light water reactor fuel analysis code femaxi-7

JAEA-DataCode

2013-005

July 2013

Japan Atomic Energy Agency 日本原子力研究開発機構

Motoe SUZUKI Hiroaki SAITOU Yutaka UDAGAWA and Fumihisa NAGASE

Light Water Reactor Fuel Analysis Code

FEMAXI-7 Model and Structure

Reactor Safety Research UnitNuclear Safety Research Center

本レポートは独立行政法人日本原子力研究開発機構が不定期に発行する成果報告書です

本レポートの入手並びに著作権利用に関するお問い合わせは下記あてにお問い合わせ下さい

なお本レポートの全文は日本原子力研究開発機構ホームページ（httpwwwjaeagojp）

より発信されています

独立行政法人日本原子力研究開発機構研究技術情報部研究技術情報課

319-1195 茨城県那珂郡東海村白方白根 2 番地 4

電話 029-282-6387 Fax 029-282-5920 E-mailird-supportjaeagojp

This report is issued irregularly by Japan Atomic Energy Agency

Inquiries about availability andor copyright of this report should be addressed to

Intellectual Resources Section Intellectual Resources Department

Japan Atomic Energy Agency

2-4 Shirakata Shirane Tokai-mura Naka-gun Ibaraki-ken 319-1195 Japan

Tel +81-29-282-6387 Fax +81-29-282-5920 E-mailird-supportjaeagojp

copy Japan Atomic Energy Agency 2013

i

JAEA-DataCode 2013 - 005




Reactor Safety Research Unit

Nuclear Safety Research Center


Tokai-mura Naka-gun Ibaraki-ken

(Received March 14 2013)

A light water reactor fuel analysis code FEMAXI-7 has been developed for the purpose

of analyzing the fuel behavior in both normal conditions and anticipated transient conditions

This code is an advanced version which has been produced by incorporating the former

version FEMAXI-6 with numerous functional improvements and extensions In FEMAXI-7

many new models have been added and parameters have been clearly arranged Also to

facilitate effective maintenance and accessibility of the code modularization of subroutines

and functions have been attained and quality comment descriptions of variables or physical

quantities have been incorporated in the source code With these advancements the

FEMAXI-7 code has been upgraded to a versatile analytical tool for high burnup fuel behavior

analyses This report describes in detail the design basic theory and structure models and

numerical method of FEMAXI-7 and its improvements and extensions

Keywords LWR Fuel FEM Analysis Transient Pellet Cladding Fission Gas Release

PCMI Burn-up

ITOCHU Techno-Solutions Corporation Tokyo Japan

ii


軽水炉燃料解析コード

FEMAXI-7 のモデルと構造

日本原子力研究開発機構安全研究センター

原子炉安全研究ユニット

鈴木元衛斎藤裕明宇田川豊永瀬文久

（2013 年 3 月 14 日受理）

FEMAXI-7 は軽水炉燃料の通常運転時及び過渡条件下のふるまい解析を目的と

するコードとして前バージョン FEMAXI-6 に対して多くの機能の追加改良を実施

した高度化バージョンである特にソースコードの整備及び解読の効率化を図るた

めにサブルーチンやファンクションのモジュール化とコメント記述の充実を図りコ

ードのさらなる拡張を容易にしたまた新しいモデルを追加するとともにユーザ

ーの使いやすさにも考慮して多くのモデルのパラメータを整理したこれらにより

FEMAXI-7は高燃焼度燃料の通常時のみならず過渡時ふるまいの解析に対する強力な

ツールとなった

本報告はFEMAXI-7 の設計基本理論と構造モデルと数値解法改良と拡張

採用した物性値等を詳述したものである

原子力科学研究所（駐在）319－1195 茨城県那珂郡東海村白方白根 2－4

伊藤忠テクノソリューションズ株式会社東京

iii

Contents

1Introduction 1

11 Characteristics of Fuel Performance Code 1

111 Features and roles of fuel code 1

112 Prediction of fuel behavior by FEMAXI 2

113 History of FEMAXI code development 3

114 Usage of FEMAXI 4

12 Features of Structure and Models of FEMAXI 5

121 What is model 5

122 What is fuel modeling 6

123 Analytical targets 7

13 Whole Structure 8

14 Features of Numerical Method and Modeling 11

15 Interfacing with Other Codes etc 14

151 Interfacing with burning analysis codes RODBURN and PLUTON 14

152 Re-start function from base-irradiation to test-irradiation periods 14

153 Execution system environments 14

References 1 15

2 Thermal Analysis Models 17

21 Heat Transfer to Coolant and Thermal-hydraulics Model 17

211 Coolant enthalpy increase model 20

212 Determination of cladding surface heat transfer coefficient 23

213 Flow of thermal-hydraulics calculation 33

214 Transition process of coolant condition 35

215 Equivalent diameter and cross-sectional area of flow channel 38

22 Cladding Waterside Corrosion Model 39

23 One-Dimensional Temperature Calculation Model 41

231 Selection of number of elements in the radial direction 41

232 Determination of cladding surface temperature 43

233 Solution of thermal conduction equation 45

JAEA-DataCode 2013-005

iv

234 Fuel pellet thermal conductivity 50

24 Determination of Heat Generation Density Profile 53

241 Use of RODBURN output 53

242 Use of PLUTON output 54

243 Robertsonrsquos formula 54

244 Accuracy confirmation of pellet temperature calculation 55

25 Gap Thermal Conductance Model 60

251 Modified Ross amp Stoute model 60

252 Bonding model for gap thermal conductance 63

253 Bonding model for mechanical analysis 65

254 Pellet relocation model 66

255 Swelling and densification models 68

26 Model of Dry-out in a Test Reactor Capsule 68

261 Modelling the dry-out experiment 69

262 Thermal and materials properties 73

263 Numerical solution method 74

27 Generation and Release of Fission Gas 78

271 Fission gas atoms generation rate 79

272 Concept of thermally activated fission gas release in FEMAXI 79

273 Thermal diffusion accompanied by trapping 81

274 White+Tucker model of intra-grain gas bubble radius and its number

density (GBFIS=0) 85

275 Radiation re-dissolution and number density model of intra-granular gas

bubbles (GBFIS=1) 88

276 Pekka Loumlsoumlnen model for intra-granular gas bubbles (GBFIS=2) 92

277 Galerkinrsquos solution method for partial differential diffusion equation

(common in IGASP=0 and 2) 94

278 Fuel grains and boundaries (common in IGASP=0 and 2) 99

279 Amount of fission gas atoms migrating to grain boundary


2710 Amount of fission gas accumulated in grain boundary 101

2711 Amount of released fission gas 103


v

2712 Sweeping gas atoms to grain boundary by grain growth 105

2713 Re-dissolution of fission gas (common in IGASP=0 and 2) 110

2714 Threshold density of fission gas atoms in grain boundary

(Equilibrium model IGASP=0) 111

2715 In-grain diffusion coefficient of fission gas atoms

(common to IGASP=0 and 2) 114

28 Rate- Law Model (IGASP=2) 117

281 Assumptions shared with the equilibrium model 117

282 Growth rate equation of grain boundary bubble 117

29 Swelling by Grain-boundary Gas Bubble Growth in the Equilibrium

Model (IGASP=0) 125

291 Grain boundary gas bubble growth 125

292 Bubble swelling 128

210 Swelling by Grain-boundary Gas Bubble Growth in the Rate-law Model

(IGASP=2 IFSWEL=1) 129

2101 Coalescence and coarsening of bubbles 129

2102 Swelling 132

211 Thermal Stress Restraint Model 132

2111 Option designated by IPEXT 133

2112 Comparative discussion on the gap gas conductance increase during

power down 134

212 High Burnup Rim Structure Model 135

2121 Basic concept 135

2122 Density decrease of fission gas atoms in solid phase of rim structure 136

2123 Effect of rim structure formation on fuel behavior 140

213 Selection of Models by Input Parameters 148

2131 Number of elements in the radial direction of pellet 148

2132 Calculation of thermal conductivity of fuel pellet 149

2133 Selecting swelling model 150

2134 Selecting fission gas release model 151

2135 Bubble growth and swelling 152

2136 Options for the rim structure formation model 153


vi

214 Gap Gas Diffusion and Flow Model 154

2141 Assumptions and methods of diffusion calculation 155

215 Method to Obtain Free-space Volumes in a Fuel Rod 163

2151 Definition of free space 163

216 Internal Gas Pressure 168

2161 Method of assigning plenum gas temperature and cladding temperature 169

2162 Calculation of the variation in internal gas condition during irradiation 169

217 Time Step Control 170

2171 Automatic control 170

2172 Time step increment determination in FGR model 172

2173 Time step increment determination in temperature calculation 173

References 2 174

3 Mechanical Analysis Model 178

31 Solutions of Basic Equations and Non-linear Equations 178

311 Basic equations 178

312 Initial stress method initial strain method and changed stiffness method

for non-linear strain 184

313 Solution of the basic equations in FEMAXI (non-linear strain) 187

32 Mechanical Analysis of Entire Rod Length 194

321 Finite element model for entire rod length analysis (ERL) 195

322 Determination of finite element matrix 199

323 Derivation of stress-strain (stiffness) matrix of pellet and cladding 201

324 Pellet cracking 208

325 Crack expression in matrix 211

326 Definition equations of equivalent stress and strain in FEMAXI 214

327 Hot-pressing of pellet 216

328 Supplementary explanation of incremental method 219

33 Method to Calculate Non-linear Strain 223

331 Creep of pellet and cladding 223

332 Method of creep strain calculation 224

333 Plasticity of pellet and cladding 233


vii

334 Derivation of stiffness equation 242

34 Formulation of Total Matrix and External Force 244

341 Stiffness equation and total matrix 244

342 Upper plenum and lower plenum boundary conditions 247

343 Pellet-cladding contact model 254

344 Axial force generation at the P-C contact surfaces 256

345 Algorithm of axial force generation at PCMI contact surface 257

346 Evaluation of axial force generated at the P-C contact surface Option A 260

347 Evaluation of Axial force generation Option B 265

35 2-D Local PCMI Mechanical Analysis 271

351 Element geometry 272


353 Stiffness equation 278

354 Boundary conditions 284

355 PCMI and contact state problem 289

356 Axial loading - Interaction between target pellet and upper pellet - 298

36 Mechanical Properties Model of Creep and Stress-strain 301

361 Creep 301

362 Model of cladding stress-strain relationship 305

363 Ohta model 307

364 Method to determine yield stress in FRAPCON and MATPRO

models (ICPLAS=3 6) 313

37 Skyline Method 315

References 3 322

4 Materials Properties and Models 323

41 UO2 Pellet 323

42 MOX Pellet 343

43 Zirconium Alloy Cladding 353

431 Cladding properties 353

432 Cladding oxide properties 368

44 SUS304 Stainless Steel 370


viii

45 Other Materials Properties and Models 372

46 Burnup Calculation of Gd2O3-Containing Fuel 377

References 4 380

Acknowledgement 382


ix

目次

1序言 1

11 燃料解析コードの性格 1

111 燃料解析コードの特徴と性格 1

112 FEMAXI による燃料挙動の予測 2

113 FEMAXI コード開発の歴史 3

114 FEMAXI コードの利用 4

12 FEMAXI の構造とモデルの特徴 5

121 モデルとは何か 5

122 燃料モデリングとは何か 6

123 解析対象 7

13 全体構造 8

14 数値計算とモデルの特徴 11

15 他コードとの関係その他 14

151 燃焼解析コード RODBURN および PLUTON との接続 14

152 ベース照射から Re-start による試験照射への接続 14

153 実行環境など 14

参考文献 1 15

2 熱的解析モデル 17

21 冷却材への伝熱と熱水力モデル 17

211 冷却材のエンタルピー上昇のモデル 20

212 被覆管表面熱伝達率の決定 23

213 熱水力計算の流れ 33

214 冷却材状態の遷移 34

215 流路相当直径および流路断面積 38

22 被覆管水側腐食モデル 39


x

23 一次元温度計算モデル 41

231 半径方向リング要素数の選択 41

232 被覆管表面温度の決定 43

233 熱伝導方程式の解法 45

234 ペレット熱伝導率 50

24 発熱密度プロファイルの決定 53

241 RODBURN 出力の利用 53

242 PLUTON 出力の利用 54

243 Robertson の式 54

244 ペレット温度計算の精度確認 55

25 ギャップ熱伝達モデル 60

251 修正 Ross amp Stoute モデル 60

252 熱伝達に対するギャップボンディングモデル 63

253 力学計算に対するギャップボンディングモデル 65

254 ペレットリロケーションモデル 66

255 スエリングと焼きしまりモデル 68

26 試験炉キャプセル内ドライアウトモデル 68

261 ドライアウト試験のモデル化 69

262 熱的材料的物性値 73

263 数値解法 74

27 FP ガス生成放出モデル 78

271 FP ガス原子生成速度 79

272 熱活性拡散放出モデルの概念 79

273 トラッピングを伴う熱拡散 81

274 粒内バブル半径と数密度の White+Tucker モデル(GBFIS=0) 85

275 粒内バブル半径と数密度の照射溶解モデル(GBFIS=1) 88

276 粒内バブルの Pekka Loumlsoumlnen モデル(GBFIS=2) 92

277 偏微分拡散方程式の Galerkin 解法(IGASP=0 と 2 で共通) 94

278 結晶粒と粒界(IGASP=0 と 2 で共通) 99


xi

279 粒界へ移動する FP ガス原子の量(IGASP=0 と 2 で共通) 100

2710 粒界に溜まる FP ガス量 101

2711 放出されるガス量 103

2712 粒成長に伴う粒界へのガスの掃き出し 105

2713 FP ガスの再溶解(IGASP=0 と 2 で共通) 110

2714 粒界 FP ガス原子濃度の飽和値（平衡論 IGASP=0） 111

2715 FP ガス原子の粒内拡散定数(IGASP=0 と 2 で共通) 114

28 速度論モデル（IGASP=2） 117

281 平衡論モデルとの共通の仮定 117

282 粒界バブル成長速度式 117

29 粒界バブル成長スエリング平衡論（IGASP=0） 125

291 粒界バブル成長 125

292 バブルスエリング 128

210 バブル成長スエリング－速度論(IGASP=2 IFSWEL=1) 129

2101 バブルの合体粗大化 129

2102 スエリング 132

211 圧力抑制モデル 132

2111 IPEXT で指定されるオプション 133

2112 出力降下時の軸方向ガスコンダクタンス増加との比較考察 134

212 高燃焼度リム組織モデル 135

2121 基本概念 135

2122 リム組織固相における FP ガス原子の濃度減少 136

2123 リム組織の燃料ふるまいへの影響 140

213 パラメータによる計算モデルの選択 148

2131 ペレット径方向リング要素数 148

2132 ペレット熱伝導率計算 149

2133 スエリングモデルの選択 150

2134 FGR モデルの選択 151

2135 バブル成長とスエリング 152


xii

2136 リム組織生成モデルのオプション 153

214 ギャップ内ガスの拡散流動モデル 154

2141 拡散計算の仮定と方法 155

215 燃料棒内空間体積の求め方 163

2151 自由空間の定義 163

216 内部ガス圧力 168

2161 プレナムのガス温度と被覆管温度の設定方法 169

2162 照射途中における内部ガス状態変更の計算 169

217 タイムステップの制御 170

2171 自動制御 170

2172 FP ガス放出モデルにおけるタイムステップ幅の決定 172

2173 温度計算におけるタイムステップ幅の決定 173

参考文献 2 174

3 力学解析モデル 178

31 基本式と非線形力学の解法 178

311 基本式 178

312 非線形歪みに対する初期応力法初期歪み法剛性変化法 184

313 FEMAXI コードでの基本式の解法（非線形歪み） 187

32 燃料棒全長の力学解析 194

321 全長力学解析の有限要素モデル 195

322 有限要素マトリックスの決定 199

323 ペレットと被覆管の応力歪み（剛性）マトリックスの導出 201

324 ペレットのクラック 208

325 マトリックスにおけるクラック表現方法 211

326 FEMAXI における相当応力相当歪みの定義式 214

327 ペレットのホットプレス 216

328 増分法の補足説明 219


xiii

33 非線形歪みの計算方法 223

331 ペレットと被覆管のクリープ 223

332 クリープ歪みの計算方式 224

333 ペレットと被覆管の塑性 233

334 剛性方程式の導出 242

34 全体マトリックスの定式化と外力 244

341 剛性方程式と全体マトリックス 244

342 上下プレナム境界条件 247

343 ペレット被覆管接触モデル 254

344 ペレット被覆管接触面における軸力発生 256

345 PCMI 接触面における軸力発生のアルゴリズム 257

346 PCMI 接触面における発生軸力の評価オプション A 260

347 PCMI 接触面における発生軸力の評価オプション B 265

35 2 次元局所 PCMI 力学解析 271

351 要素体系 272

352 基本式 276

353 剛性方程式 278

354 境界条件 284

355 PCMI と接触問題 289

356 軸方向荷重―上部ペレットとの軸方向相互作用― 298

36 クリープ及び応力-歪みに関する機械的性質モデル 301

361 クリープ 301

362 被覆管の応力歪み関係式のモデル 305

363 太田モデル 307

364 FRAPCON と MATPRO モデルの降伏応力の決定方法

(ICPLAS=3 6) 313

37 スカイライン法 315

参考文献 3 322


xiv

4 物性値とモデル 323

41 UO2 ペレット 323

42 MOX ペレット 343

43 ジルコニウム合金被覆管 353

431 被覆管の物性値 353

432 被覆管酸化物の物性値 368

44 SUS304 ステンレス被覆管 370

45 その他の物性値モデル 372

46 燃焼度の単位換算 377

参考文献 4 380

謝辞 382


- 1 -

1 Introduction A light water reactor has occupied the most essential part in nuclear power generation

for years With the advancement in LWR performance fuel improvement has been actively

pursued in nuclear industries worldwide In particular burnup extension has been promoted

recently and high-burnup fuel behavior has been widely known as a key issue in terms of rod

design and reliability

Under these circumstances development of a computer code which can precisely analyze

fuel behavior is required to evaluate the behavior of high-burnup fuels to assess their

reliability and to serve as a safety cross-check tool In the code introduction of new models

is inevitable departing from the conventional fuel analysis methods because it has been

observed that fuel behavior in a high-burnup region of over 40 - 50 GWdt essentially differs

from that in lower burnup region Such difference is generated by fuel thermal conductivity

degradation rim structure formation in fuel periphery and increase in FGR associated with

the structure change etc They can impose a strong influence on the overall behavior of fuel

rod

On the basis of these considerations an integrated fuel performance code FEMAXI-7

has been developed for analyzing the behavior of fuels under normal operation and

anticipated transients (excluding accident conditions) This code has integrated and

extended the capabilities of the predecessor FEMAXI-6 by incorporating new models and

functions

11 Characteristics of Fuel Performance Code 111 Features and roles of fuel code

Fuel rods are the objects in which radioactive substances are highly accumulated and

concentrated in a reactor core Thermal and mechanical behavior of fuel rod is resulted from

complicated interactions among a number of such factors as temperature neutron flux

nuclear fission accumulation of fission products etc which outspread from atomistic level

phenomena to engineering-controllable macroscopic entities

A fuel assembly has to attain concurrently the two roles ie safety and power

generation efficiency Realizing this purpose needs a various RampD efforts based on

irradiation experiences which gives a fundamental cause for the research of fuel behavior in

terms of engineering and safety

On the other hand that fuel behavior is a result of complicated interactions implies that

fuel research is a field where actual irradiation experiences have inevitably a marked


- 2 -

significance In other words fuel research is a field where theoretical and deterministic

models cannot always have a precise predictability some materials properties or behavior

cannot be expressed by such theoretical model as those of reactor physics or cannot be

clearly defined by established physical laws and there are so many unknown factors such as

the effect of small quantity of impurities crystal grain structure dependence on fabrication

process etc Furthermore so many factors are involved in fuel behavior that it is sometimes

difficult to evaluate the contribution of each factor to whole behavior by simply summing up

each factor-wise calculation Accordingly irradiation experiences have an indispensable role

in simulation design and understanding fuel behavior

However irradiation of fuels ie irradiation tests and post-irradiation examinations

needs in general much cost and long period of time and quantity and quality of measured

data are limited for the cost and time The irradiation test cannot give all that are

necessary to understand the inside processes of fuel rod As a result in some cases we are

obliged to depend on a few specific measurable physical quantities such as fuel center

temperature and internal pressure or on the indirectly evaluated quantities such as burnup

distribution in assessing the whole behavior of fuel rod

In these situations making use of a fuel performance code is an important measure to

recognize and reproduce the interlinked structures inside fuel rod to grasp the controlling

parameters for fuel behavior and to predict the fuel behavior on the basis of the ldquodirectly or

indirectly measured physical quantitiesrdquo as if solving an inverse problem in an exploratory

manner FEMAXI is an engineering tool serving for this purpose In other words

development of FEMAXI is a process which constitutes reproduces and predicts the

interaction structure hidden behind observed fuel behavior on the basis of the limited data

or experiences of irradiation

112 Prediction of fuel behavior by FEMAXI A fuel design code of fuel manufacturer often uses empirical correlations which have

been derived from irradiation test data base not depending on the mechanistic models such

as those in FEMAXI and the calculated results are compared with measured data for

validation Therefore its predictions are reliable at least within the scope of the test data

base and its small external region and it is considered that the prediction is effective for fuel

design

On the other hand many mechanistic and complementary empirical models and

parameters to control the models have been implemented in FEMAXI-7 For prediction of


- 3 -

unknown behavior of fuel FEMAXI-7 could bring about so different calculation results

depending on the combination of its models and parameters that the final prediction would

be capable of having a considerable uncertainty

Accordingly it is necessary to find models and data set combinations which are as

generally applicable as possible to the prediction in the process of verification calculations If

a code can reproduce by a set of models and parameters the behavior of fuel A of which

irradiated results are known and if the same set is applied to another fuel rod B and can

result in a successful reproduction of the irradiated state of the rod B predictability of the

set is verified with respect to at least these fuel rods It depends on the number of analyzed

cases on analyzed fuel designs and irradiation conditions how far such verification has been

made ie how many types of models and parameter sets have been tested and how much

generalization of predictability is guaranteed in FEMAXI In this sense verification of

FEMAXI-7 is on-going As some of verification activities test analyses have been

performed in the international benchmark test FUMEX-III(11) and by using the results of

the Halden test irradiations Through these activities refinement of models and

parameters set are being followed up

At the same time FEMAXI-7 can allow by the combination of its various models and

controlling parameters an attempt to reproduce such fuel behavior (irradiation data) that

the conventional models and parameters cannot obtain This is one of the means which

enables us to estimate the unknown linkage structure of factors inside fuel rod

Furthermore the code can be applied to the prediction of fuel behavior in the conditions

which is planned in test irradiation or new design of fuels or even in the conditions beyond

the experimentally feasible scope

113 History of FEMAXI code development History of development up to the latest version FEMAXI-7 is briefly described It is one

of features that up to FEMAXI-7 a number of researchers have been engaged in the

development and their knowledge and experiences have been accumulated in the code The

code can be considered to be based on the combination of wisdom and experiences of

Japanese fuel researchers and engineers

(1) The first version of FEMAXI was developed by Ichikawa (JAERI Japan Atomic

Energy Research Institute) in the Halden Project It featured a 2-D local elasto-plastic

PCMI analysis function (1974)


- 4 -

(2) Creep model was implemented in FEMAXI-II by Kinoshita (CRIEPI The Central

Research Institute of Electric Power Industry) (1977)

(3) FEMAXI-III(12) was developed by an extensive cooperation of JAERI universities

Hitachi Toshiba NFD(Nuclear Fuel Development Co) CRIEPI and CRC(Century

Research Center Co) Development of this version received an award from Japan

Society of Nuclear Science and Technology It exerted a predominant capability in

international benchmarking calculation Released to NEA-DataBank(13) (1984)

(4) FEMAXI-IV(14)-(112) was a version improved by Nakajima et al It was used in

JNES(113) ( former NUPECNuclear Power Engineering Corporation) as one of cross

check codes This version was taken over to MSuzuki in JAERI Ver2 was developed

and released to NEA-DataBank (1996)

(5) FEMAXI-V (114) was developed by Suzuki for the analysis of high burnup fuels

released to NUPEC and NEA-DataBank (2001)

(6) The code was further extended to FEMAXI-6 by Suzuki et al(JAEA) released to

JNES and to NEA DataBank (2003-2010)

114 Usage of FEMAXI

An extended usage of FEMAXI code can be summarized as follows

(1) Cross check code for safety licensing

The former version FEMAXI-6 is a main part and function of the cross-check code used in

JNES for safety licensing of LWR fuel rod

(2) Release to external users

A code package of FEMAXI-7 (source binary program related codes description and

manual) has been released to domestic users without charge through RIST (Research

Organization for Information Science amp Technology httpwwwristorjp ) The package

will be registered in NEA DataBank as a representative fuel performance code of Japan to

be released to foreign users The former versions FEMAXI-VI FEMAXI-V and FEMAXI-6

have been released from NEA DataBank to foreign users without charge and the total

number of the releases is over thirty Also we are positively acquiring external users

responding to questions from users In effect the FEMAXI code has been a

frame-of-reference of Japanese LWR fuel performance code


- 5 -

(3) One of the means for experimental analysis prediction of experiments and

training of fuel behavior understanding

The FEMAXI code can be used for exploratory analysis of irradiation tests and for the

prediction of state of fuel rods just prior to test-irradiation in accident conditions

Also the code is used for the planning of irradiation test for checking the adequacy of

fuel design Furthermore it is used as a training tool by performing simulations with

various input conditions for a better understanding of fuel behavior

(4) Platform to evaluate irradiation test data and new models

Newly obtained materials properties of irradiated fuels are in itself sometimes not clearly

evaluated in terms of their contribution to fuel behavior change or improvements However

incorporating them in a fuel code putting them in a various interactions of a number of

factors and performing simulations with them will enable us to evaluate their contribution

to or effects on the whole behavior of fuel rods more clearly or quantitatively The FEMAXI

code can function for such purposes

12 Features of Structure and Models of FEMAXI 121 What is model

Before discussing specific models a fundamental question is addressed what is model

as a code component What is the purpose of model development Here if the answer is

that the purpose is to reproduce behavior of a target object as precisely as possible this

answer is insufficient because it does not address the question that ldquowhy is it necessary to

reproduce preciselyrdquo and ldquowhat is the preciseness in this caserdquo To begin with since

model development starts in the situation where knowledge about the object to be modeled is

insufficient the preciseness has to own its limit In other words a model can be a model

because it is an approximation of a real object or an ideational reflection which has

quantitative nature to a considerable extent Therefore model development is upgrading in

the following steps

1) The first stage is a one which gives insight into the underlying mechanism of phenomena

as to what will appear when watching the phenomena through the prototypical model

2) The second stage allows us to estimate the underlying mechanism of phenomena by

incorporating the model in a code and performing calculation


- 6 -

3) The third stage enables us to predict the fuel behavior by changing model parameters

which will lead to the improvement of fuel performance and be used as a measure to solve

the ldquoinverse problemrdquo

However the most essential role of modeling is considered to make a reflection on

human notion and deepen the insight into the mechanism of phenomena by realizing an

interaction or mutual permeation between the human notion and the real entity which exists

independently from the human notion

122 What is fuel modeling As for the fuel modeling it is quite usual to have discussion in terms of individual

modeling of major phenomena which dominate fuel behavior such as heat generation and

conduction fission gas release Pellet-Clad Mechanical Interaction(PCMI) etc However in

high burnup fuel behavior as interactions become stronger among various phenomena the

modeling is inevitably extended to not only each phenomenon but also a linkage structure to

combine various factors That is a linkage structure of each model itself is required to be

designed This means that to what extent the interaction among the factors is explicitly

modeled

For example coupling of thermal and mechanical analyses or interlinkage among

fission gas bubble growth pellet swelling and thermal conductivity degradation etc are to

be addressed in the code design

However the interlinkage structure cannot be too complicated at the beginning and it is

an usual strategy to start with a simplified assumption or model For example porosity

increase by fission gas bubble growth has an additional decrease effect on the pellet thermal

conductivity though it is set independent from the thermal properties of pellet

On the other hand as nuclear reaction calculation has a weak interaction with thermal

and mechanical behavior the reaction calculation is carried out in a relative isolation from

fuel behavior calculation in fuel performance code That is no direct interaction with fission

reaction is incorporated in the code but it is considered implicitly as the change of pellet

power density profile with burnup or fast neutron flux corresponding to linear heat rate

These quantities are often calculated by a sub-code or an independent burning analysis code

(reactor core physics code) and the results are fed to the main part of fuel code In this

sense the burning analysis codes which give results to FEMAXI are important (Refer to

Refs(115) and (116))

Based on the above discussion features of FEMAXI models can be summarized as


- 7 -

follows

(1) The analytical object is limited to a single fuel pin and its surrounding coolant being

isolated from the interactions among the fuel pins set in an assembly or thermal and

nuclear behavior of whole reactor core

(2) In designing models and code as engineering tools the highest priority is placed on the

prediction of fuel behavior in a macroscopic scale Accordingly in this sense it is not a

preferential task to make each model based on a microscopic physical principle or theory

The code adopts a combination of mechanistic model and empirical model If a

mechanistic model cannot be designed the code begins with an empirical model

(3) The code pursues fast calculation Nevertheless a low priority is given to the modeling

efforts to carry on a much reduced calculation time for the purpose of a statistical

analysis which handles a number of fuel pins in a reactor core

123 Analytical targets Analytical objects of FEMAXI-7 is quite identical to those of FEMAXI-6 fuel behavior

in not only steady-state normal operation conditions but also such transient conditions as

fast power ramp load-following power-coolant mismatch etc Nevertheless the code does

not cover the accident conditions such as LOCA (Loss-of-coolant accident) and RIA

(Reactivity- initiated accident)

Table 11 shows the phenomena which FEMAXI-7 deals with As for the material

properties including those for MOX fuels and Gd-containing UO2 fuels those available in

open literatures have been adopted as much as possible Major functional differences

among FEMAXI-V FEMAXI-6 and FEMAXI-7are listed in Table 12 Description for each

item will be presented in chapters 2 3 and 4

Table 11 Phenomena analyzed by FEMAXI-7

Thermal process determiningtemperature distribution

Process with mechanical displacement

Pellet

Thermal conduction (heat flux distribution ) Fission gas release Swelling

Thermal expansion elasticity plasticity creep cracking relocation densification swelling hot-press

Cladding Thermal conduction Waterside corrosion

Thermal expansion elasticity plasticity creep irradiation growth

Fuel rod

Gap thermal conduction (mixedgas contact radiation) cladding surface heat transfer gap gas flow

Mechanical interaction between pellet and cladding friction and bonding between pellet and cladding


- 8 -

Table 12 Upgrades of main functions of versions up to FEMAXI-7

FEMAXI-Vreleased in 1999

FEMAXI-6 released in 2002

FEMAXI-7

Coupling solution of thermal and mechanical analyses

times

Pellet thermal conductivity degradation with burnup

High burnup structure formation in pelletperiphery

Gas bubble swelling model times

Rate-law model of grain boundary gas bubble growth

times times

Transient thermal hydraulics model

Extension of inputoutput and materials properties

Restart function to have a link between base-irradiation and test-irradiation

times times

Modularization of source codes detailed comments embedded selection of the number of elements

times times

13 Whole Structure

FEMAXI-7 consists of two main parts one for calculating the temperature distribution

of fuel rod thermally induced deformation and fission gas release etc (hereafter called

ldquothermal analysis partrdquo) and the other for calculating the mechanical behavior of fuel rod

(hereafter called ldquomechanical analysis partrdquo) Figure 11 outlines the entire code structure

and Fig12 shows the analytical geometry

In the thermal analysis part calculation always covers an entire rod length which is

divided into the maximum of 40 axial segments Namely the temperature distribution is

calculated at each axial segment as one-dimensional axis-symmetrical problem in the radial

direction and with this temperature such temperature-dependent values as fission gas bubble

growth gas release gap gas flow in the axial direction and their feedback effects on gap

thermal conduction are also calculated

Also to take into account thermal feedback among the axial segments which may arise

due to uneven power distributions in the axial direction iteration is performed until

convergence is attained for the entire rod length


- 9 -

In the mechanical analysis part users can select one of the two modes analysis of the

entire rod length 1-D ERL mechanical analysis only or the ERL mechanical analysis +

2-D local PCMI analysis in one pellet length

In the former the 1-D axis-symmetrical finite element method (FEM) is applied to each

axial segment of the entire rod length in the latter in addition to the 1-D FEM the

axis-symmetrical FEM is applied to half a pellet length for a symmetrical reason and

mechanical interaction between pellet and cladding ie local PCMI is analyzed only in one

axial segment

In the mechanical analysis the magnitude of pellet strain caused by thermal expansion

densification swelling and relocation is calculated first and a stiffness equation is formulated

with consideration given to cracking elasticityplasticity and creep of pellet

Then stress and strain of pellet and cladding are calculated by solving the stiffness

equation with boundary conditions which correspond to the pellet-cladding contact mode

When PCMI occurs and pellet-cladding contact states change calculation is re-started with

the new boundary conditions of contact from the time when the change occurs

In the ERL mechanical analysis the axial force acting on adjacent axial segment is

evaluated and as a result analysis of axial deformation of entire rod length is performed

Also the finite element method is simplified to reduce the total number of degrees of

freedom

The thermal analysis and the ERL mechanical analysis are coupled convergence

between temperature and deformation is obtained in every time step by iteration


- 10 -

Next Time Step

No

Output

Temperature Internal pressure

Input

＜Time step control＞

Mechanical analysis of entire rod length

Elasto-plastic creep PCMI axial force deformation(stress-strain)

End of Time Step

Yes

Iteration

Local PCMI analysis [half a pellet length]

Elasto-plastic creep ridging stress-strain distribution

Thermal analysis Temperature fission gas

diffusion and release

Fig11 Overview of analytical flow of FEMAXI


- 11 -

Plenum

Segment

LHR

Axial Power Profile

Zircaloy metal

ZrO2 generated bywaterside corrosion

Temperature

Power andBurnupprofiles

FEM mechanical analysisfor pellet and cladding

Nest ofcoaxialrings

Gas pressure

14 Features of Numerical Method and Modeling

(1) FEM element characteristics

Storage region and calculation time are reduced by introduction of rectangular

3-degrees-of-freedom elements into the FEM analysis in order to perform mechanical

calculation for the entire length of the fuel rod

(2) Creep solution An implicit procedure is applied to obtain numerical solution stability for high creep rate

(3) Contact problem Three contact states are dealt with as the contact conditions between pellet and cladding

open gap state clogged gap state and sliding state In the FEM contact conditions at each

node pair of pellet and cladding are determined and the boundary conditions are set in

accordance with the conditions

Fig12 Geometry of FEMAXI for the analysis of one single fuel rod


- 12 -

(4) Matrix solution The coefficient matrix of simultaneous equations of

the FEM (stiffness equations) is a diagonal symmetrical

matrix having a large number of zero elements as

shown in the figure aside Therefore the skyline

method has been adopted to solve the equations and the

memory method of non-zero elements in the matrix as

well as the calculation procedure has been improved

in order to reduce the size of the storage region and

calculation time

In the skyline method only the elements under the

solid line in the figure are stored In the fission gas

release model the procedure has been improved so that the number of computing steps for the

matrix becomes a minimum

(5) Non-equilibrium residue In order to avoid accumulation of non-equilibrium residue generated at each time step

during solution of a nonlinear problem equilibrium condition equations are formulated not in

an incremental manner but in a form which maintains the total balance of loads and stresses

(6) Behavior of cracked pellet Cracking of a pellet is modeled using a decreased stiffness approximation method similar

to the case for FEMAXI-III In order to describe behavior of the cracked pellet which is

relocated under PCMI recovery of the stiffness during compression is expressed as a function

of the amount of relocation

(7) Fission gas release model and bubble swelling model

Fission gas diffusion in fuel grain is calculated by a combination of the least residual

method in equivalent sphere model and gas bubble formation In addition a model of fission

gas release restraint by thermal stress is incorporated Gas bubble growth can be calculated

either by rate-law model or pressure equilibrium model In the both models option is

incorporated to calculate swelling by intra-grain bubble growth and by grain boundary

bubble growth

Nonzero element

Storage region in the skyline method


- 13 -

(8) Gap gas diffusion-flow model Gap gas-diffusion flow is modeled and effects of the released fission gas on gap thermal

conductance are carefully evaluated

(9) Model of oxidation of cladding Cladding waterside oxidation is modeled and change in the thermal conductivity due to

oxidation is calculated

(10) Introduction of thermal-hydraulics model

With respect to heat transfer between a fuel rod and coolant RELAP5-MOD1 and some

other models have been introduced to cover a wide range of conditions of heat transfer to

coolant including a boiling transition

(11) Non-steady phenomena analysis The fuel behavior in a non-steady state can be analyzed by mechanistic treatment of a

non-steady heat transfer model gap gas flow model and fission gas release model Also the

accuracy of prediction is improved and calculation time is reduced through use of

independent time-step controls in each model

(12) Simplification of calculation In the fission gas release model when the same type of calculations are performed both in

a low-temperature region and a high-temperature region calculation results obtained for

elements in one region are used for calculation in another region in order to avoid carrying out

similar calculations twice Also physical properties and other values which are frequently

referred to are stored in a data table and use of special mathematical functions is avoided

By means of the above-mentioned procedures FEMAXI can give highly accurate

solutions within a shorter calculation time

(13) Empirical models of rim structure formation

Empirical models are incorporated for simulating the rim structure formed in the

peripheral region of high burnup pellet


- 14 -

15 Interfacing with Other Codes etc

151 Interfacing with burning analysis codes RODBURN and PLUTON Power profiles in the radial and axial directions of a pellet and the burnup profile both

change with burnup from BOL These profile changes differ depending on the reactor type

(PWR BWR and heavy-water reactor etc) in which fuel rods are irradiated These profiles

naturally affect the temperature distribution of pellets and also affect the material

properties the performance of which depends on the burnup Accordingly FEMAXI-7

similar to FEMAX-6 includes a function which reads the output files of the burning analysis

codes RODBURN(115) and PLUTON(116) and uses them for calculation RODBURN uses

part of ORIGEN(117) and RABBLE(118) and calculates the power profiles burnup and fission

products in pellets and pellet stacks in accordance with a given power history and fuel rod

specifications and then produces files of these results

PLUTON produces isotope distributions and heat generation density profiles accurately

at high speed using an original morphological function (Refer to section 24)

152 Re-start function from base-irradiation to test-irradiation periods

FEMAXI-7 has a function to generate a result file which records the fuel rod states at

EOL ie at the end of base-irradiation This re-start file can be read by FEMAXI-7 to

restart the calculation of fuel rod in test-irradiation which has often a number of power

changes in a relatively short time In the former version FEMAXI-6 in the case where

base-irradiation is conducted with a long fuel rod and test-irradiation is conducted with a

short rod re-fabricated from the long rod users have no other choice but to perform analysis

with the short refabricated geometry of rod from the beginning of base-irradiation to the end

of test-irradiation This may give some inaccuracy in the final results and also

ineffectiveness of doing the whole calculation process every time In FEMAXI-7 these

problems have been eliminated

Furthermore this file can be read by RANNS to be used as one of the initial conditions for

the fuel rod analysis in accident conditions

153 Execution system environments (1) Machine source code and compilers

FEMAXI -V -6 and -7 have been developed to be operated usually in Windows PC while

Linux version of FEMAXI-7 has been developed The source code is written mainly in


- 15 -

Fortran-77 and partly in Fortran-90

Until now Compaq Digital Visual Fortran v6 or upper (Compaq DVF) has been used for a

compiler though this compiler has been discontinued From now Intelreg Visual Fortran

Compiler for Windows version 10xx or upper (Intel VF) (119) are recommended which

includes every function of Compaq DVF

In other words totally equivalent executable programs of the FEMAXI-7 source code can

be built by the three compilers Compaq DVF Intel VF and Linux-GNU Fortran g77

(2) Plotted output

Numerical results of FEMAXI-7 calculation can be obtained in three types of outputs

direct numerical figures plotted figures by using plotting program EXPLOT and Excel

tables corresponding to the plotted figures

(3) Release of package

A code package of FEMAXI-7 consists of the source code executable program compile

option files related programs such as plotting program code model description document

and InputOutput manual The package can be delivered to users in foreign countries

through NEA DataBank

References 1 (11) Improvement of Computer Codes Used for Fuel Behaviour Simulation - FUMEX III

httpwww-nfcisiaeaorgCRPCRPMainaspRightP=CRPDescriptionampCRPID=16

(12) Nakajima T Ichikawa M et al FEMAXI-III A Computer Code for the Analysis of

Thermal and Mechanical Behavior of Fuel Rods JAERI 1298 (1985)

(13) OECD Nuclear Energy Agency DataBank httpwwwneafrdatabank

(14)

(15)

(16)

Nakajima T FEMAXI-IV A Computer Code for the Analysis of Fuel Rod Behavior

under Transient Conditions Nucl Eng Design 88 pp69-84 (1985)

Nakajima T and Saitou H A Comparison between Fission Gas Release Data and

FEMAXI-IV Code Calculations Nucl Eng Design 101 pp267-279 (1987)

Nakajima T Saitou H and Osaka T Analysis of Fission Gas Release from UO2 Fuel

during Power Transients by FEMAXI-IV Code IWGFPT-27259 pp140-162 (1987)


- 16 -

(17)

(18)

(19)

(110)

(111)

Nakajima T and Ki-Seob Sim Analysis of Fuel Behavior in Power-Ramp Tests by

FEMAXI-IV Code Res Mechanica 25 pp101-128 (1988)

Nakajima T Saitou H and Osaka T Fuel Behavior Modelling Code FEMAXI-IV and

Its Application IAEA-T1-TC-659 Paper presented at IAEA Technical Committee on

Water Reactor Fuel Element Computer Modelling in Steady-State Transient and

Accident Conditions Preston England Sept19-22 (1988)

Uchida M Saito H Benchmarking of FEMAXI-IV Code with Fuel Irradiation Data

in Power Reactors JAERI-M 90-002 (in Japanese) (1990)

Nakajima T Saitou H and Osaka T FEMAXI-IVA Computer Code for the Analysis

of Thermal and Mechanical Behavior of Light Water Reactor Fuel Rods

Trans11th Int Conf on SMIRT (Tokyo Japan) 6297 PVC-D pp1-6 (1991)



Nucl Eng Design 148 pp41-52 (1994)

(112) Suzuki M and Saitou H Light Water Reactor Fuel Analysis Code FEMAXI-IV

(Ver2) -Detailed Structure and Userrsquos Manual- JAERI-DataCode 97-043 (1997)

(113) Japan Nuclear Energy Safety Organization httpwwwjnesgojp

(114) Suzuki M Light Water Reactor Fuel Analysis Code FEMAXI-V(Ver1)

JAERI-DataCode 2000-030 (2000)

(115) Uchida M and Saitou H RODBURN A Code for Calculating Power Distribution in

Fuel Rods JAERI-M 93-108 (in Japanese) (1993)

(116)

Lemehov SE and Suzuki M PLUTON ndash Three-Group Neutronic Code for Burnup

Analysis of Isotope Generation and Depletion in Highly Irradiated LWR Fuel Rods


(117) Bell MJ ORIGEN ndash The ORNL Isotope Generation and Depletion Code

ORNL-4628 (1973)

(118) Kier PH and Robba AA RABBLE A Program for Computation of Resonance

Absorption in Multi-region Reactor Cells ANL-7326 (1967)

(119) httpsoftwareintelcomen-usintel-compilers


- 17 -

2 Thermal Analysis Models In this chapter theory and models in the thermal analysis of FEMAXI-7 are explained

21 Heat Transfer to Coolant and Thermal-hydraulics Model

In the thermal analysis part temperature distribution at each axial segment of a fuel rod

is calculated by one-dimensional cylindrical geometry with the boundary conditions

determined by coolant temperature and pressure The basic assumption and calculation

procedure are as follows

(1) Temperatures and states of coolant for all axial segments depend on the axial distribution

of coolant enthalpy in a previous time step and are determined by temperature pressure and

flow velocity of coolant at the inlet and surface heat flux from cladding at each segment in

the current time step The thermal-hydraulics model of FEMAXI-7 can be applied not only

to steady states but also to transient changes such as increasedecrease in power and

load-following condition under normal operation and in the rapid change of coolant flow

rate

However in the case in which an active (stack) length of rod is modeled by a small

number of axial segments if the flow rate change occurs in a very short period of time some

possibility would be foreseen that gives an unstable or false solution of axial distribution of

coolant temperature (enthalpy) depending on the length of the axial segment flow rate and

time step increment etc

To eliminate this unfavorable possibility the stack length of rod is always divided into

equal length 100 sub-segments in a coolant enthalpy calculation and enthalpy distribution

calculated at these sub-segments is interpolated to obtain the enthalpy at representative

location of each segment to determine axial distribution of coolant temperature This method

enables users to designate the number and length of segments without concern for the

numerical instability or uncertainty in coolant temperatures at each segment elevation

(2) Enthalpy of coolant at each axial segment is calculated on the basis of the enthalpy

distribution at the end of the former time step with the enthalpy of the coolant at the inlet as

the initial value using the increase in the enthalpy obtained from flow velocity and the

amount of heat conducted from each segment of rod Distribution of the coolant temperature

in the axial direction is determined by thus-obtained distribution of the coolant enthalpy

Here as stated in (1) axial distribution of coolant enthalpy is derived from the


- 18 -

interpolation of values calculated at the sub-segments

(3) Next the heat transfer coefficient in each coolant mode is determined Surface temperature

of cladding at each axial segment is then calculated with heat flux and coolant temperature

The temperature profile in the radial direction from cladding surface to pellet center is

calculated in one dimensional geometry with the result of the term (2) above ie coolant

temperature as boundary condition

However heat transfer in the axial direction of a fuel rod is not taken into consideration

assuming that heat transfer in the axial direction due to the power distribution in the axial

direction is suppressed by such thermal resistance as dishes and gap at the end surface of

pellet Actually the temperature gradient in the axial direction is substantially smaller than

that in the radial direction and can be neglected Therefore inclusion of thermal calculation in

the z (axial) direction is not efficient since it increases calculation time markedly

(4) In addition to this normal mode in which cladding surface temperature is calculated by the

heat conduction from inner region of pellet and to the coolant at the cladding surface another

mode is possible in FEMAXI In this latter mode cladding surface temperature history is

specified by input and temperature calculation inside the fuel rod is performed by adopting

this surface temperature as boundary condition This is so-called the Dirichlet problem

solution

The following items will be explained in sections 23 to 27 calculation of

one-dimensional temperature profile in the radial direction gap thermal conductance model

and fission gas release model The flow of thermal analysis is shown in Fig211


- 19 -

Time Step Start

Calculation of burnup power density and fission gas atoms generation

Calc of gas flow in the axial direction

Calc of heat conduction coefficient between cladding surface and coolant

Calc of fission gas release and grain growth

Fuel center temperature has converged

Calc of temperature distribution in the radial direction of rod

Calc of gap thermal conductance

Fission gas release calculation has converged

Calc for one axial segment has been completed

Fuel center temperatures of all the axial segments have converged

Calc of internal pressure and gap gas composition

Next time step

Fig211 Calculation flow of thermal analysis

Iteration with mechanical analysis of the entire length of rod Calc of gap size and contact pressure at all the axial segments


- 20 -

211 Coolant enthalpy increase model

In this model mass flux and pressure during each time step are assumed to be

constant(21) Enthalpy of the coolant (water or steam) H(z t)(Jkg) at an axial position z (m)

and time t (s) can be given by the integral of its history beginning from an inlet of the coolant

as follows

τdzS

zQzVtHtzH

t

t primeprimeprime

+=in )(

)()()0()( in (211)

Here H(0 tin) enthalpy of coolant at the inlet (Jkg)

tm the time (s) when a coolant located at an axial position z at time t passes

through the inlet

( )τzprime axial position (m) of the coolant at time τ which was positioned at z at

time t

V specific volume (m3kg)

Q heat influx at unit length (Wm)

S flow cross section (m2)

τ time

Focusing on the time step between time t0 and t0+Δt Eq (211) can be rewritten as

H z t t H z tV z Q z

S zd

t

t t( ) ( )

( ) ( )

( )0 0 00

0+ = + prime primeprime

+

ΔΔ

τ (212)

where z0 axial coordinate (m) of a coolant at time t0 which is positioned at z at time t0+Δt

Next the active length of fuel rod is divided equally into 100 sub-segments irrespective of

the segment lengths designated by input data Fig212 shows the enthalpy increase model

for this 100-sub-segment in which the axial elevation z is set to be a representative point of

sub-segment j and a sub-segment containing the axial location z0 is set to be k

When a coolant which is positioned at z at time t0+Δt moved from z0 to z if we designate

the time required for the coolant to pass through the segment i to be δti a time step increment

Δt can be expressed as follows

Δt t t t tk k j j= + + + ++ minusδ δ δ δ1 1helliphellip (213)

By rewriting the integral in Eq(212) using the average values of V Q and S of the

segment i Vi Qi and Si and also δti enthalpy of the segment j Hj at t0+Δt can be obtained as


- 21 -

H t t H tV Q

Stj

i i

ii k

j

i( ) ( )0 0 0+ = +=Δ δ (214)

Δti can be obtained as follows using the flow velocity at the segment i as ui (ms)

δ tL

u

L

G V

S L

WVii

i

i

i i

i i

i

= = =Δ Δ Δ

(215)

where

ΔLi i-th sub-segment length (m)

Gi mass flow rate per unit cross sectional area (kgm2s)

W mass flow rate (kgs)

And ΔLi can be obtained as

node j+1

node j

node j-1

node k+1

node k

segm j

segm j-1

segm k

z

zk+1

z0

δ t j

δ t j minus1

δ tk

z

zk

Fig212 Coolant enthalpy increase model


- 22 -

ΔL

z zi j

z z k i j

z z i ki

j j

i i

k

=

minus=

minus ne neminus =

+

+

+

1

1

1 0

2( )

( )

( )

(216)

H0(t0) can be obtained by interpolating the enthalpies at zk and zk+1 as

( ) ( ) ( ) ( )( )H t H t H t H tz z

z zk k kk

k k0 0 0 1 0 0

0

1

= + minusminusminus+

+

(217)

Fig213 shows the calculation sequence Here the enthalpy at a representative point

on a segment is the average value of enthalpies at the upper and lower nodes of the segment

Here FEMAXI does not accommodate single-steam-phase inlet conditions with respect

to the relationship between temperature and pressure When inlet temperature and pressure of

coolant are given by input the steam table incorporated in FEMAXI-7 is capable of

identifying the inlet temperature as a temperature which exceeds the saturation temperature

due to the rounding of figures and numerical errors in the inlet temperature data In such cases

input enthalpy cannot be determined Thus when inlet temperature is determined to be the

saturation temperature or higher at the coolant pressure the inlet temperature is assumed as

the saturation temperature for a single liquid phase in the code and the inlet enthalpy may be

determined (Refer to name-list parameter SUBCL)

Based on the enthalpy distribution obtained at the 100 equal-length sub-segments by the

above method the enthalpy distribution for the input axial segments is derived by

interpolating Eq(217)

A flow of the calculation is shown in Fig213


- 23 -

Inlet enthalpy

Inlet temperature

Pressure

Mass flow rate

Heat flux at a segment

Inlet flow

velocity

A part of a flowing coolant X ascends and when it reaches a position of a node of an axial segment the enthalpy at the part X in the previous time step is calculated using axial distribution of the enthalpy in the previous step and the axial position of the part X When the part X is judged to be positioned below the coolant inlet the enthalpy of the position X in the previous step is set to be equal to the enthalpy of the coolant at its inlet

Calc of enthalpies at the nodes of 100 sub-segments

Enthalpies of the 100 sub-segments at the former time step

Calc of enthalpy at the representative point of

each node of the axial segments by

interpolating the enthalpies of nodes of the 100

sub-segments

Fig213 Process of enthalpy increase calculation

212 Determination of cladding surface heat transfer coefficient

In FEMAX cladding surface heat transfer coefficient is determined mainly by

RELAP5MOD1 model(22)

(1) Classification of boiling state Boiling is classified into saturation boiling and surface boiling (sub-cool boiling)

depending on the liquid temperature into in-tube boiling and pool boiling depending on the

presence of flow and into nucleate boiling transition boiling and film boiling depending on

the mode of phenomenon Figure 214 qualitatively shows the boiling modes and heat


- 24 -

transfer characteristics in terms of the heat flux and magnitude of super-heating of the heat

transfer surface The magnitude of super-heating of the heat transfer surface ΔTsat shown on

the horizontal axis in Fig 214 can be given by

satwsa TTT t minus=Δ (218)

where Tsat is obtained from coolant pressure and temperature using the steam table

Fig 214 Classification of boiling and heat-transfer characteristic curve

(2) Single-phase heat transfer In the single-phase region (non-boiling region) shown in Fig 214 the Dittus-Boelter

equation is used for the FEMAXI calculation of in-tube forced convection heat transfer

coefficient

Dittus-Boelter equation(23)

( )Bw TThq minus= (219)

Here

8040

e

RePr0230D

Kh =

Hea

t flu

x l

og q

D

A Boiling initiation pointB Critical heat flux pointC Local minimum heat

flux point

Single phase

Nucleate

boiling

Transient

boiling

Film boiling

Magnitude of super-heating of heat transfer surface logΔTsat

A

E

B

C


- 25 -

μeRe DG=

where k coolant thermal conductivity (WmK)

De equivalent thermal hydraulic diameter (m)

V coolant velocity (ms)

ρ coolant density (kgm3)

μ coolant viscosity (kgms)

Pr Prandtlrsquos number

TB coolant temperature (K)

K thermal conductivity (WmK)

Re Reynoldrsquos number(－)

G mass flow rate(kgm2s)

(3) Nucleate boiling heat transfer In nucleate boiling temperature of a heat-transfer surface decreases periodically near the

saturation temperature due to rapid vaporization of thin liquid films remaining at the bottom

of bubbles which leads to the realization of extremely high thermal conductivity

In FEMAXI-7 nucleate heat transfer is given by the sum of heat transfer from surface to

water in sub-cooled state and heat transfer from surface to water in nucleate-boiling state

The Dittus-Boelter equation given by Eq(219) is applied to calculate the heat transfer

from heated surface to sub-cooled water and either Chenrsquos equation or Jens-Lottes equation is

used to calculate the heat transfer from surface to nucleate boiling water

The equation for the nucleate heat transfer is

( ) ( ) ( )satWBBWWBW TThTThTThq minus+minus=minus= (2110)

where q heat flux (Wm2)

h cladding surface heat transfer coefficient (Wm2K)

hw heat transfer coefficient to single phase (non-boiling sub-cooled) water (Wm2K) determined by using the Dittus-Boelter equation

hB heat transfer coefficient to boiling water (Wm2K) determined by using either Chenrsquos equation or Jens-Lottes equation

Tw cladding surface temperature (K)

TB coolant temperature (K) and

Tsat saturation temperature (K)


- 26 -

In Chenrsquos equation heat flux of the nucleate boiling is expressed using the overheating of

surface ΔTsat and mainly applied to the coolant with a large void ratio (cross-sectional area

of a flow channel occupied by gas phasetotal cross-sectional area of the flow channel) in

power transient or in the early phase of LOCA

The Jens-Lottes equation can be applied to regions with a low void ratio and is applied

to mainly sub-cool boiling of water or in BWR condition

Chenrsquos equation (24)

SPTH

CKh 750240

sat240g

240fg

290f

50

490f

450

fP790

fw 00120 ΔΔ=

ρμσρ

(2111)

hw cladding surface heat transfer coefficient (Wm2K)


CP Specific heat at constant pressure (Jkg)

ρ density (kgm3)

σ surface tension(Nm)

μ dynamic viscosity(kgms)

Hfg latent heat (Jkg)

satTΔ magnitude of super-hearing of surface (K)

coolsat PPP minus=Δ

Psat Pressure at which cladding surface temperature is the saturation

temperature(Nm2)

Pcool coolant pressure (Nm2)

S Chenrsquos suppression factor (see Table21)

Here suffices f and g represent liquid phase and gas phase respectively

Jens-Lottes empirical equation(25)

hP

WW

N= sdottimes

012636 201 106

0 75 exp

φ (2112)

hW surface heat transfer coefficient (Wcm2K)

PW coolant water pressure (Nm2)

φN cladding outer surface heat flux (Wcm2)


- 27 -

Table21(1) CHENrsquoS REYNOLDS NUMBER FACTOR F

10

f

g

50

g

f

90

e

e

1

minus

=μμ

ρρ

x

xr

ex quality (-)

p density (kgm3)


R F

00 100

01 107

02 121

03 142

04 163

06 202

10 275

20 430

30 560

40 675

60 910

100 1210

200 2200

500 4470

1000 7600

4000 20000

Fitting parameter AKFAC The cladding surface heat transfer coefficient is adjusted using the following equation

(4) Burn-out and critical heat flux A phenomenon in which the heat transfer characteristic deteriorates due to rapid change

in the boiling mode is generally called boiling crisis and the heat flux with which this boiling

crisis occurs is called critical heat flux There are two types of boiling crisis one generated

Table21(2) CHENrsquoS SUPPRESSION

FACTOR S 251

fRe F S

103 1000

104 0893

2x104 0793

3x104 0703

4x104 0629

6x104 0513

105 0375

2x105 0213

3x105 0142

4x105 0115

6x105 0093

106 0083

108 0000

Selection of model by JL The equation to calculate the nucleate boiling heat transfer coefficient is selected using the input parameter JL JL=0 Chenrsquos equation JL=1 Jens-Lottes equation Default=1


- 28 -

from the transition from nucleate boiling to film boiling and another due to the breakage of

ring-type liquid films etc The former boiling crisis is generated in the case of pool boiling

or forced convection boiling in a low-quality or sub-cool region which is defined as departure

from nucleate boiling (DNB) and the critical heat flux (CHF) in this case is generally called

burn-out heat flux In contrast the latter boiling crisis is generated in the case of high-quality

forced convection boiling which is generally called dry-out

Since so-called burn-out is accompanied by the rapid damage of heat-generating body by

burning it is also called high-speed burn-out In contrast increase in the temperature of

heat-generation body after the boiling crisis is generally mild in the dry-out and therefore

dry-out is called low-speed burn-out In FEMAX Bjornard and Griffithrsquos model(26) is used

to distinguish high-speed burn-out generated by the transition from nucleate boiling to film

boiling from low-speed burn-out ie breakage of ring-like liquid film In the model to

distinguish whether the boiling crisis is high-speed burn-out generated in low-quality regions

or low-speed burn-out generated in high-quality regions the flow velocity ratio of saturated

steam Jg and the flow velocity ratio of saturated water Jf are used as a flow velocity to

normalize the generation rate of bubbles

The procedure is as follows

( ) 21gfgee

g

minusminus= ρρρDGxJ g (2113)

( ) ( ) 21gffee

f 1 minusminusminus= ρρρDxGJ g (2114)

gJ velocity ratio of saturated steam(－)

fJ velocity ratio of saturated water(－)

ex quality(－)

G mass flow rate (kgm2s)

g gravity acceleration (ms2) gρ density of saturated steam (kgm3)

fρ density of saturated water (kgm3)

The following conditions are used to identify the high-speed and low-speed conditions

(2115)

12 12g f crit

12 12g f crit

12 12crit g f crit

(low speed)

12 (highspeed)

12 (interpolation between low speedand highspeed)

J J J

J J J

J J J J

+ le + ge times lt + lt times


- 29 -

Here 136 is used for jcrit under normal flow condition

The quality (gas-phase mass flowtotal mass flow) is given by

( ) fgse HHHx minus= quality (2116)

H enthalpy (Jkg)

Hs saturation enthalpy (Jkg)


Next calculation to obtain critical heat flux (CHF) is explained Tong derived the W-3

equation to obtain an optimally fitted curve of experimental values of DNB heat flux qDNB in a

pressurized water channel of uniform heat generation This is used to evaluate CHF in

high-speed burn-out(27)

( ) ( )( ) ( )[ ]( )

( )

( ) inf7

e

e

eee

e78

863DNB

1041292382580

05487124exp8357026640

86901571

03713563117290596114840

1098705517718exp104268117220

10237960222101543

HH

D

x

Gxxx

xPP

PqqDNBW

minustimes+timesminus+times

minustimes++minustimes

timesminustimesminus+

timesminustimestimes==

minus

minusminus

minus

(2117)

Here

qDNB DNB heat flux(Wm2)

P pressure(Nm2)

Hf saturated enthalpy (liquid phase) (Jkg)

Hin inlet enthalpy (Jkg)

The applicable range of Eq(2117) is

69times106≦P≦158times106 (Pa) 14≦G≦68(kgm2s)

ex ≦01 Hin≧93times105(Jkg)

When xegt01 a correction equation of W-3 equation was presented by Hsu and Beckner(28) as

( )inf

e 03DNB 960761HH

xW DNBqq

==minustimes= α (2118)

α void fraction(－)

Meanwhile for evaluation of CHF in low-speed burn-out a modified version of Zuberrsquos

equation(29) is used


- 30 -

( ) ( )[ ]21

gf

gf250gffgDNB 960231640

+minusminus=

ρρρρ

ρρσα gHq (2119)


ρ density(kgm3)

σ surface tension (Nm)

g gravity acceleration (ms2)


(5) Heat transfer in transition-boiling and film-boiling When heat flux q is increased in a nucleate boiling region the boiling mode rapidly

changes and heat transfer deteriorates When heat flux exceeds the critical heat flux (point B

in Fig 214) the heat transfer surface is covered with a steam film and the heat flux value

moves rightward from point B in Fig 214 along the dotted line and jumps onto point D

Point D is in a film-boiling region nucleate boiling suddenly transits to film boiling

Conversely when the heat flux gradually decreases starting from point D it reaches point C

along the curve However when it passes point C the phenomenon suddenly changes into

that of the nucleate boiling where heat flux moves to point E (on A-B) along the dotted line in

Fig 214 Point C is the local minimum heat flux In the region between B-C heat flux

decreases as ΔTsat increases and extremely unstable phenomena are observed This region is

called the transition boiling region

Among these transitions of boiling rapid changes in the boiling mode from nucleate

boiling to film boiling are incorporated in the boiling transition analysis in FEMAXI As the

evaluation equation of heat transfer coefficients in the transition boiling and film boiling after

high-speed burn-out a modified evaluation equation by Condie-Bengston(22) is used that is

( ) satFBTBtotal Thhq Δ+= (2120)

qtotal heat flux(Wm2)

( )DNB

satsatDNBFBDNBTB 22

expDNB T

TTTqqh

T ΔΔ

ΔminusΔminus=Δ

(2121)

hTB transition boiling heat transfer coefficient (Wm2K)

qDNB critical heat flux(Wm2) refer to eqs(2117) and (2118)

DNB

FB Tq

Δ heat flux by film boiling at DNB(Wm2)

DNBFBFBDNB

ThqT

Δ=Δ

DNBTΔ satTΔ at DNB (K)


- 31 -

( )[ ]

( ) [ ]w7

590282e

78420e

1ln2456060040g

30702w

43760g

FB Pr1045041exp89471131

RePr0810330

e

PxD

Kh

xminus

++

timesminus+

=

(2122) hFB film boiling heat transfer coefficient (Wm2K)

K thermal conductivity(WmK)

Pr Prandtlersquos number(－)


P pressure(Nm2)

Here suffix w represents wall temperature and suffix g represents the value obtained

using saturation temperature

On the other hand as an evaluation equation for the heat transfer coefficient at transition

boiling or film boiling after low-heat burn-out the following equation(22) is used

vaporliquidtotal qqq += (2123)

( ) ( )fwc TThTThq minus=minus= αfwvvapor (2124)

( )

( )

=330

filmfilm

250film

film

c

PrGr738170

PrGr151443max

e

e

D

KD

K

h (2125)

( ) ( )( )2filmfwfilm

3e 2Gr μρβ TTD minus= g (2126)

qvapor heat flux of steam(Wm2)

hv thermal conductance of steam(Wm2K)

Tw cladding surface temperature(K)

Tf coolant temperature(K)


Gr Grasshof number(－)

β thermal expansion coefficient(K-1)

ρ density(kgm3)


Kfilm is obtained using a value at 2

fwfilm

TTT

+=


- 32 -

( ) ( )

( )

ge

leΔ

gtΔminusΔ+

=

9600

1

1960

satDNB

satsatFBTB

liquid

α

α

BTq

BTThh

q (2127)

( ) ( ) satTB

BTTB ehqqBh Δminus

=Δ

minusminusminus= 1FBvaporDNB

sat960712 α (2128)

( ) 250

satge

fggfg3

g

1720

c

eFB 526775

Δminus

=

TD

HKDh

μρρρ

λg

(2129)

( )50

gfc

minus=

ρρσλ

g (2130)

( ) ( )( ) ( )

1 2

1 2

001368 K at 4137000 N m

001476 K at 6205500 N m

B PB

B P

minus

minus

= == = =

B is obtained by interpolation on a log-log graph with respect to P

qliquid heat flux of the liquid phase (Wm2)

hTB heat transfer coefficient at transition boiling (Wm2K)

hFB heat transfer coefficient at film boiling (Wm2K)

Hfg latent heat(Jkg)


Here suffixes f and g represent liquid phase and gas phase respectively

(6) Heat transfer at steam single-phase When the quality xe becomes 1 the coolant is in a steam single-phase (superheated

steam) accordingly Dittus-Boelterrsquos equation (Eq(219)) is applied again

(7) Condensation heat transfer For the heat transfer at condensation two-phase flow Coillierrsquos evaluation equation(210)

is used Here suffixes f and g represent liquid phase and gas phase respectively


- 33 -

( )q h T Tw sat= minus (2131)

h = max(hlaminar hturbulent) (2132)

( )

( )

2503f

min 2960

minusminus

=wsatfe

fggffarla TTD

KHh

μρρρ g

(2133)

ef

ggpfff

e

gg

f

pff

f

ff

turbulent

D

VCK

D

V

K

CKh

μμρ

μμμ

ρ

310857749

02300650

minustimes=

=

(2134)

q heat flux(Wm2)


Tsat saturation temperature(K)

Cp specific heat(Jkg)

V velocity(ms)

213 Flow of thermal-hydraulics calculation The logical flow of calculations of coolant temperature and cladding surface heat transfer

coefficient in FEMAXI is shown in Fig215 As the figure indicates coolant temperature at

each axial segment is calculated from the coolant pressure and enthalpy at each segment by

using the steam table Also thermal conductivity viscosity density and Prandtle number etc

are calculated by referring to the steam table For this purpose FEMAXI has a built-in steam

table


- 34 -

Fig215 Flow of thermal-hydraulics calculation

Thermal conductivity viscosity density Prandtle number etc

Calculation of enthalpy at a point positioned at each

node of the sub-segments in the previous step

Inlet enthalpy

Inlet temperature

Pressure

Heat flux at each segment

Inlet flow

velocity

Enthalpy at the segment

Mass flow rate

Enthalpy distribution of sub-segments at previous time step

Pressure

Temperature at the segment

Low-flow

high-flow

NO

NO

YES

YES

NO

Water single-phase

Is heat flux equal to or less than critical heat flux

Is the quality 099 or higher

Coolant flow rate

Either Chenrsquos or Jens-Lottesrsquos eq+

Dittus-Boelterrsquos eq

Dittus-Boelterrsquos equation RETURN

Dittus-Boelterrsquos eq (steam single phase)

Equation for transition boiling + film boiling (modified

Condie-Bengstonrsquos equation)

RETURN

RETURN

RETURN

Equation for transition boiling + film boiling (RELAP-5)

(

RETURN

YES

Is the magnitude of super-heating of heat transfer surface negative

and the quality positive

Heat transfer equation for the

condensation two-phase flow

(RELAP-5)

RETURN


- 35 -

214 Transition process of coolant condition

Transient boiling phenomenon modeled in FEMAXI is not the transient boiling

accompanying the heat flux change but the transient boiling accompanying the flow rate

change Therefore heat transfer characteristics are obtained from the relationship between

the quality and magnitude of super-heating of the heat transfer surface rather than from the

relationship between the heat flux and magnitude of super-heating of heat transfer

Figure 216 shows the classification of the heat transfer characteristics in FEMAXI

modeling Table 22 lists the conditions set for each heat transfer mode Here using Fig

216 outline of the calculation of the boiling transition performed in FEMAXI is explained

In the figure ΔTsat on the vertical axis shows the magnitude of super-heating of the heat

transfer surface namely the amount of increase from the saturation temperature The quality

xe on the horizontal axis is defined by Eq(2116)

Fig216 Classification of heat transfer characteristics based on quality magnitude

of super-heating magnitude of sub-cooling and magnitude of super-heating

of the heat transfer surface

[ ]E prime Forced

convection to wall

[A] Forced convection to water

(TwgtTB)

(TwltTB)

(TwltTB)

[E] Forced convection to steam

[ ]Aprime Forced convection to wall

satTΔ

1

[C] transition boiling

[Β] Nucealte boiling

[D] film boiling

0

[F] Condensation

two-phase flow

xe quality

(TwgtTB)

Tw cladding surface temperature

TB coolant bulk temperature

x

Superheated steam region

Sub-cool region


- 36 -

Here since xe is quality (= gas phase mass flow ratetotal mass flow rate) 0lexele1

should hold However we obtain the following results of the calculation using Eq (2116)

when HltHs (in a case of sub-cooling water) xelt0

when HgtHs+Hfg (in a case of super-heated steam) xegt1

Accordingly output of the calculation is expressed as QUALITY when 0ltxe lt1

magnitude of super-heating of the steam (MSUPHTG) when xe ge1 magnitude of sub-cooling

(MSBUCOOL) when xe le 0

(1) Normal operation In normal operation the coolant states belong to either region [A] or [B] In other

words when the cladding surface temperature is equal to or less than the saturation

temperature (ΔTsat le 0) the coolant state is in a region of forced convection heat transfer from

the cladding surface to water phase (region [A]) When the cladding surface temperature

exceeds the saturation temperature (ΔTsat gt 0) it is in a region of heat transfer of nucleate

boiling (including sub-cool boiling) (region [B])

(2) Decrease in the coolant flow rate When the heat flux remains constant and the coolant flow rate is decreased the quality

increases and gradually reaches the critical heat flux Consequently the coolant state is

transited to region [C] or [D] In the high-speed burn-out in a low-quality region the

cladding wall temperature rapidly increases whereas in the low-speed burn-out in a

high-quality region the cladding wall temperature gradually increases These behaviors are

determined by the heat flux at the time of change in the flow rate and the value of flow rate

after the decrease in the flow rate

In addition when the coolant flow rate is lowered significantly the coolant is transited to

a single-phase steam region (super-heating) of [E] In region [E] ΔTsat increases with

increasing enthalpy (temperature) of the coolant Namely the cladding surface temperature is

determined by the axial position of coolant and coolant pressure at the time when the coolant

enters region [E]

(3) Recovery of coolant flow rate When the coolant flow rate recovers enthalpy of the coolant rapidly decreases with

decreasing specific volume V (see Eq(211)) In the film-boiling region [D] ΔTsat becomes

large as the quality decreases Accordingly when ΔTsat is constant if the enthalpy decreases


- 37 -

the quality decreases (see Eq(2116)) and the coolant state transits from the film-boiling

region [D] to the nucleate boiling region [B] In region [B] since the heat transfer rate is

very high the cladding wall temperature rapidly decreases and the normal operation condition

is recovered

When the coolant flow rate is recovered in the steam single-phase region [E] the coolant

temperature decreases due to the decrease in enthalpy With this ΔTsat decreases

Consequently the cladding surface temperature decreases However since it is the

temperature decrease under the condition of the quality being 1 the coolant state transits to

the nucleate boiling region [B] under very low ΔTsat value which rapidly decreases the quality

in region [B] That is the water which first enters is consumed to decrease ΔTsat and the

water which enters thereafter decreases the quality When the quality decreases sufficiently

normal operation is recovered

In region [F] (0ltxelt1) there is a flow of droplets and stream which is not related to the

boiling transition

Table 22 Set conditions for each heat transfer mode

Tw cladding surface temperature TB coolant temperature Heat transfer mode Model or eq Set conditions Convection heat transfer (single phase)

Dittus-Boelter

TW

ltTB

When cladding surface temperature le saturation temperature (water single-phase) when quality ge 099 (steam single-phase)

TW

gtTB

When quality = 0 (water single-phase) when cladding surface temperature ge saturation temperature (steam single phase)

Nucleate boiling heat transfer (two phase)

Either Jens-Lottes or Chen1 ＋Dittus-Boelter2

TW

gtTB

When cladding surface temperature ge saturation temperature (two phase) and heat flux le critical heat flux

Transition-boiling + film-boiling heat transfer (two phase high-flow)

Modified Condie- Bengston

TW

gtTB

When cladding surface temperature ge saturation temperature and heat flux ge critical heat flux and the quality le 099 (two phase) and coolant flow rate is high

Transition-boiling + film-boiling heat transfer (two phase low-flow)

RELAP-5 MOD-1model

TW gtTB

When cladding surface temperature ge saturation temperature and heat flux ge critical heat flux and quality le 099 (two phase) and coolant flow rate is low

Condensation (two phase liquefaction Collier

TW

ltTB

quality is positive and cladding surface temperature le saturation temperature (two phase)

1 0w satT Tminus gt Even if the coolant enthalpy does not reach the saturation enthalpy ie

2 0sat BT Tminus gt if the coolant is unsaturated the flow in the vicinity of cladding surface is

assumed to be a two-phase flow


- 38 -

215 Equivalent diameter and cross-sectional area of flow channel

Equivalent thermal-hydraulics diameter and cross-sectional area of the coolant flow

channel are the parameters required for obtaining cladding surface temperature The

equivalent diameter De is the diameter of a virtual column when the shape of a coolant

channel around the fuel rod is converted to a column De is used in Eq(219) to determine

the cladding surface heat transfer coefficient and the cross-sectional area of the channel is

used in Eq(211) to calculate the enthalpy of each axial node

Usually these two parameters are directly set by input in the case that one of them is not

set or none of them is set a method of converting one value to the other or of calculating both

values using the rod pitch is incorporated in FEMAXI These methods are explained below

(1) Equivalent diameter of the channel De A) When De is specified by input data the input value is used

B) When De is not input but the cross-sectional area of the channel S is input De is obtained

using

DS

re = 4

2π (2135)

where S cross-sectional area of the channel

Name-list input parameter FAREA

r fuel rod outer radius

eD Name-list input parameter DE

C) When neither De nor S is input these values are

calculated using the pitch between fuel rods Assuming

channel III in Fig217 S is given by

S l r= minus4 2 2π (2136)

where l half of the pitch between fuel rods

Rod pitch input name-list parameter PITCH

De is calculated using Eq(2135) after obtaining S

S

Symmetrical line

l1

Fig217 Channel model


- 39 -

(2) Flow cross section area

A) When flow cross sectional area S is specified by input the input value is used

B) When flow cross sectional area S is not specified but the equivalent diameter De is

specified by input the flow cross sectional area S is determined by

4

2 eDrS

π= (2137)

22 Cladding Waterside Corrosion Model

In the high burnup region waterside corrosion (oxidation) of a cladding advances and its

effect on heat transfer and mechanical behavior of the cladding cannot be neglected Growth

rate of oxide layers depends on temperature and it is calculated in FEMAXI as a function of

temperature at the boundary between the metal substrate and oxide layer This boundary

temperature is obtained by the thermal calculation explained in the following section

Thermal conductivity of oxide layers is specified by name-list input IZOX Specific heat

of the oxide layer is given by MATPRO-A(211)（Refer to section 42）

A typical one is the MATPRO-A model as follows(211)

CN = 0835 + 181times10-4T (221)

Specific heat of oxide layers is given by MATPRO-A(211) as

272 1014110116565 minusminus timesminustimes+= TTCP (222)

(300 1478 )T Kle le

The oxide growth (thickness increase) model is an empirical one and designated by input

In this initial thickness of oxide can be specified segment-wise by input In any models of the

oxide growth the corrosion rate is given as a function of temperature at the metal-oxide

interface This temperature is calculated in the thermal analysis

In Eq(221) since the value of the temperature-dependent term is approximately one

order of magnitude lower than the constant term the temperature gradient across the oxide

layer can be approximated by a straight line (Fig222) Therefore in FEMAXI thermal

conductivity is assumed to be uniform inside the oxide layer and to determine the thermal

conductivity in Eq(221) the average value of the oxide surface temperature and the

boundary temperature is designated as the temperature of the oxide layer

Decrease in the thickness of cladding metal phase with progress of corrosion is


- 40 -

quantitatively related to the volume of oxide layer In Zircaloy volume of the oxide expands

approximately 156 times from that of the volume of the metal phase consumed for oxide

generation This ratio is called the Piling-Bedworth ratio (PBR) (Fig 221)

Using PBR decrease in the thickness of cladding metal part and increase in the cladding

outer diameter can be given as SPBR and S(PBR-1)PBR respectively where S is the

oxide layer thickness Both the thermal calculation and the mechanical calculation are

performed with this changing thickness of radial ring element by re-setting the element

thickness at each time step

Name-list input 1) Option of oxidation modelICORRO =0 No oxidation =1 EPRI model(212)

=2MATPRO-09 (PWR) model(213) =3 MATPRO-09 (BWR) model (Default=1)

2) Oxidation rate tuning factorRCORRO (Default=10) Oxidation rate is multiplied by

RCORRO times

3) Oxide thermal conductivity factor OXFAC (Default=10)

thermal conductivity of oxide NC is adjusted by OXFACCC NN times=

4) PX radial fraction of volumetric increase in X=PBR-10 by oxide growth ()(default=99)

5) PBR volumetric increase ratio by oxide growth ie Piling-Bedworth ratio (default=156)

[Input value initial oxide layer thickness OXTH(40)] Initial oxide layer thickness is designated for each segment using OXTH (microm) Standard value is 01 microm

Temperature

Cladding meta l part

Meta l par t a f ter oxidat ion

Oxide layer Cladding

Meta l part

Oxide

Radius

( )Stimes PBR -1S

PBR PBR

Co

ola

nt

Fig 221 Volume increase due to cladding corrosion

Fig222 Temperature distribution inside cladding


- 41 -

23 One-Dimensional Temperature Calculation Model 231 Selection of number of elements in the radial direction

In performing the thermal and mechanical analyses in FEMAXI it is possible to select a

mesh system ie number of ring elements in the pellet stack and cladding of one axial

segment in one-dimensional geometry as follows (this is explained in the mechanical analysis

chapter again)

(1) Number of ring element divisions in pellet stack

As shown in Table 231 a name-list parameter MESH can designate the number of ring

elements of pellet stack The default is ldquoMESH=3rdquo In this iso-volume ring elements of pellet

stack thickness of the ring decreases as a function of radial location of the ring as shown in

Fig231 In the peripheral region of pellet there is a peaking of heat generation density and

burnup and the rim structure is formed so that a fine digitizing of this steep change region

will markedly contribute to the accuracy and preciseness of calculation

However when the HBS model (rim structure model) which will be later described is

activated by input it is necessary to always input MESHgt0 If the HBS model is activated

MESH=3 is automatically set even if MESH=0 is designated by input

Table 231 Number of radial ring elements of pellet stack in thermal and mechanical analyses Parameter value

1-D thermal analysis 1-D mechanical analysis

2-D local PCMI analysis

MESH=0 Iso-thickness 10 ring elements

Iso-thickness 10 ring elements


MESH=1 Iso-volumetric 36 ring elements Iso-volumetric 18

ring elements Iso-volumetric 9 ring elements

MESH=2 Iso-volumetric 72 ring elements




[Adjustment parameter FCOPRO]

Corrosion rate can be adjusted by the following parameters ( )FCORROdt

dS

dt

dS += 1


- 42 -

When name-list parameter ISHAPE=1 is selected by input the number of ring elements

in MESHgt0 is doubled which allows a finer calculation

(2) Number of ring element divisions in cladding wall Ring element division in cladding is common to both the 1-D thermal analysis and 1-D

ERL mechanical analysis irrespective of the MESH value designation shown above That is

A) Cladding with no Zr-liner

8 iso-thickness rings for metallic wall + 2 iso-thickness rings for outer surface oxide layer

B) Cladding with Zr-liner

2 iso-thickness rings for Zr-liner + 8 iso-thickness rings for metallic wall + 2 iso-thickness

rings for outer surface oxide layer

In the 2-D local mechanical analysis

C) Cladding with no Zr-liner

4 iso-thickness rings for metallic wall + 1 ring for outer surface oxide layer

D) Cladding with Zr-liner

One ring for Zr-liner + 4 iso-thickness rings for metallic wall + 1 ring for outer surface

oxide layer

However when ISHAPE=1 the number of rings is doubled from the above in the 2-D

local mechanical analysis

Plot output of cladding deformation In the ERL mechanical analysis in outputting the plotted

figures of cladding diameter or radial displacement IDNO=174 refers to the position of metal-oxide

interface and IDNO=7 and 11refer to the oxide outer surface

In the local 2-D mechanical analysis mechanical properties of oxide layer is not considered


- 43 -

The geometry with these 36 or 72 iso-volume ring elements in pellet stack is effective in

performing a precise calculation of thermal conduction by using a detailed power density

profile in the radial direction This holds for example in loading the result file of the

RODBURN code (36 ring elements) or PLUTON code (36 72 or more ring elements) or

even more detailed profile into FEMAXI Furthermore the 36 or 72 ring elements

geometry is indispensable in conducting the HBS (rim structure) model of pellet which is

described in section 2121 However the 36- or 72-ring elements geometry itself can be used

when the rim structure model is not applied The 10-ring elements model can be still used

as it was in FEMAXI-7 In this case the PLUTON file and the HBS model cannot be used

In chapter 3 the ring element number is explained again

232 Determination of cladding surface temperature

In FEMAXI Eq(231) is obtained from the finite difference equation of thermal

conduction inside a fuel rod (see section 233)

φN = ATN + B (231)

where φNcladding surface heat flux (Wm2)

ＴNcladding surface temperature (K)

N cladding surface mesh number

Fig231 Ring elements and analytical geometry of pellet stack and cladding

of one axial segment

00 02 04 06 08 10

ZrO2

1 2 3 4 5 6 7 8

Cladding

ring element number

Pellet stack 36 equal-volume ring elements

Relative radius of pellet

Gap or Bonding layerCenterline

Relative power density


- 44 -

Here A and B are the coefficients determined by the finite difference equation of thermal

conduction Surface temperature TN can be obtained from the heat transfer coefficient of a

cladding surface determined using Eq(231) That is when coolant temperature is

designated as TB (K) heat flux at cladding surface can be given as

φN = hW(TN ndashTB) (232)

where hW surface heat transfer coefficient (Wm2K)

φN cladding surface heat flux (Wm2)

TN cladding surface temperature (oxide surface ) (K)


TN can be obtained by eliminating φN from Eqs(231) and (232)

However as stated in the previous section FEMAXI has an optional mode in which

cladding surface temperature is not calculated but given by input This mode has two options

option-1 is to specify the cladding surface (oxide surface) temperature and option-2 is to

specify the temperature at metal-oxide interface In the option-1 name-list parameter ITCP=0

is set and rod temperature calculation is performed with the oxide surface temperature as

boundary condition In the option-2 ITCP=1 is set and the temperature at the oxide surface is

obtained by

ln(2 )N

N OXOX OX

rqT T K rπ

= minus sdot

(233)

where

TN cladding surface(oxide surface) temperature (K)

TOX metal-oxide interface temperature given by input (K)

q linear heat rate (Wm)

KOX oxide thermal conductivity (WmK)

rN cladding (oxide) surface radius (m)

rox radius at metal-oxide interface (m)


- 45 -

233 Solution of thermal conduction equation

In the analysis of temperature gradient in the radial direction at each axial segment

one-dimensional thermal conduction equations are used while the heat conduction in the axial

direction is neglected It is assumed that the thermal properties of the fuel at each ring

element are dependent on temperature Based on these assumptions thermal conduction at

one axial segment is described as follows

( ) ( )[ ] ( ) ( ) ( )partpart t

c T r T r t k T r T r t q r tv = nabla sdot nabla + (234)

Here

T temperature (K)

r coordinate in the radial direction (m)

Cν volume specific heat (Jm3)

k thermal conductivity (WmsdotK)

q heat generation per unit volume (Jm3sdots)

n-1 n n+1

kpn

Cvpn

qpn

ksn

Cvsn

qsn

hpn hsn

hpn 2 hsn 2

Fig 232 Ring element model

Name-list Parameter IS ICTP Setting either IS=3 or =4 enters the mode in which cladding surface temperature history is given by input In this mode ICTP=1 specifies the temperature at metal-oxide interface (default) ICTP=0 specifies the temperature at oxide surface When either IS=3 or =4 thermal-hydraulic calculation is essentially unnecessary Therefore if additionally IS3P=0 is set thermal-hydraulic calculation is omitted and calculation time is reduced IS3P=1 thermal-hydraulic calculation is On (default) IS3P=0 Off To plot out the cladding surface heat flux (IDNO=62) and surface heat transfer coefficient (IDNO=63) IS3P=1 should be set


- 46 -

Volume integration is performed with Eq(234) Here the volume to be integrated is

that enclosed by the dotted lines in Fig 232

( ) ( ) ( ) ( ) ( ) V v V VC T r T r t dV k T r T r t dV q r t dVt

part = nabla sdotnabla +

part (235)

Since this is a one-dimensional problem dimensions of the volume are set as unity except

for the radial direction Eliminating the common factor 2π and using the forward finite

difference method for time differentiation the first term of Eq(235) is expressed as

( ) ( ) ( ) ( )

( )1

2 4 2 4

V v V v

m mn n pn pn sn sn

v pn n v sn npn

TC T r T r t dV C T r r t dV

t t

T T h h h hC r C r

h

+

part part asymp part part

minus asymp minus + +

(236)

where Tnm is the temperature at coordinate rn and time tm and 1+m

nT is the temperature at

coordinate rn and time tm+1 The second term of Eq(235) is expressed as

( ) ( ) ( ) ( )( ) ( )1 1 1 1

1 1

[ ]

2 2

V s

m m m mpn n n sn n npn sn

n npn sn

k T r T r t dV k T r T r t d S

k T T k T Th hr r

h h

+ + + +minus +

nabla sdotnabla = nabla sdot

minus minus asymp minus minus + +

(237)

Here continuity conditions of the heat flux at the inner boundary are applied for

evaluation of the plane integration along the boundary surface

The following terms are defined for convenience sake

hh

rh

h rh

hh

rh

hh

rh

pnV pn

npn

snV

nsn

pns

pnn

pnsns

snn

sn

= minus

= +

= minus

= +

2 4 4

1

2

1

2

(238)

D C h C hn v pn pnV

v sn snV= +

Then the heat generation term q(r t) of the third term in Eq(235) is separated with

variables as

( ) ( ) ( )q r t Q r P t= (239)

Here Q(r) is the relative power generation distribution in the radial direction and P(t) is a

standard value of the heat generation density as a function of time Then the third term of


- 47 -

Eq(234) is expressed as

( ) ( ) ( ) V VV pn pn sn sn mq r t dV Q h Q h P t θ+ asymp + (2310)

By adding the right sides of Eqs(236) (237) and (2310) which are approximate

expressions of the terms in Eq(235) Eq(235) is converted into the difference

approximation equation for the n-th mesh point as follows

( ) ( )( )( ) ( )

1

1

1

nm n n m m s

n n pn pn

m m sn n sn sn

V Vpn pn sn sn n

T T DT T k h

t

T T k h

Q h Q h P t

θ θ

θ θ

θ

+ + +minus

+ ++

+

minus= minus minus

Δ+ minus

+ +

(2311)

Upon application of the Crank-Nicholson method a fully implicit method ( 10θ = ) to

Eq(2310) the following equation holds

( ) ( ) ( )

( ) ( ) ( )

T T D

t

T T T Tk h

T T T Tk h

Q h Q h PP t P t

nm

nm

n nm

nm

nm

nm

pn pns

nm

nm

nm

nm

sn sns

pn pnV

sn snV

fm m

+ +minus

+minus

++

++

+

minus=

+minus

+

+ + minus +

+ ++

1 11

11

11

11

1

2 2

2 2

2

Δ

(2312)

Here thermal conductivity kpn and ksn and Tm+1 and Tm are derived from the average of

temperatures at n-1-th and n-th elements and temperatures at element nodes n and n+1

respectively

Through modification of Eq(2312) the differential equation at ring point n in the

medium region is given as

a T b T c T dn nm

n nm

n nm

nm

minus+ +

+++ + =1

1 11

1 (2313)

where

( )

( ) ( ) ( )

ak h t

ck h t

b D a c

d a T D a c T c T

t Q h Q h PP t P t

npn pn

s

nsn sn

s

n n n n

nm

n nm

n n n nm

n nm

pn pnV

sn snV

fm m

= minus = minus

= minus minus

= minus + + + minus

+ ++

minus +

+

Δ Δ

Δ

2 2

2

1 1

1

(2314)


- 48 -

For the pellet-clad gap region the following equations hold as in the case of Eq (2312)

Next the P-C gap region is shown in Fig233

Pellet CladdingGap

n-1 n n+1 n+2

Tn-1 Tn Tn+1 Tn+2

hpn hsn+1

Fig 233 Gap thermal conductance model

For the gap region similarly to Eq(2312)

( ) ( ) ( )( )1

111

11

11

+

+++

+minus

++

+

minus+minusminus=Δ

primeminus

mfVpnpn

gm

nm

nspnpn

mn

mn

nm

nm

n

tPPhQ

hTThkTTt

DTT

(2315)

( ) ( ) ( )( )11

111

11

211

111

11

++

+++

++

+++

+++

++

+

minus+minusminus=Δ

primeprimeminus

mfns

snsns

mn

mng

mn

mn

nm

nm

n

tPPQ

hkTThTTt

DTT (2316)

Holds where 1 1 1V Vn v pn pn n v s n s nD C h D C h+ + +prime primeprime= =

and

gh is the gap thermal conductance

( ) ( )( )

( )

11 1 1 1 1

1 1 1

1 12 1 1 1

1 1 1

m mn n n m m s

n n g p n p n

m m sn n s n s n

Vs n s n m

T T DT T h h h

t

T T k h

Q h P t

++ + + + +

+ + +

+ ++ + + +

+ + +

primeprimeminus= minus minus

Δ+ minus

+

(2317)

Here prime = primeprime =+ + +D C h D C hn v pn pnV

n v s n s nV

1 1 1 1sn p nh h += = gap width and hg is gap thermal

conductance The gap region is shown in Fig 233

The center region of the fuel rod is shown in Fig 234(a) and the gap and cladding

surface region is shown in Fig 234(b)


- 49 -

Fig234(a) Fuel rod center model Fig 234(b) Cladding surface model

Here the oxide layer on the outer surface of cladding has a lower thermal conductivity

than that of the metal phase

For the center of the fuel rod Eq (2312) is transformed as follows

( ) ( ) ( )10000

10

110

01

0+

+++

+minus=Δminus

mfVss

sss

mmsv

mm

tPPhQhkTTCt

TT (2312a)

The boundary equation for the cladding surface region ( Nr r= ) is

( ) ( ) ( )1

11

1

1

++

minus+

+

+minusminusminus=Δminus

mfVpnpnnn

spnpn

mn

mn

Vpnpnv

mn

mn tPPhQrhkTThC

t

TT φ (2318)

The difference approximation equation (2318) is expressed as

a T b T dn nm

n nm

nminus+ ++ =1

1 1 (2319)

Then the coefficients nn ba and nd are given as

a

k h tn

pn pns

= minusΔ

2 (2320)

b C h an v pn pnV

n= minus (2321)

( )( ) ( )

d a T C h a T

t r tQ h PP t P t

d d

n n nm

V pn pnV

n nm

nnm

nm

pn pnV

fm m

n n nm

= minus + minus

minus +

+

+

= prime + primeprime

minus

++

+

1

11

1

2 2

Δ Δφ φ

φ

(2322)

However in the normal operation period the heat flux mnφ at time tm in Eq(2322) is

Pellet center

0 1

T0 T1

hs0

Pellet Surface Cladding outer surface

n-1 n=N

Tn-1 Tn=TN

hpn

φ φn N=

Gap


- 50 -

given by

φ πnm m

nq r= 2 (2323)

qlinear heat rate (Wm)

This is to avoid the accumulation of numerical difference errors which may occur

through repeated re-setting of ring element on the calculated heat flux

Here φN is the cladding surface heat flux Now unknown variables in Eqs(2318) to

(2321) are 1 11 m m

n nT T+ +minus and 1mφ +

Next using coefficients expressed as E and F which are obtained using the forward

elimination of the Gaussian elimination method starting from the first point to the (N-1)th

point the following equation is derived

T E T Fnm

n nm

nminus+

minus+

minus= minus +11

11

1 (2324)

By substituting Eqs(2319) to (2322) into Eq(2318) we obtain

A T Bnm m

11

11+ ++ = φ (2325)

where

( )A b a E dn n n n1 1= minus primeprimeminus (2326)

( )B a F d dn n n n1 1= minus prime primeprimeminus (2327)

However Eq(2325) is equivalent to Eq(231) and in Eq(231) A1 and B1 are known

values Therefore 1m

nT + can be determined as stated in section 232 Hereafter using the

backward substitution of the Gaussian elimination method Tm+1 is sequentially obtained

234 Fuel pellet thermal conductivity (1) Thermal conductivity models UO2 and MOX fuel thermal conductivity decreases with burnup Typical models ie the

NFI model(214) and Halden model(215) are shown in Fig235 The thermal conductivity model

can be chosen by name-list input parameter IPTHCN (See section 41)

Details of equations and literatures are described in section 41


- 51 -

500 1000 1500 20001

2

3

4

5

6

80 GWdtHM

40 GWdtHM

Ther

mal

Con

duct

ivity

(Wm

K)

Temperature (K)

NFI UO2

NFI MOX Pu=10 Halden UO2

Halden MOX0 GWdtHM

(2) Effect of porosity The thermal conductivity is also affected by porosity and density as shown in Chapter 4

Here between porosity p and theoretical density ratio D the following equation holds

Option for fuel pellet thermal conductivity IPTHCN Detailed in Chap4 =1 MATPRO-09 =2 Washington =3 Hirai =4 Halden (Wiesenack) =5 Modified Hirai =6 Forsberg =7 Kjaer-Pedersen =8 BaronampCouty =10 Lucta Matzke ampHastings

=11 Tverberg Amano Wiesenack (UO2 Gadlinia-containing fuel) =13Fukushima (UO2 Gadlinia-containing fuel) =15Daniel Baron (UO2 Gadlinia-containing fuel) =16 CRIEPI (Kitajima amp Kinosita)=17Halden new (Wiesenack) (UO2) =18 PNNL-modified Halden model(UO2) =90 Ohira amp Itagaki (UO2) =91 Ohira amp Itagaki (NFI) latest Model for 95TD UO2 and MOX pellets =92 Ohira amp Itagaki Modified in FRAPCON 33 for 95TD UO2 pellet =30Martin (MOX) =31MATPRO-11(MOX) =32Martin+Philipponneau (MOX) =33Duriez et al(MOX) =34Philipponneau (MOX) =35 Halden (new MOX) =36Daniel Baron (MOX UO2 Gadlinia-containing fuel) =38 FRAPCON-3 PNNL Modified model =39 FRAPCON-3 PNNL-modified Halden model

Fig235 Comparison of fuel thermal conductivity degradation with burnup


- 52 -

p(I)＝10 - D(I) (2328)

where I pellet ring element number

D(i) is defined as

D(I)=Di-(3ΔLr) (2329)

where Di initial theoretical density ratio

ΔL radial displacement due to volume change generated

by irradiation

r pellet radius

In FEMAXI it is possible to specify the definition of porosity p (or D) in Eq(2328) by

designating the parameter IPRO (default is 0) as follows

IPRO =0 Initial porosity 0p is used when D(I)=Di initial theoretical density ratio

=1 00

swgVp p

V

Δ= + is used

=2 0

0 0

dens hotV Vp p

V V

Δ Δ= + + is used

=3 00 0 0

swg dens hotV V V

p pV V V

Δ Δ Δ= + + + is used

where 0

swgV

V

Δ volumetric swelling of pellet induced by fission gas bubble growth

0

densV

V

Δ volumetric densification of pellet

0

hotV

V

Δ volumetric shrinkage of pellet due to hot-pressing

Name-list input parameter IPRO (Default=0)

However this calculation is not performed in every ring element but is performed at

each axial segment to derive the average porosity over the segment Plotting output is done by

IDNO=57 pellet density =1 pminus

Another related name-list parameter MPORO is an option to designate the assumed

effect of porosity on pellet thermal conductivity

That is

MPORO=0 pellet thermal conductivity is calculated using the segment-wise average

porosity which is calculated on the basis of IPRO (default setting)

MPORO=1 pellet thermal conductivity is calculated using the ring-element-wise

calculated porosity 00 0

swg densV V

p pV V

Δ Δ= + +


- 53 -

Here definitions of swelling and densification are equivalent to that of IPRO However

the element-wise change of pellet thermal conductivity induced by the activation of HBS

model is independent from designation of MPORO so that effect of element-wise porosity

change by the HBS model can be included in calculation irrespective of the MPORO value

The corresponding plotting outputs are done by IDNO=257 total porosity and

IDNO=260 fission gas bubble porosity

(3) Relationship between porosity and thermal conductivity models The thermal conductivity models explained in section (1) implicitly adopt p=005 as a

default value When the porosity changes from 005 with extension of burnup pellet thermal

conductivity would change To take into account of this effect IPRO is used as above It is to

be noted that some models of pellet thermal conductivity have already included this effect

For example the Halden model (IPTHCN=4 17) has been derived from the measured

data of linear heat rate and cladding temperature of test rod under irradiation Accordingly it

can be considered that this model implicitly includes the effect of porosity change If IPRO=1

2 or 3 is designated for this model calculation takes into account of the effect doubly by the

model and by the FEMAXI algorithm Thus the calculated results will be either over- or

under estimation Therefore to reflect the porosity change in thermal conductivity change it is

necessary to confirm the condition in which thermal conductivity model has been derived

24 Determination of Heat Generation Density Profile

Heat generation density profile in the radial direction of a pellet can be obtained by any

one of the following four methods

1 Designation by input as a function of burnup

2 Using the output of RODBURN(1155)

3 Using the output of PLUTON(116) and

4 Using a distribution function Q(r) of the Robertsonrsquos formula (Halden experimental

equation) (216)

241 Use of RODBURN output

The burning analysis code RODBURN calculates the power density profile burnup

profile and fast neutron flux in the axial and radial directions of a fuel rod from data on fuel


- 54 -

size shape materials temperature power history and thermal neutron flux history and then to

produce an output file FEMAXI can read this file as one of the calculation conditions

242 Use of PLUTON output

The burning analysis code PLUTON(116) calculates accurately and at high speed the

power density profile burnup profile fast neutron flux distribution of Pu isotope production

and other factors such as fission product element production distribution in the axial and

radial directions of a fuel rod from data on the fuel size shape materials temperature and

power history and produces output files FEMAXI can read these files as one of the

calculation conditions

243 Robertsonrsquos formula

Robertson proposed the following equation regarding heat generation density profile of a

pellet in the Halden reactor(216)

( ) ( ) ( )( ) ( )RkK

RkK

RkIRkIRQ sdotsdot

sdotsdot+sdot= 0

11

1101 (241)

Here

I first transformed Bessel function

R1 pellet center hole radius

K second transformed Bessel function

R radial coordinate in pellet

k Inverse of the neutron diffusion distance (cm-1)

Here k is calculated using

( ) ( )k E DR

E Dp

= sdot +

sdot sdot0328 054

050 8

0 82

0 19

(242)

where

E 235U enrichment ()

Name-list input parameter IFLX =－2 input from RODBURN results

=－1 input from PLUTON results =0 using the Robertsonrsquos formula

IFLXgt1 Designated by input as a function of burnup

The heat generation density profile in between the input burnup points is

determined by interpolation in terms of burnup


- 55 -

D pellet theoretical density ratio (minus)

Rp pellet diameter (cm)

The distribution function (241) can be approximated by the following function

( ) ( )( ) ( ) ( )[ ]( )( ) ( )( )22

2

2 1121

112

ahhhahb

harahbrQ

bb

bb

minusminus+minusminusminus+minusminus+= (243)

where

r normalized pellet radius )10( lele r

a heat generation density ratio of inner surface to outer surface of the pellet (minus)

b shape factor of the heat generation density function Q(r) (minus) and

h normalized inner radius of the pellet (minus)

In the case of Halden reactor the shape factor in Eq (243) can be set as b=2 since Eq

(243) can be well approximated with a quadratic function The heat generation density

ratio a is expressed as follows using Eq (241)

( )( )pRQ

RQa

1

11= (244)

244 Accuracy confirmation of pellet temperature calculation

To evaluate the accuracy of the numerical calculation of the temperature distribution in

the radial direction in a pellet in FEMAXI a comparison between analytical solutions and

numerical results by FEMAXI is shown below

(1) Power profile in the radial direction in a pellet The temperature distribution in a pellet can be given as

1po

po pi

T r r

pT r rdT r q dr dr

rλ = sdot sdot (245)

where pλ pellet thermal conductivity (W(cmK))

r location in the radial direction in the pellet (cm)

T temperature (K) at location r

rpi radius of the center hole in the pellet (cm) and

q heat generation density at radial location r（Wcm3）

By normalizing Eq(245) with respect to location r the heat generation density q is


- 56 -

written as a function of location r as q(r)

1 1( )

po

T r

pT r hdT r q r dr dr


Here r normalized radial location in the pellet

h normalized radius of the central hole in the pellet

The function of generated heat density distribution Eq (246) q(r) is given by

( ) )(1

)(2

rh

Prq LHR ϕ

πsdot

minus= (247)

where )(rϕ is the heat generation density distribution function and PLHR is the radiation

power Using Eqs(246) and (247) when the pellet thermal conductivity pλ is constant and

the heat generation density distribution function )(rϕ is uniform the temperature

distribution in a pellet can be easily obtained For example for the normalized radius h=0 of

the center hole in the pellet the linear power PLHR=400 (Wcm) the heat generation density distribution function )(rϕ =1 and pellet thermal conductivity pλ =003（W(cmk)） by

substituting Eq(247) into Eq(246) the following equation is obtained

( )

21 1

00

1212

1 1

2

12 2 2 4

po

rT r

LHR LHRpT r r

LHR LHR LHR

rr

P P rdT r dr dr dr

r r

P P r Prdr r

λπ π

π π π

= sdot sdot = sdot

= = = minus

(248)

Accordingly [ ] ( )214

rP

T LHRTTp po

minus=π

λ and analytical solutions for the temperature in the

pellet can be obtained by providing the values of pλ PLHR and the outer-surface temperature

of pellet TPO When TPO＝450 (oC) the temperature T at relative radius r is given as

( ) )1(0304

4004501

422 rr

PTT

p

LHRPO minus

sdot+=minus+=

ππλ (249)

Then using Robertsonrsquos formula assuming that the heat generation density ratio of

internal to external surfaces of pellet is a and the form exponent of the heat generation density

function is b the heat generation density distribution function )(rϕ is

( )( ) ( ) ( )[ ]( )( ) ( )( )22

2

1121

112)(

ahhhahb

harahbr

bb

bb

minusminus+minusminusminus+minusminus+=ϕ (2410)

By substituting Eq(2410) into Eq(247) and substituting Eq(247) into Eq(246) the


- 57 -

following equation is obtained

( )4po

TLHR

pT

PdT F rλ

π= sdot (2411)

Here

( ) [ ][ ])1)(1(2))(1(

ln)2)(()1(22)1)()(2(2

)1)(1(4

22

2222

ahhhahb

rhbhaharhabb

rarF

bb

bbbb

minusminus+minusminus

+minus+minus+minusminus+++

minusminus= ++

(2412)

When h=0 Eqs(249) and (2410) are simplified to

( ) ( )[ ]

2

12)(

++minus+=

ba

arabr

b

ϕ (2413)

( ) )2(

)1()2(2

)1)(1(4 22

+

minus+++

minusminus

=

+

ba

rabb

ra

rF

b

(2414)

For example when a=07 and b=2

( )24 03 07( )

34

rrϕ

+= ( ) 4 203(1 ) 28(1 )

34r r

F rminus + minus

=

Accordingly by obtaining F(0) F(0)=3134=0912 The value obtained using

Eq(2412) is 0912-fold that under uniform power distribution

Moreover with respect to )(rϕ according to [ 7030 2 +r ] in which the central value is

07 and the surface value is 1 to set the integral value at 1 it is necessary to multiply )(rϕ

by 434=1176 Here when the pellet radius is divided into 10 equal-thickness rings the

relative power density distributions applied to the rings are 08244 08315 08456 08668

08950 09303 09727 10221 10785 and 11421 from the inside toward the outside of the

pellet These are relative power density distributions obtained by volume averaging the

solution for each ring using Eq(2413)

Because F(r) is obtained using Eq(2412) when λp is constant at pλ =003 (Wcm-K)

the analytical solution for the pellet temperature distribution is obtained from Eq(2410)

Table 241 shows a comparison between the values of analytical and numerical solutions

The latter is obtained using FEMAXI by discretizing the radial distance into 10 and 100

segments The numerical solutions in Table 241 in the case when the power distribution in


- 58 -

the radial direction is given are the results obtained using a table for the relative power

density distribution obtained by volume averaging the solution for each ring as mentioned

above In the 100 mesh system results were obtained by calculating values for 100 points by

interpolating the above 10 data points using the following equations

f(x)＝arb＋C (2415)

( ) (1 ) PFACf x CSFR r CSFR= minus sdot + (2416)

(Here strictly ( ) ( )2 1( )

2

bb a x af x

ba

+ minus + =+

where a=CSFR and b=PFAC)

Furthermore the numerical solution in the case in which CSFR=07 and PFAC=20 are

given for both the10- and 100-mesh systems is the value obtained by calculating the central

coordinate of r (r is normalized by setting the radius of a pellet to 10) and calculating the

value of )(rϕ corresponding to r using Eq(2411)

As a result as is apparent from Table 241 it has been confirmed that the difference in

pellet temperatures between the analytical solution and numerical solution for the same power

density distribution is 01 or smaller In addition compatibility with the analytical solution

is better when simply the relative power density distribution at the coordinate center (the

numerical solution obtained by providing CSFR=07 and PFAC=20) is used than when the

relative power density distribution obtained by volume averaging (the numerical solution

when the power distribution in the radial direction is given) is used

Table 241 Comparison between values from the analytical and numerical

solutions obtained by FEMAXI

Relative radius

of pellet

Analytical solution of

integral equation

Numerical solutions

obtained by providing power

profile in the radial direction

Same as left (number of rings in the

radial direction of pellet = 100)

Numerical solution when CSFR=07 and PFAC=20 are

given

Same as left (number of rings

in the radial direction of

pellet = 100)

00 01 02 03 04 05 06 07 08 09 10

14174114086713823113380112752111931110907196678819846482245000

141873 140997 138358 133922 127633 119412 109158 96747 82033 64848 45000

141758 140883 138246 133814 127530 119317 109075

96679 81984 64822 45000

141787 140912 138276 133844 127560 119346 109101

96701 82001 64831 45000

141758 140883 138246 133814 127530 119317 109075

96679 81984 64822 45000


- 59 -

This is because the calculation of temperature distribution in a pellet is one dimensional

in the radial direction the effect of relative power density obtained by volume averaging on

the temperature distribution is greater on the peripheral side (the side with higher relative

power density) where the volume is larger than on the inner side

(2) Thermal conductivity of pellet Since the thermal conductivity of pellet depends on temperature burnup and density the

heat conduction equation for pellet is nonlinear and it is impossible to obtain an analytical

solution Factors such as burnup and density dependences are effectively equivalent to a

difference in sensitivity to temperature dependence temperature dependence alone should be

taken into consideration in the examination of programs and the accuracy of numerical

solutions The remaining issue is a proper coding of physical models in the program The

thermal conductivity equation of Wiesenack(43) was used in the comparison of simulated

solutions with the numerical solutions based on analytical solutions

Ignoring the burnup dependence of the Wiesenack (Halden) equation(43) gives the

following equation

( )41 01148 2475 10 00132exp(000188 ) (WmK)K T Tminus= + times sdot + sdot (2417)

T Temperature (degC)

This equation is applied to FEMAXI calculations In the calculation based on analytical

solutions the error in pellet center temperature on the left-hand side of Eq(2410) was set to

01degC Because the thermal conductivity equation is temperature-dependent the conductivity

cannot be determined accurately unless the temperature distribution is determined

Furthermore unless thermal conductivity is determined temperature distribution cannot be

obtained thus iterative calculations are required In iterative calculations a pellet is divided

into 10 ring elements and the pellet surface temperature is adopted as a boundary condition

In determining the temperature distribution for the first iteration the temperature of the inner

side of the pellet ring element is obtained by calculating thermal conductivity using the

temperature of the outer side of each element In the determination of temperature distribution

for the second iteration and thereafter the average temperatures of the inner side and outer

side of each element are calculated using the results of temperature calculations in the

preceding stage Using these results thermal conductivity is obtained and the temperature of

the inner side is determined As a result of these iterations when the differences in

temperature at the center of the pellet reach 01degC or smaller it is assumed that convergence is


- 60 -

attained

According to Table 242 which was obtained from these results it was confirmed that

the difference between the analytical solution and numerical solution was within 02

Table 242 Comparison of analytical solutions with numerical solutions by FEMAXI

Relative radius of

pellet


integral equation

Numerical solution

obtained from power profile in the radial

direction


radial direction= 100)


given


in the radial direction = 100)

00 137512 137794 137535 137686 137507 01 136408 136688 136430 136581 136403 02 133109 133381 133129 133278 133106 03 127661 127919 127677 127823 127658 04 120151 120388 120163 120302 120149 05 110726 110937 110734 110863 110725 06 99607 99784 99611 99725 99606 07 87088 87227 87090 87182 87087 08 73533 73628 73534 73599 73533

09 59357 59405 59357 59391 59357 10 45000 45000 45000 45000 45000

25 Gap Thermal Conductance Model

The gap model can be selected from 7 options In standard cases Eq(251) of the

modified Ross amp Stoute model (IGAPCN=0) is used while in pellet-clad bonding case

IGAPCN=5 or 6 (bonding gap model) is used

251 Modified Ross Stoute model

Gap thermal conductance is expressed by the following equation which is modified from

the Ross and Stoute model(217)

( ) ( ) 1 22 1 2 05

gas m cr

eff

Ph h

R HC R R g g GAP

λ λ sdot= + +sdot sdot+ + + +

(251)

The first term on the right side of the equation is the gas conduction component the

second term is solid contact conduction and the third term radiation conduction

[Selection of option IGAPCN] IGAPCN=0 modified Ross amp Stoute model =1 MATPRO-09 model =2 Ross amp Stoute model =3 Modified Dean model =5bonding model 1 =6 bonding model 2


- 61 -

Here

C Pc= minus times sdotminus2 77 2 55 10 8

( ) ( )

( ) ( )λ

λ λλ λm

p po c ci

p po c ci

T T

T T=

sdot+

2 (WcmK)

hT T

T Trp c

po ci

po ci

= + minus

sdot sdotminusminus

minus1 1

1

1 4 4

ε εσ

Tpo pellet outer surface temperature (K)

Tci cladding inner surface temperature (K) λ p pellet thermal conductivity (WcmK)

λc cladding thermal conductivity (WcmK)

λ gas thermal conductivity of the mixed gas (WcmK)

Pc pellet-cladding contact pressure (Pa) [described in section 26(4)]

Reff apparent surface roughness of pellet (cm)

R2 cladding surface roughness (cm)

g g1 2+ temperature jump distance between solid phase and liquid phase (cm)

GAP radial gap width (cm)

R R Reff

22

2

2

+

H cladding Mayerrsquos hardness (Pa) ( H Y= 2 8 σ (213))

σ Y yield stress (Pa) (See section 342)

σ Stefan-Boltzmann constant (WcmK4) ( )σ = times minus5 67 10 12

ε p pellet emissivity (minus)(See chapter 4)

ε c cladding black oxide emissivity (minus)(See chapter 4)

Equations for describing λgas and g1 + g2 are as follows

(1) Thermal conductivity of mixed gas This value can be represented as follows using the MATPRO-09 model (213)

1

1 1ji j

ni

gas nji

iji

x

x

λλφ

=ne

=

= minus Σ

(252)

The roughness Reff and R2 can be designated by the input parameters R1 and R2 (μm) respectively The default values are both 10 μm


- 62 -

Here the equation

( )( )

( )φ

λλ

ij

i

j

i

j

i

j

i j i j

i j

M

M

M

M

M M M M

M M=

+

+

+minus minus

+

1

2 1

12 41 0142

1

2

1

4

2

3

2

1

2

2

(253)

holds where xi molar ratio of gas i

λ i thermal conductivity of gas i (WcmK)

M i molar mass of gas i

The gases of this model are Helium Nitrogen Krypton and Xenon Naturally gas

composition depends on fission gas release which will be explained in section 274

Thermal conductivities of gases (213) are shown

3 0668

4 0701

5 0872

5 0923

4 08462

3366 10

3421 10

40288 10 (WmK) (K)

4726 10

2091 10

He

Ar

Xe

Kr

N

k T

k T

k T k T

k T

k T

minus

minus

minus

minus

minus

= times

= times = times = times = times

(254)

Here it is noted that the thermal conductivity of He increases by only a few even if

pressure of He gas increases from 1MPa to 10MPa(218) (219)

(2) Temperature jump distance at solid-gas inetrface The temperature jump distance is given by

( )5

1 2 1 21

10n

iigas

g g g gP=

+ = Σ + sdot (255)

Here

(g1 + g2) jump distance of gas i between solid phase and liquid phase

(cm)

Pgas gas pressure (Pa)

Name-list input Parameter IAR Gas can be specified among He N Kr and Xe

When IAR=rsquoARrsquo is designated the gas can be specified into Argon instead of Nitrogen


- 63 -

According to the Ross amp Stoute model the values of (g1+g2) for Helium Nitrogen

Krypton and Xenon are 10times10-4 5times10-4 1times10-4 and 1times10-4(cm) respectively

Calculation of Pgas is explained later in section 2167 using Eq(2171)

(3) Pellet- Cladding contact pressure To calculate the gap thermal conductance in Eq(251) it is necessary to calculate the

contact pressure cP between the pellet and cladding In FEMAXI-7 the contact pressure

which is calculated by the entire length mechanical analysis is directly used to calculate the

gap thermal conductance

252 Bonding model for gap thermal conductance

(1) Phenomena In LWR fuel rods the gap between pellet and cladding is closed after average burnup

reaches approximately 40 GWdtU and bonding layer is formed(217)as a result of a chemical

reaction between pellet and cladding After the formation of bonding layer the heat

conductance significantly increases from the values predicted by Eq(251) In addition

regardless of the magnitude of the contact pressure biaxial stress-state analysis is required in

which effect of bonding (anchored) gap states on PCMI is taken into consideration

Therefore modeling of bonding is inevitable in the situation where the importance of

PCMI has revived in high-burnup-fuel behavior By assuming that a bonding layer formation

advances as a function of burnup contact pressure and time and that the resulting changes in

the gap thermal conductance occur the following model is considered

(2) Calculation and assumption of the development of bonding The following is considered based on the mechanical analysis of the entire length of fuel

rod When pellet and cladding are in contact in PCMI the bonding reaction begins and

advances During this period the thermal conductance of the gap changes (increases)

depending on the development of bonding layer

Here bonding development (BD) is defined as

start

t

CtBD P dt X= (0 10)BDle le (256)

where Pc contact pressure between pellet and cladding

tstart time at which contact is initiated


- 64 -

X empirically determined parameter (hourMPa)

Equation (256) is integrated for the period of contact in the entire length mechanical

calculation If contact is interrupted during the course of the analysis the integration is

terminated and when the contact redevelops the integration is restarted the total integration

is monitored from the time of initial contact If the value of BD reaches 10 the bonding layer

is considered to be completed and BD is assumed not to increase from this point

Value of X Name-list input parameter BDX=1000000 hr-MPa (Default)

() Note Regardless of the usage of the above bonding model Eq(256) is always calculated

(3) Gap thermal conductance If IGAPCN=5 or 6 is selected ldquogap conductance (GC) rdquo is defined as shown below

A) When bonding does not develop the GC1=Ross amp Stoute model is used as is usual

B) When bonding has already begun and the gaps are determined to be closed assuming

that the GC given by Eq (251) is OpenGC the gap conductance GC2 under such

conditions is given

GC2=(1－BD)OpenGC + BDBondGC (Wcm2K) (257)

BondGC= ( ) (1 ) ( )p p ox cF T Tθ λ θ λsdot + minus sdot (0 1 0 )Fθle le lt (258)

The terms )( pp Tλ and )( cox Tλ denote the thermal conductivities of pellet and oxide film

of cladding respectively at the outer surface of pellet and inner surface of cladding Here the

thermal conductivity of the cladding oxide layer can be selected and adjusted by using

name-list parameters IZOX and CNOX (See section 43)

Furthermore the thermal conductivity of pellet is set to a value corresponding to burnup

at the outer surface of pellet The terms F and θ are input parameters determined empirically

and have the following default values

Name-list input parameter F FBONDG=100 (Default) θ SBONDG=001 (Default)

C) In a segment at which bonding has already begun and when the mechanical calculations

determine that the gap is open due to power decrease or high internal pressure if

IGAPCN=5 the following GC3 is used

GC3=OpenGC(10+αBD) α adjustment parameter (gt0) (259)

The term α is an input parameter which can be determined empirically and has the


- 65 -

following default values

Name-list input parameter α ALBD=07 (default)

D) In a segment in which bonding has already begun when the mechanical calculations

determine that the gap becomes open due to power decrease or high internal pressure and if

IGAPCN=6 BondGC in Eq(258) is used as it is

That is the mechanical model calculates the gap size on the assumption that the pellet

stack is a solid continuum However in an actual rod pellet fragments are likely to remain

bonded to the inner surface of cladding In this situation the ldquonumerically resulted

re-openingrdquo of gap due to the cladding outward creep deformation or pellet shrinkage can

be considered to correspond to the increase of crack space among the pellet fragments In

this case in FEMAXI gap thermal conductance is given the value which is determined in

the bonded gap state This is the physical implication of IGAPCN=6

253 Bonding model for mechanical analysis

In the entire rod length (ERL) mechanical analysis independent from IGAPCN which

specifies the gap thermal conductance model a name-list input parameter IBOND can specify

the bonding state between pellet and cladding as follows (This is explained in chapter 3

again)

IBOND=0 bonding state is not applied (default)

IBONDge1 bonding state is applied

(1) In case of IBOND=1 A) After the bonding reaction begins it is assumed that contact between pellet and cladding is

kept depending on its progress irrespective of the magnitudes of power or internal pressure

Accordingly in a segment in which bonding has occurred (or is progressing) the clogged

(bonded) state is assumed after a certain value of bonding development (above a certain value

of BD) irrespective of the calculated existence of contact or magnitude of contact pressure

A certain value of BD name-list input parameter BDTR=05 (default)

B) In a segment in which clogged gap has occurred due to bonding the mechanical calculation

is performed with the displacements of pellet outer surface and cladding inner surface always

being kept equal in the axial direction even if the gap re-opens


- 66 -

(2) In case of IBOND=2 After the bonding reaction begins it is assumed that contact between pellet and cladding is



(bonded) state is assumed after a certain value of bonding reaction progress (above a certain

value of BD) while if the contact pressure drops under PCPRES the bonding (clogged) state

is cancelled and the gap recovers to a normal state before the bonding reaction

PCPRESName-list input parameter PCPRES =001(MPa) (default)




(bonded) state is assumed after a certain value of bonding reaction progress(above a certain

value of BD) while if the gap re-opens more than GWD (mm) in the radial direction the

bonding (clogged) state is cancelled and the gap recovers to a normal state before the bonding

reaction

GWDName-list input parameter GWD =01(mm) (default)

254 Pellet relocation model As one of the factors dominating the pellet temperature FEMAXI model of fuel pellet

relocation having an effect on the gap size is explained

(1) Basic concept of relocation model In the start-up period of a fresh fuel temperature of fuel pellet increases with power and

thermal stress is generated to give rise to radial cracks and pellet is broken into some pieces of

fragments These fragments pushes each other due to thermal expansion and they bring

themselves outward to partially fill the gap space This is referred to as relocation This

relocation magnitude cannot be measured in situ It can be estimated only indirectly from the

temperature and linear power of rod

The relocation model of FEMAXI has been designed to input a constant value in both the

radial and axial directions in the beginning of irradiation Name-list parameter FRELOC

(=035 as default) specifies the decrease ratio of initial hot gap size which is determined by


- 67 -

pellet diameter thermal expansions of pellet and cladding and elastic deformation of cladding

immediately after initial start-up

The gap size decreases with extension of burnup while when gap is closed and PCMI

contact pressure is generated pellet fragments are pushed back by cladding and the initial

relocation is decreasing This process is explained in the mechanical analysis part sections

33 and 34

Irradiation in a test reactor such as the Halden reactor a test rod is subjected to frequent

power changes and cycles This induces a thermal cycle in pellet fragments and at every cycle

the relocation amount is likely to change finely To reproduce this situation FEMAXI has

the following controlling function

(2) Function to decouple thermal relocation from mechanical relocation

In this function the mechanical analysis uses FRELOC as per usual while in the thermal

analysis it is possible to add a modification term DELTAR into the radial gap size GAP in

Eq(251) which is resulted from the mechanical analysis

A) Users can specify by input DELTAR ie the pellet thermal relocation modification amount

(μm) of axial segments at any history point Namely suppose that GAPi is the gap size

resulted in the mechanical analysis at i-step the thermal gap size GAP1i is expressed as

GAP1i=GAPi - Σi(DELTAR) (2510)

where Σi(DELTAR) is the accumulated sum of DELTAR until i-step

B) Input parameters are set as DELTAR(IZ IN) (Default DELTAR(IZ IN)=0

(1 le IZ leNAX 1 le IN leNHIST)) DELTAR is the increment for Σi-1(DELTAR) of the

previous time step For the history points with no designations DELTAR=0 is assumed

C) When GAP1i= GAPi - Σi(DELTAR) = 0 GAP1=0 is applied to calculation even if

DELTARgt0 is specified in the following history points

D) At the instant when PCMI occurs in the mechanical analysis Σi(DELTAR) or the

accumulated sum of DELTAR until then is reset to be null If the gap re-opens after this

Σi(DELTAR)=0 is applied to calculation even if DELTARgt0 is specified in the following

history points


- 68 -

E) Value of Σi(DELTAR) is given as a modification term to the radial gap size in the gap

thermal conductance model Namely the first term of Eq(251) ie GAP value is modified

Here it is noted that DELTAR is effective only when no PCMI occurs In this situation the

second term (solid contact term) is 0 and is not affected by DELTAR at all

255 Swelling and densification models With burnup extension fission products accumulate in a fuel pellet and pellet volumetric

increase occurs This is referred to as swelling This swelling consists of solid fission product

swelling which is proportional to burnup and fission gas bubble swelling which is dependent

on both burnup and fuel temperature The swelling model of FEMAXI is broadly classified

into an empirical model and gas bubble swelling model which are described in detail in

section 27 - 213 in connection with fission gas bubble growth model

26 Model of Dry-out in a Test Reactor Capsule

FEMAXI-7 has implemented a function to perform a predictive analysis of dry-out

experiment in JMTR (Japan Materials Testing Reactor) The analytical geometry is shown in

Fig261 A typical set of analytical conditions is as follows

Fuel rod BWR-9x9A type LHR=about 100Wcm

Heated steam phase inlet conditions= 72MPa quality 10 flow velocity parameter

Flow channel tube double-layer thermal insulated tube made of SUS316

Circulating water 72MPa 50oC 001ms returning condensed water from heating steam

Capsule outer mantle material SUS316

Coolant water 04MPa inlet temp40 oC flow velocity 19mh


- 69 -

Fig261 Geometry of cladding temperature regulation system

261 Modelling the dry-out experiment

(1) Basic assumption A) The boundary condition temperature is fixed either at the inner surface of internal tube or

outer surface of the capsule outer tube Basically it is assumed that the outer tune surface

temperature is fixed at 40oC

B) With respect to the heat exchanging flows only the overheated steam and circulating water

are taken into account and the thermal removal by Helium flow in the Helium insulation layer

is neglected

C) Therefore heat flow in the axial direction is taken into account only by the overheated

steam and circulating water

Fuel pellet

96

Cladding

978

113 Internal tube

Outer tube

Capsule outer m

antle

Heated steam

layer

Helium

layer

Circulating w

ater

Coolant w

ater

180

200

210

230

270

320

Diameter (mm)Center line


- 70 -

(2) Basic equations Fig262 shows a cladding temperature controlling system In this figure j denotes one

segment j of the axial segments

jl

Fixed at 40

Fuel pellet

T2j T3j T4j T5j T6j T7j T1j

R1 Heated steam layer

R2 R3 R4 R5 R6 R7

Cladding

Internal tube

He layerCapsule

outer mantle

Circulating w

ater

Outer tube

Fig262 Cladding temperature controlling system

In the heated steam layer shown in Fig262 assuming that the steam enthalpy is H2j-1 at

the boundary of segment j-1and its upper segment j

( ) )(2

112220

22

1222

2222 jjjj

j

jjjj qRqR

RRHH

V

W

t

TC minus

minus=minus+ minuspart

partρ (261)

holds where

ρ density (kgm3)

C specific heat (JkgK)

T temperature (K)


V volume (m3)

H steam enthalpy (Jkg)

R radius (m)

q heat flux by heat conduction (Wm2)

By transforming Eq(261) neglecting the mass flow rate change in the axial direction


- 71 -

)(2

)(

112221

22

112

122

1111 jCjC

jj

jjj qRqR

RRz

TC

RR

W

t

TC minus

minus=

minus+

partpart

πpartpart

ρ (262)

is obtained

For the internal tube neglecting the heat flow in the axial direction

)(2

223322

23

222 jCjC

jjj qRqR

RRt

TC minus

minus=

partpart

ρ (263)

is obtained Similarly for the helium thermal insulating layer neglecting the heat flow in the

axial direction

)(2

334423

24

333 jCjC

jjj qRqR

RRt

TC minus

minus=

partpart

ρ (264)

is obtained Also similarly for the outer tube neglecting the heat flow in the axial direction

)(

244552

425

444 jCjC

jjj qRqR

RRt

TC minus

minus=

partpart

ρ (265)

is obtained For the circulating water taking into account of the heat flow in the axial

direction

)(2

)(55662

526

552

526

5555 jCjC

jj

jjj qRqR

RRz

TC

RR

W

t

TC minus

minus=

minus+

partpart

πpartpart

ρ (266)

is obtained For the capsule outer mantle neglecting the heat flow in the axial direction

)(2

66

7726

27

666 jCjC

jjj qRqR

RRt

TC minus

minus=

partpart

ρ (267)

is obtained

7

1 jCjC qq in Eqs(262) to (267) are

( )jNjjjjC BTAq 0

1 )( φminus=minussdotminus= (268)

Here Aj and Bj are the coefficients determined by the difference equations of thermal

conduction as follows

jN φ heat flux at the outer surface of cladding (Wm2K)

jT 0 temperature at the outer surface of cladding (K)

)( 1222 jjjC TThq minus= (269)





- 72 -


)( 677

7 jjjC TThq minus= (2614)

where h thermal conductance (Wm2K)

The thermal conductance h in Eqs(268) to (2614) are given respectively as follows

2

32

2

2

1

2

2ln

11

R

RRR

h

h+

+prime

=

λ

(2615)

1hprime forced convection thermal conductance including radiation in

the superheated steam layer (Wm2K)

2λ thermal conductivity of the inner tube (WmK)

3

43

3

3

32

3

2

33

2ln

2ln

1

R

RRR

RR

RRh

++

+

=

λλ

(2616)

3λ equivalent thermal conductivity of the Helium thermal insulating

layer (WmK)

4

54

4

4

43

4

3

44

2ln

2ln

1

R

RRR

RR

RRh

++

+

=

λλ

(2617)

4λ thermal conductivity of the outer tube (WmK)

554

5

4

55 12

ln

1

hRR

RRh

prime+

+

=

λ

(2618)


the circulating water superheated steam layer (Wm2K)

6

76

6

6

5

6

2ln

11

R

RRR

h

h+

+prime

=

λ

(2619)

λ6 thermal conductivity of the capsule outer mantle (WmK)

76

7

6

77 2

ln

1

RR

RRh

+

=

λ

(2620)


- 73 -

262 Thermal and materials properties

(1) Materials properties

The necessary properties in the present model are listed below

A) SUS316

thermal conductivity (non-disclosure data) density ρ=7830 (kgm3)

Isobaric specific heat (non-disclosure data)

B) Helium

thermal conductivity(211) 3 069293 10 (WmK) (K)T Tλ ηminus= times sdot

where η adjustment factor

Density(221) 2 30 (kgm )a b cρ ρ+ + =

( ) ( )( )

4

3

54245 10 1

1890

(K) bar 00207723 bar kg K

Pa b c

T RT

T P R

minus= times + = = minus+

= sdot

Isobaric specific heat (222) ( )( )3519 10 at 773K J kg KPC = times sdot

C) Heated steam and circulating water

Calculated by using the steam table at 50oC for the circulating water

Thermal conductivity 0653 (WmK)λ = Densityρ=73645 (kgm3)

Isobaric specific heat for heated steam ( )( )5456 J kgtimesKPC =

Thermal conductivity

(290 C) (400 C) (600 C) (800 C)0064 0065 0088 0100 (WmK)λ deg deg deg deg=

Density ρ=3766 (kgm3)


(2) Thermal conductance

Thermal conductance hw is given by the sum of convection term and radiation term For

the convection term the heated steam layer is a single phase steam region (quality=1 [mass

flow rate of gas ][total mass flow rate]) and the circulating water is a single phase liquid

region so that the Dittus-Boelter equation(23) is applied to both the terms


- 74 -

Dittus-Boelter equation 8040

e

RePr0230D

KhW = (2621)

where μeRe DG=


De equivalent diameter (m)

Pr Prandtle number (－)

Re Reynoldrsquos number (－)

G Mass flow flux (kgm2s)

μ dynamic viscosity coefficient (kgms)

Also the radiative thermal conductance is

21

42

41

1

21

111

TT

TThr minus

minussdotsdot

minus+=

minus

σεε

(2622)

where σ Stephan-Boltzmann constant（Wm2K4） ( )810675 minustimes=σ

1ε emissivity of cladding outer surface or inner tube outer surface (-)

2ε emissivity of cladding inner surface or outer tube inner surface

263 Numerical solution method Based on the modeling method described in section 22 equations are converted into

difference formulae as follows though in this conversion an outlet-representation format is

adopted This is to make a difference in equations for the heat flow in the axial direction from

upper stream with respect to the flow direction and for the heat flow in the radial direction

with respect to the outlet temperature at the corresponding axial segment

(1) Converting the constitutive equation to difference equation

For heated steam layer

( )

))()((2

)(

011

11

22221

22

111

111112

122

111

111

jn

jjnj

nj

njj

njj

j

nj

nj

jj

BTARTThRRR

TCTClRR

W

t

TTC

minussdot+minusminus

=

sdotminussdotminus

+Δminus

++

++minusminus

+

πρ

(2623)


- 75 -

))(2

2

)()(

011111

221

22

22

112

122

222

122

11111112

122

111

jn

jjnj

jjnj

nj

j

jjjnj

j

j

BTARTt

CT

RR

hR

TRR

hR

lRR

CW

t

CT

lRR

CW

minussdot+Δ

=minus

minus

minus+

minusminus

Δ+

minus

+

++minus

minus

ρ

πρ

π (2624)

For inner tube

)()(2 1

11

2221

21

33322

23

21

222

+++++

minusminusminusminus

=Δminus

nj

nj

nj

nj

nj

nj

jj TThRTThRRRt

TTCρ (2625)

( ) nj

jjnj

nj

jjnj T

t

CT

RR

hRT

RR

hRhR

t

CT

RR

hR2

22132

223

33122

223

223322112

223

22 222

Δ=

minusminus

minus+

+Δ

+minus

minus +++ ρρ(2626)

For Heium gas layer

)()(2 1

21

3331

31

44423

24

31

333

+++++


=Δminus

nj

nj

nj

nj

nj

nj

jj TThRTThRRRt

TTCρ (2627)

( ) nj

jjnj

nj

jjnj T

t

CT

RR

hRT

RR

hRhR

t

CT

RR

hR3

33142

324

44132

324

334433122

324

33 222

Δ=

minusminus

minus+

+Δ

+minus

minus +++ ρρ (2628)

For outer tube

)()(2 1

31

4441

41

55524

25

41

444

+++++


=Δminus

nj

nj

nj

nj

nj

nj

jj TThRTThRRRt

TTCρ (2629)

( ) nj

jjnj

nj

jjnj T

t

CT

RR

hRT

RR

hRhR

t

CT

RR

hR4

44152

425

55142

425

445544132

425

44 222

Δ=

minusminus

minus+

+Δ

+minus

minus +++ ρρ(2630)

For circulating water

( )

))()((2

)(

14

1555

15

16662

526

155

115152

526

551

555

++++

++++

+


=

sdotminussdotminus

+Δminus

nj

nj

nj

nj

njj

njj

j

nj

nj

jj

TThRTThRRR

TCTClRR

W

t

TTC

πρ

(2631)

( )

nj

jjnj

nj

j

j

nj

j

jjjnj

Tt

CT

RR

hRT

lRR

CW

TRR

hRhR

lRR

CW

t

CT

RR

hR

5551

625

26

661152

526

155

152

526

556625

26

5555142

526

55

2

)(

2

)(

2

Δ=

minusminus

minus+

minus+

+minus

minusΔ

+minus

minus

+++

+

++

ρπ

πρ

(2632)


- 76 -

For capsule outer mantle

)()(2 1

51

6661

67772

627

61

666

++++


=Δminus

nj

nj

njj

nj

nj

jj TThRTThRRRt

TTCρ (2633)

( ) 72

627

776

66162

627

667766152

627

66 222j

nj

jjnj

jjnj T

RR

hRT

t

CT

RR

hRhR

t

CT

RR

hR

minus+

Δ=

minus+

+Δ

+minus

minus ++ ρρ (2634)

where 7 jT =40 oC is the temperature at the outer boundary of capsule outer mantle

At the inlet boundary of heated steam layer

( )

))()((2

)(

110111

111

122221

22

11111

111

12

122

1111

111111

BTARTThRRR

TCTClRR

W

t

TTC

nnn

nninin

nn


=

sdotminussdotminus

+Δminus

++

+++

πρ

(2635)

where 11

+ninT is the heated steam temperature at inlet

At the inlet boundary of circulating water

( )

))()((2

)(

14

1555

15

16662

526

155

1552

526

551

555

++++

+++


=

sdotminussdotminus

+Δminus

nN

nN

nN

nN

nNN

ninin

N

nN

nN

NN

TThRTThRRR

TCTClRR

W

t

TTC

πρ

(2636)

where 15+ninT is the water temperature at inlet and N is the number of segments

in the axial direction

(2) Method of solution

The algebraic equations ordered into n+1 steps of temperatures are solved

The unknown quantities are 6timesN ie 1 111 61 n nT T+ + 1

1nNT + and 1 1

1 6 n nN NT T+ +

By set out these unknowns from Eqs(2623) (2626) (2628) (2630) (2632) (2634)

and (2636) the coefficient matrix is obtained as Eq(2637) where Eq(2637) is a case with

four axial segments


- 77 -

(2637)

By solving the above matrix by the Gaussian elimination method n+1 steps temperatures

are obtained By this the axial distribution of heated steam and cladding surface heat transfer

coefficients are determined Thus temperatures of cladding outer surface and fuel can be

obtained


- 78 -

27 Generation and Release of Fission Gas FEMAXI calculates the diffusion of fission gas atoms bubble formation and growth

and release of fission gas on the basis of the temperature obtained in the one-dimensional

thermal analysis at each ring element of axial segment Fig271 shows the flow of analysis

Time Step Start

Calc of grain growth

Updating element nodal coordinate with grain growth

Calc of amount of grain-growth-induced migration and diffusion of fission gas atoms to grain boundary grain boundary gas bubble growth and grain boundary gas density

Solution of the total simultaneous equations to determine the distribution of in-grain fission gas atoms

Re-distribution of density of intra-grain fission gas atoms at updated grain size

Calc of gas release threshold (grain boundary bubble radius grain boundary saturation density of gas etc)

Intra-grain gas atom density calculation converges

Calc of the total amount of released gas by the current time step

Calc for all the ring elements completed Calc for all the axial segments completed

Calc of in-grain gas bubble (size and gas amount) and bubble density

Calc of amount of fission gas release during time step

Calc of element matrix and element vector Formation of entire matrix and entire vector (Eq277-22)

Fig271 Process of fission gas generation and release in the equilibrium and rate-law

models in FEMAXI


- 79 -

271 Fission gas atoms generation rate

Composition of fission gas consists of Xe and Kr while generation of He can be

designated by input The fission gas generation rate in the axial segment j and radial element i

is represented by

PY f q

E Nij

ij j

f A

= sdot sdotsdot

(271)

Here

Pij fission gas generation rate per unit length in element ij (molcm3s)

( )+

minus=

50

502

i

i

ij rdrrf φπ

φ(r) heat generation density profile function in the radial direction Q(r) in Eq (239)

q j average heat generation density of axial segment j (Wcm3) = ( )2 21

jLHR

pout

P

r hπ minus

PLHRj linear heat rate of axial segment j (Wcm)

h normalized radius of the center hole of pellet (minus)

Ef energy generated per one fission 3204times10minus11 J (=200 MeV)

Y fission yield of fission gas (Xe+Kr) = 03

NA Avogadros number 602times1023

【Note】 Iodine release calculation in FEMAXI To evaluate the Iodine-SCC of cladding under PCMI condition FEMAXI has a function

to estimate the release of Iodine In this function Iodine yield ratio is assumed as 005 (0025

for I2) with respect to the yield of Xe and using the atomic weight of Iodine = 1269045gmol

released amount (g) of Iodine is calculated as per the unit axial length (cm) of rod It is

considered that inside a fuel pellet Iodine is not present in the form of single element but

substantially in the form of CsI However as a reference value Iodine concentration (gcm2)

on the inner surface of cladding is determined by dividing the single element Iodine quantity

by the inner surface area (cm2) of unit axial length of cladding

272 Concept of thermally activated fission gas release in FEMAXI 272-1 Equilibrium model and Rate-law model With respect to the calculation method of grain boundary gas bubble growth in the fission

gas diffusion and release model one of the following two types are selected an equilibrium


- 80 -

model assuming that the grain boundary bubble is growing with the gas pressure being kept in

equilibrium with the external compressive pressure of surrounding solid matrix and a

rate-law model assuming that the grain boundary bubble radius is changing in accordance with

the differential pressure across the bubble boundary and temperature

272-2 Assumptions of fission gas release mechanism Assumption of FEMAXI model for fission gas release is as follows

A) The fission gas atoms (Xe and Kr) produced in fuel (UO2 or MOX) grains transfer to

grain boundaries through the two mechanisms described below and that there it

accumulates and forms grain boundary gas bubbles

i) Gas atom diffusion within grains

ii) Sweeping of gas atoms to grain boundaries by grain growth

B) When the amount of fission gas atoms in the gas bubbles reaches a certain saturation

value a number of bubbles connect at grain boundary and a tunnel which leads to a free

surface of pellet is formed

C) After the tunnel is formed additional gas atoms which have diffused from inside of

grains to the grain boundaries enter the tunnel and are released to the free surface The

gas accumulated at the grain boundaries also re-dissolves into the grains at a certain rate

D) Evaluation of formation of the intra-granular gas bubbles includes trapping of gas atoms

by the intra-granular bubbles and re-dissolution of gas from the bubbles into grain

matrix

(1) Gas release in the equilibrium model The equilibrium model in FEMAXI assumes that the fission gas atoms continuously

accumulate in the grain boundary bubbles while the pressure of gas in the bubble is always in

equilibrium with external compressive pressure + surface tension and that when the density

of gas atoms per unit grain boundary area exceeds a threshold value

N fmax

which is

determined by the combination of temperature bubble radius and bubble coverage fraction on

grain boundary etc the excess fission gas over the threshold value is released to free space

outside the fuel pellet That is in a situation in which tunneling has occurred ie grain

boundary accumulation of fission gas atoms has exceeded the value N fmax

the fission gas

atoms enter the tunnel and are immediately released However if the accumulated amount

of gas atoms becomes lower than the value N fmax

due to temperature decrease etc the release

is assumed to cease


- 81 -

(2) Rate-law model of FGR FEMAXI-7 has introduced a new model of FGR rate-law model This model is based

on the same understanding of that of the equilibrium model for fission gas atoms diffusion

accumulation in grain boundary and release However this model calculates the growth or

shrinkage rate of radius of the grain boundary bubble and assumes that if either the bubble

size or grain boundary coverage reaches a certain value gas release begins

(3) Contents shared by the two FGR models A) Calculation of fission gas atoms diffusion from inside of grain to grain boundary is

common in both the models The diffusion calculation addresses the gas atoms trapping by the

intra-grain bubbles and re-dissolution to solid matrix Also the method to calculate the

intra-grain bubble size is different between the two models

B) Direct release of gas by recoil and knock-out from pellet surface assumes 05 of the total

fission gas atoms generation That is even if no FGR occurs from fission gas bubbles 05

release is assumed to take place However this 05 is a default value This can be specified

by name-list parameter FRMIN in input file For example when FRMIN=0 FGR is 0 if

no release occurs from the bubbles

C) FGR from fission gas pores in the high burnup structure (HBS) ie rim structure and a

direct FGR from the HBS are evaluated by an empirical model as an additional amount of gas

release apart from the above two models

273 Thermal diffusion accompanied by trapping (1) Diffusion accompanied by trapping Figure 272 shows states of intra-granular gas bubbles and grain boundary gas bubbles in

a pellet and Fig 273 shows an ideal model of crystal grains of pellet


- 82 -

INTRAGRANULAR

GAS BUBBLE INTERGRANULAR

GAS BUBBLE

Fig 272 Schematics of intra-granular and grain boundary gas bubbles in pellet

Spherical shell containing half the amount of FP gas retained on a grain boundary

Fig273 Model of ideal spherical grains of pellet

The grain shown in Fig273 is illustrated as if it were covered with a spherical blanket

An actual grain boundary is a thin amorphous layer separating adjacent crystal grains All

the descriptions hereafter address one single crystal grain and the parameters accompanying

titles are name-list parameters and its values to control models

For example ldquo (IGASP=0 2)rdquo indicates that calculation is performed in a mode which is

in accordance with either IGASP=0 or 2 A specific explanation of parameters is given in

section 214 and chapter 4

273-1 Partition and diffusion equation of fission gas atoms in intra-grain bubbles (common in IGASP=0 and 2)

It is assumed that the relative amounts of gas atoms dissolved in the matrix and present in

intra-granular bubbles depend on the equilibrium between trapping and re-dissolution The


- 83 -

gas atoms dissolved in the matrix moves toward the grain boundary by concentration-gradient

driven diffusion The gas atoms diffusion rate in a spherical solid matrix with a radius of a is

given by Speightrsquos equation in which trapping and re-dissolution are taken into

consideration(223)

β+prime+minus

partpart+

partpart=

partpart

mbgcr

c

rr

cD

t

c 22

2

(273-1)

Here c number of gas atoms (Xe and Kr) dissolved per unit volume of solid matrix

(atomscm3) D diffusion coefficient of gas atoms (cm2s)

g rate of trapping of gas atoms by intra-granular bubbles (sminus1)

bprime rate of re-dissolution of gas atoms into solid phase (sminus1)

m number of gas atoms per unit bubble volume (atomscm3)

β generation rate of the gas atoms per unit volume of pellet (atomscm3s)

The gas atoms generation rate ijβ (atomscm3s) in a unit volume of one pellet region i j

which is shown in Eq(273-1) is using ijP in Eq(271)

1 12 2

2 2 ( )ij ij A i j i jP N r rβ π + minus= minus (273-2)

where AN the Avogadro constant

12 i jr+ outer radius of pellet region i j(cm)

12 i jrminus inner radius of region i j(cm)

The difference between trapping and re-dissolution is

0m

mt gc b mt

part primeΔ = = minus nepart

(273-3)

It is assumed that calculation at the end of every time step gives the following equilibrium

relation

0gc b mprimeminus = (273-4)

Thus addition of Eq(273-1) to (273-4) gives

( ) 2

2

2c m c cD

t r r r

part part part βpart part part+

= + +

(273-5)


- 84 -

Transforming Eq(273-3) gives

1 0mt

gc b mmb

Δ primeminus minus = prime (273-6)

Here 1

1

1 ii

i

mtb

m bα minus

minus

Δprime= minus prime (273-7)

is defined to determine the initial value of iα at current time step by using the value at the

previous time step i-1 Then

( )i

i

c c mg

αα

= ++

(273-8)

holds

Substituting the above equation into Eq(273-5)

( ) ( ) ( )2

2

2i

i

c m c m c mD

t g r r r

part part partα βpart α part part

+ + += + + +

(273-9)

is obtained Setting the sum of the number of gas atoms in the solid matrix and the number of

those in the intra-granular bubbles as

Ψ = +c m (273-10)

the diffusion equation in the intra-granular spherical coordinate system including trapping is

2

2

2D

t r r r

part part part βpart part part

Ψ Ψ Ψprime= + +

(273-11)

Here

i

i

DD

g

αα

prime =+

(273-12)

and Ψ represents the apparent average number of gas atoms per unit intra-granular volume

(atomscm3) regardless of the location of gas atoms (ie whether it is present in the matrix or

in the intra-granular bubbles) Dprime is the apparent diffusion coefficient which includes the

effects of trapping re-solution and partial destruction of bubbles by fission fragments

Thus the FEM numerical solution of Eq(273-11) and intra-granular bubble radius R

and its number density N are interlinked

Next to solve Eq(273-11) g brsquo R and D are determined assuming that there are N

intra-granular bubbles with R radius in an unit volume of pellet It is likely that the actual

radius R has a size distribution though its weighted average value (representative value) is


- 85 -

considered This R and N are determined by one of the methods described in sections 274 -

276 below

274 White+Tucker model of intra-grain gas bubble radius and its

number density (GBFIS=0)

Assuming White amp Tuckerrsquos model that N intra-granular gas bubbles with an average

radius of R are present per unit volume R and N are related as follows in an equilibrium

state during irradiation(224)

( )2152 oN R Zα πλ= + (274-1)

Here

N intra-granular bubble density (bubblescm3)

R average radius of intra-granular bubbles (cm)

Z0 affected zone (=10-7cm)

α bubble generation rate (= 24 bubblesfission fragment)

λ range (6 times 10minus4 cm)

The initial value of R is determined by a name-list parameter APORE (default=1 nm)

The rate of trapping by intra-granular bubbles g is given by (225)

4g DRNπ= (274-2)

The number density of bubbles N in Eq(274-2) is obtained by Eq(274-1)

The rate of re-dissolution of the gas into the matrix is given by (224)

( )2

0303b F R Zπλprime = + (274-3)

Here

Ffission rate (fissionscm3s)

λ fission distance（=6times10-4cm）

Zoaffected range (cm)

In addition the lowest limit of 1013

(fissionscm3s) is set for fission rate F in Eq(2714) The value is set to suppress the

occurrence of unnatural results when b prime changes suddenly due to a decrease in power

Namelist input BFCT

Re-solution ratio to solid phase can be adjusted by

BFCT prime = prime timesb b BFCT (Default=10)

The rate of re-dissolution brsquo is controlled using

BFCT bprime = bprime times BFCT (Default value 10)

Fitting parameter BFCT


- 86 -

Diffusion coefficient D of fission gas atoms inside a pellet grain is described in section

2715 In the meantime if average radius R of the intra-grain bubbles is given by Eqs(274-1)

(274-2) and (274-3) g brsquo and D are determined and D is derived from Eq(273-12)

Therefore by solving Eq(273-11) Ψ can be obtained Solution method of the partial

differential equation Eq(273-11) is presented in section 277

Here the method used to obtain R in the current time step is explained R is obtained

using van der Waalsrsquo state equation which is

( )PR

V m B m kT0

2+

minus =γ (274-4)

( )

gasP V m B m kTsdot minus =

(274-5) where

R average intra-granular bubble radius (cm)

P0 external force applied on intra-granular bubbles (dyncm2)

γ surface tension (626 ergcm2)

V intra-granular bubble volume ( = 43

3πR cm3)

B van der Waals constant ( = times minus8 5 10 23 cm3atom)

m number of gas atoms per unit bubble volume (atomsbubble)

k Boltzmanns constant ( = times minus1 38 10 16 ergK)

T temperature (K)

As the model assumes that the gas pressure inside the grain boundary bubble is in

balance with the sum of compressive static pressure extP exerted by surrounding solid and

surface tension

2

gas extP PR

γ= + (274-6)

holds Here Pext usually takes a negative value in the mechanical analysis because Pext is

usually a compressive stress Nevertheless in the calculation of Eq(274-6) and related

equations absolute value of Pext is taken

By solving Eq (274-4) with respect to m using 343V Rπ=

( )

3 24

3 2ext

ext

P RRm

B P R kTR

γπγ

+= sdot+ +

(274-7)


- 87 -

is obtained

On the other hand by Eq(273-3) ie the equilibrium relationship between trapping of

gas atoms by intra-gain bubbles and re-dissolution from the bubbles

0gc b m mtprimeminus minus Δ = (274-8)

holds Here assuming 0mtΔ = gives

gc b m= prime (274-9)

By transforming this equation into an equation which appears m by using m=mN and

c m= minusΨ

( )prime + =b g m N g Ψ (274-10)

is obtained

Substituting Eqs(274-2) and (274-3) into Eq(274-10) gives

( )( )2 0303 4 4F R Z DRN m DRπλ π π+ + = Ψ (274-11)

By substituting Eq(274-7) into the above equation

( )( ) ( )3

2

0

( 2 )303 4 3

2ext

ext

R ZR P R

F DRN DB P R kTR

γπλ πγ

+++ sdot = Ψ

+ + (274-12)

is obtained Further by substituting Eq(274-1) to the above equation and arranging with

respect to R

( )( ) ( )

32

0 2

608 ( 2 )303 3

2ext

exto

DR R P RF R Z D

B P R kTRR Z

α γπλγλ

+ + + sdot = Ψ + ++ (274-13)

is obtained Solution of this equation determines the bubble radius R

Eq(274-13) indicates that R is dependent on Ψ The number density N is determined

by Eq(274-1) once R is determined Trapping rate and re-dissolution rate are determined

by substituting R and N into Eqs(274-2) and (274-3)

In actual numerical calculations by designating the m obtained from

Eq(274-7) as 1m and by designating the m obtained from Eq(274-9) as

2m

R which satisfies 2

1 mm = is obtained The range of R is assumed to be 10-15 (cm)

to 10-5 (cm) R is calculated by applying the dichotomy to R10log


- 88 -

275 Radiation re-dissolution and number density model of Intra-granular gas bubbles (GBFIS=1)

Similarly to the method in the previous section 274 it is assumed that per unit volume

of fuel pellet there are N intra-granular bubbles with average radius of R The following

assumption is set to obtain the changing rate of the number density N (bubblescm3) of

intra-granular bubbles

1) α bubble nucleus is generated from one fission

2) Fission fragment generates the bubble nucleus in the solid volume except the bubble

volume in the solid matrix That is it generates bubble nucleus in the solid volume which

has a fraction of 341

3N Rπ minus sdot

of the total solid volume

3) The generated bubble nucleus is immediately destroyed by punching of other fragments

This destruction rate is assumed as

22 ( ) ( )F R Z f R Nπ λsdot + (275-1)

where ( )f R implies the effect that the bubble is not destroyed by fission fragments when

the bubble has grown to reach a certain size That is if the bubble radius R is under R0

destruction rate is 10 if over R0 it is a partial destruction rate which depends on R as

follows

0

00

10 ( )

( )( )

k

R R

f R RR R

R

le= ge

(275-2)

Namelist Parameters k= KFIS (default=20) R0= RFIS (default= 20 nm)

Based on the above assumption changing rate of the number density N（bubblescm3）

during irradiation is obtained by incorporating ( )f R into the White and Tuckerrsquos equation as

3 20

42 1 2 ( ) ( )

3

dNF N R F R Z f R N

dtα π π λ = minus sdot minus sdot +

(275-3)

where F fission rate (fissionscm3s)

α bubble nucleus generation rate（=24 bubblesfission fragment）

R average radius of intra-granular bubble（cm）

λ fission track distance（＝ 46 10 cmminustimes ）

0Z affected range of fission (=10-7cm)

Then


- 89 -

3 20

3 20

42 2 ( ) ( )

3

2 2

4( ) ( )

3

dNF F R R Z f R N

dt

F FKN

K R R Z f R

α πα πλ

α

πα πλ

= minus + + sdot

= minus

= + +

(275-4)

holds Thus the numerical solution of Eq(275-4) is

1( 2 )exp( 2 ) 2i iN N FK F t FKαminus= minus minus sdotΔ + (275-5)

Assuming that the total number of gas atoms contained in an intra-granular bubble is m

by the number of gas atoms m (atomscm3) in bubbles in unit volume and by the number

density of bubbles N (bubblescm3)

m m N= (275-6)

is obtained Changes of m and m are determined by the difference between the number of

gas atoms entering bubbles by trapping and the number of gas atoms sputtered out by fission

fragment (re-dissolution)

The trapping rate g (s-1) of fission gas atoms by the intra-granular bubbles is given by

4g DRNπ= (275-7)

where D diffusion coefficient of fission gas atoms (cm2s)

Further setting that c is an average number of fission gas per unit volume of fuel pellet

containing the gas bubbles irrespective of whether the number of gas atoms Ψ (atomscm3)

which are dissolved in unit volume of matrix is in a solid phase of the matrix or in


Ψ = +c m (275-8)

holds Assuming that the re-dissolution rate to solid phase is

20303 ( )b F R Zπλprime = + (275-9)

2

04 303 ( )

( )

dmDRNc F R Z m

dtgc b m g g b m

π πλ= minus + sdot

prime prime= minus = Ψ minus + (275-10)

is obtained Accordingly the numerical solution of Eq(275-10) is

1( )exp( ( ) )i im m g g b t gminus prime= minus Ψ minus + Δ + Ψ (275-11)

At the same time assuming that the state equation can hold for the bubble gas

32 4

3

RP m B m kT

R

γ π + sdot minus sdot =

(275-12)


- 90 -

is obtained However in the present model the intra-granular bubble gas pressure is in balance

with the static pressure extP from external matrix and surface tension then

2gas extP P

R

γ= + (275-13)

holds

Here to determine the values of andm m R N at the i-step a numerical solution is

obtained which simultaneously fulfills Eqs(275-5) (275-6) (275-11) and (275-12)

At the same time since the bubbles increase their volume by influx of vacancies setting

that the vacancy concentration is nv and vacancy diffusion coefficient is Dv

4 2v v P

H gas

dn D RP

dt kT R

π γσ = + minus

(275-14)

gas

v v

m kTP

n=

Ω (275-15)

3

4

3

v vV n m B

Rπ

= Ω +

= (275-16)

are obtained(226 ) where

vΩ vacancy volume ( 29 3409 10 mminussdot )

Pγ UO2 pore surface energy (Jm2)

4 2

041

085 140 10 (Jm )

P SV SV

SV T

γ γ γγ minus

= sdot =

= minus times

GSV (0ltTlt2850C)

Default GSV=041 (227)

Also Turnbull(228) presents

2v V vD k S j C= sdot (275-17)

where S atom jump distance=1

3Ω =34454E-10 m

Ωatomic volume = ( )29 3409 10 mminustimes

( )13 410 exp 552 10 j RTν = minus times

Rgas constant =1987 calmoleK

Ttemperature (K)

Vk adjusting factor (KV=10 default)

C lattice defects concentration to which vacancy jumps

is approximated as 10(228)


- 91 -

------------------------------- 【Explanation of numerical solution】 -----------------------------------

To determine the values of andm m R N at the i-time step a numerical solution is

obtained which simultaneously fulfills Eqs(275-5) (275-6) (275-11) (275-12)

(275-13) and (275-14) For this purpose to determine a new value of ivn in Eq(275-12)

at the time step -i- the followings are set

1i i vv v

dnn n t

dtminus= + sdot Δ (275-18)

1

11

4 2i ii iv vH gasi i

dn D RP

dt kT R

π γσminus

minusminus

= + minus

(275-19)

The determined value of ivn by using these formulations is put into Eq(275-14) and

by using the value of m a new value of iR is determined

Here it is noted that ivn cannot exceed the value of vn which is obtained by 0vdn

dt=

because vn is a value at an infinite time elapse When calculating ivn by using Eqs(275-18)

and (275-19) it is not guaranteed that ivn will not exceed vn Therefore when calculating

ivn by using Eqs(275-18) and (275-19) it is necessary to set a sufficiently small time step

increment tΔ though setting of tΔ size is much difficult because the size depends on state

variables

Accordingly after determining the limiting value of vn paying a special attention not to

exceed this limiting value of vn Eq(275-19) is transformed by the Crank-Nicholson method

That is Eq(275-20) is the transformed equation of Eq(275-19) by the Crank-Nicholson

method

1

1

1

422

22

i ii

i iv

gas gasi ivi iH gasi

R RD P Pdn

PR Rdt kT

π γσ

minus

+

minus

+ +

= + minus +

(275-20)

Naturally when using Eq(275-20) iteration is required to obtain convergence

First calculation of vn derived from 0vdn

dt= is explained

Setting that R obtained from vn is R and gasP is gasP

4 20

iiv vH gasi

dn D RP

dt kT R

π γσ = + minus =

(275-21)


- 92 -

is obtained From this

( ) 2 0iH gasP Rσ γ+ minus = (275-22)

holds Also gasP and R in Eqs(275-21) and (275-22) are defined as follows

gasv

AP

n= (275-23)

3

vR nα β= + (275-24)

To substitute Eqs(275-20) and (275-21) into Eq(275-19) both sides of Eq(275-22)

are cubed

33 3( ) 8H gasP Rσ γ+ = (275-25)

By substituting Eqs(275-23) and (275-24) into Eq(275-25) and rearranging

( ) ( ) ( )3 4 2 3 3 3 2 2 2 3 2 33 8 3 3 3 0H v H H v H H v H vn A n A A n A A n Aασ ασ βσ γ ασ βσ α βσ β+ + minus + + + + + =

(275-26)

is obtained By using Eq(275-23) the solution in 0vn gt is obtained by the N-R method

-------------------------------------------------------------------------------------------------

276 Pekka Loumlsoumlnen model for intra-granular gas bubbles (GBFIS=2)

This model is an option in the rate-law model In this model the intra-granular bubble

radius R and its number density ( )oN N= are predicted by using the Loumlsoumlnenrsquos empirical

correlation(229) by assuming the relationship expressed by Eqs(274-2) and (274-3)

(1) Initial bubble radius is assumed as 0 001R = nm

Then the number density is determined

Number density 0 0

00762 Bu 104( ) ( )

98N Bu T N T

minus sdot += (cm-3) (276-1)

170 ( ) (157 000578 ) 10N T T= minus sdot times (cm-3) (276-2)

Bu burn up (MWdkgU) T Kelvin

However values of Eqs(276-1) and (276-2) are fixed at those at T=2173K above 2173K

(Cut-off)

(2) Average radius of intra-granular bubbles Next an average radius of growing bubbles is determined


- 93 -

1 3

0

3

4

m MR

Nπρ prime

=

(276-3)

where M is molecular weight of gas atoms and gas density 34 (gcm )ρ = is assumed

GROU=40 (default)

Next ldquogas concentration mrsquo (molecm3)rdquo is calculated which is defined as the total gas

(mole) in all the bubbles in one fuel grain divided by the grain volume (cm3)

04 ( ) ( ) ( )g g g

dmRDc r t N r t Fbm r t

dtπprime prime= minus (276-4)

173 10b = times =BBC (cm3) BBC= 173 10minustimes (default)

3 fission rate (1cm s)F sdot

However the initial gas concentration 0mprime is calculated by 0 001R nm= and Eq(276-3)

(3) Coalescence and growth of intra-granular bubbles Coalescence of bubbles is predicted by the following empirical equation when the

temperature exceeds the coalescence threshold mgT

8000

273159

ln0005

mgTBu

= +minus

(276-5)

( )4 4( ) 1 exp( 83 10 ) ( )exp( 83 10 )g F ini gR r t R t R r t tminus minus= minus minus times + minus times (276-6)

100 nmFR = coarsened radius RADMG=100nm (default)

iniR radius before coarsening


t elapsed time from the start of bubble coalescence (s)

In Eq(276-5) if 91Bu le Bu is fixed at 91 By Eq(276-6) the coalescence

terminates after about 15 hour and the bubbles become a single bubble with radius of

RAMDG (nm) Then in the process of this change the number density 0N and bubble gas

concentration mrsquo change though as an approximation mrsquo is assumed to hold the

pre-coalescence value as a kind of equilibrium state That is after coalescence Eq(276-4)

is no longer used and gas atoms influx to and re-dissolution from the bubbles are assumed to

maintain the equilibrium

Here the number density of bubbles at the start and end of coalescence and after that is


- 94 -

expressed as

0 3

3 ( )( )

4g

g

m r t MN r t

Rπ ρprime

= (276-7)

1) R Rρlt 0ρ ρ= Rρ RROU (Default=10 nm)

2) R Rρge 0

N

iniR

R

ρ

ρ ρ =

N ρ NROU (default=10)

When the temperature once exceeds mgT and then falls below mgT the bubble radius is

assumed to remain at 100 nm Shrinkage of the intra-granular bubbles is associated with so

many factors and there are only scarce observation data Consideration of the bubble

shrinkage remains a future task

277 Galerkinrsquos solution method for partial differential diffusion equation (common in IGASP=0 and 2)

Solving the partial differential diffusion equation (273-11) for the diffusion of fission

gas atoms uses the FEM in spherical geometry assuming that the grain is spherical and inside

the elements formulation is made by using the Galerkinrsquos weighted residual method

The Galerkinrsquos method as one of the FEM allows a precise solution with a small number

of elements and accordingly in a short calculation time Also it is easy to treat the

re-dissolution and this method is theoretically adaptable to transient situation

Eq (273-11) can be rewritten as

02 =minusΨnablaprimeminuspartΨpart βDt

(277-1)

where an approximation function of the function Ψ(t r) is set as follows

( ) ( ) ( ) ( )Ψ Ψ Ψt r t r t rj jj

asymp = φ

Here φj(r) is an arbitrary known function called the basic function and Ψ j is an

unknown coefficient where the residuals are represented as

( ) βminusΨnablaprimeminuspartΨpart=Ψ

ˆ 2Dt

rtR (277-2)

The coefficient Ψ j which provides the minimum residuals in Eq(277-2) can be

determined by requiring a condition in which weighted integration becomes zero When the

basic function is selected as a weight function this requirement can be satisfied by the


- 95 -

equation

( )R t r dviV Ψ φ = 0 (277-3)

To satisfy Eq(277-3) for an arbitrary weight function of φ j RΨ(t r)=0 must be

satisfied Therefore Eq(277-3) is equivalent to Eq(277-1)

Integrating Eq(277-3)

2ˆ

ˆ 0iVD dv

tβ φ

partΨ primeminus nabla Ψ minus = part

2ˆ

ˆ 0i iV Vdv D dv

tβ φ φ

partΨ primeminus minus nabla Ψ = part (277-4)

are obtained The second term of Eq(277-4) can also be expressed as

( ) ( ) ( )

( ) ( ) ˆˆ

ˆˆˆ2

nablasdotΨnablaprime+sdotΨnablaprimeminus=

nablasdotΨnablaminusΨnablasdotnablaprimeminus=Ψnablaprimeminus

nabla V ii

V iV iV i

dvDsdD

dvdvDdvD

φφ

φφφ

part

(277-5)

Here when a weight function which satisfies iφ =0 at the grain boundary Vpart is selected the

surface integration term is eliminated and Eq(277-4) is expressed by

( ) ( )ˆ

ˆ 0i iV Vdv D dv

tβ φ φ

partΨ primeminus + nablaΨ sdot nabla = part (277-6)

Then in Eq(277-6) ˆj jφΨ = Ψ is set in a certain finite elementＶｅ Here

jΨ is the

fission gas atoms concentration at three nodes of the element eV and jφ is an interpolating

function inside the element Originally the previously defined iφ is an arbitrary function

group though derivation from Eq(277-3) can hold as-is if this iφ is defined as the

interpolating function as above Accordingly this function is treated as the same function

ie iφ ＝jφ

Then in Eq(277-6) fission gas generation term βis a constant inside the element so

that

( ) ( ) 0jV j i V j j idV D dV

t

partφ β φ φ φ

partΨ primeminus + nabla Ψ sdot nabla =

(277-7)

can be written In Eq(277-7) concentration at node jΨ and generation term β can be


- 96 -

put out from the volume integral and

0jV j i V i V j i jdV dV D dV

t

partφ φ φ β φ φ

partΨ

primesdot minus sdot + nabla sdot nabla sdot Ψ = (277-8)

is obtained Consequently for a certain element eV Eq(277-9) and the following three

equations are derived or FEM divides Eq(277-7) into the following finite elements

( contents inside sum of Eq(277-9))

( )Ψ Ψj ji j ji ij

E D A H+ prime minus = β 0 (277-9)

where

Veji j i

j iji Ve

i Ve i

E dV

A dVr r

H dV

φ φ

partφ partφpart part

φ

=

=

=

(277-10)

Then the elements in the present geometry is set as quadratic and the following

quadratic functions are adopted as basic functions φ j

( ) ( )

( )

( ) ( )ξξξφ

ξξφ

ξξξφ

+=

minus=

minusminus=

12

1

1

12

1

3

22

1

(277-11)

In introducing FEM a direct use of real coordinates is often inconvenient in

calculation To facilitate the following development of equations an actual

element region is converted into -1 to +1 region in a relative coordinate system

ξ A quadratic element has three nodes In this relative coordinate system the

nodes are atξ= -1 0 and 1 In this case an arbitrary extensive variable ie

fission gas atoms concentration is given by an approximate function and this

function values can be determined by the values at the nodes This means that

the function must be an approximate function that guarantees the values at

nodes Eq(277-11) which is the basic function (interpolating function) in the

quadratic elements takes the following values to guarantee the continuity at

each node


- 97 -

( ) ( )1

11

2φ ξ ξ ξ= minus minus

gives 1 atξ= -1 and 0 at ξ=0

( ) 22 1φ ξ ξ= minus gives 1 atξ=0 and 0 at ξ= -1 and 1

( ) ( )3

11

2φ ξ ξ ξ= + gives 1 atξ=1 and 0 at ξ= -1 and 0

Therefore a basic function within the element ie the interpolating function to express

the gas atoms concentration profile in the radial direction is expressed as

( ) ( ) ( ) ( )21 2 3

1 1( 1 ) 1 ( 1 )

2 2ξ ξ ξ ξ ξ ξΨ = minus minus Ψ + minus Ψ + + Ψ

(277-12)

and

( ) 11Ψ minus = Ψ ( ) 20Ψ = Ψ ( ) 31Ψ = Ψ (277-13)

holds

Figure 274 schematically shows Eq(277-11) by the red broken curve inside the grain

and Eq(277-12) by the red solid line

Using these basic functions Eji Aji and Hi in Eq(277-11) are represented using Rk ie

coordinates of the midpoint of element Ve and using half a element width ΔR as follows

gas atom concentration

boundary layer

Fig274 Gas concentration profiles in FEM element of a grain which is

approximated by sphere


- 98 -

( )

minus

Δ+

Δminus

Δ+Δ=

==

1

1

2

2

4

4

ξξφφπ

φφπφφ

dRRR

drrdVE

kij

RR

RR ijVe ijji

K

K (277-14)

( )

2

1 2

1

4

4

k

Ve k

R Rj ji iji R R

j ik

A dV r drr r r r

R R dR

partφ partφpartφ partφπpart part part partpartφπ partφ ξ ξpartξ partξ

+Δ

minusΔ

minus

= =

= + ΔΔ

(277-15)

( )1 22

14 4

k

Ve k

R R

i i i i kR RH dV r dr R R R dφ π φ π φ ξ ξ

+Δ

minusΔ minus= = = Δ + Δ (277-16)

Next Eq(277-10) is represented by an implicit formula

( )( )1

1 1 0n nj j n n

ji j j ji ij

E D A Ht

θ θ β+

+ Ψ minus Ψ + minus Ψ + Ψ minus = Δ (277-17)

Here

( )( )

D D Dn n

n n

= minus prime + prime

prime = minus +

+

+

1

1

1

1

θ θβ θ β θβ

(277-18)

The reason to use an implicit method here is as follows The diffusion equation is

included in the FGR model and the FGR model is set in a temperature convergence

calculation process Accordingly there are two temperatures Tn at the n-th (previous)

step and Tn+1 at the n+1th (current) step so that the current value is brought into the

convergence process by using the implicit method Also there are njΨ and 1n

j+Ψ for

fission gas atoms concentration The implicit method eg θ =1 allows fission gas

atoms to diffuse by concentration gradient in 1nj+Ψ The diffusion equation in

Eq(277-17) enables us to obtain 1nj+Ψ on the basis of the concentration gradient

derived from the unknown quantity 1nj+Ψ by setting θ =1 in matrix calculation

(implicit method with respect to jΨ ) Introduction of the implicit method allows a

larger time step increment which beneficially reduces computing time

Through rearrangement of the unknown and known values of Eq(277-17)


- 99 -

( ) 1 1 11n n

ji ji j ji ji j ij j

E D A E D A Ht t

θ θ β+ + Ψ = minus minus Ψ + Δ Δ (277-19)

is obtained Therefore the element matrix Ｗij and the element vector Qi are

1ij ji jiW E D A

kθ= +

Δ (277-20)

( ) 11 n

i ji ji j ij

Q E D A Ht

θ β = minus minus Ψ + Δ (277-21)

Through superimposition of the equations for all elements and setting of the condition of

Ψn+1 = 0 on the surface according to the assumption that the gas atoms concentration is zero at

grain boundary Ψjn+1 can be obtained by solving the simultaneous equations below

[ ] W Qij jn

iΨ + =1 (277-22)

278 Fuel grains and boundaries (common in IGASP=0 and 2)

It should be noted that the diffusion explanation described until now and the logic to

determine ldquoboundary accumulation amount nfn rdquowhich will be discussed hereafter assumes a

situation that there is one single grain exists in space and grain boundary surrounds its surface

In an actual case fuel pellet is poly-crystal

and the ring element of pellet stack consists of a

number of crystal grains Since two neighboring

crystal grains share a boundary it is necessary to

take into account of such actual conditions when

considering the concentration and amount of

fission gas atoms at grain boundary

Here attention is focused on one grain

boundary Since concentration of gas atoms at

the boundary increases by the influxes diffused

from the two adjacent grains the actual

concentration of gas atoms at boundary is 2

times that of the result of diffusion calculation

Now in Fig275A which illustrates the grains by two-dimensional hexagonal shape for

Fig275A Schematic of grain

boundary (1)

A B


- 100 -

simplicity the area of a partial boundary PB of the central grain A with radius ldquoardquo is

one-sixth of the total boundary area The diffusion influx from the adjacent grain B comes into

this PB similarly to the influx from the grain A The rest five-sixth boundary has an identical

situation After all the actual concentration of fission gas atoms accumulated at the boundary

including grain boundary gas bubbles is equivalent to 2-times

that of the diffusion calculation result as shown in Fig275B

However in the boundary gas inventory Eq(279-1)

which is explained in the section 279 below this 2-times

multiplication is not necessary because the amount of gas

atoms which diffused from each grain to boundary is

multiplied by the number of grains contained in one ring

element of pellet Also in the calculation of released gas

amount described in section 2711 the number of grains per

one ring element is taken into account

Furthermore because the gas release threshold N fmax

which is described in section 2714 is simply derived by applying the state equation to the

gas in a representative bubble in which gas pressure is in balance with the compressive stress

from surrounding matrix and surface tension this N fmax

itself does not have to be changed

In other words FEMAXI models that grains and boundaries in each ring element are

represented by one single grain and its surrounding boundary which eliminates the

consideration of the ldquodouble amountrdquo in actual grain boundaries 279 Amount of fission gas atoms migrating to grain boundary (common in IGASP=0 and 2)

Amount of gas atoms 1nfn +Δ (atoms) diffusing to grain boundary from inside of a grain

during time step tΔ is calculated on the basis of the change of fission gas atoms inside grain

as a baseline Assuming that

total amount of fission gas atoms inside grain at time step n ntotalΨ (atoms)

amount of fission gas atoms generation during tΔ V

tdVβΔ (atoms)

total amount of fission gas atoms in grain after tΔ

(at time step n+1)1n

total+Ψ (atoms)

Fig275B Schematic of grain

boundary (2)


- 101 -

1 1n n nf total totalV

n tdVβ+ +Δ = Ψ + Δ minus Ψ (279-1)

is obtained where ntotalΨ and 1n

total+Ψ are determined as follows by using the results of section

277 in the case in which the grain has three elements

( )

( ) ( )

32

1

3

1

ˆ4

ˆ

1

e

k ktotal V

e

k kj j

j

r r dr

r r

k n n

π

φ

=

=

Ψ = Ψ Ψ = Ψ

= +

(279-2)

Here if grain growth occurs the amount of gas atoms 1nfn +Δ which move to grain

boundary during tΔ is determined in the following way setting that the grain radius before

growth is na and the radius after the growth is 1na + which is a baseline radius (refer to the

next section 2710) and using the effective fission gas atoms generation rate β which is

defined by Eq(2712-2) later

Amount of fission gas atoms inside grain at time step n is

3

11 nntotal

n

a

a++

sdot Ψ

(279-3)

where radius 1na + is a baseline and amount of fission gas atoms generated during tΔ is

( )31

4

3 n Vt a tdVβ π β+ sdot Δ = Δ

(279-4)

consequently 3

1 11n n nnf total totalV

n

an tdV

aβ+ ++

Δ = Ψ + Δ minus Ψ

(279-5)

is obtained

2710 Amount of fission gas accumulated in grain boundary

In the fission gas atoms diffusion calculation thus far it has been assumed that the gas is

in an atomic molecule except in the intra-granular gas bubbles However in the discussion

of grain boundary bubbles a gas state is assumed

A value is considered which is obtained by dividing the total amount of fission gas atoms

ie grain boundary inventory invG accumulated at grain boundary of unit volume of pellet

ring element at time step -n- with the total number of grains in the unit volume of ring


- 102 -

element Here all the grain boundaries in the ring element -i- having a volume of iringV is

considered Assuming that the unit volume of pellet consists of igrainN grains with radius

ofga

3

143

igrain

g

Naπ

= (2710-1)

holds

Determination of grain boundary accumulation amount nfn (or 1n

fn + )(atoms) assumes a

situation in which two kinds of movements of gas atoms are present atomic diffusion from

inside of grains and gaseous release to external free space Therefore a relationship n i

inv f grainG n N= (2710-2)

holds Thus invG times the volume of each ring element is the inventory in the element

(1) Equilibrium model (IGASP=0)

If we assume that the amount of fission gas retained in the grain boundary at time step n

is nfn (atoms) amount of gas in the boundary after tΔ will be the smaller one of

1n nf fn n ++ Δ or max

fn ie

1 1 max 1min( )n n n nf f f fn n n n+ + += + Δ (2710-3)

Here maxfn is the maximum amount of gas that can be retained in the grain boundary

Using maxfN (atomscm2) as the saturation value of number of gas atoms in the unit area of

grain boundary maxfn is expressed by

max 1 2 max 114n n

f n fn a Nπ+ ++= sdot (atoms) (2710-4)

The present model assumes that when amount of fission gas at the boundary exceeds maxfn ( max

fN ) excess amount of gas is immediately released to free space of rod and after this

gas transported by diffusion from inner region of grain is released as long as the amount of gas

in the bubble is above maxfn ( max

fN )

In the case of grain growth [see the next section 2712] amount of fission gas retained at

the boundary at time step n is

3

1 nnf

n

an

a+

sdot

(2710-5)


- 103 -

and amount of gas transported to the boundary during tΔ is 1nfn +Δ

Then amount of gas retained at the boundary after tΔ is given by

3

1 1 max 11min n n n nnf f f f

n

an n n n

a+ + ++

= sdot + Δ

(2710-6)

(2) Rate-law model (IGASP=2)

Amount of fission gas retained in the grain boundary after tΔ is a value which is

obtained by subtracting the released amount rlsfn from

1n nf fn n ++ Δ This value is a grain

boundary inventory of one grain when gas release occurs It is

1 1 1n n n rls nf f f fn n n n+ + += + Δ minus

(2710-7)

where 1rls n

f releasen n t+ = timesΔ (2710-8)

and releasen is the gas release rate which is explained in section 28 Total quantity of gas

release from a ring element is given by

1i rls ngrain fN n +times

(2710-9)

In the case of grain growth [see the next section 2712] amount of fission gas retained

in the grain boundary at time step n is

3

1 nnf

n

an

a+

sdot

(2710-10)


Then amount of gas retained at the boundary after tΔ is given as

3

1 1 11n n n rls nnf f f f

n

an n n n

a+ + ++

= sdot + Δ minus

(2710-11)

2711 Amount of released fission gas


Assuming that the amount of fission gas per one grain released to outer free space from

pellet during tΔ is 1nrn +Δ (atoms) in the case of no grain growth we have


- 104 -

( )1 1 max 10n n n nr f f fn n n n+ + +Δ = + Δ le (2711-1)

( )1 1 max 1 1 max 1n n n n n n nr f f f f f fn n n n n n n+ + + + +Δ = + Δ minus + Δ gt (2711-2)

In the case of grain growth [see next section] we have

3

1 1 max 110n n n nnr f f f

n

an n n n

a+ + ++

Δ = sdot + Δ le

(2711-3)

3 3

1 1 max 1 1 max 11 1n n n n n n nn nr f f f f f f

n n

a an n n n n n n

a a+ + + + ++ +

Δ = sdot + Δ minus sdot + Δ gt

(2711-4)

Also decrease in gas concentration 1nrN +Δ (atomscm3) of grain boundary by the release

during tΔ is given by

1

12

1

nn rr

n

nN

aπ

++

+

ΔΔ = (2711-5)

Total quantity of gas release from a ring element is given by

1i ngrain rN n +timesΔ (2711-6)


Refer to the previous section 2710(2)

(3) Fission gas release rate FGR (common to IGASP=0 and 2)

Total quantity of gas released from one grain during tΔ from the 1-th to n-th time steps

is given by

1

nkr

k

n=

Δ (2711-7)

and the gas release rate is assuming that the total amount of fission gas atoms generation in

one grain corresponding to the burnup at n-th time step is ntotalFPG given by

1

nkr

kntotal

nFGR

FPG=

Δ=

(2711-8)


- 105 -

2712 Sweeping gas atoms to grain boundary by grain growth


During grain growth the gas in grains is swept into the grain boundary The rate of gas

sweeping into the grain boundary is given by

3

1 1ng

n

af

a+

= minus

(2712-1)

Here

1na + grain radius after grain growth (cm)

na grain radius before grain growth (cm)

Figure 276 shows a schematic of the grain growth model The gas present in the

region with hatched lines is released at the grain boundary during grain growth

an+1

an

Fig276 Conceptual diagram of grain growth

The grain growth model uses the following Itohrsquos model as a representative one(230) (231)

da

dtK

a

N N

af f

m

= minus+

1 1 max

(μmh) (2712-2)

( )K RT= times minus times5 24 10 2 67 107 5 exp (μm2h)

a grain radius at current time step (μm)

am maximum grain radius ( )( )= times minus2 23 10 76203 exp T (μm)

R gas constant (=8314 JmolK)

Nf gas atom density at grain boundary (atomscm2)

Nfmax saturation value of the gas atom density at grain boundary (atomscm2)

[Fitting Parameter FGG] The rate of gas sweepings is controlled with FGG as fg = fg times FGG


- 106 -

Fitting parameters AG and GRWFThe grain growth rate can be controlled using AG and GRWF as follows 1) Control of the gas atom density Nf = Nf times AG (Default value AG=15)

2) Grain growth rate GRWFda da

dt dt= times

(Default value GRWF=10)

The option of the grain growth model can be selected using IGRAIN as follows IGRAIN=0 Itohrsquos model(default) =1 Ainscough model =2 MacEwan model =3 Lyons model =4 MATPRO-09 model

Selection of option IGRAIN

In the diffusion calculation using FEM each grain consists of 3 elements as shown in

Fig277 but can have 5 elements at the maximum

Boundary

layer

Node

Fig277 Element division of a pellet grain

The outermost element is the boundary layer which is a special element for

consideration of re-dissolution When grain growth occurs during time step from n to n+1

elements (mesh) in the grains are re-meshed

Setting the boundary layer width as 2ΔR coordinates r at the node of the first and second

elements are given as

11

2

2n

n nn

a Rr r

a R+

+minus Δ=

minus Δ (2712-3)

where

rn+1 coordinates at the node point after grain growth and

rn coordinates at the node point before grain growth

Coordinates at the three node points in the outermost boundary layer are given as

1 2na R+ minus Δ 1na R+ minus Δ and 1na +

The generation term β is changed by taking account of the dissolution only at the

interface element These elements are of ldquothree-node quadratic elementrdquo with the assumed

boundary condition of zero concentration of gas atoms at the grain boundary


- 107 -

The fission gas atoms concentration Ψ at a given set of coordinates after re-meshing is

given by

Ψn+10(r) = Ψn(r) (0 le r le an) (2712-4)

Ψn+10(r) = 0 (an le r le an+1)

where

Ψn+10(r) initial concentration distribution of fission gas after grain growth

Ψn(r) fission gas concentration distribution before grain growth

Namely it is assumed that sweeping of the gas into the grain boundary due to grain

growth occurs instantaneously

Next fission gas atoms concentration at unit area of grain boundary fN is obtained

Expressing Eq (2710-3) again


Here in the case where no grain growth occurs using 1n na a +=

( )

11 max 1

21

1 max 1

min 4

min

n nf fn n

f fn

n n nf f f

n nN N

a

N N N

π

++ +

+

+ +

+ Δ=

= + Δ

(atomscm2) (27-12-5)

holds However since grain growth is assumed in this section

Δ+sdot=

Δ+sdot

=

+++

+

+

++

+

1max11

1max2

1

1

3

1

1

min

4

min

nf

nf

nf

n

n

nf

n

nf

nf

n

n

nf

NNNa

a

Na

nna

a

Nπ

(2712-6)

holds where24

nfn

fn

nN

aπ= and

11

214

nfn

fn

nN

aπ

++

+

ΔΔ =

Model parameters NODEG and RREL The number of elements in grains is designated by

NODEG The default value is 3 and the maximum value is 5 The width of layers other than

the outermost layer namely the elements of NODEG-1 is designated by the ratio RREL The

default value is 5 1 which means that the ratio of the width of the first to second layers is 51


- 108 -


In the case where the grain growth occurs

3

11

12

1

3

2 11

21

11

4

4

4

n nnf f

nnf

n

n nnn f f

n

n

n nnf f

n

an n

aN

a

aa N n

a

a

aN N

a

π

π

π

++

+

+

++

+

++

sdot + Δ

=

sdot + Δ

=

= sdot + Δ

(2712-7)

holds where 24

nfn

fn

nN

aπ= and

11

214

nfn

fn

nN

aπ

++

+

ΔΔ =

If no grain growth occurs 1n na a += is set in the above equations

(3) How to calculate 1n

fN + (common in IGASP=0 and 2)

Here how to calculate the change nfNrarr

1nfN +

during tΔ ie time step nrarrn+1 when

grain growth occurs is explained To determine 1nfN + it is necessary to take into account of

the transport of gas atoms from inside of grain to the boundary and their re-dissolution during

grain growth period tΔ so that iteration calculation is required Assuming that the initial

value of this iteration is N fn+1 0

( )3

2 2 10114 4n n n nn

total n f total n fn

aa N a N

aπ π ++

+

Ψ + sdot = Ψ + sdot

(2712-8)

holds where Ψtotaln total number of fission gas atoms inside the grain

before grain growth (atoms)

Using Eq(2712-1) and solving Eq(2712-8) with respect to N f

n+1 0 give

10 12

14

ng totaln nn

f fn n

f aN N

a aπ+ +

+

Ψ= + (2712-9)

where 2

14

ng total

n

f

aπ +

Ψ is the amount by grain-growth sweeping and n

fN is the grain boundary

concentration at the beginning of time step


- 109 -

Here if i is supposed to be a count index of iteration

1

1 12

1

4

n

ng totaln i n in

f fr an n

f aN N D t tdV

a a rβ

π+

+ +

=+

Ψ partΨ= + minus sdot Δ minus Δpart (2712-10)

holds The third term of the right hand side of Eq(2712-10) is the amount of fission gas

atoms migrated from inside of the grain to the boundary the fourth term is the amount of fission gas atoms that return to the inside of grain by re-dissolution and iβ is the

re-dissolution rate defined by

( ) ( )1 1 11

2 2i n n i

f f

bN N ADDF

Rβ θ θ + minusprime = minus +

Δ (2712-11)

Eq(2712-11) will be explained in the next section

Now the first and second terms of the right hand side of Eq(2712-10) are known by

Eq(2712-9) Consequently the following process holds

A) Substituting iβ given by Eq(2712-11) into Eq(274-11) and solving the

equation 1 11 1( ) ( )n n

n na a r+ ++ +Ψ Ψ minus Δ and 1

1( 2 )nna r+

+Ψ minus Δ are obtained

B) Using these values the third term of the right hand side of eq(2712-10)

1nr a

D tr

+=

partΨminus sdot Δpart

(2712-12)

is determined

C) Substituting Eqs(27112-11) and (2712-12) into Eq(2712-10) proceeds the

iteration count and 1n ifN + is obtained By this value β in Eq(2712-11) is

updated

Then the steps A) to C) are repeated until convergence is attained ie

1 1 1

1

n i n if f

n if

N N

Nε

+ + minus

+

minuslt is satisfied

However in the actual numerical calculation of N f

n+1 the following equation which is

equivalent to eq (2711-10) is used

( ) 1

1 12

1

1

4

n ng total total Vnn n

f fn n

f tdvaN N

a a

βπ

++ +

+

+ Ψ minus Ψ + Δ= + (2712-13)

Here Ψtotaln +1number of fission gas atoms in grains at the end of time step (atoms)


- 110 -

β Δtdv amount of fission gas atoms (atoms) generated excluding the

amount of re-dissolved quantity during time step tΔ where 3

1

4

3V ndv aπ + = holds

2713 Re-dissolution of fission gas (common in IGASP=0 and 2)

In the model of fission gas atom re-dissolution from grain boundary into grain it is

considered that the boundary separates adjacent grains In sections 288 to 2810 amount of fission gas atoms ( fn fN ) which are accumulated at the boundary of one grain is

considered

As explained in section 288 in the actual grains the amount of gas atoms accumulated at boundary is two times larger than ( fn fN ) due to influxes of gas atoms from the adjacent

two grains However as the re-dissolution occurs for the two adjacent grains concurrently for

each one grain the re-dissolution rate can be evaluated with respect to the amount of fission gas atoms fN per unit area at boundary Namely re-dissolution amount is given by

ADDFfbN sdot (atomscm2s) (2713-1)

where b re-dissolution rate (10-6s)

The re-dissolution means physical transfer of fission gas atoms from the grain boundary

into grains which is virtually equivalent to an additional generation of fission gas atoms in

grains Accordingly when the thickness of regions where fission gas re-dissolves is assumed

to be equal to the boundary layer thickness (200 Aring) the apparent rate of fission gas atoms

generation primeβ in the boundary layer is determined

Therefore primeβ is given by the following equation

( ) ( )1 11 ADDF

2 2n nf f

bN N

Rβ θ θ +prime = minus +

Δ (atomscm3s) (2713-2)

Here Nf

n concentration of fission gas atoms in the grain boundary at time step n (atomscm2)

N fn+1 concentration of fission gas atoms in the grain boundary at time step(n+1)

(atomscm2)

θ interpolation parameter between time steps in the implicit solution

where values of Nfn and N f

n+1 do not depend on the occurrence of grain growth

Accordingly β in Eq (2713-2) for the boundary layer is expressed as


- 111 -

Model parameter RB This parameter controls the thickness of boundary

layer in which fission gas atoms are re-dissolved RB specifies the

thickness Default value is 2times10-6cm

~β β β= + prime (2713-3)

and ~β is defined as the effective fission gas generation rate

2714 Threshold density of fission gas atoms in grain boundary (Equilibrium model IGASP=0)

The gas release model described in the previous section 281 ie White+Tucker model

assumes that when number of fission gas atoms at grain boundary (= area density of fission

gas atoms) exceeds a threshold value the isolated babbles are connected and tunneling path to

free space is formed and the gas atoms diffused from inside of grain are instantaneously

released to free space In other words in the model described in section 271 the diffusion

equation is solved but the growth of grain boundary bubbles is only implicitly considered and

not directly calculated Furthermore the model does not contain the idea of number density of

bubbles in the boundary Nevertheless the model calculates the threshold ie upper limit of

gas atoms concentration in the boundary for the onset of gas release by definition equation

including the grain boundary bubble radius

The method to determine this threshold (saturation) concentration maxfN (atomscm2) is

explained in the followings(224)

(1) Radius of grain boundary gas bubble

(common to equilibrium model and rate-law model)

As shown in Fig278 fr is defined as a radius of curvature of lenticular bubble shape

and br is the radius of the orthogonal projection of the bubble on the grain boundary surface

We have

sinb fr r θ= (2714-1)

The amount of gas re-dissolved is controlled

using ADDF The default value of ADDF is 90

Fitting parameter ADDF


- 112 -

(2) Average concentration and release threshold of grain boundary gas atoms

By applying the state equation of gas to the bubble gas

1gas bubbleF P V m kT (2714-2)

holds where 1m is the number of gas atoms considered in one grain geometry It is the total

amount of gas atoms diffused from inside of one grain to its boundary and the bubble size

corresponds to the condition in which this amount of gas is concentrated in a hypothetical

single gas bubble Here by using the van der Waals constant B of fission gas (Xe+Kr) a

modification factor F is used to take into account of the slight deviation from an ideal gas

1

1 gas

FP B

kT

=

+

(2714-3)

Volume of a bubble is

34( )

3bubble f fV r f (2714-4 )

By the equilibrium assumption

2

gas extf

P Pr

(2714-5)

holds Obtaining the average gas atom density fN by dividing the number of gas atoms m1

by the area which a bubble occupies in the boundary we have

Fig278 Curvature radius of lenticular bubble and its orthogonal projection radius

on the boundary

Grain boundary


- 113 -

12

2

( sin )

4 ( ) 2

3 sin

ff

fext

f

mN

r

r fP F

kT r

π θ

θ γθ

=

= +

(2714-6)

The threshold or upper limit maxfN considered as a quantity for total grain boundary area is

assuming that fb is the covering fraction of lenticular bubbles over grain boundary surface

maxf f bN N f= sdot (2714-7)

From Eqs(2714-6) and (2714-7) we obtain (224)

( )max

2

4 2

3 sinf f

f b extf

r fN f P F

kT r

θ γθ

= +

( )2atomscm (2714-8)

Here N fmax saturation value of the number of gas atoms per unit area at the

grain boundary (atomscm2)

rf intergranular gas bubble radius _=RF (default=05μm)

ff(θ) volume ratio of lenticular bubble to sphere bubble

θ lenticular bubble angle (=50deg)

γ surface tension (626 ergcm2) (220)

k Boltzmanns constant (=138times10minus16 ergK)

fb covering fraction of lenticular bubbles over grain boundary surface

_=FBCOV (default=025)

Pext external force (dynecm2)

The external pressure Pext is obtained using a procedure described below taking into

account the plenum pressure contact pressure between pellet and cladding and pellet internal

thermal stress

Here instead of considering directly the number of grain boundary bubbles or their

actual size the amount of gas moved to grain boundary is considered first and the average gas

atoms concentration in the boundary at the onset of gas release is defined by Eq(2714-8)

In Ref(224) as the specific parameter values to determine N fmax 05μmfr = o50θ =

and 025bf = are presented However PIE observation data suggests that

05 10μmfr = minus and 05bf cong are considered to give more realistic prediction of FGR


- 114 -

2715 In-grain diffusion coefficient of fission gas atoms (common to IGASP=0 and 2) (1) Diffusion coefficient models Details of diffusion coefficients models which are incorporated in FEMAXI-7 are

presented in Chap4 The temperature dependent curves calculated by using some

representative models are shown in Fig279

(2) Fitting function of temperature dependency of coefficient

FEMAXI-7has a function to adjust the temperature dependency of diffusion coefficient

by multiplying a factor FGDIF However when FGDIFX=0 (default) this function is

ineffective and FGDIFX=1 effective

Input parameters OPORO FBSAT ALHOT Fraction of inter-granular bubbles which have always conduit path to free surface is specified by OPORO (default=00) FBSAT is a multiplication factor for the saturation value of inter-granular gas bubbles (Default=10)

500 1000 1500 200010-22

10-21

10-20

10-19

10-18

10-17

10-16

Diff

usio

n C

onst

ant

(m2 s

)

Temperature (K)

Turnbull original Modified Turnbull White 100Wcm White 200Wcm 300Wcm 400Wcm Kogai original Kogai case D White+Tucker Kitajima Kinoshita

Fig279 Comparison of temperature dependency of representative diffusion coefficient models for fission gas atoms in grain


- 115 -

4 5 6 7 8 9 10 11 12 13 14 15 161E-22

1E-21

1E-20

1E-19

1E-18

1E-17

1E-16

1E-15

Diff

usio

n co

nsta

nt (

m2 s

)

10000T (1K)

Turnbull Kitajima K White+Tucker modified Turnbull

Enhanced Turnbulby magnifying factor

(FGDIF0-1)

FGDIF FGDIF010000

9 exp 7

a

a a aT

sdot= minus + sdot sdot minus minus (2715-1)

6a = (default) T (Kelvin)

where name-list parameter a = EFA=6 (default) FGDIF0=10 (default) and FGDIFX =0

(default)

The temperature dependency of this factor is shown in Fig2710 In this case the

diffusion coefficient increases in high temperature region Accordingly as indicated in

Fig2711 in the default condition the original Turnbullrsquos curve is 10-times enhanced above

about 1250K

However this is only an example An optimization on the basis of experimental data is

indispensable

Fig2710 Examples of the values of temperature dependency adjusting factors


- 116 -

5 6 7 8 9 100

2

4

6

8

10

FGDIF0=5 a=6

FGDIF0=10 a=4

FGDIF0=10 a=8

FGD

IF M

agni

fyin

g fa

ctor

10000T (1K)

FGDIF0=10 a=6 default

Fig2711 An example of enhanced diffusion coefficient in high temperature region

of the Turnbull model by applying the temperature dependency adjusting

factor FGDIF


- 117 -

28 Rate-Law Model (IGASP=2)

The rate-law model has been explained only partially in sections 272-2 (2)

2710(2) and 2712(2) Here in this section an essential explanation is described

281 Assumptions shared with the equilibrium model

(1) Calculations of diffusion and intra-grain gas bubble growth

The rate-law model has common model and algorithm to those of the equilibrium

model for the diffusion of gas atoms from inside of grain to grain boundary trapping by

intra-grain bubbles and influence on the effective diffusion rate by growth of intra-grain

bubble growth etc In other words the rate-law model is applied to only the grain boundary

gas bubble growth gas release and coalescencecoarsening of grain boundary bubbles

Also either the irradiation re-dissolution model or Pekka Loumlsoumlnen model which is

explained in section 276 is applied to the intra-grain bubble growth model

(2) Criteria for gas release Gas release from grain boundary gas bubbles is theoretically dependent on the

conductance of gas pass connecting the bubbles and pellet free surface and on pressure

difference between the free space and gas bubbles and on temperature However in the

present model considering that the quantitative evaluation of pass conductance is difficult

and that the known assumption of ldquogas release by tunneling of bubblesrdquo has worked

successfully in predicting FGR behavior as described in section 2714 and below it is

assumed that when either the radius of grain boundary bubble fr or grain boundary

coverage fraction bf reaches a certain threshold level gas in the bubbles begins to be

released to free space

282 Growth rate equation of grain boundary bubble

(1) Grain boundary bubble radius its number density and grain boundary inventory Based on the discussion in section 278 a hypothetical situation is first supposed in

which one isolated crystal grain exists in a space and ldquograin boundaryrdquo wraps its surface

and grain boundary gas bubbles of the grain are generated by the gas atoms diffusion from

inside of the grain

Then bbN the number of grain boundary bubbles of a single grain is defined as


- 118 -

24bb g densN a Nπ= (282-1)

where ga grain radius densN number density of grain boundary bubbles (bubblescm2)

Here it is assumed that

1 densN d= (μm2) d=183x183 (μm2) (282-2)

Namely default value of a name-list parameter NDENS corresponding to densN = is

311x10-2 Also when bubbles coalesce bbN ( densN ) decreases in inverse proportion to

the times of coalescence MGB (See section 282 (3))

Assuming that the bubble radius is fr (cm) internal pressure bP (MPa) of one bubble is

determined by the state equation as follows

34( )

3b fP r f mB mkTπ θ sdot minus =

(282-3)

where

m number of gas atoms per one grain boundary bubble

(atomsbubble) refer to Eq(282-24)

B van der Waals constant for (Xe+Kr) (85x10-23 cm3atoms)

Also grain boundary coverage fraction bf is defined as

2

2 2 2 22

sin 4 sin4

fb f bb g bb

g

rf r N a N

aπ θ π θ= sdot = (282-4)

Here by setting

2 2

2

sinfc

rF

s

π θ= (282-5)

(where s is defined in the next section) ( )c bF f= = FBCOV (282-6)

( )2 2

2 2 2 2 22 2

sin 4 sin sin4 4

f fb bb g dens f dens

g g

r rf N a N r N

a aθ π θ π θ= = = sdot sdot and (282-7)

2 2sin

bdens

f

fN

rπ θ=

sdot (282-8)

hold

Since the value of bf is calculated by Eq(282-4) it varies with fr and ga Here the

present model assumes that gas release from bubbles begins when either one of the


- 119 -

following conditions is satisfied bf becomes equal to FBCOV (input name-list parameter

default=025) by the changes of fr and ga or the bubble radius fr reaches RF (input

name-list parameter default=5E-5 cm) Accordingly FBCOV can be used as a threshold

value for gas release ( Refer to the later section (4))

(2) Rate-law equation to determine grain boundary bubble radius

Radius of grain boundary gas bubble is determined by integrating the growth rate fdr

dt

with time based on the model of Kogai et al(232)

1

nf

f kk

drr t

dt=

= Δ (282-9)

2

22

2FRAD

4 ( )kT

Vf bd gb

b ext ff f

dr s DP P k

dt r f r

δ γθ

Ω= sdot minus minus sdot

(282-10)

where extP gap gas pressure of the axial segment calculated It is set equal to

plenum gas pressure

fr radius of bubble defined in section 2714 (1)

FRADadjustment factor

When 0fdr

dtgt or in a bubble-growing period vacancy moves by grain

boundary diffusion so that FRAD= HP (=20 default) is set

Reference Calculation 1) Assuming 2 11 21 d μm 3 10 (bubblesm )densN = = times (d=183x183μm2) and setting

605μm 05 10 mfr minus= = times and o50θ = we have 2 2(sin50 ) (0766044) 058682= =

and 6 2 11 2(05 10 m) 3 10 m 058682 01383bf π minus minus= times sdot times sdot =

2) Inversely setting 025bf = 605μm 05 10 mfr minus= = times and o50θ = Eq(282-8)

gives 11 25424 10 (bubblesm )densN = times

3) Obtaining the number of gas bubbles K per one grain boundary by referring to

section 291 (1) we have 2

2 2

4

sing b

f

a fK

r θsdot

=

Substituting 025bf = and grain radius 5 μmga = gives K =1704 (bubblesgrain

boundary)


- 120 -

When 0fdr

dtle or in a bubble-shrinking period vacancy moves by lattice-

diffusion so that FRAD＝ HN (=10 default) is set

Vacancy sink strength of grain boundary bubble is(233)

Vacancy sink strength

22

8(1 )

( 1)(3 ) 2 lnfks

φφ φ φ

minus=minus minus minus

(282-11)

2 2 2

2

sin sinf f br r f

s s

θ θφ

π = = =

(282-12)

3

z rextP θσ σ σ+ += (282-13)

2s (cm2)bubble spacing area or territory area of

one bubble including bubble itself on grain

boundary

2 1densN ssdot = (282-14)

2 2

2 sinf

b

rs

f

π θ= (282-15)

bδ grain boundary thickness = name-list parameter GBTHIK (default=5x10-10 m) V

gbD grain boundary diffusion coefficient of vacancy (=assumed

to be equal to that of Xe atom)

6 4138 10 exp( 2876 10 )VgbD Tminus= times minus times (m2s)

Ω atomic volume of Xe = 409x10-29 m3

Be careful in unit conversion Here since

2 2 2 2 22

sin sin sindens f ff

HN r r H

rφ θ θ θ= = =

φ = 007465x058682=00438 is obtained That is in Eq(282-11) 2 2fs ksdot has a

constant value


- 121 -

(3) Changes of the number density and size of bubbles

The assumption that densN is inversely proportional to the square of bubble curvature

radius fr =RF (default=05E-6m) as shown in Eq(282-8) is supported by observations(226)

That is 2densf

HN

r= holds where H is a proportional constant Also when 025bf = and

05 μmfr = H 01356= is obtained for 11 22

5424 10 (bubblesm )densf

HN

r= = times (Refer

to the reference calculation in the previous section (1) As the bubble is tiny in its embryo

stage it is necessary to circumvent ldquoabnormal numerical resultsrdquo during the embryo stage

For this purpose value of densN is determined in the following method

First the initial value of fr is set as RINIT (default= 001μm=1E-6 cm) Namely

RINITfr ge holds

A) In case of RF (default=5E-5 cm)fr lt

A fixed vale is set 115424 10densN = times (Refer to the above reference calculation 2) In

accordance with the equations of 24bb g densN a Nπ= and inv

bb

Gm

N= not only fr but also bbN

and m change

B) In case of RFfr ge gas release occurs though if the diffusion influx of gas atoms from

inside of grain exceeds the release amount the bubble grows In this situation in a case in

which no coalescence of bubbles is assumed (NGB=1) bf changes in accordance with

Eq(282-8) but the number density is fixed at 115424 10densN = times

In other case in which the coalescence is assumed (NGB≧2) the above two variables

change in inverse proportion to the coalescence number NGB That is after coalescence

the number density becomes

1new

dens densN N = times

MGB

NGB (282-16)

where MGB is the times of coalescence (Refer to section 2101)


- 122 -

(4) Gas release rate from grain boundary bubbles Release of fission gas in the grain boundary bubbles begins in accordance with the van Uffelen model(234) if either 05μmfr = (RF sinb fr r θ= ) or 025bf = (FBCOV) is

【Note】As fr has a very strong dependence on fr numerical instability happens to

occur To circumvent this the following approximate method is adopted in the

calculation

1) For the calculation of Eq(282-9)1

nf

f kk

drr t

dt=

= Δ it is assumed that when the radius

growth rate becomes 0 at some time step ie the bubble grows to reach an equilibrium

0fdr

dt= This is equivalent to obtain the equilibrium size fr when

2b ext

f

P Pr

γ= + holds

In other words

f fr r= is obtained by solving

342

( ) 3

f

f

rP f m b m kT

r

πγ θ

+ sdot minus sdot = (a)

However in the actual situation which the rate-law model addresses f fr rlt

2) Next time required for the growth of bubble from present fr to fr is calculated by

using Eq(282-10) As a first step of calculation the process of change fr rarr fr is

divided equally into 100 steps That is

2 (100 1)100 100 100 100f f f f f f f f

f f f f f f

r r r r r r r rr r r r r r

minus minus minus minusrarr + + rarr + + minus rarr (b)

Then at each divided step time to change ie time to realize the growth amount in

each step is calculated For example if the change at i-th step is expressed as

1f i f ir r +rarr and if change rate of fr is small time itΔ required to the growth from

f ir to 1f ir + ( 1f i f ir r +rarr ) can be obtained by using Eq(282-10) as follows

1

f if i f i i

drr r t

dt+minus = Δ (c)

3) By comparing the time =

Δ=i

jji tt

1

obtained from itΔ with the actual time step

increment tΔ fr is determined as follows

( )( ) ( )iiiifififf ttttrrrr minusminusΔminus+= ++ 11 (d)

Namely tΔ lies between the i-th time it and i+1-th time 1it + of the 100-division

steps so that setting that fr is f ir at it and 1f ir + at 1it + fr を is determined by

Eq(d) If 100tt gtΔ ff rr = holds


- 123 -

satisfied with the following rate

2

FGCND c bb b gbrelease

V N P Sn

kTη= sdot (282-17)

22669

MTης

= gas viscosity (282-18)

4 2 2 2( ) 16b plenumS a P P Lπ η= minus (282-19)

(principle equation)(232)

24gb gS aπ= grain boundary surface area

a capillary inner radius

FGCND adjustment factor of gas conductance in capillary (default=1000)

24bb g densN a Nπ=

Here it is assumed that when fr =RF the number density of grain boundary bubbles is

115424 10densN = times (bubblesm2) and this is proportional to the square of bubble radius

(refer to the previous section (3))

Furthermore the followings are set

Atomic mass of Xe M=131

Its diameter as a solid sphere approximation ς = 4047x10-10 (m)

Capillary volume 0 ( ) ( )c c c eV V f F g σ= (282-20)

4

0 12 32 210 (m )

16 3c

aV

L

π minus= = sdot (282-21)

0( ) 1 exp[1 ( ) ]NFCc c cf F F F= minus minus NFC= 10 default (282-22)

0 0784cFπ= = a baseline value

0( ) 1 exp[1 ( ) ]NSCe e eg σ σ σ= minus minus NSC= 10default (282-23)

0 10 MPaeσ = a baseline value SIG0=10 default

Here it is predicted that after the bubble coalescence the baseline value of 0cF increases

In other words bubble spacing increases and probability of tunneling between bubbles

becomes less Accordingly

After one coalescence of bubbles 0 085cF = is set FC1=085 default


- 124 -

After twice coalescences 0 09cF = is set FC2=09 default

After three times coalescences 0 095cF = is set FC3=095 default

After four times coalescences 0 099cF = is set FC4=099 default

These name-list parameters are given provisional values They should be optimized by

measured data taking into consideration of various factors

(5) Gas amount accumulated in grain boundary (grain boundary inventory)

Here invG is set as fission gas inventory (accumulated amount) at boundary of grains

which are contained in a unit volume of fuel pellet gas amount per a bubble is set as m

(atoms) and the number of grain boundary bubbles is similarly to the previous section (1) set as bbN

As explained in section 278 an actual grain boundary is shared by the two adjacent

grains and the gas in the boundary is also shared However we suppose a situation that one

single grain is isolated in a space and its surface is wrapped by ldquograin boundaryrdquo Assuming that the pellet unit volume contains i

grainN grains with radius ga

iinv bb grainG mN N= sdot (282-24)

holds Therefore gas amount per one bubble m (atoms) is

invi

bb grain

Gm

N N=

sdot (282-25)

Accordingly invG times the volume of each ring element is an inventory value in each

element

(6) Effective stress acting the grain boundary

2 2 2

2 2 2

sin

sin 1h f b c b

ef c

s r P F P

s r F

σ π θ σσπ θ

+ sdot += =minus minus

(282-26)

h ext gapP Pσ = + (282-27)

In these Eqs(282-26) and (282-27) solely the sign is reversed with respect to that in

usual formulation ie compressive stress has a positive value And

2 2 2 2

2 2 2

sin sin

sinf b f

c bf

r f rF f

s r

π θ π θπ θ

= = = (282-28)

is obtained


- 125 -

29 Swelling by Grain-boundary Gas Bubble Growth in the Equilibrium Model (IGASP=0)

291 Grain boundary gas bubble growth In this section a model of gas bubble swelling is explained in the calculation process of

the grain boundary gas bubble growth on the basis of temperature equilibrium pressure and

number of atoms

(1) Calculation of number density of grain boundary bubbles Instead of calculating the growth rate of gas bubbles at grain boundaries the grain radius

which balances with the [external pressure + surface-tension] ie equilibrium model is

calculated by assuming a constant number density at each time step

Here the number density of bubbles K per grain boundary is defined as follows

The orthogonal projection radius of a gas bubble radius fr on the grain boundary

surface is sinfr θ and assuming that the grain radius is gra the total grain boundary

surface area is 24 graπ (Refer to section 2710)

Furthermore assuming that the area number density of bubbles at the moment of gas

release (in the saturated (threshold) condition) is K (though this is a non-dimensional

quantity its physical implication is bubblescm2) the bubble area covering the grain

boundary surface in the saturated condition is given by 2 2sinfK rπ θsdot The fraction of this

area to the total grain boundary surface area is bf thus the following equation holds

2 2 2sin 4f gr bK r a fπ θ πsdot = sdot (291-1)

In other words 2

2 2

4

singr b

f

a fK

r θsdot

= (291-2)

Assuming that the number of bubbles on the grain boundary is inversely proportional to the square of the gas bubble radius fr on the grain boundary and calculating Eq(291-2) by

setting fr=05 μm and θ =50 (223) the following is obtained

2

2

6816 b gr

f

f aK

r

sdot= (291-3)

Namely the number density of bubbles in the saturated condition is given by

Eq(291-3) at the same time the number density of bubbles in the condition prior to the


- 126 -

saturated condition can be obtained by substituting gra bf and fr into Eq(291-3)

Hence the number density of grain boundary bubbles is given by Eq(291-3) K in

Eq(291-3) is adjusted simultaneously with N fmax using the name-list parameter

bf =FBCOV (Refer to 2714 (2) and 282(1))

(2) Growth of bubbles at grain boundaries Assuming that the volume of fission gas remaining on the grain boundary of a grain at

time step n is nfn (atoms) the following holds

nfn K n= sdot (291-4)

Here the term n is the number of fission gas atoms in a bubble and nfn is obtained as

a numerical solution of the diffusion equation More specifically nfn is obtained from

Eqs(279-5) and (2710-3) as

1 maxmin( )n n n nf f f fn n n nminus= + Δ (291-5)

3

1

1

n n nnf total totalV

n

an tdV

aβminus

minus


(291-6)

Fission gas atoms flow into the micro-bubbles generated at the grain boundary by

diffusion bubble gas pressure increases and bubble growth starts

It is assumed that the gas pressure gasP within the bubble is always equal to the [external

pressure + surface tension pressure]

That is 2

gas extf

P Pr

γ= + (291-7)

The gas pressure within the bubble Pgas during growth is determined by the following equation using fr n B and T

34( )

3f

gas f

rP f n B n kT

πθ

sdot sdot minus sdot = sdot

(291-8)

By solving Eqs(291-3) (291-4) (291-7) and (291-8) simultaneously gasP and fr

are obtained From Eqs(291-3) and (291-4)

2

26816

nf f

b gr

r nn

f a

sdot=

sdot (291-9)


- 127 -

By substituting this into Eq(291-8) and using Eq(291-7)

3 2 2

2 2

42( )

3 6816 6816

n nf f f f f

ext ff b gr b gr

r r n r nP f B kT

r f a f a

πγ θ sdot sdot

+ sdot sdot minus sdot = sdot sdot sdot (291-10)

By dividing both sides by 2fr and letting 26816

nf

b gr

nQ

f a=

sdot

42( )

3f

ext ff

rP f QB QkT

r

πγ θ

+ sdot sdot minus = (291-11)

From Eq(291-11)

( ) ( ) 0223

4

3

4 2 =minus

minusminussdotsdot+sdot QBrQKTQBPfrPf fextffextf γθγπθπ (291-12)

By solving this quadratic equation rf is obtained

In the above equation let

( ) ( ) QBcQKTQBPfbPfa extfextf γθγπθπ 223

4

3

4 minus=minusminussdotsdot== (291-13)

where agt0 and clt0 two real roots are obtained

The larger of the two solutions is the solution sought ie rf

a

acbbrf 2

42 minus+minus= (291-14)

Then using Eq(291-7) Pgas is obtained

As explained the calculation of bubble size on the basis of the equilibrium model is

performed until fr grows to 05μm when it reaches 05μm gas is released as in the

conventional model which is discussed later Bubbles which released gas do not grow beyond

fr =05 μm

(3) In the case grain growth is accompanied

abbreviated Identical to section 2712(1)

(4) Limit of bubble growth due to fission gas release

abbreviated Identical to section 2714


- 128 -

292 Bubble swelling

(1) Modeling of grain boundary bubble swelling

A) Number of grain boundary bubbles of one crystal grain

Writing again Eq(291-3) 2

2

6816 b gr

f

f aK

r

sdot= (291-3)

Here the number of bubbles in the condition prior to the saturated condition can be obtained by substituting gra bf and fr into Eq(291-3)

B) Bubble volume and swelling

The gas bubble volume Vbb and the volumetric bubble swelling are defined for bubble size fr obtained by the above procedure First the volume of one bubble is

34( )

3f

bb f

rV f

πθ= sdot (292-1)

In terms of the grain radius agr the grain volume Vgr is given as

34

3gr

gr

aV

π= (292-2)

Then volumetric gas swelling is

0 2

Bubble

bb

gr

V V K

V V

Δ sdot=

(292-3)

A coefficient of 2 in the denominator on the right-hand side results because one bubble

is shared by two grains (In reality a bubble at a corner of a grain is shared by two or more

grains and in that case the coefficient should be slightly larger than 2 on an average)

In this context 0

001436Bubble

V

V

Δ =

for 5μm 05μmgr fa r= = and 025bf = the

volumetric swelling is 144

(2) Intra-grain bubble swelling Assuming that the number density of intra-grain bubbles is 3(bubbles cm )N and

radius of the intra-grain bubble is intraR (cm) the volumetric swelling is similarly to the

derivation of Eq(292-3)


- 129 -

3intra

0

4

3

BubbleV

R NV

π Δ = sdot

(292-4)

(3) Changes of porosity and density Pellet porosity change pΔ and related density change ρΔ are associated with the total

bubble swelling in the following formula

Total bubble swelling is 3intra

0

4

3

Bubble

bb

grTotal

V V KR N

V Vπ

Δ sdot= sdot +

(292-5)

and 0

Bubble

Total

Vp

V

ΔΔ =

(292-6)

and pρΔ = minusΔ (292-7)

hold

210 Swelling by Grain-boundary Gas Bubble Growth in the Rate-law Model (IGASP=2 IFSWEL=1)

Concerning the bubble growth associated number density of grain boundary bubbles

and gas release are described in sections 281 and 282 Also the bubble growth and gas

release in the case in which grain growth occurs are described in section 2712

Hereafter as part of the model of grain boundary bubble swelling a model of bubble

coalescence and coarsening are described This model is only tentative and has no

verification Here only some basic assumption is presented

[Note] In the equilibrium model (IGASP=0) coalescence and coarsening of grain

boundary bubbles are not dealt with because the equilibrium model assumes that [bubble gas

pressure + surface tension] is always in balance with the compressive force from surrounding

solid matrix However the coalescence and coarsening of grain boundary bubbles are driven

by the excess pressure from the equilibrium pressure Therefore they contradict with the

assumption of the model

2101 Coalescence and coarsening of bubbles

(1) Definition and condition of bubble coalescence Coalescence of grain boundary bubbles cannot occur only by the contact of two isolated

but growing bubbles By contact they are connected and give rise to tunneling leading to gas


- 130 -

release Then by gas release gas pressure decreases inside the connected bubbles and

when the surrounding pressure ie compressive force from solid matrix exceeds the gas

pressure the tunneling path will close by creep down That is tunneling open path of the

connected bubbles is closed After this the connected bubbles will become an isolated and

coalesced bubble by surface tension and the Ostwald growth This process is expressed as

ldquocoalescencerdquo

The above process can be realized when the following two conditions are fulfilled one

is the pellet temperature should rise above the threshold level MGT TMG=1200K (default)

and the other is that the external compressive pressure exceeds the bubble gas pressure This

pressure condition can be expressed by

2b ext gapb

P P Pr

γle + + (2101-1)

03c bF f= ge = FCON (2101-2)

FCON=03 (default)

Here gapP is the gap gas pressure in the axial segment This is approximated

as gap plenumP P= In a future extension of model gapP can be derived by the calculation of gap

gas flow using gap gas flow cross section gapS In this case gapS will vary depending on the

gap state ie open gap closed gap and bonded gap

(2) Number of coalesced bubbles It is assumed that the coalescence is such that M-bubbles merges at one coalescence

and after that the merged bubbles become an isolated single bubble and that this new bubble

grows or shrinks without gas release The bubble number density at grain boundary becomes

1M Growth of bubble would be affected by the Ostwald growth so that the growing rate

should be to some extent deviated from the previous model of ldquodiffusion control driven by

flux from the inner region of grainrdquo

ldquoIGASP=2rdquo condition is controlled by the following parameter

Number of coalescent bubbles by name-list input for ldquoMrdquo NGB default=4

Number of coalescence times by name-list input for ldquoMGBrdquo MGB default=10

In this case when NGB=0 is designated no coalescence occur Also MGB 1ge is

assumed


- 131 -

(3) Geometry of coalescence The above described coalescence is considered

Suppose that M bubbles 1 with curvature radius 1fr r= merge to form a larger single

bubble 2 with radius r2 Here the grain boundary area S which is occupied by the M

bubbles 1 is

2 21 sinS M rπ θ= sdot (2101-3)

Assuming that M does not change and the M bubbles 1 are replaced hypothetically by a

single bubble an equivalent radius r1 of this hypothetical bubble is

1 1r M r= sdot (2101-4)

(4) Change of bubble radius

Next when M bubbles 1 merge into a single bubble 2 with radius r2 it is assumed that

gas pressure surface energy γ and the external pressure extP will remain unchanged before

and after the coalescence and that the coalescence occurs instantaneously so that no influx of

fission gas atoms are diffused into the bubble during coalescence

At this process the lenticular shape is held (exactly speaking the bubbles become a tunnel shape once but this is ignored) increase of radius lessens the surface tension from

1

2γ

r

to2

2γ

r The bubble gas equilibrium pressure would be in a decreasing trend but this change is

so small that it is neglected

Here volume of the bubble 2 is M-times that of the bubble 1 while the radius r2 falls to 3 M times as follows

3 3

1 2

32 1

4 4

3 3r M r

r M r

π πtimes =

= sdot (2101-5)

Therefore the area on the grain boundary occupied by this single bubble is 23 M times

That is the radius of bubble 2 formed by coalescence decreases from the effective radius r1

after coalescence by

31 2 1 1r r r M r M rΔ = minus = minus (2101-6)

Then the number of bubbles on the grain boundary has decreased to 1M from that which is

expressed by K


- 132 -

2

2 2

4

singr b

f

a fK

r θsdot

= (2101-7)

And the fraction of grain boundary coverage has decreased to

2

2 3

1

bb

r ff

r M

sdot =

(2101-8)

(5) Growth of coalesced single bubble

Let us consider the condition of gas re-release by growth of coalesced bubble For the

bubble 2 with radius r2 to grow to the hypothetical bubble with the original effective radius

r1 or to reach the former value of bf the bubble 2 should grow by the rate-law model to

have a radius

2 1 1 sinr r M r θ= = sdot (2101-9)

That is a radius increment is

32 1 1( ) sinr r r M M r θΔ = minus = minus sdot (2101-10)

In this condition the bubble volume becomes M -times larger as the following equation

indicates

32

1

1 rM

M r

=

(2101-11)

2102 Swelling Bubble swelling by the rate-law model is identical to that which is described in the

equilibrium model presented in section 292 Namely if the volumes (radius) of bubbles at

grain boundary and inside grain and their number density are once calculated the swelling is

derived in a similar way to the equilibrium model Relationship between porosity and density

is also identical

211 Thermal Stress Restraint Model

In the equilibrium model of FGR (IGASP=0) a restraint effect on gas release imposed

by the external pressure Pext acting on the grain boundary bubbles can be evaluated by

selection of name-list parameter IPEXT which takes into account of the plenum pressure

PCMI contact pressure and pellet internal thermal stress


- 133 -

For example in Eq(2714-6) increase in Pext will increase the threshold value maxfN

which decreases gas release and vice versa However in the mode ldquoFBSAT=1rdquo if

ldquoFBSATS= maxfN (atomscm3) (default=5E15)rdquo is designated by input max

fN remains

unchanged irrespective of the IPEXT value

In the rate-law model of FGR (IGASP= 2) the gas release threshold is determined by

either the bubble radius fr or grain boundary coverage fraction rf which are indirectly

dependent on extP

However since the gas release rate is a function of bubble pressure plenum pressure and

pellet internal stress as expressed by Eqs(282-16) to (282-22) the restraint effect by Pext

on gas release is implicitly included in the model Also pellet thermal stress is affected by

the swelling model which is selected by IFSWEL When IFSWEL=1 the thermal stress is

affected by FGR which in turn depends on the FGR model selected by IGASP

Thus evaluation of Pext has an indispensable role in the prediction of the amount of

released fission gas Some options in this evaluation are described in the next two sections

It is noted here that Pext has normally a negative value because it is usually a

compressive stress nevertheless

2111 Option designated by IPEXT

Value of name-list parameter IPEXT designates the following mode of external

pressure Pext as the restraint on gas release

=0 Pext =0 =1 Pext =Plenum gas pressure only

=2 Pext =P-C contact pressure only =3 Pext = Max(plenum pressure contact pressure)

=13 Pext = plenum gas pressure + contact pressure

Here the above contact pressure is the PCMI contact pressure which is calculated by the

ERL mechanical analysis

=14(default) Pext = Sav+ plenum gas pressure

Here Sav is the thermal stress (average hydrostatic pressure) calculated by the ERL

mechanical analysis It is defined by Eq(3254) 3

r zST

θσ σ σσ + += The PCMI contact

pressure is always taken into account in the calculation of Sav


- 134 -

Only when IPEXT=14(default) Pext is set as follows

(1) When IGASP=0 Pext = |Sav|+ Plenum gas pressure Here if Sav is positive (tensile

stress) Sav=0 is set

(2) When IGASP=2 Pext =|Sav|+Plenum gas pressure When Savgt0 ie Sav is tensile

stress and 2

0extPr

γ+ = Pext is fixed at Pext =2

08r

γminus times

2112 Comparative discussion on the gap gas conductance increase during power down

In some cases of a test fuel exceeding a certain level of burnup during irradiation tests in

a test reactor when linear power decreases fast concurrently with this power dip or in the

power increase after this power dip a temporary increase of fission gas release has been

observed More exactly an increase of reading of the internal pressure sensor installed in the

plenum region is observed This temporary increase of gas pressure ie substantial increase

in fission gas release can be interpreted by the following two ways

(1) Gas conductance increase in P-C gap During high power period of irradiation pellet temperature is elevated diffusion of

fission gas atoms is enhanced tunneling of grain boundary gas bubbles to outer free space

easily takes place resulting in FGR from grain boundary gas bubbles In this situation due to

thermal expansion of pellets the P-C gasp is either closed or has narrowed markedly or

cracks in the pellet are almost healed As a result the released fission gas from pellet is

retained in a very small space remaining among pellet fragments and cannot flow to reach the

plenum space In other words gas conductance in the axial direction of rod is much limited

Contrarily when linear power decreases pellet thermal expansion mitigates and P-C gap or

crack space are re-opened recovering the gas conductance in the axial direction which

allows the locally retained gas to flow into the plenum space and is observed as an increase in

internal pressure

(2) Change from containment to release of grain boundary bubble gas Similarly to the explanation in the previous section (1) the grain boundary gas

bubbles are growing due to diffusion influx from inside of the grain while the bubbles are

under compressive thermal stress from surrounding solid fuel matrix so that it is difficult to

form a tunnel to release gas to outer free space On the contrary when linear power falls

pellet thermal expansion decreases which facilitates tunneling formation and as a result gas


- 135 -

is easily released to free space and flows to the plenum space through the opened P-C gap and

crack spaces This process is observed as increase in internal pressure This situation

corresponds to the relationship in Eq(2714-8) in which decrease in Pext will decrease the threshold value max

fN leading to gas release This implies a physical meaning of

Eq(2714-8)

【Discussion】 Comparison of the above two mechanisms (1) and (2) suggests that both

have the possibility or that both takes place concurrently in actual fuel rod It cannot be

justified to confine the mechanism interpretation to only one of the two The FEMAXI model

has assumed the second interpretation while modeling on the basis of the first mechanism is

also possible However in fact the FEMAXI model based on the second interpretation has

well reproduced the internal pressure of plenum part successfully at the power dip and

recovery

212 High Burnup Rim Structure Model Formation of the rim structure (the rim phase) affects fuel behavior(226)-(228) In

FEMAXI change in the volume of the rim structure produced (high-burnup structure=HBS)

and its effects on the thermal properties of the pellet are modeled

2121 Basic concept

There has been a number of researches and reports which address such phenomena

associated with the rim structure formation as the porosity FGR density of fission gas atoms

in solid phase swelling and thermal conductivity deterioration and recovery(236)-(243)

Nevertheless up to now no models have been successfully developed which describe

comprehensively the effect of rim structure formation on the macroscopic ie in terms of

engineering level fuel behavior change In addition for the formation of rim structure

expressing a notion that fuel is a ldquocomplex systemrdquo can by itself give no further

understanding of fuel behavior and as such totally meaningless pedantry These situations

are induced by the facts that the rim formation is intrinsically associated with a process in

which many factors are intertwined such as pellet fabrication process burnup temperature

grain size enrichment power history fast neutron flux bonding with cladding etc and also

difficulties in treating high burnup fuels in experiments

Therefore in modeling the rim structure formation in a fuel performance code it costs

too much effort to compose known observations and interactions in a mechanistic manner and

to adjust a number of parameters consistently to obtain at least a reasonable agreement to

several observed data Furthermore it is likely that the resulted predictability could not be

balanced with such tough effort in terms of engineering level


- 136 -

Here noting that the rim structure formation itself gives a relatively independent and

additional effects on the whole aspects of fuel behavior such as a certain degree of fuel

temperature rise or swelling amount the modeling which is simple and based on empirical

correlations and covers the complicated phenomena with some degree of uncertainty can be a

rational strategy in terms of the prediction of whole aspects of fuel behavior

Of course such empirical correlations and model parameters can be easily modified in

combination with the accumulation of observed data

Consequently the present model assumes the followings

A) Rim structure formation ratio

The volumetric ratio Fv of rim structure formation in each ring element of pellet is a

function of burnup Bu and temperature T

B) Porosity and fission gas inventory in rim structure

The porosity is a function of Fv and Bu and takes a fixed value for Fv=100

Gas pore size distribution and gas pressure are not calculated On the other hand fission

gas inventories inside the gas pores and in the solid matrix of fuel and additional FGR by

athermal diffusion etc are determined by empirical correlations

C) Effects of porosity and swelling on thermal conductivity of fuel pellet

Swelling amount is calculated directly from porosity The thermal conductivity

degradation is calculated by Ikatsursquos equation

D) Conditions for diffusion calculation of fission gas atoms

Fission gas atoms diffusion rate is very low at the low temperature peripheral region of

pellet Thus to avoid a redundant calculation the code has an option to duplicate the

calculated results of diffusion in a region at temperature TDIFF (default =600C) to adopt

them to the region with temperature lower than TDIFF However in the case in which the

local temperature in rim region is considerably high adoption of this option would not be

appropriate and need some attention

2122 Density decrease of fission gas atoms in solid phase of rim structure

Although not directly related to the fuel behavior calculation a method is presented first

as the simplest empirical model to evaluate the depletion of fission gas atoms concentration

due to the formation of rim structure This depletion amount is evaluated by the following

three methods while these method are not activated when IRIM=0


- 137 -

(1) Battelle model(236) (HBS=0 IRIM=1) In the model proposed in the Battelle Experimental Program for high-burnup effects

fission gas is represented by Xenon and the following calculation is performed

( )F B B MWd kgUr u u= minus gt0 00625 62 2 62 2 ( ) (2121)

Here Bu is a local burnup at the peripheral (rim region) of pellet Development of the

rim region with burnup is given as a function of the burnup as

( )W Bu= minus2 19 48 8 (2122)

where W is the thickness of the rim region (μm) Equation (2122) shows that even at

local burnup of 150 MWdkg-U the rim region has 220 μm thickness at most and can be

comparable with the thickness of the outermost ring element of pellet Therefore the

athermal fission gas release is calculated only in the outermost ring region Vr

Using Eqs(2121) and (2122) the ratio Rr of xenon which is athermally released in

addition to the release by the thermally activated diffusion from the rim region to that

generated in the outermost region is given by

R V Fr r r= sdot (2123)

( ) ( )V r r W r rr o o o i= minus minus minusπ π2 2 2 2 (2124)

where ri is the inner radius of the outermost region and ro is the pellet outer radius

Assuming that the fission gas detached from the solid matrix of rim structure is released

to outer free space the additional gas release at the outermost ring element induced by rim

structure formation is approximated to occur from grain boundaries to have a compatibility

with the grain boundary gas bubble release model Then the concentration of fission gas

atoms at the grain boundary in the n+1 time step of the grain growth is given by the following

equation rather than Eq(272-13)

( ) ( )1 1 1

1 12

1

1

4

n n n n n ng total total V r rn nn

f fn n

f tdv R M R MaN N

a a

βπ

+ + ++ +

+

+ Ψ minus Ψ + Δ + minus= + (2125)

Here Mn+1 number of fission gas atoms generated in the grains with a radius of an+1 up

to the time tn+1 (atoms)

Mn number of fission gas atoms generated in the crystalline grains with a radius

of an+1 up to the time tn (atoms)


- 138 -

Rrn+1 fission gas release rate due to rim region formation up to the time tn+1

Rrn fission gas release rate due to rim region formation up to the time tn

(2) Cunningham model(225) (HBS=0 IRIM=2) In the Cunningham model fission gas is also represented by xenon and the amount of

additional release is calculated by the following equation

R F Pr r r= (2126)

Here Rr ratio of the amount of fission gas released from the rim region to the

amount of fission gas generated in the entire pellet

Fr (cross-section area ratio of rim to pellet) x (amount of fission gas released

from rim regionamount of fission gas generated in the rim region)

Pr Ratio of the average burnup at the rim region to the average burnup of

pellet ( assumed to be 13)

In addition the following equation holds

F B Br u u= times minus times sdot + times sdotminus minus minus790 10 698 10 834 103 4 6 2 ( Bu gt65MWdkgU) (2127)

where Bu is the local burnup at the pellet peripheral region ie the rim structure region

Fig2121 shows the change of Fr with burnup Thus the amount of additional fission gas

release can be obtained as a ratio to the total amount of fission gas generation which is

calculated by the FGR model

By setting Fr = 0 in Eq (2127) Bu = 1349 and 7020 are obtained as solutions

Accordingly Fr has positive values in the regions Bult1349 and Bugt7020 Hence

because the range of positive Fr values is 7020 MWdKgU or greater in the domain of

Bugt65 MWdKgU the domain of Bu in Eq (2127) is set at Bugt7020 MWdKgU

In addition because fission gas release rate in an entire pellet is obtained by integrating

the release rate of gas in each ring element of pellet formulation of Eq(2126) does not fit

the constraints of FEMAXI Under these circumstances formulation of Eq(2126) is altered

as described below

Eq(2126) gives the ratio of the fission gas released from the rim region to the fission

gas produced from the entire pellet On the basis of this equation calculation of the release

rate Frim of fission gas from the rim structure region is defined as follows

Assuming that the amount of fission gas production is proportional to the burnup


- 139 -

50 60 70 80 90 100 110 120000

001

002

003

004

005

Frac

tion

of ri

m a

rea

Local burnup (GWdt)

Cunningham model

Fig2121 Change of fraction of the rim structure region as a function of pellet local burnup in the Cunningham model

(Pellet cross-section arearim area)

(Pellet average burnupRim average burnup)rim rF R= times

times (2128)

holds

(3) Modified Cunningham model (HBS=0 IRIM=3)

Next in the Cunningham model ratio Pr for the average burnup in the rim region to the

average burnup of the entire pellet is treated as a corrected rate in Eq (2126) thus this

corrected approach is adapted as an option

In other words there are two cases the case in which Pr=13 is adopted and the case in

which Pr calculated in FEMAXI is adopted

A) When Pr=13 is adopted (IRIM=2) Frim is obtained using Eq (2128) In this case the

ratio of the average burnup of pellets to rim phase average burnup calculated in FEMAXI is

used for the value (average pellet burnup)(rim phase average burnup) in Eq (2128)

therefore in general the ratio is different from 113

B) Contrary to this when Pr calculated in the code is used (IRIM=3) the product of Pr in Eq

(2128) and (average pellet burnup)(rim phase average burnup) equaling 1 is assumed

(Pellet cross-section arearim area)rim rF F= times (2129)

In both the cases the area of rim phase is required however the Cunningham model

does not consider the area of rim phase Thus one should determine the area in order to

obtain reasonable solutions Accordingly the area of rim phase is calculated expediently

using Eq(2122)


- 140 -

2123 Effect of rim structure formation on fuel behavior

The FEMAXI has systematically modelled the effects of rim structure formation on

pellet behavior These models are effective when name-list parameter HBS=1 and =2 though

in these cases IRIM=0 is automatically set and as a result the previous depletion model

cannot be applied When HBS=0 designation of IRIM is effective but these models of the

effects are not applied

2123-1 Burnup of rim structure (HBS=1 2)

In considering the rim structure formation one of the important points for modelling is

that the onset local burnup for the rim structure formation and change of the nature of

structure with burnup extension In this two options are set with respect to the burnup of

rim structure formation ie a local burnup and effective burnup to calculate the volume of

rim structure and porosity etc by designating the name-list parameter HBS

(1) Fraction of rim structure transformation Xｖ（HBS=1 2)

It is assumed that fraction of transformation of a normal high burnup fuel matrix into the

rim structure is Xv This Xv has nothing to do with the models described in the next

sections 2123-2 and 2123-3 though it is a physical quantity to indicate how much the rim

structure has been formed and gives a reference to consider the correlation between model

prediction and observation (plotting output index is IDNO=256)

A) When HBS=1 the transformation fraction Xv is Xv=10 for the ring elements of pellet in

which element local burnup have exceeded the name-list parameter BKONA (default=65

GWdtU) and Xv=00 for the elements in which local burnup is under BKONA

When in the two adjacent ring elements to a certain ring element R local burnup in the

outer element R+1 has exceeded BNONA while local burnup in the inner element R-1 does

[Fitting parameter RFGFAC] The amount of additional fission gas release due to rim region formation can be adjusted using a factor RFGFAC ( default = 10 ) F F R F G F A Cr r= times

[Option Parameter IRIM] A model for additional fission gas release from rim region can be selected from the following three models IRIM=1 Battelle model (default) =2 Cunninghum model =3 Modified Cunninghum model


- 141 -

not Xv in the ring element R is determined by the interpolated area fraction of Xv(gt0) in

R+1 element and Xv(=00) in R-1 element

B) When HBS=2 Xv is determined by ( )1

1tanXv

effB Buα δ

π

minus minus= + (2123-1)

where α=105 (ARIM default=105) δ=052 (DRIM default=052)

effB effective burnup explained later in section (3) Eq (2123-3)

1Bu BURIMS (default=600 GWdtU)

(2) Local burnup (HBS=1) In this option (HBS=1) local burnup localBu in every ring element of pellet is directly

used to the models of rim structure formation and properties change

(3) Concept of effective burnup (JAEA model HBS=2) Formation of rim structure and its volume increase depend not only on local burnup but

also local temperature Because of this a model concept of ldquoeffective burnuprdquo is introduced

on the basis of the following discussion and assumptions

A) The rim structure is generated in the region with a certain level of high burnup eg 50 - 60

GWdt and with time its volumetric fraction increases(236) - (241) This is because the high

burnup region has augmented the density of lattice defects and resulted in enhanced strain

energy which promotes the lattice structure transformation into the rim structure

B) On the other hand the distorted lattice in high burnup region can be annealed by thermal

activation process and recover its lattice to a normal condition With temperature rise this

recovery process is accelerated and as a result the thermal conductivity degradation also

mitigates (242) and the density of lattice defects and strain energy decrease which counteracts

the rim structure formation(243)

C) For these reasons modelling to predict the rim structure formation should be conditioned

by the two terms burnup and temperature history Therefore the radial direction profile of

burnup resulted from a burning analysis code is modified into an ldquoeffective burnup profilerdquo by

assuming the lattice recovery effect due to temperature rise and it is assumed that if in a

region of pellet this effective burnup profile exceeds some threshold level the rim structure is


- 142 -

generated and developed

D) The effective burnup profile is derived from a combination of local burnup extension and

temperature history

First it is assumed that at the n-th time step the effective burnup 1neffB minus up to the n-1 th

time step decreases by the recovery during the current time step increment tΔ That is

recovery process is a thermally activated process so that it decreases as a function of the

local temperature nT at the n-th time step

11 0exp ( )n

eff nB k T T tminus sdot minus minus sdot Δ (2123-2)

On the other hand at the n-th time step the burnup increases by nBΔ so that the

effective burnup at the n-th time step is

11 0exp ( )n n n

eff eff nB B k T T t Bminus= sdot minus minus sdot Δ + Δ (2123-3)

where neffB effective burnup at the n-th time step (GWdtU)

nT local temperature of fuel at the n-th time step (K)

here if 0nT Tle 0nT T= is assumed

0T baseline temperature TSTD=1000K (default)

1k constant 1 KON1= 1E-8 (default)

tΔ time step increment (s)

nBΔ local burnup increment during the n-th time step

Fig2122 shows a comparison of burnup profiles at different average burnups of fuel

rod

00 01 02 03 04 050

10

20

30

40

50

60

70

80

90

Burn

up (G

WdtU

)

Pellet Radius (cm)

Fig2122 Comparison of profiles of burnup and the effective burnup at different

average burnups of fuel rod


- 143 -

The present model assumes that the rim structure is developed in the pellet region in

which the local effective burnup calculated by Eq(2123-3) exceeds a certain threshold

value

2123-2 Fission gas release from gas pores (HBS=1 2)

(1) Model of fission gas atoms migration from rim structure to gas pores In the previous section 2122 depletion of fission gas atoms concentration in rim

structure ie migration into gas pores is expressed by the Battelle and Cunningham models

(HBS=0) The present new model (HBS=1 =2) adopts the Lassmann empirical model(243)

instead of these two models and expresses the fraction of fission gas atoms retained in solid

phase to the total fission gas atoms generated in the rim structure as FSOLID and also

expresses the fraction FPOR of gas which moves from solid phase and gas pores to the free

space outside pellets as a function of burnup Bu (either localBu or effB ) Namely the model

evaluates the partition and released amount to the solid phase and gas phase of fission gas

atoms generated in the solid phase of rim structure

The Lassmann model represents the fission gas atoms in the rim structure by Xe and the

Xe concentration in solid phase 1Xe is

( )1 0 0(wt) FPINF (GEN1 FPINF)exp GEN2 ( )Xe Bu Bu Bu= + sdot minus minus sdot minus (2123-4)

Therefore the fraction FSOLID of gas retained in solid phase is

[ ]1FSOLID Total Xe generation(wt)Xe= (2123-5)

Also the fraction FPOR of gas moved from the solid phase is

[ ]1

FPOR 10 FSOLID

10 Total Xe generation(wt)Xe

= minus= minus

(2123-6)

However when 0Bu Bule all the gas is assumed to be retained in the solid phase

Here

[ ]Total generation(wt) GEN1 (GWdt)Xe Bu= sdot (2123-7)

Default FPINF=025(wt )

GEN1=00145 GEN2=01 0Bu =BURIMXE=600

are defined Fig2123 shows change of Xe concentration in solid phase with burnup

Also assuming that the fraction of gas which migrates directly from the solid phase to


- 144 -

the free space by athermal release is ATHMR (default=00) 0 ATHMR FPORle le holds

Here as an inter-conversion between A wt and B atomscm3 of Xenon

3

3

23

001 [pellet density(gcm )]B (atomscm )=A(wt)

131 (gmol)

6022 10 (atomsmol)

ρtimestimes

times sdot (2123-8)

is used

Here it is better to put it in order though repetition that the gas content GS0 existing in

the rim structure is defined by a result of subtractions of the following three terms from total

amount of gas generated in the rim structure GB1 for the gas atoms moving to the grain

boundaries by thermal diffusion GB2 for the gas atoms moving into the intra-granular

bubbles in grain and GR for the gas atoms migrating from the rim structure at the fraction of

FPOR As a result

GS0 = GG ndash GB1 ndash GB2 ndash GR

holds Therefore when HBS is either =1 or = 2 the following procedure is performed

A) At a certain time step in a certain ring element once the rim structure is formed the

thermal diffusion calculation is ceased at the element At this moment the amounts of gas

atoms in solid phase GS0a and the gas accumulated in the bubbles GB1 and GB2 are

determined by the result of diffusion calculation by that time step

B) After this time step with formation of rim structure neither GB1 nor GB2 changes and

the amount of gas atoms retained in the solid phase is calculated by subtracting GR from

GS0a

0 50 100 150 20000

05

10

15 BURIM=60GWdt at 65 GWdt at 70 GWdt at 75 GWdt

Xe

conc

(w

t)

Local burnup (GWdt)

Total generation rate of Xe

Fig2123 Change of Xe concentration in solid phase by the Lassmann model


- 145 -

(2) FGR from rim structure (HBS=1 2 RMOGR=012) An additional release of fission gas from the rim structure to outer free space is

predicted by using empirical equations while in the case of RMOGR=0 no release is

assumed from the gas pores

Of the total gas generated in the rim structure the Lassmanrsquos model defines the fraction

ldquoFPORrdquo of gas which migrates from solid phase and gas pores of the rim structure into free

space That is FPOR is the fraction of amount of gas detached from the solid phase of rim

structure However the present model described in this section (2) assumes that the gas

which migrates to the pores is all accumulated in the pores and that no release of gas occurs

directly from the solid phase to free space In addition it is assumed that out of all the gas

atoms contained in all the pores only a fraction OPR is released to outer free space Then

fission gas release rate FGRIM from the pores out of the total generation of fission gas atoms

in the rim structure is

FGRIM=OPRFPOR (2123-9)

A fraction of fission gas atoms retained in the matrix of rim structure except pores is

(1-FPOR) of the total generation Therefore FGR of the ring element is

0(Bu) FGRIMFGR FGR= + (2123-10)

Bu local burnup (either localBu or effB ) (GWdt)

where FGR by the thermal diffusion is 0FGR

The open pore fraction OPR is determined by either the A) or B) in the followings

A) Option A to specify the release rate (RMOGR=1 ) By using ATHMR which is explained in section 21232 (1) the open pore fraction is

defines as

OPR=ATHMR (2123-11)

where 0 OPR ATHMR FPORle = le That is after the rim structure is formed a part of gas

out of the generated gas in the rim structure is always released to outer free space at the rate

of (ATHMR x 100) via pores in addition to the FGR induced by the thermal diffusion

B) Option B to specify the open pore fraction (RMOGR=2) The open pore fraction OPR to the porosity of rim structure porosity rimp is defined as a

function of rimp as follows(241)

003OPR (0 023)023 rim rimp p= sdot le le

(2123-12)

OPR 003 (015 003)(100 23) (023 024)rim rimp p= + minus sdot minus le le (2123-13)


- 146 -

20 40 60 80 100 120 1400

5

10

15

20

Min

Por

osity

(

)

Local burnup (GWdt)

Max

Fig2124 Change of porosity with burnup in the NFD correlation equation


where the rim structure porosity rimp is determined by the next section 2123-3

2123-3 Porosity thermal conductivity and swelling (HBS=1 and 2 RMPST=01 and 2)

The porosity change induced by fission gas pore generation in the rim structure its

associated degradation of thermal conductivity and the effect of swelling by the rim structure

are modeled The porosity is specified by a name-list parameter RMPST When RMPST=0

rimp =005 is set not calculating the porosity change

(1) Empirical correlation equation of NFD data (RMPST=1)

The porosity rimp is calculated by the following empirical equation(240) which is shown

in Fig2124

2

2

PROMAX 0005 ( 40) 25

PROMIN 00013 ( 40) 25

PROMAX (10 06) 06 PROMIN

= sdot minus += sdot minus +

= sdot minus + sdotrim

Bu

Bu

p

(2123-15)

Bu bunrup (either localBu or effB ) (GWdt)

Here it is assumed that the upper limit of rimp is set as PMX=0254 which is common to

RMPRS=2 mode


- 147 -

(2) Billaux model ( RMPST=2)

Not calculating the amount of fission gas in pores the rim porosity rimp is obtained by

an empirical correlation with respect to burnup That is by applying partially the Billauxrsquos

correlation equation (246)

as-fabricated porosity=p

AP1 AP2 BP1 BP2 GWdtM= sdot minus le lerimp Bu Bu (2123-16)

PMX BP2 GWdtM= gtrimp Bu (2123-17)

Default AP1=00024 AP2=0106

BP1=65 (GWdt) BP2=150 PMX=0254 Bu local burnup (either localBu or effB ) (GWdt)

are assumed Here it is assumed that after rimp increases and attains a temporary highest

value maxrimp rimp remains at its highest value maxrimp even if temperature rises markedly

(3) Effect on thermal conductivity of fuel pellet (RMPRO=0 1) As one of important effects of the rim structure effect of porosity on the pellet thermal

conductivity is calculated by the equation modifying the thermal conductivity of ring

elements which contain the rim structure as follows here suppose that the thermal

conductivity is rimλ of the elements in which the rim structure has developed one of the

following two equations is selected by input

Ikatsu(247) and Bekker(248) 16(1 )rim Bx rimpλ λ= sdot minus (2123-18)

Billauxrsquos equation 25

1

095rim

rim Bx

pλ λ minus = sdot

(2123-19)

where rimp is the porosity of the ring elements defined by Eqs(2123-15) to (2123-17)

RMPRO=0 Ikatsu model (Default) =1 Billaux model

Here Bxλ is a thermal conductivity of fuel matrix at the effective burnup Bx on the basis

of the consideration that the burnup of matrix would virtually decrease ie effects of thermal

conductivity degradation due to burnup extension is mitigated because development of the

rim structure will induce the strain relief of matrix and depletion of fission gas atoms In

other words the relation can be

( )Bx T Bxλ λ= (2123-20)

where Bx effective burnup BXEQ=40 (GWdtU) (Default)


- 148 -

T local temperature at element (K)

Here RMPRO is independent from the designation of RMPST

(4) Swelling When RIMSWL=1 irrespective of the IFSWEL value the volumetric swelling rate of

ring element is set equal to [porosity + solid swelling rate] (default value of RIMSWL is 0)

213 Selection of Models by Input Parameters

The thermal analysis of FEMAXI determines analytical conditions and model options by

a combination of several input parameters ie name-list input parameters The inter-linkage

among these parameters is explained below in an organized manner Here this explanation

of options designated by name-list parameters is in principle given by the order of analysis methodrarrgeometryrarrmaterial properties (models) Further the following explanation is

on the premise of IFEMOP=2 ie a completely coupled solution of thermal and mechanical

analyses When IFEMOP ne 2 some selections below would not be available

2131 Number of elements in the radial direction of pellet

The number of ring elements of a pellet stack is linked to the selection of burnup

calculation code (heat generation density profile model) and the HBS model Here the Table

231 shown in section 231 is re-written for importance

However when the HBS (rim structure) model is designated to use MESHgt0 is required

Even if MESH=0 is input the calculation automatically sets MESH=3

Here when ISHAPE=1 is designated the number of ring elements doubles for the value

of MESHgt0 which allows a more detailed calculation














- 149 -

2132 Calculation of thermal conductivity of fuel pellet

(1) The case in which rim structure formation is not considered (HBS=0) Regardless of the value of parameter MESH let us assume that the thermal conductivity

of pellet is

100( ) ( )TDF p T Buλ λ= sdot (2131)

where ( )F p is porosity factor and this function is described in Chapter 4

The variable ldquoporosity p rdquo is calculated by following the name-list parameter IPRO and

put into the function ( )F p

IPRO=0 (Default) (refer to section 234)

1) IPRO=0 0p p= (2132)

2) IPRO=1 00

swgVp p

V

Δ= + (2133)

3) IPRO=2 00 0

dens hotV Vp p

V V

Δ Δ= + + (2134)

4) IPRO=3 00 0 0

swg dens hotV V V

p pV V V

Δ Δ Δ= + + + (2135)

where P0 initial porosity

0

swgV

V

Δ fission gas bubble swelling

0

densV

V

Δ densification

0

hotV

V

Δ hot press

Here the content of 0

swgV

V

Δ is determined by the selection of name-list input parameter

IFSWEL

(2) The case in which formation of a rim phase is considered (HBS=1 2)

The thermal conductivity of ring element which have been transformed into the rim

structure is defined by Eqs(2123-18) and (2123-19) previously in section 2123-3 (3)

Ikatsu Bekker equation 16(1 )rim Bx rimpλ λ= sdot minus (2123-18)


- 150 -

Billaux equation

251

095rim

rim Bx

pλ λ minus = sdot

(2123-19)

For the ring element having no rim structure Eq(2131) is applied to the thermal

conductivity

(3) Effects of the decrease in density

Regardless of values of MESH HBS and IPRO when IFSWEL=0 or 1 assuming that

the solid swelling is 0

swsV

V

Δ and the density decrease factor due to solid selling is Df

the corrected thermal conductivity λ is

100 ( ) ( )D D TDf f F p T Buλ λ λ= sdot = sdot sdot (2136)

D HBSfλ λ= sdot (2137)

Also when name-list parameter FDENSF=0 (default)

10Df =

and when FDENSF=1 0

10 swsD

Vf

V

Δ= minus (2138)

For IFSWEL see section 41 Chubb and Zimmermann + FEMAXI-III model or

FEMAXI model

2133 Selecting swelling model Swelling depends on the temperature of pellet At the same time it is a phenomenon

which is directly related to deformation and PCMI it has both thermal and mechanical

characteristics Here option is explained when swelling is taken into consideration in terms

of its relationship to fission gas bubbles

(1) Definitions and Classification a) Swelling due to intra-granular gas bubble formation is defined as intraS

b) Swelling due to gas bubble growth at the grain boundary is defined as interS

c) Total gas bubble swelling is defined as sg intra interS S S= +

d) Solid-state fission product swelling is defined as solid ssS S=

e) Swelling due to increase in porosity caused by the rim structure development is

defined as rimS

These are defined by the swelling model which is selected by the name-list input


- 151 -

parameters combined and calculated as described below

(2) Selection of options of swelling model by IFSWEL

A) IFSWEL=0 sgS = Chubb and Zimmermann model(249) + FEMAXI-III(12)

00025ss

solid

VS SWSLD

V

Δ = = sdot

per 1020 fissionscm3 (2139)

SWSLD=10 (Default)

B) IFSWEL=1 FEAMXI-6 model 1) sg intra interS S S= + is calculated from fission gas bubbles

2) 00025ss

solid

VS SWSLD

V

Δ = = sdot


SWSLD=10 (Default)

3) IFSWEL=2 MATPRO-9 model

4) IFSWEL=3 Kosaka model(250)

5) IFSWEL=4 Studzvik model(251)

6) IFSWEL=5 Hollowell model(252) See Chapter 4 for details of each model

7) DENSWEL=1 one of the swelling models and one of the densification models are

merged to calculate the volume change of pellet at every time step

(3) Relation to FGR models

When IFSWEL=0 2 3 4 and 5 swelling is determined independently from the gas

bubbles growth calculation

When IGASP=0 and IFSWEL=1 swelling is determined by the equilibrium bubble

growth model

When IGASP=2 and IFSWEL=1 swelling is determined by the rate-law bubble growth

model

2134 Selecting fission gas release model (1) Selection of model either the equilibrium model or rate-law model is selected

(2) Calculation of intra-granular bubble radius and its number density (IGASP=0 2) A) A strict analytical solution of the White+Tucker equilibrium state-equation model

GBFIS=0


- 152 -

B) Model of re-dissolution by irradiation including bubble destruction and vacancy

migration GBFIS=1

C) Pekka Loumlsoumlnen model GBFIS=2

2135 Bubble growth and swelling

(1) Swelling by grain boundary bubble growth and intragranular bubble growth IFSWEL=1 enables users to calculate the swelling when either IGASP=0 or 2

(2) Porosity (density) Porosity (density) is controlled by IPRO and IDENSWEL They can affect the pellet

thermal conductivity in calculation

(3) Coalescence and coarsening of grain boundary bubbles By IGASP=2 the coalescence and coarsening of grain boundary bubbles are calculated

while if NGB=1 is specified no such calculations are performed


- 153 -

2136 Options for the rim structure formation model

When HBS=0 irrespective of IGASP additional FGR from the rim structure can be

specified by IRIM and RFGFAC

When HBS=1 or 2 Burnup in the rim structure is designated Designation of IRIM

and RFGFAC is ineffective though the rim structure model can be used irrespective of

IGASP values However in the cases of both HBS=0 and 2 designation of MESH=2 or 3 or

4 is required Types of combination of FGR models and swelling models A1 A2 B1 and

B2 are summarized in Table 2131

Table 2131 List of combination of FGR models and swelling models

IGASP IFSWEL Grain boundary bubble Intra-granular bubble Method of calculation GBFIS

A1 0

Equili-

brium

model

0 2

3 4 5

Bubble growth is calculated by the White-Speight- Tucker model with the state equation of gas Not taking into account of the swelling by grain boundary and intra-granular bubbles

Bubble radius is calculated by the state equation

0 default

Bubble radius is calculated by irradiation-re-dissolution model

1

Bubble radius is calculated by Loumlsoumlnen model 2

A2 1

Bubble growth is calculated by the White-Speight- Tucker model with the state equation of gas Taking into account of the swelling by grain boundary and intra-granular bubbles


0 default


1


B1 2

Rate-

Law

model

0 2

3 4 5

Bubble growth is calculated by the rate-law model Taking into account of the coalescence and coarsening of bubbles but not considering the swelling by grain boundary and intra-granular bubbles


0 default


1


B2 1

Bubble growth is calculated by the rate-law model Taking into account of the coalescence and coarsening of bubbles and swelling by grain boundary and intra-granular bubbles


0 default


1



- 154 -

214 Gap Gas Diffusion and Flow Model

Calculation method for the flow of gap gas is described in Fig2141

Time Step Start

Diffused amount of gas between adjacent (lower and upper) segments by concentration gradient

Concentration and composition of gas of each segment after diffusion

Calculated at all the axial segments

Concentration and composition of gap gas at each segment after diffusion

Equilibrium pressure in total space inside fuel rod

Axial transport of gas under pressure gradient (adjustment from lower segment)

Total amount and composition of gas at each axial segment after pressure adjustment

Gas concentration at gap space of each segment

Mutual-diffusion constant of gas

Fission gas release in each axial segment during one time step

Total amount and composition of released gas

Fig 2141 Calculation process of gap gas diffusion and flow


- 155 -

2141 Assumptions and methods of diffusion calculation

In the case of (name-list parameter) IST=0 (default) complete and instantaneous mixing

of gases is assumed while in the case of IST=1 calculation of diffusion and flow of gap gas

is performed on the basis of the following assumptions

A) Helium and Nitrogen (or Argon) as well as released Xenon and Krypton are completely

mixed instantaneously therefore behavior of Nitrogen (or Argon) is represented by that

of Helium and also behavior of Krypton is represented by that of Xenon Thus

calculation is performed for a 2-component mixed gas ie mixture of Helium and

Xenon

B) Counter-diffusion occurs with Helium and Xenon which pass through the gap

C) The counter-diffusion calculation is performed using an implicit solution based on Ficks

first law The diffusion is independent of flow caused by pressure difference in the fuel

rod

D) Pressure in the axial segments of fuel rod is uniform during the diffusion calculation

E) Pressures at different regions inside fuel rod are assumed to be equilibrated

instantaneously Therefore gas flow across the axial segments induced by the

segment-wise pressure difference also occurs instantaneously

In the following sections the method in IST=1 case is described

(1) Calculation method for the gap space The following is a list of symbols used

N gas molar number (mol)

n molar density (molcm3)

D12 counter-diffusion constant (m2s) (D12 = D21)

T temperature (K)

V volume (m3)

C molar fraction ( )( )C n n n1 1 1 2= +

Subscripts are

I kind of gas (1 = Xenon 2 = Helium) j axial segment ( )j P N PL U= 1 2 3 (PL lower plenum PU upper plenum)

In particular subscripts for molar density are


- 156 -

nij density of gas i at axial segment j

nj density of all gases at axial segment j

Figure 2142 shows the model

PlenumSegment j Segment j+1

TpTj+1Tj

2lj 2lj+1

Fig 2142 Gas diffusion and flow model

The total amount of gas i at axial segment j is

N n V

C n VP

RC

V

T

i j i j j

i j j j i jj

j

=

= = (2141)

Here each segment has a channel cross-sectional area of Sj and length of 2l j also the

boundary of segments j and j+1 is indicated by

The amount of gas 1 (Xe) ( )G j j1 1rarr + flowing from segment j to segment j+1 over the

boundary is obtained on the basis of Ficks law

( )G j S Dn n

lj jj

j1

12 1 1rarr =minus

(2142)

Similarly the amount of gas 1 (Xe) flowing into segment j+1 over is represented as

( )G j S Dn n

lj jj

j1 1 1

12 1 1 1

1

1 rarr + =minus

+ ++

+ (2143)

From the continuity condition

( ) ( )G j G j1 1 1rarr = rarr +

holds and

S Dn n

lS D

n n

lj jj

jj j

j

j

12 1 11 1

12 1 1 1

1

minus=

minus+ +

+

+ (2144)

can be obtained By solving this equation for n1 we obtain

nS D l n S D l n

S D l S D lj j j j j j j j

j j j j j j1

121 1 1 1

121 1

1 112 12

1

=+

++ + + +

+ + + (2145)

Using Eq (2142) we obtain the following equation


- 157 -

( ) ( )

( )G j j G j

S D S D

S D l S D ln nj j j j

j j j j j jj j

1 1

121 1

12

1 112 12

11 1 1

1rarr + = rarr

=+

minus+ +

+ + ++

(2146)

Here by setting

FS D S D

S D l S D lj jj j j j

j j j j j j +

+ +

+ + +

=minus1

121 1

12

1 112 12

1

(2147)

Eq(2146) is expressed as ( ) ( )G j j F n nj j j j1 1 1 1 11rarr + = minus+ +

(2148)

Meanwhile regarding gas 2 (He) ( ) ( )G j j F n nj j j j2 1 2 2 11rarr + = minus+ +

(2149)

is obtained

Through addition of Eqs(2148) and (2149) the total amount of inflow mixed gas can

be calculated as ( ) ( )G j j F n nj j j jrarr + = minus+ +1 1 1

(21410)

Here the next equation

n n n n n nj j j j j j= + = ++ + +1 2 1 1 1 2 1

holds

Since in general the gap temperatures at segments j and j+1 differ the following equation

holds in the equilibrium state njnen j+1 (21411)

According to Eq(21410) however even if segments j and j+1 are in equilibrium

imaginary flow from segment j to j+1 is present in the numerical calculation

This is because Fickrsquos first law is applied to a field with a temperature gradient To

eliminate this flow a second term representing the effect of temperature difference must be

added to both Eqs(2149) and (21410)

The second term to be added to Eq (21410) is

( ) ( ) ( ) G j j F n n n nj j j j j jrarr + = minus minus minus+ + +1 1 1 1 (21412)

where n nj jand +1 are given by the ideal gas law as follows

nP

R Tn

P

R Tjj

jj

= =++

1 11

1

(21413)

The amount of flow of each component namely Eqs(2148) and (2149) is given

respectively as follows


- 158 -

( ) ( ) ( ) G j j F n n n nj j j j j j1 1 1 1 1 1 1 11rarr + = minus minus minus+ + + (21414)


Here

nP

R

C

Tn

P

R

C

Tjj

jj

11

1 11

1 = =+

+ (21416)

CN N

V V

n V n V

V Vj j

j j

j j j j

j j1

1 1 1

1

1 1 1 1

1

=++

=++

+

+

+ +

+

(21417)

nP

R

C

Tn

P

R

C

Tjj

jj

22

2 12

1 = =+

+ (21418)

CN N

V V

n V n V

V Vj j

j j

j j j j

j j2

2 2 1

1

2 2 1 1

1

=++

=++

+

+

+ +

+

(21419)

C C1 2 1+ = (21420)

(2) Implicit solution of He－Xe mutual diffusion The flux J R1 of gas 1 (Xe) that diffuses through the boundary surface R (boundary

between segments jminus1 and j) into segment j is given by

( ) ( ) J F n n n nR j j j j j j1 1 1 1 1 1 1 1 = minus minus minusminus minus minus (21421)

The flux J S1 of gas 1 (Xe) that diffuses through the boundary surface S (boundary

between segments j and j+1) into segment j+1 is given by

( ) ( ) J F n n n nS j j j j j j1 1 1 1 1 1 1 1 = minus minus minus+ + + (21422)

Then the change in the amount of Xenon in segment j is given by

Vn

tJ Jj

jR S

partpart

11 1

= minus (21423)

Through differentiation of Eq(21423) using Eqs(21421) and (21422) and the implicit

method the following Equation is obtained Here k denotes time-step number


- 159 -

( ) ( )[ ]

( ) ( )[ ]n n

t VF n n n n

F n n n n

jk

jk

jj j j j j j

j j j j j j

11

1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1

11 2 2 1 1

1 2 2 1 1

+

minus minus minus

minus + +

minus= minus minus minus

minus minus minus minusΔ θ

θ θ θ θ θ

θ θ θ θ θ (21424)

Here

θ 1= interpolation with respect to temperature volume pressure and

equilibrium molar ratio

θ 2= interpolation with respect to molarity

θ 1 and θ 2

are the interpolation parameters in a converging loop in the axial direction and

in the counter-diffusion calculation respectively θ 1 changes in the converging loop in the

axial direction but remains constant in the counter-diffusion calculation

Tuning parameter THG1THG2 θ1 and θ 2 can be specified by THG1 and THG2

respectively Default=10

Therefore an interpolation equation for θ 2

( )n n nj jk

jk

12

2 1 2 111

θ θ θ= minus + + (21425)

is substituted into Eq (21424) and the following equation is obtained

( )[ ( )

( )]

n n

t

F

Vn

F F

Vn

F

Vn

VF n n

n n F

jk

jk

j j

jj

k j j j j

jj

k

j j

jj

k

jj j j

kj

k

j j j j

11

1 2 11 1

1 2 1 1

11

2 11 1

11 2 1 1 1

1 1 1 1

1

1

1 1

1

1

1 1

1

1 1 1

11

1

+minus

minus+ + minus +

minus+

+minus minus

minus +

minus= minus

+

+ + minus minus

minus minus minus minus

Δθ θ

θθ

θ

θ

θ

θ θ

θ

θ

θ θθ

θ θ θ ( )[ ( ) ( )]2 1 1 1 1 1 11 1n n n nj

kj

kj j minus minus minus+ +

θ θ

(21426)

Through rearrangement of Eq (21426)

A n B n C n Dj jk

j jk

j jk

j1 11

11

1 11

minus+ +

+++ + sdot = (21427)

Here

AF

Vtj

j i

j

= minus minusθ θ

θ2 1

1

1

Δ (21428)

B

F F

Vtj

j i j j

j

= + minus+minus +

12 1 1

1 1

1

θ θ θ

θ Δ (21429)


- 160 -

CF

Vtj

j j

j

= minus +θ θ

θ2 1

1

1

Δ (21430)

( ) ( )( )

( ) ( )

DV

F t nV

F F t n

VF t n

VF n n F n n

jj

j j jk

jj j j j j

k

jj j j

k

jj j j j j j j j

= minus + minus + minus

+ minus + minus minus minus

minus minus minus +

+ + + + minus minus

11 1

11

11

1

1

1

1

1 1

1

1

1

1 1 1 1 1

1 2 1 1 1 1 2 1

1 2 1 1 1 1 1 1 1 1 1 1

θθ

θθ θ

θθ

θθ θ θ θ θ

θ θ

θ

Δ Δ

Δ ( )θ1 Δt

(21431)

First the upper and lower axial segments including plenum are considered

Since there is no boundary surface R for the first segment

( ) ( )[ ]V

n

tJ

F n n n n

S11 1

1

1 2 1 1 1 2 1 1 1 2

partpart

= minus

= minus minus minus minus

(21432)

holds therefore the following equation holds

( )

( ) ( )

n n

t

F

Vn

F

Vn

F

Vn

F

Vn

F

Vn n

k kk k k

k

1 11

1 1 2 1 2

11 1

1 2 1 2

11 2

1 2 1 2

11 1

2 1 2

11 2

1 2

11 1 1 2

1

1

1

1

1

1

1

1

1

1

1 1

1

1

++ +minus

= minus + minusminus

+minus

+ minus

Δθ θ θ

θ

θ

θ

θ

θ

θ

θθ

θ

θ

θθ θ

(21433)

This equation is rearranged using

( ) ( ) ( )

BF

Vt C

F

Vt

DF

Vt n

F

Vtn

F

Vn n tk k

12 1 2

11

2 1 2

1

12 1 2

11 1

2 1 2

11 2

1 2

11 1 1 2

1

11 1

1

1

1

1

1

1

1

1

1

1

1 1

= + = minus

= minusminus

+

minus+ minus

θ θ

θ θ

θ

θ

θ

θ

θ

θ

θ

θ

θ

θθ θ

Δ Δ

Δ Δ Δ (21434)

and we obtain

B n C n Dk k1 1 1

11 2

11sdot + sdot =+ +

(21435)

Similarly there is no boundary surface S for the uppermost segment n and thus the

following equation holds

( ) ( )[ ]V

n

tJ

F n n n n

nn

R

n n n n n n

partpart

11

1 1 1 1 1 1 1

= minus

= minus minus minus minusminus minus minus

(21436)

Therefore by setting

A

F

Vt B

F

Vtn

n n

nn

n n

n

= minus = +minus minusθ θθ

θ

θ

θ2 1 2 1

1

1

1

11 Δ Δ


- 161 -

( ) ( )

( ) tnnV

F

ntV

Fnt

V

FD

nnn

nn

knn

n

nnkn

n

nnn

Δminusminus

Δ

minusminus+Δ

minus=

minusminus

minusminus

minus

11

1

1

1

1

1

1

1111

112

1112 1

11

θθθ

θ

θ

θ

θ

θ θθ

(21437)

we obtain

A n B n Dn nk

n nk

nsdot + sdot =minus+ +

1 11

11

(21438)

Combining the above we obtain the following equation for the diffusion of gas 1

B CA B C

A B C

A B CA B C

A B

n

n

n

n

n

n

n n n

n n n

n n

k

k

k

nk

nk

nk

1 1

2 2 2

3 3 3

2 2 2

1 1 1

1 11

1 21

1 31

1 21

1 11

11

sdotsdotsdotsdotsdot


minus minus minus

minus minus minus

+

+

+

minus+

minus+

+

=


minus

minus

DDD

DDD

n

n

n

1

2

3

2

1

(21439)

Gas concentration ni jk+1 of each axial segment is obtained by solving Eq (21439)

(3) Mutual diffusion constant

The gas counter-diffusion constant adopts Presents constant(231)

DkT

m n d12

1

2

122

3

8 2

1=

ππ

(21440)

where

mm m

m m =

+1 2

1 2

(m weight of one molecule)

( )d d d1212 1 2= + (d diameter of one molecule)

n n n= +1 2 (molar density)

k Boltzmanns constant T temperature (K)

(4) Pressure adjustment calculation

Setting the total gas molar number as Nj for each axial segment at the end of time step

N Nj iji

= (21441)


- 162 -

holds where i represents kind of gas For the entire volume of the fuel rod inner space the

gas molar number is given by

jj

N N= (21442)

Upperpl enum

Segment

5

Segment

Xe injectionPressure adjustment

4

Counter-diffusion

Segment3

Segment2

Segment1

Lower

pl enum

Previous step

Current step

He Xe He Xe He Xe

After pressure adjustment

Fig 2143 Calculation of pressure adjustment (vertical axis gas molar number)

Since

P N RV

T jj

= sdot sdot

1 (21443)

holds on the basis of the ideal gas law the total molar number of gas primeN j in segment j after

the calculation of pressure adjustment is given by

prime =

NP

R

V

Tjj

(21444)


- 163 -

The discrepancy between Nj and Njrsquo is adjusted by balancing of the input and output flows

between the upper and lower adjacent segments beginning from the lowest part of pellet stack

Figure 2143 shows the pressure adjustment process which consists of three steps

Namely if the following are set

Nirarrj molar number of mixed gas transported from segment i to segment j

fij fraction of segment i in segment j before transport

the gas molar number i for segment j after transport is given by

prime = minus

+rarr rarr N f N N f Nij ij j j kk

il l jl

(21445)

Model parameter IST When using the above model of gas flow and diffusion in gap designate IST=1 Default is IST=0 which gives an instantaneous complete mixture of gas is assumed

215 Method to Obtain Free-space Volumes in a Fuel Rod 2151 Definition of free space

The gap volume in a fuel rod can be classified into the following four components A)

plenum volume B) P-C gap volume C) pellet center hole volume and D) free-gas volume in

the pellet stack This classification is required because the pressure reference temperature for

each volume is different in calculation The free gas volume in the pellet stack consists of dish

volume chamfer volume inter-gap volume between pellet ends (pellet-to-pellet aperture) and

crack gap space inside pellet

When IFEMRD=1 ie in the entire rod length mechanical analysis quantities of the A)

B) and C) components change following the calculated results while changes of ldquoD) free-gas

volume in the pellet stackrdquo ie volume changes of dish space chamfer space and inter-gap

between pellet ends are not taken into account and crack space calculation depends on the

option value of name-list parameter IRELCV (default=0) while when IFEMRD=0 ie in

the local PCMI mechanical analysis it is assumed that of the four components the volumes

of the center hole and free gas in the stack do not change during irradiation

(1) Center hole volume

The center hole volume is given using the initial inner radius of pellet rpi and the length


- 164 -

lZj of axial segment j as

V r lhj

pi Zj= π 2 (2151)

The free gas volume inside the stack includes dish volume chamfer volume

pellet-to-pellet free space volume and void volume in cracks

When the length of one pellet is set as lp dish volume per pellet as Vdish chamfer volume

as Vchem inter-gap volume between pellet ends as Vp and void volume in crack as Vcrack the

equations

V r lp po= π 2 2Δ (2152)

V r u lcrack porel

p= 2π (2153)

hold where

rpo outer radius of pellet

Δl pellet-to-pellet aperture width

urel radial displacement of pellet caused by relocation

The free gas volume V jint in the stack in axial segment j in Fig2151 is given by

( )V V V r l r u ll

lj

dish cham po porel

pZj

pint = + + + sdotπ π2 2 2Δ (2154)

(2) Plenum volume and pellet-cladding gap volume

In contrast to the above plenum volume and gas volume are assumed to change during

irradiation The plenum volume change is obtained from the change in axial segment length

The pellet-to-pellet aperture

lΔ is defined as the distance

between two edges of

adjacent pellets as described

by the right-side figure lΔ

It specifies the tilting of pellets

It is designated by parameter AY

The default value is 20 μm

Parameter AY

pellet

Fig2151 Definition of pellet-to-pellet aperture


- 165 -

inside the stack In segment j setting the thermal expansion strain averaged by pellet

volume as ε thj densification strain as ε den

j and solid swelling strain as ε ssj the segment

length ~

lZ pj including the changes by these strains is

( )~l lZ pj

thj

denj

ssj

Zj= + + +1 ε ε ε (2155)

However when IFEMOPgt0 j

Z pl is obtained by the calculated deformation of

mechanical model In cladding setting the thermal strain as Δε c thj and strain by irradiation

growth as Δε c irrj the segment length

~lZ Cj is

( )~ l lZ Cj

c thj

c irrj

Zj= + +1 ε ε (2156)

Similarly the plenum length Z pll is given by the calculated deformation of mechanical

model as follows (Refer to section 33 ldquoMechanical analysis of entire rod lengthrdquo for details)

( ) 1 pl plZ pl c th c irr Z pll lε ε= + + (2157)

The change in plenum volume ΔVpl is given by

( )2

1

NAXIj j

pl ci Z C Z p Z pl Z plj

V r l l l lπ=

Δ = minus + minus

(2158)

where NAXI represents the number of axial segments and rci the cladding inner radius

Accordingly when the initial plenum volume is set as Vpl the plenum volume is

~V V Vpl pl pl= + Δ (2159)

The gap volume of axial segment j is given as follows using the gap width δ Here δ

= 0 is assumed when numerically δ lt0

~ ~ V r lgap j ci Z p

j= sdot sdot2π δ (21510)

(3) Volumes of dish chamfer and pellet-to-pellet gap space

Space volume of chamfer which is calculated by Fig2152 is

2 3chamfer

Dp aV abπ = minus

(21511)


- 166 -

Dish space volume which is calculated by Fig2153 is

2 2( 3 )6dish

dV d r

π= + (21512)

Fig2153 Volume of dish space

Fig2152 Volume of chamfer

a

b

Dp

Dp

r

d


- 167 -

Space of tilting end faces of pellet which is calculated by Fig2154 is

2 21tan ( )

8tilting inV Dp Dp Dπ θ= sdot minus (21513)

where tanθ＝0002 is assumed

(4) Volume of cracks inside pellet Change of crack space volume inside pellet is not taken into account when IRELCV=0

When IRELCV=1 this change is taken into account on the basis of the relocation strain

change inside pellet Here setting as

A= P-C gap space volume change induced by relocation strain and

B=Space volume of cracks inside pellet produced by relocation

a model of ldquocrack space volume inside a pelletrdquo is introduced which assumes that A+B can

remain unchanged irrespective of the change of relocation strain However when

IRELCV=0 is applied A+B will not remain unchanged with the change of relocation strain

Accordingly for the internal pressure calculation IRELCV=1 has to be applied

When IRELCV=1 is applied and change of crack space is taken into account on the basis

of the relocation strain the crack space approaches zero when cladding creeps down inward

P-C gap is closed pellet fragments are pushed back relocation displacement is decreased and

compressive stress is generated inside pellet This is linked with the model of pellet Youngrsquos

modulus The pellet stiffness change model is applied to the pellet volume change by

Fig2154 Space volume by tilting of pellet end face

θ

Din

Center hole

Dp


- 168 -

relocation Details are described in section 323

216 Internal Gas Pressure It is assumed that the gas in a fuel rod behaves as an ideal gas and the pressure in the

fuel rod is uniform in every part of rod Then the gas pressure is calculated as

int

1

Tgas j j jM

pl L pl U gap hj j j

jpl L pl U gap pi av

n RP

V V V V V

T T T T T=

sdot=

+ + + +

(2161)

where

Pgas gas pressure inside fuel rod (Pa)

nT total molar number of all gases (mol)

R gas constant 8314 JK sdot mol

VplL lower plenum volume (m3)

TplL lower plenum gas temperature (K)

VplU upper plenum volume (m3)

TplU upper plenum gas temperature (K)

Tw coolant temperature (K)

Vgapj

gap volume in axial segment j (m3)

Vhj pellet center hole volume in axial segment j (m3)

V jint free gas volume in pellet stack (m3)

Tgapj gap temperature in axial segment j (K) = 05 ( )T Tps

jci

j+

Tpij pellet center temperature in axial segment j (K)

Tpij volumetric average temperature of pellet in axial segment j (K)

Here temperatures used to calculate the gas pressure inside fuel rod are

A) Plenum gas temperature is a sum of temperature of coolant at the plenum elevation and

DTPL (Refer to the next section 2161)

B) Gap gas temperature is an average of pellet outer surface temperature and cladding inner

surface temperature

C) Pellet center hole gas temperature is the temperature of pellet center or pellet center hole

inner wall temperature

D) Free gas temperature inside pellet stack is a volumetric average temperature of pellet


- 169 -

Here the free gas volume V jint includes dish volume chamfer volume inter-pellet free

space volume and void volume in crack V jint is already known by Eq(2154)

2161 Method of assigning plenum gas temperature and cladding temperature

Temperatures at plenums are assigned by two parameters DTPL and ITPLEN

(1) ITPLEN=0 (default) specifies the followings

The lower plenum gas temperature is given as the [coolant inlet temperature + DTPL]

and the upper plenum temperature is given as the [coolant outlet temperature + DTPL] The

value of DTPL can be specified by input parameter DTPL (default value is 25C) Here one

must be cautious with the plenum gas temperature because the plenum gas pressure is

calculated using the [coolant temperature + DTPL] even under a cold state

In addition the upper plenum cladding temperature is set equal to the calculated coolant

outlet temperature The lower plenum cladding temperature is set equal to the coolant inlet

temperature These cladding temperatures are not dependent on the calculation procedures

whether the cladding surface temperature is given by input or by calculation Here the outlet

coolant temperature at the upper plenum section is determined by the heat conduction

calculation at the segment immediately below the upper plenum section of the fuel rod

(2) ITPLEN=1 specifies the followings

The plenum gas temperature is set as the [coolant inlet temperature + DTPL] In this

option the cladding temperatures of the upper and lower plenum sections are given by the

coolant inlet temperature When there is a lower plenum section pl L pl UT T= =[coolant inlet

temperature + DTPL] is adopted for Eq (2171)

2162 Calculation of the variation in internal gas condition during irradiation

In FEMAXI it is possible to change the composition pressure of the internal gas and

plenum volume up to 20 times during irradiation by specifying the name-list input parameters

This option enables users to perform analysis of experiments on fuel rod which is

base-irradiated in a commercial reactor and then test-irradiated in a test reactor after

instrumentation or analysis of creep deformation experiment of cladding by increased internal

pressure during irradiation The name-list input parameters for these purposes are ITIME(20)


- 170 -

GASPRN(20) PLENM(20) and GMIXN(4 20)

Here when a rod is refabricated and instrumented it is assumed that the instrumentation

is carried out at room temperature and with no power Accordingly this input condition has to

designate zero linear power at the irradiation history point which is specified by ITIME

written in input file In this zero power point the rod temperature becomes equal to the

coolant temperature at this history point For example if coolant temperature is 300K and

even if DTPLgt0 temperature of the entire rod including plenum becomes 300K Therefore

the pressure and volume of internal gas should be considered as quantities at 300K

Tuning parameter DTPL Default value of DTPL is 25K As the plenum gas

temperature is always set as [coolant temperature +DTPL] it is to be noted that the

plenum gas pressure is calculated by using the temperature which has additional

DTPL even in the cold state after irradiation

Option parameter ITPLEN Default value is ITPLEN=0

217 Time Step Control The time-step width is automatically controlled in FEMAXI In the entire calculation

the width is determined by the restriction Δt1 while in the thermal analysis it is subdivided

into Δt2 and in the mechanical analysis it is subdivided into Δt3

2171 Automatic control

(1) Restriction for entire calculation Δt1 is the minimum among the values determined by the following four conditions

1) Changes in linear heat rate is within 10 Wcm during one time-step

2) Changes in burnup is within 100 MWdt during one time-step

3) Creep strain rate is limited so that creep strain increment does not exceed elastic

strain

4) Time step width is within 15 times the former time step width

5) Change of coolant flow rate is within 10 of that of the former time step

The condition 1) is to stabilize calculation when power changes more than 10Wcm the

condition 2) is determined by subroutine PHIST to make the burnup increment less than


- 171 -

100MWdtUO2 The condition 3) is a restriction to avoid divergence in creep calculation for

all elements and Δt1 is determined by

EFCOEF1 sdotsdot

leΔcE

tε

σ (2171)

where EFCOEF is a tuning parameter

The condition 4) restricts the expansion of Δt1 within a factor of 15 when rod power

increases at less than 10Wcm during one time step The condition 5) is for the fast change of

coolant flow rate However in the thermal analysis a time step width Δt2 which is smaller

than Δt1 can be used in the following manner

(2) Restriction for sub-division of time step width for thermal analysis Δt2 is used for the gas diffusion time-step width which is the subdivision of Δt1 and

determined as the shortest time among the time required to reach the equilibrium state of gas

flow in each axial segment and the time determined from the maximum amount of transport

of gas between the segments as well as from pressure adjustment conditions with its upper

limit value set as either 100 ms or Δt1

The time required to reach the equilibrium state in each axial segment is derived from

Eq(21423) as

( )

Δtn n V

J Jji j j j

R S

=minus

minus1 1

1 1

(2172)

where the superscript i represents axial segment number

Next the time determined from the maximum amount of transport of gas between

segments is obtained as follows Using the maximum molar number of the gas transported

These factors are related to 1t Δ The term DPXX

is the variation size of linear power within a time step its default value is DPXX=10

(Wcm) The term DPBU is the burnup increase within a time step its default value is

DPBU=1 00 (MWdt O2 ) This default value occasionally causes an extremely long

computation In many cases adopting DPBU=5000 is effective in reducing computation

time Setting DPXX value is effective in reducing the extremely long calculation time

For example it can be set to DPXX=90

Tuning parameter DPXX DPBU


- 172 -

between segments Nmax as standard since ( ) ( ) J JR j S j1 1 1= minus we represent Δt2 as follows

using ( )J R j1

( )ΔtN

Jj

R j

2

1

= max

(2173)

Then the time determined from the pressure adjustment conditions in gas flow is obtained

as follows Due to the assumption of instantaneous equilibrium of pressure inside the fuel

rod numerical instability resulting from gas flow between segments may occur To avoid

this instability we must impose a restriction on the amount of gas transport at each segment

Setting the gas transport rate in each segment as β (mols) the restriction is given by

ΔtN

jj

3 = max

β (2174)

Thus under the conditions

( )Δ Δ Δ Δt t t tn21 11

21 1

2= sdotmin η (2175)

( )Δ Δ Δ Δt t t tn22 12

22 2= min (2176)

( )Δ Δ Δ Δt t t tn23 13

23 3= min (2177)

Δt2 is given by

( )Δ Δ Δ Δt t t t2 21 22 23= min (2178)

Nmax in Eqs(2173) and (2174) and η 2 are the control parameters programmed in the

code

Tuning parameters EFCOEF AMLMX2 AMLX3 and DTPR

Default value of EFCOEF in Eq(2171) is 10 Nmax in Eq(2173) can be specified by AMLMX2 Default value is 10-6(mole)

Nmax in Eq(2174) can be specified by AMLMX3 Default value is 2times10-6(mole)

η2 in Eq(2175) can be specified by DTPR Default value is 001

2172 Time step increment determination in FGR model In the fission gas release model the time step increment

2tΔ given by section 2171 is

further divided by a particular method which uses the equivalent diffusion constant Drsquo and


- 173 -

nodal interval 2RΔ of the second zone of spherical FEM system to give sub-divided time step

as

22

3 005 FMULTR

tD

ΔΔ = times timesprime

(2179)

Adjusting input parameter FMULT Default=10

2173 Time step increment determination in temperature calculation In a non-steady temperature calculation the time step increment 2tΔ given by section

2181 is further divided by a particular method which uses the thermal diffusivity Ki and ring

element thickness rΔ to give sub-divided time step as

( )2

64

min

10i

rt

K

Δ Δ = sdot

(21710)

Here the factor 106 is very large because it is determined by an extremely thin oxide

thickness rΔ and has been empirically determined through sensitivity analysis

In some cases 4tΔ is determined as a very small value with respect to 2tΔ If 4tΔ is less

than 1100 of 2tΔ to proceed a time step of transient thermal calculation the time step period

2tΔ is not divided with equal intervals of 4tΔ but the time step begins to increase with the

initial value of 4tΔ 分 and doubles the former time step tΔ to circumvent the too large

number of time steps

On the other hand as a result of non-steady thermal calculation using 4tΔ if the obtained

temperature values are abnormal or if the iteration does not reach convergence after more

than 20 times iterations which is set as the maximum value in the process of determining a

temperature-dependent materials properties by iteration calculation this result is discarded

and re-calculation is performed with the time step width of 230 times 4tΔ as a new 4tΔ This

230 times is the value determined empirically through sensitivity analysis This procedure

has proved to suppress the numerical instability of temperature calculation


- 174 -

References 2

(21) Uchida M Otsubo N Models of Multi-rod Code FRETA-B for Transient Fuel Behavior Analysis(Final Version) JAERI 1293(1984)

(22) Ransom VH et al RELAP5MOD1 Code Manual Volume 1 System Models and

Numerical Methods NUREGCR-1826(EGG-2070) (1980)

(23) Dittus FW Boelter LMK Univ Calif Pubs Eng2 443(1930)

(24) Chen J A Correlation for Boiling Heat Transfer to Saturated Fluids in Convective Flow Process Design Developments 5(1966)

(25) Jens WH Lottes PA Analysis of Heat Transfer Burnout Pressure Drop and

Density Data for High Pressure Water ANL-4627(1951)

(26) Bjornard TA Griffith P PWR Blowdown Heat Transfer ASME Symposium on the Thermal and Hydraulic Aspects of Nuclear Reactor Safety Vol1(1977)

(27) Tong LS Weisman J Thermal Analysis of Pressurized Water Reactors American

Nuclear Society (1970) 76-HT-9(1976)

(28) Hsu YY Beckner WD A Correlation for the Onset of Transient CHF cited in LS Tong GLBennett NRC Water Reactor Safety Research Program Nuclear Safety 18 1 JanuaryFebruary(1977)

(29) Smith RA Griffith P A Simple Model for Estimating Time to CHF in a PWR

LOCA Transactions of American Society of Mechanical Engineers paper No76-HT-9(1976)

(210) Collier JG Convection Boiling and Condensation London McGraw-Hill Book

Company Inc(1972)

(211) SCDAPRELAP5MOD31 Code Manual Vol4 MATPRO-A A Library of Materials Properties for Light-Water-Reactor Accident Analysis NUREGCR-6150(1995)

(212) Gazarolli F Garde AM et al Waterside Corrosion of Zircaloy Fuel Rods

EPRI-NP 2789 (1982)

(213) MATPRO-09 A Handbook of Materials Properties for Use in the Analysis of Light Water Reactor Fuel Rod Behavior USNRC TREE NUREG-1005 (1976)

(214) Ohira K and Itagaki N Thermal Conductivity Measurements of High Burnup UO2

Pellet and a Benchmark Calculation of Fuel Center Temperaturerdquo Proc ANS Topical Meeting on Light Water Reactor Fuel Performance pp541-549 Portland USA March (1997)


- 175 -

(215) Wiesenack W and Tverberg T Thermal Performance of High Burnup Fuel ndash In-pile Temperature Data and Analysisndash Proc 2000 Int Topical Mtg on LWR Fuel Performance Park City USA (2000)

(216) Robertson JAL kdθ in Fuel Irradiation CRFD-835 (1959)

(217) Ross AM and Stoute RL Heat Transfer Coefficient between UO2 and Zircaloy-2

CRFD-1075 (1962)

(218) Wilson MP Jr GA-1355 (1960)

(219) Hasegawa and Mishima Handbook of Materials for Nuclear Reactor p466 Nikkan Kougyou Shinbunsha (1977) [in Japanese]

(220) Une K Nogita K Kashibe S Toyonaga T and Amaya M Effect of Irradiation-

Induced Microstructural Evolution on High Burnup Fuel Behavior Proceedings to Int Topical Meeting on LWR Fuel Performance Portland USA pp478-489 (1997)

(221) Fujimoto N Nakagawa S Tuyuzaki N et al Verification of Plant Dynamics

Analysis Code for HTTR ldquoASURArdquo JAERI-M 89-195 (1989) [in Japanese]

(222) The Japan Society of Mechanical Engineers -JSME Data Book Heat Transfer 5th Edition- (2009)

(223) Speight MV A Calculation on the Migration of Fission Gas in Material Exhibiting

Precipitation and Re-solution of Gas Atoms under Irradiation Nucl Sci Eng 37p180 (1969)

(224) White RJ and Tucker MO A New Fission Gas Release Model

J Nucl Mater 118 pp1-38 (1983)

(225) Ham FS J Phys Chem Solids 6 p335 (1958)

(226) White RJ Fission Gas Release HWR-632 (2000)

(227) Hall ROA and Mortimer MJ Surface Energy Measurements on UO2 - A Critical Review JNuclMater148 pp237-256(1987)

(228) Turnbull JA Friskney CA Findlay FR Johnson FA and Walter AJ The

Diffusion Coefficients of Gaseous and Volatile Species during the Irradiation of Uranium Dioxide J Nucl Mater 107 pp168-184 (1982)

(229) Loumlsoumlnen P Modelling intragranular fission gas release in irradiation of sintered LWR

UO2 fuel JNuclMater 304 pp29-49 (2002)

(230) Itoh K Iwasaki R and Iwano Y Finite Element Model for Analysis of Fission Gas


- 176 -

Release from UO2 Fuel JNuclSciTechnol 22 pp129-138 (1985) (231) Ainscough JB and Oldfield BW Ware JO Isothermal Grain Growth Kinetics in

Sintered UO2 Pellets J Nucl Mater49 pp117-128 (197374)

(232) Kogai T Modelling of fission gas release and gaseous swelling of light water reactor fuels JNuclMater 244 pp131-140 (1997)

(233) Matthews JR and Wood MH A simple Operational Gas Release and Swelling

Model IIGrain boundary gas JNucleMater91 pp241-256(1980)

(234) Van Uffelen P Parametric Study of a Fission Gas release Model for Light Water Reactor Fuel Proc 2000 Int Topical Mtg on LWR Fuel Performance Park City USA (2000)

(235) White RJ The Development of Grain Face Porosity in Irradiated Oxide Fuel

Seminar on Fission Gas Behavior in Water Reactor Fuels Cadarache France NEANSCDOC-20 (2000)

(236) Central Research Institute of Electric Power Industry Battelle High Burnup Effects

Program Task 3 Report Komae Research Laboratory Report T90802 (1990) [in Japanese]

(237) Walker CT Kameyama T Kitajima S and Kinoshita M Concerning the

microstructure changes that occur at the surface of UO2 pellets on irradiation to high burnup JNuclMater 188 pp73-79 (1992)

(238) Kitajima S and Kinoshita M Development of a Code and Models for High Burnup

Fuel Performance Analysis IAEA TCM on Water Reactor Fuel Element Modelling at

High Burnup and its Experimental Support Windermere UK Sept(1994)

(239) Nogita K and Une K Irradiation-induced recrystallization in high burnup UO2 fuel JNuclMater 226 pp302-310 (1995)

(240) Une K Nogita K Suzawa Y Hayashi K Ito K and Etoh Y Effects of grain size

and PCI restraint on the rim structure formation of UO2 fuels Proc IAEA Technical

Committee Meeting on Nuclear Fuel Behavior Modelling at High Burnup and its

Experimental Support Windermere UK (2000)

(241) Yan-Hyun Koo Je-Yong Oh Byung-Ho Lee Dong-Seong Sohn Three-dimensional

simulation of threshold porosity for fission gas release in the rim region of LWR UO2

fuel JNuclMater 321 pp 249-255 (2003)

(242) Cunningham ME Freshley MD and Lanning DD Development and Characteristics of the Rim Region in High Burnup UO2 Fuel Pellets JNuclMater 188 pp19-27 (1992)


- 177 -

(243) Lassmann K Walker CT van de Laar J and Lindestroem F Modelling the high burnup UO2 structure in LWR fuel J NuclMater 226 pp 1-8(1995)



fuel JNM 321 pp 249-255(2003)

(245) Kinoshita M Ultra High Burnup Characteristics of UO2 Fuel Proceedings to KNSAESJ Joint Seminar YongPyung Korea (2002)

(246) Billaux MR Sontheimer F Arimescu VI and Landskron H Fuel and Cladding

Properties at High Burnup Seminar on Fission Gas Behaviour in Water Reactor Fuels- Executive Summary Cadarache France NEANSCDOC(2000)20 (2000)

(247) Ikatsu N Itagaki N Ohira K and Bekker K Influence of rim effect on fuel center

temperature Proc IAEA Meeting on High Burn-up Fuel Specially Oriented to Fuel Chemistry and Pellet-Clad Interaction Sweden Sept (1998)

(248) Bakker K Kwast H and Cordfunke EHP Determination of a Porosity Correction

Factor for the Thermal Conductivity of Irradiated UO2 Fuel by Means of the Finite Element Method JNuclMater 226 pp 128-143 (1995)

(249) Chubb W Storhok VW and Keller DL Factors Affecting the Swelling of Nuclear

Fuel at High Temperatures Nucl Technol 18 pp231-255 (1973)

(250) Kosaka Y Thermal Conductivity Degradation Analysis of the Ultra High Burnup Experiment (IFA-562) HWR-341 (1993)

(251) Schrire D Kindlund A and Ekberg P Solid Swelling of LWR UO2 Fuel

HPR-34922 Enlarged HPG Meeting Lillehammer (1998)

(252) Hollowell TE The Development of an Improved UO2 Fuel Swelling Model and Comparison between Predicted Pellet Cladding Gaps and PIE Measured Gaps (Gap Meter Measurements) HPR-229 Enlarged Halden Programme Group Meeting on Fuel Performance Experiment and Evaluation Hanko vol1 (1979)

(253) Present RD Kinetic Theory of Gases Mc-Graw Hill NY p55 (1958)


- 178 -

x y z yz zx xyσ σ σ τ τ τ

ε ε ε γ γ γx y z yz zx xy

3 Mechanical Analysis Model First the analysis based on the finite element method used in FEMAXI is explained

31 Solutions of Basic Equations and Non-linear Equations

The governing equations of elastic solid mechanics in analyzing the behavior of objects

are linear This implies the followings

(a) The equation relating strain to displacement is linear

(b) The equation relating stress to strain is linear

However there are many nonlinear phenomena among actual objects Accordingly to

address these phenomena it is necessary to expand the method of numerical calculation to

accommodate nonlinearity In this manner phenomena such as plasticity and creep or other

complex constitutive equations (the equation which expresses the relationship between stress

and strain) which encompass the entire range of solid mechanics can be replaced by simple

and linear elastic relationships This approximation is explained below

311 Basic equations

In FMAXI because constitutive equations are constructed on the basis of infinitesimal

deformation theory (model of small increment) the relational expression of

strain-displacement is linear The problem of nonlinearity occurs in the relational expression

of stress and strain To analyze this problem a general method of replacing the nonlinear

relationship between stress and strain with a linear relationship is discussed

(1) Basic assumption and formulation of mechanical analysis

In general a mechanical analysis of materials comes down to solve the problem to obtain

the following set of 15 components

1) three components of displacement u v w

2) six components of strains

3) six components of stress

Here equations and conditions to obtain these quantities are

1) boundary condition

2) equations of equilibrium in the three principal directions ( x y z or rθ z )

3) six equations of strain-displacement relation and


- 179 -

10r rr rz R

r r z rθ θτ σ σσ τ

θpart minuspart part+ + + + =

part part part

10zrz z rz Z

r r z rθττ σ τθ

partpart part+ + + + =part part part

210r z r

r r z rθ θ θ θτ σ τ τ

θpart part part+ + + + Θ =part part part

0rr

r rθσ σσ minuspart + =

part

rr

u

rε part=

part1r uu

r rθ

θεθ

part= +part

zz

u

zε part=

part

[ ]1( ) pl cr sw

r r z TE θε σ ν σ σ α ε ε ε= minus + + + + +

[ ]1( ) pl cr sw

r z TEθ θε σ ν σ σ α ε ε ε= minus + + + + +

[ ]1( ) pl cr sw

z z r TE θε σ ν σ σ α ε ε ε= minus + + + + +

0r zθε ε ε+ + =

4) six stress-strain constitutive equations

In theory solution can be obtained by numerical calculation which satisfies all these

equations and conditions

These equations are shown below with respect to a cylindrical coordinate system

A group equations of equilibrium in the three directions

Here R Z and Θ are body force such as gravity inertia electromagnetic force etc

B group three equations of strain-displacement relation omitting another three equations

for shear strain

C group four constitutive equations of stress-strain relation omitting another four equations

for shear components


- 180 -

The last equation implies the conservation (invariance) of volume of material Therefore

it holds in elastic plastic and creep deformations It does not hold in phenomenon which

includes irreversible volumetric change or concurrent expansion or shrinkage in the three

directions such as thermal expansion irradiation growth swelling and hot-pressing These

phenomena are dealt with by another method in FEMAXI

Here the approximation and hypothesis in numerical calculation by FEMAXI are as

follows

1) within a small region homogeneity of material and uniformity of temperature are

assumed

2) the body forces such as gravity inertia and electromagnetic force are neglected

3) geometrical symmetry is taken into consideration In a fuel rod axis-symmetry is

assumed

4) Either plane strain condition or plane stress condition is adopted case by case

(2) Incremental method First the idea of a typical [incremental method] is explained below In an incremental

interval linearity is assumed for the relational expression of stress and strain (relational

equation for stress increment-strain increment)

The equation for creep is an example of a nonlinear stress-strain relational equation The

general equation for creep cannot express the stress level explicitly by strain however strain

(or its increment) can be expressed explicitly by stress

In other words the equation

( ) tt

f Δsdotpartpart=Δ σε (311)

holds and in general ( )xf is nonlinear equation

Here Eq(311) is expressed as shown below in an incremental interval

tε αΔ = sdot Δ (312)

Here ( )nt

f σαpartpart=

nσ initial stress in the current increment interval (the stress

obtained in the preceding time step)

In the interval between time steps ( )σt

f

partpart

changes with the stress σ change and if

the incremental time tΔ is sufficiently small resulting change in stress is so small that


- 181 -

approximation ( ) ( )nt

f

t

f σσpartpart=

partpart

can be used and the increase in strain can be determined

independent of change in stress using Eq(312) As shown linearity can be assumed between

the stress increment and strain increment in an incremental interval

Next the analysis of FEM using the incremental method is explained

(3) Outline of a solution using the incremental method First an outline of a FEM solution of the basic equation is presented when a linear

relationship holds approximately over an incremental interval This is a basic method in

elastic solid mechanics when linearity can be assumed

In the incremental method which is explained using the creep strain all strain increments

other than increments of elastic strain are represented as known strain and these are

represented by initial strain incremental vectors 0εΔ

Also denoting external force as F stress as σ and a matrix which connects the

strain ε in an FEM element to the displacement of node u as [ ]B

Then

[ ] B uε = (313)

holds This is equivalent to the ldquoB group three equations of strain-displacement relationrdquo

The nodal external force is set as F which is statically equivalent to the element

boundary stress In FEM element force exerts only at nodal points To formulate the FEM

matrix it is necessary to obtain F For this purpose an arbitrary virtual nodal displacement

is given and for this displacement the equilibrium is set between the work done to the

element by the force and the work done by the stress In other words the work by force and

the work by stress are set as equivalent Such a virtual displacement is set as ν

The virtual work FW done to the element node by the external force F is equal to the

sum of the products of each component of force and corresponding displacement

T

FW Fν= (314)

Similarly a virtual internal work per unit volume done by the stress is

[ ] TT

INW Bν σ= (315)

Assuming that the external work determined by integrating these two equations over entire

element volume V is equal to the internal work (the principle of virtual work) the next

equation is obtained


- 182 -

[ ]

[ ]

TT T

V

TT

V

F B dV

B dV

ν ν σ

ν σ

=

=

(316)

Consequently by using the principle of virtual work the equations of equilibrium which

correspond to the above ldquoA group equations of equilibrium in the three directionsrdquo

[ ] T

VF B dVσ= (317)

is obtained Here since Eq(317) can hold at any stress-strain relationship irrespective of

the type of stress-strain relationship it allows FEM to be applied to not only the elastic

deformation of material but also plastic and creep deformation That is theoretically the

principle of virtual work is equivalent to both the equilibrium equations and mechanical

boundary conditions In other words in FEM work done to elements by the external force is

equal to the strain energy change inside elements In summary FEM assumes the ldquoprinciple of

virtual workrdquo which assumes that external work is equal to internal work

However any type of body forces inside elements are taken into account in the equation

formulation process in the code in terms of strains which are associated with volumetric

expansion or shrinkage induced by temperature or structure changes An initial strain and

initial stress are similarly dealt with in the code

Here considering the equilibrium condition (equilibrium equation) at (n+1) th time step

ie at time tn+1

[ ] ++ = dVBF nT

n 11 σ (318)

holds where

Fn+1 external force vector at the n+1 step and

σ n+1 stress vector at the n+1 step

To clearly distinguish an unknown quantity from a known quantity by separating σ n+1

into known stress σ n and unknown stress increments Δσ n+1 Eq (318) is changed to

[ ] [ ] F B dV B dVn

T

n

T

n+ += + 1 1σ σΔ (319)

[ ] [ ] B dV F B dVT

n n

T

nΔσ σ+ += minus 1 1 (3110)

In Eq(319) the right-hand side is a known value and the left-hand side is an unknown

quantity

Here the relationship between stress and strain (which correspond to the above ldquoC


- 183 -

group four constitutive equations of stress-strain relation omitting another four equations for

shear componentsrdquo) is given by

[ ] 1e

n nDσ ε+ +Δ = Δ (3111)

Even if there are other strain components this relationship always holds Besides as a

result of calculation covering all strain components eg if plastic strain increment plasticεΔ

exists σΔ decreases and elastic strain eεΔ decreases also Plastic strain and creep strain are

quantities determined by σΔ andσ and they have no direct dependences on eεΔ

In Eq(3111) [ ]D is called a strain-stress matrix (stiffness matrix) and shows the

stiffness of an element This consists of Youngrsquos modulus and Poisonrsquos ratio indicating a

stiffness of element ie mechanical resistance against deformation

Here the following equation holds

Δσ n+1 stress increment vector from tn to tn+1

Δε ne

+1 elastic strain increment vector from tn to tn+1

[ ] ( )[ ] [ ]D D Dn n n+ += minus + =

θ θ θ θ1

1

21 and

[ ]Dn Stress-strain (stiffness) matrix at tn

Defining [ ]C matrix as an inverse matrix of [ D ] makes Eq(3111) become

[ ] Δ Δε σθne

n nC+ + +=1 1 (3112)

By substituting Eq(3111) into Eq(3110)

[ ] [ ] [ ] B D dV F B dVT

ne

n

T

nΔε σ+ += minus 1 1 (3113)

Since the following relationship holds among the elastic strain increment Δεne+1 total

strain increment Δεn+1 and initial strain increment as known quantity 01+Δ nε

10

11 +++ ΔminusΔ=Δ nnen εεε (3114)


[ ] [ ] [ ] minus=ΔminusΔ +++ dVBFdVDB nT

nnnT σεε 1

011 (3115)

Then transferring the known term [ ] [ ] +Δ dVDB nT 0

1ε to the right-hand side gives

[ ] [ ] [ ] [ ] [ ] minusΔ+=Δ +++ dVBdVDBFdVDB nT

nT

nnT σεε 0

111 (3116)

By substituting the strain-stress incremental relationship [ ] B uεΔ = sdot Δ into Eq(3116)


- 184 -

[ ] [ ][ ] [ ] [ ] [ ]

1

01 1

T

n

T T

n n n

B D B dV u

F B D dV B dVε σ

+

+ +

Δ

= + Δ minus

(3117)

Hence the unknown quantity Δun+1 can be bundled out from the integral Eq(3117)

is the stiffness equation ie an equation presenting the relationship between displacement and

external force in the incremental method Also this equation can be expressed as

[ ] [ ] [ ] [ ] [ ]

T

T TH rel swl den crk c p HOT

B D B dV u F

B D dVε ε ε ε ε ε ε ε

Δ = Δ

+ Δ + Δ + Δ + Δ + Δ + Δ + Δ + Δ

(3118)

Finally by discretizing Eq(3117) into a sum of M elements

[ ] [ ][ ]

[ ] [ ] [ ]

11

01 1

1 1

MT

i i i i ni

M MT T i

n i i n i i n ii i

B D B V u

F B D V B Vε σ

+=

+ += =

Δ Δ

= + Δ Δ minus Δ

(3119)

( ΔVi volume of elementｉ M total number of elements)

Equation (3112) is a simultaneous linear equation By representing Eq(3112) in the

form of a matrix the unknown quantity of the displacement incremental vector Δun+1 can

be obtained

The above method is applicable when a governing equation can be linearly approximated

in incremental intervals However as explained in the case of creep the assumption (linear

approximation) of ( ) ( )nt

f

t

f σσpartpart=

partpart

is only applicable when stress variations are small

enough ie only when the stress increment σΔ is sufficiently small Therefore in creep

calculation restriction on the size of time increment becomes strong To resolve this problem

use of a more general constitutive equation containing nonlinear strain in FEMAXI is

explained in the next section and thereafter

312 Initial stress method initial strain method and changed stiffness method for non-linear strain

(1) Equation for stiffness As discussed in the preceding section 311-(1) non-linear problems can be converted

into linear elastic problems of small strain which may be formulated using the displacement

method by assuming linearity between stress increments and strain increments in intervals An

equation for stiffness which is expressed by Eq(3119) is obtained

[ ] 0=ΔminusΔ FuK (3120)

In linear elastic problems Eq(3120) is solved and usually the final form of the


- 185 -


In formulating the above equation it is assumed that not only the linear equation of

strain-displacement continuity of displacement and balance of force but also the linear elastic

equation is satisfied [ ] ( ) 00 σεεσ Δ+ΔminusΔ=Δ D (3121)

Here 0εΔ initial strain increment

0σΔ initial stress increment

On the other hand in stress-strain equations which pose problems due to nonlinearity in

general assuming that the function ( )xF is nonlinear

( )σε F=Δ (3122)

Therefore by adjusting one or more of the parameters of [ ] 0 εΔD and 0σΔ

contained in Eq(3121) if Eqs(3121) and (3122) provide the same stress and strain as

solutions of Eq(3120) a nonlinear solution can be obtained To realize this iterative

calculation is required

(2) Iterative calculation In iterative calculation which of the three quantities discussed in (1) above should be

adjusted is determined by the following two conditions

A) solution used in equivalent linear elastic problems (method of solving nonlinear

problems by replacing them with equivalent linear elastic problems)

B) characteristics of physical principles defining the stress-strain relationship

When iterative calculation is performed after adjustment of the [ ]D matrix the process

is called the stiffness variation method When either 0εΔ or 0σΔ is adjusted the method

is called either the initial strain method or the initial stress method

In the stiffness variation method with respect to the specific behavior of materials in

terms of the relationship between stress and strain (for example creep behavior and plastic

behavior) the stiffness matrix [ ]D rewritten in the form of Eq(3112) is given as a function

of stress or strain level attained In other words considering the stiffness variation

[ ] [ ] [ ]εσ DDD == (3123)

the process can be applied In this case change in stress or strain alters the stiffness in the

stiffness matrix Therefore in the iteration to obtain convergence of the changes in stress and


- 186 -

strain solution of stress-strain of deformation is affected by the stiffness matrix thus it is

necessary to iterate until changes in the stiffness matrix converge

In other words the stiffness calculation method is a process which can be explained as

follows It is the method to obtain a deformation solution by adjusting (renewing) the material

constants which satisfy the nonlinear constitutive equation with a new material constant in the

final step after the adjustment in the calculation if it is possible to obtain a solution for a

nonlinear problem iteratively In the stiffness variation method in FEMAXI a method of

adjusting the stress-related material constants by iterative calculation is used

On the other hand in the initial strain method elastic strain increments obtained at each

step in the calculation are compared with strain increments corresponding to the constitutive

equation Eq(3119) the difference is used to calculate residual non-equilibrium force For

example in creep calculation by expressing the increment of creep strain and the increment of

elastic strain in separate forms the correction term for the initial strain increment in each step

can be directly obtained

The difference between the initial stress and initial strain methods is explained using Fig

311 In this figure the relationship between stress and strain obtained by the initial

approximation solution is shown by a Point 1 The solid line in the figure represents the real

behavior of the object described by an empirical equation and material equation In the initial

stress method the initial stress 01σΔ shown in the figure is introduced and the stress is

corrected to an appropriate level This is advantageous when strain sharply increases with

stress because the relationship of stress and strain is corrected using the small stress changes

01σΔ

εΔ

σ

εε

ε

σ

σ

)(a

)(b

1

Fig311 Initail-Strain method and Initial-Stress method (a)Hardening rate decreases with strain (b) Fastening material increases hardening rate with strain


- 187 -

However in the initial strain method strain is adjusted by the correction represented

by 01εΔ This method is advantageous for fastening materials in which stress sharply increases

with strain For the creep and plasticity of fuel pellets and claddings discussed in FEMAXI

the initial stress method proves better by comparing the initial strain method with the initial

stress method because strain changes sharply with increasing stress

However when the initial strain and initial stress methods are compared with the

stiffness variation method the latter method proves better in reducing the number of iterations

until convergence because the stress-strain relationship can be incorporated in the preparation

of the stiffness matrix In other words in the initial strain and initial stress methods the

convergence is calculated by renewing the load vectors using calculated stress and strain

On the other hand the stiffness variation method predicts changes in stiffness prior to

calculation thus it is equivalent in that the two processes carried out in the initial strain and

initial stress methods are completed in one calculation

Because of this basic equations are formulated using the stiffness variation method in

which the stiffness matrix [ ]D is given as a function of stress level attained (ie

[ ] [ ]σDD = ) in FEMAXI An outline of formulation in FEMAXI is described in the next

section

313 Solution of the basic equations in FEMAXI (non-linear strain) As described in the previous section (2) the stiffness variation method utilizing the stress

for convergence is adopted in the treatment of nonlinear strain in FEMAXI This method is

based on the formulation by the incremental method described in section (1) For example the

equation for creep is given by a relational equation between stress and strain increments as

shown in Eq (3114) which fits for the incremental method

(1) Classification and treatment of strain components Next strain used in FEMAXI code is examined The strain discussed in FEMAXI

includes the strain defined by an stress-strain relationship equation and the strain

independent of stress When the strain is given by the stress-strain equation if strain

changes stress changes and vice versa On the other hand strain independent of stress is the

component in which stress changes by the development of strain however even if stress

changes strain does not

First the strain discussed in FEMAXI is explained


- 188 -

cn

Pn

HPn

swn

denn

crkn

reln

thn

enn

111

1111111

+++

+++++++

Δ+Δ+Δ+

Δ+Δ+Δ+Δ+Δ+Δ=Δ

εεεεεεεεεε

(3124)

where Δε n+1 total strain incremental vector

Δε ne+1 elastic strain incremental vector

Δε nth+1 thermal strain incremental vector

Δε nrel+1 pellet relocation strain incremental vector

Δεncrk+1 pellet crack strain incremental vector

Δε nden

+1 pellet densification strain incremental vector

Δε nsw

+1 pellet swelling strain incremental vector

Δε nHP+1 pellet hot-press strain incremental vector

Δε nP+1 plastic strain incremental vector and

Δε nc+1 creep strain incremental vector

Of these the strain increments given by the stress-strain equation are en 1+Δε Δεn

crk+1

Δε nHP+1 Δε n

P+1 and Δε n

c+1 the strain-stress equation ( )σε f= is given by an

independent relational equation (material equation empirical equation and models) Of these the elastic strain increment is linear because ( ) σασε Δ==Δ fe The strain

incremental vectors which are independent of stress are denn

reln

thn 111 +++ ΔΔΔ εεε and sw

n 1+Δε

On the basis of this background let us write the relational equation between the

increment of elastic strain and the stress increment at time interval ( )Δt t tn n n+ += minus1 1

considering that ［C］ depends on temperature (depends on time)

[ ] e Cε σΔ = Δ (3125)

[ ] Δ Δε σθne

n nC+ + +=1 1 (3126)

where [ ] ( )[ ] [ ]

=+minus= ++ 2

11 1 θθθθ nnn CCC


[ ]Cn+θ compliance (stress-strain) matrix

Δσ n+1 stress incremental vector

θ parameter for implicit solution

In addition to obtain stress increments from Eq(3126) it is necessary to obtain elastic

strain increments The elastic strain increment can be written as


- 189 -

1 1 1 1 1 1 1

1 1 1

e th rel crk den swn n n n n n n

HP P cn n n

ε ε ε ε ε ε ε

ε ε ε+ + + + + + +

+ + +

Δ = Δ minus Δ minus Δ minus Δ minus Δ minus Δ

minus Δ minus Δ minus Δ (3127)

The strain incremental vector which is independent of stress can be summarized as

follows considering it as the initial strain incremental vector

swn

denn

reln

thnn 1111

01 +++++ Δ+Δ+Δ+Δ=Δ εεεεε (3128)

Here 01+Δ nε initial strain incremental vector

Furthermore as will be discussed in section 317 in detail because the strain

incremental vector from the hot-press of a pellet can be incorporated into the plastic and creep

strain incremental vectors of pellet Eq(3127) can be written as follows

cHn

PHn

crknnn

en 111

0111 ++++++ ΔminusΔminusΔminusΔminusΔ=Δ εεεεεε (3129)

where

PHn 1+Δε plastic strain incremental vector including hot-press strain

cHn 1+Δε creep strain incremental vector including hot-press strain

It should be noted here that all PHn

crkn 11 ++ ΔΔ εε and cH

n 1+Δε are given by ( )σε f=Δ

and the function f is nonlinear

On the other hand the total strain can be related to nodal displacement as follows

[ ] 11 ++ Δ=Δ nn uBε (3130)

where Δun+1 is an incremental vector for nodal displacement

By substituting Eqs(3126) and (3130) into Eq(3129)

[ ] [ ] 01110

111 =Δ+Δ+Δ+Δ+ΔminusΔ +++++++cHn

PHn

crknnnnn uBC εεεεσθ (3131)

In Eq(3131) the treatment of PHn

crkn 11 ++ εΔεΔ and cH

n 1+εΔ which are the nonlinear strain

increments is explained below

Let us assume that the following equations are given to the nonlinear strain increment

1 1 1 2 1 3 crk PH cHn n nf f fε σ ε σ ε σ+ + +Δ = Δ = Δ = (3132)

Here σσ 21 ff and σ3f are known functions obtained from material parameters

empirical equations or tensile tests the relationship between stress and strain is known


- 190 -

(2) Determining stress by iteration method Here a method of determining σ by solving Eq(3132) is considered In this method

without obtaining an accurate value of stress σ the strain increment of PHn

crkn 11 ++ εΔεΔ and

cHn 1+εΔ cannot be calculated

For this reason the stiffness variation method is applied to obtain numerical solutions

The Newton-Raphson method (NR method) is applied to the iterative calculation to solve this

nonlinear equation

Here the general concept of the NR method is explained It is applied to solve the

following general nonlinear equation having a single variable x

ψ( x)=0 (3133)

In the NR method assuming that ix which is sufficiently close to the exact solution is

obtained while 0)( nexψ still holds the improved test solution 1+ix is obtained in the

following manner

1+ix = ix + Δ 1+ix

Here

i

ii

dx

dxx

minus=Δ + ψψ )(

1

Figure 312 illustrates an iterative calculation using a changing gradient as described above

In the case of applying of the NR method to Eq (3130) x=σ and each )(xψ is

0)()( 111 =Δminus= +crknf εσσψ (3134)

0)()( 122 =Δminus= +PHnf εσσψ (3135)

)(xψ

idx

d

ψ

ix1+ix

)( ixψ

１

Fig 312 Iteration method of the Newton-Raphson method using a changing gradient


- 191 -

0)()( 133 =Δminus= +cHnf εσσψ (3136)

also

)( 11 σσψ fdd = (3137)

)( 22 σσψ fdd = (3138)

)( 33 σσψ fdd = (3139)

As is apparent the three equations from Eqs (3134) - (3136) include the unknown

values 1 1 1 crk PH cHn n nε ε ε+ + +Δ Δ Δ other than stress and the number of unknowns exceeds that

of the equations hence it is impossible to solve these equations simultaneously This implies

that the conditions of 0)( =σψ cannot be obtained The reason is that Eq(3131) is not in a

final form which allows the application of the NR method Therefore it is necessary to

proceed with the formulation

The following discussion is given on the basis of the application of the NR method to

iterative calculations for nonlinear analyses Here i represents the count of iterative

calculations in the NR method When the ith iterative calculation in the NR method is

completed (the variable value in the ith iteration is known ) and the i+1th iteration is being

performed strain displacement and stress other than 01+Δ nε and nσ change in response

to the iterative calculation Therefore Eq(3131) which is the equation of the NR method is

represented using superscripts to indicate the number of iterations i and i+1 as

[ ] ( )

1 11 1

0 1 1 11 1 1 1 1 0

i i in n n

crk i PH i cH i i in n n n n n n

C d B u

C

θ

θ

σ

ε ε ε ε σ σ

+ ++ + +

+ + ++ + + + + +

minus Δ

+ Δ + Δ + Δ + Δ + minus =

(3140)

Here d ni

ni

niσ σ σ+

+++

+= minus11

11

1

In addition the equilibrium condition Eq (3110) is rewritten as

[ ] [ ] V

T

ni

V

T

ni

nB dV B d dV F + ++

++ =σ σ1 11

1 (3141)

In Eqs(3140) and (3141) because the unknown values are the values at the i+1th

iteration they are d niσ ++

11

Δεncrk i+

+1

1 1

1+

+Δ iPHnε 1

1+

+Δ icHnε and Δun

i++

11

Here the

conditional equations to obtain 0)( =σψ in the NR method Eqs(3134) (3135) and

(3136) are given below

0)()( 1111 =Δminus= +

+icrk

nf εσσψ (3142)

0)()( 1122 =Δminus= +

+iPH

nf εσσψ (3143)

0)()( 1133 =Δminus= +

+icH

nf εσσψ (3144)


- 192 -

By this approach because five equations Eqs(3140)-(3144) are obtained the

unknown values 11

++

indσ Δε n

crk i+

+1

1 1

1+

+Δ iPHnε 1

1+

+Δ icHnε and Δun

i++

11

can be determined To

complete the convergence process by the NR method the addition of an equilibrium

conditional equation (balanced equation between external force and internal stress) described

by Eq(3141) is necessary

Incidentally with respect to the method of correcting the elastic matrix by the stiffness

variation method in FEMAXI suitable formulation algorithms for each characteristic of the

material behavior model are explained after section 32 Fig 313 shows the algorithm for

the treatment of the entire calculation of deformation Alteration of the contact boundary

conditions and conditions required for evaluation of yield and unloading will be explained

again in a later section The evaluation of yield and unloading is performed after completion

of the convergence calculation by the NR method in the algorithm presented in Fig 313

The number of iterative calculations in the

Newton-Raphson method is given by LMAX (The standard value is 20)

However after three or more iterations when the maximum amount of total correction

(the difference in magnitude between the preceding and current results) is 001 or less

iterations are terminated by assuming that the convergence is complete

Model parameter LMAX


- 193 -

Preparation of nonlinear strain increments

iPHn

1ˆ +Δε and icH

n

1ˆ +Δε

Preparation of elastoplastic creep coefficient matrix [ ]pD

The intercept and contact planes are obtained in the Newton-Raphson method on the basis of the quantity of the state (stress strain) at stage i

For each element Preparation of residualcorrection load inF 1+Δ

from stress and strain

Preparation of stiffness matrix [ ]θ+nK from coefficient matrix

Preparation of entire stiffness matrix and load by summing the contribution from each element The residual load is added to the total load

Solving the stiffness equation

[ ] in

in

in FuK 1

11 +

+++ Δ=Δθ

To confirm whether contradictions developed in the contact boundary conditions (anchoringslipping) based on displacement and load obtained (if there is a contradiction the contradiction should be corrected and recalculations carried out)

The corrections for strain and stress increments 11

++

indε and 1

1++

indσ are

obtained from the solution 11

++Δ i

nu and the estimated values 11

++

inε and

11

++

inσ at the i＋1th iteration are obtained

The parameters of the nonlinear process (hardening rule of creeping) are renewed

Evaluation of convergence by the Newton-Raphson method

Evaluation of contact boundary conditions (contact or noncontact) based on calculated displacement and load and evaluation of yield and unloading based on stress and plastic strain increments obtained

Fig313 Algorithm and flow of mechanical analysis in FEMAXI


- 194 -

32 Mechanical Analysis of Entire Rod Length

First an overview of the total flow is shown in Fig321 before each step is described

Fig321 Flow chart of the Global FEM mechanical analysis

Time Step Start

Contact state judgment at contact node settled

Displacement converged

Break the present time step at the instant of state change Stress and strain are interpolated into the values at the instant

Yielding unloading or contact-non-contact state change occurred

Calc of stress-strain distribution

Proceed to next Time Step

Set pellet and cladding contact state and boundary conditions at nodal points

Calculation of matrix for all the elements completed

Calc of nodal displacement by solving the total stiffness equation

Set initial strains at each element such as densification swelling and thermal expansion

Calc of material matrix of each element

Incorporation of stiffness matrix and nodal vector in each element


- 195 -

(Note) The 1-D analysis over the entire rod length (ERL) is performed in the

routine FEMROD while the analysis of 2-D local PCMI deformation of rod is

performed in other routines The 2-D analysis can be performed concurrently

with the 1-D analysis by using the input parameter IFEMRD

IFEMRD=0 1-D ERL and 2-D local analyses and IFEMRD=11-D ERL only

321 Finite element model for entire rod length analysis (ERL)

(1) Geometry

As shown in Fig322 the analysis model includes a 1-dimensional axis-symmetrical

system in which the entire length of the fuel rod is divided into axial segments and each

segment is further divided into concentric ring elements in the radial direction In this

system the stress-strain analysis is performed using the finite element method with

quadrangular elements with four degrees of freedom as shown in Fig324

Fig322 Geometry of FEMAXI model

U P PER PL ENUM

S E GMENT( M)

S E GMENT( M- 1 )

S E GMENT( 2 )

S E GMENT( 1 )

L O WER PL ENUM

Pellet

Cladding

Clad Pellet


- 196 -

(2) Ring elements Fig323 shows the ring element formation of one axial segment in ldquoMESH=3rdquomode

This is the same as Fig231

To help understand this geometry a simplified FEM element structure is shown in

Fig324 in which the number of ring element of pellet stack is 10 and the number of ring

elements in cladding is 4 In this figure dummy space elements for dish and chamfer are

shown which is described in section 322 The inner region three elements 13 14 and 15 of

cladding are metallic part and the outer one element 16 is oxide layer (ZrO2) The outer

surface oxide element is re-meshed to increase in accordance with the growth of oxide layer

(refer to chapter 2 section 22)

Also nodal coordinates in one ring element are shown in Fig325

Uz2

Lz9Lz3

Lz2Lz1

PELLET(10ring-elements) CLADDING

r12

Uz3Uz1

Uz10

r1 r2 r3 r10 r11

Lz10

r13 r14 r15 r16

Lz11Lz12

Lz13Lz14

Uz11Uz12

Uz13 Uz14Uz9

Mz1Mz2

Mz3

Mz9Mz10

Fig324 Ring element of FEM for one axial segment

00 02 04 06 08 10

ZrO2

1 2 3 4 5 6 7 8

Cladding

ring element number



Bonding layerCenterline

Fig323 Ring elements geometry for one axial segment


- 197 -

(3) Degree of freedom of elements

As shown in Fig 324 and Fig325 each node has one degree of freedom in either the

radial or axial direction On the assumption that the axial displacements of all nodes are

identical which is the generalized plane strain condition a degree of freedom of one element

is used as a representative value for degrees of freedom in the axial direction of pellet stack

1 10L Lz z - 1 10

U Uz z- and in the axial direction of cladding 11 14L Lz z- and 11 14

U Uz z - as shown

below

( )( )

( )( )U

CUUUU

LC

LLLL

Up

UUU

Lp

LLL

zzzzz

zzzzz

zzzz

zzzz

====

====

====

====

14131211

14131211

1021

1021

(321)

That is it is assumed that the bottom plane at each segment keeps the same elevation for

both pellet stack and cladding and that the top plane of pellet stack and that of cladding have

a freedom to displace independently from each other in the axial direction with the two top

planes being kept flat

This is required by the characteristics of FEM where ri and ri+1 have a freedom in the

radial direction solely and ZiL and ZiU have a freedom in the axial direction solely It should

be noted that this applies only to the mechanical analysis over the entire length of a fuel rod

Generally in FEM displacement and strain are continuous at nodes and boundaries of each

element while stress is not continuous However derivatives of displacement and strain are

not always continuous

(4) Number of digitized elements of pellet stack in the radial direction

Although this is already explained in Table 231 in chapter 2 it is presented here again

for its importance

ri ri+1

ZiU

ZiL

Fig325 Quadrangle four nodes element with

degree of freedom (ri ri+1 ZiL ZiU)


- 198 -

As Table 231 has shown the name-list parameter MESH can specify the number of ring

elements in the pellet stack Default value is MESH=3 In the iso-volumetric elements

thickness of ring decreases insomuch as the outer region of pellet stack When ISHAPE=1

the number of elements doubles in the case of MESHgt0 which allows a finer calculation

(5) Number of ring elements of cladding

The 1-D thermal analysis shares the ring elements of cladding with the 1-D ERL

mechanical analysis irrespective of the designated value of ldquoMESHrdquo

That is in the 1-D thermal and mechanical analyses

A) Cladding without Zr liner 8 metallic elements + 2 outer surface oxide elements

B) Cladding with Zr liner 2 Zr-liner elements + 8 metallic elements + 2 outer surface

oxide elements

Also in the 2-D local PCMI mechanical analysis which is later described

C) Cladding without Zr liner 4 metallic elements + 1 outer surface oxide elements

D) Cladding with Zr liner

1 Zr-liner element+ 4 metallic elements + 1 outer surface oxide elements

(6) Cladding oxide layer The volume expansion of the oxide during each time step is added as an axial strain

increment to the oxide layer element The fraction of volume increment X is

X=PBR ndash 10 (322)







MESH=1 Iso-volumetric 36 ring elements Iso-volumetric

18 ring elementsIso-volumetric 9 ring elements






- 199 -

and X is assumed to be distributed in the radial direction by PX () and in the axial direction

by 100-PX () respectively PX is given as an input parameter and PBR is the Piling

Bedworth ratio

Accordingly the metal phase thickness decreases and is re-meshed with the increase in

thickness of the oxide layer Mechanical characteristics such as Youngrsquos modulus thermal

expansion coefficient creep rate and irradiation growth rate naturally differ between the metal

phase and oxide layer generating an internal stress in the metallic substrate of cladding

mainly in the axial direction affecting the mechanical behavior of the entire cladding This

stress is included in the mechanical analysis However since no measurement data on the

creep rate and irradiation growth rate of ZrO2 has been obtained creep in the oxide layer is

neglected and the irradiation growth rate is tentatively set as that of Zircaloy multiplied by a

factor RX (input parameter default value is 10) in the calculation

322 Determination of finite element matrix Here various matrices of the finite element method will be determined

(1) Derivation of strain-displacement matrix ［B］

Denoting the radial displacement as u and axial displacement as ν and setting ir 1ir+ Miz

Uiz and

Liz as respectively pre-determined values of radial coordinate and axial

coordinate of the nodes of ring element ie fixed values given by input displacement (u v) at

an arbitrary point inside an element shown in Fig324 can be written as

111

1+

++

+

minusminus

+minusminus

= iii

ii

ii

i urr

rru

rr

rru (323)

( )L

M Lii iM L

i i

z z

z zν ν νminus= minus

minus

or ( )M

U Mii iU M

i i

z z


minus (324)

In axis-symmetric geometry the strain component inside the element is

[Input Parameter PBR and PX] Piling-Bedworth ratio is given by PBR Default value is 156 Default value of PX is 800


- 200 -

[ ] r

z

ur

u B ur

z

θ

partpartε

ε εε partν

part

= = equiv

(325)

Using Eq(323) and Eq(324)

( ) ( ) 111

1

111

11

+++

+

+++

minusminus

+minusminus

=

minus+

minusminus=

iii

ii

ii

i

iii

iii

urrr

rru

rrr

rr

r

u

urr

urrr

u

partpart

(326)

Here by putting 1

2i ir r

r ++= as central-point coordinate of the element we obtain

( ) ( )

1 11

11 1

1 1

11 1

2 2

2 21 1

i i i ii i

i ii i i i

i i i i

i ii i i i

r r r rr ru

u ur r r rr r r r r

u ur r r r

+ ++

++ +

+ +

++ +

+ +minus minus= ++ +minus minus

= ++ +

(327)

M Li iM Li iz z z

partν ν νpart

minus=minus

or U Mi iU Mi iz z z

partν ν νpart

minus=minus

(328)

By substituting Eqs(326) (327) and (328) into Eq(325) we obtain

1 1

1

1 1

1 10 0

1 10 0

1 10 0

ii i i i

r i

i

ii i i i L

i

z iM

M L M L ii i i i

ur r r r

u

r r r r

z z z z

θ

ε

εν

εν

+ +

+

+ +

minus minus minus = + +

minus minus minus

Or

1 1

1

1 1

1 10 0

1 10 0

1 10 0

ii i i i

r i

i

ii i i i M

i

z iU

U M U M ii i i i

ur r r r

u

r r r r

z z z z

θ

ε

εν

εν

+ +

+

+ +


minus minus minus

(329)

Consequently we obtain the following as components of the matrix［B］(shape function)


- 201 -

[ ]

minusminusminus

++

minusminusminus

=++

++

Li

Mi

Li

Mi

iiii

iiii

zzzz

rrrr

rrrr

B

1100

0011

0011

11

11

or [ ]

minusminusminus

++

minusminusminus

=++

++

Mi

Ui

Mi

Ui

iiii

iiii

zzzz

rrrr

rrrr

B

1100

0011

0011

11

11

(3210)

Eqs(329) and (3210) are used in all the FEM solutions of the ERL (entire rod length)

mechanical analysis for both the linear and non-linear deformations

The volume of each element is

( )( )Li

Miiii zzrrV minusminus= +

221π

or ( )( )Mi

Uiiii zzrrV minusminus= +

221π (3211)

323 Derivation of stress-strain (stiffness) matrix of pellet and cladding

We obtain strain-stress matrix [C] The stress-strain relationship is

[ ] Δ Δσ ε= D e (3212)

where [D] is the stress-strain matrix that represents the stiffness of an element

When matrix [C] is defined as an inverse matrix of [D] Eq(3212) is expressed as

[ ] Δ Δε σe C= (3213)

Accordingly matrix [C] is given by

[ ]

minusminusminusminusminusminus

=1

1

11

νννννν

EC (3214)


- 202 -

where E is Youngs modulus and ν is the Poisson ratio

Here solution of the basic constitutive equations for stress-strain analysis is outlined

The external force and initial strain vector are set as F and 0εΔ respectively When

the calculation has proceeded to the n-th time step and calculation for the next (n+1)-th time

step is performed the equilibrium equation

[ ] F B dVn

T

n+ += 1 1σ (3215)

is given using the principle of virtual work

Here

Fn+1 external force vector at (n+1)th time step

σ n+1 stress vector at (n+1)th time step

To distinguish unknown variables from the known we divide σ n+1 into known stress

σ n and unknown stress Δσ n+1 then we obtain transformed equations of Eq(3215) as


T

n

T

n+ += + 1 1σ σΔ (3216)


n n

T

nΔσ σ+ += minus 1 1 (3217)

The right-hand side of Eq(3217) consists of known values and the left side consists of

unknown values Substituting Eq (3212) into Eq (3217) we obtain

[ ] [ ] [ ] B D dV F B dVT

ne

n

T

nΔε σ+ += minus 1 1 (3218)

Here the following equation holds among elastic strain increment Δε ne+1 total strain

increment Δε n+1 and the known value initial strain increment 01+Δ nε

0111 +++ ΔminusΔ=Δ nn

en εεε (3219)

Accordingly substituting Eq (3219) into Eq (3218) we obtain


nnnT σεε 1

011 (3220)

Transposing the known value term [ ] [ ] +Δ dVDB nT 0

1ε to the right-hand side


nT

nnT σεε 0

111 (3221)


- 203 -

is obtained Substituting Eq (325) into Eq (3221) we obtain

[ ] [ ][ ]

[ ] [ ] [ ] 1

01 1

T

n

T T

n n n

B D B dV u

F B D dV B dVε σ

+

+ +

Δ

= + Δ minus

(3222)

and the unknown value Δun+1 can be calculated by integration

Namely expressing Eq(3222) in a discrete form with summation of M elements

[ ] [ ] [ ]

[ ] [ ] [ ]

11

01 1

1 1

MT

i i i i ni

M MT T i

n i i n i i n ii i

B D B V u

F B D V B Vε σ

+=

+ += =

Δ Δ

= + Δ Δ minus Δ

(3223)

( ΔVi represents the volume of the element i and M represents total number of elements)

Thus the unknown displacement increment vector Δun+1 is expressed using known values

The above is the basic calculation method In actual calculations a more complex

solution is required due to the use of implicit solutions for creep and plasticity

In the next section modification of the basic equations for the purpose of implicit

solutions will be described

(1) Details of pellet stack elements

In a cylindrical shape pellet solid or hollow thermal stress is induced by temperature

gradient in the radial direction which generates a strong thermal stress inside pellet and

sometimes a ridging deformation of cladding occurs by pellet hour-glass shape expansion

(refer to chapter 35) In actual fuel rods these thermal deformations are markedly mitigated

by the dish and chamfer spaces However a simple solid cylinder stack model which neglects

the dish and chamfer spaces would generate numerically a far stronger thermal compressive

stress in the central region of pellet stack and contrarily a stronger tensile stress in the outer

region This is caused by the generalized plane strain condition adopted in FEMAXI in which

pellet upper plane keeps a flat plane in displacing in the axial direction (Refer to section

321 (2))

To avoid these virtual thermal stresses which would be induced by the model

characteristics and at the same time to keep the generalized plane strain condition dish or

chamfer space can be specified by input in the upper part of the pellet stack by designating a

parameter IDSELM=1 Namely a one pellet stack of segment can consist of three elements


- 204 -

This is illustrated in

Figs326 and 327 Fig326

shows a pellet stack geometry

of one segment The number of

ring elements is set 10 for

simplicity of explanation

Here while actually a

pellet stack of ldquoone segmentrdquo

length consists of several or a

few tens of pellets in the

present model the yellow

region represents the pellet

solid part and the other region

corresponds to the dish

chamfer and inter-pellet

aperture spaces of all the pellets

that consist of this segment

In other words Fig326 is

a geometry which takes into

account of the effects of dish and chamfer spaces on the

solid cylinder model These ldquovoid space elementsrdquo are given

a very weak stiffness which is shown in the figure to act as

a buffer for the thermal expansion or swelling of pellet

Fig327 is the 7th element which has neither dish nor

chamfer out of the elements in Fig326 Here the upper

non-hatched part is the inter-pellet (IP) space and lower

hatched part is the pellet solid element The division ratio of

this element is that the upper part is BUFSP() of the

segment length and the lower part (100 - BUFSP) of the

segment length and BUFSP is given by input Namely in

this element the segment is divided at BUFSP (100 - BUFSP) ratio

The IP space element is presented in Fig326 as a uniform thickness zl -spaced element

over the dish and chamfer spaces and zl is initially given by input as BUFSP () of the

Fig327 Pellet ring element

without dish and chamfer

MiZ

ir

UiZ

LiZ

1ir +

centerline

Outer surface

Fig326 FEM element division corresponding to the pellet stack of one axial segment

M3Ζ

M7Ζ

Chamfer space

Dish space

U1Ζ U

2Ζ U3Ζ

U4Ζ

U5Ζ U

6Ζ U7Ζ U

8Ζ U9Ζ U

10Ζ

M10Ζ

M1Ζ

M4Ζ

M5Ζ

M2Ζ

L1Ζ L

2Ζ L3Ζ L

4Ζ L5Ζ L

6Ζ L7Ζ L

8Ζ L9Ζ L

10Ζ

M6Ζ

M8Ζ

M9Ζ

Pellet-pellet gap space which is assumed to have a fixed thickness z lz ie BUFSP () of stack length of one axial segment


- 205 -

segment length That is the pellet shape matrix in the ERL mechanical analysis is composed

with reference to the initial coordinate value of the segment baseline and the IP thickness is

fixed at zl in the axial direction

Here for the pellet stack of one segment space element depth of dish (or chamfer) in the

axial direction is specified in accordance with the ratio of depths of dish and chamfer In

Fig326 depth of dish element is calculated by [ one pellet dish depthtimes2timesnumber of pellets

in one segment] which determines Uiz

Miz and

Liz as initial conditions

The axial length of the chamfer space is similarly determined The concave region and

gap shown in Fig326 indicate the spaces which take into account of this shape feature

They are ldquobuffer spaces with weak spring stiffnessrdquo which allows to calculate a realistic

thermal stress and strain

It should be noted that the axial displacement of the upper coordinate of pellet solid part Miz is calculated in each ring element while the axial shear stress exerting between the

adjacent elements is not taken into account in the present 1-D mechanical model Also the

displacement of the top plane of segment follows the generalized plane strain condition which

is defined by the uniform displacement as U Uiz vΔ = and ( )1 2

U U U Ui pz z z z= = = = =

(Refer to Eq(321))

That is in the calculation of the total matrix the displacement of each solid ring element

of pellet in the axial direction is determined as ( )M Li iv vminus where L

iv is equal to Uv of the

lower adjacent segment Accordingly it is the same in all the ring elements of a segment

(Refer to section (1)) Further

U Uiz vΔ = is determined by the force which is obtained by summing up the

spring forces of each ring element calculated by the concave space depth and axial gap

stiffness (or spring constant) of all ring elements for the values of ( )M Li iv vminus of each ring

element

Also Uv includes implicitly the effect of axial forces which are induced by the PCMI

apart from the above spring forces That is Uv is determined by incorporating all these

interactions among the forces into the matrix

Default value of BUFSP is 10() It is assumed as an uniform thickness of space elementwith BUFSP of the stack length ofone axial segment

Adjustment parameter

Default value of IDSELM is 0 Model parameter IDSELM (used only in mechanical


- 206 -

(2) Assumption of mechanical properties of elements consisting of pellet stack The calculation method by this element geometry is specifically described In Fig326

when the pellet solid element is expanding by thermal expansion or swelling this element can

increase its volume almost freely in the axial direction until the dish space (concave void) is

fully filled and during this expansion the deformation calculation can be performed with the

original mechanical properties of pellet in both the tensile and compressive stress state

Next when the dish space shown in Fig326 is filled the subsequent expansion in the

axial direction fills the inter-pellet gap space elements In this situation stiffness of the gap

space elements is recovering to the original property of pellet so that the solid elements are no

longer able to expand or shrink freely in the axial direction For the chamfer elements an

identical numerical process is conducted

However there is some situation where the spaces of dish and chamfer elements can be

recovered when power is falling and pellet is shrinking During this process stiffness of the

spring is decreasing reversely in accordance with the depth recovery of the inter-pellet gap

space elements

(3) Change of the stiffness (spring constant) in the axial direction The above stiffness change of the dish (chamfer) and inter-pellet gap spaces is modeled

by a single spring in the axial direction for the [dish + gap space] [chamfer + gap space] and

[gap space] However the stiffness recovery of pellet stack element by crack healing is

modeled in another way which will be described in section 324 Here the spring stiffness

in the axial direction is treated by the spring constant change as follows

When the gap space is assumed to be completely replaced with the solid pellet the spring

constant zk of the gap space element is given as

z

zz l

SEk = (3224)

where zE is Youngrsquos modulus of pellet S is initial cross sectional area normal to the center

axis of pellet of the dish or chamfer elements and if the pellet has neither dish nor chamfer

S is initial cross sectional area normal to the center axis of pellet zl is the depth (thickness)

of the gap space element which is uniform in the radial direction as explained in the previous

section Since value of zl remains constant zk is affected only by temperature dependence

of zE and if zE remains unchanged zk keeps a constant value

Next the spring constant zdk which is shared by the dish element and gap space


- 207 -

element (or the chamfer element and gap space element) in calculation is required to take into

account of the stiffness change or recovery which is caused by the decrease or increase of the

dish (chamfer) space elements due to pellet expansion or shrinkage

Here depth (thickness) value of dish (or chamfer) in the axial direction of a certain ring

element i at a certain time step is set as zd il with the upper surface level of inter-pellet gap

space element being a baseline and downward direction is set as positive

A) The spring constant is set as

5 10zd zk k FAC FAC minus= sdot = (3225)

during the period in which dish (chamfer) space is not filled (ie z zd il llt ) This means that the

spring constant zdk through the dish (chamfer) + gap space elements has a very small value

Here it is noted that the dish could become deeper than the initial state depending on

irradiation condition

B) When expansion still continues ie zd i zl lle after the dish or chamfer space is filled

completely the thickness of gap space is decreasing and the spring constant zsk of the dish

(+chamfer) + gap space is approaching zk in accordance with Eq(3226) as the expansion

amount z zd il l lΔ = minus entering the gap space is coming closer to zl

zs zz

lk k

l

Δ= sdot (3226)

C) For the ring elements of pellet which have no dish (chamfer) as described in Figs326 and

327 the elements are entering directly the gap space Accordingly a similar equation to

Eq(3226) expresses the change of spring constant as follows

( )zzs z zd z

z

lk k k k FAC

l

primeΔ= sdot ge = sdot (3227)

where ( )z zl lprimeΔ le is the amount of expansion in the axial direction In the case of shrinkage

zs zd zk k k FAC= = sdot holds

(4) Derivation of stress-strain matrix (stiffness matrix)［D］

Stress-elastic strain relation of each ring element of pellet is expressed as

[ ] enn D 11 ++ = εσ (3228)

[ ]D is as stated in section 31 called a strain-stress matrix ie stiffness matrix which

indicates the mechanical stiffness of elements Defining [ ]C matrix as an inverse matrix of

this [ ]D matrix gives

[ ] 111 +++ = nnen C σε (3229)


- 208 -

Here [ ]C matrix is presented by Eq(3214)

324 Pellet cracking

(1) Model of pellet cracking and stiffness recovery The concept of crack model in the mechanical analysis is described with the schematic

shown in Fig328(14)(110) When a pellet is in a tensile stress in the i direction (axial

circumferential or radial direction) due to thermal dilatation it is assumed that cracks are

generated in the orientation perpendicular to the i direction and the apparent Youngrsquos modulus

in the i direction decreases to approximately 1100 of the original material property value

(Since the Youngrsquos modulus occupies a diagonal position in the FEM matrix of the

detailed mechanical analysis the value cannot be 0)

When pellet is in a compressive stress due to PCMI it is assumed that the elasticity of

the pellet partly recovers in accordance with the decrease in relocation strain That is in the

initial state ie burnup=0 power=0 andε εi iominus = 0 Youngrsquos modulus Ei of pellet is a small

value represented by Ec

StrongInteraction

W eakInteraction

NoInteraction

Ec

E

ε eminusε rel

Fig 328 Stiffness model of pellet with cracks

However FEMAXI does not deal with the pellet cracking in an explicit way It is

assumed that a pellet stays in a cylindrical continuum and when i direction (axial hoop or

radial direction) is in a tensile stress state crack occurs in the plane normal to the stress and

the Youngrsquos modulus in the i direction becomes very small ie about 1100 of the original


- 209 -

value

Also in a compressive stress state it is assumed that the elastic stiffness recovers

depending on the relocation strain That is in an initial state ie no burnup no power and

ε εi iominus = 0 the Youngrsquos modulus Ei of pellet takes a small value Ec because pellet fragments

can move freely in the radial direction

Change of pellet stiffness with increasing power is modeled as follows

A) When pellet is not in contact with cladding thermal stress in pellet is significantly

relieved by the cracks accordingly the pellet can expand almost at its original expansion rate

and the gap is narrowed

B) When pellet is in contact with cladding the compressive strain inside the pellet increases

by the restraint of cladding Then the crack void space generated by relocation is

compressed and stiffness of the pellet increases This process is modeled by linearly changing

the Youngrsquos modulus Namely it is the process in which a mechanical interaction is gradually

enhanced between a cracked and relocated pellet and cladding

C) Next elastic stiffness of pellet completely recovers to the original value ie compressive

stiffness when fragments of cracked pellet are compressed by cladding and fill the void space

which was originally generated by relocation In this situation a strong mechanical interaction

between pellet and cladding is supposed

Youngrsquos modulus of a pellet with cracks is set by ECRAC3 The default value of ECRAC3 is 2times109 (Nm2) which is approximately 1100 of original Youngrsquos modulus of UO2

Fitting parameter ECRAC3

The pellet stiffness model shown in Fig 326 is for the case

IYNG = 1 (default) When IYNG=1 the region between minusεrel and 0 is described by a straight line When IYNG=0 this region is described by a concave quadratic function

Model parameter IYNG

(2) Relationship between relocation and stress-strain Based on the above-described model the relationship between stress and strain of pellet

concerning the relocation is described

The apparent Youngrsquos modulus in each direction Ei（Er Ez Eθ）is defined as follows


- 210 -

( )

minus=lt

ltltminus+minusminus

=lt

=

reli

ei

ei

reliccrel

i

ei

eic

i

E

EEE

E

E

εε

εεεε

ε

　

　　

　　

0

0

(3230)

where eiε elastic strain in i direction

ε irel initial relocation strain in i direction (input data)

Ec effective Youngrsquos modulus of pellet with cracks in a tensile state

Then sum of the elastic strain eiε in i direction and relocation strain r

iε in i direction is

introduced as

ri

eii εεε += (3231)

In Eq(3231) when 0=eiε rel

iri εε = holds and when rel

iei εε minus= 0=r

iε That is the

relocation strain is ε irel when stress is null ( 0=e

iε ) and when reli

ei εε minus= pellet stiffness

completely recovers and the relocation strain becomes null Accordingly in the elastic strain

range 0ltltminus ei

reli εε the relocation strain changes in the range rel

iireli εεε ltltminus

Fig329 Schematic of relationship between relocation strain and stress in cracked pellet

Elastic region

Elastic strain

Relocation strain

-εrel

εrel

σ

εi

minusσR


- 211 -

In the range relii

reli εεε ltltminus shown in Fig329 since change of the apparent Youngrsquos

modulus is expressed by a linear equation in Eq(3230) and given by εσ dd the

stress-strain relationship derived by integrating the equation of εσ dd is a quadratic

function

Therefore the stress change in relii

reli εεε ltltminus is set as a quadratic function of strain

cba iii ++= εεσ 2

(3232)

Differentiating Eq(3232) gives

bad

dE i

i

ii +== ε

εσ

2 (3233)

Here coefficients a b and c have to be determined Since the boundary condition is that at relii εε = 0=iσ at rel

ii εε = ci EE = and at relii εε minus= EEi = Eq(3232) gives

02 =++ cba rel

ireli εε (3234)

From Eq(3233)

creli Eba =+ε2 (3235)

Eba reli =+minus ε2 (3236)

holds Solving Eqs(3234) (3235) and (3236) gives

reli

ccreli

c EEc

EEb

EEa ε

ε 4

3

2

4

+minus=

+=

minusminus= (3237)

Therefore Eq(3232) becomes

reli

ci

cirel

i

ci

EEEEEE εεεε

σ4

3

242 +

minus+

+minus

minus= (3238)

In Fig329 the Y-axis intercept value of the curve of Eq(3234) is reli

ci

EE εσ4

3+minus=

and it is found that the stress at reli iε ε= minus is rel

ici EE εσ )( +minus= when the Youngrsquos modulus

recovers completely

Characteristics of the present model is to derive the stress-strain relationship including

the pellet stiffness change due to cracking by introducing the equation of stress vs [elastic

strain + relocation strain] not by introducing the equation of stress vs elastic strain

325 Crack expression in matrix As described previously the FEMAXI mechanical model consistently assumes a

cylindrical continuum for pellet and the crack formation is expressed approximately by the

decrease in stiffness of the ring elements of pellet stack That is further direct division of ring

elements of the cracked pellet is not performed


- 212 -

A crack strain increment Δε ncrk+1 which is proportional to the change in Δσ is given

by transforming the elements of [C] matrix which is in Eq(3213) from Eq(3214)

The strain-stress matrix of the pellet [ ]~C is represented as

[ ]~C

E E E

E E E

E E E

r

z

=

minus minus

minus minus

minus minus

1

1

1

ν ν

ν ν

ν νθ

(3239)

Here Er Youngrsquos modulus in the radial direction (Pa)

Ez Youngrsquos modulus in the axial direction (Pa)

Eθ Youngrsquos modulus in the circumferential direction (Pa)

E Youngrsquos modulus of pellet (material property) (Pa)

ν Poissonrsquos ratio (minus)

The apparent Youngrsquos modulus in each direction Ei (Er Ez Eθ) is defined as follows

( )

0

00

0

0

0

c i i

reli ii c c i i irel

i

reli i i

E

E E E E

E

ε ε

ε ε ε ε εε

ε ε ε

lt minus= minus= minus minus + minus lt minus lt minus ltminus =

(3240)

Here ε i strain in i direction

0iε initial strain in i direction



From Eq (3213) [ ] Δ Δε σθn

en nC+ + +=1 1 (3241)

and setting the sum of elastic strain increment and crack strain increment as

[ ] Δ Δ Δε ε σθne

ncrk i

n niC+ +

++ +

++ =1 11

11 ~

(3242)

we obtain the crack strain increment vector using Eqs (3241) and (3242) as


- 213 -

[ ] [ ] Δ Δε σ

ν ν

ν ν

ν ν

ν ν

ν ν

ν ν

θ θ

θ

ncrk i

n n ni

r

z

C C

E E E

E E E

E E E

E E E

E E E

E E E

++

+ + ++= minus

=

minus minus

minus minus

minus minus

minus

minus minus

minus minus

minus minus

11

11

1

1

1

1

1

1

~

=

minus

minus

minus

+

minus

++

++

++

+

+

+

Δ

Δ

Δ

Δ

Δ

Δ

σσσ

σ

σ

σ

θ

θθ

r ni

z ni

ni

rr ni

zz ni

ni

rE E

E E

E E

E E

11

11

11

1

1

1

1 1

1 1

1 1

1 1

minus

minus

++

++

++

d

E Ed

E Ed

r ni

zz ni

ni

σ

σ

σθ

θ

11

11

11

1 1

1 1

(3243)

By this process the unknown quantity Δε ncrk i+

+1

1 in Eq(3140) ie crack strain

increment is expressed by the unknown quantities 1 1

1 1i ir n z nσ σ+ +

+ +Δ Δ and 1

1i

nθσ ++Δ and one

unknown quantity is eliminated Here Eq(3243) indicates the process of obtaining the crack

strain increment and is incorporated into the total matrix by Eq(3242) That is it is

formulated by the relationship between the elastic and crack strain increments and stress

increment as shown in Eq(3242)

Substituting Eq(3243) which is resulted as the elimination process above into

Eq(3140) gives

[ ] ( ) ( )

1 1 01 1 1

11 1

1 11 1 1 0

i i in n n n

i i i in n n n n

P i c i i in n n n n

C d B u

C C d

C

θ

θ θ

θ

σ ε

σ σ σ

ε ε σ σ

+ ++ + + +

++ + + +

+ ++ + + +

minus Δ + Δ

+ minus + minus

+ Δ + Δ + minus =

and finally it is converted into

[ ] [ ] [ ] ( )~ ~ C d B u Cni

ni

ni

n nP i

nc i

ni

ni

n+ ++

++

+ ++

++

+ +minus + + + + minus =θ θσ ε ε ε σ σ11

11

10

11

11

1 0Δ Δ Δ Δ (3244)

Fitting parameters FACR and FACZEi = E holds when rel

iii εεε minus=minus 0 however these values are adjusted using

parameters FACR in the circumferential direction and FACZ in the axial direction as ε ε εi i i

relminus = minus sdot0 FACR and ε ε εi i irelminus = minus sdot0 FACZ respectively

The default values of FACR and FACZ are both 1


- 214 -

Fitting parameters FRELOC and EPSRLZ

Relocation strain εrel in the radial direction is given by the value which is calculated from the initial radial gap width multiplied by RELOC and divided by the pellet radius

In the axial direction relocation strain εrel is directly set by EPSRLZ The default values of FRELOC and EPSRLZ are 05 and 0003 respectively

Youngrsquos modulus of cracked pellet is specified by

ECRAC3 Default value is 2 109 2times N m about 1100 of the original Youngrsquos modulus of UO2


Default value of IYNG is 10 When

IYNG=10 in the pellet stiffness change model shown in Fig327 minusε rel and 0 is connected by

a straight line When IYNG=0 minusε rel and 0 is connected by a quadratic curve which is convex downward

Model parameter IYNG (used only in mechanical analysis)

It is assumed that a pellet can recover its stiffness up to EtimesEFAC if the pellet has an original Youngrsquos modulus E EFAC is used in only ERL mechanical analysis In the 2-D local PCMI analysis described in chapter 35 EFAC is not a fittting parameter It is fixed at 10

Fitting parameter EFAC

326 Definition equations of equivalent stress and strain in FEMAXI Definitions are shown of equivalent stress and strain which have an important role in

understanding the description of non-linear mechanics after this

(1) Equivalent stress A general form of the equivalent stress or effective stress σ or yield function h of

pellet and anisotropic cladding where shear stress components are assumed to exist in the

axial and radial directions only and three principal stresses are always identical to the hoop

axial and radial stresses

( ) ( ) ( ) ( )

( )

2 22 2

12 2

36

2

3

r z z r zr

r z

h H F GF G H θ θ

θ

σ σ σ σ σ σ σ τ

α σ σ σ

= = minus + minus + minus + + +

+ + +

(3245)

where H F G anisotropy coefficients (in the pellet H=F=G=1 is assumed)

α pellet hot-press parameter of pellet (α =0 is assumed for cladding)

Here α is the hot-press parameter which is explained in section 327 In Eq(3245) if

isotropy is assumed 10F G H= = = so that in ERL mechanical analysis (IFEMRD=1) we

have


- 215 -

( ) ( ) ( ) ( )2222 32

1zrrzzr σσσασσσσσσσ θθθ +++minus+minus+minus= (3246)

and in LCL (local PCMI) mechanical analysis (IFEMRD=0)

( ) ( ) ( ) ( )22222 362

1zrzrrzzr σσσατσσσσσσσ θθθ ++++minus+minus+minus=

(3247)

(2) Equivalent elastic strain in isotropic material In ERL mechanical analysis (IFEMRD=1)

( ) ( ) ( )05

2 2 22 1

3 2e r z z rθ θε ε ε ε ε ε ε = minus + minus + minus (3248)

In local PCMI analysis (IFEMRD=0)

( ) ( ) ( )05

2 2 2 22 1 3

3 2 4e r z z r rzθ θε ε ε ε ε ε ε γ = minus + minus + minus + (3249)

(3) Plastic strain increment (identical formulation for creep strain increment) In ERL mechanical analysis (IFEMRD=1)

2 2 22( ) ( ) ( )

3pl pl pl pl

r zd d d dθε ε ε ε = + + (3250)


( )2 2 2 2 22 1( ) ( ) ( ) ( ) ( )

3 2pl pl pl pl pl pl

r z r zd d d d d dθε ε ε ε γ γ = + + + + (3251)

Also total equivalent plastic strain is the sum of the increments from the one at the first time

step to that at the n-th time step as

1

npl pl

ii

dε ε=

= (3252)

Anisotropy parameters H F G for

pellet cladding (SUS) pure Zr and ZrO2 are respectively H0(1) F0(1) G0(1)H0(2)

F0(2) G0(2)H0(3) F0(3) G0(3) and H0(4) F0(4) G0(4) Default values are all 10

Anisotropy parameter H0(4) F0(4) G0(4)


- 216 -

327 Hot-pressing of pellet (1) Phenomenon and equation

Volume compression of pellet by hydrostatic pressure ie hot-pressing is a result of

increasing density due to crushing the void space inside pellet by hydrostatic pressure(31) The

following description is applied to both the ERL analysis and local PCMI analysis

The equivalent stress and yield function of pellet is as follows neglecting the shear stress

components as they are ineffective for hot-pressing in Eq(3235) for the ERL analysis

(IFEMRD=1)

( ) ( ) ( ) ( )

( )

2 22

12 2

3

2

3

r z z r

r z

h H F GF G H θ θ

θ

σ σ σ σ σ σ

α σ σ σ σ

= minus + minus + minus + +

+ + + equiv

(3253)

holds where

hyield function (von Mises function)

σ equivalent stress

H F G anisotropy factors though a pellet is isotropy and H=F=G=1

α hot-pressing parameter (in cladding α=0)

Eq(3243) is added by the term

( )2

23 27

3r z

r zθ

θσ σ σα σ σ σ α + + + + =

In FEMAXI it is assumed that the hot-pressing strain is proportional to plastic and creep

strains Accordingly it is convenient to formulate the hot-pressing strain as being included in

those of plastic and creep strains Thus by giving the equivalent stress with Eq(3253) the

creep strain increment including hot-pressing is obtained In this primeσ is a deviatoric stress

vector including hot-pressing which is given by

( )( )( )

( )

( )

( )

22

322

2 23

2 22

3

r zr z

r ST r z

z rz ST r z r z

ST r zr z

r z

θθ

θ

θθ θ

θ θ θθ

σ σ σ α σ σ σσ σ α σ σ σ

σ σ σσ σ σ α σ σ σ α σ σ σσ σ α σ σ σ σ σ σ α σ σ σ

minus minus + + + minus + + + minus minus prime = minus + + + = + + + minus + + + minus minus + + +

(3254)

Here the term 3r z

STθσ σ σσ + += in Eq(3254) represents a hydrostatic stress


- 217 -

If the material is metal ldquoα=0rdquo in Eq(3254) has and the deviatoric stress is

2

32

03

2

3

r z

r ST

z rz ST

STr z

θ

θ

θθ

σ σ σ

σ σσ σ σσ σ σ

σ σ σ σ σ

minus minus

minus minus minus prime = minus = = minus minus minus

(3255)

In this condition the equivalent stress σ =0 holds and no plastic strain is generated

However in the solid material such as pellet which has gas pores and voids situation is

different For example if there remain voids in pellet after sintering process even in the

condition that the deviation stress is null the voids are pressed to collapse by a compressive

static pressure ( 0)STσ lt which results in pellet shrinkage and volume change ie permanent

strain generation The rate of this shrinkage can be assumed proportional to the creep strain

rate and plastic strain rate Also the shrinkage rate is considered proportional to the static

pressure ( 0)STσ lt so that as shown in Eq(3253)

2

273

r z θσ σ σα + +

is added inside the

square root of equivalent stress formula (31)

Here by considering the plastic flow rule in the iterative calculation of the nth time step

1 1P Pn n

n θ

σε εσ+ +

+

part Δ = Δ part (3256)

holds Assuming the isotropy of pellet material ( )H F G N= = = =1 3 from Eq(3253)

( )3 22

2 3r z

r zr

θθ

σ σ σ σ α σ σ σσ σ

part minus minus = + + + part (3257)

holds Accordingly

1 3 05 3 05 3

11 3 05 3

1 3n

n n Symθ

θ θ

α α ασ α α σσ σ

α+

+ +

+ minus + minus + part = + minus + part

+

(3258)

By substituting Eq (3258) into Eq (3256) and rearranging the result

( ) 1

11 3

1

1

P P

PP P Pnn n

n P P

v v

v v

v vθ θ

θ

α εε σσ

++ +

+

minus minus+ Δ Δ = minus minus

minus minus

(3259)

where v P = minus+

0 5 3

1 3

αα

(3260)


- 218 -

This equation holds in the 2-D local PCMI analysis discussed later

The volume strain increment Δε h nP +1 due to the hot-pressing during the plastic

deformation process is formulated by

( ) ( )( )

( )

Δ Δ Δ Δ

Δ

Δ

ε ε ε ε

α εσ

σ σ σ

α εσ

σ σ σ

θ

θ

θ

h nP

nP r

nP z

nP

nP

Pr z

nP

r z

v

+ + + +

+

+

= + +

=+

minus + +

= + +

1 1 1 1

1

1

1 31 2

9

(3261)

Furthermore Eq (3261) is also applied to the volume strain increment Δε h nc

+1 due to

the hot-pressing during the creep advancement so that similarly

( )Δ Δε α εσ

σ σ σθh nc n

c

r z ++= + +119

(3262)

Accordingly the overall volume change (=hot-pressing strain) Δε h n +1 accompanied by

plastic and creep deformation is given by

( )( )

Δ Δ Δ

Δ Δ

ε ε εα

σε ε σ σ σ θ

h n h nP

h nc

nP

nc

r z

+ + +

+ +

= +

= + + +

1 1 1

1 1

9 (3263)

The method to determine the plastic and creep strains is explained in the next part

(2) Parameters for hot-pressing The name-list parameters to control the hot-pressing calculation are explained

A) BETAX

BETAX is a coefficient correlating the theoretical density ratio and the hot-press

parameter α Value of α depends on the optional parameters IHOT and IHPOP

1) For IHOT=0 α is fixed as α = BETAX (Applied to the ERL mechanical analysis)

2) For IHPOP=0 α is fixed as follows

0=α (non-contact period) α = BETAX (contact period)

(Applied to the LCL mechanical analysis)

3) For IHOT=1 (or IHPOP=1) α at the n+1 time step is defined as a

function of porosity as follows

11

1

BETAX ( )

0 ( )

nn

i

n

D DD D

D D

D D

α

α

++

+

minus= ltminus

= ge (3264)


- 219 -

Here D theoretical density ratio (-) when the hot pressing is completed

1 11n nD p+ += minus

pn+1 porosity inside the pellet

iD initial theoretical density ratio (-)

In this case α advances asymptotically to 0 if hot-pressing and densification advance and

porosity decreases however it does not assume a negative value

B) IHOT An option used to adjust α in the ERL mechanical analysis Default value=0

C) IHPOP An option used to adjust α in the local PCMI mechanical analysis Default

value=0

D) FDENH Theoretical density ratio D（－）when hot-pressing is completed This is applied

to cases when IHOT=1 or IHPOP=1 Default value=10

328 Supplementary explanation of incremental method

When speaking of ldquoelastic strain incrementrdquo does it mean ldquoa strain which includes

non-linear strains such as plastic and creep strains but can be approximated as a linear strain

in an incremental methodrdquo For some users this ldquoelastic strain incrementrdquo might imply that

the matrixes such as Eqs(328) (329) and (3217) are shown first and after that these

matrixes are also applied to plastic and creep calculation

However equations such as Eq(3217) which use the Poissonrsquos ratio cannot hold for

plastic and creep deformation Therefore a method is explained below how to treat plastic and

creep strain increments apart from the elastic deformation

Here an additional explanation is given first with respect to the calculation of elastic

strain increment

(1) Stiffness equation in incremental method The stiffness equation in incremental method is as described in chapter 31

[ ] [ ] [ ]

[ ] [ ]

T

TH rel swl den

T

crk c p HOT

B D B dV u F

B D dVε ε ε ε

ε ε ε ε

Δ = Δ

Δ + Δ + Δ + Δ + + Δ + Δ + Δ + Δ

(3265)

Eq(3264) can be transformed into


- 220 -

[ ] [ ]

[ ] [ ]

T

TH rel swl den

T

crk c p HOT

B D dV F

B D dV

ε

ε ε ε ε

ε ε ε ε

Δ = Δ

Δ + Δ + Δ + Δ + + Δ + Δ + Δ + Δ

(3266)

where the terms in the right-hand side are

THεΔ thermal strain increment

relεΔ relocation strain increment

swlεΔ swelling strain increment

denεΔ densification strain increment

crkεΔ cracking strain increment

cεΔ creep strain increment

pεΔ plastic strain increment

HOTεΔ hot-pressing strain increment

The strain in the left-hand side is total strain increment

e TH rel swl den

crk c p HOT

ε ε ε ε ε ε

ε ε ε ε

Δ = Δ + Δ + Δ + Δ + Δ

+ Δ + Δ + Δ + Δ (3267)


[ ] [ ] T eB D dV FεΔ = Δ (3268)

is obtained In Eq(3267) according to the equilibrium equation

[ ] TF B dVσΔ = Δ (3269)

holds so that we obtain

[ ] [ ] [ ] T TeB D dV B dVε σΔ = Δ (3270)

and [ ] eDσ εΔ = Δ finally Eq(3111) is obtained

Here in Eq(3268) if 0FΔ = 0eεΔ = holds

In an actual application when no differences are generated among non-linear strain

increments

0FΔ = (3271)

holds and

0eεΔ = (3272)

holds

However when some difference is generated among non-linear strain increments


- 221 -

loading is produced depending on the difference of strains among elements That is 0FΔ ne

and elastic strain increment takes place Because this FΔ value is not known it is necessary

to solve the stiffness matrix of Eq(3266)

Here in the following relationship

[ ] eDσ εΔ = Δ (3273)

a strain determining the stress is called an elastic strain irrespective of taking account of the

other strain components

(2) Initial strain increment In the initial strain incremental method the following eight components are given at the

start of every time step to solve Eq(3266)

TH rel swl den crk cε ε ε ε ε εΔ Δ Δ Δ Δ Δ p HOTε εΔ Δ

These components are referred to as initial strain increments However the strain

increments which can be defined at the start of time step ie strain change only as a function

of temperature and burnup etc are limited to the following four components

TH rel swl denε ε ε εΔ Δ Δ Δ (3274)

Accordingly since the strain increments which can be defined at the start of time step are the initial strain increments the initial strain increment 0εΔ is set as

0 TH rel swl denε ε ε ε εΔ = Δ + Δ + Δ + Δ (3275)

For the unknown strain increments at the start of time step ie stress-dependent

components crk c p HOTε ε ε εΔ Δ Δ Δ (3276)

determination of these components needs some numerical device so that in the following text

its solution method is explained sequentially

By substituting Eq(3275) into Eq(3267) we obtain an elastic strain increment vector

as Δ Δ Δ Δ Δ Δε ε ε ε ε εn

en n n

crknc

np

+ + + + + += minus minus minus minus1 1 10

1 1 1 (3277)

where HOTεΔ is included in cεΔ and pεΔ

Δε n+1 total strain increment vector

Δε n+10 initial strain increment vector


- 222 -

Δε ncrk+1 pellet cracking strain increment vector

Δε nc

+1 creep strain increment vector

Δε np+1 plastic strain increment vector

The strains by thermal expansion densification swelling and relocation are all collectively

treated as the initial strain In the next section 33 the numerical solution method for these

non-linear strains is explained


- 223 -

33 Method to Calculate Non-linear Strain 331 Creep of pellet and cladding

The creep equation of cladding and pellet is generally represented as

( ) ε σ ε φc Hf T F= (331)

Here

ε c equivalent creep strain rate (1s)

σ equivalent stress (Pa)

ε H creep hardening parameter (minus)

T temperature (K)

φ fast neutron flux (nm2sdots)

F fission rate (fissionm3sdots)

The variation rate of creep hardening parameter ε H is represented as

( ) ε σ ε φH Hf T F= (332)

Eq(331) is usually an equation of creep under uniaxial stress When this equation is

generalized for multi-axial stress state the creep strain rate vector ε c is expressed as a

vector function of stress and creep hardening parameter When this vector function is set as

β ε c is represented as

( ) ε β σ εc H= (333)

where Tφ and F are omitted because they can be dealt with as known parameters

When a calculation at time tn is finished and calculation in the next time increment

Δtn+1 is being performed the creep strain increment vector is represented as

Δ Δε ε β σ εθ θ θnc

n nc

n nHt+ + + + += =1 1

(334)

where

( )

( )σ θ σ θ σ

ε θ ε θεθ

θ

n n n

nH

nH

nH

+ +

+ +

= minus +

= minus +

1

1

1

1

(here 0≦ θ ≦1)

When θ = 0 is adopted the above equation represents the initial strain method This

method adopts the creep rate which is determined by state quantity which has known value at

the starting point of an increment In this case the calculation procedure is simple but the

solution tends to be numerically unstable therefore it is known that the time increment must


- 224 -

be extremely small at high creep rate

In contrast when θ ne 0 is adopted the calculation procedure becomes complex since the

equation contains unknown values however as θ approaches 1 stability of the numerical

solution improves this method of setting θ ne 0 is called an implicit solution

In FEMAXI θ = 1 (complete implicit method) is set to put emphasis on the importance

of numerical stability

Now in a calculation performed from tn to tn+1 when the (i+1)th iteration by the

Newton-Raphson method is being performed after completion of the i-th iteration (see

Eqs(3134) to (3136)) the creep rate vector can be expressed as follows

Since the function form of each principal direction component is identical the creep

Equation (331) is represented as follows using a flow rule and Eq(333)

ε ε σσ σ

σc c d

d

f=

= prime3

2 (335)

Here σ is an equivalent stress which is defined by Eq(3245)

332 Method of creep strain calculation Before developing equations following Eq(335) this section gives an additional

explanation on the relationship between creep strainplastic strain and elastic strain in the

FEMAXI calculation

(1) Principal stress directions and creep strain increment The creep and plastic strain increments in each principal stress direction are determined

in proportion to the magnitude of deviatoric stress following the Prandtle-Reuss flow-rule

An absolute magnitude of the equivalent strain rate is determined by stress temperature creep

strain hardening etc In other words a positive creep strain relative to average stress occurs in

the direction of tensile stress and a negative creep strain relative to average stress in the

direction of compressive stress

At the same time a positive elastic strain is generated in the tensile stress direction and a

negative elastic strain in the compressive stress direction so that the sign of elastic strain

coincides with its principal stress direction The elastic strain occurs in the direction which

【Note】The Prandtle-Reuss flow rule also holds for the creep strain components

(increments) Since creep strain increment is c ct tε εΔ = Δ the flow rule can be

applied to creep strain rate too Refer to the next section 332


- 225 -

enables the strain to accommodate the difference from other strain components and to

decrease the total strain difference which is induced by the displacement difference between

adjacent elements Here ldquodifference from other strain componentsrdquo denotes the difference of

non-elastic strains between adjacent elements in FEM

(2) Difference of strain components Displacement which corresponds to total strain is provided by materials stiffness That is

the total strain differences among all the elements that constitute an entire system are

controlled by the stiffness of each element The component which directly depends on

stiffness is elastic strain whereas the statement ldquothe total strain differences are providedrdquo

holds because the total strain difference includes elastic strain In FEM stress analysis of

FEMAXI displacement increment is pursued and therefore the total displacement increment

is pursued

The elastic strain increment in this total strain increment is provided by the stiffness of

element It follows that in the adjacent elements the element which has a larger non-elastic

strain components than the other element generates a negative elastic strain (compressive

strain) component while the element which has a smaller non-elastic strain components than

the other element generates a positive elastic strain (tensile strain) component Here extent of

elastic strain is determined by the external force imposed on element boundary and by the

magnitude relation of non-elastic strains between adjacent elements The ldquoexternal force

imposed on boundaryrdquo is represented by a loading (pressure difference across the cladding or

contact force in PCMI) in cladding Namely when this loading is exerted on some nodes of

element as external force stress is generated within the element and elastic strain is induced

Also ldquothe magnitude relation of non-elastic strains between adjacent elementsrdquo indicates

a magnitude relation obtained in comparing the element deformations For example when an

inner element has a larger thermal strain (one of the non-elastic strain components) and an

outer element has a smaller thermal strain the inner element expansion is restrained by the

outer element resulting in a negative stress (compressive elastic strain) in the inner element

and vice versa in the outer element

(3) Stress relaxation by creep and strain difference When creep strain occurs as a result that the creep strain is generated in the same

direction of that of the elastic strain ie the two kinds of strains are made to have the same

sign by the flow rule stress is relieved and elastic strain is mitigated Nevertheless the


- 226 -

difference of initial strain components which are other than the elastic and plastic strains

between the adjacent elements remains unchanged

For example four types of strains in cladding elastic creep plastic and initial (=

thermal expansion and radiation growth) strains are considered When creep occurs the elastic

strain is mitigated and stress induced by external force and thermal expansion is relieved This

means that the strain difference of elements consisting of cladding decreases On the other

hand plastic strain will not decrease in general when once generated unless the direction of

stress is reversed and also such stress direction reversal is hardly occurs in fuel rod

Meanwhile during the calculation in one time step whether plastic strain increment is

generated or not is determined by not only the elastic strain increment but also the relation

between creep strain increment and stress The other types of strains are determined by

temperature and burnup etc independently from the stress Therefore during the calculation

in one time step in some cases creep strain mitigates elastic strain relieves stress and

decreases plastic strain increment while the creep strain does not decreases the initial strains

For example the elastic strain increase induced by PCMI stress is mitigated by the creep in

the radial direction of cladding though the plastic strain which has been generated by the

previous time step or thermal strain are not decreased

(4) Method of creep calculation An important point in calculation is how much the creep deformation is generated ie to

evaluate justly the equivalent creep strain rate The equivalent creep strain rate has a high

sensitivity to stress Accordingly to evaluate the rate precisely the code has sufficiently

elaborated its numerical process to predict unknown stress change In the present section

332 and in the next section ldquo333 Plasticityrdquo explanation is given particularly for the

numerical prediction method of unknown stress change Also another explanation is given on

the concept and numerical elaboration method of ldquocreep strain hardeningrdquo for this concept is

one of the restraining factors on the equivalent creep rate change However even if the creep

equation does not include the strain hardening term no problem arises in the FEMAXI

calculation

The essential of creep calculation is to perform a partition evaluation of the strain

induced by stress change between the newly generated creep strain and elastic strain (and

plastic strain if yielding occurs) Needless to say this new creep strain is dependent not only

on strain hardening but also on stress change


- 227 -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Here we return to the explanation of numerical process By Eq(335) in the (i+1)

iteration during the n-th time step the following equation holds

( ) ( )( )

( ) ( )

ε β σ θ σ ε θ ε

σ σ θ σ ε θ ε

σ σ θ σ

σ σ θ σ

θ θ θ

θ θ

θ

θ

nc i

ni

ni

nH i

nH i

ni

ni

nH i

nH i

ni

ni

ni

ni

d d

f d d

d

d

++

+ ++

+ ++

+ ++

+ ++

+ ++

+ ++

= + +

=+ +

+

times prime +

111

11

11

11

11

11

3

2 (336)

Creep strain rate depends on stress temperature fission density fast neutron flux and

creep strain hardening etc Among these factors the rate equation is formulated assuming that

temperature fission density and fast neutron flux are known

( ) 1 1 11 1c i i i H i H i

n n n n nd dθ θ θε β σ θ σ ε θ ε+ + ++ + + + += + + (337)

is obtained Here the unknown quantities are 11

indσ +

+ and 11

H indε +

+ Since the creep strain rate

obeys the flow rule

( )( ) ( ) ( )

1 11 1 1 1

111

1 11

3

2

3

2

i i H i H in n n nc i i i

n n ni in n

i in ni

n

f d dd

d

f

θ θθ θ

θ

θ θθ

σ σ θ σ ε θ εε σ σ θ σ

σ σ θ σ

σσ

+ ++ + + ++ +

+ + +++ +

+ ++ ++

+

+ +prime= times +

+

prime=

(338)

holds

Eq(338) can be approximated through the first-order Taylor expansion with respect to the

unknown 11

indσ +

+ and 11

H indε +

+ as follows

( ) ( )

( )

( )( )

( ) 11

11

11

11

111

2

3

2

3

2

3

2

3

2

3

2

3

+++

++

++++

+

+++++

+

++++

+++

+

++

+++

+

prime

part

part+

primepartpart+

primepartpartprime+

prime

partpart+prime=prime

iHn

in

i

n

H

H

in

in

in

Hini

n

in

in

in

Hini

n

in

in

Hini

n

in

Hini

n

in

ini

n

df

df

df

dfff

εθσε

εσσ

σθσεσσσ

σθσσ

σεσσ

σθσεσσσ

σεσσ

σσ

θθθ

θθθ

θθθθ

θθθ

θθθ

θθθ

(339)

Then


- 228 -

( )

1 112

1 11 1

11

3 3

2 2

3 3

2 2

3

2

ic i i i i i in n n n n ni i

nn n

i i ii i i i

n n n ni in n nn n

ii H i

n ni Hnn

f f d

ff d d

fd

θ θ θ θ θθθ θ

θ θθ θ θθ θ

θθθ

partσε σ σ θ σσ partσσ

partσ part partσθ σ σ θ σσ partσ σ partσ partσ

partσ θ εσ partε

+ ++ + + + + +

++ +

+ ++ + + +

+ + ++ +

++ +

++

prime prime= minus

prime prime+ +

prime+

is derived and re-arranging this gives

1 11

1 11 1

3

2

3 3

2 2

i iic i c i inn n ni i

n nn n

i iii i H inn n ni i H

n nn n

f fd

f fd d

θθ θ

θ θθ θ

θθ

θ θθ θ

part partσε ε σ θ σσ partσ σ partσ

partσ partθ σ σ θ εσ partσ σ partε

+ +++ + +

+ ++ +

+ +++ + +

+ ++ +

prime= + minus

prime prime+ + prime

(3310)

The above equation is rewritten by using the relation of partσpartσ σ

σ

= prime3

2

( ) [ ]

ε ε

σpartpartσ σ

σ σ θ σ

σpartσpartσ

θ σσ

σ partpartε

θ ε

θ θ

θ θ

θ

θθ

θ

θ θ θθ

θ

nc i

nc i

ni

n

i

ni

ni i j n

i

ni

ni

ni

n

i

ni

ni n

iH

n

i

nH

f fd

fd

fd

++

+

+ +

+

++ +

+

+

+ +++

++

++

=

+

minus

prime prime

+ prime

+ prime

1

2 11

11

1

9

4

3

2

3

2i+1

(3311)

is obtained

Eq(3311) is a result of the first-order Taylor expansion and the process of equation

development up to here is a generally approved process irrespective of the creep equation

type

In the development of equations after this a calculation method is explained concerning

the creep strain hardening which is another factor to evaluate the equivalent creep strain rate

The creep strain hardening is usually observed in a primary creep region (thermally activated

creep) where stress is a function which is strongly affected by creep strain Here the

numerical process is explained by classifying the creep equations in terms of ldquowith or without

strain hardeningrdquo into A) B) and C) as follows

A) Stress is an increasing function of creep strain as in the primary creep equation This case assumes that hardening of material occurs with increasing creep strain in the

primary creep region This primary creep component is expressed as ( )1 1i H in ng θ θσ ε+ +

+ + by


- 229 -

referring to Eq(337) Then the corresponding creep strain hardening increment is expressed

by using Eq(337) as follows

( ) ( )( )

1 1 11 1

1 11 1 1

H i i H in n n n

i i H i H in n n n n

t g

t g d d

θ θ

θ θ

ε σ ε


+ + ++ + + +

+ ++ + + + +

Δ = Δ

= Δ + + (3312)

The above equation is approximated by the first-order Taylor expansion

1 11 1 1

11 1 1

11 1

H i H i H in n n

i ii i

n n n nn n

iH i

n nHn

d

gt g t d

gt d

θθ θ

θ

ε ε ε

part σθ σpartσ σ

partθ εpartε

+ ++ + +

++ + + +

+ +

++ +

+

Δ = Δ +

part = Δ + Δ part

+ Δ

(3313)

From Eq(3313) 11

H indε +

+ is expressed as

( ) 1

1 1 1 1 11

11

i ii H i i

n n n n nH i n n

n i

n Hn

gt g t d

dg

t

θθ θ

θ

part partσε θ σpartσ partσεpartθ

partε

++ + + + +

+ + ++

++

Δ minus Δ + Δ = minus Δ

(3314)

Thus formulation of the unknown quantity 11

H indε +

+ is completed which permits the

elimination of the unknown quantity 11

H indε +

+ of two unknowns 11

++

indσ and 1

1H i

ndε ++ in the

right hand side of Eq(3311)

By substituting Eq(3314) into Eq(3311) and re-arranging

1 1 11 1 2 1 3 1

1

iic i c i i i i

n n i j n n nnn

F d F d Fθ θ θ

partσε ε θ σ σ σ θ σ θ σpartσ

+ + ++ + + + ++

+

prime prime prime prime = + + + (3315)

is obtained where

( )

1

1 2

1

9

4 1

i i

i ni Hn n n

iiin nn

n Hn

f gt

ffF

gt

θ θ θ

θ θθ

θ

part partθpart partε partσpartσ σ partσ θ

partε

++ + +

+ +++

+

Δ = minus + minus Δ

Ffn

i

ni2

3

2= +

+

θ

θσ (3316)

( )1 1

3

1

3

21

ii H i

n n nHn

iin

n Hn

ft g

Fg

t

θθ

θ

θ

part εpartε

σ partθpartε

+ + ++

++

+

Δ minus Δ =

minus Δ


- 230 -

B) Creep rate is gradually decreasing under constant stress as in the secondary creep equation

This case implies that to maintain a positive strain rate of creep stress has to be

monotonously increased In the secondary creep region ie steady creep region no

time-hardening effect exists and strain hardening does not occur with time and strain Here

ldquotime-hardeningrdquo means that stress needed to maintain a certain magnitude of creep strain rate

is increasing with time By replacing this time-hardening effect term with the strain hardening

term creep equation is expressed in FEMAXI

However in the case in which creep equation is made by a combination of primary creep

and secondary creep which has the time-hardening a whole creep region is subjected to strain

hardening eg Zircaloy creep equation of MATPRO-09 Accordingly if the creep equation

is given by the following form

1 2( ) ( ) nc f T t f T tε σ σ= sdot + sdot （nlt1） (3317)

the creep equation is formulated by taking account of the strain hardening In other words

1( )s f Tε σ= in Eq(3317) is a steady rate of creep strain term and this term does not

contribute to strain-hardening so that the hardening term is limited

2 ( ) nH f T tε σ= sdot (3318)

Consequently by differentiating Eq(3318)

12 ( ) n

H n f T tε σ minus= sdot (3319)

is obtained By eliminating time t from Eqs(3318) and (3319)

[ ] ( )1 1

2( ) ( )n

n nH H Hg n f Tε σ ε σ ε

minus

= = (3320)

is derived then Eq(3320) is an equation to define ( )Hg σ ε Accordingly in Eq(3318) if

1( ) 0s f Tε σ= = a whole region is subjected to strain hardening

C) None of power function term of time t is included in creep equation In the case of a steady creep equation (no time-hardening) which has no strain hardening

effect the method of FEMAXI is as follows

The creep strain rate is given by Eq(3315) and 0=partpart

H

f

ε and g=0 in no strain

hardening state Then

( )1 2

9

4

i inii

n nn

f fF θ

θ θθ

partpartσ σσ

+

+ ++

= minus


- 231 -

2

3

2

inin

fF θ

θσ+

+

=

(3321)

3 0F =

are obtained

Since the unknown quantity of the right-hand side of Eq(3315) is only 11

indσ +

+

dividing the known and unknown quantities to express the unknown creep strain increment

gives 1 1 1

1 1 1 1c i c i c i c i in n n n n nt C dθ θε ε ε σ+ + +

+ + + + + + Δ = Δ = Δ + (3322)

where

1 1 3 1 1

3

2c i i i in n n n n ni

n

f t F tθ θθ

ε σ θ σσ+ + + + + +

+

prime primeΔ = Δ + Δ

1 1 2

iic i

n n i j nn

C t F Fθ θ θ

partσθ σ σpartσ+ + + +

prime prime prime = Δ +

(3323)

Here in the formulation of Eq(3322) the hot-pressing is included if the material is fuel

pellet as shown in Eq(3259)

Next to eliminate 11

c inε +

+Δ in Eq(3131) which is transformed from Eq(3244)

[ ] ( )

1 1 0 1 11 1 1 1 1

1 0

i i i P i c in n n n n n

i in n n

C d B u

C

θ

θ

σ ε ε ε

σ σ

+ + + ++ + + + + +

+ +

minus Δ + Δ + Δ + Δ

+ minus =

(3244)

substituting Eq(3322) and re-arranging give the following equation

( ) [ ] ( )

1 1 0 1 1 1 1 1 1

1 0

i c i i i P i c in n n n n n n

i in n n

C C d B u

C

θ θ

θ

σ ε ε ε

σ σ

+ + ++ + + + + + +

+ +

+ minus Δ + Δ + Δ + Δ

+ minus =

(3324)

[ ]( ( )

)

1 11 1 1

0 11 1 1

î i i i in n n n n n

c i P in n n

d D B u Cθ θσ σ σ

ε ε ε

+ ++ + + + +

++ + +

= Δ minus minus

minus Δ minus Δ minus Δ

（3325)

where

[ ] [ ] [ ]( ) 1111

~ˆ minus

+++ += icn

in

in CCD

ldquo＾rdquomeans an apparent stiffness 1inC + corresponds to elasticity and [ ]ic

nC 1+ to creep

Eq(3325) is a sum of terms related to the unknown quantity 11

indσ +

+ In this equation

11

c inε +

+Δ has been eliminated and application of the Newton-Raphson iteration to this is due

to the approximation by first-order Taylor expansion That is by performing a first-order


- 232 -

Taylor expansion of the creep strain rate equation with respect to the magnitude of changes of

stress and creep strain hardening a formulation is carried out to predict the change of creep

strain rate which is induced by changes of stress and creep strain hardening and an accurate

prediction of creep strain rate is obtained

Here the creep reversal calculation by the Pughrsquos method(32) is explained When stress

direction is reversed primary creep occurs again Pugh proposed a simple and convenient

method to simulate this reversal as follows

1) Creep hardening induced by creep strain progress is expressed by ε H That is

( ) ε partσpartσ

σ εcHf T=

(3326)

2) When stress is reversed ε εH gt holds and therefore

ˆ 0HHε ε ε= = (3327)

is set

3) When stress is reversed again or reversed after some time elapsed

If ε εH ge ε ε ε= = =H H0 0 is set and

If ε εH lt ε ε εH H= minus is set

Fig331 illustrates these relationships

ε c ε

ε

ε

εH =0

εH =0 εH =0

ε ε εH H= minus

①

②

③

④

Fig331 Creep strain reversal induced by stress direction reversal

Here the stress direction reversal is judged by the following way

Since creep equation is

( ) ε partσpartσ

σ εcHf T f=

ge 0 (3328)

time


- 233 -

irrespective of the occurrence of sign change of ε c between previous time step and current

time step the current time step status is used to judge the stress direction reversal In

multi-axial stress state by using the scalar product of the ε c components the judgment of

sign is performed

In the creep equation of Zircaloy the following tuning parameters can be used

Parameter name Content Default value

CRFAC Creep rate is multiplied with CRFAC for the stress-relieved material

13

IPUGH IPUGH2

PUGH=0 Pughrsquos reversal is not applied PUGH=1 Pugh s reversal is applied

1

333 Plasticity of pellet and cladding The yield condition of pellet and cladding is commonly expressed by

( ) ( )h K TPσ ε= (3329)

where

ε P equivalent plastic strain

T temperature

K yield surface area

h yield function expressed by Eq(3245)

【Note】When the yield function is h a curved surface which satisfies h=K is called

yield surface Yield is referred to as the stress state on this surface The yield function

or yield surface will expand or shift with the progress of yield state The size of yield

surface implies the change of yield surface

However the outer oxide layer is assumed to have no effect on the cladding yield

condition In other words mechanical strength of the oxide layer is neglected The yield

function h is expressed by re-writing Eq(3245) into Eq(3245A) Eq(3245A) takes into

account of the anisotropy of cladding and compressive nature of pellet (hot-pressing) though

the shear components are neglected in the ERL mechanical model

( ) ( ) ( ) ( )

( )

2 22

12 2

3

2

3

r z z r

r z

h H F GF G H θ θ

θ

σ σ σ σ σ σ σ

α σ σ σ

= = minus + minus + minus + +

+ + +

(3245A)

where


- 234 -

H F G anisotropy factor (in pellet isotropy is assumed so that H=F=G=1)

α pellet hot-pressing factor (in cladding α=0)

By this equation the hot-pressing strain increment which depends on plasticity can be

included in the plastic strain increment

Anisotropic factors of cladding H F and G

are designated by H0(2) F0(2) and G0(2) The default values for these are 10

Fitting parameters H0(2) F0(2) and G0(2)

The plastic strain increment vector from time tn to tn+1 is represented by the flow rule as

Δ Δε ε partpart σ θ

nP

nP

n

h+ +

+

=

1 1 (3330)

In the formulation of plastic strain in yielded state it is unknown at the beginning of time

step how much the equivalent plastic strain will occur during the time step and it is also

unknown if the stress will be relieved or not so that calculation in time step should be

progressed by predicting the stress and equivalent plastic strain at the end of time step For

this prediction method the code adopts the method which formulates the plastic strain with

the prediction of changes of stress and equivalent plastic strain from current state by taking the

first-order Taylor expansion of the unknown quantities of stress and equivalent plastic strain

Accordingly iteration calculation is effective to have a better precision of prediction In the

following formulation this method is explained of calculating the plastic strain during this

iteration

During the (i+1)th iteration that follows the (i)th iteration the yield condition and flow

rule are expressed as

( ) ( )h K T Tn ni

nP

nP i

n nσ σ ε ε+ = + +++

++

+Δ Δ Δ11

11

1 (3331)

Here the left-hand side term h is the equivalent stress expressed by Eq(3245) and

changing term is the stress so that ( )11

in nh σ σ +

++ Δ is presented while in the right-hand side

term K the changing terms are equivalent plastic strain and temperature so

that ( ) 11 1P P i

n n n nK T Tε ε ++ ++ Δ + Δ is presented

Then a method is described to determine the unknown quantities 11

inσ +

+Δ and 11

P inε +

+Δ so

as to allow Eq(3331) to hold assuming that 1nT +Δ is known For this purpose hereafter 11

P inε +

+Δ is expressed by 11

inσ +

+Δ in the following manner


- 235 -

Δ Δε ε partpartσ θ

nP i

nP i

n

ih

++

++

+

=

11

11 (3332)

Performing the first-order Taylor expansion for Eq (3331) with respect to Δσ ni++

11

ΔεnP i+

+1

1 and ΔTn+1 we obtain the following approximation

( ) 11

111 +

+

++

+

++

+

Δ

+Δ

+=Δ

+ n

n

iPn

i

nPn

Pn

in

i

nn T

T

KKTK

hh

θθθ partpartε

εpartpartεσ

partσpartσ (3333)

By substituting

nin

in

in d σσσσ minus+=Δ +

++++

111

11 (3334)

into Eq (3333) and rearranging the equation we obtain

( ) ( ) ( )

Δ

Δ

ε partpartσ

σ σ σ ε

partpart

partpartσ

σ

θ θ

θ θ

nP i

ni

n

i

ni

n n nP

n

nn

n

i

ni

H

hh K T

K

TT

hd

++

+ ++

++

+++

=prime

minus + minus

minus

+

11

1

1 11

1

(3335)

Eq(3335) describes the unknown 11

P inε +

+Δ by using 11

inσ +

+Δ instead of 11

inσ +

+Δ

Here

prime =

+

+

HK

ni

P n

i

θθ

partpartε

(3336)

Next since the right-hand side of Eq(3335) contains another unknown quantity 11

indσ +

+

This unknown quantity is eliminated by simultaneous solution with Eq(3335)

Substituting Eq (3325) into Eq (3336) we obtain

( ) ( )

( )

[ ] [ ] [ ] ( )

ΔminusΔminusΔminusminusminusΔ

+

minus

=

+minusΔ

+Δprime

+

++++++

+++

+

++

++

+++

i

n

iPn

icnnn

in

in

in

in

i

n

nin

i

n

nP

nnnn

iPn

in

hCuBD

h

TKhTT

KH

θθθ

θ

θ

θ

partσpartεεεσσ

partσpart

σσpartσpartσ

εσpartpartε

11

1

011

11

1

11

11

~ˆ

(3337)

Rearranging Eq (3337) with respect to Δε nP i

++1

1 we obtain


- 236 -

[ ]( ( ) )1 0 1 1 1 1

11

ˆ

ˆ

ii i i i c in n n n n n n

P i nn i i

i in n

n n

hD B u C

h hH D

θ θθ

θ θθ θ

part σ σ ε εpartσε

part partpartσ partσ

++ + + + + +

+ ++

+ ++ +

Δ minus minus minus Δ minus Δ Δ = prime +

( ) ( ) ( )1 1

ˆ

ii Pn n n n n n

n ni i

i in n

n n

h KT h K T

T

h hH D

θ θ

θ θθ θ

part partσ σ σ εpartσ part


+ ++ +

+ ++ +

minus minus Δ + minus + prime +

(3338)

Here 1

1indσ +

+ is eliminated Eq(3338) is an equation to calculate the equivalent plastic

strain increment To eliminate finally the unknown 11

P inε +

+Δ from Eq(3325) substituting

Eq(3338) into Eq(3332) we obtain

[ ] [ ]( [ ] ( ) )

[ ]

( ) ( ) ( )

[ ]

i

n

i

n

in

i

n

in

nP

nnnn

nin

i

n

i

n

in

i

n

in

icnnn

in

in

in

in

i

n

i

n

iPn

iPH

h

hD

hH

TKhTT

Kh

hD

hH

CuBDh

h

1

11

1

011

11

1

11

1

ˆ

ˆ

~ˆ

+

++

++

++

++

++

++

+++++++

+

+

++

+

+prime

minus+Δ

minusminus

+

+prime

ΔminusΔminusminusminusΔ

=

Δ=Δ

partσpart

partσpart

partσpart

εσpartpartσσ

partσpart

partσpart

partσpart

εεσσpartσpart

partσpartεε

θθ

θθ

θθ

θθ

θθ

θθθ

(3339)

Substituting this equation into Eq(3325) and eliminating 11

P inε +

+Δ the following


[ ]( ( ) )

[ ]( ( )

1 1 0 1 1 1 1 1

1 01 1 1 1

ˆ

ˆ ˆ

ˆ

i i i i i c in n n n n n n n

i ii in n

i i i cnnn n n n n nii

i in n

n n

d D B u C

h hD D

B u Ch h

H D

θ θ

θ θθθ

θ

θ θθ θ

σ σ σ ε ε

part partpartσ partσ σ σ ε εpart partpartσ partσ

+ ++ + + + + + +

+ +++++ + + + +

+ ++ +

= Δ minus minus minus Δ minus Δ

minus times Δ minus minus minus Δ minus Δ prime +

)i

( ) ( ) ( )1 1

ˆ

ˆ

i ii i Pn n n n n n n

n nn

iii i

n nn n

h h KD T h K T

T

h hH D

θθ θθ

θ θθ θ

part part partσ σ σ εpartσ partσ part


+ + ++ ++

+ ++ +

minus minus Δ + minus minus prime +

(3340)


- 237 -

This equation is rearranged as follows

[ ][ ]( [ ] ( ) )

211

1

011

11

11

ˆ

ZZTS

CuBDd

nin

icnnn

in

in

in

ipn

in

minusminusΔ+

ΔminusΔminusminusminusΔ=

++

+++++++

++

θ

θθ εεσσσ (3341)

By this procedure iteration subscript i+1 with respect to plasticity has been eliminated in

Eq(3341) so that a relationship satisfying Eq(3325) has been derived by using the known

quantity of the previous step i As this formulation is approximated by a first-order Taylor

expansion iteration calculation is required

Here the following equations hold

[ ] [ ][ ] [ ]

[ ]

D D

Dh h

D

Hh

Dh

nP i

ni

ni

n

i

n

i

ni

ni

n

i

ni

n

i+ +

++ +

+

++

++

= minus

prime +

θ θ

θθ θ

θ

θθ

θθ

partpartσ

partpartσ

partpartσ

partpartσ

(3342)

[ ]

[ ]S

Dh K

T

Hh

Dh

ni

ni

n

i

n

ni

n

i

ni

n

i+

++ +

++

++

=

prime +

θ

θθ θ

θθ

θθ

partpartσ

partpart

partpartσ

partpartσ

(3343)

[ ] ( )

[ ]Z

Dh h

Hh

Dh

ni

n

i

n

i

ni

n

ni

n

i

ni

n

i1

1

=

minus

prime +

++ +

+

++

++

θθ θ

θθ

θθ

partpartσ

partpartσ

σ σ

partpartσ

partpartσ

(3344)

[ ] ( ) ( )( )

[ ]Z

Dh h

h K T

Hh

Dh

ni

n

i

n

i

n nP

n

ni

n

i

ni

n

i21 =

minus

prime +

++ +

++

++

θθ θ

θθ

θθ

partpartσ

partpartσ

σ ε

partpartσ

partpartσ

(3345)

Thus the unknown value Δεnp i+

+1

1 in Eq(3325) was eliminated

Accordingly Eq(3341) is a constitutive equation in which all the unknown quantities

regarding pellet cracks as well as creep and plasticity of pellet and cladding are eliminated

【Supplementary explanation】

Some additional explanation is given concerning the treatment of creep and plasticity by

taking cladding as an example To simplify and make clear the description below iteration by


- 238 -

stress change is abbreviated and derivation of equations is condensed

In cladding strain components consist of elastic strain thermal strain creep strain

plastic strain and irradiation growth strain in the axial direction Since calculation is

performed at every time step in FEMAXI they are expressed in an incremental form

1 1 1 1 1 1e th P c irr

n n n n n nε ε ε ε ε ε+ + + + + +Δ = Δ + Δ + Δ + Δ + Δ (3346)

where 1nε +Δ Total strain increment vector

1enε +Δ elastic strain increment vector

1thnε +Δ thermal strain incremental vector

1Pnε +Δ plastic strain incremental vector

1cnε +Δ creep strain incremental vector and

1irrnε +Δ irradiation growth incremental vector in the axial direction

Since the thermal analysis treats the thermal strain increment and irradiation growth

increment as the known quantities in the mechanical analysis the unknown quantities are

elastic strain increment creep strain increment and plastic strain increment

FEMAXI uses a method to convert the constitutive equation by setting the unknown

quantity as only stress increment 1nσ +Δ In this a relationship equation is derived between

stress increment and those increments as elastic strain increment creep strain increment and

plastic strain increment and formulation is performed by eliminating elastic creep and plastic

strain increments from the constitutive equation to express the stress increment Namely these

strain components are expresses in the form

1 1X Xn n nC θε σ+ + + Δ = Δ (3347)

Eq(3347) expresses the relationship among elastic strain increment creep strain

increment and plastic strain increment induced by stress increment ie external force change

This equation shows the sharing amount of strain induced by the stress change with elastic

creep and plastic strain increments

By substituting Eq(3347) relationship into Eq(3346)

1 1 1 1 1 1e th P c irr

n n n n n n n n nC C Cθ θ θε σ ε σ σ ε+ + + + + + + + + Δ = Δ + Δ + Δ + Δ + Δ (3348)

is derived Once Eq(3348) is obtained


- 239 -

[ ] ( ) 1 1 1 1e P c th irr

n n n n n n nB u C C Cθ θ θ σ ε ε+ + + + + + + Δ = + + Δ + Δ + Δ (3349)

is obtained and setting e P c

n n n nC C C Cθ θ θ θ+ + + + = + + (3350)

1

n nD Cθ θ

minus

+ + = (3351)

then Eq(3349) is transformed into

[ ] ( )1 1 1 1th irr

n n n n n nD B u Dθ θσ ε ε+ + + + + + Δ = Δ + Δ + Δ (3352)

where

[ ] 1 1

T

n nF B dVσ+ +Δ = Δ (3353)

Consequently substituting Eq(3352) into Eq(3353) gives

[ ] [ ] ( )( )1 1 1 1

T th irrn n n n n nF B D B u D dVθ θ ε ε+ + + + + + Δ = Δ minus Δ + Δ (3354)

and finally

[ ] [ ] [ ] ( )1 1 1 1

T T th irrn n n n n nB D B dV u F B D dVθ θ ε ε+ + + + + + Δ = Δ + Δ + Δ (3355)

is obtained By solving this equation the displacement increment is obtained

For this solution it is necessary to express specifically every strain component in

Eq(3355) and to obtain

1

n nD Cθ θ

minus

+ + = and 1nF +Δ

For elastic strain increment Eq(3348) is

1 1en n nC θε σ+ + + Δ = Δ (3356)

Namely enC θ+ of Eq(3349) is nC θ+

For creep strain increment as it is not a function of stress increment but a function of

stress creep strain increment is also a function of stress Then creep strain increment is

expressed

( )

1 1 1 1

1 1 1 1 1

1 1 1 1

1 1

( )

( ) ( ) ( )

( ) ( ) ( )

( )

c Hn n n n

H H Hn n n n n n n n n

HH H H

n n n n n n n n n nH

H cn n n n

f T t

f T t f T t f T t

df df df T t T T t

d d d

f T t C

ε σ ε

σ ε σ ε σ ε

εσ ε σ ε σ ε σσ ε σ

σ ε σ

+ + + +

+ + + + +

+ + + +

+ +

Δ = Δ

= Δ + Δ minus Δ

= Δ + + sdot Δ Δ = Δ + Δ

(3357)


- 240 -

where ( )Hf T σ ε creep strain rate

T temperature

σ stress Hε creep hardening strain

As the first term of the right-hand side of Eq(3357) uses σ and Hε which are at the

beginning of time step they are strain terms independent from stress and the known strain

increment terms similarly to thermal strain increment The second term of the right-hand side

is the creep strain increment term which changes with stress The creep strain hardening is one

of the factors such as temperature and stress which change the creep rate and creep strain

hardening changes with stress That is Eq(3355) can be expressed by dividing the known

and unknown components for creep strain increment as

1 1 0 1c c cn n n nC θε σ ε+ + + + Δ = Δ + Δ (3358)

Here cnC θ+ of Eq(3348) is given by Eq(3358) and Eq(3354) is expressed in an

expanded form as

[ ] [ ] [ ] ( )

1

1 1 1 0 1

T

n n

T th irr cn n n n n

B D B dV u

F B D dV

θ

θ ε ε ε

+ +

+ + + + +

Δ =

Δ + Δ + Δ + Δ

(3359)

As the plastic strain increment is dominated by the size of yield surface its formulation is

different from that of the creep strain increment

Plastic strain is required to satisfy the relations

( ) ( )h K TPσ ε= (3360)


T temperature

K size of yield surface

h yield function

Here the yield function ( )nh σ is the equivalent stress equation and size of yield

surface ( )PK Tε is the yield stress determined by materials properties or the

stress ( )Py K Tσ ε= For example yield stress of Zircaloy can be formulated with

temperature plastic strain fast neutron flux etc which are experimentally measured

To satisfy Eq(3360) it is necessary for ( ) ( )1 1 1Pn n nh K Tσ ε+ + += to hold after the


- 241 -

previous time step provided that ( ) ( )Pn n nh K Tσ ε= holds at the previous time step

Therefore calculation of plastic strain increment 1Pnε +Δ is performed as follows

Setting

( ) ( )1 1 1P Pn n n n n nh K T Tσ σ ε ε+ + ++ Δ = + Δ + Δ (3361)

and

( )1 1 1P Pn n n n n nP

n n n

h K Kh K T T

Tθ θ θ

part part partσ σ ε εpartσ partε part+ + +

+ + +

+ Δ = + Δ + Δ (3362)

holds

By ( ) ( )Pn n nh K Tσ ε= the second term of the left-hand side of Eq(3362) is

1 1 1P

n n nPn n n

h K KT

Tθ θ θ

part part partσ εpartσ partε part+ + +

+ + +

Δ = Δ + Δ (3363)

Here setting nP n

KH θ

θ

partpartε+

+

prime =

and using the relation Δ Δε ε partpart σ θ

nP

nP

n

h+ +

+

=

1 1 Eq(3363)

becomes

1

1 1 1

n

Pn n n n

n n

h h KH T

Tθ

θθ θ

part part partσ εpartσ partσ part

+

minus

+ + + ++ +

primeΔ = Δ + Δ

(3364)

Expressing Eq(3363) in the form of Eq(3338) we obtain

1 1 1

1

1

1 0 1

1

n n

n

n

Pn n n

n nn

nn n

nn n

P Pn n n

h h K hT

H T

h h KT

hT

H H

C

θ θ

θ

θ

θ θθ

θ θ

θ θ

θ

part part part partε σpartσ partσ part partσ

part part partpartσ partσ partpartσ

partσ

σ ε

+ +

+

+

+ + ++ ++

++ +

++ +

+ + +

Δ = Δ minus Δ prime

Δ = Δ minus prime prime

= Δ minus Δ

(3365)

That is PnC θ+ of Eq(3350) is given by Eq(3365) and Eq(3359) is expanded to the

form

[ ] [ ] [ ] ( )

1

1 1 1 0 1 0 1

T

n n

T th irr c Pn n n n n n

B D B dV u

F B D dV

θ

θ ε ε ε ε

+ +

+ + + + + +

Δ

= Δ + Δ + Δ + Δ + Δ

(3366)

That is plastic strain term 0 1Pnε +Δ is added as the last term

In the actual development of plasticity equations in FEMAXI [D] matrix conversion

method is used This is almost identical to the method of Eq(3366) in which residual


- 242 -

processing for ( ) ( )Pn n nh K Tσ ε= in the previous time step is performed concurrently

Here residual processing for ( ) ( )Pn n nh K Tσ ε= is as follows

Taylor expansion as described in Eq(3363) will inevitably result in

( )1nh σ + asymp ( )1 1Pn nK Tε + + Then in the next time step instead of Eq(3363) Eq(3362) is

directly used for formulation In other words considering the correction of the

quantity ( ) ( )Pn n nK T hε σminus

( ) 1

0 1

n

Pn n n n

nPn

n

KT K T h

T h

Hθ

θ

θ

part ε σpart partε

partσ+

++

++

Δ + minus Δ = prime (3367)

is used

334 Derivation of stiffness equation The constitutive Eq(3341) in which pellet cracks creep and plasticity are expressed with

only known values 11

indσ +

+ 11

inu +

+Δ and constant terms is rewritten

[ ] ( ) ( )

1 1 0 1 1 1 1 1

1 1 2

î P i i i i c in n n n n n n n

in n

d D B u C

S T Z Z

θ θ

θ

σ σ σ ε ε+ ++ + + + + + +

+ +


+ Δ minus minus

(3368)

The strain components which do not depend on stress are presented by Eq(3128) as initial

strain amounts

Substituting Eq(3368) into the equilibrium condition Eq(3141) and rearranging the

equation we obtain the following stiffness equation

[ ] K u Fni

ni

ni

+ ++

+=θ Δ Δ11

1 (3369)

To apply the stiffness equation Eq(3369) to each element each term of the equation is

determined by integrating each term with element volume as shown in the following

equations

[ ] [ ] [ ][ ]K B D B rdrdzni T

nP i

+ += θ θπ2 (3370)


- 243 -

[ ] ( )

[ ] ( ) [ ]

1 1

0 1 1

1 1 2

1 1

ˆ ˆ2

2

2

Ti P i i in n n n n

c in n

T in n

T in n

F B D C

rdrdz

B S T Z Z rdrdz

F B rdrdz

θ θ

θ

π σ σ

ε ε

π

π σ

+ + + +

+ +

+ +

+ +

Δ = minus

+ Δ + Δ

minus Δ minus minus

+ minus

(3371)

Here the right-hand side of the stiffness equation of elements is expressed with Δ Fni+1

because this value represents the load increment vector since the solution obtained from

stiffness equation Eq(3369) is a displacement increment Δuni++

11

Here the symbol ^ in

Δ Fni+1 means that this term is a tentative value of ΔFn+1 during iteration calculation


- 244 -

34 Formulation of Total Matrix and External Force

341 Stiffness equation and total matrix

The stiffness equations of each element (3369) and (3371) are rewritten below

[ ] K u Fni

ni

ni

+ ++

+=θ Δ Δ11

1 (3369)

[ ] ( )

[ ] ( ) [ ]

1 1

0 1 1

1 1 2

1 1

ˆ ˆ2

2

2

Ti P i i in n n n n

c in n

T in n

T in n

F B D C

rdrdz

B S T Z Z rdrdz

F B rdrdz

θ θ

θ

π σ σ

ε ε

π

π σ

+ + + +

+ +

+ +

+ +

Δ = minus

+ Δ + Δ


+ minus

(3371)

When the Newton-Raphson iteration is completed these equations can be expressed as

follows with omission of subscripts for iteration i

[ ] K u Fn n n+ + +=θ Δ Δ1 1 (341)

Eq(341) can be expressed as follows for convenience in the following discussion

[ ] B u F Bee n n et nΔ + + += +1 1 1 (342)

Here [ ] [ ] B K

B F F

ee n

et n n n

=

= minus+

+ + +

θ

1 1 1Δ

Thus the element matrices have been formulated the formulation of the total matrix is

explained next

(1) Method to formulate total matrix The present analyses is performed for the system shown in Fig341 however since

component expression of the matrix is too complex the explanation is given for a simplified

example with three pellet layers

and two cladding layers as

shown in Fig 341

Fig 341 Simplified example of the analytical model

v2U v1

U

① ② ③ ④ ⑤

lz u2u1 u3 u4 u5 u6 u7

v2L v1

L


- 245 -

In Fig 341 u represents radial displacement and v represents axial displacement The

total matrix of this segment can be written as

ΔΔΔΔΔΔΔΔΔΔΔ

+

+++

++

+

++

+

+++

+

+++

++++

++

++++

U

U

L

L

v

v

vv

u

u

u

u

u

u

u

BB

BB

BBB

BBB

B

BB

B

B

BB

BB

B

BB

BB

BBB

B

B

BB

BB

B

B

BBBBBB

BBBBBBBBB

BBB

BBBBBB

BBB

BBB

BBBBBB

BBBBBB

BBB

2

1

2

1

7

6

5

4

3

2

1

544

444

534

434

344

244

144

334

234

134

524

514

424

414

324

314

224

214

124

114

543

443

533

433

343

243

143

542

532

541

442

531

432

441

431

342

341

242

241

142

141

333

233

133

333

332

233

231

132

131

523

522

521

513

423

512

511

422

421

413

412

411

323

322

321

213

223

312

311

222

221

213

123

212

211

122

121

113

112

111

0

0

00

0

0

0

0

0

0

0

0

0

0

000

0

0

0

0

0

0

0

00000

000000

00000

000000

000000

00000

00000

000000

+

++

+++

+

++

+

minus

=

54

44

34

24

14

53

43

33

23

13

52

51

42

41

32

31

22

21

12

11

0

0

00

0

0

0

0

0

etet

etetet

etet

etetet

et

etet

et

et

etet

etet

et

pcb

pca

BB

BBB

BB

BBB

B

BB

B

B

BB

BB

B

F

F (343)

where only Fpca (internal gas pressure) and Fpcb (coolant pressure) in Fig342 (later) are

considered in Eq(343) regarding external force vector Fn+1

By rewriting Eq(343) we obtain

times

+++++++++

++

++++

+++++++++

times

544

444

542

541

442

441

543

443

344

244

144

342

341

242

241

142

141

343

243

143

524

523

514

424

513

423

414

413

22477

313

314

224

313

223

214

124

213

123

114

113

534

434

532

531

432

431

533

433

334

234

134

332

331

232

231

132

131

333

233

133

000000

0000

00

00

00

00

00

00

00

000000

00000

BBBBBBBB

BBBBBBBBBBBB

BB

BBBB

BB

BBB

BBBB

BBBB

BB

BBBBBBBB

BBBBBBBBBBBB


- 246 -

+++

+

++

+++

+

minus=


54

44

34

24

14

52

51

42

41

32

31

22

21

12

11

53

43

33

23

13

2

1

7

6

5

4

3

2

1

2

1

0

0

0

0

0

0

0

0

0

etet

etetet

et

etet

et

et

etet

etet

et

etet

etetet

pcb

pca

U

U

L

L

BB

BBB

B

BB

B

B

BB

BB

B

BB

BBB

F

F

v

vu

uu

u

u

u

u

v

v

(344)

Next formulation of a total stiffness matrix using a stiffness matrix of each segment

given by Eq(344) is shown

(2) Formulation of total stiffness matrix of entire rod length When a total stiffness matrix is created using a stiffness matrix of each segment the

degree of axial freedom of a common node of adjacent segment is superimposed Namely

at the boundary between j and j+1 in the axial direction degrees of freedom of a cladding are

set as

L

jU

j

Lj

Uj

vv

vv

112

111

+

+

=

= (345)

For example when the number of axial segments is three

+

=

ΔΔΔ

ΔΔ=ΔΔ=Δ

Δ

ΔΔ=ΔΔ=Δ

Δ

ΔΔΔ

++ 11

32

31

37

31

3222

3111

27

21

2212

2111

17

11

12

11

3

2

1

netn

U

U

LU

LU

LU

LU

L

L

ee

ee

ee

BF

v

v

u

u

vv

vv

u

u

vv

vv

u

u

v

v

B

B

B

(346)


- 247 -

For simplicity in the following discussion Eq(346) is rewritten in the form of

Eqs(341) and (342) as

[ ] FuBee Δ=Δ (347)

Since no upper and lower boundary conditions are set in Eq(346) a method of setting

boundary conditions for the upper plenum and lower plenum is described in the next section

342 Upper plenum and lower plenum boundary conditions

Presence of upper andor lower plenums creates one of boundary conditions in

mechanical analyses FEMAXI-7 assumes the plenum section as a normal spring element

The following is an explanation of the effect of this condition on the total matrix

(1) External force and contacting force acting on pellets and cladding Figure 342 shows the forces acting on the pellet and cladding

z

rrca

PELLET CLADDINGF3

lz

F3

Fpca Fpcb

F2

F2

rcb

Fig 342 Forces acting on the fuel and cladding

where F2 pellet-cladding contact force

F3 axial force

Fpca plenum gas pressure

Fpcb coolant pressure

Since the gas enters the crack spaces of pellet stack it is assumed that the effect of

deformation of the pellet due to plenum gas pressure can be ignored


- 248 -

(2) Constitutive equation for plenum section

Upper andor lower plenums assign one of the boundary conditions in the mechanical

analysis How this boundary condition is reflected in the total matrix is explained below

A) Upper plenum

Figure 343 illustrates the modeling of upper plenum section as one of boundary

conditions Assuming that plenum gas pressure is Ppca (Pa)coolant pressure is Ppcb (Pa)

inner radius of cladding is rca and outer radius is rcb the external force Fpcz at the boundary

surface (I) is

pcbcbpcacapcz PrPrF 22 ππ minus= (348)

In addition the reaction force F12z acts on the boundary (I) F12z is resultant from the plenum

spring force F1z and cladding force F2z at plenum section

That is assuming that the nodal force at the boundary (I) is FI

( )zzpczzpczI FFFFFF 2112 +minus=+= (349)

holds Here assuming that the upper plenum spring constant is ksU zF1 is expressed by

( ) ( )onUszUsUs

UNszUsz TTlkvvkF minusminusminus= +111 α (3410)

where

szv axial displacement at the boundary (I)

Fig343 Boundary condition of the upper plenum section

Upper plenum spring

Ppcb

Ppca lsz

rca

rcb

(I)

(II)

PELLET CLADDING

lsz

(I)

(II)

Ppca－Ppcb


- 249 -

UNv 1 axial displacement of pellet stack at the boundary (II) (N is the

segment number of the top segment)

αsU thermal expansion coefficient of the spring in the upper

plenum

lszU length of the upper plenum

Tn+1 plenum temperature at the current (n+1 th) time step

To initial temperature of plenum

Expressing Eq(3410) in an incremental form gives

( ) )( 111 nnUszUsUs

UNszUsz TTlkvvkF minusminusΔminusΔ=Δ +α (3411)

where

szvΔ axial displacement increment at the boundary (I) U

Nv 1Δ axial displacement increment of pellet stack at the boundary (II)(N is the

number of the top segment)

Tuning parameters XKSU XKSL ALSU ALSL

Plenum spring constant (Nm) is given by XKSU for the upper plenum and by

XKSL for the lower plenum Default value are XKSU=15times103(Nm) and

XKSL=10times107(Nm) Plenum spring thermal expansion coefficient is given by

ALSU for the upper plenum and by ALSL for the lower plenum Default values are

ALSU=15times10-5(1K) and ALSL=15times10-5(1K)

Expressing Eq(3411) in matrix form gives

( )( )

ΔΔ

minus

minusminus

minusminusminus

=

ΔΔminus

+

+U

N

sz

UsUs

UsUs

nnUszUsUs

nnUszUsUs

z

z

v

v

kk

kk

TTlk

TTlk

F

F

1

1

1

1

1

αα

(3412)

Next zF2 which is the force from the plenum section cladding is calculated by the

formulation equivalent to that of the active length cladding The stiffness equation of each

element of the plenum section cladding is given by Eqs(3413) and (3414)

[ ] K u Fni

ni

ni

+ ++

+=θ Δ Δ11

1 (3413)


- 250 -

[ ] ( )

[ ] ( ) [ ]

1 1

0 1 1

1 1 2

1 1

ˆ ˆ2

2

2

Ti P i i in n n n n

c in n

T in n

T in n

F B D C

rdrdz

B S T Z Z rdrdz

F B rdrdz

θ θ

θ

π σ σ

ε ε

π

π σ

+ + + +

+ +

+ +

+ +

Δ = minus

+ Δ + Δ


+ minus

(3414)

Here it is noted that the temperature referred to by the plenum spring element is [coolant

temperature at inlet + DTPL] for the lower plenum and [coolant temperature at outlet +

DTPL] for the upper plenum (See section 2161)

When ITPLEN=1 is adopted regardless of the plenum locations (upper or lower

section) the temperature at the coolant inlet is used as the plenum section cladding

temperature

The fast neutron flux used for the calculation of creep strain and irradiation growth strain

of the plenum section cladding is given by FPLFAC(2) times the fast neutron flux of segment

1 (the lowest segment of fuel stack region) for the lower plenum section and also given by

FPLFAC(1) times the fast-neutron flux of segment NAXI (the segment at the upper-end of

stack region) for the upper plenum section

Parameter XLSZU XLSZL The upper plenum section length is given by

XLSZU and the lower plenum section length is given by XLSZL When XLSZU and

XLSZL are not specified the value obtained by dividing the upper plenum volume PLENUM(2) and the lower plenum volume PLENUM(1) by 2

cirπ ( cir inner radius of

the cladding) are given as plenum lengths The presence of the lower plenum does not

depend on XLSZL but is determined on the basis of the selection of PLENUM(1) ie

(gt0) or 0 The unit for XLSZU and XLSZL is cm

Adjusting parameter FPLFAC The fast-neutron flux at the lower plenum is given

by FPLFAC(2) times the fast neutron flux at segment 1 (the lowest segment of fuel

stack region ) and the fast-neutron flux at the upper plenum is given by FPLFAC(1)

times the fast neutron flux at segment NAXI (the top segment of the stack)

The default values are FPLFAC(1)=05 and FPLFAC(2)=05

By applying Eq(3413) to the example geometry described in Fig341 Eq(3415) is

obtained Here u is the radial displacement and v is the axial displacement


- 251 -

++

++

minus=

ΔΔΔΔΔ

++++++

+++

54

44

53

43

52

51

42

41

2

7

6

5

544

444

543

443

542

541

442

441

534

434

533

433

532

531

432

431

524

523

522

521

514

424

513

423

512

511

422

421

414

413

412

411

0

0

0

0

etet

etet

et

etet

et

pcz

pcb

pca

sz

UN

BB

BB

B

BB

B

F

F

F

v

v

u

u

u

BBBBBBBB

BBBBBBBB

BBBB

BBBBBBBB

BBBB

(3415)

where

szv axial displacement at the boundary (I) U

Nv 2 axial displacement of cladding at the boundary (II) (N is the number of

the top segment)

Then expressing the matrix of the upper plenum section by using Eqs(3412) and

(3415)

ΔΔΔΔΔΔ

+++minus++++

minus

+++

sz

UN

UN

UsUs

UsUs

v

v

v

u

u

u

kBBBBkBBBB

BBBBBBBB

kk

BBBB

BBBBBBBB

BBBB

2

1

7

6

5

544

444

543

443

542

541

442

441

534

434

533

433

532

531

432

431

524

523

522

521

514

424

513

423

512

511

422

421

414

413

412

411

0

0000

00

0

00

( )

( )

minus+++

minusminus

+

+

minus=

+

+

nnUszUsUsetet

etet

nnUszUsUs

et

etet

et

pcz

pcb

pca

TTlkBB

BB

TTlk

B

BB

B

F

F

F

15

44

4

53

43

1

52

51

42

41

0

0

0

α

α (3416)

is obtained

By incorporating the matrix of upper plenum section obtained by Eq(3416) into the

total matrix given by Eq(346) the boundary condition for the upper plenum section is set

B) Lower plenum

For a rod which has a lower plenum the boundary (I) in Fig344 makes a fixed

boundary


- 252 -

Accordingly for the respective axial displacements Lv 11 and

Lv 12 of the bottom of pellet

stack and cladding of the lowest segment

00 1211 neΔne LL vv

holds That is for the nodes corresponding to them the force from lower plenum spring F1z is

given by ( ) ( )01111 TTlkvvkF n

LszLsLs

LLszLsz minus+minus= +α (3417)

where

Lszv displacement at the boundary (I) ( 0 =Lszv )

lszL length of lower plenum section

Lsk spring constant of the lower plenum spring

Lsα thermal expansion coefficient of the lower plenum spring

Converting Eq(3417) into an incremental form gives

( ) ( )nnLszLsLs

LLszLsz TTlkvvkF minus+ΔminusΔ=Δ +1111 α (3418)

Further expressing Eq(3418) in a matrix formulation

( )

( )

ΔΔ

minus

minusminus

minusminusminus

=

ΔΔminus

+

+L

Lsz

LsLs

LsLs

nnLszLsLs

nnLszLsLs

z

z

v

v

kk

kk

TTlk

TTlk

F

F

11

1

1

1

1

αα

(3419)

holds

For the cladding of the lower plenum section a similar formulation to that of the

cladding of the upper plenum section is possible This formulation is given by Eq(3415)

when applied to the geometry shown in Fig341 The matrix of the lower plenum section is

expressed by using Eqs(3415) and (3419) as

Lower plenum spring

Ppcb

Ppca lsz

rca

rcb

(I)

(II)

PELLET CLADDING

lsz

(I) (Δv=0)Fixed boundary

(II)

Fig344 Boundary condition of the lower plenum section


- 253 -

ΔΔΔΔΔΔ

+++minus++++

minus

+++

Lsz

L

L

LsLs

LsLs

v

v

v

u

u

u

kBBBBkBBBB

BBBBBBBB

kk

BBBB

BBBBBBBB

BBBB

12

11

7

6

5

544

444

543

443

542

541

442

441

534

434

533

433

532

531

432

431

524

523

522

521

514

424

513

423

512

511

422

421

414

413

412

411

0

0000

00

0

00

( )

( )

minusminus++

minus

+

+

minus

minus=

+

+

nnLszLsLsetet

etet

nnLszLsLs

et

etet

et

pcz

pcb

pca

TTlkBB

BB

TTlk

B

BB

B

F

F

F

15

44

4

53

43

1

52

51

42

41

0

0

0

α

α (3420)

Here the boundary condition 0 =Δ Lszv is added to Eq(3420) This is done by

deleting the row and column corresponding to Lszv Δ to satisfy the condition 0 =Δ Lszv

( )

+minus

++

minus=

ΔΔΔΔΔ

++

++

+5

34

3

1

52

51

42

41

12

11

7

6

5

533

433

532

531

432

431

523

522

521

513

423

512

511

422

421

413

412

411

0

0

0

0

0000

00

0

00

etet

nnLszLsLs

et

etet

et

pcb

pca

L

L

Ls

BB

TTlk

B

BB

B

F

F

v

v

u

u

u

BBBBBB

k

BBB

BBBBBB

BBB

α (3421)

Thus by incorporating the lower plenum section matrix shown by Eq(3421) into the

total matrix shown in Eq(346) the boundary condition of the lower plenum section is set

Incorporating the plenum section matrices obtained by Eqs(3116) and (3121) into the

total matrix gives the total matrix as follows

+

=

Δ

+++

111

1

netnn

UPee

nee

ee

LPee

BFu

B

B

B

B

(3422)

Here LPeeB is the matrix of the lower plenum section shown by Eq(3421) UP

eeB is the

matrix of the upper plenum section shown by Eq(3116) and 1 nee eeB Bsdot sdot sdot sdot are the matrices of

the segments of active length of rod Also for the loading vectors 1nF + and 1et nB + the

loading vectors given by Eqs(3421) and (3116) are added The resulted Eq(3422) is the


- 254 -

total (symmetrical) matrix in the non-contact state

If a rod has no lower plenum the boundary (II) in Fig344 makes a fixed boundary and

the nodes of the bottom end of pellet stack and cladding of the lowest segment are fixed in the axial direction ie 01211 =Δ=Δ LL vv is set In this case also the corresponding freedom is

deleted from the matrix That is by deleting the row and column corresponding to 11LvΔ and

21LvΔ the boundary condition is satisfied

343 Pellet-cladding contact model

In this section the method for obtaining the contact force between pellet and cladding

ie contact model is explained In FEMAXI two modes bonded and sliding modes are

considered as contact states Purpose of the contact model is to obtain the contact force F an

important element in the calculation of PCMI

The following improvement has been made to stabilize the calculation during the contact

period in the ERL mechanical analysis model

In the finite elements of pellet and cladding belonging to one segment j the calculation is

performed for a nodal point Pnj－Cnj (= node pair) at which the outer surface of pellet and inner surface of cladding face each other Here 401 lele j

(1) Determination of contact and non-contact A) Calculation up to trarr t t+ Δ for time step n (which starts at time t) is performed and

the state of contactnoncontact is evaluated from a point located at the top section to a point

located at the lower section of node pair Pni－Cni and alternates between noncontacthArr

contact states are checked for all node pairs If there are any changes the point of time tprime

which is determined by the following conditions is obtained for a node pair Pn1－Cn1 at

which changes first occurred

Then tracing back the time all time-dependent calculation results are linearly

interpolated with time and the values are corrected to those corresponding to tprime Furthermore

by altering the state of contact for node pair Pn1－Cn1 at tprime and reducing the time-step width tΔ in the recalculation from tprime to t t+ Δ only mechanical analysis is carried out

B) With respect to the results of A) evaluation and recalculation are performed similarly

to A) for the changes tprimerarr t t+ Δ at node pairs Pni－Cni and Pn1－Cn1

non-contact rarr contact change The time at which the gap width becomes zero

(The coordinate values in the radial direction of nodal points Pn1 and Cn1 are

equal)

contact rarr non-contact change The time at which the contact pressure becomes

zero


- 255 -

C) The presence and absence of changes between noncontact hArr contact are rechecked

for all node pairs at the stage when all the processes of B) are completed If there is any

change the point of time tprimeprime at which the state changed is obtained similarly to A) By

tracing back the time the values for tprimeprime are obtained after linearly interpolating all

time-dependent results with respect to time Furthermore the contact state is altered at tprimeprime

and recalculation from tprimeprime to t t+ Δ is carried out only for mechanical analysis

(2) Judgment of slidingbonded state in P-C contact condition One of the purposes of the 1-D ERL mechanical analysis is to predict elongation of

entire rod or active part of rod macroscopically by calculating the elongations induced by

PCMI of pellet stack and cladding of each segment Accordingly the model predicts the

three modes of contact state open gap state sliding state and bonded state

The 1-D ERL model gives the contact force to the radial and axial freedoms of nodes

which represent the contact surface of each segment The radial freedom of node

representing the contact surface is given to both the pellet outer surface and cladding inner

surface The axial freedom of node representing the contact surface is given to the upper end

planes of pellet stack and cladding permitting the axial force to be generated at each segment

Criteria for the judgment of bondedsliding state of the contact surface is set by the

following relationship sliding state when

bonded state when

where U and V are the contact force in the radial and axial directions respectively and μ is the

friction coefficient defined below and explained later U and V are obtained by assuming that

the contact surface is in the bonded state

In the sliding state since Ult0 sliding direction is first determined by the sign of V For

the node couple (contact pair) which satisfies the sliding condition the boundary condition of UV μplusmn= is given to the node couple and iteration calculation is performed for all the axial

segments until either bonded state or sliding state is determined This method is identical to

that of the 2-D local PCMI model though designation of friction coefficient is different

The coefficient μ used in the 1-D ERL model is set as follows

( )

22

2μ

ππμ sdot

minus=

lr

rr

i

io (3423)

where l segment axial length cladding inner radius(=pellet outer radius) and

cladding outer radius

And is the original frictional coefficient (AMU2 default=04) Eq(3423) is explained

later in section 345(2)

UV μge

UV μlt

ir

orμ


- 256 -

(3) Judgment of contactnon-contact of pellet and cladding Judgment of contactnon-contact is made in the condition that the bonding or sliding

modes at all axial segments in contact state have been determined However when IBOND=1 is set the slidingbonded state judgment is not performed only the bonded state is assumed

Further with respect to the bonded state an axial force due to the restriction in the axial

displacement is considered even in the non-contact state In this case method to set boundary

conditions is discussed in item (5) At the beginning of a new time step the bonded state is assumed at the axial segments

which have a contact state of pellet and cladding If the contact force and friction force at

the axial segment j which is assumed to have a bonded state is

jrjz FF μgt

the contact mode is changed ldquobonded staterdquo to ldquosliding staterdquo At this process sliding direction is determined by the direction of jzF Then for the segments which are judged as being in a ldquosliding staterdquo the contact

boundary condition is changed into ldquoslidingrdquo For the segments which are judged as being in a

ldquobonded staterdquo the contact boundary condition is not changed This procedure is repeated

until the ldquobondedslidingrdquo assumption agrees to the judgment result for all the axial segments

which are in contact state When the assumption agrees to the result the judgment of

ldquobondedslidingrdquo contact state is completed

Here the axial forces of each segment of a rod which has more than nine axial segments

are listed in Table 341 in section 346-2 For the rod which has more than ten segments

an identical treatment applies

344 Axial force generation at the P-C contact surfaces 344-1 Common conditions PCMI analysis by the ERL mechanical model in FEMAXI-7 for the rod having several

axial segments is considered The following discussion supposes a rod with three axial

segments though this discussion can be generalized to the situation of rod with four or more

segments

In PCMI state pellet stack and cladding mutually exert axial force to the other In the

extreme situation of this PCMI ldquono frictionrdquo state and ldquobonded gaprdquo state exist In these

situations one of the most important issues in mechanical analysis is the evaluation of axial

force which is acted as a reaction force between pellet stack and cladding

The common conditions of the mechanical analysis including the options described

below are as follows 1) generalized plane strain condition is applied 2) displacement and

strain are calculated with respect to the initial coordinate prior to deformation Therefore


- 257 -

Pellet stacksegment

Initial state

Irradiated anddeformed state

A B

Fig345 Comparison of the two rod states

the initial state A vs irradiated and

deformed state B

displacement of each nodal point is defined as the displacement from the location in the initial

coordinate 3) fixed boundary condition applies which assumes that in one segment the

bottom plane is shared by pellet stack and cladding And 4) weights of pellet stack and

cladding are always ignored in the model

Determining the displacement and strain of

one segment with reference to the initial coordinate

effects that the contact surfaces of pellet stack and

cladding which constitute one segment are not

affected by the axial displacements of the pellet

stack and cladding of the other segments That is

the contact condition is always set at the initial

elevation of contact surface and calculation of

contact force and frictional force is performed at

the surface This is explained in Fig345 where a

rod is supposed to consist of four axial segments

ldquoArdquo is the initial state having no deformation ldquoBrdquo

is the rod which is considerably irradiated In B

P-C gap is closed and cladding is axially elongated

by the frictional force due to the elongation of

pellet stack in the axial direction ie PCMI Since

the magnitude of axial elongation is larger in pellet

stack than in cladding a relative difference is

given rise to between pellet stack and cladding in

the corresponding segments of A and B However in PCMI (contact) calculation this relative

departure is ignored and it is assumed that the pellet stack faces cladding in a manner as

described in A

345 Algorithm of axial force generation at PCMI contact surface

In the 1-D ERL FEM analysis geometry it is assumed that a straight cylinder of pellet

stack makes a contact with cladding and generate a contact force 2 jF in the radial direction

and that at the same time a friction force in the axial direction is induced by the difference of

thermal expansion magnitudes between pellet stack and cladding The friction force is in

proportion to the contact force This assumption is discussed in detail below

(1) Local ridging and macroscopic friction force It should be noted here that the frictional force generated by the thermal expansion

difference between the two cylinders in segment-wise geometry should be much smaller than

the frictional force which would be actually generated in PCMI if the force is evaluated by


- 258 -

the formula 3 2μj axial jF F= with an ordinary friction coefficient μaxial because which is

explained below in each pellet of actual fuel rod the ridging deformation occurs and PCMI is

in such situation as pellet edge is locally digging into cladding as shown in Fig346

Another problem in calculating PCMI mechanical calculation by such friction formula as 3 2μj axial jF F= is

that when thermal expansion difference between the straight

cylinders exceeds a limit imposed by the static frictional

force generated by this difference this model is no longer

able to express the dynamic frictional force over this limit

ie a frictional force in sliding state

In other words using simply the relationship of

3 2μj axial jF F= will lead to evaluate only the static frictional

force thus sliding state is not taken into consideration and

the analysis is limited to only ldquoeither open gap or bonded

statesrdquo

The above consideration implies that in the ERL

mechanical analysis it is impossible to obtain a significant

frictional axial force which can express the actual situation

including sliding state by multiplying the ordinary frictional

coefficient on the force which is derived by an

average of calculated contact force between pellet

and cladding

(2) Expression for macroscopic friction force It is necessary to consider the pellet ridging deformation and related cladding

deformation in evaluating the frictional force between pellet stack and cladding As shown

in Fig345 concavity and convexity interlock at the ridging region which would enhance the

local mechanical resistance against the independent deformation of pellet stack and cladding

in the axial direction Particularly in power ramp a reaction force which consists of

resistance force and moment force to suppress the pellet ridging is produced by cladding as a

result the mechanical resistance exerting between pellet and cladding against the axial

displacement includes a complicated mechanical behavior

In the 1-D model it is necessary to elaborate a model which expresses a macroscopic

axial frictional force including all these resistance

On the basis of the above consideration in the ERL model instead of the axial frictional

force lrF ijj πμσμ 222 sdot= (3424)

Cladding

Pellet

Fig346 Mechanical resistance generation induced by concavity and convexity interlock at ridging part


- 259 -

( )2 22 2j j o iF r rμ μσ π= sdot minus (3425)

is given where l the segment length ri inner radius of cladding (metal part) ro outer radius

of pellet and j2σ contact pressure in the radial direction evaluated at a segment which is an

averaged contact forces over those generated at ridging part and at the other part

In other words the physical condition assumed here is not one in which the friction force

contributing to the axial force is simply proportional to the contact area 2 irlπ of the cladding

but is one in which only part of the contact area between the pellet and cladding strongly

contributes to the generation of friction ie axial force

The cladding cross sectional area is included in Eq(3425) because it can be considered

that the mechanical stiffness of cladding is correlated with the reaction force against the pellet

ridging and contributes as a main factor to the resistance against the axial displacement of

cladding at the concavity and convexity interlock at ridging Namely the axial force is not a

function of contact area but a function of cladding mechanical strength which is proportional

to the cross sectional area

It should be noted here that it is physically natural to consider that the frictional force

increases with segment length Therefore the frictional force should have a form which is

proportional to the segment length l However at the same time it is also noted that users

will specify the length l without consideration to the mode change of contact state That is

if a long l is set in input file the frictional force increases in proportion to l which leads to

generation of a large axial force in a long segment in the sliding state In this case judgment

of ldquobondingrdquo and ldquoslidingrdquo states would be coarse as a result the sliding state rarely occurs

and bonded state is foreseen to dominate If this is the case it can be considered that

eliminating the dependence on l in local region would allow the judgment to be realistically

performed Therefore it is assumed that the adjusted contact stress per unit area is expressed

as 3 2AMU2j jσ σ= sdot (3426)

and the axial force jF 3 transmitted to the lower adjacent segment is

( )2233 iojj rrF minussdot= πσ (3427)

Then contact force jF 2 is obtained by multiplying the contact force with contact area

lrF ijj πσ 222 sdot= (3428)

By transforming the equation

( )3 2

2 2AMU2

2j j

io i

F F

rlr r ππ= sdot

minus (3429)

is obtained This is expressed as jj FF 23 sdot= μ and we obtain

( )2 2

AMU22

o i

i

r r

rl

πμ

πminus

= sdot (3430)


- 260 -

cf

pf+cf

cf

cf

cf

3

2

1

cf

cf

cf

cf

CladdingPellet stack

Plenum space

In the calculation axial force generation by frictional force is given by incorporating the frictional force jF 2μ obtained at each axial segment into the total matrix

346 Evaluation of axial force generated at the P-C contact surface Option A

346-1 Axial force in sliding-gap state A state is supposed in which a fuel rod is

divided into three axial segments segment 1

(bottom) segment 2(middle) and segment

3(top) as schematically shown in Fig347 In

this case every segment has frictional force

exerting between pellet stack and cladding As a

result axial force is generated in both pellet stack

and cladding

A) At the top segment 3 axial elongation of pellet

stack emerges though the stack is subjected to the

axial reaction force cf=cr3 from cladding due to

static friction In addition plenum spring force

pf exerts on the stack Thus the axial elongation

is restricted to some extent The force ldquocr3 + pfrdquo is the axial force imposed on the top end of

the pellet stack This force ldquocr3+pfrdquo is transferred from the bottom plane of the stack of

segment 3 to the top plane of the stack of segment 2 Thus static frictional force generated

by the elongation of stack is given to cladding as its axial force cx3= -cr3

B) Similarly in the segment 2 the pellet stack has reaction force cr2 due to the friction with

cladding and the cladding has axial force cx2= -cr2 Further axial force cr3+pf from the

segment 3 is added to the top plane of stack of segment 2 and axial force cx3 is added to

the cladding top plane of segment 2 Accordingly the axial force in segment 2 is

cr2+cr3+pf for the stack and cx2+cx3 for cladding

C) Similarly in segment 1 the reaction force cr1 and the axial force cx1= -cr1 Further axial

force cr2+cr3+pf from the segments 2 and 3 is added to the top plane of stack of segment

2 and axial force cx2+cx3 is added to the cladding top plane of segment 1 Accordingly

the axial force in segment 1 is ccr1+r2+cr3+pf for the stack and cx1+cx2+cx3 for cladding

Fig347 Transfer of axial force from one segment to another

Friction coefficient μ is controlled by name-list input parameter AMU2 (default=04) in the ERL analysis axf is designated by AXF (default=10) Parameter AMU2 and AXF


- 261 -

346-2 Calculation method of Axial force in sliding-gap state

When the pellet and cladding are determined to be in contact and in a sliding state the

following two boundary conditions are set A) the boundary condition which directly

determines the axial contact force by multiplying the radial contact force by the friction

coefficient and B) the boundary condition which equates the radial displacement increment

of the outer surface of the pellet at the contact point to the radial displacement increment of

the inner surface of the cladding as indicated by Eq(3431)

When segment j makes a contact in the sliding state the following conditions are set in

the matrix for the axial segment in which the sliding state is assumed

11 12j ju uΔ = Δ (3431)

jj FF 23 μ= (3432)

where μ friction coefficient

Equation (3420) shows the matrix for segment j when segment j is sliding

Δ=

ΔΔ

ΔΔ

minus

minus

minus

00001100

0

0

0

0

1

1

0

0

2

12

11

F

F

u

u

B

j

Ujc

Ujp

j

j

ee

νν

μμ

(3433)

In Eq (3433) letting the rows corresponding to ju 11Δ ju 12Δ U

jP νΔ UjC νΔ and F2 j be

l1 l2 l3 l4 and l5 the following boundary conditions are set for Eq (3433)

Row l1 =

Δ=+ΔL

lljjll FFuK

1211 11

(3434)

Row l2 =

Δ=minusΔL

lljjll FFuK

1212 22

(3435)

Row l3 =

Δ=+ΔL

llj

UjPll FFvK

12 33

μ (3436)

Row l4 =

Δ=minusΔL

llj

UjCll FFvK

12 34

μ (3437)


- 262 -

Row l5 01211 =ΔminusΔ jj uu (3438)

In Eq (3434) the contact force F2j in the radial direction of segment j is added to the

surface of the pellet and in Eq (3435) the contact force -F2j in the axial direction of segment j is added to the internal surface of the cladding In Eq (3436) jF 2μminus is added to the upper cross section of segment j of the pellet In Eq (3437) jF 2μ is added to the upper

cross section of segment j of the cladding Equation (3438) defines the conditions for

generating contact force F2 j under the conditions specified by Eq (3431)

To solve Eq (3433) the axial force from the upper part of segment is required At the

uppermost segment the axial force is known as the coolant pressure and plenum spring force

The boundary condition at the bottom of the lowest segment depends on the existence of

lower plenum If there is no lower plenum the axial displacements of both pellet stack and

cladding are fixed at zero at the bottom of the segment

If there is a lower plenum the displacement of cladding at the bottom is fixed at zero in

the axial direction while at the bottom of the pellet stack spring force is taken into account

ie the axial force is acted on the bottom plane of the pellet stack

vkF lz sdot= (3439)

where v displacement of the bottom plane of pellet stack

zF lower plenum spring force and

lk spring constant of the lower plenum

Consequently Eq(3433) is calculated in series from the upper segments and by

exerting the axial force in accordance with Table 341 and Eq(3433) is solved in the next

lower adjacent segment Here the force to be determined by Eq(3433) is the friction force

jF 2μ obtained in sliding condition This force is an axial force generated in each segment

but not the axial load locally generated due to friction at a contact section between pellet stack

and cladding Its physical concept is that it is a sum of all the forces in the axial direction

which are generated by local frictional contact at the contact surface of each pellet and

cladding This has been explained in the above sections 345(1) and (2) Here the axial forces of each segment of a rod which has more than nine axial segments

are listed in Table 341 For the rod which has more than ten segments an identical

treatment applies


- 263 -

Table 341 Determination of axial force for a rod having ten axial segments

Segment No

Contact state

Load derived from sliding condition

Load derived from bonding condition

Axial force at each segment

Plenum － 0 0 uF

9 Non-contact 0 0 uF


7 Sliding 7Fμ 0 7FFu μ+

6 Sliding 6Fμ 0 7 6uF F Fμ μ+ +

5 Bonding 0 5f 7 6 5uF F F fμ μ+ + +

4 Bonding 0 4f 7 6 5 4uF F F f fμ μ+ + + +

3 Bonding 0 3f 7 6 5 4 3uF F F f f fμ μ+ + + + +

2 Non-contact 0 0 7 6 5 4 3uF F F f f fμ μ+ + + + +


The term uF represents the force of a plenum spring pushing down the pellet top

The terms 6F and 7F represent the contact forces in the radial direction and 3f 4f and 5f

represent the contact forces in the axial direction under the anchoring boundary

conditions

D) These axial forces are incorporated in the PCMI calculation as one of the loading factors

and mechanical equilibrium is attained at every time step during PCMI

In this way the ldquoOption Ardquo model does not calculate the axial force in an independent

manner in each segment but the axial force generated in upper segment is sequentially

summed up in lower segments Therefore the lowermost segment happens to be subjected

to an unreasonably large axial force In this situation the axial elongations of pellet stack and

cladding are constricted That is stress and strain in the axial direction deviate to a

compressive side As a result strain in the radial direction is affected due to Poissonrsquos effect

which in turn imposes some effect on PCMI state Further this effect is recurred to the

calculation of frictional force ie reaction force in cladding

346-3 Calculation method of Axial force in bonded gap state A) It is supposed that in all segments pellet stack is bonded to cladding and the stack and

cladding are displaced in a unit in the axial direction In this situation calculation is

performed simply in the condition that with respect to the axial and radial directions

displacement of outer surface of pellet stack is forced to be equal to that of inner surface of

cladding That is the calculation is dominated by displacement and gives no unnatural

results


- 264 -

B) It is assumed that some segments have bonded state and other segments frictional sliding

state In the bonded segments the axial force is calculated in the way described in the next

section 346-4 and thus-calculated axial force is transferred additionally to the adjacent

lower segment if the lower segment is in the sliding state If the lower segment has bonded

state too the displacement-dominating calculation is performed independently in the lower

segment and gives no unnatural results

346-4 Calculation method of axial force in bonded-gap state

In the (assumed) bonded state the boundary condition is given which sets the same

displacement increments to the pellet outer surface and cladding inner surface at contact

This is expressed by the followings which are set in the total matrix

11 12j ju uΔ = Δ (3431) U

jCU

jP vv Δ=Δ (3440)

where 11 juΔ radial displacement increment of pellet stack outer surface of

segment j 12 juΔ radial displacement increment of cladding inner surface of

segment j

U

jPv Δ axial displacement increment of upper end plane of pellet stack

of segment j

UjCv Δ axial displacement increment of upper end plane of cladding of

segment j

The total matrix [K] for the case where the segment j is in the bonded state is

Δ=

Δ=Δ

Δ=Δ

F

u

B

jCU

jP

jj

ee

1211

νν

μ

(3441)

The matrix [K] is the matrix of left-hand side of Eq(3441) This is a LxL square matrix

with rank L Here for the lines corresponding to 11 juΔ= 12 juΔ and

UjPv Δ=

UjCv Δ a common


- 265 -

freedom is shared by 11 juΔ and 12 juΔ and by UjPv Δ and U

jCv Δ in the matrix so that the

matrix becomes a square matrix with (L-2) rank

346-5 The case assuming the bonded state IBOND=1 by bonding layer formation including gap re-opening state

In the calculation with IBOND=1 after BD=1 is attained by generation of bonding layer

the boundary condition is given which forces the axial displacement to be always shared by

the upper end of pellet stack and cladding at contact point This is applied to the calculation

even if the gap re-opens in the mechanical calculation when fuel temperature decreases and

pellet stack shrinks ie a non-contact state is resulted again The following condition is set in

the total matrix U

jCU

jP vv Δ=Δ (3440)

where UjPv Δ axial displacement increment of upper end plane of pellet stack

of segment j U

jCv Δ axial displacement increment of upper end plane of cladding

of segment j

The total matrix when segment j is in a bonded state is shown by Eq(3442)

Δ=

Δ=Δ

F

B

jCU

jP

ee

νν

(3442)

Similarly to the bonded state U

jPv Δ is set to share a common freedom with UjCv Δ

347 Evaluation of Axial force generation Option B This model assumes that axial force generated in each segment is processed only within

the segment independently from other segments


- 266 -

347-1 Axial force in sliding-gap state

A) Take segment 3 for example Axial elongation of pellet stack will occur though the

elongation is restricted by the reaction force cf3 from friction at cladding and spring force pf

from plenum spring This reaction force cr3+pf is NOT transferred to segment 2 but

processed within the segment 3 only The frictional force generated by the elongation of

pellet stack is equal to the axial force of cladding ie cx3 = -cr3

B) Similarly in segments 2 and 1 the reaction force and axial force are generated from

friction though the calculation is performed for the balance of force within one segment only

independently from the axial force generated in the upper and lower adjacent segments

This is a model which assumes an axial displacement occurs in each independent segments

C) In the sliding state the axial force is determined as follows by giving the differential

pressure across the cladding and downward-direction frictional force to pellet stack and by

giving the differential pressure across the cladding and upward-direction frictional force to

cladding

Frictional force in a sliding state= plusmn frictional coefficient x contact force

In the condition which gives the differential pressure and upward spring force the frictional

force in a sliding state is defined by

[frictional force]= plusmn [frictional coefficient]x[contact force] (3443)

where the direction of frictional force ie sign of the force has been determined

The frictional force which is obtained by Eq(3443) is the axial force in a sliding state

When the contact state is judged as a sliding state such boundary conditions are set as

that which determines the axial frictional force directly by multiplying the radial contact force

with frictional coefficient and that which sets equal radial displacement increments as shown

in Eq(3431) for the outer surface of pellet stack and inner surface of cladding at the contact

node instead of the boundary conditions which as shown in Eq(3440) set the equal axial

displacement increments for the outer surface of pellet stack and inner surface of cladding


In the case which assumes a sliding state contact in the segment j the following

conditions are set in the total matrix

11 12j ju uΔ = Δ (3431)

jj FF 23 μ= (3444)

The total matrix when segment j has a sliding state is shown by Eq(3445)


- 267 -

Δ=

ΔΔ

ΔΔ

minus

minus

minus

00001100

0

0

0

0

1

1

0

0

2

12

11

F

F

u

u

B

j

Ujc

Ujp

j

j

ee

νν

μμ

(3445)

Since Eq(3445) is identical to Eq(3433) Eqs(3434) to (3438) are the boundary

conditions for Eq(3445)

347-3 Axial force in bonded-gap state

A) In all the segments pellet stack is assumed to be completely bonded to cladding so that

the pellet stack and cladding displace in the axial direction as one solid unit In this case the

calculation is performed simply with each displacement of pellet stack and cladding being

forced to be the same Namely this is a displacement-dominating calculation and no

unnatural results are obtained

B) It is assumed that some segments have a bonded gap and other segments have a frictional

sliding state In the bonded gap segments the axial force is calculated by the above method

A) while this axial force is not transferred to the lower adjacent segment even if the lower

segment is in the sliding state If the lower adjacent segment is also in the bonded state the

displacement-dominating calculation is performed independently in this segment too and no


C) Here it is noted that the displacement increment resulted from the calculation by the

subroutine FEMROD is the value which is obtained on the assumption that each axial

segment makes one continuum ie a fuel rod Accordingly the bonded state is defined as the

condition that the axial displacement increment of the upper end node PZU of pellet stack is

equal to the axial displacement increment of the counterpart upper end node CZU of cladding

and that the axial displacement increment of the lower end node PZL of pellet stack is equal

to the axial displacement increment of the counterpart lower end node CZL of cladding

Accordingly in this boundary condition the axial force induced by the boundary

condition of the equal axial displacement increment in segment A is transferred to the lower

adjacent segment B Namely this is contradictory to the ldquocondition that axial force of segment


- 268 -

A is not transferred to the segment Brdquo ie ldquocondition of independent displacement in each

segmentrdquo That is in a bonded segment setting the bonded state boundary condition that the

axial displacement increments of PZU and PZL are equal to those of the counterpart CZU and

CZL allows the axial force to be transferred to the lower adjacent segment This means that

the ldquocondition of independent displacement in each segmentrdquo of the present option B prohibits

the boundary condition to set an identical displacement increment of pellet stack and

cladding

Consequently the option B adopts the following method such axial force is calculated as

that in which the axial force calculated for the P-C contact on the basis of bondedsliding state

of each segment is added to the sum of the differential pressure across the inside and outside

of fuel rod and spring force and the resulted axial force is given to each axial segment as an

external force

D) Namely the axial force given to one segment is considered as follows

In a bonded state the top planes of pellet stack and cladding have the same axial

displacement increments even if their z-axis coordinate values are different Here the two

planes are considered to be a common plane which gives the boundary condition that has the

same displacement increments putting the relative difference of their z-axis coordinate values

on the shelf The frictional force at the P-C contact surface obtained in the condition of the

same displacement increments and of the imposed differential pressure on the common plane

is the axial force in the bonded state A specific method of calculation is explained next

347-4 Calculation method of Axial force in bonded-gap state

When a segment is (assumed to be) in the anchored state the boundary condition at

which the increment of displacement at the pellet outer surface of the contact point and the

increment of displacement of the inner surface of the cladding are equal is given The

following conditions are imposed for the entire matrix where subscript ldquopordquo denotes pellet

outer surface and ldquocirdquo cladding inner surface

po j ci ju uΔ = Δ (=Eq(3431)) (3446)

UjC

UjP vv Δ=Δ (3440)

Here

po juΔ incremental displacement in the radial direction of the outer surface of

segment j of the pellet

ci juΔ incremental displacement in the radial direction of the inner surface of segment

j of the cladding

UjPv Δ incremental displacement in the axial direction at the upper boundary of


- 269 -

segment j of the pellet U

jCv Δ incremental displacement in the axial direction at the upper boundary surface

of segment j of the cladding

The total matrix for the bonded state in segment j is expressed by Eq(3447) where

pFΔ and cFΔ are the contact forces acted on the mode of pellet stack outer surface and cladding

inner surface respectively

(3447)

In Eq(3447) letting the rows corresponding to 11 juΔ 12 juΔ U

jPv Δ UjCv Δ F2j and

F3j be respectively l1 l2 l3 l4 l5 and l6 the following conditions are set

Row l1 =

Δ=+ΔL

lljlll FFuK

12 11

(3448)

Row l2 =

Δ=minusΔL

lljlll FFuK

12 22

(3449)

Row l3 =

Δ=+ΔL

lljlll FFuK

13 33

(3450)

Row l4 =

Δ=minusΔL

lljlll FFuK

13 44

(3451)

Row l5 11 12 0j ju uΔ minus Δ = (3452)

Row l6 0 =ΔminusΔ UjC

UjP vv (3453)

where F2j radial contact force in segment j

F3j axial contact force in segment j

Eq(3548) denotes that the radial contact force F2j in segment j is added to the pellet

stack surface and Eq(3449) denotes that the radial contact force －F2j in segment j is

added to the cladding inner surface Eq(3450) denotes that the axial contact force －F3j in

segment j is added to the top plane of pellet stack and Eq(3451) denotes that the axial

contact force F3j in segment j is added to the top plane of cladding Eqs(3452) and

2

3

0 0

1

1

0

0

1

1

0 0 1 1 0 0 0 0

0 1 1 0 0 0

0 0

p j

c j

po j

ci jee

Up

Uc

j

j

u

uBF

Fv

Fv

F

F

Δ Δminus Δ = ΔΔ Δminus Δ minus minus


- 270 -

(3453) are equivalent to Eqs(3446) (=Eq(3431)) and (3440) determining the conditions

for the generation of contact forces F2j and F3j

Here the sign of contact force is set as follows F2j gt0 when pellet stack is pushed by

cladding ie contact force to thrust cladding back is generated and F3jgt0 when axial

compressive force is imposed on the pellet stack ie cladding is under tensile force in the

axial direction

The force F3j obtained through the above process is the frictional force in bonded state

However the force given by the plenum spring is imposed in the downward direction to the

pellet stack and in the upward direction to the cladding As a result the spring force is

cancelled out in the axial force calculation The spring force is an internal force

The stiffness matrix and loading vector of each axial segment are made before the total

matrix formations as described above to calculate the axial force in each segment in both the

sliding state and bonded state

The derived axial force in each segment is added to the external force vector component

in the total loading vector This component is the one that is imposed on the top end planes of

pellet stack and cladding Here the total loading vector is the loading vector to cover all the

analytical geometry composed by superposing the loading vectors obtained in each segment

before the addition of the axial force and has the loading components corresponding to the

degree of freedom of node

After all the displacement vector is calculated from the obtained total loading vector and

total stiffness matrix

In this manner in the new option axial force and axial displacement (stress and strain) are

calculated in each segment independently Therefore lower segments cannot be subjected to

unreasonably large axial force Radial strain due to Poissonrsquos effect and some impact on

PCMI behavior are included only in the equilibrium calculation within each segment


- 271 -

Time Step Start

Contact state in contact nodes converged

Displacements converged

Current time step is divided at the state change instant and stress-strain is interpolated at the instant

Yield or unloading occurs Or state change ldquonon-contact contactrdquo occurs

Calc of distribution of stress and strain

Going to next time step

Calc of materials matrix of each element

Calc of stiffness matrix and nodal vector of each element

Incorporation of element matrixes and nodal vectors into total stiffness matrix

Calculation completed for all the elements

Solving the total stiffness matrix and calculate nodal displacements

Calc of densification swelling and thermal expansion etc in every element

Setting the contact state between pellet and cladding and boundary conditions of nodes

35 2-D Local PCMI Mechanical Analysis

Fig351 Flow of 2-D local PCMI mechanical calculation

Figure 351 summarizes the flow of steps of the 2-D local PCMI analysis explained below


- 272 -

351 Element geometry

The objective of the present 2-D local PCMI mechanical analysis is to analyze the local

stress-strain of one pellet and its counterpart of cladding in PCMI conditions independently

from the rod-wise PCMI state Namely the objective is to perform a detailed stress-strain

analysis for the ridging deformation For this the model does not regard the contact state

between pellet and cladding of a whole length of rod The model addresses exclusively the

contact force and axial force which are generated in PCMI in a local geometry

(1) Element division in 2-D axis-symmetric FEM

In the model 2-D axis-symmetric FEM is applied to half a pellet length for symmetrical

reason We call this part an objective segment Fig352 shows a simplified example of

element division This figure illustrates half a pellet length of the objective segment which is

divided into coaxial ring elements

In the present model a 8-node iso-parametric ring element is used and displacement

inside the element is approximated by a quadratic function of node displacement An example

of this analytical target ie elements and nodes in half a pellet length is shown in Fig353

Fig353 illustrates a case of ldquoMESH=0rdquo in Table 351 below with respect to the radial

element division Here a pellet is divided into five ring elements and the half pellet is eg

divided into three elements with 111 ratio from the upper end plane The cladding metallic

Fuel Rod

Gap

Cladding

R

Pellet

Fig352 Local PCMI model

1 segment


- 273 -

Lower half of a pellet Cladding

Upper pellet

Centerline of rod

1

1

1

Gap

Fig353 Elements and nodes in half a pellet length

Half a pellet length

Oxide layer

Upper half of a pellet

part has four ring elements and the outer oxide layer is treated as one layer element The outer

oxide layer growth and resulted decrease of cladding metallic part thickness is updated at

every time step by using the results of the thermal analysis and ERL mechanical analysis Also

when inner surface Zr liner of cladding is designated by input the innermost ring element is

treated as the liner element Two nodes each located at pellet and cladding at each side of a

gap form a nodal pair which is described again in section 355

As stated in section 321(4) number of elements in the radial direction of pellet stack

can be specified by input name-list parameter MESH In Table 351 below the control of

number of elements in the 2-D local PCMI analysis is indicated by specifying ldquoMESHrdquo value

which has a default value of MESH=3

When the name-list parameter ISHAPE=1 is designated in input number of ring

elements in the case of MESHgt0 doubles which allows a more detailed calculation

Table 351 The number of pellet stack ring elements

Parameter value Number of pellet ring elements MESH=0 Iso-thickness 5 ring elements MESH=1

Iso-volume 9 ring elements MESH=2 MESH=3

Iso-volume 18 ring elements MESH=4


- 274 -

00 02 04 06 08 10

R(ridge)M(mid)


Half a pellet consisting of 18 equal-volume ring elements

Cladding

1 2 3 4

B(body)

2n-1 2n

n

2n-1 and 2n elements in 1-D thermal

and mechanical analyses

n elements in 2-D local mechanical

analysis

XStrain evaluation point (Gaussian point) Node

Number of cladding ring elements are irrespective of the ldquoMESHrdquo designation above as

follows

1) In case of no Zr-liner cladding 4 ring elements for metallic part + one ring element for

outer oxide layer

2) In case of Zr-liner cladding one ring element for Zr-liner + 4 elements for metallic part

+ one element for outer oxide layer

However when ISHAPE=1 these numbers of elements double

Fig354 shows schematically the FEM elements and nodes of ridging deformation of pellet

and cladding in the 2-D local PCMI analysis in the case of ldquoMESH=3 or 4rdquo The analysis is

applied to half a pellet length making use of the upper-lower symmetrical assumption This

figure indicates that the geometry has a ridge (R) body (B) and mid(M)

Fig354 Correspondence between FEM elements and pellet in the 2-D local PCMI analysis

Fig355 Correspondence of pellet ring elements in the two mechanical models


- 275 -

As the correspondence relation between the 1-D thermal and ERL mechanical analyses

and the 2-D local PCMI mechanical analysis the relation in pellet elements is indicated in

Fig355 and the relation in cladding elements is indicated in Fig356 which means that

after obtaining the temperatures of elements by 1-D elements the average temperature of the

two elements is applied to one element of the 2-D local PCMI analysis geometry

However for the thicknesses of metallic part and outer oxide layer of cladding values

determined in the 1-D ERL analysis are applied to the 2-D analysis The mechanical analysis

itself has no direct relationship between the 1-D and 2-D analyses except the outer surface

oxide growth and the two types of analyses perform stress-strain calculation independently

from each other

(2) Division in the axial direction and contact node pair

The number of element divisions and its ratio in the axial direction of half a pellet length

can be specified by the name-list input parameters K1 and AZ1 Here the default is K1=3

and AZ1=510 the largest value of two parameters are 5 Fig353 shows the case with

K1=3 and AZ1=10 10 10 For example if K1=4 and AZ1=10 1015 20 are input the

geometry has four elements in the axial direction and its relative length ratio is 1 15 2

The nodes facing with each other across the P-C gap make a contact node pair as described in

section 355 In this section the strain component within an element is defined as

Fig356 Correspondence of cladding ring elements in the 1-D and 2-D mechanical analyses where X mark indicates Gaussian point

1-D analysis

2-D analysis

Growing oxide at outer surface

8 elements of iso-thickness metallic part of cladding



- 276 -

ε

ε

ε

ε

γpartpart

partpart

θ

=

=

+

r

z

rz

du

dr

dv

dz

u

r

v

r

u

z

(351）

where εr εz εθ and γ rz are strains in the radial axial circumferential and shearing (r

z) directions respectively The shearing component was not considered in the previous

analysis however since one of the objectives here is stress evaluation in ridging deformation

part of a pellet under strong PCMI the shearing component is taken into account in this

analysis

352 Basic equations

Although the FEM method in the 2-D local PCMI model is identical to that of the 1-D

model described in sections 31-34 this 2-D model method is presented below again because

of its importance

The equilibrium condition at time tn+1 is presented based on the principle of virtual

work as

[ ] B dV FT

n n + +minus =σ 1 1 0 (352)

Here

[B]T transposed matrix of a strain-displacement matrix [B]

σn+1 stress vector at time tn+1

Fn+1 nodal point loading vector at time tn+1

The strain-displacement matrix [B] is assumed to be determined from a dimension prior to

deformation [B] matrix is given by Eqs(3524) and (3525) presented later

The constitutive equation in an incremental form (stress-strain relationship equation) is

given by

[ ] σ εn neD+ +=1 1Δ (353)

where

Δσ n+1 stress increment vector from time tn to tn+1

Δε ne

+1 elastic increment vector from time tn to tn+1


- 277 -

[ ] ( )[ ] [ ]D D Dn n n+ += minus + =

θ θ θ θ1

1

21

[ ]Dn stress-strain matrix (stiffness) at time tn

This relationship can always hold even if there are other components of strains

Defining matrix [C] as an inverse matrix of this D matrix Eq(353) is transformed into

[ ] Δ Δε σθne

n nC+ + +=1 1 (354)

Matrix [C] in Eq(354) is given as

[ ]CE

=

minus minus

minus minus

minus minus

+

1

1 0

1 0

1 0

0 0 0 2 1

ν ν

ν ν

ν ν

ν( )

(355)

where E is Youngrsquos modulus and ν is Poissonrsquos ratio

The elastic strain increment vector is expressed as follows

Δ Δ Δ Δ Δ Δε ε ε ε ε εne

n n ncrk

nc

np


1 1 1 (356)

where

Δε n+1 total elastic strain increment vector

Δε n+10 initial elastic strain increment vector

Δεncrk+1 pellet crack strain increment vector

Δε nc

+1 creep strain increment vector including pellet hot pressing

Δε np+1 plastic strain increment vector including pellet hot pressing

and strains by thermal expansion densification swelling and relocation are treated as initial

strain on the block

The strain-displacement relationship is expressed as

[ ] ( )Δε n n nB u u+ += minus1 1 (357)

where

un nodal point displacement vector at time tn+1

The following equation is obtained from Eqs(353) - (357)

[ ] ( ) [ ] ( ) C B u un n n n n n np

nc

+ + + + + +minus minus minus + + + =θ σ σ ε ε ε1 1 10

1 1 0Δ Δ Δ (358)

Here

[ ] [ ] [ ]

[ ] [ ]

1

1

1(1 )

2n n n

n n

C C C

C D

θ θ θ θ+ +

minus

= minus + =

=


- 278 -

Note Eq(3341) is to be substituted into Eq(3512) In formulation of Eqs(3342) to

(3345) which should be incorporated in Eq(3341) shear stress components are taken

into account Therefore the yield function h（＝equivalent stress σ ) shown by

Eq(3245) is modified as follows (refer to section 326 and 333)

( ) ( ) ( ) ( ) ( ) 052 22 2 232 3

2 r z z r rz r zh H F G NF G H θ θ θσ σ σ σ σ σ σ τ α σ σ σ

= = minus + minus + minus + + + + + +

Eq(352) and (358) are the basic equations for obtaining unknown values σ n+1

and un+1 Eq(358) can be solved by iteration using the Newton-Raphson method since

Δε np+1 and Δε n

c+1 are functions of unknown value σ n+1 (refer to section 32 for

details) Pellet cracks creep and plasticity are treated similarly to the case of 1-D ERL

mechanical analysis described in sections 328 and 33

At the (n+1)th stage (time-step) calculation when the i-th iteration is completed and the

(i+1)th iteration is being performed Eq(358) is expressed as

[ ] ( ) [ ] ( ) C d B u du un ni

ni

n ni

ni

n n np

nc

+ + ++

+ ++

+ + ++ minus minus + minus + + + =θ σ σ σ ε ε ε1 11

1 11

10

1 1 0Δ Δ Δ

(359)

Here

d ni

ni

niσ σ σ+

+++

+= minus11

11

1 (3510)

du u uni

ni

ni

++

++

+= minus11

11

1 (3511)

The equilibrium condition of Eq(352) is given by substituting Eq(3510) into Eq(352)

as

[ ] [ ] B dV B d dV FT

ni T

ni

nσ σ+ ++

+ + =1 11

1 (3512)

353 Stiffness equation The formulation method of the stiffness equation is the same for both the 2-D mechanical

analysis and the 1-D ERL mechanical analysis Therefore by substituting Eq(3341) which

was obtained in the ERL mechanical analysis into the equilibrium condition Equation(3512)

we obtain the equilibrium condition at the (i+1)th iteration

That is


- 279 -

[ ] [ ] [ ][ ] [ ] [ ][ ] ( )[ ] [ ] ( )[ ] ( ) 12111

1

01

1

11

1

ˆ

ˆ

ˆ

+++

+++

+++

++++

=minusminusΔ+

Δ+Δminus

minusminus

+

nnin

T

icnn

iPn

T

nin

ii

iPn

T

in

iPn

Tin

T

FdVZZTSB

dVDB

dVCDB

dVduBDBdVB

εε

σσ

σ

θ

θθ

θ

(3513)

is obtained where

[ ]( )

~C

E E E

E E E

E E E

E E

r

z

r z

=

minus minus

minus minus

minus minus

++

1 10

10

10

0 0 04 1

ν

ν ν

ν ν

νθ

(3514)

Eq(3514) is rearranged as follows

[ ] K u Fni

ni

ni

+ ++

+=θ Δ Δ11

1 (3515)

This is the stiffness equation for elements

Here

[ ] [ ] [ ][ ]K B D B dVni T

nP i

+ += θ θ (3516)

and

[ ]

[ ] ( ) [ ] ( )

1 1 1

0 1 1 1

1 1 1 2

ˆ

ˆ

Ti in n n

T P i i i c in i n n n n

T in n

F F B dV

B D C dV

B S T Z Z dV

θ θ

σ

σ σ ε ε

+ + +

+ + + + +

+ +

Δ = minus

+ minus + Δ + Δ


(3517)

hold and represent stiffness matrix and loading vector respectively

Assuming axis-symmetry in deformation and performing the integration with the

circumferential coordinates θ in Eqs(3516) and (3517) we obtain

[ ] [ ] [ ][ ]K B D B rdrdzni T

nP i

+ += θ θπ2 (3518)

[ ] [ ] [ ][ ] ( ) [ ] ( ) 2

ˆ2

2ˆ

2111

1

011

111

minusminusΔminus

Δ+Δ+minus+

minus=Δ

++

+++++

+++

rdrdzZZTSB

rdrdzCDB

rdrdzBFF

nin

T

icnnn

in

ii

iPn

T

in

Tn

in

π

εεσσπ

σπ

θθ (3519)


- 280 -

On the right-hand side of Eq(3519) the first term is the non-equilibrium residue the

second term is an apparent load due to stress and initial strain and the third term is a

compensation term for yield stress change

The stiffness equation for one element is shown below

Displacement of an arbitrary point of an element in the axial and circumferential directions

uv and displacement u1v1u2v2u8v8 is related as follows

[ ]u

vN

u

v

u

v

u

v

=

1

1

2

2

8

8

(3520)

Here [N] is called a shape function which is a matrix as follows

[ ] [ ]1 2 3 8

3 81 2

3 81 2

0 00 0

0 00 0

N N I N I N I N I

N NN N

N NN N

=

=

(3521)

Here I is an unit matrix 1 0

0 1I

=

That is

1 1 2 2 3 3 8 8u N u N u N u N u= + + + +

1 1 2 2 3 3 8 8v N v N v N v N v= + + + +

hold

Using a local coordinate system shown in Fig 357 (shown later) Ni is expressed as

( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )

N

N

N

N

N

N

N

N

1

2

3

4

5

6

7

8

1 1 1 4

1 1 1 4

1 1 1 4

1 1 1 4

1 1 1 2

1 1 1 2

1 1 1 2

1 1 1 2

= minus minus minus minus minus

= + minus + minus minus



= + minus minus

= + + minus

= minus + minus

= minus + minus

ξ η ξ ηξ η ξ ηξ η ξ ηξ η ξ ηξ ξ ηξ η ηξ η ηξ η η

(3522)

Strain in the element of the axis-symmetrical system is given by the following equation


- 281 -

which includes shear components This is different from the equation of the ERL mechanical

analysis

[ ]

r

z

rz

u r

v z B u

u z v rθ

εε

εεε

part part = = part part = part part + part part

(3523)

Therefore matrix [B] is given by

[ ] [ ]B = B B B B1 2 3 8 (3524)

Here sub-matrix Bi is given by

B i

i

i

i

i i

N

rN

zN

rN

z

N

r

=

partpart

partpart

partpart

partpart

0

0

0 (3525)

In Eq(3522) Ni is defined using a local coordinate system For representation of

matrix [B] in Eq(3523) in concrete form the following coordinate transformation is

performed

The relationship between the local coordinate system and cylindrical coordinate system

r z is given by

[ ]partpartξ

partpartη

partpartξ

partpartξ

partpartη

partpartη

partpart

partpart

partpart

partpart

N

N

r z

r z

N

rN

z

N

rN

z

i

i

i

i

i

i

=

=

J (3526)

or

[ ]partpart

partpart

partpartξ

partpartη

N

rN

z

N

N

i

i

i

i

=

minusJ 1 (3527)

Here [J] is a Jacobian matrix The location(r z) in an element is represented using a nodal

coordinate (ri zi i=1～8) as


- 282 -

r N r

z N z

i ii

i ii

=

=

=

=

1

8

1

8

Using this formulation the Jacobian matrix [J] is expressed as

[ ]J =

= =

= =

partpartξ

partpartξ

partpartη

partpartη

Nr

Nz

Nr

Nz

ii

i

ii

i

ii

i

ii

i

1

8

1

8

1

8

1

8 (3528)

Therefore [J]-1 is given by

[ ] [ ]JJ

minus = =

= =

=minus

minus

1 1

8

1

8

1

8

1

8

1

det

partpartη

partpartξ

partpartη

partpartξ

Nz

Nz

Nr

Nr

ii

i

ii

i

ii

i

ii

i

(3529)

Here

[ ]det J =

minus

= = = = part

partξpartpartη

partpartη

partpartξ

Nr

Nz

Nr

Nzi

ii

ii

i

ii

i

ii

i1

8

1

8

1

8

1

8

(3530)

Components in Eq(3525) are transformed into the local coordinate system using the

above relationships and we obtain

[ ]

[ ]

partpart

partpartη

partpartξ

partpartξ

partpartη

partpart

partpartξ

partpartη

partpartη

partpartξ

N

r

Nz

N Nz

N

N

z

Nr

N Nr

N

N

r

N

N r

i jj

j

i jj

j

i

i jj

j

i jj

j

i

i i

i ij

=

minus

=

minus

=

= =

= =

=

1

1

1

8

1

8

1

8

1

8

1

8

det

det

J

J

(3531)

where drdz is transformed by the equation

[ ]drdz d d= det J ξ η (3532)

Using the above transformation equation we express Eqs(3518) and (3519) in the

local coordinate system as follows

[ ] [ ] [ ][ ] [ ]K B D B N r d dni T

nP i

j jj

+ +=

minusminus=

θ θπ ξ η2

1

8

1

1

1

1 det J (3533)


- 283 -

[ ] [ ]

[ ] ( ) [ ]

[ ] ( ) [ ]

81 1

1 1 11 11

81 1 0

1 1 11 11

81 1

1 1 21 11

ˆ 2 det J

ˆ2 det J

2 det J

Ti in n n j j

j

T P i i i c in n n n n n j j

j

T in n j j

j

F F B N r d d

B D C N r d d

B S T Z Z N r d d

θ θ

θ

π σ ξ η

π σ σ ε ε ξ η

π ξ η

+ + +minus minus=

+ + + + +minus minus=

+ +minus minus=

Δ = minus

minus minus + Δ + Δ


(3534)

In FEMAXI this integration is approximated by the Gaussian integration method as

f d d H H fi j i jj

m

i

n

( ) ( )ξ η ξ η ξ ηminusminus

== =

1

1

1

1

11

(3535)

In the case of rectangular 8-node elements the number of integration points is known to

be sufficient with n = m = 2 In this case since Hi = Hj = 1 the following equation is

applied

f d d f i jji


== =

1

1

1

1

1

2

1

2

(3536)

Here

n number of integration points in the ξ direction

m number of integration points in the η direction

ξ I coordinates of integration point in the ξ direction

ηj coordinates of integration point in the η direction

Hi Hj weight coefficients

Fig 357 Integration points for rectangular 8-node element

The number of integration points is 4 the coordinates of each point are (a a) (a -a)

(-a a) and (-a -a) as shown in Fig357 and a=05773502691 (Note the Gaussrsquos

integral formula for 2-point approximation ＝ ( ) ( ) ( )( )f x dx f a f aminus cong + minus1

1

By calculating the stress vector strain vector and compensation term for each of the 4

integration points the integral terms in Eqs(3533) and (3534) can be obtained

Node

ξ

η

1-1

-1

1

Integrationpointa

a-a

-a


- 284 -

354 Boundary conditions

The boundary conditions which are shown in Fig358 are explained

(1) Geometry for boundary condition of the 2-D local PCMI mechanical analysis Figure 358 shows the definition of the boundary planes in this system There are four

boundaries (1)-(4) on the pellet side and four boundaries (5)-(8) on the cladding side At these

boundaries conditions corresponding to the symmetry of deformation cooling water pressure

gas pressure and contact mode are given

(4)

(1)

(2) (3) (6) (7)

(5)

(8)

r

z

Pellet

CladdingGap

Fig 358 Boundaries of the 2-D model

Setting of degree of freedom in the geometry is shown in Fig359 Details of

boundary conditions with respect to PCMI are explained by Fig3510 in the next section

Fig359 Degree of freedom in the geometry of 2-D local PCMI analysis


- 285 -

(2) Boundary conditions in no contact state between pellet and cladding

As shown in A of Fig3510 the following state of PCMI is assumed

A) The upper end plane (or part of it) of a pellet in the objective segment makes a contact

with the end plane (or part of it) of an upper adjacent pellet

B) Freedom of only 1-D displacement in the axial direction is allowed to the pellet center

1-D freedom of displacement in the radial direction is allowed to the mid-height plane of

pellet (hypothetical bottom plane ie symmetrical plane)

C) Freedom of 1-D displacement in the radial direction is allowed to the cladding upper

plane of the objective segment and concurrently the cladding upper plane can displace in

the axial direction having interaction with the cladding axial force and keeping a flat

plane Freedom of 1-D displacement in the radial direction is allowed to the mid-height

plane of cladding which corresponds to the mid-height plane of pellet

(3) Boundary conditions in contact state between pellet and cladding As shown in B of Fig3510


with the end plane (or part of it) of an upper pellet


1-D freedom of displacement in the radial direction is allowed to the mid-height plane

of pellet (hypothetical bottom plane ie symmetrical plane) However the center axis

of pellet is fixed

Fig3510 Boundary conditions A no contact between pellet and cladding

B contact occurs between pellet and cladding

Cladding

Mid-heightplane

Center

Pellet (half height)

Cladding

Center


A B


- 286 -


plane of the objective segment and concurrently the cladding upper plane can displace

in the axial direction having interaction with the cladding axial force and keeping a flat

plane which has the same elevation of that of the pellet center height Freedom of 1-D

displacement in the radial direction is allowed to the mid-height plane of cladding

which corresponds to the mid-height plane of pellet

(4) Loading vector and displacement at boundaries

The loading vector in Eq(352) is expressed as the sum of loadings acting on nodes

located on the 8 boundaries in Fig358 as follows

F F F F F F F F Fn n n n n n n n n+ + + + + + + + += + + + + + + +1 11

12

13

14

15

16

17

18( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (3537)

Here Fni+1( ) is the vector of loading which acts on the node at boundary (i)

It is assumed that boundaries (1) and (5) which are shown in Fig358 are symmetrical

planes and nodes on these boundaries do not displace in the axial direction that is

Δv ( )1 0= (3538)

Δv ( )5 0= (3539)

hold Here ( ) Δv 1

and ( ) Δv 5

are the displacement increment vectors of nodes on

boundaries (1) and (5) in the axial direction respectively The constraint force generated at

boundary (1) due to the constraint condition of Eq (3538) is Fn+11( )

and the constraint force

generated at boundary (5) due to the condition of Eq (3539) is Fn+15( )

In the case of a solid pellet nodes located on boundary (2) are not displaced in the radial

direction since axis-symmetry is assumed for deformation That is for the radial displacement

increment vector Δu( )2of the node located on boundary (2)

Δu( )2 0= (3540)

The constraint force generated at boundary (2) (axis of rotational symmetry) by the

condition of Eq (3540) is Fn+12( )

Here in the case of a hollow pellet Fn+ =12 0( )

holds

since radial displacement is not restricted

The surface force given by coolant pressure acts on boundary (7) or Fn+17( )

represents

the vector of equivalent node point loading due to this action

At boundary (3) contact force exists due to the interaction (PCI) of boundaries (3) and

(6) Gas pressure inside the fuel rod also acts on boundary (3) however this is not taken into


- 287 -

account since it is canceled by gas pressure present in a crack void inside the pellet

Fn+13( ) is a vector of contact force generated by the interaction of boundary (3) with the

pellet (boundary (6))

At boundary (6) actions of contact force produced by PCMI and gas pressure are

considered Fn+16( )

represents the sum of the contact force vector and the vector of equivalent

node point loading by gas pressure

At boundary (8) the number of degrees of freedom in the axial direction of a node

group on the boundary is set as 1 based on the assumption of symmetry of deformation

Δ Δv v( ) ( )8 8= (3541)

Here Δv ( )8 is the axial displacement increment of nodes on boundary (8) At

boundary (8) in addition to the restraint force generated from the condition of Eq (3572)

such loads as axial tensile load produced by the PCMI in an upper region of a fuel rod and

axial loads given by gas pressure and coolant pressure are considered The axial tensile load

due to PCMI in the upper region of the fuel rod which acts on boundary (8) is determined in

an empirical manner This will be explained later in section 356

At boundary (4) the above-described loads in the axial direction which is determined

empirically acts as a compressive force The contact force distribution which can be

obtained by solving the pelletpellet contact problem on the assumption of plane symmetry

under this axial load is the load acting on the boundary

As described above the kinds of load which act on the 8 boundaries are of the following

6 types reaction due to constraint of displacement equivalent nodal force due to coolant

pressure equivalent nodal force due to gas pressure pelletcladding contact force axial tensile

load and pelletpellet contact force balanced with the axial tensile load

For clarity of expression these forces are represented by subscripts R W G PC AX and

PP respectively and Eq (3537) is expressed as

F F F F F F F

F F F F F F

n nR

nR

nPC

nPP

nR

nPC

nG

nW

nR

nAX

nW

nG

+ + + + + + +

+ + + + + +

= + + + + +

+ + + + + +

1 11

12

13

14

15

16

16

17

18

18

18

18

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) (3542)

These forces are classified into three types according to the method of treatment of forces

as

F F F Fn n n n+ + + += prime + primeprime + primeprimeprime1 1 1 1 (3543)


- 288 -

where

prime = + + ++ + + + +F F F F Fn nG

nW

nW

nG

1 16

17

18

18( ) ( ) ( ) ( ) (3544)

primeprime = + + ++ + + + +F F F F Fn nR

nR

nR

nR

1 11

12

15

18( ) ( ) ( ) ( ) (3545)

primeprimeprime = + + ++ + + + +F F F F Fn nPC

nPP

nPC

nAX

1 13

14

16

18( ) ( ) ( ) ( ) (3546)

prime+Fn 1 is treated as those in which values are known at time tn+1 and primeprime+Fn 1 is the

constraint force due to the known boundary condition (designated displacement) primeprimeprime+Fn 1 is

determined by boundary conditions which depend on unknown values for obtaining primeprimeprime+Fn 1

the interaction between pellet and cladding as well as pellet and pellet must be solved

(5) Obtaining method of primeprimeprime+Fn 1 in Eq(3546)

When iteration of the Newton-Raphson method is completed Eqs(3515) (3516) and

(3517) are expressed as follows with omission of subscript i for iteration Here ^ indicates

that the value is a tentative one during the iteration calculation

[ ] K u Fn n n+ + +=θ Δ Δ1 1 (3547)

[ ] [ ] [ ][ ]K B D B dVn

T

nP

+ += θ θ (3548)

[ ] [ ] [ ][ ] ( ) [ ] ( )

minusminusΔminus

Δ+Δ+minus+

minus=Δ

++

+++++

+++

dVZZTSB

dVCDB

dVBFF

nnT

cnnn

inn

Pn

T

nT

nn

211

10

11

111

~ˆ

ˆ

θ

θθ εεσσ

σ

(3549)

Here with substitution of Eq (3543) into the right-hand side of Eq (3549) the load

term can be given by

[ ] [ ] ( ) [ ] ( )

1 1 1 1 1

01 1 1

1 1 2

1 1

ˆ

ˆ

ˆ

T

n n n n n

T P i cn n n n n n

T

n n

n n

F F F F B dV

B D C dV

B S T Z Z dV

F F

θ θ

θ

σ

σ σ ε ε

+ + + + +

+ + + + +

+ +

+ +

prime primeprime primeprimeprimeΔ = + + minus

+ minus + Δ + Δ


prime primeprimeprime= Δ +

(3550)

where


- 289 -

[ ] [ ] ( ) [ ] ( )

1 1 1 1

01 1 1

1 1 2

ˆ

ˆ

T

n n n n

T P i cn n n n n n

T

n n

F F F B dV

B D C dV

B S T Z Z dV

θ θ

θ

σ

σ σ ε ε

+ + + +

+ + + + +

+ +

prime prime primeprimeΔ = + minus



(3551)

Here Δ prime+Fn 1 is the node point loading increment vector under the condition primeprime+Fn 1

(non-contact condition)


[ ] K u F Fn n n n+ + + += prime + primeprimeprimeθ Δ Δ1 1 1 (3552)

1nF +primeΔ of the right-hand side of Eq(3552) is a loading vector when no contact loading occur

and 1nF +primeprimeprime is a contact loading vector Eq(3552) expresses the contact loading and the other

loading vector in a separate manner to make clear the treatment of loading vectors in the

stiffness equation

355 PCMI and contact state problem In the 2-D local PCMI analysis the two planes contacting each other are expressed by 7

to 11 contact nodes ie contact node couples Here the pellet which is targeted by the 2-D

PCMI analysis in the pellet stack of the objective segment is referred to as the target pellet

(1) Sliding state and bonded state in contact-mode calculation

A) In the contact state analysis in the 2-D geometry the nodes on the cladding inner surface

are set as the node counterparts of the target pellet outer surface (see Fig3511) and

judgment of bondedsliding state is performed on these nodal pairs Here the bonded state is

the condition in which no relative displacement occurs between the two nodes of one nodal

pair and the sliding state is the condition in which some relative displacement can occur

parallel to the contact plane

The criteria to judge the contact state is as follows When UV μge the sliding state occurs

When UV μlt the bonded state occurs

where μ is the friction coefficient (AMU default=04) U is the contact pressure in the

radial direction and V is the frictional force in the axial direction

Here the U and V which are obtained on the assumption that at each node is in the bonded state are used in the judgment of UV μge or UV μlt If a node pair is judged as being in

a sliding state ie UV μge sliding direction is determined by the sign of V since Ult0


- 290 -

Then the node pair is given a boundary condition of UV μplusmn= Thus iteration calculation

is carried out for all the nodal pairs until bonded sliding state is determined at all the nodal

pairs

B) This sliding state is described in the following process the end edges of thermally

expanded pellet are pushing cladding to cause a local ridging deformation At the same time

the end edges are being shifted upward relative to cladding and mitigate the ridging

deformation of end edges by the reaction force from cladding

This process is given rise to because the bending moment received by the pellet ridging

part from the cladding ridging part is larger than that received by the cladding from the pellet

ridging part and because the end edges of pellet make contact with cladding in the early

period of PCMI so that they are imposed by the reaction force induced by contact from the

early phase of PCMI

That is bondedsliding state of P-C contact has an effect on the height of ridging and

local stress of cladding

C) As one of the basic assumptions of the 2-D local PCMI analysis it is assumed that a

pellet which has the identical PCMI state to that of the target pellet exists as an upper adjacent

pellet just above the target pellet and that the contact force generated across the end planes of

the two pellets acts on the target pellet as axial force

In the ridging analysis which is performed with this axial force being as one of the

boundary conditions the condition has been found which can most successfully reproduce the

observed ridging deformation data This condition is realized by assuming that pellet and

cladding are in a nearly locking state Here ldquolockingrdquo denotes that axial displacement

increment of pellet upper end is equal to that of cladding upper end in the 2-D analytical

geometry This situation is explained below

First a situation is considered in which no axial force is assumed on the pellet upper end

and the displacement increment of the pellet upper end is larger than that of the cladding

Then when the cladding top end is forced to be pulled up to the level of pellet upper end

height a tensile axial force is generated in the cladding In this situation displacement

increment of the pellet is larger than that of the cladding Then a downward axial force on the

pellet is augmented until the displacement increment of pellet upper end is equal to that of

cladding In this process a downward axial force acts on the pellet upper end and an

upward axial force acts on the cladding upper end because cladding has less thermal


- 291 -

expansion than pellet When pellet is shrinking the directions of the axial forces are reversed

D) To realize these conditions a name-list variable ILOCK has been introduced When

ILOCK=0 (default) the locking state is assumed as explained above If the locking is not assumed ie ILOCK=2 the target pellet and its upper adjacent pellet

will not make a contact That is this boundary condition allows the objective pellet to expand

freely in the axial direction Therefore in this situation no axial force from the upper adjacent

pellet is generated on the target pellet

Also since the plenum gas pressure is considered balanced with the gas pressure inside

the cracks of pellet the plenum gas is assumed to have no mechanical effect on pellet surface

An important point of view here is that the ldquolockingrdquo boundary condition in the 2-D local

PCMI analysis assumes that the pellet and cladding are in the bonded state during the locking

period and that this is the optimum boundary condition to predict a local strain and stress of

cladding

The above A) to D) explanations are elucidated more specifically in the following

sections (2) to (7) and section 357

(2) Contact node pair and process of calculation Fig3511 illustrates an example in which contact occurs in the geometry with three

elements in the axial direction As contact conditions three states are considered non-contact

state bonded gap state and sliding state The bonded state which is treated in the 1-D ERL

analysis is not explicitly covered though it can be represented by the bonded state of contact

pairs

Pellet Surface Boundary

Cladding Surface Boundary

Node Couple

Pellet Gap

Fig 3511 Pellet-cladding contact node couple


- 292 -

Thermal expansion rate is different between pellet and cladding so that as the boundary

condition indicated by Fig3511 a slight difference can exist between the axial elevations of

the contact nodes which face each other in the radial direction across the gap shown in

Fig3511 irrespective of the contact (PCMI) or non-contact state However in FEM since

force is necessarily transmitted at the nodes in the case where mechanical interaction is caused

with each two components this method to perform PCMI analysis by using the ldquonode-pairrdquo is

a kind of approximation calculation

In FEMAXI this pelletcladding contact problem is solved in the following procedure

1) A contact state (either contact or non-contact) is assumed for each couple of nodes which

are possibly in contact with each other Initially this assumption is made based on the

results for the previous time step

2) The stiffness equation is solved using the boundary conditions corresponding to the

assumed contact state (contact or non-contact)

3) Based on the solutions of nodal displacement and contact force obtained in 2) judgment is

made on whether or not the assumption made in 1) for each node couple is correct If at one

node couple the assumptions is contradictory to the solution obtained by calculation the

procedure is returned to 1) the assumption is corrected for that node couple and procedures

2) and 3) are performed again

4) Thus the calculation procedures from 1) to 3) are repeated until no discrepancy is observed

between the assumption and calculated results When a change in state of non-contact or

contact (bonded gap state or sliding state) occurs interpolation is carried out for the calculated

results into the time when the change in state occurred the assumption is corrected after that

time and the procedure is returned to 1) for continuous calculation

The calculation method of this process is presented below

(3) Non-contact state between pellet and cladding

In Eq(3542) FnPC

+13( ) and Fn

PC+16( ) are zero As described later Fn

PP+14( ) and

FnAX

+18( ) are also set as zero and accordingly primeprimeprime =+Fn 1 0 Therefore from Eq(3552) we

obtain

[ ] K u Fn n n+ + += primeθ Δ Δ1 1 (3553)


- 293 -

(4) When one contact node couple is in a sliding state and the remaining contact node couples are in a non-contact state

In this case FnPC

+13( ) and Fn

PC+16( ) are represented as

FU

VnPC n

n+

+

+

=

13 1

1

0

0

0

0

( )

(3554)

FU

VnPC n

n+

+

+

=minusminus

16 1

1

0

0

0

0

( )

(3555)

where

V Un n+ += plusmn1 1μ (3556)

Vn+1 contact force in the radial direction at time step tn+1

Un+1 contact force in the axial direction at time step tn+1

μ pelletcladding friction coefficient

Therefore Eq (3553) is represented as

i

j

k

l

m

K

u

v

u

v

U

Fn

p n

p n

c n

c n

n

n

0

0

1

1

0

0

0 0 1 0 1 0 0 0 0 0

1

1

1

1

1

1

minusplusmn

plusmn

minus

= prime

+

+

+

+

+

+

+

μ

μθ

ΔΔΔΔ

Δ

(3557)


- 294 -

Here Δup n +1radial displacement increment of the node on the pellet side of a sliding-state

contact node couple at time tn+1

Δvp n +1axial displacement increment of the node on the pellet side of a sliding-state


Δuc n +1radial displacement increment of the node on the cladding side of a sliding-state


Δvc n +1axial displacement increment of the node on the cladding side of a sliding-state


Equations expressing degrees of freedom of contact nodes i j k and l in the above

equations are

K u F Uni

n n n+ + + += prime +θ( ) ( )Δ Δ1 1

11 (3558)

K u F Unj

n nj

n+ + + += prime plusmnθ μ( ) ( )Δ Δ1 1 1 (3559)

1)(

11)( ˆ

++++ minusprimeΔ=Δ nk

nnk

n UFuK θ (3560)

1)(

11)( ˆ

++++ plusmnprimeΔ=Δ nl

nnl

n UFuK μθ (3561)

Also the m-th row linear equation is minus + =+ +Δ Δu up n c n 1 1 0 (3562)

(5) The stiffness equation when one contact node couple is in a bonded gap state

and all the other couples are in a non-contact state

The stiffness equation is represented as

i

j

k

l

m

n

K

u

v

u

v

U

V

n

p n

p n

c n

c n

n

n

0 0

0 0

1 0

0 1

1 0

0 1

0 0

0 0

0 0 1 0 1 0 0 0 0 0

0 0 0 1 0 1 0 0 0 0

1

1

1

1

1

1

minusminus

minusminus

+

+

+

+

+

+

+

θ

ΔΔΔΔ

=prime

+Δ Fn 1

0

0

(3563)


- 295 -

Here Δup n +1radial displacement increment of the node on the pellet side of a bonded-gap-state


Δvp n +1axial displacement increment of the node on the pellet side of a bonded-gap-state


Δuc n +1radial displacement increment of the node on the cladding side of a bonded-gap-

state contact node couple at time tn+1

Δvc n +1 axial displacement increment of the node on the cladding side of a

bonded-gap-state contact node couple at time tn+1

Stiffness equations expressing degrees of freedom of contact nodes i j k and l in the

above equations are

1)1(

11)( ˆ

++++ +primeΔ=Δ nnni

n UFuK θ (3564)

K u F Vnj

n nj

n+ + + += prime +θ( ) ( )Δ Δ1 1 1 (3565)

K u F Unk

n nk

n+ + + += prime minusθ( ) ( )Δ Δ1 1 1 (3566)

1)(

11)( ˆ

++++ minusprimeΔ=Δ nl

nnl

n VFuK θ (3567)

Also the m-th and n-th rows linear equations are

minus + =+ +Δ Δu up n c n 1 1 0 (3568)

minus + =+ +Δ Δv vp n c n 1 1 0 (3569)

(6) Judgment of contact state and post-processing

A) In cases of a non-contact state at the previous time step Presence of overlap of displacement is checked for Namely

g g u un n p n c n+ + += minus +1 1 1Δ Δ (3570)

where gn+1 gap width at time step tn+1

When gn+1gt 0 Eq (3570) is judged as being converged when gn+1le 0 the time step is

returned to tn+α using the equation

tt t

g gg tn

n n

n nn n+

+

+

= minusminus

+α1

1

(3571)

Then the calculation from time tn+α to tn+1 is performed again after changing of the contact

state to a bonded gap state


- 296 -

B) In cases of a bonded gap state at the previous time step

The state change is judged in terms of a contact force vector Namely when

U p n + lt1 0 and V Up n p n + +lt1 1μ (3572)

are satisfied the contact state is judged as being converged

when

U p n + lt1 0 and V Up n p n + +ge1 1μ (3573)

are satisfied the assumed contact state is changed to a sliding state and the calculation is

performed again Also when

U p n + ge1 0 (3574)

the contact state is changed from a bonded gap state to a non-contact state therefore the time

step is returned to tn+α using

tt t

U UU tn

n n

p n p np n n+

+

+

= minusminus

+α1

1 (3575)

and the calculation from time tn+α to tn+1 is performed again after changing of the contact

state to a non-contact state

C) In cases of a sliding state at the previous time step



are satisfied the contact state is judged as being converged However when


hold the contact state is changed to a bonded gap state

Also when

U p n + lt1 0 and v vp n c n

+ +le1 1 (3578)

are satisfied the assumed contact state is changed from a sliding state to a bonded gap state

Here v p n

+1 axial displacement of a node on the pellet side after the start of contact

vc n

+1 axial displacement of a node on the cladding side after the start of contact

In contrast when


- 297 -

U p n + ge1 0 (3579)

is satisfied the contact state is judged as being changed from a sliding state to a non-contact

state the time step is returned to tn+α using Eq (3575) and the variable values are

interpolated for the time step tn+α then the calculation from time tn+α to tn+1 is performed

again after changing of the contact state to a non-contact state

In a system having several contact node couples as shown in Fig 315 if we try to

converge the contact state for each node couple independently a large number of iterations are

required for convergence of the entire system Therefore the following conditions are added

in order to reduce the number of iterations

1) For node couples in the contact state all of the states are initially assumed to be

bonded gap states

2) Among them if some node couples satisfy the condition V Up n p n + +ge1 1μ the

assumed contact state for such node couples and contact states of node couples

located above them are changed to sliding states

3) At this time directions of sliding are aligned That is friction force is exerted

downward on the pellet surface and upward on the inner surface of cladding

However if all axial components of the contact force of the node couples

which were judged to be in a sliding state when they were assumed to be in a

bonded gap state are in the reverse directions the direction of sliding should be

reversed

With the above method convergence of the entire system can be achieved within less than

3 iterations in general cases

(7) In treating a situation having plural contact pairs In treating the situation having plural contact node pairs as shown in Fig3511 if

contact state convergence is tried at one of the contact pairs independently from the other

contact pairs it is found that a number of iteration calculation is required to obtain an eventual

convergence for all the contact pairs within one time step Then the following additional

condition has been given to obtain convergence in 3 iterations normally

A) For the contact state pairs all the pairs are once assumed to have a bonded state and

equations are solved

B) In these pairs if a node exists which satisfies the condition V Up n p n + +ge1 1μ

this


- 298 -

node pair and its upper node pair are judged as being in a sliding state

C) In this process to have a fast convergence sliding direction is aligned That is it is

assumed that frictional forces are all directing downward at pellet outer surface and upward

at cladding inner surface However if the axial component of contact forces of the contact

pairs which are judged in a sliding state are all directing inversely the sliding direction is

reversed

356 Axial loading - Interaction between target pellet and upper pellet -

The objective axial segment contains normally several pellets and the present 2-D local

PCMI model assumes that every one of the pellets has the same mechanical behavior eg

PCMI Therefore this implies that in calculating deformation of half a pellet length of the

target pellet the upper and lower pellets adjacent to the target pellet are deformed in a

completely identical manner and condition to those of the target pellet Here when the gap

between pellet and cladding falls below a pre-determined (specified) value it is assumed that

the target pellet and its upper adjacent pellet have a bonded state with cladding and the

displacement of the target pellet in the axial direction is given some restraint In other words

ldquolockingrdquo is generated

To model the locking condition among the nodes which are distributed on the upper

surface of the target pellet a displacement increment of node i with the maximum coordinate

value in the axial direction is set to be equal to the displacement increment of the boundary

surface of cladding ie boundary (8) shown in Fig 358 By this setting axial load is

generated for the pellet and cladding This condition is called a locking condition

The procedures for the judgment of contact condition between the target pellet and upper

adjacent pellet as well as the procedures following the judgment are as follows

When the axial coordinate of another node j on the upper surface of the target pellet

exceeds that of node i the axial displacement increment of node j is also set to be equal to the

axial displacement increment of the boundary surface (8) as in the case of the node i That

is the locking condition is set on node j also

At this time the time is returned to the state change point tn+α using the equation

( ) ( )tZ Z

Z Z Z Zt t tn

i n j n

j n j n i n i nn n n+

+ ++=

minusminus minus minus

minus +α

1 11 (3580)

Here

Z Zi n n +1 axial coordinates of node i in the locking condition at time tn and tn+1

Z Zj n j n +1 axial coordinates of node j at time tn and tn+1


- 299 -

From time tn+α to tn+1 the degree of freedom in the axial direction of node j is set as

the same degree of freedom in the axial direction of nodes i which is also the same degree of

freedom in the axial direction of the upper edge surface of the cladding (boundary (8))

namely the common degree of freedom then the calculation is performed again

By this method the locking condition is generally realized by setting of a boundary

condition in which axial displacement increments of a pellet-node group which supports the

axial load on boundary (4) in Fig358 and those of a cladding-node group on boundary (8)

are set to be equal In this procedure the number of degrees of freedom of the pellet-node

group in the axial direction is assumed to be one and those of the cladding-node group is also

assumed to be one That is they are assumed to have common degrees of freedom

The contact force in the axial direction at an arbitrary time of an arbitrary node in the

contact state is given by Eq (352)

That is in

[ ] F B dVn

T

n+ += 1 1σ (352)

since the only load which acts on boundary (4) is the contact force FnPP

+14( ) in the axial

direction

[ ] F F B dVnPP

n

T

n+ + += = 14

1 1( ) σ (3581)

holds

The contact force in the axial direction of a node obtained from Eq (3581) is represented

by Vn+1 When Vn+1lt 0 the locking condition is maintained on this node When Vn+1ge0

the locking condition is judged to be disengaged and the time is returned to tn+α using the

equation

( )tV

V Vt t tn

n

n nn n n+

++=

minusminus +α

11 (3582)

From time tn+α to tn+1 the degree of freedom in the axial direction of this node is set to be

different from the common degree of freedom

Here when the sign of V is positive at all nodes on the edge surface of the pellet the

locking condition is disengaged

The locking condition is set for the second time or later when a node on the edge surface

of the pellet reaches the axial coordinates at the time when the previous locking condition was disengaged Here the load F ( )8 which acts on the upper edge surface of the cladding


- 300 -

(boundary (8)) at the time of the locking condition setting is expressed as

F F F F FR AX W G( ) ( ) ( ) ( ) ( )8 8 8 8 8= + + + (3583)

where F ( )8 axial load which acts on boundary (8)

F R( )8 nodal force generated by setting of the number of degrees of

freedom in the axial direction as 1 at boundary (8) F AX( )8 axial tensile load due to the locking condition at boundary (8)

F W( )8 axial load due to coolant pressure

F G( )8 axial load due to plenum gas pressure

The radial gap width with which the locking condition is

given is adjusted by GAPLK The default value of GAPLK is 3 (μm)

Fitting parameter GAPLK


- 301 -

36 Mechanical Properties Model of Creep and Stress-strain Some important points of the models of cladding creep and stress-strain relationship and

pellet creep are explained below

361 Creep

(1) Pellet The creep strain rate of a pellet can be selected from several models The upper

temperature limit used in the creep calculation can be specified by a name-list Input

Parameter TCS Above this temperature TCS pellet temperature is set equal to TCS in creep

calculation The default value is 127315K Refer to chapter 4

(2) Zirconium-based alloy cladding The creep function f c= ε of cladding in FEMAXI does not contain time t in an explicit

way Therefore if a creep rate equation in literature is used the equation needs to be

transformed into a form in which time t is eliminated from the formulation of creep rate ε c

The creep rate equation can be selected from several model options Refer to chapter 4

(3) SUS alloy cladding creep model A creep equation has been derived from SUS cladding creep data The equation is

constituted by approximate derivation of the dependencies of creep strain on fast neutron flux

and time In this case the equation processing is to a certain degree different from that of the

zirconium-based cladding so that some explanation is described below

1) Creep equation of SUS304(33) Irradiation creep equation of SUS304 in FEMAXI is given by

( ) ( ) 3221 exp CCtECECE +minussdot= ϕϕσϕε (361)

where ε creep strain rate (mmhour)

E = fast neutron energy (=10 MeV)

ϕ = fast neutron flux (ncm2s)

σ = equivalent stress (psi)

t time (hour)

C 1= 17times10-23 C 2= 55times1015 C 3= 27times10-26


- 302 -

Eq(361) is a Black burn type creep equation This type of creep equation consists of the

sum of a primary creep strain ( ) ( )( )( ) 1 exp p f T r T tε σ σ= minus minus and the secondary creep

strain s cs tε ε= and generally expresses as Eq(362)

( ) ( )( )( ) ttTrTf cs

c εσσε +minusminus= exp1 (362)

It is necessary to express this equation in the form of strain-hardening law the following

transformation is made

By differentiating Eq(362)

( ) ( ) ( )( ) cs

c tTrTrTf εσσσε +minussdot= exp (363)

is obtained Here time t cannot be eliminated from Eqs(362) and (363) Then assuming

that ldquocreep hardening depends only on the primary creep amountrdquo we set

( ) ( )( )( )tTrTfH exp1 σσε minusminus= (364)

and substituting this equation into Eq(363)

( ) ( )( )( ) ( )( )

minus=

+minus=HH

cs

Hc

TfTr

TfTr

εσσεεεσσε

(365)

is obtained That is we have a simultaneous differential equation for creep strain cε and

strain-hardening Hε

This is a general solution method of the Black burn type creep equation

By applying this method to Eq(361) and comparing Eqs(363) and (361)

( ) 1f T C Eσ ϕσ= (366)

( ) 2 r T E Cσ ϕ= (367)

3cs C Eε ϕσ= (368)

are obtained

Consequently ( )c FFUNCε = is derived from Eq(365)

( )2 1 3c HE C C E C Eε ϕ ϕσ ε ϕσ= minus + (369)

( )c

FSεσ

part =part

is obtained by differentiating Eq(369) with respect to σ

2 1 3c

E C C E C Eε ϕ ϕ ϕσpart = sdot +part

(3610)

( )c

H FHεε

part =part

is obtained by differentiating Eq(369) with respect to Hε


- 303 -

2c

H E Cε ϕεpart = minuspart

(3611)

Also ( )H GFUNCε = is derived from Eq(365)

( )2 1H HE C C Eε ϕ ϕσ ε= minus (3612)

( )H

GSεσ

part =part

is obtained by differentiating Eq(3612) with respect toσ

2 1H

E C C Eε ϕ ϕσpart = sdotpart

(3613)

( )H

H GHεε

part =part


2H


(3614)

From Eqs(369) to (3614)

( )c FFUNCε = ( )c

FSεσ

part =part

( )c

H FHεε

part =part

( )H GFUNCε =

( )H

GSεσ

part =part

( )H

H GHεε

part =part

are obtained so that the creep strain increment can be obtained from the following equations

Creep strain increment is

1 1 11 1 1 1

c i c i c i c i in n n n n nt C dθ θε ε ε σ+ + +

+ + + + + + Δ = Δ = Δ + (3615)

where

1 1 3 1 1

3


n

f t F tθ θθ

ε σ θ σσ+ + + + + +

+

prime primeΔ = Δ + Δ (3616)

[ ] [ ]C t F Fnc i

n i j n

i

n

i

+ + + += prime + prime

θ θ θ

θ σ σ partσpartσ

Δ 1 1 2 (3617)

and F1 F2 and F3 are

( )1

1 2

1

9

4 1

i i

i ni Hn n n

iiin nn

n Hn

f gt

ffF

gt

θ θ θ

θ θθ

θ


partε

++ + +

+ +++

+


(3618)

Ffn

i

ni2

3

2= +

+

θ

θσ (3619)

( )1 1

3

1

3

21

ii H i

n n nHn

iin

n Hn

ft g

Fg

t

θθ

θ

θ

part εpartε

σ partθpartε

+ + ++

++

+

Δ minus Δ =

minus Δ

(3620)


- 304 -

Since in Eqs(3615) to (3620)

( )f FFUNC= ( )f FSσpart =part ( )H

f FHεpart =part ( )g GFUNC= ( )g GSσ

part =part

( )Hg GHε

part =part 1

c inε +Δ and

c inC θ+ can be determined

(Example of coding the SUS304 creep equation)

SUBROUTINE CCPEQ7 1 (FFUNC F1 F2 F3 GFUNC GS GH EPSH 2 DEPSH TEMP FAI FAIT SIGE CRFAC THETC DTIME 3 NST IFLG ) C----------------------------------- C MATETRIAL SUS304 C SIGE=STRESS(Nm2) C----------------------------------- IMPLICIT DOUBLE PRECISION (A-HO-Z) COMMONTCCR TCCS DATA EBAR10 C117D-23 C255D15 C327D-26 C IF(TEMPGTTCCS) TEMP = TCCS C SIGPSI = SIGE1450389D-4

FFUNC=(EBARFAIC2(C1EBARFAISIGPSI-EPSH) 1 +C3EBARFAISIGPSI) CRFAC

IF(IFLGEQ3) RETURN C FS = (EBARFAIC2C1EBARFAI+C3EBARFAI)1450389D-4CRFAC FH = -EBARFAIC2CRFAC GFUNC = EBARFAIC2(C1EBARFAISIGPSI-EPSH)CRFAC GS = EBARFAIC2C1EBARFAI1450389D-4CRFAC GH = -EBARFAIC2CRFAC IF(IFLGEQ2) RETURN C F1=9(4SIGE2)

1 (FS-FFUNCSIGE+FHTHETCDTIMEGS(1-THETCDTIMEGH)) F2=15FFUNCSIGE F3=15FHSIGE(GFUNCDTIME-DEPSH)(1-THETCDTIMEGH) C RETURN END

【Some cautions in coding】

1) As Eqs(3616) and (3618) to (3620) indicate each equation is divided by the equivalent

stress (σ ) which could cause ldquozero-divisionrdquo when 0σ = Accordingly 0σ ne has to be

designated In the coding example if σ lt10(MPa) σ =10(MPa) is set

2) In the formulation of Eqs(369) and (3612) unit of cε and Hε is 1s In FEMAXI unit


- 305 -

of cε and Hε is given by 1hr so that the unit in the equations has to be multiplied 3600

times (d(hour)d(s)) The same is applied to cε

σpart

part

c

Hε

εpart

part

Hεσ

partpart

and

H

Hε

εpart

part

3) In the formulation of Eqs(3611) and (3613) unit of cε

σpart

part and

Hεσ

partpart

is 1(sMPa)

In FEMAXI unit of cε and Hε is given by 1hr so that the unit in the equations has to

be multiplied by

( )3 3 6( ) ( )36 10 36 10 ( ) 10 ( )( ) ( )

d hr d Pad s d MPa

minus minustimes = times times

362 Model of cladding stress-strain relationship (1) FEMAXI-III model(12)

A basic option (ICPLAS=1 Default) is explained

σ ε= K n

where σ true stress（kgmm2）

ε true strain

n strain hardening exponent

( )( ) ( )1

2202 3344 1 exp yK B kg mmσ= + minus minus Φ (3621)

( )21 14292 10 exp 16 10B φminus minus= times minus times

Φ fast neutron fluence (ncm2) E gt1 MeV

φ fast neutron flux (ncm2s) E gt1 MeV

σ y = Yield stress (kgmm2)

=2160-00213T (recrystalized material)

=3132-00213T (stress-relieved material)

n=00504+00001435T

T temperature (oC) 220 450le leT oC

(2) MATPRO-11 model(34)

ICPLAS=2 specifies the original model of MATPRO-11 and ICPLAS=3 a modified

model of MATPRO-11 Here modification in option ICPLAS=3 is that the yield stress

levels in high burnup region (refer to Fig361) is lowered in accordance with the actual data


- 306 -

00 05 10 15 200

200

400

600

800

1000

1200

1400300K

600K

350K

900K

800K

True

Stre

ss

(MP

a)

True Strain ()

100 s-1

1 s-1

0001 s-1

MATPRO No fluence300K

MATPRO-11 Fluence=13x1026 nm2

FEMAXI 300KFlux=10x1014 ncm2

00 05 10 15 200

200

400

600

800

1000

0001s

0001s

0001s

100s

100s

900K

600K

400K

True

Stre

ss

(MP

a)

True Strain ()

352K100s

(Refer to Fig362) Since the original model of MATPRO-11 constitutes its empirical

equation on the basis of low-burnup cladding data the model predicts unreasonably higher

yield stress of cladding in high burnup region This problem is eliminated by the option

ICPLAS=3 Refer to chapter 4

Fig361 Comparison of stress-strain curves of Zircaloy between MATPRO-11 model and FEMAXI model

Fig362 Modified MATPRO-11 model The yield stress levels are modified on the basis of measured data of irradiated cladding


- 307 -

363 Ohta model(35)

The stress-strain relationship

and parameters of this model are

derived from the uni-axaial

tensile test data of highly

irradiated claddings The basic

form of equation is identical to

that of the MATPRO-11 model

though this model is featured by

incorporation of an intermediate

region connecting the elastic and

plastic regions In this

ldquoconnecting regionrdquo mixture of

elastic and plastic deformations

occurs

The connecting point of

elastic region and intermediate

region is expressed as 1 1( )ε σ

and the connecting point of

intermediate region and plastic


which is shown in Fig363

where the vertical axis is true

stress and horizontal axis is true strain

(1) Stress-strain relationship equation

Elastic region Eσ ε= (3622)

E Youngrsquos modulus

Intermediate region m

nmid

midK

= minus310

εεσ (3623)

Plastic region 310

mnK

εσ ε minus =

(3624)

K strength coefficient

0

300

600

900

1200

1500

0 001 002 003 004True s train [frac]

True

stre

ss [M

pa]

Predict InterfacePlas tic-fit Meas data

00E+0

50E+4

10E+5

15E+5

0 001 002 003 004

True s train [frac]

dσ

dε [M

Pafr

ac]

Predict Interface

Interimregion

Fig363 Connection points among elastic interim and plastic regions and yield point of MATPRO model

( )y yε σ

1 1( )ε σ

2 2( )ε σTru

e s

tress

(M

Pa)


- 308 -

n strain hardening coefficient

ε true strain rate Ifε＜10-5 s then ε＝10-5 s

m strain rate sensitivity coefficient

Hereafter 310

εminus

m

K is expressed as K prime for simplicity

The MATPRO-11model lacks a part indicated by Eq(3623) In the present model for

the cross point ( )y yε σ of Eqs(3622) and (3624) where

1

1ε

minusprime =

n

y

K

E and y yEσ ε= (3625)

if yε εle Eq(3622) is used and if yε εgt Eq(3624) is used

The present model has to define the bound of the intermediate region For this typical

values of 1 02yε εminus = and 2 03yε εminus = are adopted which have been obtained by

summarizing the uniaxial tensile test data In addition the present model adopts a restriction

to avoid that 1ε becomes negative when yε is less than 02 and to have a balance between

the magnitudes of 2 yε εminus and 1yε εminus when 1yε εminus is small as follows

1 max( 0002 05)y yε ε ε= minus times

2 min( 0003 175)y yε ε ε= + times (3626)

Then midK and midn which enable the curved yield surface to include the two points

( )11σε and ( )22 σε are obtained The condition is as follows 0=pε holds at 1ε ε= and

Ep2

2

σεε minus= holds at 2ε ε= Also a relationship 2 2

nKσ εprime= holds for 2ε and 2σ

Thus the yield function for the intermediate region is obtained from Eq(3623)

m

nmid

midK

= minus311 10

εεσ

(3627)

m

nmid

midK

= minus322 10

εεσ

(3628)

Dividing the both sides of Eq(3628) by Eq(3627) to obtain midn gives

midn

=

1

2

1

2

εε

σσ

(3629)

Consequently

( ) ( )( ) ( )12

12

lnln

lnln

εεσσ

minusminus=midn (3630)


- 309 -

mn

mid

mid

K

=

minus32

2

10

εε

σ

(3631)

are obtained Namely this process defines the curved yield surface which includes the two

points ( )11σε and ( )22 σε The interchange between the intermediate region and yield region

is carried out by comparing the magnitude relation between ( )ε σ and ( )22 σε

The above process permits the determination of the true stress by using Eqs(3622)

(3623) and (3624) in accordance with the relationship 1ε εle 1 2ε ε εlt le and 2ε εlt

when true strain ε is given Inversely when true stress is given and true strain is to be

determined Eqs(3622) (3623) and (3624) are used in accordance with the relationship

1σ σle 1 2σ σ σlt le and 2σ σlt

(2) Strength factor K (MPa)

A) K of stress-relieved (SR) cladding of PWR rod K is assumed to depend on temperature T and fast neutron fluence Φ（gt10MeV） and

given as follows

T≦900 K （62685）

1 2

253 4

( 27315)

( 27315) 1 exp( 1751 10 )

K L L T

L L T minus

= minus minus

+ minus minus sdot minus minus times Φ

1 2 3 41745 2517 526 0748L L L L= = = = (3632)

K MPa T Kelvin Φ neutronsm2 (gt1MeV)

900＜T＜1000 K (62685～72685 oC)

1 2 3 4

05251 2 3 4

(

( ) 1 exp( 1751 10 )

K A T A T A TA

B T B T B TB minus

= + + +

+ + + + times minus minus sdot Φ (3633)

A1＝ 3041233371times104 B1＝ 33282432 times102

A2＝－8285933869times101 B2＝ 298027653

A3＝ 7490958541times10－2 B3＝－681286681times10－3

A4＝－2242589096times10－5 B4＝ 351229783times10－6

1000 K≦T （72685oC~） 0525

1 2 3 4exp( ) 1 1 exp( 1751 10 )K H H H T H minus = + sdot + minus minus sdot Φ

61 2 3 41 10 8775 8663 0341566H H H Hminus= sdot = = = (3634)

【Note 1】The MATPRO-11models the strength factor K as a value for non-irradiated

non-cold worked material Effects of irradiation and cold-work are taken into account by the


- 310 -

additional factor DK and sum of the two K DK+ is equivalent to K in Eq(3624)

On the other hand K of the present model includes the effects of irradiation and

cold-work Particularly for the effect of cold-work when fluence is null the present model

gives K value which corresponds to the manufactured condition of SR material ie cladding

with 8 to 10 cold-work Furthermore the fluence-dependent term includes the property that

the effect of cold-work diminishes with fluence

In the present model the followings are assumed

K value over 1000K is the same as that of high temperature region（gt900K） of the

MATPRO model However to have a continuity of Eq(3632) K value increases with

irradiation at the same rate of fluence dependence at 900K

In 900 to 1000 K the factor is determined so as to have continuities of both K value and its

temperature-differentialdK

dT at 900 and 1000K This is identical to the method of

MATPRO-11model which interpolates the 730 to900K range

B) K of re-crystallized (RX) cladding of BWR rod

Dependencies on temperature T and fast neutron fluence Φ (gt10MeV) is given as follows

T≦700 K （42685）

1 2

05263 4

( 27315)

( 27315) 1 exp( 9699 10 )

K L L t

L L T minus

= minus minus

+ minus minus times minus minus sdot Φ

(3635)

L1＝1201 L2＝2508 L3＝1119 L4＝0627

700＜T≦800 K （42685～52685）

1 2 3 4

05265 6 3 4

(

( ) 1 exp( 9699 10 )

K A T A T A TA

A T A T A TA minus

= + + +

+ + minus + times minus minus sdot Φ (3636)

A1＝ 3656432405times104 A2＝－1361626514times102

A3＝ 1695545314times10－1 A4＝－7055897451times10－5

A5＝－3338799880times104 A6＝ 1330276514times102

800＜T＜950 K （52685～67685）

For non-irradiated cladding

1 2 3exp( )non irradiatedK H H H Tminus = + (3637)

61 2 31 10 4772 9740H H Hminus= sdot = =


- 311 -

For cladding with fluenceΦ

0526

2 3 4( 1 exp( 9699 10 )

non irradiated

non irradiated

K K

B T B T B TB K

minus

minusminus

=

+ + + + minus times minus minus sdot Φ (3638)

B1＝－3617377246times104 B2＝ 1361156965times102

B3＝－1636731582times10－1 B4＝ 6386789409times10－5

950K≦T （67685oC≦T）

0526

1 2 3 4exp( ) 1 1 exp( 9699 10 )K H H H T H minus = + times + minus minus sdot Φ (3639)

61 2 3 41 10 4772 9740 5267783H H H Hminus= sdot = = =

【Note 2】Assumption with respect to temperature dependence of K value is it is ldquoa linear

function of temperaturerdquo in low temperature region proportional to exp(-QT) at the highest

temperature region and these two regions are connected by a cubic function of temperature

This is similar to the case in the SR material

However in non-irradiated material RX material has a smaller K value than SR material

so that if K of the linear function of RX material is extrapolated with temperature K=0 is

obtained at 400 to 500oC To circumvent this it is connected to the function of exp(-QT) at

lower temperature than the equation of SR material

In the above description for non-irradiated material Eq(3636) is a connecting point by

the cubic function while increment term by irradiation has a connecting point indicated by

Eq(3638) For the region in which K value is proportional to exp(QT) Q is adjusted so as

to obtain a continuity with the extrapolation of the linear function

(3) Strain hardening exponent n ldquonrdquo is calculated by the following equations assuming that it is dependent only on

temperature T but has no dependence on irradiation

T≦67315 K (400oC)

1 2 ( 27315)n C C T= minus minus (3640)

67315ltTlt1000 K （400～72685）

3 4 5( )n C T C C T= + + (3641)

C1＝0213 C2＝1808times10－4 C3＝34708518times10－2

C4＝49565251times10－4 C5＝－50245302times10－7


- 312 -

1000 K≦T （72685～）

n = C6 = 0027908 (3642)

【Note 3】

A) n value above 1000K is set equal to that of high temperature region (gt850K) of the

MATPRO model

B) In the intermediate temperature range 67315 to 1000 K the coefficients are determined to

satisfy the condition that at the both bounds of temperature the ldquonrdquo value be continuous and at

the same time dndT be continuous at 67315K Since if this range is narrow ldquonrdquo value

changes too fast with temperature and results in an unusual fluctuation in the yield stress ie

function of temperature the temperature to settle the n value is shifted to a higher region than

the MATPRO model

(4) Strain rate sensitivity exponent m

ldquomrdquo has not been changed from the MATPRO model and is calculated by the following

equations assuming that it is dependent only on temperature but has no dependence on

irradiation

T≦730 K （45685） m＝002 (3643)

730ltT＜900 K （45685～62685）

5 6 7 8( )m A T A T A A T= + + + (3644)

900 K ≦T （62685～）

9 10m A A T= + m＝－647times10－2＋2203times10－4T (3645)

A5＝ 2063172161times101 A6＝－7704552983times10－2

A7＝ 9504843067times10－5 A8＝－3860960716times10－8

A9＝ -067times10－2 A10＝－2203times10－4

Name-list Input Parameter ICPLAS Option for cladding stress-strain (plasticity) equations ICPLAS=1 FEMAXI-III model(Default) =2 MATPRO-11 model

=3 modified MATPRO-11 model =4 Ohta model =41SUS304


- 313 -

364 Method to determine yield stress in FRAPCON and MATPRO models (ICPLAS=3 6)

When temperature changes and concurrently stress increases and shifts from elastic

region to plastic region the yield stress level varies from one moment to another In this

situation the FEMAXI calculation determines the yield stress by calculating the intersection

point of the line of elastic stress-strain relationship and the curve of plastic stress-strain

relationship In the case where plastic strain is already generated the yield stress is obtained

by the relation shown in Fig364

Fig364 illustrates the relationship between Eq (3646)

310

mnK

s

εσ ε minus =

(3646)

where σ true stress (Pa)

ε true strain (-)

ε true strain rate (1s)

and Eq(3647) of the line of elastic strain-stress

eEεσ = (3647)

The stress at the intersection point of these two lines is the yield stress sought

In Eq(3646) representing 310

εminus

m

K collectively by K prime allows to express Eq(3646)

as

nK εσ prime= (3648)

Using Eqs(3648) and (3649) the following derivation is performed for the

Newton-Raphson method

( ) σεσ minusprime= nKf (3649)

( )

11 minusprime

= minusn

E

Kn

d

df εσσ

(3650)

and ( )( )σσσσσ

d

dff

n

nnn minus=+1 (3651)

At each time step iteration calculations are done until convergence nn σσ asymp+1 is obtained

As the iteration process illustrated in Fig364 the yield stress Yσ is obtained as a converged

solution


- 314 -

σ=Kε nσ=Eεσ=Eε+cσ=σγ

ε

σ σ0

σ1σ2 σ3

εp

nK εσ =

eEεσ =

Yσσ =cE e += εσ


ε

σ σ0

σ1σ2 σ3

εp

nK εσ =

eEεσ =

Yσσ =cE e += εσ

Fig364 Determination of yield stress by iteration


- 315 -

37 Skyline Method

A structural analysis by FEM eventually comes down to the solution of simultaneous

linear equations such as Eqs(347) and (3547) Except the equilibrium condition equation

for the contact force degree of freedom of the simultaneous equations is equal to the degree

of freedom of nodes so that matrix is symmetrical Only in the case in which ldquocontactrdquo

boundary condition is added as shown in Eqs(347) and (3447) non-symmetrical

components appear In FEMAXI Skyline method is adopted to solve the simultaneous

equations This method is the one which reduces the working area and computing number of

times by improving the Gaussian elimination algorithm

The equations to be solved are generally written as follows

(0) (0) (0) (0)11 1 12 2 1 1

(0) (0) (0) (0)21 1 22 2 2 2

(0) (0) (0) (0)31 1 32 2 3 3

(0) (0) (0)1 1 2 2

(1)

(2)

(3)

n n

n n

n n

n n nn n

a x a x a x b

a x a x a x b

a x a x a x b

a x a x a x

+ + sdotsdot sdot sdot sdot sdot sdot sdot + =


+ + sdotsdot sdot sdot sdot sdot sdot sdot + =sdotsdotsdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot

+ + sdotsdot sdot sdot sdot sdot sdot sdot + = (0) ( )nb n

(371)

The suffixes at the upper right of each coefficient ija and known loading vector increment

ib are added in order to clarify the elimination process

The first step of elimination is to make the terms 0 suffix having 1x in equations(2) (3)

hellip (n) or to make

(0) (0) (0)21 1 31 1 1 1 na x a x a xsdot sdot sdotsdotsdot sdot (372)

0

To eliminate 1)0(

21 xa by multiplying Eq(372) with )0(11

)0(21 aa and subtract it from

Eq(372) we obtain

(0) (0) (0)

(0) (0) (0) (0) (0) (0)21 21 2122 12 2 2 1 2 1(0) (0) (0)

11 11 11n n n

a a aa a x a a x b b

a a a

minus + sdotsdot sdot sdot sdot + minus = minus

(373)

Then to eliminate 1)0(

31 xa multiplying eq(1) with )0(11

)0(31 aa and subtracting this from

Eq(373) are performed This process is repeated until it reaches the elimination of 1)0(

1 xan

then we have the following form of equations


- 316 -

(0) (0) (0) (0)11 1 12 2 1 1

(1) (1) (1)22 2 2 2

(1) (1) (1)32 2 3 3

(1) (1) (1)2 2

(1)

(2)

(3)

( )

n n

n n

n n

n nn n n

a x a x a x b

a x a x b

a x a x b

a x a x b n

prime + + sdotsdot sdot sdot sdot sdot sdot sdot + =

prime+ sdot sdot sdot sdot sdot sdot sdot sdot + = prime+ sdot sdot sdot sdot sdot sdot sdot sdot + = sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot prime+ sdot sdot sdot sdot sdot sdot sdot sdot + =

(374)

where (0) (0) (0) (0)

(1) (0) (1) (0)21 12 31 1222 22 32 32(0) (0)

11 11

(0) (0) (0) (0)(1) (0) (1) (0)21 1 31 12 2 3 3(0) (0)

11 11

a a a aa a a a

a a

a b a bb b b b

a a

= minus = minus sdotsdot sdotsdot sdot sdot sdot

= minus = minus sdotsdot sdot sdotsdot sdot sdot

In the next step of elimination terms )1(32a 2x helliphellip )1(

2na 2x having x in Eqs(373)rsquo

(374)rsquo helliphellip (n)rsquo are made 0 This procedure is quite similar to that of the first step except

the first line (1) if the lines (2)helliphellip (n) are considered newly given equations

This sequential elimination will finally come down to the following simultaneous

equations which have a triangle in its upper right region

=

=+

=+++

=++++

minusminus

minusminusminus

minusminus

minusminus

)1()1(

)2()2()2(

)1(2

)1(21

)1(122

)1(22

)0()0(11

)0(112

)0(121

)0(11

nnn

nnn

nnn

nnnn

nnn

nnnn

nnnnn

bxa

bxaxa

bxaxaxa

bxaxaxaxa

　　　　　　　　　　　　

　　　　　　　　

　　　　

　　　

(375)

Here each coefficient can be generally written as follows

Nkj

Nki

a

aaaa k

kk

kkj

kikk

ijk

ij

1)1(

)1()1()1()(

=+=

minus= minus

minusminusminus (376)

Nkiba

abb k

kkkk

kikk

ik

i 1)1()1(

)1()1()( +=minus= minus

minus

minusminus 　 (377)

It is easy to solve Eq(375) From the last equation we can directly obtain

)1(

)1(

minus

minus

minusn

nn

nn

n a

bx (378)

By substituting this into the second equation from the last 1minusnx is obtained In general


- 317 -

by using the following the equation

)1(2

)1(2

)1(1

)1(

minus+

minus++

minus+

minus minusminus= k

kk

kkkkk

kkk

kk

k a

xaxabx

(379)

and sequentially substituting the already obtained solution all the solutions up to 1x are

obtained

The above procedure is called the Gaussian elimination method In this method every

time elimination is performed re-writing the coefficients is required

However by using the following process ie Crout reduction )(kija and )(k

ib of

Eq(375) can be directly derived from their initial values )0(ija and )0(

ib

Now Eqs(376) and(377) are expressed as follows by using the relationship 1minus= ik

which is clearly seen in Eq(375)

)2(11

)2(1

)2(1)2()1(

minusminusminus

minusminus

minusminusminusminus minus= i

ii

iji

iiii

iji

ij a

aaaa

)2(1)2(

11

)2(1)2()1( minus



iiii

iiii

ii

i ba

abb

For example the followings can be written

)0(11

)0(1

)0(21)0(

2)1(

2 a

aaaa j

jj minus=

=

minus

minusminus

minus=

minusminus=minus=

2

1)1(

)1()1()0(

3

)1(22

)1(2

)1(32

)0(11

)0(1

)0(21)0(

3)1(22

)1(2

)1(32)1(

3)2(

3

kk

kk

kkj

kik

j

jjj

jjj

a

aaa

a

aa

a

aaa

a

aaaa

　　

so that in general

minus

=minus

minusminusminus minus=

1

1)1(

)1()1()0()1(

i

kk

kk

kkj

kik

iji

ij a

aaaa (3710)

holds Similarly

minus

=

minusminus

minusminus minus=

1

1

)1()1(

)1()1(

i

k

kkk

kk

kik

ii

i ba

abb (3711)

holds

By substituting nji 32 = sequentially all the values up to )1( minusnnna )1( minusn

nb are

obtained


- 318 -

The coefficient matrix of simultaneous equations which is to be solved in FEMAXI contains partially asymmetrical components For the symmetrical components )1( minusi

ija and )1( minusi

ib are determined by the following equations

ji

mj

a

aaaa

i

kk

kk

kkj

kik

iji

ij

2

21

1)1(

)1()1()0()1(

==

minus= minus

=minus


miba

abb

i

k

kkk

kk

kik

ii

i 21

1

)1()1(

)1()1( =

minus= minus

=

minusminus

minusminus (3713)

where =m degree of freedom of symmetrical part of the matrix

For the asymmetrical region the following equations are applied

( 1) ( 1)1

( 1) (0)( 1)

1

for 2

for 2

k kiik kji

ij ij kk kk

i m j na aa a

j m i na

minus minusminusminus

minus=

gt == minus gt =

　 (3714)

nmiba

abb

i

k

kkk

kk

kik

ii

i 1

1

)1(

)1(

)1()1( =

minus= minus

=

minusminus

minusminus (3715)

where =n all the degree of freedom of matrix

In this procedure also for the process of Eq(375) a backward substitution is performed

similarly to Eqs (378) and (379)

Here the coefficient matrix is supposed to be schematically shown in Fig371 and

memorizing the coefficients surrounded by solid lines in a 1-dimensional array is considered

Fig371 Schematic coefficient matrix

Assuming that the line number is jL in which non-zero component appears for the first

times timestimestimestimes０timestimes００００times０times００timestimes００００

timestimes００times０００００timestimes０timestimes００００

timestimestimestimestimestimes０timestimes０timestimestimestimestimes

times０timestimes０timestimestimestimestimestimes

timestimestimestimestimes０times

timestimestimes

i ｊ

Symmetry

-Li

-Lj


- 319 -

time in the j -th column the coefficient ija in the j -th column is memorized in the following

number position of 1-D array

( )=

+++minusj

kkLkji

1

1 (3716)

because in the k -th column the number of coefficients to be memorized is 1++ kLk and

memory address of jja can be written as ( )=

++j

kkLk

1

1

This memory method and accompanied solution method of simultaneous equations is

called Skyline method In the followings solution method of this memorizing system is

described

We consider to apply the process ie Grout Reduction expressed by Eqs(3712) and (3713) to the j -th column jL and iL are set as the line numbers in which non-zero

coefficient appear for the first time respectively in the i -th and j -th columns Then since

ija is not modified if either ika or kia is 0 the summation Σ of these equations does not have

to begin with 1=k but can adopt a larger one of jL or iL Furthermore because

0=rja when jLr lt the first non-zero coefficient of the j -th column remains unmodified

Therefore evaluation can be done from 1+= jLi This process ie modified Crout

reduction is expressed as follows

minus

=

minusminusminus minus=1

)1()1()0()1(

0

î

kk

kkj

kkiij

iij aaaa (3717)

where )1(

)1()1(ˆ minus

minusminus =

kkk

kkik

ki a

aa

mj 2=

11 minus+= jLi j

( )ji LLk max0 =

mjaaaaj

Lk

kkj

kkjjj

ijj

j

2ˆ1

)1()1()0()1( =

minus= minus

=


and

( )bi

bj

kk

kk

kkii

ii LLk

mLibabb

max

1ˆ

0

1)1()1()0()1(

0=

+=

minus= minus

=


=bL line number of a first non-zero component of vector b


- 320 -

Here memorized components are set as

mLia

bb bi

ii

iii )1(

)0()1( =

= minusminus (3720)

The backward substitution process is as follows

( 1) 1kk kx b k mminus= = (3721)

( 1) 1 2ˆ

1 1k

k k ki ii i

i m mx x a x

k L L iminus = minus

= minus= + minus

(3722)

Here every time Eq(3717) is applied )1( minuskkia in the above equations is converted into

)1()1()1(ˆ minusminusminus = kkk

kki

kki aaa and memorized

Next it is considered that the upper limit of the number of procedure repetitions in the

above calculation of )1( minusi

ija )1( minusi

ib is set equal to the number of equations to be eliminated This

procedure ie partial reduction is expressed as follows

Coefficient matrix in the symmetrical part of matrix is

minus

=


)1()1()0()1(

0

î

kk

kkj

kkiij

iij aaaa (3723)

where

)1(

)1()1(ˆ minus

minusminus =

kkk

kkik

ki a

aa (3724)

mj 2=

( )jpLi j min1 +=

( )ji LLk max0 =

=p number of equations eliminated in the symmetrical

part of matrix

and

pjaaaaj

Lk

kkj

kkjjj

ijj

j

2ˆ1

)1()1()0()1( =

minus= minus

=


where

( )jpLk

mi

a

aa

jk

kk

kkik

ki min

2ˆ

)1(

)1()1(

==

= minus

minusminus (3726)

Eq(3726) is evaluated concurrently with Eq(3725) from 2=j to pj =


- 321 -

From 1+= pj to mj = evaluation of Eq(3726) is conducted after the procedure of

Eq(3725)

On the other hand the known vectors are

minus

=


)1()1()0()1(

0

î

kk

kk

kkii

ii babb (3727)

where pLi b 1+=

( )bi LLk max0 =

and also

pLia

bb bi

ii

iii

i )1(

)1()1( =

= minus

minusminus (3728)

is set

In the asymmetrical part of matrix

minus

=


)1()1()0()1(

0

î

kk

kkj

kikij

iij aaaa (3729)

holds where

)1(

)1()1(ˆ minus

minusminus =

kkk

kikk

ki a

aa (3730)

in nmj 1+=

( )jpLi nj min1 +=

in nmi 1+=

( )jpLj ni min1 +=

and

n

j

Lk

kkj

kjkjj

ijj ppjaaaa

j

1ˆ1

)1()1()0()1( +=

minus= minus

=


where

( )jpLk

nmj

a

aa

njk

kk

kjkk

ki min

1ˆ

)1(

)1()1(

=+=

= minus

minusminus (3732)

=np total number of equations eliminated in both the

symmetrical and asymmetrical parts of matrix

The known vectors are


- 322 -

n

i

kk

kk

kkii

ii ppibabb 1ˆ

1)1()1()0()1(

0

+=

minus= minus

=


where niii

iii ppi

a

bb 1)1(

)0()1( +=

= minusminus (3734)

is set

The alternate substitution in the above procedures is conducted by the following

sequence

( 1) 1kk k nx b k pminus= = (3735)

( )( 1)

1 2ˆ

min 1k

k k ki ii n

i n nx x a x

k L p iminus = minus

= minus= minus

(3736)

The procedure expressed by Eq(3723) to Eq(3736) is the Skyline solution method of

simultaneous equations in FEMAXI

References 3

(31) Rashid YR Tang HR and Johansson EB Mathematical Treatment of Hot

Pressing of Reactor Fuel NuclEngDes29 pp1-6 (1974)

(32) Pugh Current Recommended Constitution Equations for Inelastic Analysis of FFTF

Components ORNL-TM-3602 (1972)

(33) Refer to the Chap4 Section 4 materials properties of SUS304

(34) Hagrman DL and Reyman GA MATPRO-Version11 A Handbook of Materials

properties for use in the analysis of light water reactor fuel rod behavior

NUREGCR-0497 TREE-1280 Rev3 (1979)

(35) Ohta A Development of irradiated cladding stress-strain model for LWR fuel analysis

code Fuel Safety Research Meeting Tokai April 20-21 (2006)


- 323 -

4 Materials Properties and Models

Material properties empirical equations and models usable in FEMAXI-7 are listed in

the tables below by categories ie pellets claddings and others Items listed in the column of

ldquoConditionrdquo are name-list parameters it is possible to designate material properties and

models etc by assigning the parameter values For items labeled ldquoNo restrictionsrdquo there is no

selection by means of name-list parameter ie materials properties are uniquely fixed

41 UO2 Pellet (1) Thermal conductivity UO2 thermal conductivity (1) MATPRO-Version 09 Condition Ref

for a temperature range of 0minus1650degC

( ) ( )K

D

TT=

minus minusminus

sdot+

+ times times

minus minus1 1

1 0 05

40 4

4641216 10 1867 104 3

ββ

exp


( ) ( ) K

DT=

minus minusminus

sdot + times timesminus minus1 1

1 0 050 0191 1216 10 1867 104 3β

β exp

Kthermal conductivity of UO2 pellet (WcmK)

DUO2 theoretical density ratio

Ttemperature (degC)

βporosity coefficient(213) )(10580582 3 CTTminustimesminus=β

Default

value

IPTHCN=1

213

UO2 thermal conductivity (2) Washington

IPTHCN=2

41 4 1 12 3(0040 257 10 ) 726 10 (WmK)K KK F T Tminus minus minus= sdot + times sdot + times sdot

F P= minus sdot10 2 5

P porosity=10 ndash (actual UO2 densitytheoretical density) TK temperature（K）

UO2 thermal conductivity (3) Hirai

IPTHCN=3

42 95 (1 ) (1 005 ) (WmK)TDK K pβ β= sdot minus minus sdot

K T TTD K K952 4 1 12 32 35 10 2 55 10 357 10= times + times sdot + times sdotminus minus minus minus( )

)(10580582 3 CTTminustimesminus=β rarrRef(213)

P porosity TK temperature（K）


- 324 -

UO2 thermal conductivity (4) Wiesenack (Halden) Condition Ref

4 7 01148 00035 (2475 10 8 10 )

00132exp(000188 ) (WmK)

K F B B T

F T

minus minus= + sdot + times minus times sdot

+ sdot sdot

F p= minus minus sdot( ) ( )1 1 0 05β β


P porosity Bburnup（MWdkgUO2) Ttemperature (C)

IPTHCN=4 43

UO2 thermal conductivity (4A- Revised) Wiesenack (Halden)

IPTHCN=17 215

4 01148 00040 (2475 10 000333 )

00132exp(000188 ) ( )

K F B B

F T W mK

minus= + sdot + times minus sdot Θ

+ sdot sdot


β = minus times minus2 58 0 58 10 3 T rarrRef(213)

P porosity Bburnup (MWdkgUO2) min( 1650)TΘ =

T(C)

UO2 thermal conductivity (5) Modified Hirai

IPTHCN=5 42

95 (1 ) (1 005 ) (WmK)TDK K pβ β= sdot minus minus sdot

12 3 4 795

312

(235 10 35 10 ) (255 10 80 10 )

357 10

TD K

K

K B B T

T

minusminus minus minus minus

minus

= times + times sdot + times minus times sdot sdot

+ times sdot

3258 058 10 ( )T T Cβ minus= minus times rarrRef(213)

P porosity TK temperature (K) Bburnup (MWdkg-U)

A provisional model in which burnup dependence similar to the Halden

model(43)(44) is added to the original Hirairsquos model(42)

UO2 thermal conductivity (6) Forsberg

IPTHCN=6 44

13 4 80047 173 10 (25 10 541 10 )

(WmK)

KK F B B Tminusminus minus minus= + times sdot + times minus times sdot

F p= minus sdot10 β



UO2 thermal conductivity (7) Kjaer-Pedersen

IPTHCN=7 45

K T P G B TK K= sdot + + sdot + sdot + sdot sdotminus sdot93 0 0224 11 56 3 2 5 10 14 4 ( ( ))

2

05

10 05( 10)

10 (( 75) 10)

B G B

B G B

le =ge = minus

P porosity TKtemperature (K) Bburnup (MWdkg-U)


- 325 -

UO2 thermal conductivity (8) Baron amp Couty Condition Ref

( ) ( )[ ] ( )TDCTBTATK sdot++= minus 31 (WmK)

23210 gAgAxAAA +++=

2210 gBgBBB ++=

A0=00254 A1=40 A2=03079 A3=122031

B0=2533x10-4 B1=8606x10-4 B2=-00154

C=10x10-12

g Gd content (weight ) x absolute value of (2-OM) ( ) ( )( ) ( )( )TTPTD αα 05011 minussdotminus=

( )( )

327384 058 10 ( 127315 )

20 ( 127315 )

T T T K

T T K

αα

minus= minus times sdot le

= gt

IPTHCN=8 46

UO2 thermal conductivity (9) Ohira amp Itagaki (NFI)

IPTHCN=9 214

( ) ( ) ( )ThBugBufTBA sdot++sdot+=λ1 (mKW)

( ) BuBuf sdottimes= minus310871

( ) 2800380 BuBug sdot= Bu Burnup (GWdtU)

( ) ( )

2 4

1 temperature(K)

1 396exp 6380

452 10 (mKW) 246 10 (mKW)

h T TT

A Bminus minus

=+ minus

= times = times

Latest Model for 95TD UO2 and MOX pellets (10)

Ohira amp Itagaki (NFI)

IPTHCN

=91 410

Latest model

( ) ( ) ( )2 4

1

Gd P

C D

A B T f Bu g Bu h T

T T

λα β

=+ sdot + sdot + sdot + + sdot

minus + (WmK)


( ) ( )2 4(807 10 651 10 ) 1 exp( 0142 )g Bu Bu Buminus minus= sdot + sdot sdot times minus minus sdot

Bu Burnup (GWdtHM) T Kelvin

( ) ( )

2 4

1

1 396exp 6380

453 10 (mKW) 246 10 (mKW)

h TT

A Bminus minus

=+ minus

= times = times

C=547E-9 (Wm-K3) D=229E-14 (Wm-K5)

3812 10α minus= sdot Gd Gd2O3 concentration (wt) 3124 10β minus= sdot P PuO2 concentration (wt)

【Note】This equation is applicable to 95TD pellet In general

apply the followings )0501()1(95 ββ minusminussdot= pKK TD 3 o258 058 10 ( C)T Tβ minus= minus sdot


- 326 -

UO2 thermal conductivity (10) Lucuta Matzke amp Hastings Condition Ref

( )

( )

( ) ( )

( ) ( )( )

( )

1 1 2 3 4 0

9

0 4 2

1 3265 3265

1

2

W mK

1 4715 10 16361exp W mK

00375 2165 10

109 00643 1arctan

109 00643

0019 11

3 0019 1 exp 1200 100

1 15 for s

1 1

d P P x r

d

P

P

K K K K K

T T T

K TT

KT

PK

P

λ λ

λ

ββ β β β

ββ

σσ

minus

=

times = + minus + times sdot = + +

= + sdotminus + minus minus

minus= =+ minus

( )

( )( )3 4

4

pherical pore porosity

021 1

1 exp 900 80

temperature (K)

burnup by atom 1067 10 MWdtU

x r

P

K KT

T

Bu Buβ minus

= = minus+ minus

= times sdot

IPTHCN=10 47

UO2 + Gadolinia thermal conductivity (13) Fukushima

IPTHCN=13 48

KaT b

=+

1 thermal conductivity (WmK)

Ttemperature (degC)

a bfunction of Gd2O3 concentration (see table below)

Gd2O3 (wtfraction) a b

00 232733times10minus4 1180229times10minus1




In order to interpolate a and b for Gd2O3 concentrations the values of a and b in the above table are substituted into K rAGd= to obtain r and A so that the two values of K may coincide


- 327 -

UO2 thermal conductivity (14) Baron Condition Ref

( )( )95 22 20 1 2 3 0 1 2 3

1exp (WmK)

1TD

C Dg WK

T kTA A x A g A g B B q B g B g T

+ = + minus + + + + + + +

T = temperature (K)

x = absolute value of the stoichiometric deviation abs[2-OM]) =Y

q = plutonium weight content (-) = PU

g = Gadolinium weight content (-) 0-12 =GD

k = Boltzmann constant (138E-23 JK)

W= 141MeV = 1411610-19 J

A0 = 0044819 + 0005BU+20(G - 54702) (mKW)

B0 = (24544 - 00125BU - 70(G - 54702))10-4 (mW)

A1 = 4 (-) B1 = 08 (-)

A2 = 0611 (mKW) B2 = 960310-4 (mW)

A3 = 11081 (mKW) B3 = -176810-2 (mW)

C = 5516109 (WKm) D = - 43021010 (WKm)

BU = burn-up (GWdtU)

G = the lattice parameter in

A =LT

This equation is for 95TD pellet In general case K is given by )0501()1(95 ββ minusminussdot= pKK TD

)(10580582 3 CTT minussdotminus=β

[Note 1 Baronrsquos equation can be used also in MOX fuel by MOX=2]

[Note 2 Y PU GD and LT belong to the name-list parameters

designated by input Default value of LT is 54702]

IPTHCN=15 49

In the case of IPTHCN=13 or 15 the following value is input

Variable name Content Default value

GD Gd2O3 concentration (Weight fraction) 00

Even if MOX=0 (=Default value) designating the parameter ldquoPUrdquo in Baronrsquos model

(IPTHCN=15) permits the effect of Pu in calculation


- 328 -

UO2 thermal conductivity (15) Kitajima and Kinoshita Condition Ref

1

mat ek

K KA B T

= ++ sdot

(kWmK) for 100TD

where ( )5 31216 10 exp 1867 10eK Tminus minus= times times

Ke is the electronic conduction term using MATPRO-11 model For a pellet with less than 100TD

matK F K= sdot F p= minus sdot10 β

β = minus times minus2 58 0 58 10 3 T rarr Ref(213)

1)

12731540 412 ( )022 ( 50)0447 000453 (50 80)0084 (80 )

kT KA Bu Bu MWd kgUB Bu

Bu BuBu

lt= + sdot= le= minus sdot lt le= lt

2) 147315

40 174022 00003

kT KA BuB Bu

gt= + sdot= + sdot

3) 127315 147315kT Kle le Kmat is determined by a linear

interporation between the two values at the two terminal temperatures Bu is burnup assuming fission power =207MeVfission

IPTHCN=16 238

UO2 熱伝導率 in FRAPCON-3 PNNL modified Halden model

IPTHCN =18 411

( )95 1 ( ) 00132exp 000188 (WmK)K Q x T= + for 95TD pellet 4( ) 01148 gad 11599 00040B 2475 10 (1 00033B)Q x a x minus= + sdot + + + sdot minus Θ

11599a = ldquogadrdquo= weight fraction of gadolinia

x = 200-OM (oxygen to metal ratio) =Y (Default=00) B burn-up (MWdkgUO2) min( 1650)TΘ = T degree C

(2) Pellet thermal expansion

UO2 thermal expansion (1) MATPRO-11 Condition Ref

ΔL

LT T T= minus times + times + times + timesminus minus minus minus4 972 10 7107 10 2 581 10 1140 104 6 9 2 13 3

ΔL

Llinear expansion rate T temperature (degC)

IPTHEX=0 (default)

IPTHEX=1

34


- 329 -

UO2 thermal expansion (2) Burdick and Parker Condition Ref

ΔL

LT T= minus times + times + timesminus minus minus30289 10 8 4217 10 21481 104 6 9 2

T temperature (degC) for the range 27minus1260degC

ΔL L linear thermal expansion normalized by the length at 27degC

IPTHEX=2

UO2 thermal expansion (3) Halden

ΔL

LT T T= times + times + timesminus minus minus6 0 10 2 0 10 17 106 9 2 12 3

Ttemperature (degC) for the range 900minus2800degC (melting point)

IPTHEX=3

UO2 thermal expansion (4) Conway and Fincel

ΔL

LT T= minus times + times + timesminus minus minus1723 10 6 797 10 2 896 104 6 9 2

Ttemperature (degC) for the range 1000minus2250degC

IPTHEX=4

UO2 thermal expansion (5) MATPRO-A

IPTHEX=5 211

Thermal strain(=0 at 300 K) 1 2 30

exp DL EK T K K

L kT

Δ = minus + minus

T temperature (K) k Boltzmann constant 23(138 10 )J Kminustimes

ED =69x10-20 energy of formation of a defect (J)

K1 = 10x10-5 K-1 K2 = 30x10-3 K3 = 40x10-2

UO2 thermal expansion (6) Martin

IPTHEX=6 426

273 923T Kle le 3 6 10 2

13 3

( ) (273) 266 10 9802 10 2705 10

4391 10

L T L T T

T

minus minus minus

minus

Δ = minus sdot + sdot minus sdot+ sdot

1 6

10 2 13 3

( ) (273) (99734 10 9802 10

2705 10 4391 10 )

L T L T

T T

minus minus

minus minus

= sdot sdot + sdotminus sdot + sdot

973 3120T Kle le

3 5 12 2

12 3

( ) (273) 328 10 1179 10 2429 10

1219 10

L T L T T

T

minus minus minus

minus


1 5

9 2 12 3

( ) (273) (99672 10 1179 10

2429 10 1219 10 )

L T L T

T T

minus minus

minus minus


Non-stoichiometry of UO2+x has no effects

source literatures are not found


- 330 -

(3) UO2 density

UO2 density MATPRO-Version 09 Condition Ref ρ ＝1096 ρ UO2 theoretical density (gcm3) No restrictions 213

(4) UO2 Youngrsquos modulus

UO2 Youngrsquos modulus (1) MATPRO-Version 09 Condition Ref ( ) ( )[ ]E T D= times minus times minus minusminus2 26 10 1 1131 10 1 2 62 111 4

EYoungrsquos modulus (Pa) Ttemperature (degC) Dtheoretical density ratio

No

restrictions

213

UO2 Youngrsquos modulus (2) MATPRO-11

IPLYG=2 Default

34 ( )[ ][ ]TDE 411 10091511175221103342 minustimesminusminusminustimes=

E Youngrsquos modulus (Nm2) Dtheoretical density ratio (-) T temperature (K)

(5) UO2 Poissonrsquos ratio

Poissonrsquos ratio of UO2 MATPRO-Version 09 Condition Ref

ν＝0316 No restrictions 213

(6) UO2 Emissivity UO2 Emissivity MATPRO-11 Condition Ref

Tlt1000 K e = 08707 Tgt2050 K e = 04083

1000≦T≦2050 K Te 40140443111 minustimesminus=

No

restrictions

34

(7) UO2 melting point

Temperature dependency of melting point is assumed as 32 degrees decrease per 10 GWdt

according to the Christensenrsquos equation(424)

Pellet melting point (Solidus) MATPRO-Version 11+Christensen Condition Ref T T P P Bum u= minus + minus5 414 0 007468 0 00322

Tm melting point (K)

Tumelting point of un-irradiated UO2 pellet (=3113K)

P molar fraction of PuO2 () Buburn-up (MWdt)

No

restrictions

34

413


- 331 -

(8) UO2 creep

UO2 Creep (1) MATPRO-Version 09 Condition Ref

( ) ( )( )

( )( ) ( )ε

σ σσ=

+ minus

++

minus+

+ minusA A F Q RT

A D G

A Q RT

A DA F Q RT

1 2 1

32

44 5

2

56 3

exp exp exp

ε steady-state creep rate (1h)

6 121 2

43 4

285 6

1 2 3

9728 10 324 10

877 1376 10

905 924 10

90000 132000 5200 (calmol)

A A

A A

A A

Q Q Q

minus

minus

minus

= times = times

= minus = times

= minus = times= = =

Ffission rate ( )8 4 10 118 1017 20 ~ sectimes times sdotfissm3

σstress 1000-16000 (psi)

Ttemperature 713-2073 (K)

Ddensity 92-98 (TD)

Ggrain size 4-35 (μm)

Rgas constant 1987 (calmol sdot K)

No restrictions

34


- 332 -

UO2 Creep (2) MATPRO-11 Condition Ref

( ) ( ) ( )

( )

451 2 1 1 4 8 2 2

23 6

7 3

exp ( ) exp

( )

exp

c

A A F Q RT A A F Q RT

A D G A D

A F Q RT

σ σε

σ

+ minus + minus= +

+ +

+ minus

(s-1)

A1=03919 A2=13100times10-19 A3=-877 A4=20391times10-25

A6=-905 A7=372264times10-35A8=00 1

1

20178848 exp 80 10 7212423

log( 199999)Q

x

minus minus= minus + + minus

1

2

20198720 exp 80 10 1115435

log( 199999)Q

x


Q1 Q2 activation energy (calmol)

x OxygenMetal ratio[normally 200]

F = fission rate (fissionm3)s

σ =Stress(Pa) T temperature (K)

D theoretical density ratio G=grain diameter(μm)

Q3=26167times103RR=19872 (calmolK)

Where σ equivalent stress（Nm-2）

When 05714

165474168

Gσ le 1σ σ= and 002 =σ

When 05714

165474168

Gσ gt 1 05714

165474168

Gσ = and 2σ σ=

IPCRP=2

(Default) 412

The following fitting parameters can be used for the UO2 creep rate equation


FCRFAC Magnification factor for the creep rate equation 10

TCS Cut-off value at the reference temperature in the creep

calculation (K) 207315


- 333 -

UO2 creep (3) SKI model for UO2 Condition Ref

Converted equation of Steady creep rate from original SKI equation

45 21 1 22

( )( ) exp( ) exp( )

A FA F Q RT Q RT C F

G

σε σ σ= minus + minus +

(1s) ( MPa)σ =

4 17 610

1

14 33

6 5 63

2

4 13

138 10 46 10( ) 1877 10

905

259 10 863 10(1s)

905

973 10 324 106895 10

877

671 10 223 10(1s)

877

FA F

F

FA

F

ρ

ρ

ρ

ρ

minus minus minusminus

minus minus

minus minusminus

minus

times + times= times timesminus +

times + times=minus +



5 51 25523 10 (Jmol) 3766 10 (Jmol)Q Q= sdot = sdot

23 6 3 317 10 6895 10 4827 10C minus minus minus minus= times times sdot = sdot

F fission rate (fissionsm3s) [1 fissionscm3s=10-6 fissionsm3s]

σ uni-axial stress (MPa) [1 psi= 6895E-3 MPa]

G =10 grain size (assumed constant 10μm) ρ density (TD) above 92 If it is less than 92 assume 92

T temperature (K) R gas constant (831441 Jmol-K)

IPCRP=4

(Default)

414

(4) Halden UO2 Creep model derived from Petitprezrsquos report (HWR-714) Condition Ref5

4 303505 108577 10 exp 148 10 f

RTε σminus times= sdot minus + sdot sdot

σ uni-axial stress (MPa) 3 fissionsm -sf

R 831441 (Jmol-K) T temperature (K)

Derived from the correlation in Fig26 assuming f=12E19m-3s-1

and σ=24 MPa

IPCRP=5 415


- 334 -

(9) UO2 swelling

Shown below are the equations representing UO2 swelling There are two types of UO2

swelling solid FP swelling and fission gas-bubble swelling

Using the options one of the following two cases can be selected a case in which the

solid FP swelling and gas-bubble swelling are treated independently and a case in which both

types of swelling are given by a single material property equation

UO2 swelling (1) Chubb Condition Ref

(gas-bubble swelling)

ΔV

V T

gs

S

= times minus times

4 396 10

1645 1024

exp T TS = minus 100

ΔV

V

gs

volume fraction of gas-bubble swelling per 1020

fissionscm3

Ttemperature (K) IFSWEL=0

(default )

249

UO2 swelling (1a) FEMAXI-III

11 (Solid FP swelling)

ΔV

V

ss

= 00025 volume fraction of FP swelling per 1020

fissionscm3 cong 062 vol(10GWdt)

In the case of ISWEL = 0 the following fitting parameters can be used


STFCP Suppress generation of gas-bubble swelling average contact pressure (Pa) (used in the deformation model)

minus3times107

SPCON Suppress generation of gas-bubble swelling average contact pressure (Pa) (used in the thermal model)

minus106

UO2 swelling （2） FEMAXI model Condition Ref

Fission gas bubble swelling see chapter 2

(Solid fission products swelling= FEMAXI-III model) ΔV

V

ss

= 00025 volmetric swelling per 1020 fissionscm3

IFSWEL=1 12


- 335 -

UO2 swelling (3) Hollowel model Condition Ref

The Hollowel model defines total swelling as a function of burnup

It is given by Eq(1) having no distinction between solid fission

products swelling and gas bubble swelling However Eq(1) assumes the

followings the swelling occurs in no restriction conditions it does not

take into account the effect of restriction the total quantity of swelling

does not fall below the swelling rate obtained by solid fission products

swelling rate eg 025 per 1020fissionscm3

Swelling in no restriction conditions 5 7

1 1

5 51 1

2413 10 10634 10 011509

286 10 ln 20184 10 ln

VT T Bu

V

T BU T GR

minus minus

minus minus

Δ = times + times +

+ times minus times (1)

where

V

VΔ volumetric swelling rate T1＝T－900

BU burnup (GWDtUO2) T temperature (degC)

GR grain size (μm)

IFSWEL=5 252

UO2 swelling (4) MATPRO-Version 09

IFSWEL=2 213

Temperature at pellet center (degC) Swelling ratio (1020fisscm3)

T le 1400 028 1400 le T le 1800 ( )[ ]0 28 1 0 00575 1 400 + minusT

1800 le T le 2200 ( )[ ]0 28 33 0 004 1800 minus minusT

2200 lt T 0476

UO2 swelling (5) Kosaka

IFSWEL=3

250 Volumetric swelling ratio

205 (10MWdkgUO )V

V

Δ =


- 336 -

UO2 swelling (6) Studsvik Condition Ref

( ) ( )

( ) ( )BuBuBuBuAV

V

V

V

BuBuBuFGRCAV

V

SWSW

SW

ltminus+

Δ=

Δ

lelesdotsdotminus=

Δ

1121

111 0

whereSW

V

V

1

Δ

is volumetric swelling() at Bu=Bu1

A1=0095 C1=0043 Bu1=57 A2=00317

SW

V

V

Δ

volumetric swelling ()

FGR FP gas release rate()

Bu Bu1 Burn-up(MWdkgU)

IFSWEL=4 251

(10) Densification of UO2

UO2 densification (1) FEMAXI-III Condition Ref

( )BuCeV

V

V

V sdotminusminusΔ=Δ1

max

ΔV

V volume contraction due to densification

V

V maxΔmaximum volume contraction due to

densification

Bu burnup (MWdt-UO2)

IDENSF=0

(default value)

11

The following fitting parameters are used in the case of IDENSF=0


DMAX(40) Maximum volume contraction due to densification

ΔV

V

max

() 10

SBU

Burnup of 90 completion of densification

given by CSBU

= 2 3025 (MWdt-UO2)

where exp (－23025) =01

25000


- 337 -

UO2 densification (2) Rolstad Condition Ref

The upper limit of the dimension change in the axial direction due to

densification is given by

( )( )

ΔL

L

TD

TS= times

minusminus

22 2100

1180

TDtheoretical density ratio of UO2 (TD)

TSsintering temperature (degC)

Burnup dependence ΔL

Le eBU BU= minus + +minus minus30 0 93 2 07 35

BUburnup (GWdt-U)

IDENSF=1 416

Here the following fitting parameter is used for IDENSF


TDNSF Sintering temperature (K) 20000

UO2 densification (3) NRC Condition Ref Density increment due to densification Δρ (gcm3)

( )

( ) ( )( )

ΔΔΔ Δ

ρρρ ρ

= le= + lt lt= ge

0 1

1 2

2

BU BU

m BU b BU BU BU

BU BUsn

log

Δρ sndensity increment (gcm3) due to out-of-pile heat treatment

(1700degCtimes24h)

BUburnup (MWdt)

BU1 BU2constants for determining the burnup range dependent on Δρ sn

m bconstants for connecting the three equations for each

burnup range

IDENSF=2 417

The following fitting parameter is used in the case of IDENSF=2


DMAX(40) Relative density increase () due to out-of-pile heat

treatment (1700degCtimes24h) 10


- 338 -

UO2 densification (4) Wiesenack (Halden) Condition ref

Volumetric decrease by densification is ( )

( )ΔV

V

BU

BUADST= minus

minus minus sdot

+ minus minus sdot

sdot0 01420 1 6 7943

0 00793 1 11434

exp( )

exp( )

Here BU Burnup(MWdkgUO2)

ADST adjusting factor(=06)

IDENSF=4 43

UO2 densification (5) Marlowe

Δρ = +

M

A

AD Ft

Girr irrln

10

03

Δρ amount of in-pile densification (TD)

M densification rate constant (TD-cm)

A grain growth rate constant (cm)

Dirr0 irradiation diffusion constant=127times10minus29 (m3fiss)

F fission rate (fisscm3 sdot s)

tirr irradiation time (s)

Ftirr burnup (fisscm3)

G0 initial crystal grain size (cm)

The constants M and A are calculated from the following equation

using the results of the out-of-pile heat treatment test in which a

fuel produced in the same batch as the fuel for irradiation is used

AG G

Dt=

minus30

3

MA

InADt

G

th=

+

Δρ

10

3

( )D D Q RT= sdot minus0 exp

G crystal grain size after heat treatment (cm)

t heat treatment time (s)

Δρ th density change due to heat treatment (TD)

D0 thermal diffusion constant 232times10minus6 (cm2s)

Q 82000 calmol

R gas constant 1987 (calmol)

T treatment temperature (K)

IDENSF=3 418



- 339 -


DMAX(40) degree of in-pile densification (TD) 10

GG crystal grain size after heat treatment (m) 10minus5

GGO initial crystal grain size (m) 8times10minus6

SITIM heat treatment time (h) 240

UO2 densification (6) Schlemmer and Ichikawa Condition Ref

bD

D sdot=Δ150 (b≦15 GWdt)

bD

D sdotminus=Δ100 (15 GWdt ≦b)

where D density (TD) b burnup (GWdt)

IDENSF=30 419

420

(11) Pellet relocation

Pellet fragments which result from cracking at the start-up stage expand thermally and

push with each other and consequently the gap is narrowed This phenomenon is called

relocation In FEMAXI the initial relocation is assumed to be proportional to the gap width

at hot stand-by where burnup is 0 and linear heat rate is 0 as

( )Relocation c cci pir rγ= sdot minus (411)

where rcic and rpi

c are the cladding inner radius and pellet radius (cm) at hot stand-by

respectively γ is an empirical fitting parameter which is determined by comparison with

experimental data and is called a relocation parameter The relocation parameter is as

follows


FRELOC Relocation parameter α in the radial direction 05

EPSRLZ Axial relocation strain 0003


- 340 -

(12) UO2 plasticity

Equations representing yield stress and the strain hardening parameter (tangent stiffness)

are selected from among the following

UO2 plasticity (1) Tachibana Condition Ref

( )( )

( )( )σ

εεY

p

p

T T

T T

T C

T C=

minus sdot + minus sdot sdotminus sdot + minus sdot sdot

legt

66 9 0 0397 520 0 0 386

36 6 0 0144 139 5 0 06875

1200

1200

( )( )prime =

minus sdotminus sdot

legt

HT

T

T C

T C

520 0 0 386

139 5 0 06875

1200

1200

σ Y yield stress (kgmm2)

primeH tangent stiffness (kgmm2)

T temperature (degC)

IFY=1

(default value) 421

UO2 plasticity (2) Rodford

422

σ Y T T

T

= minus sdot + times sdot

minus times sdot

minus

minus

1176 1 1 688 8 179 10

1 293 10

4 2

7 3

primeH =00tangential stiffness σ Y yield stress (MPa)


IFY=0

(13) UO2 Specific heat

UO2 specific heat (1) MATPRO-11 Condition Ref

( )( )

( )2

1 322 22

exp exp

2exp 1D

P D

K T O M K EC K T E RT

RTT T

θ θθ

= + + minus minus

CP (JkgK) T temperature (K)

OM OM ratio (－) R=83143(JmolK)

UO2 PuO2 Unit

K1 2967 3474 JkgK

K2 243times10-2 395times10-4 JkgK2

K3 8745times107 3860times107 Jkg

θ 535285 571000 K

ED 1577times105 1967times105 Jmol

No

restrictions 34


- 341 -

(14) UO2 grain growth

Equations representing UO2 grain growth are selected from among the followings

UO2 grain growth (1) Itoh (modified Ainscough ) Condition Ref

sdot+minus=

m

ff

D

NNE

DK

dt

dDmax11

Dgrain size at time t (μm)

Krate constant (μm2h)

( )K RT= times minus times524 10 2 67 107 5 exp

R=8314 (JmolK) Ttemperature (K)

Dmlimit grain size (μm)

Dm=223times103 exp (minus7620T)

Efitting parameter

N f fission gas atom density at grain boundary (atomscm2)

N fmax saturated value of FP gas atom density at grain

boundary (atomscm2)

IGRAIN = 0

(default value) 230

The fitting parameters in the case of IGRAIN=0 are as follows


AG Fitting parameter E 10

GRWF The grain growth rate is multiplied by GRWF 15

UO2 grain growth (2) Ainscough Condition Ref

dD

dtk

D

B

Dm

= minus +

1 1 0 002

Dgrain size at time t (μm) t time (h) k rate constant (μm2h) ( )k RT= times minus times5 24 10 2 67 107 5 exp R=8314 (JmolK) T temperature (K) for a range 1300minus1500degC Bburnup (MWdtU)

IGRAIN=1 231


- 342 -

UO2 grain growth (3) MacEwan Condition Ref

( )D D k t RT20

20

0 8 87 000minus = minus exp D grain size at time t (μm) D0initial grain size (μm) k0 constant of proportionality (ko=39times1010) t time (h) Rgas constant 1987 (calmolK) T temperature (K) for the range1555minus2440degC

IGRAIN=2 423

UO2 grain growth (4) Lyons

IGRAIN=3 424

D D k e tQ RT30

30minus = sdotminus

D grain size at time t (μm) D0initial grain size (μm) k0rate constant (μm3h) Q activation energy required for grain growth (calmol) R =1987 (calmolK) T temperature (K) t time (h)

( )K Q045 096 2 476 10= + times sdotminusexp

For the experimental data obtained by MacEwan Q=119000(calmol)

UO2 grain growth (5) MATPRO-Version 09

IGRAIN=4 213

D D e tRT40

4 13 92 400618 10minus = times sdotminus

D grain size at time t (μm)

D0 initial grain size (μm)

R =1987 (calmol sdot K) T temperature (K)

t time (h)

UO2 grain growth (6) MATPRO-11

IGRAIN=5 34

Equiaxed grain growth

tRTDD sdottimesminustimes=minus )10873exp(107171 51040

4

D grain size (μm) Do initial diameter (μm)

R=8314 JmolK t time(s)


- 343 -

42 MOX Pellet Various material properties for MOX pellets can be used in FEMAXI By eliminating the

use of the name-list variable ldquoMOXrdquo consistency is maintained with the name-list parameters

for UO2 fuel For MOX fuel use the name-list parameters and the material properties listed in

the followings

(1) Optional parameters for fission gas release model Fission gas release rate Condition

The value obtained by multiplying the release rate given by the fission gas release model in FEMAXI by FPMOX is used

10 (Default)

Fission gas composition assumption Condition Ref

Average XeKr ratio is assumed to be 87 13 IXEKR = 0 (Default)

425 Average XeKr ratio is assumed to be 86 14 IXEKR = 1

Average XeKr ratio is assumed to b 16 1 IXEKR = 2

Data from PLUTON are used for the average ratios of XeKr of a

pellet Here when PLUTON is not used (when IFLX=-1 is not used)

IXRKR=3 is ignored and the same calculation is performed as in the

case with IXEKR=0

IXEKR = 3

(2) Thermal conductivity of MOX

MOX pellet thermal conductivity (1) Martin Condition Ref f molar fraction 21 O)PuU( ffminus

1) 012 ≦ f ≦030

( ) 1 34 120037 333 237 10 789 10MOX K KK y T Tminusminus minus= + + times + times

2) 0 ≦ f ≦012 ( ) ( )K f K f K fu MOX= sdot minus + sdot1 012 012

where

Kthermal conductivity (WmK) KT temperature (K)

ｙMO2 - y Ku UO2 thermal conductivity (WmK)

( ) ( )

( )

1 34 12

2

0035 225 10 830 10 1

0 005

u K K

z

K T T z

z UO z

minusminus minus

minus

= + sdot sdot + sdot sdot sdot + le le

IPTHCN=30 426


- 344 -

MOX pellet thermal conductivity (2) MATPRO-11 Condition Ref

( )13 4

2

exp C

KK P K K T

K T

= + +

( )0 1550CTdeg lt le (WcmK)

( )5 3 4exp CK P K K K T= + ( )1550 CTlt (WcmK)

( )( )P

D=

minus minusminus minus1 1

1 1 0 95

ββ

porosity factor

D TD β =258－058times10-3Tc

TC temperature() K1 = 330 K2 = 375 K3 = 154times10-4 K4 = 171times10-3

K5 = 00171 K is applicable for 005 ≦ f ≦030 where f weight fraction

IPTHCN=31 34

MOX pellet thermal conductivity (3) Martin + Philipponneau

IPTHCN=32 426 427

( ) ( )[ ])()20()(201 2 pfKpFfKpfK MOXUO αsdotsdot+sdotminussdot=

where K (f p) thermal conductivity (WmK) f 21 O)PuU( ffminus 2000 lele f

p

ppF

21

1)(

+minus=

p

pp

21

1

8610

1)(

+minussdot=α

p porosity

z OM ratio (MO2 - z)

τ burnup factor (01 for 10 at-burnup)

TK temperature (K)

KMOX thermal conductivity of MOX fuel ( f = 020) (WmK)

( )

sdotsdot+

sdotsdot+sdot+minus+=minus

minusminus

312

14

103876

108852440105500093105281

K

KMOX

T

TzK

τ

KUO2 thermal conductivity of UO2 (WmK)

( ) )( zTTK KKUO +sdot

sdotsdot+sdotsdot+sdot+= minusminusminus 110083102524400350 31214

2 τ


- 345 -

MOX pellet thermal conductivity (4) Duriez et al Condition Ref

95

1 1 005

1 2 1 2 005

p

pλ λ minus minus = times + + times

(WmK)

( )2951 exp( )K KK

C DA BT TT

λ = + minus +

)(0350852)( WmKxxAA +==

)(10)862157()( 4 WmxxBB minustimes+minus==

)(106891 9 mWKC times= 13520D K=

Temperature range 700 2600KK T Kle le

Applicable for 003 ≦ f ≦015 f molar fraction 21 O)PuU( ffminus

IPTHCN=33

428

MOX pellet thermal conductivity (5) Philipponneau

IPTHCN=34

427

( ) 14

312

1528 000931 01055 044 2885 10( )

7638 10

KMOX

K

z TK p

T

τα

minusminus

minus

+ minus + sdot + sdot sdot = sdot + sdot sdot

KMOX thermal conductivity ( f = 020) (WmK) ( 2000 lele f f molar fraction )

p

pp

21

1

8610

1)(

+minussdot=α p porosity



TK temperature (K)

MOX pellet thermal conductivity (6) Halden Model

IPTHCN=35

429

1

2 3 7 4 5 6

( )(1 ) ( )

moxfuel

C

K aK F p

a a BU a BU a Gd a a BU T

sdot= sdot+ minus + + minus

(WmK)

where 3 7(1 )a BU a BUminus =184 (1 0003 )BU BUminus sdot

1 1 005( )

1 2 1 2 005

pF p

p

minus minus = + + times p porosity

1 2 3

4 5 6

7

404 464 184

116 10 00032

0003

a a a

a a a

a

= = == = ==

CT C deg

210 for UO

092 for MOXmoxK =

= MWdkgMOXBU

TD=095 is assumed as a standard condition Name-list parameter Kmox = KMOX = 092 (Default)


- 346 -

MOX pellet thermal conductivity (7) Baron Condition Ref

( )( )95 22 20 1 2 3 0 1 2 3

1exp

1TD

C Dg WK


+ = + minus + + + + + + +

K WmK T = Kelvin x = abs[2-OM]) =Y q = Pu weight fraction (-) = PU g = Dg weight fraction (-) 0-12 =GD k = Boltzmann constant (13810-23 JK) W= 141MeV = 141x1610-19 J A0 = 0044819 + 0005BU+20(G - 54702) (mKW) B0 = (24544 - 00125BU - 70(G - 54702))10-4 (mW)

A1 = 4 (-) B1 = 08 (-) A2 = 0611 (mKW) B2 = 960310-4 (mW) A3 = 11081 (mKW) B3 = -176810-2 (mW) C = 5516109 (WKm) D = - 43021010 (WKm)

BU = burn-up (GWdtU) G = lattice constant(

A ) =LT This equation is applied to 95TD pellet In general )0501()1(95 ββ minusminussdot= pKK TD

)(10580582 3 CTT minussdotminus=β holds

IPTHCN=15 49

[Note] In Baronrsquos model Y PU GD and LT are name-list input parameters Default values

are Y of MO2 ndash y =00 PU=00 GD=00 and LT = 54702 PU is PuO2 weight fraction

which is converted into a molar fraction in FEMAXI-6

MOX pellet thermal conductivity (8) FRAPCON-3 PNNL-modified Halden model

Condition Ref

( )95 092 ( ) 00132exp 000188 (WmK)K Q x T= +

4( ) 01148 gad 11599 00040B 2475 10 (1 00033B)Q x a x minus= + sdot + + + sdot minus Θ

11599a = ldquogadrdquo= weight fraction of gadolinia x = 200-OM (oxygen to metal ratio) =Y (Default=00)

B burn-up (MWdkgUO2) min( 1650)TΘ = T degree C

Porosity p correction factor 95

1( ) 10789

1 05

pK p K

p

minus=+

IPTHCN=37

430


- 347 -

(10) Ohira amp Itagaki (NFI) The latest Model for 95TD UO2 and MOX pellets

IPTHCN =91

410

Latest model

( ) ( ) ( )2 4

1

Gd P

C D

A B T f Bu g Bu h T

T T

λα β


minus + (WmK)




( ) ( )

2 4

1

1 396exp 6380

453 10 (mKW) 246 10 (mKW)

h TT

A Bminus minus

=+ minus

= times = times

C=547E-9 (Wm-K3) D=229E-14 (Wm-K5) 3812 10α minus= sdot Gd Gd2O3 concentration (wt) 3124 10β minus= sdot P PuO2 concentration (wt)

【Note】This equation is applicable to 95TD pellet In general apply

the followings )0501()1(95 ββ minusminussdot= pKK TD 3 o258 058 10 ( C)T Tβ minus= minus sdot

MOX pellet thermal conductivity (9) PNNL model in FRAPCON-3 modified on the basis of the Duriez model

Condition Ref

mod95 2

1 ( ) expC D

K P xT T

= + minus

( ) ( ) gad ( ) ( )

(1 09exp( 004 )) ( ) ( )

P x A x a B x T f Bu

Bu g Bu h T

= + sdot + ++ minus minus

( ) 285 0035 (mKW)A x x= + x=200-OM (oxygen to metal ratio

=Y default=00)

11599a = ldquogadrdquo=weight fraction gadolinia (not expected in MOX) 4( ) (286 715 ) 10 (mW)B x x minus= minus sdot

f(Bu)= 000187Bu effect of fission products in crystal matrix g(Bu)= 0038Bu028 effect of irradiation defects

Bu burnup in GWdtHM

h(T)= 1 1 396expQ

T

+ minus temperature dependence of annealing

on irradiation defects

Q ldquoQRrdquo 6380K Cmod=15E9 W-Km D=13520 K


1( ) 10789

1 05

pK p K

p

minus=+

IPTHCN =38

430


- 348 -

(3) MOX pellet creep

Creep rate of MOX pellet (1) MATPRO-11 Condition Ref

( ) ( )[ ]( ) ( )[ ]fBDBTQFBB

fBDBTQG

FBBc

47454

265

4332121

1exp

1exp

+minus+minus++

+minus+minus+=

σ

σε

B1=01007 B2=757times10-20 B3=333 B4=0014

B5=6469times10-25 B6=00 B7=103

Q3=503271 Q4=704580 where ε c creep rate (s-1) σ 1 equivalent stress(Nm2)

G grain diameter (μm) F fission rate (fissionsm3s)

D theoretical density ratio(－)

p porosity f weight fraction of Pu T (K)

σ equivalent stress (Nm-2) Here

When 05714

165474168


when 05714

165474168

Gσ gt

571401816547416

G=σ and 2σ σ=

IPCRP=30 34

Creep rate of MOX pellet (2) SKI-based MOX model Condition Ref

1) fH=1 222

( )exp( )

A FQ RT C F

G

σε σ= minus +

5 13

2

263 10 893 10(1s)

877

FA

ρ

minustimes + times=minus +

52 3776 10Q = times 30278 10C minus= times

σ uni-axial stress (MPa)

F fission rate (fissionsm3s)

G =10 grain size (assumed constant 10μm) ρ density (TD) above 92 if less than 92 92

T temperature (K)

R gas constant (831441 Jmol-K) 2) fH=2 30556 10C minus= times Others are the same 3) fH=3 30973 10C minus= times Others are the same

fH=1 IPCRP=11 fH=2 IPCRP=12 fH=3 IPCRP=13

414


- 349 -

4 5 6 7 8 9 10 11 12 13 14 15 161E-10

1E-9

1E-8

1E-7

1E-6

1E-5

fH= 3 fH= 2

SKI MOX fH=1

SKI UO2

UO2Halden

MOXHalden

12x1019 fissionsm3s 24MPa

Fuel

cre

ep ra

te

(1s

)

10000T (1K)

Figure 411 shows a comparison of fuel creep rates of Halden model and SKI model

(4) Poissonrsquos ratio MOX pellet Poissonrsquos ratio (1) MATPRO-11 Condition Ref

μ = 0276 IPOIS=30 34

MOX pellet Poissonrsquos ratio (2) Nutt + Yamada

IPOIS=31 431 432

( ) pf sdotminus+sdotminus= 3203240230 μ where p porosity f molar fraction 21 O)PuU( ffminus

No temperature dependence is assumed

Creep rate of MOX pellet (3) Halden model Condition Ref

54 303505 10

8577 10 exp 973 10 fRT

ε σminus times= sdot minus + sdot sdot

σ uni-axial stress (MPa) 3 fissionsm -sf R 831441 (Jmol-K) T temperature (K)

IPCRP=15 415

Fig411 Comparison of creep rate models


- 350 -

(5) Youngrsquos modulus MOX pellet Youngrsquos modulus (1) MATPRO-11 Condition Ref

( )[ ][ ]( )E D T f= times minus minus minus times +minus2 334 10 1 2 752 1 1 10915 10 1 01511 4

E Youngrsquos modulus(Nm2) D Theoretical density ratio T (K)

f weight fraction of Pu

IPLYG=30 34

(6) Specific heat

MOX pellet specific heat (1) MATPRO-11 Condition Ref

CPmox=CP UO2 (1－f )＋CP PuO2f

( )( )[ ]

CK T

T TK T

O M K E

RT

E

RTPD D=

minus+ +

minus

12

2 2 23

21 2

θ θ

θ

exp

exp

exp


CP specific heat(JKgK) T (K) OM OM ratio(－) R=83143(JmolK)

UO2 PuO2 Unit K1 2967 3474 JKgK K2 243times10-2 395times10-4 JKgK2

ISPH=30 34

K3 8745times107 3860times107 JKg θ 535285 571000 K ED 1577times105 1967times105 Jmol

(7) Melting point MOX pellet melting point (Solidus) (1)

MATPRO-11+Christensen Condition Ref

ummum BffTT 0032000746804145 2 minus+minus=

Tm melting point(K)

Tu melting point of un-irradiated UO2(=3113K)

fm weight fraction of Pu Buburnup (MWdt)

Note On the basis of Christensenrsquos equation it is assumed that the

melting point falls by 32 C per 10GWdt

No restrictions

(shared with that of UO2 )

34 413


- 351 -

(8) Thermal expansion MOX pellet thermal expansion (1) MATPRO-11 Condition Ref

( ) fL

Lf

L

L

L

L

PuOUO 22

1Δ+minusΔ=Δ

ΔL

L linear expansion(－)

ΔL

LT T T

UO2

4 972 10 7107 10 2581 10 1140 104 6 9 2 13 3= minus times + times + times + timesminus minus minus minus

2

4 6 9 2 16 339735 10 84955 10 21513 10 37143 10PuO

LT T T

Lminus minus minus minusΔ = minus times + times + times + times

T temperature (C) f weight fraction of Pu

IPTHEX=30 34

MOX pellet thermal expansion (2) Martin + Tokar

( ) ( )xfL

Lf

L

L

L

L

PuOUOMOXsdot+sdot

sdotΔ+minussdotΔ=Δ9311

22

where MOXL

LΔ thermal expansion of MOX fuel

2UOL

LΔ thermal expansion of UO2

2PuOL

LΔ thermal expansion of PuO2

x OM ratio defined by MO2-x f molar fraction 21 O)PuU( ffminus

Also thermal expansions of UO2 and PuO2 are

(279 K≦T≦923 K)

( ) 11039141070521080291097349 31321061

2minussdottimes+sdottimesminussdottimes+times=Δ minusminusminusminus TTT

L

L

UO

(923 K≦T≦3120 K)

( ) 11021911042921017911096729 3122951


L

L

UO

(279 K≦T≦1693 K)

( ) ( )( )

24 6 9

2312

12232 10 75866 10 273 56948 10 273

59768 10 273PuO

LT T

LT

minus minus minus

minus

Δ = minus times + times sdot minus + times sdot minus

minus times sdot minus

( ) ( )518415 27310289712731040924 minussdottimesminusminussdottimes+ minusminus TT T temperature (K)

IPTHEX=31 426433


- 352 -

MOX pellet thermal expansion (3) MATPRO-A Condition Ref

Thermal strain (=0 at 300 K) 1 2 30

exp DL EK T K K

L kT

Δ = minus + minus

T (K) ED = 70x10-20 lattice defect generation energy (J)

k Boltzmann constant 23(138 10 )J Kminustimes

K1 = 90x10-6 K-1 K2 = 27x10-3 K3 = 70x10-2

IPTHEX=32 213

(9) Densification MOX pellet densification Schlemmer + Ichikawa Condition Ref

bD


bD


D density (TD) b burnup (GWdt)

IDENSF=30 419 420


- 353 -

43 Zirconium Alloy Cladding 431 Cladding properties (1) Cladding thermal conductivity Zircaloy thermal conductivity MATPRO-Version 09 Condition Ref

From room temp to melting point

k T T T= + times minus times + timesminus minus minus7 51 2 09 10 145 10 7 67 102 5 2 9 3 kthermal conductivity (WmK)

Ttemperature (K)

No

restrictions 213

(2) Zircaloy thermal expansion

Zircaloy thermal expansion (1) MATPRO-Version 09 Condition Ref

For 27minus800degC

thermal expansion in the axial direction

ΔL L T 05 62 506 10 4 441 10= minus times + timesminus minus

Therefore thermal expansion coefficient is

α = times minus4 441 10 6

thermal expansion in the radial direction

ΔD D T 04 62 373 10 6 721 10= minus times + timesminus minus

therefore α = times minus6 721 10 6

Ttemperature (degC)

ICATHX=0

(default value)

213

Zircaloy thermal expansion (2) Scott

ICATHX=1

For 21minus429degC


ΔL L To = +42 53230 3 96753 ( in in )times10minus6

Ttemperature (degC)

Zircaloy thermal expansion (3) MATPRO-11

ICATHX=3 34

300ltTlt1073 K

6 3

0

444 10 124 10L

TL

minus minusΔ = times minus times axial expansion

6 3

0

672 10 207 10D

TD

minus minusΔ = times minus times diameter expansion



- 354 -

In the case of ICATHX=0 the following parameters can be designated

Variable name

Content Default value

ATHEX Expansion coefficient in the axial direction is given by ATHEX Default value is that of MATPRO-09 model 4 441 10 6 times minus

RTHEX Expansion coefficient in the radial direction is given by RTHEX Default value is that of MATPRO-09 model 6 721 10 6 times minus

(3) Youngrsquos modulus of cladding Youngrsquos modulus of Zircaloy (1) Fisher Condition Ref

( )[ ]E T= times minus times minus times times9 900 10 566 9 27315 9 8067 105 4

EYoungrsquos modulus (Pa) Ttemperature (K) IZYG=1 434

Youngrsquos modulus of Zircaloy (2) MATPRO-A -11

IZYG=2

Default

213

34

1) α phase 11 7

1 2 3(1088 10 5475 10 ) Y T K K K= times minus times + + (Pa)

2) β phase 10 7921 10 405 10Y T= times minus times (Pa)

T cladding temperature (K)

K1 effect of oxidation (Pa)

K2 effect of cold work (Pa)

K3 effect of fast neutron fluence (-)

11 81 (661 10 5912 10 )K T= times + times Δ

102 26 10K C= minus times 3 25

K =088+012exp10

Φ minus

Δ average oxygen concentration minus oxygen concentration

of as-received cladding (kg Okg Zircaloy) As-received

oxygen concentrations are so small (12gOkg Zircaloy) that the

exact magnitude of the as-received concentration will not affect

the correlation predictions

C cold work (-) Φ fast neutron fluence (nm2) Youngrsquos modulus of ZrO2 MATPRO-A No

restrictions 213

E T= times minus times1367 10 3 77 1011 7 (Pa) T(K)


- 355 -

(4) Poissonrsquos ratio of cladding Poissonrsquos ratio of Zircaloy Fisher Condition Ref

( )v T= + times minusminus0 3303 8 376 10 273155 Ttemperature (K) No restrictions 434

Poissonrsquos ratio of ZrO2 MATPRO-A No restrictions 213

v = 0 30

(5) Creep of Zircaloy Zircaloy creep (1) MATPRO-Version 09 Condition Ref

In-pile creep equation ( ) ( ) 5000010exp minusminus+= tRTBeK Cσσφε (1)

ε biaxial creep strain rate (mms)

K = 5129times10minus29 B = 7252times102

C = 4967times10minus8 R = 1987 (calmolK)

Ttemperature (K) ttime (s)

φ fast neutron flux (nm2s) E gt10 MeV

σ circumferential stress (Nm2)

[Method of numerical analysis]

Integrating Eq(1) gives

( )( ) ( )ε φ σ σ= + minus2 10000 0 5K B C RT texp exp (2)

Since Eq(2) is a primary creep and power index of t is not unity ε εH = (total creep strain) holds

Eliminating t from Eqs(1) and (2)

( )( ) ( ) exp exp ε φ σ σ ε= + minus2 10000 2K B C RT

Accordingly

( )( ) ( )f K B C RTc c= = + minus exp exp ε φ σ σ ε2 10000 2

is obtained where

ε c equivalent creep strain rate of cladding (1s)

ε c equivalent creep strain of cladding and


CRPEQ=0

213


- 356 -

Zircaloy creep (2) Franklin model Condition Ref

expm p ncreep

D QA t

D Tθε φ σΔ = = sdot sdot sdot sdot minus

13111 10

0682

0550

0579

1173

A

m

p

n

Q

minus= times====

2

hours

ncm s (gt1MeV)

MPa

Kelvin

t

Tθ

φσ

sdot

CRPEQ=2 435

Zircaloy creep (3) NENANSEN

CRPEQ=1 436

In-pile creep equation for Zircaloy-4

ε ε εcthc

irrc= + (1hr)

1) thermal creep rate setting thermal creep strain as ε thc mK t= sdot

we have the creep rate as

( )

ε εthc m

HmmK=

minus11

1

where ε ε εH thc

th kc

kk

n

t( ) = = sdot= Δ

1

thermal creep strain in the

previous time step(n-1)

( ) K T Teq= minus + minus + times + timesminus minusexp 133 0 416 8 22 10 6 59 104 3σ

( )m T Teq= minus + minus + times + timesminus minus107 0 00343 7 27 10 195 105 3 σ

σ eq equivalent stress (kgfmm2)

T temperature （Ｋ）

2) irradiation creep rate(ref source is unknown) ε φ σirr

ceq= times minus6 64 10 25 1 23 1 34

φ fast neutron flux (ncm2s)

[Application in numerical calculations]

Based on the form of the equation instead of eliminating time t f c= ε is used The creep strain increments in the radial and axial

directions are given by

( )d dcz

cεσ

σ σ εθ θ= minus sdot1

22

( )d dzc

zcε

σσ σ εθ= minus sdot1

22

where ε c in the above is given as

ε ε θ ε θcnc cd= + sdot =

1

2

and ε c is obtained using repetitive convergence calculations


- 357 -

Zircaloy creep (4) Halden (McGrath) model Condition Ref

expm p ncreep

D QA t


1216 10

0450

0550

0579

1173

A

m

p

n

Q

minus= times====

2

hours

ncm s (gt1MeV)

MPa

Kelvin

t

Tθ

φσ

sdot

CRPEQ=3 437

High temperature Zircaloy creep (H1) Rosinger Condition Ref

expε σα = minus

2000342205 32

T ( )730 1073K T Kle le

εα Creep rate (s-1) σ equivalent stress (MPa)

HTCRP=1

(Default) 438

940≦T≦1073 K 5322000 exp( 284600 )kTαε σ= minus

1073ltTlt1273 K

4 4

1073 107310

200 200

T Tα β α βε ε ε+

minus minus = minus +

1273≦T K 37981 exp( 142300 )kTβε σ= minus k 8314 JmolK

High temperature Zircaloy creep (H2) Donaldson

HTCRP=

20

21

22

439

0 10 1 exp

nG Q

AT G RT

σε = minus

10 7 3 23326 10 2244 10 ( 273) 2161 10 ( 273)G T T= times minus times minus + times minus

(Nm-2)

( 973 1273T Kle le )

A1（N-1m2Ks-1） n Q1（kJmol-1） Westinghouse tube 443x1016 531plusmn006 2668plusmn140 20

Wolverine tube 183x1017 543plusmn002 2722plusmn120 21 Sandvik(NRU) tube 654x1016 494plusmn009 2940plusmn160 22

In normal operation and transient analyses the models (H1) and (H2) for

high-temperature creep are used only when the creep rate obtained from the models (H1)

and (H2) exceeds the creep rate obtained from the creep models (1)-(4) for normal

operation temperature In this case CRPEQ=0 1 2 or 3 and HTCRP=1 or 20-22


- 358 -

The following fitting parameters are used for the Zircaloy creep equation Variable name Content Default value

CRFAC For Stress-Relieved material the creep strain rate is multiplied by CRFAC

13

IPUGH PUGH=0 inversion of Pugh is not taken into account PUGH=1 inversion of Pugh is taken into account

0

Here Pughrsquos method(444) of cladding creep inversion caused by stress reversal is explained

When the direction of stress is reversed during creep the primary creep occurs again

Pugh proposed a simplified method to simulate this phenomena An outline of Pughrsquos

method is given as follows

1) The creep hardening is represented by ε H and

( ) ε partσpartσ

σ εcHf T=

(431)

2) At the time of the first stress inversion ε εH gt and

ε ε ε= =HH 0 (432)

is set

3) At the time of the second or later inversions

when ε εH ge then ε ε ε= = =H H0 0 is set and

when ε εH lt then ε ε εH H= minus is set

Figure 431 shows the above states

ε c

ε

ε

ε

εH =0

εH =0εH =0

ε ε εH H= minus

①

②

③

④

Fig 431 Inversion of creep strain due to stress inversion


- 359 -

Here the stress inversion is judged as follows Since the equation for the creep is

( ) ε partσpartσ

σ εcHf T f=

ge 0 (433)

whether or not the sign of ε c is reversed between the previous time step and the current

time step is used for the judgment In a multi-axial state the judgment of sign reversal is

performed using a scalar product of the component of ε c

(6) Cladding stress-strain relationship Cladding stress-strain relationship (1) FEMAXI-III Condition Ref

σ ε= K n

σtrue stress (kgmm2) εtrue strain

n strain-hardening exponent

n

E

KKK minus+= )0020( 20

20

( )( )1

202 3344 1 expyK B tσ φ= + minus minus sdot (kgmm2)

( )φ1421 1061exp10922 minusminus timesminustimes=B

φtfast neutron fluence (ncm2s) E gt1 MeV σ y =yield stress (kgmm2)

=2160minus00213T (recrystallized material) =3132minus00213T (stress-relieved material)

n=00504+00001435T

Ttemperature (degC) 220 450le leT degC

ICPLAS=1 11

The following parameter is used for the Zircaloy strain-hardening curve


ISTR In the case of ISTR ne 0 K y= +σ 3344 is set 0


- 360 -

Cladding stress-strain relationship (2) MATPRO-11 Condition Ref

In the following equations the strain rate ε is determined in the

prior time step

310

mnK

s

εσ ε minus =

σ true stress (Pa)

ε true strain rate If ε lt10-5s ε =10-5s

Tlt850 K 2 4 7186 10 7110 10 7721 10n T Tminus minus minus = minus times + times minus times

Tgt=850 K 0027908n =

Adjustment factor n for RIC reflects both the effect of irradiation and

cold work 2

13

7 8

0847exp( 392 ) 0153 ( 916 10 0229 )

exp373 10 2 10

RIC CWN CWN CWN

CWN

φ

minus = minus + + minus times + minustimes

times + times sdot

CWN effective cold work (area reduction) for strain-hardening

exponent n

φ effective fast neutron fluence (gt10MeV nm2)

Strength coefficient K

Tlt=730 K 9 610884 10 10571 10K T= times minus times

730ltTlt900 K [ ]1 2 3 4( )K A T A T A A T= + + + sdot

9 71 28152540534 10 3368940331 10A A= minus times = times


900lt=T 8663exp 8755K

T = +

180546 554 10DK CWK K φminus= sdot sdot + times strength coefficient for

irradiated and cold-worked material minus the equations (Pa)

CWK effective cold work for strength coefficient (area reduction)

Strain rate sensitivity exponent m

T le 730 K m = 002 730ltTlt900 K [ ]5 6 7 8( )m A T A T A A T= + + +

1 25 62063172161 10 7704552983 10A A minus= times = minus times

5 87 89504843067 10 3860960716 10A Aminus minus= times = minus times

900 le T le 1090 K 2 4647 10 2203 10m Tminus minus= minus times + times

ICPLAS=2 34


- 361 -

Cladding stress-strain relationship (3) Modified MATPRO-11 Condition


prior time step

310

mnK

s

εσ ε minus =

(original)

σ true stress (Pa)


n the same as those in the original MATPRO-11

model

Strength coefficient K the same as those in the original MATPRO-11

model

Modified DK strength coefficient for irradiated and cold-worked

material minus the equations (Pa)

18 250546 554 10 ( 26 10 )DK CWK K φ φminus= sdot sdot + times sdot le timesKF

18 25

8 25

0546 554 10 26 10

0546 144 10 (26 10 )

DK CWK K

CWK K φ

minus= sdot sdot + times times times sdot= sdot sdot + times sdot times lt

KFKF

where KF is a name-list parameter KF=10 (Default)

Strain rate sensitivity exponent m the same as those in the original

MATPRO-11 model

ICPLAS=3

00 05 10 15 200

200

400

600

800

1000

0001s

0001s

0001s

100s

100s

900K

600K

400K

True

Stre

ss

(MP

a)

True Strain ()

352K100s

Fig432 Stress-strain relationship of Zircaloy predicted by the modified

MATPRO-11 model (ICPLAS=3) Dependence on strain rate and

temperature is shown


- 362 -

Cladding stress-strain relationship (4) Ohta model Condition Ref

Refer to the model description in section 362

1 Stress-strain equation Elastic region Eσ ε= E Youngrsquos modulus

Intermediate region

mn

midmidK

= minus310

εεσ

Plastic region 310

mnK

εσ ε minus =

K strength coefficient nstrain-hardening exponent

εtrue strain rate If ε lt10-5 s ε＝10-5 s

mstrain rate sensitivity exponent

2 Strength coefficient K (MPa) (1) K of stress-relieved (SR) cladding of PWR rod

K is assumed to depend on temperature T and fast neutron fluence

Φ (gt10MeV) and given as follows

T≦900 K （62685）

1 2

253 4

( 27315)

( 27315) 1 exp( 1751 10 )

K L L T

L L T minus

= minus minus


1 2 3 41745 2517 526 0748L L L L= = = =


900＜T＜1000 K (62685～72685 oC)

1 2 3 4

05251 2 3 4

(

( ) 1 exp( 1751 10 )

K A T A T A TA

B T B T B TB minus

= + + +

+ + + + times minus minus sdot Φ

A1＝ 3041233371times104 B1＝ 33282432 times102

A2＝－8285933869times101 B2＝ 298027653

A3＝ 7490958541times10－2 B3＝－681286681times10－3

A4＝－2242589096times10－5 B4＝ 351229783times10－6

1000 K≦T （72685oC -）

0525


61 2 3 41 10 8775 8663 0341566H H H Hminus= sdot = = =

ICPLAS=

4 for

PWR

ICPLAS=

5 for

BWR

35


- 363 -

(2) K of re-crystallized (RX) cladding of BWR rod

T≦700 K （42685）

1 2

05263 4

( 27315)

( 27315) 1 exp( 9699 10 )

K L L t

L L T minus

= minus minus


700＜T≦800 K （42685～52685）

1 2 3 4

05265 6 3 4

(

( ) 1 exp( 9699 10 )

K A T A T A TA

A T A T A TA minus

= + + +

+ + minus + times minus minus sdot Φ

A1＝ 3656432405times104 A2＝－1361626514times102

A3＝ 1695545314times10－1 A4＝－7055897451times10－5

A5＝－3338799880times104 A6＝ 1330276514times102

800＜T＜950 K （52685～67685）


1 2 3exp( )non irradiatedK H H H Tminus = + 6

1 2 31 10 4772 9740H H Hminus= sdot = =


05261 2 3 4( 1 exp( 9699 10 )

non irradiated

non irradiated

K K

B T B T B TB K

minus

minusminus

=

+ + + + minus times minus minus sdot Φ

950K≦T （67685oC≦T）

0526

1 2 3 4exp( ) 1 1 exp( 9699 10 )K H H H T H minus = + times + minus minus sdot Φ 6

1 2 3 41 10 4772 9740 5267783H H H Hminus= sdot = = =

3 Strain hardening exponent n T≦67315 K (400oC) 1 2( 27315)n C C T= minus minus

67315ltTlt1000 K (400-72685) 3 4 5( )n C T C C T= + +

C1＝0213 C2＝1808times10－4 C3＝34708518times10－2

C4＝49565251times10－4 C5＝－50245302times10－7

1000 K≦T (72685 - ) n = C6 = 0027908

4 Strain rate sensitivity exponent m

T≦730 K (45685) m＝002 730ltT＜900 K (45685～62685) 5 6 7 8( )m A T A T A A T= + + +

900 K ≦T (62685 - ) 9 10m A A T= + m＝－647times10－2＋2203times10－4T

B1＝－3617377246times104 B2＝ 1361156965times102

B3＝－1636731582times10－1 B4＝ 6386789409times10－5


- 364 -

A5＝

A7＝

A9＝

2063172161times101

9504843067times10－5

-067times10－2

A6＝

A8＝

A10＝

－7704552983times10－2

－3860960716times10－8

－2203times10－4

Cladding stress-strain relationship (5) FRAPCON_34 FRAPTRAN_v14

ICPLAS =6 430


prior time step

310

mnK

s

εσ ε minus =

σ true stress (Pa)


(1) Strength coefficient K

( ) ( ) ( )( ) ( )ZryKKCWKTKK φ++sdot= 1

where )(TK is

T lt 750 K ( ) 32359 1727521102818531054859410176281 TTTTK sdot+timesminustimes+times=

750 K lt T lt 1090 K

( )

timestimes=2

66 1085000272

exp105224882T

TK

1090 K lt T lt 1255 K ( ) TTK 58 1043454481108413760391 timesminustimes=

1255K lt T lt 2100K ( ) 332147 103371075793106856103304 TTTTK minustimesminustimes+timesminustimes=

( ) CWCWK sdot= 5460

)(φK is 225 1010 mntimesltφ ( ) ( ) ( )TCWfK 10464114640 25φφ minustimes+minus=

22525 1021010 mntimesltlttimes φ ( ) φφ 26109282 minustimes=K

22525 1012102 mntimesltlttimes φ ( ) φφ 271066182532360 minustimes+=K

( ) ( ) 110

550exp1min20exp252 +

minussdotsdotminus= T

CWTCWf

)(ZryK is K(Zry) =1 for Zircaloy-4

K(Zry) =1305 for Zircaloy-2

T temperature (K)

CW effective cold work for strength coefficient (unitless area

reduction)

φ effective fast neutron fluence (nm2) (gt10MeV)


- 365 -

(2) Strain hardening exponent n

( ) ( ) ( )ZrynnTnn φsdot=

where ( )Tn is

Tlt4194 K ( ) 114050=Tn

4194 K lt T lt 10990772 K

( ) 2 3

6 2 10 3

9490 10 1165 10

1992 10 9588 10

n T T

T T

minus minus

minus minus

= minus times + times

minus times + times

10990772 K lt T lt 1600 K ( ) TTn 41052226551190 minustimes+minus=

T gt 1600 K ( ) 6089531=Tn

( )φn is

φ lt 01times1025 nmsup2 ( ) φφ 25104803211 minustimes+=n

01times1025nmsup2 ltφlt 2times1025 nmsup2 ( ) φφ 251009603691 minustimes+=n

2times1025 nmsup2 ltφlt 75times1025 nmsup2 ( ) φφ 2510008727054351 minustimes+=n

φgt75times1025 nmsup2 ( ) 6089531=φn ( )Zryn is n(Zry) = 1 for Zircaloy-4 n(Zry) = 16 for Zircaloy-2

T temperature (K) φ effective fast neutron fluence (nm2gt10MeV)

(3) Starin rate sensitivity exponent m T lt 750K 0150=m

750K lt T lt 800K 5443380104587 4 minustimes= minus Tm

T gt 800K 20701010241243 4 minustimes= minus Tm

Fig433 Yield stress of Zircaloy-4 in the FRAPCON( FRAPTRAN) model (ICPLAS=6)

300 400 500 600 700 800 900

100200

300400500

600700800

900

1000

Zry-4 strain rate=1s fluence=1026 (nm2) Cold work=0

Cla

ddin

g Y

ield

Stre

ss (

MPa

)

Temperature (K)

Zry-4 strain rate=1s fluence=0 Cold work=0


- 366 -

(7) Zircaloy irradiation growth The equation representing Zircaloy irradiation growth is selected from among the

following

Zircaloy irradiation growth (1) MATPRO-Version 09 Condition Ref

( )[ ][ ] [ ] [ ]ΔL

LA T t f CWZ= minus +exp 2408 1 3 1 2 0

12φ

ΔL

L irradiation growth (minus)

A = times minus1407 10 16 (nm2)minus05 Ttemperature (K)

φ tfast neutron fluence (nm2) (E gt1 MeV)

fzfactor in the axial direction

CWcold work

ICAGRW=1

213

The following parameters are used in the case of ICAGRW=1


CATEXF Factor in the axial direction fz 005

COLDW Cold work CW 081

Zircaloy irradiation growth (2) Manzel Condition Ref

( )ΔL

Lt

n

0

2101 10= sdot φ ΔL

L0

irradiation growth ()

φ sdot t fast neutron fluence (ncm2)

n constant between 065minus067 (066 is used)

ICAGRW=2

Zircaloy irradiation growth (3) Hanerz

( )ΔL

Lc t

o

perp = times minus6 10 12 0 4φ

ΔL

Lc

o

perp growth of single crystal in the direction perpendicular to

the C axis (minus) φ sdot t fast neutron fluence (ncm2)

ICAGRW=3

Zircaloy irradiation growth (4) Hesketh

ICAGRW=4 445

ΔL

LA t

o

= sdot sdotφ ΔL

Lo

irradiation growth (minus)


Afitting constant ( )= times minus5 10 14



- 367 -

(8) Zircaloy specific heat and density Zircaloy specific heat MATPRO-Version 09 Condition Ref

Cp T T= + minus times minus24511 015558 33414 10 5 2

Cpspecific heat (JkgK)

T temperature (K) (300-1090K)

No restrictions

213

Zircaloy density MATPRO-Version 09 Condition Ref ρ = 655 (gcm3) No restrictions 213

(9) Zircaloy cladding corrosion Cladding corrosion rate (1) EPRI Condition Ref

Pre-transition corrosion rate ( ) ( )dS dt A S Q RT exp = minus21

Post-transition corrosion rate ( )dS dt C Q RT exp = minus 2

( )C C U MP= +0 φ

Oxide layer thickness at transition ( )D Q RT ETexp minus minus3

dS dt corrosion rate (μmday)

Soxide layer thickness (μm)

T temperature at oxide-metal interface (K)


Rgas constant (1987 calmol_K)

A = times6 3 109 (μm3day) Q1 32289= (calmol)

7

0 10048 times=C (μmday) U = times2 59 108 (μmday)

M = times minus7 46 10 15 (cm2sn) P = 024

Q2 27354= (calmol) D = times2 14 107 (μm)

Q3 10763= (calmol) E = times minus117 10 2 (Kminus1)

ICOPRO=1

212

Cladding corrosion rate (2) MATPRO-A ICORRO=2 (PWR)

ICORRO=3 (BWR)

213

Pre-transition ( )( )S A t T So= times sdot sdot minus +minus4 976 10 156603 31

3 exp

Post-transition ( )S A t T STRAN= sdot sdot minus +82 88 14080 exp

Oxide thickness at transition ( )S TTRAN = times minusminus7 749 10 7906 exp

SOxide thickness（m） ttime（days）

T temperature at metal-oxide interface (K)

A15（for PWR） =9（for BWR）

When ICORRO=0（default value） no oxidation of cladding is assumed to occur in the

calculation The following parameters are used in the case of ICORROgt0


- 368 -


FCORRO

Accelerating factor for corrosion

( )ds

dtT FCORRO

ds

dt= + sdot1 0 04 Δ

ΔT temperature gap between inner and outer side of oxide layer（metal-oxide interface and waterside surface）

00

Comment Piling-Bwedworth Ratio of ZrO2

ZrO2 Zr ox ZrO2X = Y + times Xρ ρ α ρ

X Oxide layer thickness (μm)

Y Zircaloy thickness converted into oxide (μm)

Zrρ Zircaloy density 655 or 649 (MATPRO-A) (gcm3)

ZrO2ρ Oxide density 56 or 582 (MATPRO-A) (gcm3 )

oxα Oxygen weight fraction in oxide layer ZrO2 = 0260

Piling-Bedworth ratio = 2

(10 026)Zr

ZrO

X

Y

ρρ

=minus

=151 or 156 or 158

432 Cladding oxide properties

(1) Cladding oxide thermal conductivity ZrO2 thermal conductivity (2) MATPRO-11 Condition Ref

310274 109511043610412961 TTTk minusminusminus timesminustimes+timesminus= (WmK)

T temperature (K)

IZOX=2

Default 34

ZrO2 thermal conductivity MATPRO-A No

restrictions

213 40835 181 10OXk Tminus= + times sdot (WmK)

Reference information Slovacek derived the followings

OXk =107 plusmn 013 at 249C 112 plusmn 012 at 252C 117 plusmn 011 at 254C

and 123 plusmn 010 (WmK)

442

The OXk equation of an oxide layer has the following fitting parameter to incorporate the

effects of the thickness of oxide layer into the thermal conductivity of oxide layer

Variable name Contents Default value

CNOX

The thermal conductivity equation of ZrO2 is given by

( )K T f= + times sdot sdotminus0 835 181 10 4

f CNOX S= minus sdot1 SOxide thickness (μm)

00


- 369 -

(2) ZrO2 thermal expansion emissivity specific heat and density ZrO2 thermal expansion MATPRO-A Condition Ref

36 103421087 minusminusΔ sdotminussdotsdot= TL

L ( )KT 1478300 lele No restriction 213

Emissivity of ZrO2 is assumed as 0809ε =

ZrO2 specific heat Couglin

No restrictions

Cp T T T= + minus times + timesminus minus46 2 134 2 936 10 1 103 2 6 348 433 (JkgK)

T temperature (K) ZrO2 density MATPRO-A

No restrictions 213 65=ρ (gcm3)


(3) Youngrsquos modulus of ZrO2

Youngrsquos modulus of ZrO2 MATPRO-A Condition Ref 117 10637110773 times+timesminus= TY 300ltTlt1478 K

117 102252100248 times+timesminus= TY 1478ltTltTsol

(Tsol=melting temperature of oxide 2970K)

1=Y TsolltT

Y youngrsquos modulus (Pa) Ttemperature (K)

No

restriction

213


- 370 -

44 SUS304 Stainless Steel source literatures are not open

SUS304 thermal conductivity Condition Ref

T temperature RT~848 K 401065 1326 10k Tminus= + times

(WcmK) ITMC=41

ldquo ITMC=1rdquo designates Zircaloy thermal conductivity as a Default

SUS304 Oxide layer thermal conductivity ZOX=41

Not examined

ldquo ZOX=1rdquo designates the thermal conductivity of oxide layer of Zircaloy as a Default

SUS304 oxide emissivity OXEMS=41

Not examined Assumed to be identical to that of oxide layer of Zircaloy

ldquo OXEMS=1rdquo designates the emissivity of oxide layer of Zircaloy as a Default

SUS304 Youngrsquos modulus Condition Ref

6218400 9326( 27315) 98067 10 (Pa)E T= minus minus times sdot

Ttemperature(K) IZYG=41

SUS304 thermal expansion Condition Ref

For range T=300 - 848 K thermal expansion coefficient (1K) is 5 9

12 2 15 3

1629 10 3285 10

2198 10 1629 10

T

T T

λ minus minus

minus minus

= times + timesminus times + times

Therefore the thermal expansion is 0LTL sdotΔsdot=Δ λ

ICATHX=41

SUS304 oxide thermal expansion OXTHM=41 Not examined Assumed to be identical to that of oxide layer of

Zircaloy

ldquo OXTHM=1rdquo designates the thermal expansion of oxide layer of Zircaloy as a Default

SUS304 irradiation growth ICAGRW=41

Assumed to be null

SUS304 Poissonrsquos ratio Condition Ref

( )50265 7875 10 27315v Tminus= + times minus Ttemperature (K) ICPOS=41


- 371 -

ldquo ICPOS=1rdquo designates the Poissonrsquos ratio of Zircaloy as a Default

SUS304 irradiation creep Condition Ref

( ) ( ) 1 2 2 3 exp E C E C E t C Cε ϕσ ϕ ϕ= sdot sdot minus sdot sdot +

ε creep strain rate (mmhr) E = neutron energy (MeV) ϕ = fast neutron flu(ncm2s)

σ =equivalent stress (psi) t time (hour)


CRPEQ =41

SUS304 stress-strain relationship Condition Ref

σ ε= K n To be obtained from material X

σ true stress (kgmm2) ε true strain


8 5 2

8 5 20 2

9 6 21 2

9 5 21 2

3945 10 225 10 ( 0 n cm ) (1)

7550 10 613 10 ( 15 10 n cm ) (2)

1378 10 179 10 ( 145 10 n cm ) (3)

1093 10 660 10 ( 605 10 n cm ) (4)

Y

Y

Y

Y

T t

T t

T t

T t

σ φ

σ φ

σ φ

σ φ

minus

minus

minus

minus

= times minus times sdot = sdot

= times minus times sdot = times sdot



(Pa) (C)Y Tσ

Eqs(1)-(4) are interpolated with respect to fluence tφ

However above the level of 21 2605 10 n cmtφ minus= times sdot Eq(4)

holds Strain hardening after yielding is not assumed so that n 0YK σ= = in nKσ ε=

ICPLAS=41

SUS304 density MATPRO-A Condition Ref

3 379 10 kgmsdot at 300K ICDENS =41 213

SUS304 specific heat MATPRO-A Condition Ref

5 2300 1558 326 0298 956 10 (JkgK )

1558 558228p

p

T K C T T

T K C

minusle lt = + sdot minus sdot

ge = ICSPH=41 213


- 372 -

45 Other Materials Properties and Models

(1) Gap thermal conductance

Gap conductance (1) Ross and Stoute Condition Ref

Refer to section 25 Here thermal conductivities of helium

nitrogen krypton and xenon are as follows(MAPRO-Version 09) K THe = times minus3366 10 3 0 668

2

4 08462091 10NK Tminus= times

K TKr = times minus4 726 10 5 0 923

5 087240288 10XeK Tminus= times

Kthermal conductivity (WmK) Ttemperature (K)

IGAPCN=0

(default value) 217

Gap conductance (2) MATPRO-Version 09

IGAPCN=1 213

( )h F h Fhgap c= minus +1 0

hgap gap conductance (Wm2K)

h0 non-contact gap conductance (Wm2K)

hc contact gap conductance (Wm2K)

F circumference ratio of pellet-cladding contact part

hK

rmix

0 =+Δ δ

Kmix thermal conductivity of mixed gas (WmK)

δ = times minus4 4 10 6 rms surface roughness of pellet and cladding (m)

Δr hot gap width (m)

h PK

cn mix= times sdot +minus5 10 4

δ

P contact pressure between pellet and cladding (Nm2)

when Flt1 P=0 when 0ltPlt1000 psi n=1

when Pgt1000 psi n=05

F

ar

RF

a=

sdot

+

+1

1001429

0 3

1

2Δ

RF pellet hot diameter (m) a a1 2 constant depending on burnup (x)

when xgt600MWdt-U a F1 100 98= minus prime a F2 4 05= minus prime

prime = minusminus

+F

x1

1

600

10001

4

when x le 600 MWdt-U a1 100= a2 4=


- 373 -


h h hgap g s= +

hgap gap conductance (WcmdegC)

hg non-contact gap conductance (Wcm2degC)

hs contact gap conductance (Wcm2degC)

( ) ( )hK

C R R g g rgg=

+ + + +1 2 1 2 Δ

K g gap gas thermal conductivity (Wcm2degC)

He00021 Ar000029 Kr00001

Cconstant for contact pressure

C = 25 for 100 kgfcm2


R1 and R2surface roughness of pellet and cladding (cm)

g1 and g2temperature jump distance (cm)

He10 10 4times minus Ar5 10 4times minus Kr1 10 4times minus

Xelt times minus1 10 4 (value of g1 + g2)

Δr hot gap width (cm)

hK P

a Hsm=

sdotsdot

KK K

K Km =sdot

+2 1 2

1 2

Pt Ys

ID= sdot2 a a R= 0

( )R R R= +1

2 12

22

K1 and K2thermal conductivities of pellet and cladding

(WcmdegC)

K

KKm

1

2

0 038

0140 05977

==

=

Pcontact pressure (kgfcm2)

t cladding thickness (cm)

IDcladding inner diameter (cm)

Yscladding yield stress (Ys = 1980 kgfcm2)

aaverage radius of contact part (cm)

a0 = 05 cm12

HMeyerrsquos hardness of soft side (kgcm2) H=3Ys

IGAPCN=2 217


- 374 -

Gap conductance (4) Dean Condition Ref h h hgap s g= +

hgapgap conductance (Btuhft2degF)

hg non-contact gap conductance (Btuhft2degF)

hs contact gap conductance (Btuhft2degF)

hP

M

K K

K Ks = sdot sdotsdot+

sdotsdot

280

1

2

1 2

1 2 1 2

12

λ λ

P contact pressure (psi)

P=267minus675psi (186minus457 kgfcm2)

M2 Meyerrsquos hardness of zircaloy (psi)

M2=120000psi (844kgcm2)


K1 = 220 Btuhft2degF (0038 Wcm2degC)

K2 = 809 Btuhft2degF (014 Wcm2degC) 1λ and λ 2 wavelengths of surface profile of pellet and

cladding (inch) λ λ1 2sdot ~ 2RUO

RZr (roughness)＝ ( )50 10510 6

minus+ minustimes rms in

hK

gg=sdot052 δ

Kg gap gas thermal conductivity (Btuhft2degF)

Ar00168 He0139 FP00088

δ gap between pellet and cladding (ft) δ ~ Ruo2

modified Deanrsquos equation

h Ps = 0 6 hK

gg=

+ times minusδ 14 4 10 6

IGAPCN=3


The following fitting parameter is used in the case of IGAPCN=0 or 2


R1 Pellet surface roughness (μm) 10


- 375 -

(2) Cladding surface heat transfer

To obtain a cladding surface heat transfer coefficient empirical equations such as those

by Chen and Dittus-Boelter etc are used as described in chapter 2 When the fitting

parameter AKFAC is designated the surface heat transfer coefficient is multiplied by

AKFAC The default value of AKFAC is 10

When the input parameter ISCNHAL = 1 the following Slovacekrsquos empirical

equation(442) for the Halden reactor can be used

2276 00110 000001152h q q= + sdot + sdot (kWm2K) (451)

q linear heat rate (kWm)

(3) Fission gas atom diffusion constant

Fission gas atom diffusion constant (1) Turnbull Condition Ref

( )D RT S j V F= times minus times + divide timesminus minus7 6 10 7 10 2 1010 4 2 40 exp υ (m2s)

Rgas constant =1987 (calmolK)

Ttemperature (K)

Satom jump distance = Ω2

3

Ωatom volume = times minus4 09 10 29 (m3) ( )j RTυ = minus times10 552 1013 4exp

( )

( )V

S ZV

Z

K Z

j S ZV

s o

s o

=+

+ prime

+

minus

part

υ part

2

2 2

1

2

21

41

part s = 1015 (mminus2 sink intensity)

Z = 2 (number of lost points) Kprime = 104 (loss ratio per atom) ( )V RT0

4552 10= minus timesexp

F = 1019 (fissionsm3s fission rate)

IDCNST=1 228

Fission gas atom diffusion constant (2) White

IDCNST=2 235 ( )TD 6614exp10091 17

1 minussdot= minus (m2s) Tgt1650 K

( )TD 22884exp10142 132 minussdot= minus (m2s) 1381ltTlt1650 K

( )TD 9508exp10511 173 minussdot= minus (m2s) Tlt1381 K


- 376 -

Fission gas atom diffusion constant (3) Kitajima amp Kinoshita Condition Ref

321 DDDD ++= (cm2s)

minussdotminussdot= minus

1280

11

9861

1001exp102

521

1 TD


1300

11

9861

10762exp102

440

2 TD

FD 403 102 minussdot=

1910=F (Fissionm3s fission rate) T temperature(K)

IDCNST=3 238

Fission gas atom diffusion constant (4) Modified Turnbull

IDCNST=4 441

( )10 4

2 40

AM1 76 10 exp 7 10

BM1 ( 2 10 )v

D RT

s j V F

minus

minus

= sdot times minus times

+ sdot + times (m2s)

where Name-list parameter AM1=10 BM1=10 (Default)

Fission gas atom diffusion constant (5) White model

IDCNST=5 226

( )( )( )

10 4

17

22

AM1 76 10 exp 7 10

15 10 exp 10600

767 10 exp 2785

D RT

R T

R T

minus

minus

minus


+ times minus

+ times sdot minus

(m2s) ( KT J sgR sdot )

Fission gas atom diffusion constant (6) Kogai model

IDCNST=6 232

a b cD D D D= + + (m2s)

( )10 976 10 684 10 40 exp( 70000 )aD Bu RTminus minus= times + times sdot minus

Bu GWdt (40 is an upper limit) though provisionally Bu is

assumed to be 0 in the code because this term gives an excessive

increase to Da in high burnup region

2 1 3 10 34454 10 (m)b vD s j V s minus= = Ω = times

13 110 exp( 55200 ) (s )vj RT minus= minus

( )( )052 2 2 2

0 0( ) 2 1 4 ( ) 1s v sV ZV Z K Z j s ZVα α = + + + minus

15 210 (m )sα minus= Z=100 0 exp( 55200 )V RT= minus

5 4 1 5 10 2 10 (s )K f minus minus= times Ω = times

40 3 192 10 fissions(m s) fixed at 10 cD f fminus= times sdot =

T Kelvin R=1987 cal(molK)

In the code ( ( ))a b cD D D D= sdot + sdot +FACD AM1 BM1


- 377 -

(4) He-Xe gas counter diffusion constant

He-Xe gas counter diffusion constant Present Condition Ref

DkT

m n d12

1

2

122

3

8 2

1=

ππ (m2s)

mm m

m m =

+1 2

1 2

(mweight of 1 molecule)

( )d d d12 1 21

2= + (ddiameter of molecule)

= times minus345 10 10 (m) n n n= +1 2 (molecular density)

kBoltzmannrsquos constant ( )= times minus1380 10 16 erg K

Ttemperature (K)

No

restrictions 253

46 Burnup Calculation of Gd2O3-Containing Fuel In FEMAXI when the weight fraction of Gd contained in a pellet is given its burnup is

calculated in the following manner

(1) Mass number and density A) Mass number

235U AU235=235 238U AU238=238 O AO=16

GdAGd=15725（average mass number）

B) Density

UO2 ( )396102

cmgUO =ρ Gd ( )3208 cmgGd =ρ

(Due to the fact that the density of Gd2O3 is unavailable the above values are

provisionally used at present)

(2) Calculation of abundance ratio of gadolinium XGｄ

Denoting the concentration of 235U and weight fraction of Gd as ENR and WGｄ

respectively XGｄ is calculated using

( ) ENRAENRAA UUU sdot+minus= 235238 1


- 378 -

( ) ( ) ( ) 5112 GdOGdGdOU

GdGdGd XAAXAA

XAW

sdot++minussdot+sdot=

(Example) When ENR=005 and WGｄ=01

85237050235950238 =times+times=UA

( ) ( ) ( )16250

511625157121685237

2515710

=times++minustimes+

=

Gd

GdGd

Gd

X

XX

X

(3) Calculation of pellet density

fWW

UO

UO

Gd

OGd

sdot+

=

2

232

1

ρρ

ρ f theoretical density ratio

A

XAW

GdGdO

OGd

sdot= 51

32 (weight fraction of gadolinia)

( )A

XAW GdUO

UO

minus=

12

2 (weight fraction of UO2)

( ) GdGdOGdUO

OGdGdO

OUUO

XAXAA

AAA

AAA

sdot+minus=

sdot+=

sdot+=

512

51

2

1

51

2

(Example) When ENR=005 WG ｄ =01 and f=095 since XGd=01625 2UOA =26985

51GdOA =18125 and A=25546 32OGdW =01153 and

2UOW =08847

Thus ( )302310

9610

88470

208

11530950

cmg=+

=ρ

(4) Calculation of burnup (Example) When ENR=005 WGd=01 and f=095 and pellet diameter is 1 cm and the pellet

is irradiated at 300 Wcm for 1000 days the burnup is given as follows

The volume per unit length of the pellet is

( )2 2 3314 05 07854 Pr cm cmπ = times =


- 379 -

Thus the weight per unit length is

( )( )6

10023 07854 7872

7872 10

g cm

t cmminus

times =

= times

The weight fraction WU of U is

( ) ( )( ) ( )

2

2

6

6

2378508847 07798

26985

300 10 100048871

7872 10 07798

UU UO

UO

AW W

A

MW cm dBu MWd tU

t cm

minus

minus

= sdot = sdot =

times times= =

times times

For the burnup of a UO2 pellet under the same conditions

( )2

31096 095 10412 UO g cmρ = times =

Therefore the weight per unit length is

( )( )6

10412 07854 8178

8178 10

g cm

t cmminus

times =

= times

88140185269

852372

2

=sdot== UOUO

UU W

A

AW

Thus

( ) ( )( ) ( )

6

6

300 10 100041620

8178 10 08814

MW cm dBu MWd tU

t cm

minus

minus

times times= =

times times

Accordingly when WGｄ=01 under the same conditions the burnup of fuel containing

gadolinia is 1174 times that of UO2 fuel


- 380 -

References 4 (41) Washington ABG Preferred Values for the Thermal Conductivity of Sintered

Ceramic Fuel for Fast Reactor Use UKAEA TRG 2236(D) (1973)

(42) Ishimoto S Hirai M Ito K and Korey Y J Nucl Sci Technol 31 pp796-802 (1994)

(43) Wiesenack W Vankeerberghen M and Thankappan R Assessment of UO2 Conductivity Degradation Based on In-pile Temperature Data HWR-469 (1996)

(44) Forsberg K Lindstrom F and Massih AR Modelling of Some High Burnup Phenomena in Nuclear Fuel IAEA TCM on Water Reactor Fuel Element Modelling at High Burnup and its Experimental Support Windermere UK Sept(1994)

(45) Kjaer-Pedersen N Rim Effect Observations from the Third Riso Fission Gas Project Fission Release and Fuel Rod Chemistry Related to Extended Burnup IAEA -TECDOC-697 (1993)

(46) Baron D and Couty JC A Proposal for a Unified Fuel Thermal Conductivity Model Available for UO2 (U-PuO)2 and UO2-Gd2O3 PWR Fuel IAEA TCM on Water Reactor Fuel Element Modelling at High Burnup and its Experimental Support Windermere UK Sept(1994)

(47) Lucuta PG Matzke Hj and Hastings IJ A Pragmatic Approach to Modelling Thermal Conductivity of Irradiated UO2 Fuel Review and Recommendations JNuclMater 232 pp166-180 (1996)

(48) Fukushima S Ohmichi T Maeda A and Watanabe H The Effect of Gadolinium Content on the Thermal Conductivity of Near-Stoichiometric (UGd)O2 Solid Solutions JNuclMater 105 pp201-210 (1982)

(49) Baron D About the Modelling of Fuel Thermal Conductivity Degradation at High-Burnup Accounting for Recovering Process with Temperature

Proceedings of the Seminar on Thermal Performance of High Burn-up LWR fuel pp129-143 Cadarache France 3-6 March (1998)

(410) Kamimura J Ide H Ohira K Itagaki N Thermal and Mechanical Behavior Modeling For High Burnup Fuel Proc of 2008 Water Reactor Fuel Performance Meeting No8118 October Seoul Korea (2008)

(411) Lanning D Beyer C Geelhood K FRAPCON-3 Updates Including Mixed- Oxide Fuel Properties NUREGCR-6534 Vol4 PNNL-11513 (2005)

(412) Martin DG A Re-Appraisal of the Thermal Conductivity of UO2 and Mixed (UPu) Oxide Fuels J Nucl Mater 110 pp73-94 (1982)

(413) Christensen JA et al Melting Point of Irradiated Uranium Dioxide WCAP-6065 (1965)

(414) Massih AR Models for MOX fuel behavior A selective review SKI Report

200610 (2006)


- 381 -

(415) Benoit Petitprez Ramp Tests with Two High Burnup MOX Fuel Rods in IFA-6293 HWR-714 (2002)

(416) Rolstad E Hanevik A and Knudsen K Measurements of the Length Changes of UO2 Fuel Pellets during Irradiation Paper 17 in Enlarged Halden Programme Group Meeting on Computer Control and Fuel Research Related to Safe and Economic Plant Operation Sandefjord Norway June 1974 HPR-188 (1975)

(417) USNRC WASH-1236 (1972)

(418) Marlowe MO Predicting In-Reactor Densification Behavior of UO2 Trans ANS 17 p166 (1974)

(419) Schlemmer FU et al IAEA Technical Committee Meeting on Recycling of Pu and U in Water Reactor Fuels Cadarache France Nov (1989)

(420) Ichikawa M et al Journal of Japan Nuclear Society vol39 p 93 (1997) [in Japanese]

(421) Tachibana T Furuya H and Koizumi M Dependence on Strain Rate and Temperature Shown by Yield Stress of Uranium Dioxide J Nucl Sci Technol 13 pp497-502 (1976)

(422) Rodford C and Terwilliger R Compressive Deformation of Polycrystalline UO2 JAmCeramSoc 58 (1975)

(423) MacEwan JR Grain Growth in Sintered Uranium Dioxide I Equiaxed Grain Growth J Am Cer Soc 45 p37 (1962)

(424) Lyons MF Boyle RF Davis JH Hazel VE and Rowland TC Nucl Eng Design 21 p167 (1972)

(425) Cadelli N et al Post Irradiation of Plutonium Bearing Fuel Elements Irradiated in LWR Power Plants - A European Community Campaign IAEA MO1 (1984)

(426) Martin DG The Thermal Expansion of Solid UO2 and (U Pu) Mixed Oxides ndash A Review and Recommendations JNuclMater152 pp94-101 (1988)

(427) Philipponneau Y Thermal Conductivity of (U Pu)O2-x Mixed Oxide Fuel JNuclMater188 pp194-197 (1992)

(428) Duriez C Alessandri JP Gervais T and Philipponneau Y Thermal conductivity of hypostoichiometric low Pu content (U Pu)O2-x mixed oxide JNuclMater 277 pp143-158 (2000)

(429) Gates GA Takano K and White BJ Thermal Performance of MOX Fuel HWR-589 (1999)



- 382 -

(431) Nutt AW Jr et al JAmCeramSoc 53 p205 (1970)

(432) Yamada K et al TechnolReps Osaka Univ 47 p181 (1997)

(433) Tokar M Nutt A and Keenan TK Linear Thermal Expansion of Plutonium Dioxide NuclTechnol17 pp147-152 (1973)

(434) Fisher EF and Renken CJ Single-Crystal Elastic Moduli and the hcp-bcc Transformation in Ti Zr and Hf Phys Rev ppA482-A494 (1954)

(435) Franklin DG Lucas GE and Bement AL Creep of Zirconium Alloys in Nuclear Reactors ASTM STP-815 pp103-105 (1983)

(436) Specialists committee for nuclear fuel safety Research on the non-elastic mechanical properties of Zircaloy cladding Report of Nuclear Safety Research Association February 1986 [in Japanese]

(437) McGrath MA In-Reactor Creep of Zircaloy Cladding Proc 2000 IntTopical Meeting on LWR Fuel Performance Park City Utah USA (2000)

(438) Rosinger HE Bera PC et al Steady-State Creep of Zircaloy-4 Fuel Cladding from 940 to 1873 K JNuclMater 82 p286(1979)

(439) Donaldson AT Horwood RA and Healey T Biaxial Creep Deformation of Zircaloy-4 in the High Alpha-phase Temperature Range Water Reactor Fuel Element Performance Computer Modelling edited by J Gittus pp81-95 APPLIED SCIENCE PUBLISHERS London and New York (1983)

(440) Hesketh RV Non-linear Growth in Zircaloy-4 JNuclMater30 pp217-222 (1969)

(441) Turnbull JA White RJ and Wise C The Diffusion Coefficient for Fission Gas Atoms in Uranium Dioxide (IAEA-TC-65935) IWGFPT32 p174

(442) Slovacek M Zirconium Oxide Conductivity Assessment with Cladding Elongation Data from IFA-5159 HWR-354 (1993)

Acknowledgement The authors thank DrJinichi Nakamura of JAEA and users of the preceding version

FEMAXI-6 for their cooperation for improvement modeling and information-gathering on

material properties and characteristics Development of FEMAXI-7 would not have been

possible without their kind cooperation


　　国国際際単単位位系系（（SSII））

乗数　接頭語記号乗数　接頭語記号

1024 ヨタＹ 10-1 デシ d1021 ゼタＺ 10-2 センチ c1018 エクサＥ 10-3 ミリ m1015 ペタＰ 10-6 マイクロ micro1012 テラＴ 10-9 ナノ n109 ギガＧ 10-12 ピコ p106 メガＭ 10-15 フェムト f103 キロｋ 10-18 アト a102 ヘクトｈ 10-21 ゼプト z101 デカ da 10-24 ヨクト y

表５SI 接頭語

名称記号 SI 単位による値

分 min 1 min=60s時 h 1h =60 min=3600 s日 d 1 d=24 h=86 400 s度 deg 1deg=(π180) rad分 rsquo 1rsquo=(160)deg=(π10800) rad秒 rdquo 1rdquo=(160)rsquo=(π648000) rad

ヘクタール ha 1ha=1hm2=104m2

リットル Ll 1L=11=1dm3=103cm3=10-3m3

トン t 1t=103 kg

表６SIに属さないがSIと併用される単位

名称記号 SI 単位で表される数値

電子ボルト eV 1eV=1602 176 53(14)times10-19Jダルトン Da 1Da=1660 538 86(28)times10-27kg統一原子質量単位 u 1u=1 Da天文単位 ua 1ua=1495 978 706 91(6)times1011m

表７SIに属さないがSIと併用される単位でSI単位で表される数値が実験的に得られるもの


キュリー Ci 1 Ci=37times1010Bqレントゲン R 1 R = 258times10-4Ckgラド rad 1 rad=1cGy=10-2Gyレム rem 1 rem=1 cSv=10-2Svガンマ γ 1γ=1 nT=10-9Tフェルミ 1フェルミ=1 fm=10-15mメートル系カラット 1メートル系カラット = 200 mg = 2times10-4kgトル Torr 1 Torr = (101 325760) Pa標準大気圧 atm 1 atm = 101 325 Pa

1cal=41858J（｢15｣カロリー）41868J（｢IT｣カロリー）4184J（｢熱化学｣カロリー）

ミクロン micro 1 micro =1microm=10-6m

表10SIに属さないその他の単位の例

カロリー cal

(a)SI接頭語は固有の名称と記号を持つ組立単位と組み合わせても使用できるしかし接頭語を付した単位はもはや　コヒーレントではない(b)ラジアンとステラジアンは数字の１に対する単位の特別な名称で量についての情報をつたえるために使われる

　実際には使用する時には記号rad及びsrが用いられるが習慣として組立単位としての記号である数字の１は明　示されない(c)測光学ではステラジアンという名称と記号srを単位の表し方の中にそのまま維持している

(d)ヘルツは周期現象についてのみベクレルは放射性核種の統計的過程についてのみ使用される

(e)セルシウス度はケルビンの特別な名称でセルシウス温度を表すために使用されるセルシウス度とケルビンの

　単位の大きさは同一であるしたがって温度差や温度間隔を表す数値はどちらの単位で表しても同じである

(f)放射性核種の放射能（activity referred to a radionuclide）はしばしば誤った用語でrdquoradioactivityrdquoと記される

(g)単位シーベルト（PV200270205）についてはCIPM勧告2（CI-2002）を参照

（c）３元系のCGS単位系とSIでは直接比較できないため等号「　　」

　　は対応関係を示すものである

（a）量濃度（amount concentration）は臨床化学の分野では物質濃度

　　（substance concentration）ともよばれる（b）これらは無次元量あるいは次元１をもつ量であるがそのこと　　を表す単位記号である数字の１は通常は表記しない

名称記号SI 基本単位による

表し方

粘度パスカル秒 Pa s m-1 kg s-1

力のモーメントニュートンメートル N m m2 kg s-2

表面張力ニュートン毎メートル Nm kg s-2

角速度ラジアン毎秒 rads m m-1 s-1=s-1

角加速度ラジアン毎秒毎秒 rads2 m m-1 s-2=s-2

熱流密度放射照度ワット毎平方メートル Wm2 kg s-3

熱容量エントロピージュール毎ケルビン JK m2 kg s-2 K-1

比熱容量比エントロピージュール毎キログラム毎ケルビン J(kg K) m2 s-2 K-1

比エネルギージュール毎キログラム Jkg m2 s-2

熱伝導率ワット毎メートル毎ケルビン W(m K) m kg s-3 K-1

体積エネルギージュール毎立方メートル Jm3 m-1 kg s-2

電界の強さボルト毎メートル Vm m kg s-3 A-1

電荷密度クーロン毎立方メートル Cm3 m-3 sA表面電荷クーロン毎平方メートル Cm2 m-2 sA電束密度電気変位クーロン毎平方メートル Cm2 m-2 sA誘電率ファラド毎メートル Fm m-3 kg-1 s4 A2

透磁率ヘンリー毎メートル Hm m kg s-2 A-2

モルエネルギージュール毎モル Jmol m2 kg s-2 mol-1

モルエントロピーモル熱容量ジュール毎モル毎ケルビン J(mol K) m2 kg s-2 K-1 mol-1

照射線量（Ｘ線及びγ線）クーロン毎キログラム Ckg kg-1 sA吸収線量率グレイ毎秒 Gys m2 s-3

放射強度ワット毎ステラジアン Wsr m4 m-2 kg s-3=m2 kg s-3

放射輝度ワット毎平方メートル毎ステラジアン W(m2 sr) m2 m-2 kg s-3=kg s-3

酵素活性濃度カタール毎立方メートル katm3 m-3 s-1 mol

表４単位の中に固有の名称と記号を含むSI組立単位の例

組立量SI 組立単位

名称記号面積平方メートル m2

体積立法メートル m3

速さ速度メートル毎秒 ms加速度メートル毎秒毎秒 ms2

波数毎メートル m-1

密度質量密度キログラム毎立方メートル kgm3

面積密度キログラム毎平方メートル kgm2

比体積立方メートル毎キログラム m3kg電流密度アンペア毎平方メートル Am2

磁界の強さアンペア毎メートル Am量濃度 (a) 濃度モル毎立方メートル molm3

質量濃度キログラム毎立法メートル kgm3

輝度カンデラ毎平方メートル cdm2

屈折率 (b) （数字の）　１ 1比透磁率 (b) （数字の）　１ 1

組立量SI 基本単位

表２基本単位を用いて表されるSI組立単位の例

名称記号他のSI単位による

表し方SI基本単位による

表し方平面角ラジアン(ｂ) rad 1（ｂ） mm立体角ステラジアン(ｂ) sr(c) 1（ｂ） m2m2

周波数ヘルツ（ｄ） Hz s-1

力ニュートン N m kg s-2

圧力応力パスカル Pa Nm2 m-1 kg s-2

エネルギー仕事熱量ジュール J N m m2 kg s-2

仕事率工率放射束ワット W Js m2 kg s-3

電荷電気量クーロン C s A電位差（電圧）起電力ボルト V WA m2 kg s-3 A-1

静電容量ファラド F CV m-2 kg-1 s4 A2

電気抵抗オーム Ω VA m2 kg s-3 A-2

コンダクタンスジーメンス S AV m-2 kg-1 s3 A2

磁束ウエーバ Wb Vs m2 kg s-2 A-1

磁束密度テスラ T Wbm2 kg s-2 A-1

インダクタンスヘンリー H WbA m2 kg s-2 A-2

セルシウス温度セルシウス度(ｅ) K光束ルーメン lm cd sr(c) cd照度ルクス lx lmm2 m-2 cd放射性核種の放射能（ｆ）ベクレル（ｄ） Bq s-1

吸収線量比エネルギー分与カーマ

グレイ Gy Jkg m2 s-2

線量当量周辺線量当量方向

性線量当量個人線量当量シーベルト（ｇ） Sv Jkg m2 s-2

酸素活性カタール kat s-1 mol

表３固有の名称と記号で表されるSI組立単位SI 組立単位

組立量


バール bar １bar=01MPa=100kPa=105Pa水銀柱ミリメートル mmHg 1mmHg=133322Paオングストローム Å １Å=01nm=100pm=10-10m海里Ｍ１M=1852mバーン b １b=100fm2=(10-12cm)2=10-28m2

ノット kn １kn=(18523600)msネーパ NpベルＢ

デジベル dB

表８SIに属さないがSIと併用されるその他の単位

SI単位との数値的な関係は　　　　対数量の定義に依存

名称記号

長さメートル m質量キログラム kg時間秒 s電流アンペア A熱力学温度ケルビン K物質量モル mol光度カンデラ cd

基本量SI 基本単位

表１SI 基本単位


エルグ erg 1 erg=10-7 Jダイン dyn 1 dyn=10-5Nポアズ P 1 P=1 dyn s cm-2=01Pa sストークス St 1 St =1cm2 s-1=10-4m2 s-1

スチルブ sb 1 sb =1cd cm-2=104cd m-2

フォト ph 1 ph=1cd sr cm-2 104lxガル Gal 1 Gal =1cm s-2=10-2ms-2

マクスウｪル Mx 1 Mx = 1G cm2=10-8Wbガウス G 1 G =1Mx cm-2 =10-4Tエルステッド（ｃ） Oe 1 Oe　 (1034π)A m-1

表９固有の名称をもつCGS組立単位

（第8版2006年改訂）

この印刷物は再生紙を使用しています

Abstract+Contents

1MAN

2M1-6

2M7_Diffusion

2M8_Rate law model

2M9--13

2M14-17_Gap_gas

3M1

3M2

3M3

3M4

3M5_2D

3M6_Creep_Plasticity

3M7_Skyline-Ref3

4M1_UO2

4M2_MOX

4M3_Zr

4M4-SUS

4M5-6+Ref

本レポートは独立行政法人日本原子力研究開発機構が不定期に発行する成果報告書です

本レポートの入手並びに著作権利用に関するお問い合わせは下記あてにお問い合わせ下さい

なお本レポートの全文は日本原子力研究開発機構ホームページ（httpwwwjaeagojp）

より発信されています

独立行政法人日本原子力研究開発機構研究技術情報部研究技術情報課

319-1195 茨城県那珂郡東海村白方白根 2 番地 4

電話 029-282-6387 Fax 029-282-5920 E-mailird-supportjaeagojp

This report is issued irregularly by Japan Atomic Energy Agency

Inquiries about availability andor copyright of this report should be addressed to

Intellectual Resources Section Intellectual Resources Department


2-4 Shirakata Shirane Tokai-mura Naka-gun Ibaraki-ken 319-1195 Japan

Tel +81-29-282-6387 Fax +81-29-282-5920 E-mailird-supportjaeagojp

copy Japan Atomic Energy Agency 2013

i






















PCMI Burn-up


ii







（2013 年 3 月 14 日受理）













iii

Contents

1Introduction 1







121 What is model 5









References 1 15














iv


































v











Model (IGASP=0) 125






2102 Swelling 132




power down 134













vi












References 2 174





















vii


















361 Creep 301


363 Ohta model 307




References 3 322


41 UO2 Pellet 323

42 MOX Pellet 343






viii



References 4 380

Acknowledgement 382


ix

目次

1序言 1









123 解析対象 7

13 全体構造 8






参考文献 1 15










x




















263 数値解法 74











xi
















2102 スエリング 132





2121 基本概念 135










xii










2171 自動制御 170



参考文献 2 174



311 基本式 178













xiii















351 要素体系 272

352 基本式 276

353 剛性方程式 278

354 境界条件 284




361 クリープ 301


363 太田モデル 307


(ICPLAS=3 6) 313


参考文献 3 322


xiv










参考文献 4 380

謝辞 382


- 1 -














rod





functions














- 2 -






























design




- 3 -

































- 4 -
















114 Usage of FEMAXI
















- 5 -

























the following steps






- 6 -















modeled
















Refs(115) and (116))



- 7 -

follows

























Pellet





Fuel rod




- 8 -




FEMAXI-7


times





times times




times times


times times

13 Whole Structure
















- 9 -








axial segment











freedom




- 10 -

Next Time Step

No

Output


Input




End of Time Step

Yes

Iteration







- 11 -

Plenum

Segment

LHR

Axial Power Profile

Zircaloy metal


Temperature



Nest ofcoaxialrings

Gas pressure













- 12 -









calculation time


















bubble growth

Nonzero element



- 13 -
























- 14 -































- 15 -








(2) Plotted output














(14)

(15)

(16)








- 16 -

(17)

(18)

(19)

(110)

(111)






















(116)





ORNL-4628 (1973)





- 17 -













rate



















- 18 -


















solution





- 19 -

Time Step Start












Next time step




- 20 -





as follows

τdzS

zQzVtHtzH

t

t primeprimeprime

+=in )(

)()()0()( in (211)



through the inlet


time t




τ time



S zd

t

t t( ) ( )

( ) ( )

( )0 0 00


+

ΔΔ

τ (212)













- 21 -

H t t H tV Q

Stj

i i

ii k

j

i( ) ( )0 0 0+ = +=Δ δ (214)


δ tL

u

L

G V

S L

WVii

i

i

i i

i i

i

= = =Δ Δ Δ

(215)

where





node j+1

node j

node j-1

node k+1

node k

segm j

segm j-1

segm k

z

zk+1

z0

δ t j

δ t j minus1

δ tk

z

zk



- 22 -

ΔL

z zi j

z z k i j

z z i ki

j j

i i

k

=

minus=

minus ne neminus =

+

+

+

1

1

1 0

2( )

( )

( )

(216)


( ) ( ) ( ) ( )( )H t H t H t H tz z

z zk k kk

k k0 0 0 1 0 0

0

1


+

(217)

















- 23 -

Inlet enthalpy

Inlet temperature

Pressure

Mass flow rate


Inlet flow

velocity







sub-segments










- 24 -









coefficient



Here

8040

e

RePr0230D

Kh =

Hea

t flu

x l

og q

D


flux point

Single phase

Nucleate

boiling

Transient

boiling

Film boiling


A

E

B

C


- 25 -

μeRe DG=





























- 26 -








SPTH

CKh 750240

sat240g

240fg

290f

50

490f

450

fP790

fw 00120 ΔΔ=

ρμσρ

(2111)




ρ density (kgm3)







temperature(Nm2)





hP

WW

N= sdottimes

012636 201 106

0 75 exp

φ (2112)





- 27 -


10

f

g

50

g

f

90

e

e

1

minus

=μμ

ρρ

x

xr

ex quality (-)

p density (kgm3)


R F

00 100

01 107

02 121

03 142

04 163

06 202

10 275

20 430

30 560

40 675

60 910

100 1210

200 2200

500 4470

1000 7600

4000 20000






FACTOR S 251

fRe F S

103 1000

104 0893

2x104 0793

3x104 0703

4x104 0629

6x104 0513

105 0375

2x105 0213

3x105 0142

4x105 0115

6x105 0093

106 0083

108 0000



- 28 -


















( ) 21gfgee

g


( ) ( ) 21gffee




ex quality(－)





(2115)

12 12g f crit

12 12g f crit

12 12crit g f crit

(low speed)

12 (highspeed)


J J J

J J J

J J J J



- 29 -




H enthalpy (Jkg)







( ) ( )( ) ( )[ ]( )

( )

( ) inf7

e

e

eee

e78

863DNB

1041292382580

05487124exp8357026640

86901571

03713563117290596114840

1098705517718exp104268117220

10237960222101543

HH

D

x

Gxxx

xPP

PqqDNBW





minus

minusminus

minus

(2117)

Here


P pressure(Nm2)







( )inf

e 03DNB 960761HH

xW DNBqq






- 30 -

( ) ( )[ ]21

gf


+minusminus=

ρρρρ

ρρσα gHq (2119)


ρ density(kgm3)





















( )DNB

satsatDNBFBDNBTB 22

expDNB T

TTTqqh

T ΔΔ

ΔminusΔminus=Δ

(2121)



DNB

FB Tq


DNBFBFBDNB

ThqT

Δ=Δ



- 31 -

( )[ ]

( ) [ ]w7

590282e

78420e

1ln2456060040g

30702w

43760g

FB Pr1045041exp89471131

RePr0810330

e

PxD

Kh

xminus

++

timesminus+

=





P pressure(Nm2)







( )

( )

=330

filmfilm

250film

film

c

PrGr738170

PrGr151443max

e

e

D

KD

K

h (2125)










ρ density(kgm3)



fwfilm

TTT

+=


- 32 -

( ) ( )

( )

ge

leΔ

gtΔminusΔ+

=

9600

1

1960

satDNB

satsatFBTB

liquid

α

α

BTq

BTThh

q (2127)

( ) ( ) satTB

BTTB ehqqBh Δminus

=Δ


sat960712 α (2128)

( ) 250

satge

fggfg3

g

1720

c

eFB 526775

Δminus

=

TD

HKDh

μρρρ

λg

(2129)

( )50

gfc

minus=

ρρσλ

g (2130)

( ) ( )( ) ( )

1 2

1 2

001368 K at 4137000 N m

001476 K at 6205500 N m

B PB

B P

minus

minus

= == = =













- 33 -



( )

( )

2503f

min 2960

minusminus

=wsatfe

fggffarla TTD

KHh

μρρρ g

(2133)

ef

ggpfff

e

gg

f

pff

f

ff

turbulent

D

VCK

D

V

K

CKh

μμρ

μμμ

ρ

310857749

02300650

minustimes=

=

(2134)

q heat flux(Wm2)




V velocity(ms)






table


- 34 -





Inlet enthalpy

Inlet temperature

Pressure


Inlet flow

velocity


Mass flow rate


Pressure


Low-flow

high-flow

NO

NO

YES

YES

NO

Water single-phase



Coolant flow rate







RETURN

RETURN

RETURN


(

RETURN

YES





(RELAP-5)

RETURN


- 35 -
















[ ]E prime Forced

convection to wall


(TwgtTB)

(TwltTB)

(TwltTB)



satTΔ

1



[D] film boiling

0

[F] Condensation

two-phase flow

xe quality

(TwgtTB)



x


Sub-cool region


- 36 -

























enters region [E]





- 37 -




is recovered










boiling transition



Dittus-Boelter

TW

ltTB


TW

gtTB




TW

gtTB




TW

gtTB



RELAP-5 MOD-1model

TW gtTB



TW

ltTB






- 38 -













using

DS

re = 4

2π (2135)








S l r= minus4 2 2π (2136)




S

Symmetrical line

l1



- 39 -





4

2 eDrS

π= (2137)










CN = 0835 + 181times10-4T (221)



(300 1478 )T Kle le













- 40 -












RCORRO times






Temperature




Meta l part

Oxide

Radius

( )Stimes PBR -1S

PBR PBR

Co

ola

nt




- 41 -





chapter again)

























dS

dt

dS += 1


- 42 -















oxide layer








- 43 -














φN = ATN + B (231)






00 02 04 06 08 10

ZrO2

1 2 3 4 5 6 7 8

Cladding

ring element number






- 44 -

















obtained by

ln(2 )N

N OXOX OX

rqT T K rπ

= minus sdot

(233)

where








- 45 -







( ) ( )[ ] ( ) ( ) ( )partpart t


Here

T temperature (K)





n-1 n n+1

kpn

Cvpn

qpn

ksn

Cvsn

qsn

hpn hsn

hpn 2 hsn 2




- 46 -





part (235)




( ) ( ) ( ) ( )

( )1

2 4 2 4

V v V v

m mn n pn pn sn sn

v pn n v sn npn


t t

T T h h h hC r C r

h

+



(236)




( ) ( ) ( ) ( )( ) ( )1 1 1 1

1 1

[ ]

2 2

V s


n npn sn


k T T k T Th hr r

h h

+ + + +minus +



(237)




hh

rh

h rh

hh

rh

hh

rh

pnV pn

npn

snV

nsn

pns

pnn

pnsns

snn

sn

= minus

= +

= minus

= +

2 4 4

1

2

1

2

(238)

D C h C hn v pn pnV

v sn snV= +


variables as

( ) ( ) ( )q r t Q r P t= (239)




- 47 -






( ) ( )( )( ) ( )

1

1

1

nm n n m m s

n n pn pn

m m sn n sn sn

V Vpn pn sn sn n

T T DT T k h

t

T T k h

Q h Q h P t

θ θ

θ θ

θ

+ + +minus

+ ++

+

minus= minus minus

Δ+ minus

+ +

(2311)



( ) ( ) ( )

( ) ( ) ( )

T T D

t

T T T Tk h

T T T Tk h

Q h Q h PP t P t

nm

nm

n nm

nm

nm

nm

pn pns

nm

nm

nm

nm

sn sns

pn pnV

sn snV

fm m

+ +minus

+minus

++

++

+

minus=

+minus

+

+ + minus +

+ ++

1 11

11

11

11

1

2 2

2 2

2

Δ

(2312)



respectively



a T b T c T dn nm

n nm

n nm

nm

minus+ +

+++ + =1

1 11

1 (2313)

where

( )

( ) ( ) ( )

ak h t

ck h t

b D a c

d a T D a c T c T

t Q h Q h PP t P t

npn pn

s

nsn sn

s

n n n n

nm

n nm

n n n nm

n nm

pn pnV

sn snV

fm m

= minus = minus

= minus minus

= minus + + + minus

+ ++

minus +

+

Δ Δ

Δ

2 2

2

1 1

1

(2314)


- 48 -



Pellet CladdingGap

n-1 n n+1 n+2

Tn-1 Tn Tn+1 Tn+2

hpn hsn+1



( ) ( ) ( )( )1

111

11

11

+

+++

+minus

++

+

minus+minusminus=Δ

primeminus

mfVpnpn

gm

nm

nspnpn

mn

mn

nm

nm

n

tPPhQ

hTThkTTt

DTT

(2315)

( ) ( ) ( )( )11

111

11

211

111

11

++

+++

++

+++

+++

++

+

minus+minusminus=Δ

primeprimeminus

mfns

snsns

mn

mng

mn

mn

nm

nm

n

tPPQ

hkTThTTt

DTT (2316)


and


( ) ( )( )

( )

11 1 1 1 1

1 1 1

1 12 1 1 1

1 1 1

m mn n n m m s

n n g p n p n

m m sn n s n s n

Vs n s n m

T T DT T h h h

t

T T k h

Q h P t

++ + + + +

+ + +

+ ++ + + +

+ + +


Δ+ minus

+

(2317)


n v s n s nV






- 49 -





( ) ( ) ( )10000

10

110

01

0+

+++

+minus=Δminus

mfVss

sss

mmsv

mm

tPPhQhkTTCt

TT (2312a)


( ) ( ) ( )1

11

1

1

++

minus+

+


mfVpnpnnn

spnpn

mn

mn

Vpnpnv

mn

mn tPPhQrhkTThC

t

TT φ (2318)


a T b T dn nm

n nm

nminus+ ++ =1

1 1 (2319)


a

k h tn

pn pns

= minusΔ

2 (2320)

b C h an v pn pnV

n= minus (2321)

( )( ) ( )

d a T C h a T

t r tQ h PP t P t

d d

n n nm

V pn pnV

n nm

nnm

nm

pn pnV

fm m

n n nm

= minus + minus

minus +

+

+


minus

++

+

1

11

1

2 2

Δ Δφ φ

φ

(2322)


Pellet center

0 1

T0 T1

hs0


n-1 n=N

Tn-1 Tn=TN

hpn

φ φn N=

Gap


- 50 -

given by

φ πnm m

nq r= 2 (2323)





(2321) are 1 11 m m





T E T Fnm

n nm

nminus+

minus+

minus= minus +11

11

1 (2324)


A T Bnm m

11

11+ ++ = φ (2325)

where




values Therefore 1m








- 51 -

500 1000 1500 20001

2

3

4

5

6

80 GWdtHM

40 GWdtHM

Ther

mal

Con

duct

ivity

(Wm

K)

Temperature (K)

NFI UO2


Halden MOX0 GWdtHM







- 52 -

p(I)＝10 - D(I) (2328)


D(i) is defined as

D(I)=Di-(3ΔLr) (2329)



by irradiation

r pellet radius




=1 00

swgVp p

V

Δ= + is used

=2 0

0 0

dens hotV Vp p

V V

Δ Δ= + + is used

=3 00 0 0

swg dens hotV V V

p pV V V


where 0

swgV

V


0

densV

V


0

hotV

V








That is





swg densV V

p pV V

Δ Δ= + +


- 53 -

























equation) (216)





- 54 -













( ) ( ) ( )( ) ( )RkK

RkK

RkIRkIRQ sdotsdot

sdotsdot+sdot= 0

11

1101 (241)

Here







( ) ( )k E DR

E Dp

= sdot +

sdot sdot0328 054

050 8

0 82

0 19

(242)

where








- 55 -




( ) ( )( ) ( ) ( )[ ]( )( ) ( )( )22

2

2 1121

112

ahhhahb

harahbrQ

bb

bb


where








( )( )pRQ

RQa

1

11= (244)






1po

po pi

T r r

pT r rdT r q dr dr









- 56 -


1 1( )

po

T r






( ) )(1

)(2

rh

Prq LHR ϕ

πsdot

minus= (247)







( )

21 1

00

1212

1 1

2

12 2 2 4

po

rT r

LHR LHRpT r r

LHR LHR LHR

rr

P P rdT r dr dr dr

r r

P P r Prdr r

λπ π

π π π

= sdot sdot = sdot

= = = minus

(248)


rP

T LHRTTp po

minus=π




( ) )1(0304

4004501

422 rr

PTT

p

LHRPO minus

sdot+=minus+=

ππλ (249)




( )( ) ( ) ( )[ ]( )( ) ( )( )22

2

1121

112)(

ahhhahb

harahbr

bb

bb




- 57 -


( )4po

TLHR

pT

PdT F rλ

π= sdot (2411)

Here

( ) [ ][ ])1)(1(2))(1(

ln)2)(()1(22)1)()(2(2

)1)(1(4

22

2222

ahhhahb

rhbhaharhabb

rarF

bb

bbbb



minusminus= ++

(2412)


( ) ( )[ ]

2

12)(

++minus+=

ba

arabr

b

ϕ (2413)

( ) )2(

)1()2(2

)1)(1(4 22

+

minus+++

minusminus

=

+

ba

rabb

ra

rF

b

(2414)


( )24 03 07( )

34

rrϕ

+= ( ) 4 203(1 ) 28(1 )

34r r

F rminus + minus

=
















- 58 -





f(x)＝arb＋C (2415)



2

bb a x af x

ba

+ minus + =+















Relative radius

of pellet


integral equation

Numerical solutions






given



pellet = 100)

00 01 02 03 04 05 06 07 08 09 10

14174114086713823113380112752111931110907196678819846482245000

141873 140997 138358 133922 127633 119412 109158 96747 82033 64848 45000

141758 140883 138246 133814 127530 119317 109075

96679 81984 64822 45000

141787 140912 138276 133844 127560 119346 109101

96701 82001 64831 45000

141758 140883 138246 133814 127530 119317 109075

96679 81984 64822 45000


- 59 -














following equation



















- 60 -

attained




Relative radius of

pellet


integral equation

Numerical solution


direction




given



00 137512 137794 137535 137686 137507 01 136408 136688 136430 136581 136403 02 133109 133381 133129 133278 133106 03 127661 127919 127677 127823 127658 04 120151 120388 120163 120302 120149 05 110726 110937 110734 110863 110725 06 99607 99784 99611 99725 99606 07 87088 87227 87090 87182 87087 08 73533 73628 73534 73599 73533

09 59357 59405 59357 59391 59357 10 45000 45000 45000 45000 45000








( ) ( ) 1 22 1 2 05

gas m cr

eff

Ph h

R HC R R g g GAP


(251)





- 61 -

Here


( ) ( )

( ) ( )λ

λ λλ λm

p po c ci

p po c ci

T T

T T=

sdot+

2 (WcmK)

hT T

T Trp c

po ci

po ci

= + minus

sdot sdotminusminus

minus1 1

1

1 4 4

ε εσ










R R Reff

22

2

2

+








1

1 1ji j

ni

gas nji

iji

x

x

λλφ

=ne

=

= minus Σ

(252)



- 62 -

Here the equation

( )( )

( )φ

λλ

ij

i

j

i

j

i

j

i j i j

i j

M

M

M

M

M M M M

M M=

+

+

+minus minus

+

1

2 1

12 41 0142

1

2

1

4

2

3

2

1

2

2

(253)







3 0668

4 0701

5 0872

5 0923

4 08462

3366 10

3421 10

40288 10 (WmK) (K)

4726 10

2091 10

He

Ar

Xe

Kr

N

k T

k T

k T k T

k T

k T

minus

minus

minus

minus

minus

= times


(254)




( )5

1 2 1 21

10n

iigas

g g g gP=

+ = Σ + sdot (255)

Here


(cm)





- 63 -
























start

t





- 64 -













conditions is given

















- 65 -

















again)













- 66 -














reaction














- 67 -






33 and 34






















history points


- 68 -






















- 69 -








is neglected



Fuel pellet

96

Cladding

978

113 Internal tube

Outer tube

Capsule outer m

antle

Heated steam

layer

Helium

layer

Circulating w

ater

Coolant w

ater

180

200

210

230

270

320



- 70 -



jl

Fixed at 40

Fuel pellet



R2 R3 R4 R5 R6 R7

Cladding

Internal tube

He layerCapsule

outer mantle

Circulating w

ater

Outer tube




( ) )(2

112220

22

1222

2222 jjjj

j

jjjj qRqR

RRHH

V

W

t

TC minus


partρ (261)

holds where

ρ density (kgm3)


T temperature (K)


V volume (m3)


R radius (m)




- 71 -

)(2

)(

112221

22

112

122

1111 jCjC

jj

jjj qRqR

RRz

TC

RR

W

t

TC minus

minus=

minus+

partpart

πpartpart

ρ (262)

is obtained


)(2

223322

23

222 jCjC

jjj qRqR

RRt

TC minus

minus=

partpart

ρ (263)


axial direction

)(2

334423

24

333 jCjC

jjj qRqR

RRt

TC minus

minus=

partpart

ρ (264)


)(

244552

425

444 jCjC

jjj qRqR

RRt

TC minus

minus=

partpart

ρ (265)


direction

)(2

)(55662

526

552

526

5555 jCjC

jj

jjj qRqR

RRz

TC

RR

W

t

TC minus

minus=

minus+

partpart

πpartpart

ρ (266)


)(2

66

7726

27

666 jCjC

jjj qRqR

RRt

TC minus

minus=

partpart

ρ (267)

is obtained

7


( )jNjjjjC BTAq 0











- 72 -


)( 677




2

32

2

2

1

2

2ln

11

R

RRR

h

h+

+prime

=

λ

(2615)




3

43

3

3

32

3

2

33

2ln

2ln

1

R

RRR

RR

RRh

++

+

=

λλ

(2616)


layer (WmK)

4

54

4

4

43

4

3

44

2ln

2ln

1

R

RRR

RR

RRh

++

+

=

λλ

(2617)


554

5

4

55 12

ln

1

hRR

RRh

prime+

+

=

λ

(2618)



6

76

6

6

5

6

2ln

11

R

RRR

h

h+

+prime

=

λ

(2619)


76

7

6

77 2

ln

1

RR

RRh

+

=

λ

(2620)


- 73 -




A) SUS316



B) Helium




( ) ( )( )

4

3

54245 10 1

1890


Pa b c

T RT

T P R


= sdot
















- 74 -


e

RePr0230D

KhW = (2621)

where μeRe DG=








21

42

41

1

21

111

TT

TThr minus

minussdotsdot

minus+=

minus

σεε

(2622)











( )

))()((2

)(

011

11

22221

22

111

111112

122

111

111

jn

jjnj

nj

njj

njj

j

nj

nj

jj

BTARTThRRR

TCTClRR

W

t

TTC


=

sdotminussdotminus

+Δminus

++

++minusminus

+

πρ

(2623)


- 75 -

))(2

2

)()(

011111

221

22

22

112

122

222

122

11111112

122

111

jn

jjnj

jjnj

nj

j

jjjnj

j

j

BTARTt

CT

RR

hR

TRR

hR

lRR

CW

t

CT

lRR

CW

minussdot+Δ

=minus

minus

minus+

minusminus

Δ+

minus

+

++minus

minus

ρ

πρ

π (2624)

For inner tube

)()(2 1

11

2221

21

33322

23

21

222

+++++


=Δminus

nj

nj

nj

nj

nj

nj

jj TThRTThRRRt

TTCρ (2625)

( ) nj

jjnj

nj

jjnj T

t

CT

RR

hRT

RR

hRhR

t

CT

RR

hR2

22132

223

33122

223

223322112

223

22 222

Δ=

minusminus

minus+

+Δ

+minus

minus +++ ρρ(2626)

For Heium gas layer

)()(2 1

21

3331

31

44423

24

31

333

+++++


=Δminus

nj

nj

nj

nj

nj

nj

jj TThRTThRRRt

TTCρ (2627)

( ) nj

jjnj

nj

jjnj T

t

CT

RR

hRT

RR

hRhR

t

CT

RR

hR3

33142

324

44132

324

334433122

324

33 222

Δ=

minusminus

minus+

+Δ

+minus

minus +++ ρρ (2628)

For outer tube

)()(2 1

31

4441

41

55524

25

41

444

+++++


=Δminus

nj

nj

nj

nj

nj

nj

jj TThRTThRRRt

TTCρ (2629)

( ) nj

jjnj

nj

jjnj T

t

CT

RR

hRT

RR

hRhR

t

CT

RR

hR4

44152

425

55142

425

445544132

425

44 222

Δ=

minusminus

minus+

+Δ

+minus

minus +++ ρρ(2630)


( )

))()((2

)(

14

1555

15

16662

526

155

115152

526

551

555

++++

++++

+


=

sdotminussdotminus

+Δminus

nj

nj

nj

nj

njj

njj

j

nj

nj

jj

TThRTThRRR

TCTClRR

W

t

TTC

πρ

(2631)

( )

nj

jjnj

nj

j

j

nj

j

jjjnj

Tt

CT

RR

hRT

lRR

CW

TRR

hRhR

lRR

CW

t

CT

RR

hR

5551

625

26

661152

526

155

152

526

556625

26

5555142

526

55

2

)(

2

)(

2

Δ=

minusminus

minus+

minus+

+minus

minusΔ

+minus

minus

+++

+

++

ρπ

πρ

(2632)


- 76 -


)()(2 1

51

6661

67772

627

61

666

++++


=Δminus

nj

nj

njj

nj

nj

jj TThRTThRRRt

TTCρ (2633)

( ) 72

627

776

66162

627

667766152

627

66 222j

nj

jjnj

jjnj T

RR

hRT

t

CT

RR

hRhR

t

CT

RR

hR

minus+

Δ=

minus+

+Δ

+minus

minus ++ ρρ (2634)



( )

))()((2

)(

110111

111

122221

22

11111

111

12

122

1111

111111

BTARTThRRR

TCTClRR

W

t

TTC

nnn

nninin

nn


=

sdotminussdotminus

+Δminus

++

+++

πρ

(2635)

where 11



( )

))()((2

)(

14

1555

15

16662

526

155

1552

526

551

555

++++

+++


=

sdotminussdotminus

+Δminus

nN

nN

nN

nN

nNN

ninin

N

nN

nN

NN

TThRTThRRR

TCTClRR

W

t

TTC

πρ

(2636)






1nNT + and 1 1

1 6 n nN NT T+ +



four axial segments


- 77 -

(2637)




obtained


- 78 -




Time Step Start














models in FEMAXI


- 79 -




is represented by

PY f q

E Nij

ij j

f A

= sdot sdotsdot

(271)

Here


( )+

minus=

50

502

i

i

ij rdrrf φπ



jLHR

pout

P

r hπ minus

















- 80 -



















matrix





N fmax

which is





the fission gas




is assumed to cease


- 81 -





















- 82 -

INTRAGRANULAR


GAS BUBBLE















- 83 -




consideration(223)

β+prime+minus

partpart+

partpart=

partpart

mbgcr

c

rr

cD

t

c 22

2

(273-1)









1 12 2






0m

mt gc b mt


(273-3)


relation



( ) 2

2

2c m c cD

t r r r


= + +

(273-5)


- 84 -


1 0mt

gc b mmb


Here 1

1

1 ii

i

mtb

m bα minus

minus




( )i

i

c c mg

αα

= ++

(273-8)

holds


( ) ( ) ( )2

2

2i

i

c m c m c mD

t g r r r


+ + += + + +

(273-9)



Ψ = +c m (273-10)


2

2

2D

t r r r


Ψ Ψ Ψprime= + +

(273-11)

Here

i

i

DD

g

αα

prime =+

(273-12)











- 85 -


276 below






( )2152 oN R Zα πλ= + (274-1)

Here








4g DRNπ= (274-2)



( )2

0303b F R Zπλprime = + (274-3)

Here







Namelist input BFCT







- 86 -








( )PR

V m B m kT0

2+

minus =γ (274-4)

( )


(274-5) where





3πR cm3)




T temperature (K)



surface tension

2

gas extP PR

γ= + (274-6)





( )

3 24

3 2ext

ext

P RRm

B P R kTR

γπγ

+= sdot+ +

(274-7)


- 87 -

is obtained







c m= minusΨ

( )prime + =b g m N g Ψ (274-10)

is obtained


( )( )2 0303 4 4F R Z DRN m DRπλ π π+ + = Ψ (274-11)


( )( ) ( )3

2

0

( 2 )303 4 3

2ext

ext

R ZR P R

F DRN DB P R kTR

γπλ πγ

+++ sdot = Ψ

+ + (274-12)


respect to R

( )( ) ( )

32

0 2

608 ( 2 )303 3

2ext

exto

DR R P RF R Z D

B P R kTRR Z

α γπλγλ

+ + + sdot = Ψ + ++ (274-13)







2m

R which satisfies 2




- 88 -










3N Rπ minus sdot




22 ( ) ( )F R Z f R Nπ λsdot + (275-1)




follows

0

00

10 ( )

( )( )

k

R R

f R RR R

R

le= ge

(275-2)




3 20

42 1 2 ( ) ( )

3

dNF N R F R Z f R N


(275-3)






Then


- 89 -

3 20

3 20

42 2 ( ) ( )

3

2 2

4( ) ( )

3

dNF F R R Z f R N

dt

F FKN

K R R Z f R

α πα πλ

α

πα πλ

= minus + + sdot

= minus

= + +

(275-4)






m m N= (275-6)





4g DRNπ= (275-7)






Ψ = +c m (275-8)


20303 ( )b F R Zπλprime = + (275-9)

2

04 303 ( )

( )

dmDRNc F R Z m

dtgc b m g g b m






32 4

3

RP m B m kT

R


(275-12)


- 90 -



2gas extP P

R

γ= + (275-13)

holds





4 2v v P

H gas

dn D RP

dt kT R

π γσ = + minus

(275-14)

gas

v v

m kTP

n=

Ω (275-15)

3

4

3

v vV n m B

Rπ

= Ω +

= (275-16)




4 2

041

085 140 10 (Jm )

P SV SV

SV T

γ γ γγ minus

= sdot =

= minus times

GSV (0ltTlt2850C)





3Ω =34454E-10 m




Ttemperature (K)





- 91 -






1i i vv v

dnn n t


1

11


dn D RP

dt kT R

π γσminus

minusminus

= + minus

(275-19)




dt=





variables




method

1

1

1

422

22

i ii

i iv


R RD P Pdn

PR Rdt kT

π γσ

minus

+

minus

+ +

= + minus +

(275-20)



dt= is explained


4 20

iiv vH gasi

dn D RP

dt kT R

π γσ = + minus =

(275-21)


- 92 -


( ) 2 0iH gasP Rσ γ+ minus = (275-22)


gasv

AP

n= (275-23)

3

vR nα β= + (275-24)


are cubed

33 3( ) 8H gasP Rσ γ+ = (275-25)



(275-26)


-------------------------------------------------------------------------------------------------







Number density 0 0

00762 Bu 104( ) ( )

98N Bu T N T

minus sdot += (cm-3) (276-1)




(Cut-off)



- 93 -

1 3

0

3

4

m MR

Nπρ prime

=

(276-3)


GROU=40 (default)



04 ( ) ( ) ( )g g g








8000

273159

ln0005

mgTBu

= +minus

(276-5)















- 94 -

expressed as

0 3

3 ( )( )

4g

g

m r t MN r t

Rπ ρprime

= (276-7)


2) R Rρge 0

N

iniR

R

ρ

ρ ρ =















(277-1)


( ) ( ) ( ) ( )Ψ Ψ Ψt r t r t rj jj

asymp = φ




ˆ 2Dt

rtR (277-2)





- 95 -

equation

( )R t r dviV Ψ φ = 0 (277-3)




2ˆ

ˆ 0iVD dv

tβ φ


2ˆ

ˆ 0i iV Vdv D dv

tβ φ φ



( ) ( ) ( )

( ) ( ) ˆˆ

ˆˆˆ2



nabla V ii

V iV iV i

dvDsdD

dvdvDdvD

φφ

φφφ

part

(277-5)



( ) ( )ˆ

ˆ 0i iV Vdv D dv

tβ φ φ



jΨ is the





ie iφ ＝jφ


that


t

partφ β φ φ φ


(277-7)



- 96 -



t


partΨ







where

Veji j i

j iji Ve

i Ve i

E dV

A dVr r

H dV

φ φ


φ

=

=

=

(277-10)



( ) ( )

( )

( ) ( )ξξξφ

ξξφ

ξξξφ

+=

minus=

minusminus=

12

1

1

12

1

3

22

1

(277-11)











each node


- 97 -

( ) ( )1

11




( ) ( )3

11




( ) ( ) ( ) ( )21 2 3

1 1( 1 ) 1 ( 1 )


(277-12)

and

( ) 11Ψ minus = Ψ ( ) 20Ψ = Ψ ( ) 31Ψ = Ψ (277-13)

holds






boundary layer




- 98 -

( )

minus

Δ+

Δminus

Δ+Δ=

==

1

1

2

2

4

4

ξξφφπ

φφπφφ

dRRR

drrdVE

kij

RR

RR ijVe ijji

K

K (277-14)

( )

2

1 2

1

4

4

k

Ve k

R Rj ji iji R R

j ik

A dV r drr r r r

R R dR


+Δ

minusΔ

minus

= =

= + ΔΔ

(277-15)

( )1 22

14 4

k

Ve k

R R


+Δ



( )( )1

1 1 0n nj j n n

ji j j ji ij

E D A Ht

θ θ β+


Here

( )( )

D D Dn n

n n


prime = minus +

+

+

1

1

1

1

θ θβ θ β θβ

(277-18)






j+Ψ for









- 99 -

( ) 1 1 11n n


E D A E D A Ht t



1ij ji jiW E D A

kθ= +

Δ (277-20)

( ) 11 n

i ji ji j ij

Q E D A Ht





[ ] W Qij jn

iΨ + =1 (277-22)




















boundary (1)

A B


- 100 -




























tdVβΔ (atoms)



total+Ψ (atoms)


boundary (2)


- 101 -






( )

( ) ( )

32

1

3

1

ˆ4

ˆ

1

e

k ktotal V

e

k kj j

j

r r dr

r r

k n n

π

φ

=

=

Ψ = Ψ Ψ = Ψ

= +

(279-2)







3

11 nntotal

n

a

a++

sdot Ψ

(279-3)


( )31

4


(279-4)

consequently 3


n

an tdV

aβ+ ++


(279-5)

is obtained









- 102 -



ofga

3

143

igrain

g

Naπ

= (2710-1)

holds











fn ie





max 1 2 max 114n n






fN )



3

1 nnf

n

an

a+

sdot

(2710-5)


- 103 -



3


n

an n n n

a+ + ++

= sdot + Δ

(2710-6)







(2710-7)

where 1rls n





(2710-9)



3

1 nnf

n

an

a+

sdot

(2710-10)



3


n

an n n n

a+ + ++

= sdot + Δ minus

(2710-11)






- 104 -




3


n

an n n n

a+ + ++

Δ = sdot + Δ le

(2711-3)

3 3


n n

a an n n n n n n

a a+ + + + ++ +


(2711-4)



1

12

1

nn rr

n

nN

aπ

++

+

ΔΔ = (2711-5)







is given by

1

nkr

k

n=

Δ (2711-7)



1

nkr

kntotal

nFGR

FPG=

Δ=

(2711-8)


- 105 -





3

1 1ng

n

af

a+

= minus

(2712-1)

Here





an+1

an



da

dtK

a

N N

af f

m

= minus+

1 1 max

(μmh) (2712-2)









- 106 -



dt dt= times






Boundary

layer

Node







11

2

2n

n nn

a Rr r

a R+

+minus Δ=

minus Δ (2712-3)

where









- 107 -


given by

Ψn+10(r) = Ψn(r) (0 le r le an) (2712-4)


where









( )

11 max 1

21

1 max 1

min 4

min

n nf fn n

f fn

n n nf f f

n nN N

a

N N N

π

++ +

+

+ +

+ Δ=

= + Δ

(atomscm2) (27-12-5)


Δ+sdot=

Δ+sdot

=

+++

+

+

++

+

1max11

1max2

1

1

3

1

1

min

4

min

nf

nf

nf

n

n

nf

n

nf

nf

n

n

nf

NNNa

a

Na

nna

a

Nπ

(2712-6)

holds where24

nfn

fn

nN

aπ= and

11

214

nfn

fn

nN

aπ

++

+

ΔΔ =






- 108 -



3

11

12

1

3

2 11

21

11

4

4

4

n nnf f

nnf

n

n nnn f f

n

n

n nnf f

n

an n

aN

a

aa N n

a

a

aN N

a

π

π

π

++

+

+

++

+

++

sdot + Δ

=

sdot + Δ

=

= sdot + Δ

(2712-7)

holds where 24

nfn

fn

nN

aπ= and

11

214

nfn

fn

nN

aπ

++

+

ΔΔ =





1nfN +






( )3

2 2 10114 4n n n nn


aa N a N

aπ π ++

+


(2712-8)




n+1 0 give

10 12

14

ng totaln nn

f fn n

f aN N

a aπ+ +

+

Ψ= + (2712-9)

where 2

14

ng total

n

f

aπ +





- 109 -


1

1 12

1

4

n

ng totaln i n in

f fr an n

f aN N D t tdV

a a rβ

π+

+ +

=+





( ) ( )1 1 11

2 2i n n i

f f

bN N ADDF


Δ (2712-11)







1( 2 )nna r+



1nr a

D tr

+=


(2712-12)

is determined



updated


1 1 1

1

n i n if f

n if

N N

Nε

+ + minus

+





( ) 1

1 12

1

1

4


f fn n

f tdvaN N

a a

βπ

++ +

+

+ Ψ minus Ψ + Δ= + (2712-13)



- 110 -



1

4





considered












( ) ( )1 11 ADDF

2 2n nf f

bN N



Here Nf



(atomscm2)






- 111 -




~β β β= + prime (2713-3)



















We have

sinb fr r θ= (2714-1)





- 112 -









1

1 gas

FP B

kT

=

+

(2714-3)


34( )



2

gas extf

P Pr

(2714-5)




on the boundary

Grain boundary


- 113 -

12

2

( sin )

4 ( ) 2

3 sin

ff

fext

f

mN

r

r fP F

kT r

π θ

θ γθ

=

= +

(2714-6)





( )max

2

4 2

3 sinf f

f b extf

r fN f P F

kT r

θ γθ

= +

( )2atomscm (2714-8)













thermal stress








- 114 -









500 1000 1500 200010-22

10-21

10-20

10-19

10-18

10-17

10-16

Diff

usio

n C

onst

ant

(m2 s

)

Temperature (K)




- 115 -

4 5 6 7 8 9 10 11 12 13 14 15 161E-22

1E-21

1E-20

1E-19

1E-18

1E-17

1E-16

1E-15

Diff

usio

n co

nsta

nt (

m2 s

)

10000T (1K)



(FGDIF0-1)

FGDIF FGDIF010000

9 exp 7

a

a a aT




(default)




about 1250K


indispensable



- 116 -

5 6 7 8 9 100

2

4

6

8

10

FGDIF0=5 a=6

FGDIF0=10 a=4

FGDIF0=10 a=8

FGD

IF M

agni

fyin

g fa

ctor

10000T (1K)




factor FGDIF


- 117 -


























inside of the grain



- 118 -




1 densN d= (μm2) d=183x183 (μm2) (282-2)






34( )


(282-3)

where





2

2 2 2 22

sin 4 sin4

fb f bb g bb

g

rf r N a N

aπ θ π θ= sdot = (282-4)

Here by setting

2 2

2

sinfc

rF

s

π θ= (282-5)


( )2 2

2 2 2 2 22 2

sin 4 sin sin4 4


g g

r rf N a N r N


2 2sin

bdens

f

fN

rπ θ=

sdot (282-8)

hold




- 119 -







dt


1

nf

f kk

drr t

dt=

= Δ (282-9)

2

22

2FRAD

4 ( )kT

Vf bd gb

b ext ff f

dr s DP P k

dt r f r

δ γθ


(282-10)


plenum gas pressure



When 0fdr










2 2

4

sing b

f

a fK

r θsdot

=


boundary)


- 120 -

When 0fdr





22

8(1 )

( 1)(3 ) 2 lnfks

φφ φ φ


(282-11)

2 2 2

2

sin sinf f br r f

s s

θ θφ

π = = =

(282-12)

3

z rextP θσ σ σ+ += (282-13)



boundary


2 2

2 sinf

b

rs

f

π θ= (282-15)







2 2 2 2 22


HN r r H

rφ θ θ θ= = =


constant value


- 121 -




That is 2densf

HN




HN

r= = times (Refer





RINITfr ge holds




bb

Gm


and m change








1new


MGB

NGB (282-16)



- 122 -




calculation


nf

f kk

drr t

dt=



0fdr


2b ext

f

P Pr

γ= + holds

In other words


342

( ) 3

f

f

rP f m b m kT

r

πγ θ






2 (100 1)100 100 100 100f f f f f f f f

f f f f f f







1

f if i f i i

drr r t

dt+minus = Δ (c)


Δ=i

jji tt

1








- 123 -


2


V N P Sn

kTη= sdot (282-17)

22669

MTης







24bb g densN a Nπ=








4

0 12 32 210 (m )

16 3c

aV

L











- 124 -
















invi

bb grain

Gm

N N=

sdot (282-25)


element


2 2 2

2 2 2

sin

sin 1h f b c b

ef c

s r P F P

s r F

σ π θ σσπ θ


(282-26)

h ext gapP Pσ = + (282-27)



2 2 2 2

2 2 2

sin sin

sinf b f

c bf

r f rF f

s r

π θ π θπ θ

= = = (282-28)

is obtained


- 125 -




number of atoms














In other words 2

2 2

4

singr b

f

a fK

r θsdot

= (291-2)



2

2

6816 b gr

f

f aK

r

sdot= (291-3)




- 126 -










Eqs(279-5) and (2710-3) as


3

1

1


n

an tdV

aβminus

minus


(291-6)





That is 2

gas extf

P Pr

γ= + (291-7)


34( )

3f

gas f

rP f n B n kT

πθ


(291-8)



2

26816

nf f

b gr

r nn

f a

sdot=

sdot (291-9)


- 127 -


3 2 2

2 2

42( )

3 6816 6816

n nf f f f f

ext ff b gr b gr

r r n r nP f B kT

r f a f a

πγ θ sdot sdot



nf

b gr

nQ

f a=

sdot

42( )

3f

ext ff

rP f QB QkT

r

πγ θ


From Eq(291-11)

( ) ( ) 0223

4

3

4 2 =minus





4

3




a

acbbrf 2






fr =05 μm






- 128 -

292 Bubble swelling




2

6816 b gr

f

f aK

r

sdot= (291-3)




34( )

3f

bb f

rV f

πθ= sdot (292-1)


34

3gr

gr

aV

π= (292-2)


0 2

Bubble

bb

gr

V V K

V V

Δ sdot=

(292-3)




In this context 0

001436Bubble

V

V

Δ =







- 129 -

3intra

0

4

3

BubbleV

R NV

π Δ = sdot

(292-4)




0

4

3

Bubble

bb

grTotal

V V KR N

V Vπ

Δ sdot= sdot +

(292-5)

and 0

Bubble

Total

Vp

V

ΔΔ =

(292-6)


hold


















- 130 -











2b ext gapb

P P Pr

γle + + (2101-1)

03c bF f= ge = FCON (2101-2)

FCON=03 (default)















assumed


- 131 -




bubbles 1 is

2 21 sinS M rπ θ= sdot (2101-3)



1 1r M r= sdot (2101-4)







1

2γ

r

to2

2γ




3 3

1 2

32 1

4 4

3 3r M r

r M r

π πtimes =

= sdot (2101-5)






expressed by K


- 132 -

2

2 2

4

singr b

f

a fK

r θsdot

= (2101-7)


2

2 3

1

bb

r ff

r M

sdot =

(2101-8)





have a radius

2 1 1 sinr r M r θ= = sdot (2101-9)




indicates

32

1

1 rM

M r

=

(2101-11)





is also identical







- 133 -




fN remains




dependent on extP





















r zST




- 134 -





stress and 2

0extPr


08r

γminus times


















internal pressure







- 135 -





Eq(2714-8)







recovery




2121 Basic concept



















- 136 -

































- 137 -






( )W Bu= minus2 19 48 8 (2122)

















( ) ( )1 1 1

1 12

1

1

4


f fn n

f tdv R M R MaN N

a a

βπ

+ + ++ +

+

+ Ψ minus Ψ + Δ + minus= + (2125)






- 138 -





R F Pr r r= (2126)




















as described below






- 139 -

50 60 70 80 90 100 110 120000

001

002

003

004

005

Frac

tion

of ri

m a

rea

Local burnup (GWdt)

Cunningham model




times (2128)

holds

















using Eq(2122)


- 140 -



























- 141 -




1tanXv

effB Buα δ

π

























- 142 -



temperature history





11 0exp ( )n




11 0exp ( )n n n










rod

00 01 02 03 04 050

10

20

30

40

50

60

70

80

90

Burn

up (G

WdtU

)

Pellet Radius (cm)




- 143 -



value

















[ ]1

FPOR 10 FSOLID


= minus= minus

(2123-6)


Here







- 144 -



3

3

23


131 (gmol)

6022 10 (atomsmol)

ρtimestimes

times sdot (2123-8)

is used






FPOR As a result









GS0a

0 50 100 150 20000

05

10


Xe

conc

(w

t)

Local burnup (GWdt)




- 145 -





















defines as

OPR=ATHMR (2123-11)







(2123-12)



- 146 -

20 40 60 80 100 120 1400

5

10

15

20

Min

Por

osity

(

)

Local burnup (GWdt)

Max











in Fig2124

2

2

PROMAX 0005 ( 40) 25

PROMIN 00013 ( 40) 25




Bu

Bu

p

(2123-15)



RMPRS=2 mode


- 147 -



















1

095rim

rim Bx

pλ λ minus = sdot

(2123-19)








( )Bx T Bxλ λ= (2123-20)



- 148 -

































- 149 -



of pellet is

100( ) ( )TDF p T Buλ λ= sdot (2131)





1) IPRO=0 0p p= (2132)

2) IPRO=1 00

swgVp p

V

Δ= + (2133)

3) IPRO=2 00 0

dens hotV Vp p

V V

Δ Δ= + + (2134)

4) IPRO=3 00 0 0

swg dens hotV V V

p pV V V

Δ Δ Δ= + + + (2135)


0

swgV

V


0

densV

V

Δ densification

0

hotV

V

Δ hot press


swgV

V


IFSWEL






- 150 -

Billaux equation

251

095rim

rim Bx

pλ λ minus = sdot

(2123-19)


conductivity




swsV

V






10Df =

and when FDENSF=1 0

10 swsD

Vf

V

Δ= minus (2138)


FEMAXI model










defined as rimS



- 151 -




00025ss

solid

VS SWSLD

V

Δ = = sdot


SWSLD=10 (Default)


2) 00025ss

solid

VS SWSLD

V

Δ = = sdot


SWSLD=10 (Default)











growth model


model



GBFIS=0


- 152 -


migration GBFIS=1









- 153 -











A1 0

Equili-

brium

model

0 2

3 4 5



0 default


1


A2 1



0 default


1


B1 2

Rate-

Law

model

0 2

3 4 5



0 default


1


B2 1



0 default


1



- 154 -



Time Step Start














- 155 -









Xenon




rod










T temperature (K)

V volume (m3)


Subscripts are




- 156 -





TpTj+1Tj

2lj 2lj+1



N n V

C n VP

RC

V

T

i j i j j

i j j j i jj

j

=

= = (2141)





( )G j S Dn n

lj jj

j1

12 1 1rarr =minus

(2142)


( )G j S Dn n

lj jj

j1 1 1

12 1 1 1

1

1 rarr + =minus

+ ++

+ (2143)


( ) ( )G j G j1 1 1rarr = rarr +

holds and

S Dn n

lS D

n n

lj jj

jj j

j

j

12 1 11 1

12 1 1 1

1

minus=

minus+ +

+

+ (2144)


nS D l n S D l n


j j j j j j1

121 1 1 1

121 1

1 112 12

1

=+

++ + + +

+ + + (2145)



- 157 -

( ) ( )

( )G j j G j

S D S D


j j j j j jj j

1 1

121 1

12

1 112 12

11 1 1

1rarr + = rarr

=+

minus+ +

+ + ++

(2146)

Here by setting

FS D S D


j j j j j j +

+ +

+ + +

=minus1

121 1

12

1 112 12

1

(2147)


(2148)


(2149)

is obtained



(21410)


n n n n n nj j j j j j= + = ++ + +1 2 1 1 1 2 1

holds











nP

R Tn

P

R Tjj

jj

= =++

1 11

1

(21413)




- 158 -



Here

nP

R

C

Tn

P

R

C

Tjj

jj

11

1 11

1 = =+

+ (21416)

CN N

V V

n V n V

V Vj j

j j

j j j j

j j1

1 1 1

1

1 1 1 1

1

=++

=++

+

+

+ +

+

(21417)

nP

R

C

Tn

P

R

C

Tjj

jj

22

2 12

1 = =+

+ (21418)

CN N

V V

n V n V

V Vj j

j j

j j j j

j j2

2 2 1

1

2 2 1 1

1

=++

=++

+

+

+ +

+

(21419)

C C1 2 1+ = (21420)








Vn

tJ Jj

jR S

partpart

11 1

= minus (21423)




- 159 -

( ) ( )[ ]

( ) ( )[ ]n n

t VF n n n n

F n n n n

jk

jk

jj j j j j j

j j j j j j

11

1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1

11 2 2 1 1

1 2 2 1 1

+

minus minus minus

minus + +



θ θ θ θ θ

θ θ θ θ θ (21424)

Here




θ 1 and θ 2







( )n n nj jk

jk

12

2 1 2 111

θ θ θ= minus + + (21425)


( )[ ( )

( )]

n n

t

F

Vn

F F

Vn

F

Vn

VF n n

n n F

jk

jk

j j

jj

k j j j j

jj

k

j j

jj

k

jj j j

kj

k

j j j j

11

1 2 11 1

1 2 1 1

11

2 11 1

11 2 1 1 1

1 1 1 1

1

1

1 1

1

1

1 1

1

1 1 1

11

1

+minus

minus+ + minus +

minus+

+minus minus

minus +

minus= minus

+

+ + minus minus


Δθ θ

θθ

θ

θ

θ

θ θ

θ

θ

θ θθ

θ θ θ ( )[ ( ) ( )]2 1 1 1 1 1 11 1n n n nj

kj


θ θ

(21426)


A n B n C n Dj jk

j jk

j jk

j1 11

11

1 11

minus+ +

+++ + sdot = (21427)

Here

AF

Vtj

j i

j

= minus minusθ θ

θ2 1

1

1

Δ (21428)

B

F F

Vtj

j i j j

j

= + minus+minus +

12 1 1

1 1

1

θ θ θ

θ Δ (21429)


- 160 -

CF

Vtj

j j

j

= minus +θ θ

θ2 1

1

1

Δ (21430)

( ) ( )( )

( ) ( )

DV

F t nV

F F t n

VF t n

VF n n F n n

jj

j j jk

jj j j j j

k

jj j j

k

jj j j j j j j j



minus minus minus +

+ + + + minus minus

11 1

11

11

1

1

1

1

1 1

1

1

1

1 1 1 1 1

1 2 1 1 1 1 2 1

1 2 1 1 1 1 1 1 1 1 1 1

θθ

θθ θ

θθ

θθ θ θ θ θ

θ θ

θ

Δ Δ

Δ ( )θ1 Δt

(21431)



( ) ( )[ ]V

n

tJ

F n n n n

S11 1

1

1 2 1 1 1 2 1 1 1 2

partpart

= minus


(21432)


( )

( ) ( )

n n

t

F

Vn

F

Vn

F

Vn

F

Vn

F

Vn n

k kk k k

k

1 11

1 1 2 1 2

11 1

1 2 1 2

11 2

1 2 1 2

11 1

2 1 2

11 2

1 2

11 1 1 2

1

1

1

1

1

1

1

1

1

1

1 1

1

1

++ +minus


+minus

+ minus

Δθ θ θ

θ

θ

θ

θ

θ

θ

θθ

θ

θ

θθ θ

(21433)


( ) ( ) ( )

BF

Vt C

F

Vt

DF

Vt n

F

Vtn

F

Vn n tk k

12 1 2

11

2 1 2

1

12 1 2

11 1

2 1 2

11 2

1 2

11 1 1 2

1

11 1

1

1

1

1

1

1

1

1

1

1

1 1

= + = minus

= minusminus

+

minus+ minus

θ θ

θ θ

θ

θ

θ

θ

θ

θ

θ

θ

θ

θθ θ

Δ Δ

Δ Δ Δ (21434)

and we obtain

B n C n Dk k1 1 1

11 2

11sdot + sdot =+ +

(21435)



( ) ( )[ ]V

n

tJ

F n n n n

nn

R

n n n n n n

partpart

11

1 1 1 1 1 1 1

= minus


(21436)


A

F

Vt B

F

Vtn

n n

nn

n n

n


θ

θ

θ2 1 2 1

1

1

1

11 Δ Δ


- 161 -

( ) ( )

( ) tnnV

F

ntV

Fnt

V

FD

nnn

nn

knn

n

nnkn

n

nnn

Δminusminus

Δ

minusminus+Δ

minus=

minusminus

minusminus

minus

11

1

1

1

1

1

1

1111

112

1112 1

11

θθθ

θ

θ

θ

θ

θ θθ

(21437)

we obtain

A n B n Dn nk

n nk


1 11

11

(21438)


B CA B C

A B C

A B CA B C

A B

n

n

n

n

n

n

n n n

n n n

n n

k

k

k

nk

nk

nk

1 1

2 2 2

3 3 3

2 2 2

1 1 1

1 11

1 21

1 31

1 21

1 11

11



minus minus minus

minus minus minus

+

+

+

minus+

minus+

+

=


minus

minus

DDD

DDD

n

n

n

1

2

3

2

1

(21439)




DkT

m n d12

1

2

122

3

8 2

1=

ππ

(21440)

where

mm m

m m =

+1 2

1 2







N Nj iji

= (21441)


- 162 -



jj

N N= (21442)

Upperpl enum

Segment

5

Segment


4

Counter-diffusion

Segment3

Segment2

Segment1

Lower

pl enum

Previous step

Current step

He Xe He Xe He Xe



Since

P N RV

T jj

= sdot sdot

1 (21443)



prime =

NP

R

V

Tjj

(21444)


- 163 -








prime = minus


il l jl

(21445)



















- 164 -


V r lhj

pi Zj= π 2 (2151)





equations

V r lp po= π 2 2Δ (2152)

V r u lcrack porel

p= 2π (2153)

hold where





( )V V V r l r u ll

lj

dish cham po porel

pZj

pint = + + + sdotπ π2 2 2Δ (2154)












Parameter AY

pellet



- 165 -




length ~


( )~l lZ pj

thj

denj

ssj

Zj= + + +1 ε ε ε (2155)





~lZ Cj is

( )~ l lZ Cj

c thj

c irrj

Zj= + +1 ε ε (2156)





( )2

1

NAXIj j


V r l l l lπ=

Δ = minus + minus

(2158)



~V V Vpl pl pl= + Δ (2159)







2 3chamfer

Dp aV abπ = minus

(21511)


- 166 -


2 2( 3 )6dish

dV d r

π= + (21512)



a

b

Dp

Dp

r

d


- 167 -


2 21tan ( )


















θ

Din

Center hole

Dp


- 168 -




int

1

Tgas j j jM



n RP

V V V V V

T T T T T=

sdot=

+ + + +

(2161)

where









Vgapj





jci

j+







surface temperature





- 169 -






























- 170 -






















strain






- 171 -



EFCOEF1 sdotsdot

leΔcE

tε

σ (2171)












Eq(21423) as

( )

Δtn n V

J Jji j j j

R S

=minus

minus1 1

1 1

(2172)













- 172 -


using ( )J R j1

( )ΔtN

Jj

R j

2

1

= max

(2173)






ΔtN

jj

3 = max

β (2174)


( )Δ Δ Δ Δt t t tn21 11

21 1


( )Δ Δ Δ Δt t t tn22 12

22 2= min (2176)

( )Δ Δ Δ Δt t t tn23 13

23 3= min (2177)

Δt2 is given by

( )Δ Δ Δ Δt t t t2 21 22 23= min (2178)


code









- 173 -


as

22

3 005 FMULTR

tD


(2179)





( )2

64

min

10i

rt

K

Δ Δ = sdot

(21710)
















- 174 -

References 2















Company Inc(1972)



EPRI-NP 2789 (1982)





- 175 -




CRFD-1075 (1962)

(218) Wilson MP Jr GA-1355 (1960)




















- 176 -


























- 177 -




fuel JNM 321 pp 249-255(2003)
















- 178 -















311 Basic equations

















- 179 -

10r rr rz R



part part part

10zrz z rz Z



210r z r



0rr


part

rr

u

rε part=

part1r uu

r rθ

θεθ

part= +part

zz

u

zε part=

part

[ ]1( ) pl cr sw


[ ]1( ) pl cr sw


[ ]1( ) pl cr sw


0r zθε ε ε+ + =








for shear strain




- 180 -







follows


assumed



assumed









( ) tt





Here ( )nt

f σαpartpart=




f

partpart




- 181 -


f

t

f σσpartpart=

partpart













Then

[ ] B uε = (313)










T

FW Fν= (314)


[ ] TT

INW Bν σ= (315)





- 182 -

[ ]

[ ]

TT T

V

TT

V

F B dV

B dV

ν ν σ

ν σ

=

=

(316)



[ ] T

VF B dVσ= (317)













ie at time tn+1

[ ] ++ = dVBF nT

n 11 σ (318)

holds where






T

n

T

n+ += + 1 1σ σΔ (319)


n n

T

nΔσ σ+ += minus 1 1 (3110)


quantity



- 183 -



[ ] 1e

n nDσ ε+ +Δ = Δ (3111)










Δε ne


[ ] ( )[ ] [ ]D D Dn n n+ += minus + =

θ θ θ θ1

1

21 and



[ ] Δ Δε σθne

n nC+ + +=1 1 (3112)


[ ] [ ] [ ] B D dV F B dVT

ne

n

T

nΔε σ+ += minus 1 1 (3113)



10




nnnT σεε 1

011 (3115)




nT

nnT σεε 0

111 (3116)



- 184 -

[ ] [ ][ ] [ ] [ ] [ ]

1

01 1

T

n

T T

n n n

B D B dV u

F B D dV B dVε σ

+

+ +

Δ

= + Δ minus

(3117)




[ ] [ ] [ ] [ ] [ ]

T


B D B dV u F


Δ = Δ

+ Δ + Δ + Δ + Δ + Δ + Δ + Δ + Δ

(3118)


[ ] [ ][ ]

[ ] [ ] [ ]

11

01 1

1 1

MT

i i i i ni

M MT T i

n i i n i i n ii i

B D B V u

F B D V B Vε σ

+=

+ += =

Δ Δ

= + Δ Δ minus Δ

(3119)




be obtained




f

t

f σσpartpart=

partpart














- 185 -









( )σε F=Δ (3122)

















[ ] [ ] [ ]εσ DDD == (3123)




- 186 -






















01σΔ

εΔ

σ

εε

ε

σ

σ

)(a

)(b

1



- 187 -

















section














- 188 -

cn

Pn

HPn

swn

denn

crkn

reln

thn

enn

111

1111111

+++

+++++++

Δ+Δ+Δ+



(3124)






Δε nden


Δε nsw






crk+1

Δε nHP+1 Δε n

P+1 and Δε n




reln


n 1+Δε




[ ] e Cε σΔ = Δ (3125)

[ ] Δ Δε σθne

n nC+ + +=1 1 (3126)

where [ ] ( )[ ] [ ]

=+minus= ++ 2









- 189 -

1 1 1 1 1 1 1

1 1 1


HP P cn n n


ε ε ε+ + + + + + +

+ + +





swn

denn

reln

thnn 1111

01 +++++ Δ+Δ+Δ+Δ=Δ εεεεε (3128)





cHn

PHn

crknnn

en 111


where








[ ] 11 ++ Δ=Δ nn uBε (3130)



[ ] [ ] 01110


PHn











- 190 -







nonlinear equation



ψ( x)=0 (3133)



following manner

1+ix = ix + Δ 1+ix

Here

i

ii

dx

dxx

minus=Δ + ψψ )(

1




0)()( 122 =Δminus= +PHnf εσσψ (3135)

)(xψ

idx

d

ψ

ix1+ix

)( ixψ

１



- 191 -

0)()( 133 =Δminus= +cHnf εσσψ (3136)

also

)( 11 σσψ fdd = (3137)

)( 22 σσψ fdd = (3138)

)( 33 σσψ fdd = (3139)














[ ] ( )

1 11 1

0 1 1 11 1 1 1 1 0

i i in n n


C d B u

C

θ

θ

σ

ε ε ε ε σ σ

+ ++ + +

+ + ++ + + + + +

minus Δ

+ Δ + Δ + Δ + Δ + minus =

(3140)

Here d ni

ni

niσ σ σ+

+++

+= minus11

11

1


[ ] [ ] V

T

ni

V

T

ni

nB dV B d dV F + ++

++ =σ σ1 11

1 (3141)



11

Δεncrk i+

+1

1 1

1+

+Δ iPHnε 1

1+

+Δ icHnε and Δun

i++

11

Here the



0)()( 1111 =Δminus= +

+icrk

nf εσσψ (3142)

0)()( 1122 =Δminus= +

+iPH

nf εσσψ (3143)

0)()( 1133 =Δminus= +

+icH

nf εσσψ (3144)


- 192 -


unknown values 11

++

indσ Δε n

crk i+

+1

1 1

1+

+Δ iPHnε 1

1+

+Δ icHnε and Δun

i++

11



















- 193 -


iPHn

1ˆ +Δε and icH

n

1ˆ +Δε








[ ] in

in

in FuK 1

11 +

+++ Δ=Δθ



++

indε and 1

1++

indσ are


++Δ i


++

inε and

11

++







- 194 -




Time Step Start














- 195 -







(1) Geometry







U P PER PL ENUM

S E GMENT( M)

S E GMENT( M- 1 )

S E GMENT( 2 )

S E GMENT( 1 )

L O WER PL ENUM

Pellet

Cladding

Clad Pellet


- 196 -











Uz2

Lz9Lz3

Lz2Lz1


r12

Uz3Uz1

Uz10

r1 r2 r3 r10 r11

Lz10

r13 r14 r15 r16

Lz11Lz12

Lz13Lz14

Uz11Uz12

Uz13 Uz14Uz9

Mz1Mz2

Mz3

Mz9Mz10


00 02 04 06 08 10

ZrO2

1 2 3 4 5 6 7 8

Cladding

ring element number






- 197 -






1 10L Lz z - 1 10


U Uz z - as shown

below

( )( )

( )( )U

CUUUU

LC

LLLL

Up

UUU

Lp

LLL

zzzzz

zzzzz

zzzz

zzzz

====

====

====

====

14131211

14131211

1021

1021

(321)













for its importance

ri ri+1

ZiU

ZiL




- 198 -











oxide elements





















- 199 -



Bedworth ratio













Uiz and




111

1+

++

+

minusminus

+minusminus

= iii

ii

ii

i urr

rru

rr

rru (323)

( )L

M Lii iM L

i i

z z


minus

or ( )M

U Mii iU M

i i

z z


minus (324)




- 200 -

[ ] r

z

ur

u B ur

z

θ

partpartε

ε εε partν

part

= = equiv

(325)


( ) ( ) 111

1

111

11

+++

+

+++

minusminus

+minusminus

=

minus+

minusminus=

iii

ii

ii

i

iii

iii

urrr

rru

rrr

rr

r

u

urr

urrr

u

partpart

(326)

Here by putting 1

2i ir r


( ) ( )

1 11

11 1

1 1

11 1

2 2

2 21 1

i i i ii i

i ii i i i

i i i i

i ii i i i

r r r rr ru

u ur r r rr r r r r

u ur r r r

+ ++

++ +

+ +

++ +


= ++ +

(327)

M Li iM Li iz z z

partν ν νpart

minus=minus


partν ν νpart

minus=minus

(328)


1 1

1

1 1

1 10 0

1 10 0

1 10 0

ii i i i

r i

i

ii i i i L

i

z iM

M L M L ii i i i

ur r r r

u

r r r r

z z z z

θ

ε

εν

εν

+ +

+

+ +


minus minus minus

Or

1 1

1

1 1

1 10 0

1 10 0

1 10 0

ii i i i

r i

i

ii i i i M

i

z iU

U M U M ii i i i

ur r r r

u

r r r r

z z z z

θ

ε

εν

εν

+ +

+

+ +


minus minus minus

(329)



- 201 -

[ ]

minusminusminus

++

minusminusminus

=++

++

Li

Mi

Li

Mi

iiii

iiii

zzzz

rrrr

rrrr

B

1100

0011

0011

11

11

or [ ]

minusminusminus

++

minusminusminus

=++

++

Mi

Ui

Mi

Ui

iiii

iiii

zzzz

rrrr

rrrr

B

1100

0011

0011

11

11

(3210)




( )( )Li


221π

or ( )( )Mi


221π (3211)



[ ] Δ Δσ ε= D e (3212)



[ ] Δ Δε σe C= (3213)


[ ]


=1

1

11

νννννν

EC (3214)


- 202 -






[ ] F B dVn

T

n+ += 1 1σ (3215)


Here






T

n

T

n+ += + 1 1σ σΔ (3216)


n n

T

nΔσ σ+ += minus 1 1 (3217)



[ ] [ ] [ ] B D dV F B dVT

ne

n

T

nΔε σ+ += minus 1 1 (3218)




en εεε (3219)



nnnT σεε 1

011 (3220)




nT

nnT σεε 0

111 (3221)


- 203 -


[ ] [ ][ ]

[ ] [ ] [ ] 1

01 1

T

n

T T

n n n

B D B dV u

F B D dV B dVε σ

+

+ +

Δ

= + Δ minus

(3222)



[ ] [ ] [ ]

[ ] [ ] [ ]

11

01 1

1 1

MT

i i i i ni

M MT T i

n i i n i i n ii i

B D B V u

F B D V B Vε σ

+=

+ += =

Δ Δ

= + Δ Δ minus Δ

(3223)

















321 (2))






- 204 -




































MiZ

ir

UiZ

LiZ

1ir +

centerline

Outer surface


M3Ζ

M7Ζ

Chamfer space

Dish space

U1Ζ U

2Ζ U3Ζ

U4Ζ

U5Ζ U

6Ζ U7Ζ U

8Ζ U9Ζ U

10Ζ

M10Ζ

M1Ζ

M4Ζ

M5Ζ

M2Ζ

L1Ζ L

2Ζ L3Ζ L

4Ζ L5Ζ L

6Ζ L7Ζ L

8Ζ L9Ζ L

10Ζ

M6Ζ

M8Ζ

M9Ζ



- 205 -








Miz and










U U U Ui pz z z z= = = = =

(Refer to Eq(321))









element








- 206 -














space elements








z

zz l

SEk = (3224)









- 207 -

















zs zz

lk k

l

Δ= sdot (3226)




( )zzs z zd z

z

lk k k k FAC

l






[ ] enn D 11 ++ = εσ (3228)




[ ] 111 +++ = nnen C σε (3229)


- 208 -


324 Pellet cracking












StrongInteraction

W eakInteraction

NoInteraction

Ec

E

ε eminusε rel







- 209 -

value



























- 210 -

( )

minus=lt


=lt

=

reli

ei

ei

reliccrel

i

ei

eic

i

E

EEE

E

E

εε

εεεε

ε

　

　　

　　

0

0

(3230)






introduced as

ri

eii εεε += (3231)



iei εε minus= 0=r

iε That is the


iε ) and when reli



range 0ltltminus ei




Elastic region

Elastic strain

Relocation strain

-εrel

εrel

σ

εi

minusσR


- 211 -

In the range relii




function




(3232)


bad

dE i

i

ii +== ε

εσ

2 (3233)



02 =++ cba rel

ireli εε (3234)

From Eq(3233)




reli

ccreli

c EEc

EEb

EEa ε

ε 4

3

2

4

+minus=

+=

minusminus= (3237)


reli

ci

cirel

i

ci

EEEEEE εεεε

σ4

3

242 +

minus+

+minus

minus= (3238)


ci

EE εσ4

3+minus=



recovers completely









- 212 -




[ ]~C

E E E

E E E

E E E

r

z

=

minus minus

minus minus

minus minus

1

1

1

ν ν

ν ν

ν νθ

(3239)







( )

0

00

0

0

0

c i i


i

reli i i

E

E E E E

E

ε ε

ε ε ε ε εε

ε ε ε


(3240)






en nC+ + +=1 1 (3241)



ncrk i

n niC+ +

++ +

++ =1 11

11 ~

(3242)



- 213 -

[ ] [ ] Δ Δε σ

ν ν

ν ν

ν ν

ν ν

ν ν

ν ν

θ θ

θ

ncrk i

n n ni

r

z

C C

E E E

E E E

E E E

E E E

E E E

E E E

++

+ + ++= minus

=

minus minus

minus minus

minus minus

minus

minus minus

minus minus

minus minus

11

11

1

1

1

1

1

1

~

=

minus

minus

minus

+

minus

++

++

++

+

+

+

Δ

Δ

Δ

Δ

Δ

Δ

σσσ

σ

σ

σ

θ

θθ

r ni

z ni

ni

rr ni

zz ni

ni

rE E

E E

E E

E E

11

11

11

1

1

1

1 1

1 1

1 1

1 1

minus

minus

++

++

++

d

E Ed

E Ed

r ni

zz ni

ni

σ

σ

σθ

θ

11

11

11

1 1

1 1

(3243)


+1




+ +Δ Δ and 1

1i

nθσ ++Δ and one






Eq(3140) gives

[ ] ( ) ( )

1 1 01 1 1

11 1

1 11 1 1 0

i i in n n n

i i i in n n n n


C d B u

C C d

C

θ

θ θ

θ

σ ε

σ σ σ

ε ε σ σ

+ ++ + + +

++ + + +

+ ++ + + +

minus Δ + Δ

+ minus + minus

+ Δ + Δ + minus =


[ ] [ ] [ ] ( )~ ~ C d B u Cni

ni

ni

n nP i

nc i

ni

ni

n+ ++

++

+ ++

++


11

10

11

11

1 0Δ Δ Δ Δ (3244)







- 214 -



















( ) ( ) ( ) ( )

( )

2 22 2

12 2

36

2

3

r z z r zr

r z

h H F GF G H θ θ

θ


α σ σ σ


+ + +

(3245)





have


- 215 -

( ) ( ) ( ) ( )2222 32



( ) ( ) ( ) ( )22222 362


(3247)


( ) ( ) ( )05

2 2 22 1



( ) ( ) ( )05

2 2 2 22 1 3



2 2 22( ) ( ) ( )

3pl pl pl pl

r zd d d dθε ε ε ε = + + (3250)


( )2 2 2 2 22 1( ) ( ) ( ) ( ) ( )





1

npl pl

ii

dε ε=

= (3252)






- 216 -







(IFEMRD=1)

( ) ( ) ( ) ( )

( )

2 22

12 2

3

2

3

r z z r

r z

h H F GF G H θ θ

θ

σ σ σ σ σ σ

α σ σ σ σ


+ + + equiv

(3253)

holds where






( )2

23 27

3r z

r zθ

θσ σ σα σ σ σ α + + + + =






( )( )( )

( )

( )

( )

22

322

2 23

2 22

3

r zr z

r ST r z

z rz ST r z r z

ST r zr z

r z

θθ

θ

θθ θ

θ θ θθ




(3254)

Here the term 3r z



- 217 -


2

32

03

2

3

r z

r ST

z rz ST

STr z

θ

θ

θθ

σ σ σ


σ σ σ σ σ

minus minus


(3255)









2

273


is added inside the



1 1P Pn n

n θ

σε εσ+ +

+



( )3 22

2 3r z

r zr

θθ



holds Accordingly

1 3 05 3 05 3

11 3 05 3

1 3n

n n Symθ

θ θ


α+

+ +


+

(3258)


( ) 1

11 3

1

1

P P

PP P Pnn n

n P P

v v

v v

v vθ θ

θ

α εε σσ

++ +

+


minus minus

(3259)

where v P = minus+

0 5 3

1 3

αα

(3260)


- 218 -




( ) ( )( )

( )

Δ Δ Δ Δ

Δ

Δ

ε ε ε ε

α εσ

σ σ σ

α εσ

σ σ σ

θ

θ

θ

h nP

nP r

nP z

nP

nP

Pr z

nP

r z

v

+ + + +

+

+

= + +

=+

minus + +

= + +

1 1 1 1

1

1

1 31 2

9

(3261)


+1 due to


( )Δ Δε α εσ

σ σ σθh nc n

c

r z ++= + +119

(3262)



( )( )

Δ Δ Δ

Δ Δ

ε ε εα

σε ε σ σ σ θ

h n h nP

h nc

nP

nc

r z

+ + +

+ +

= +

= + + +

1 1 1

1 1

9 (3263)



A) BETAX









11

1

BETAX ( )

0 ( )

nn

i

n

D DD D

D D

D D

α

α

++

+

minus= ltminus

= ge (3264)


- 219 -









value=0













strain increment


[ ] [ ] [ ]

[ ] [ ]

T

TH rel swl den

T

crk c p HOT

B D B dV u F

B D dVε ε ε ε

ε ε ε ε

Δ = Δ

Δ + Δ + Δ + Δ + + Δ + Δ + Δ + Δ

(3265)



- 220 -

[ ] [ ]

[ ] [ ]

T

TH rel swl den

T

crk c p HOT

B D dV F

B D dV

ε

ε ε ε ε

ε ε ε ε

Δ = Δ

Δ + Δ + Δ + Δ + + Δ + Δ + Δ + Δ

(3266)











e TH rel swl den

crk c p HOT

ε ε ε ε ε ε

ε ε ε ε

Δ = Δ + Δ + Δ + Δ + Δ

+ Δ + Δ + Δ + Δ (3267)


[ ] [ ] T eB D dV FεΔ = Δ (3268)


[ ] TF B dVσΔ = Δ (3269)


[ ] [ ] [ ] T TeB D dV B dVε σΔ = Δ (3270)




increments

0FΔ = (3271)

holds and

0eεΔ = (3272)

holds



- 221 -





[ ] eDσ εΔ = Δ (3273)


















en n n

crknc

np


1 1 1 (3277)





- 222 -


Δε nc







- 223 -



( ) ε σ ε φc Hf T F= (331)

Here




T temperature (K)




( ) ε σ ε φH Hf T F= (332)





( ) ε β σ εc H= (333)





n nc

n nHt+ + + + += =1 1

(334)

where

( )

( )σ θ σ θ σ

ε θ ε θεθ

θ

n n n

nH

nH

nH

+ +

+ +

= minus +

= minus +

1

1

1

1

(here 0≦ θ ≦1)






- 224 -












ε ε σσ σ

σc c d

d

f=

= prime3

2 (335)




FEMAXI calculation














- 225 -











is pursued





















- 226 -



























calculation






- 227 -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -



( ) ( )( )

( ) ( )



σ σ θ σ

σ σ θ σ

θ θ θ

θ θ

θ

θ

nc i

ni

ni

nH i

nH i

ni

ni

nH i

nH i

ni

ni

ni

ni

d d

f d d

d

d

++

+ ++

+ ++

+ ++

+ ++

+ ++

+ ++

= + +

=+ +

+

times prime +

111

11

11

11

11

11

3

2 (336)




( ) 1 1 11 1c i i i H i H i



indσ +

+ and 11

H indε +


obeys the flow rule

( )( ) ( ) ( )

1 11 1 1 1

111

1 11

3

2

3

2


n n ni in n

i in ni

n

f d dd

d

f

θ θθ θ

θ

θ θθ


σ σ θ σ

σσ

+ ++ + + ++ +

+ + +++ +

+ ++ ++

+

+ +prime= times +

+

prime=

(338)

holds


unknown 11

indσ +

+ and 11

H indε +

+ as follows

( ) ( )

( )

( )( )

( ) 11

11

11

11

111

2

3

2

3

2

3

2

3

2

3

2

3

+++

++

++++

+

+++++

+

++++

+++

+

++

+++

+

prime

part

part+

primepartpart+

primepartpartprime+

prime


iHn

in

i

n

H

H

in

in

in

Hini

n

in

in

in

Hini

n

in

in

Hini

n

in

Hini

n

in

ini

n

df

df

df

dfff

εθσε

εσσ

σθσεσσσ

σθσσ

σεσσ

σθσεσσσ

σεσσ

σσ

θθθ

θθθ

θθθθ

θθθ

θθθ

θθθ

(339)

Then


- 228 -

( )

1 112

1 11 1

11

3 3

2 2

3 3

2 2

3

2


nn n

i i ii i i i

n n n ni in n nn n

ii H i

n ni Hnn

f f d

ff d d

fd


θ θθ θ θθ θ

θθθ




+ ++ + + + + +

++ +

+ ++ + + +

+ + ++ +

++ +

++

prime prime= minus

prime prime+ +

prime+


1 11

1 11 1

3

2

3 3

2 2


n nn n


n nn n

f fd

f fd d

θθ θ

θ θθ θ

θθ

θ θθ θ



+ +++ + +

+ ++ +

+ +++ + +

+ ++ +

prime= + minus


(3310)


σ

= prime3

2

( ) [ ]

ε ε

σpartpartσ σ

σ σ θ σ

σpartσpartσ

θ σσ

σ partpartε

θ ε

θ θ

θ θ

θ

θθ

θ

θ θ θθ

θ

nc i

nc i

ni

n

i

ni

ni i j n

i

ni

ni

ni

n

i

ni

ni n

iH

n

i

nH

f fd

fd

fd

++

+

+ +

+

++ +

+

+

+ +++

++

++

=

+

minus

prime prime

+ prime

+ prime

1

2 11

11

1

9

4

3

2

3

2i+1

(3311)

is obtained



type









+ + by


- 229 -



( ) ( )( )

1 1 11 1

1 11 1 1

H i i H in n n n


t g

t g d d

θ θ

θ θ

ε σ ε


+ + ++ + + +

+ ++ + + + +

Δ = Δ

= Δ + + (3312)


1 11 1 1

11 1 1

11 1

H i H i H in n n

i ii i

n n n nn n

iH i

n nHn

d

gt g t d

gt d

θθ θ

θ

ε ε ε


partθ εpartε

+ ++ + +

++ + + +

+ +

++ +

+

Δ = Δ +

part = Δ + Δ part

+ Δ

(3313)

From Eq(3313) 11

H indε +

+ is expressed as

( ) 1

1 1 1 1 11

11

i ii H i i

n n n n nH i n n

n i

n Hn

gt g t d

dg

t

θθ θ

θ


partε

++ + + + +

+ + ++

++


(3314)


H indε +



H indε +


++

indσ and 1

1H i

ndε ++ in the



1 1 11 1 2 1 3 1

1

iic i c i i i i

n n i j n n nnn

F d F d Fθ θ θ


+ + ++ + + + ++

+


is obtained where

( )

1

1 2

1

9

4 1

i i

i ni Hn n n

iiin nn

n Hn

f gt

ffF

gt

θ θ θ

θ θθ

θ


partε

++ + +

+ +++

+


Ffn

i

ni2

3

2= +

+

θ

θσ (3316)

( )1 1

3

1

3

21

ii H i

n n nHn

iin

n Hn

ft g

Fg

t

θθ

θ

θ

part εpartε

σ partθpartε

+ + ++

++

+

Δ minus Δ =

minus Δ


- 230 -
















2 ( ) nH f T tε σ= sdot (3318)


12 ( ) n



[ ] ( )1 1

2( ) ( )n


minus

= = (3320)






H

f



( )1 2

9

4

i inii

n nn

f fF θ

θ θθ

partpartσ σσ

+

+ ++

= minus


- 231 -

2

3

2

inin

fF θ

θσ+

+

=

(3321)

3 0F =

are obtained


indσ +

+


gives 1 1 1


+ + + + + + Δ = Δ = Δ + (3322)

where

1 1 3 1 1

3


n

f t F tθ θθ

ε σ θ σσ+ + + + + +

+


1 1 2

iic i

n n i j nn

C t F Fθ θ θ



(3323)




c inε +


[ ] ( )

1 1 0 1 11 1 1 1 1

1 0


i in n n

C d B u

C

θ

θ

σ ε ε ε

σ σ

+ + + ++ + + + + +

+ +


+ minus =

(3244)


( ) [ ] ( )

1 1 0 1 1 1 1 1 1

1 0


i in n n

C C d B u

C

θ θ

θ

σ ε ε ε

σ σ

+ + ++ + + + + + +

+ +


+ minus =

(3324)

[ ]( ( )

)

1 11 1 1

0 11 1 1


c i P in n n


ε ε ε

+ ++ + + + +

++ + +

= Δ minus minus


（3325)

where

[ ] [ ] [ ]( ) 1111

~ˆ minus

+++ += icn

in

in CCD


nC 1+ to creep


indσ +

+ In this equation

11

c inε +




- 232 -









( ) ε partσpartσ

σ εcHf T=

(3326)


ˆ 0HHε ε ε= = (3327)

is set





ε c ε

ε

ε

εH =0

εH =0 εH =0

ε ε εH H= minus

①

②

③

④




( ) ε partσpartσ

σ εcHf T f=

ge 0 (3328)

time


- 233 -




sign is performed




13

IPUGH IPUGH2


1


( ) ( )h K TPσ ε= (3329)

where


T temperature












( ) ( ) ( ) ( )

( )

2 22

12 2

3

2

3

r z z r

r z

h H F GF G H θ θ

θ


α σ σ σ


+ + +

(3245A)

where


- 234 -










nP

nP

n

h+ +

+

=

1 1 (3330)










iteration



( ) ( )h K T Tn ni

nP

nP i

n nσ σ ε ε+ = + +++

++

+Δ Δ Δ11

11

1 (3331)



in nh σ σ +



that ( ) 11 1P P i



inσ +

+Δ and 11

P inε +

+Δ so


P inε +


inσ +



- 235 -


nP i

nP i

n

ih

++

++

+

=

11

11 (3332)


11

ΔεnP i+

+1


( ) 11

111 +

+

++

+

++

+

Δ

+Δ

+=Δ

+ n

n

iPn

i

nPn

Pn

in

i

nn T

T

KKTK

hh

θθθ partpartε

εpartpartεσ

partσpartσ (3333)

By substituting

nin

in


++++

111

11 (3334)


( ) ( ) ( )

Δ

Δ

ε partpartσ

σ σ σ ε

partpart

partpartσ

σ

θ θ

θ θ

nP i

ni

n

i

ni

n n nP

n

nn

n

i

ni

H

hh K T

K

TT

hd

++

+ ++

++

+++

=prime

minus + minus

minus

+

11

1

1 11

1

(3335)


P inε +

+Δ by using 11

inσ +

+Δ instead of 11

inσ +

+Δ

Here

prime =

+

+

HK

ni

P n

i

θθ

partpartε

(3336)


indσ +

+



( ) ( )

( )

[ ] [ ] [ ] ( )


+

minus

=

+minusΔ

+Δprime

+

++++++

+++

+

++

++

+++

i

n

iPn

icnnn

in

in

in

in

i

n

nin

i

n

nP

nnnn

iPn

in

hCuBD

h

TKhTT

KH

θθθ

θ

θ

θ


partσpart

σσpartσpartσ

εσpartpartε

11

1

011

11

1

11

11

~ˆ

(3337)


++1

1 we obtain


- 236 -

[ ]( ( ) )1 0 1 1 1 1

11

ˆ

ˆ


P i nn i i

i in n

n n

hD B u C

h hH D

θ θθ

θ θθ θ



++ + + + + +

+ ++

+ ++ +


( ) ( ) ( )1 1

ˆ

ii Pn n n n n n

n ni i

i in n

n n

h KT h K T

T

h hH D

θ θ

θ θθ θ



+ ++ +

+ ++ +


(3338)

Here 1

1indσ +



P inε +



[ ] [ ]( [ ] ( ) )

[ ]

( ) ( ) ( )

[ ]

i

n

i

n

in

i

n

in

nP

nnnn

nin

i

n

i

n

in

i

n

in

icnnn

in

in

in

in

i

n

i

n

iPn

iPH

h

hD

hH

TKhTT

Kh

hD

hH

CuBDh

h

1

11

1

011

11

1

11

1

ˆ

ˆ

~ˆ

+

++

++

++

++

++

++

+++++++

+

+

++

+

+prime

minus+Δ

minusminus

+

+prime


=

Δ=Δ

partσpart

partσpart

partσpart

εσpartpartσσ

partσpart

partσpart

partσpart

εεσσpartσpart

partσpartεε

θθ

θθ

θθ

θθ

θθ

θθθ

(3339)


P inε +

+Δ the following


[ ]( ( ) )

[ ]( ( )

1 1 0 1 1 1 1 1

1 01 1 1 1

ˆ

ˆ ˆ

ˆ


i ii in n


i in n

n n

d D B u C

h hD D

B u Ch h

H D

θ θ

θ θθθ

θ

θ θθ θ

σ σ σ ε ε


+ ++ + + + + + +

+ +++++ + + + +

+ ++ +



)i

( ) ( ) ( )1 1

ˆ

ˆ


n nn

iii i

n nn n

h h KD T h K T

T

h hH D

θθ θθ

θ θθ θ



+ + ++ ++

+ ++ +


(3340)


- 237 -


[ ][ ]( [ ] ( ) )

211

1

011

11

11

ˆ

ZZTS

CuBDd

nin

icnnn

in

in

in

ipn

in

minusminusΔ+


++

+++++++

++

θ







[ ] [ ][ ] [ ]

[ ]

D D

Dh h

D

Hh

Dh

nP i

ni

ni

n

i

n

i

ni

ni

n

i

ni

n

i+ +

++ +

+

++

++

= minus

prime +

θ θ

θθ θ

θ

θθ

θθ

partpartσ

partpartσ

partpartσ

partpartσ

(3342)

[ ]

[ ]S

Dh K

T

Hh

Dh

ni

ni

n

i

n

ni

n

i

ni

n

i+

++ +

++

++

=

prime +

θ

θθ θ

θθ

θθ

partpartσ

partpart

partpartσ

partpartσ

(3343)

[ ] ( )

[ ]Z

Dh h

Hh

Dh

ni

n

i

n

i

ni

n

ni

n

i

ni

n

i1

1

=

minus

prime +

++ +

+

++

++

θθ θ

θθ

θθ

partpartσ

partpartσ

σ σ

partpartσ

partpartσ

(3344)

[ ] ( ) ( )( )

[ ]Z

Dh h

h K T

Hh

Dh

ni

n

i

n

i

n nP

n

ni

n

i

ni

n

i21 =

minus

prime +

++ +

++

++

θθ θ

θθ

θθ

partpartσ

partpartσ

σ ε

partpartσ

partpartσ

(3345)


+1








- 238 -





1 1 1 1 1 1e th P c irr

















1 1X Xn n nC θε σ+ + + Δ = Δ (3347)






1 1 1 1 1 1e th P c irr




- 239 -

[ ] ( ) 1 1 1 1e P c th irr



n n n nC C C Cθ θ θ θ+ + + + = + + (3350)

1

n nD Cθ θ

minus

+ + = (3351)


[ ] ( )1 1 1 1th irr


where

[ ] 1 1

T

n nF B dVσ+ +Δ = Δ (3353)


[ ] [ ] ( )( )1 1 1 1


and finally

[ ] [ ] [ ] ( )1 1 1 1





1

n nD Cθ θ

minus

+ + = and 1nF +Δ


1 1en n nC θε σ+ + + Δ = Δ (3356)




expressed

( )

1 1 1 1

1 1 1 1 1

1 1 1 1

1 1

( )

( ) ( ) ( )

( ) ( ) ( )

( )

c Hn n n n


HH H H


H cn n n n

f T t

f T t f T t f T t

df df df T t T T t

d d d

f T t C

ε σ ε

σ ε σ ε σ ε


σ ε σ

+ + + +

+ + + + +

+ + + +

+ +

Δ = Δ

= Δ + Δ minus Δ

= Δ + + sdot Δ Δ = Δ + Δ

(3357)


- 240 -


T temperature











expanded form as

[ ] [ ] [ ] ( )

1

1 1 1 0 1

T

n n

T th irr cn n n n n

B D B dV u

F B D dV

θ

θ ε ε ε

+ +

+ + + + +

Δ =

Δ + Δ + Δ + Δ

(3359)




( ) ( )h K TPσ ε= (3360)


T temperature


h yield function







- 241 -



Setting


and

( )1 1 1P Pn n n n n nP

n n n

h K Kh K T T

Tθ θ θ


+ + +

+ Δ = + Δ + Δ (3362)

holds


1 1 1P

n n nPn n n

h K KT

Tθ θ θ


+ + +

Δ = Δ + Δ (3363)

Here setting nP n

KH θ

θ

partpartε+

+

prime =


nP

nP

n

h+ +

+

=

1 1 Eq(3363)

becomes

1

1 1 1

n

Pn n n n

n n

h h KH T

Tθ

θθ θ


+

minus

+ + + ++ +

primeΔ = Δ + Δ

(3364)


1 1 1

1

1

1 0 1

1

n n

n

n

Pn n n

n nn

nn n

nn n

P Pn n n

h h K hT

H T

h h KT

hT

H H

C

θ θ

θ

θ

θ θθ

θ θ

θ θ

θ



partσ

σ ε

+ +

+

+

+ + ++ ++

++ +

++ +

+ + +



= Δ minus Δ

(3365)


form

[ ] [ ] [ ] ( )

1

1 1 1 0 1 0 1

T

n n


B D B dV u

F B D dV

θ

θ ε ε ε ε

+ +

+ + + + + +

Δ

= Δ + Δ + Δ + Δ + Δ

(3366)





- 242 -







( ) 1

0 1

n

Pn n n n

nPn

n

KT K T h

T h

Hθ

θ

θ


partσ+

++

++


is used



indσ +

+ 11

inu +


[ ] ( ) ( )

1 1 0 1 1 1 1 1

1 1 2


in n

d D B u C

S T Z Z

θ θ

θ

σ σ σ ε ε+ ++ + + + + + +

+ +


+ Δ minus minus

(3368)


strain amounts



[ ] K u Fni

ni

ni

+ ++

+=θ Δ Δ11

1 (3369)



equations

[ ] [ ] [ ][ ]K B D B rdrdzni T

nP i

+ += θ θπ2 (3370)


- 243 -

[ ] ( )

[ ] ( ) [ ]

1 1

0 1 1

1 1 2

1 1

ˆ ˆ2

2

2

Ti P i i in n n n n

c in n

T in n

T in n

F B D C

rdrdz

B S T Z Z rdrdz

F B rdrdz

θ θ

θ

π σ σ

ε ε

π

π σ

+ + + +

+ +

+ +

+ +

Δ = minus

+ Δ + Δ


+ minus

(3371)




11




- 244 -




[ ] K u Fni

ni

ni

+ ++

+=θ Δ Δ11

1 (3369)

[ ] ( )

[ ] ( ) [ ]

1 1

0 1 1

1 1 2

1 1

ˆ ˆ2

2

2

Ti P i i in n n n n

c in n

T in n

T in n

F B D C

rdrdz

B S T Z Z rdrdz

F B rdrdz

θ θ

θ

π σ σ

ε ε

π

π σ

+ + + +

+ +

+ +

+ +

Δ = minus

+ Δ + Δ


+ minus

(3371)



[ ] K u Fn n n+ + +=θ Δ Δ1 1 (341)


[ ] B u F Bee n n et nΔ + + += +1 1 1 (342)

Here [ ] [ ] B K

B F F

ee n

et n n n

=

= minus+

+ + +

θ

1 1 1Δ


explained next





shown in Fig 341


v2U v1

U

① ② ③ ④ ⑤


v2L v1

L


- 245 -




+

+++

++

+

++

+

+++

+

+++

++++

++

++++

U

U

L

L

v

v

vv

u

u

u

u

u

u

u

BB

BB

BBB

BBB

B

BB

B

B

BB

BB

B

BB

BB

BBB

B

B

BB

BB

B

B

BBBBBB

BBBBBBBBB

BBB

BBBBBB

BBB

BBB

BBBBBB

BBBBBB

BBB

2

1

2

1

7

6

5

4

3

2

1

544

444

534

434

344

244

144

334

234

134

524

514

424

414

324

314

224

214

124

114

543

443

533

433

343

243

143

542

532

541

442

531

432

441

431

342

341

242

241

142

141

333

233

133

333

332

233

231

132

131

523

522

521

513

423

512

511

422

421

413

412

411

323

322

321

213

223

312

311

222

221

213

123

212

211

122

121

113

112

111

0

0

00

0

0

0

0

0

0

0

0

0

0

000

0

0

0

0

0

0

0

00000

000000

00000

000000

000000

00000

00000

000000

+

++

+++

+

++

+

minus

=

54

44

34

24

14

53

43

33

23

13

52

51

42

41

32

31

22

21

12

11

0

0

00

0

0

0

0

0

etet

etetet

etet

etetet

et

etet

et

et

etet

etet

et

pcb

pca

BB

BBB

BB

BBB

B

BB

B

B

BB

BB

B

F

F (343)




times

+++++++++

++

++++

+++++++++

times

544

444

542

541

442

441

543

443

344

244

144

342

341

242

241

142

141

343

243

143

524

523

514

424

513

423

414

413

22477

313

314

224

313

223

214

124

213

123

114

113

534

434

532

531

432

431

533

433

334

234

134

332

331

232

231

132

131

333

233

133

000000

0000

00

00

00

00

00

00

00

000000

00000

BBBBBBBB

BBBBBBBBBBBB

BB

BBBB

BB

BBB

BBBB

BBBB

BB

BBBBBBBB

BBBBBBBBBBBB


- 246 -

+++

+

++

+++

+

minus=


54

44

34

24

14

52

51

42

41

32

31

22

21

12

11

53

43

33

23

13

2

1

7

6

5

4

3

2

1

2

1

0

0

0

0

0

0

0

0

0

etet

etetet

et

etet

et

et

etet

etet

et

etet

etetet

pcb

pca

U

U

L

L

BB

BBB

B

BB

B

B

BB

BB

B

BB

BBB

F

F

v

vu

uu

u

u

u

u

v

v

(344)






set as

L

jU

j

Lj

Uj

vv

vv

112

111

+

+

=

= (345)


+

=

ΔΔΔ

ΔΔ=ΔΔ=Δ

Δ

ΔΔ=ΔΔ=Δ

Δ

ΔΔΔ

++ 11

32

31

37

31

3222

3111

27

21

2212

2111

17

11

12

11

3

2

1

netn

U

U

LU

LU

LU

LU

L

L

ee

ee

ee

BF

v

v

u

u

vv

vv

u

u

vv

vv

u

u

v

v

B

B

B

(346)


- 247 -


Eqs(341) and (342) as

[ ] FuBee Δ=Δ (347)








z

rrca

PELLET CLADDINGF3

lz

F3

Fpca Fpcb

F2

F2

rcb



F3 axial force






- 248 -




A) Upper plenum




surface (I) is







( ) ( )onUszUsUs


where



Upper plenum spring

Ppcb

Ppca lsz

rca

rcb

(I)

(II)

PELLET CLADDING

lsz

(I)

(II)

Ppca－Ppcb


- 249 -




plenum







where











( )( )

ΔΔ

minus

minusminus

minusminusminus

=

ΔΔminus

+

+U

N

sz

UsUs

UsUs

nnUszUsUs

nnUszUsUs

z

z

v

v

kk

kk

TTlk

TTlk

F

F

1

1

1

1

1

αα

(3412)




[ ] K u Fni

ni

ni

+ ++

+=θ Δ Δ11

1 (3413)


- 250 -

[ ] ( )

[ ] ( ) [ ]

1 1

0 1 1

1 1 2

1 1

ˆ ˆ2

2

2

Ti P i i in n n n n

c in n

T in n

T in n

F B D C

rdrdz

B S T Z Z rdrdz

F B rdrdz

θ θ

θ

π σ σ

ε ε

π

π σ

+ + + +

+ +

+ +

+ +

Δ = minus

+ Δ + Δ


+ minus

(3414)






temperature





















- 251 -

++

++

minus=

ΔΔΔΔΔ

++++++

+++

54

44

53

43

52

51

42

41

2

7

6

5

544

444

543

443

542

541

442

441

534

434

533

433

532

531

432

431

524

523

522

521

514

424

513

423

512

511

422

421

414

413

412

411

0

0

0

0

etet

etet

et

etet

et

pcz

pcb

pca

sz

UN

BB

BB

B

BB

B

F

F

F

v

v

u

u

u

BBBBBBBB

BBBBBBBB

BBBB

BBBBBBBB

BBBB

(3415)

where



the top segment)


(3415)

ΔΔΔΔΔΔ

+++minus++++

minus

+++

sz

UN

UN

UsUs

UsUs

v

v

v

u

u

u

kBBBBkBBBB

BBBBBBBB

kk

BBBB

BBBBBBBB

BBBB

2

1

7

6

5

544

444

543

443

542

541

442

441

534

434

533

433

532

531

432

431

524

523

522

521

514

424

513

423

512

511

422

421

414

413

412

411

0

0000

00

0

00

( )

( )

minus+++

minusminus

+

+

minus=

+

+

nnUszUsUsetet

etet

nnUszUsUs

et

etet

et

pcz

pcb

pca

TTlkBB

BB

TTlk

B

BB

B

F

F

F

15

44

4

53

43

1

52

51

42

41

0

0

0

α

α (3416)

is obtained



B) Lower plenum


boundary


- 252 -







LszLsLs


where






( ) ( )nnLszLsLs



( )

( )

ΔΔ

minus

minusminus

minusminusminus

=

ΔΔminus

+

+L

Lsz

LsLs

LsLs

nnLszLsLs

nnLszLsLs

z

z

v

v

kk

kk

TTlk

TTlk

F

F

11

1

1

1

1

αα

(3419)

holds





Lower plenum spring

Ppcb

Ppca lsz

rca

rcb

(I)

(II)

PELLET CLADDING

lsz


(II)



- 253 -

ΔΔΔΔΔΔ

+++minus++++

minus

+++

Lsz

L

L

LsLs

LsLs

v

v

v

u

u

u

kBBBBkBBBB

BBBBBBBB

kk

BBBB

BBBBBBBB

BBBB

12

11

7

6

5

544

444

543

443

542

541

442

441

534

434

533

433

532

531

432

431

524

523

522

521

514

424

513

423

512

511

422

421

414

413

412

411

0

0000

00

0

00

( )

( )

minusminus++

minus

+

+

minus

minus=

+

+

nnLszLsLsetet

etet

nnLszLsLs

et

etet

et

pcz

pcb

pca

TTlkBB

BB

TTlk

B

BB

B

F

F

F

15

44

4

53

43

1

52

51

42

41

0

0

0

α

α (3420)



( )

+minus

++

minus=

ΔΔΔΔΔ

++

++

+5

34

3

1

52

51

42

41

12

11

7

6

5

533

433

532

531

432

431

523

522

521

513

423

512

511

422

421

413

412

411

0

0

0

0

0000

00

0

00

etet

nnLszLsLs

et

etet

et

pcb

pca

L

L

Ls

BB

TTlk

B

BB

B

F

F

v

v

u

u

u

BBBBBB

k

BBB

BBBBBB

BBB

α (3421)





+

=

Δ

+++

111

1

netnn

UPee

nee

ee

LPee

BFu

B

B

B

B

(3422)


eeB is the





- 254 -




























equal)


zero


- 255 -


















bonded state when









( )

22

2μ

ππμ sdot

minus=

lr

rr

i

io (3423)





UV μge

UV μlt

ir

orμ


- 256 -








jrjz FF μgt













segments









- 257 -

Pellet stacksegment

Initial state


A B



deformed state B


























described in A











- 258 -














statesrdquo







and cladding













Cladding

Pellet



- 259 -































( )3 2

2 2AMU2

2j j

io i

F F

rlr r ππ= sdot

minus (3429)


( )2 2

AMU22

o i

i

r r

rl

πμ

πminus

= sdot (3430)


- 260 -

cf

pf+cf

cf

cf

cf

3

2

1

cf

cf

cf

cf


Plenum space










and cladding






















- 261 -










11 12j ju uΔ = Δ (3431)

jj FF 23 μ= (3432)



Δ=

ΔΔ

ΔΔ

minus

minus

minus

00001100

0

0

0

0

1

1

0

0

2

12

11

F

F

u

u

B

j

Ujc

Ujp

j

j

ee

νν

μμ

(3433)




Row l1 =

Δ=+ΔL

lljjll FFuK

1211 11

(3434)

Row l2 =

Δ=minusΔL

lljjll FFuK

1212 22

(3435)

Row l3 =

Δ=+ΔL

llj

UjPll FFvK

12 33

μ (3436)

Row l4 =

Δ=minusΔL

llj

UjCll FFvK

12 34

μ (3437)


- 262 -














vkF lz sdot= (3439)













treatment applies


- 263 -


Segment No

Contact state




Plenum － 0 0 uF













conditions
















results


- 264 -











11 12j ju uΔ = Δ (3431) U

jCU

jP vv Δ=Δ (3440)



segment j

U


of segment j


segment j


Δ=

Δ=Δ

Δ=Δ

F

u

B

jCU

jP

jj

ee

1211

νν

μ

(3441)



UjPv Δ=

UjCv Δ a common


- 265 -










the total matrix U

jCU

jP vv Δ=Δ (3440)


of segment j U


of segment j


Δ=

Δ=Δ

F

B

jCU

jP

ee

νν

(3442)






- 266 -














cladding
















11 12j ju uΔ = Δ (3431)

jj FF 23 μ= (3444)



- 267 -

Δ=

ΔΔ

ΔΔ

minus

minus

minus

00001100

0

0

0

0

1

1

0

0

2

12

11

F

F

u

u

B

j

Ujc

Ujp

j

j

ee

νν

μμ

(3445)


























- 268 -







cladding





external force















po j ci ju uΔ = Δ (=Eq(3431)) (3446)

UjC

UjP vv Δ=Δ (3440)

Here




j of the cladding



- 269 -







(3447)




Row l1 =

Δ=+ΔL

lljlll FFuK

12 11

(3448)

Row l2 =

Δ=minusΔL

lljlll FFuK

12 22

(3449)

Row l3 =

Δ=+ΔL

lljlll FFuK

13 33

(3450)

Row l4 =

Δ=minusΔL

lljlll FFuK

13 44

(3451)



UjP vv (3453)








2

3

0 0

1

1

0

0

1

1

0 0 1 1 0 0 0 0

0 1 1 0 0 0

0 0

p j

c j

po j

ci jee

Up

Uc

j

j

u

uBF

Fv

Fv

F

F



- 270 -






axial direction





















- 271 -

Time Step Start


















- 272 -



















Fuel Rod

Gap

Cladding

R

Pellet


1 segment


- 273 -


Upper pellet

Centerline of rod

1

1

1

Gap



Oxide layer



















- 274 -

00 02 04 06 08 10

R(ridge)M(mid)



Cladding

1 2 3 4

B(body)

2n-1 2n

n




analysis



follows


outer oxide layer











- 275 -










from each other










1-D analysis

2-D analysis





- 276 -

ε

ε

ε

ε

γpartpart

partpart

θ

=

=

+

r

z

rz

du

dr

dv

dz

u

r

v

r

u

z

(351）





analysis

352 Basic equations



of its importance


work as

[ ] B dV FT

n n + +minus =σ 1 1 0 (352)

Here







given by

[ ] σ εn neD+ +=1 1Δ (353)

where


Δε ne



- 277 -

[ ] ( )[ ] [ ]D D Dn n n+ += minus + =

θ θ θ θ1

1

21




[ ] Δ Δε σθne

n nC+ + +=1 1 (354)


[ ]CE

=

minus minus

minus minus

minus minus

+

1

1 0

1 0

1 0

0 0 0 2 1

ν ν

ν ν

ν ν

ν( )

(355)




n n ncrk

nc

np


1 1 1 (356)

where




Δε nc




strain on the block


[ ] ( )Δε n n nB u u+ += minus1 1 (357)

where



[ ] ( ) [ ] ( ) C B u un n n n n n np

nc


1 1 0Δ Δ Δ (358)

Here

[ ] [ ] [ ]

[ ] [ ]

1

1

1(1 )

2n n n

n n

C C C

C D

θ θ θ θ+ +

minus

= minus + =

=


- 278 -





( ) ( ) ( ) ( ) ( ) 052 22 2 232 3











[ ] ( ) [ ] ( ) C d B u du un ni

ni

n ni

ni

n n np

nc

+ + ++

+ ++


1 11

10

1 1 0Δ Δ Δ

(359)

Here

d ni

ni

niσ σ σ+

+++

+= minus11

11

1 (3510)

du u uni

ni

ni

++

++

+= minus11

11

1 (3511)


as


ni T

ni

nσ σ+ ++

+ + =1 11

1 (3512)





That is


- 279 -

[ ] [ ] [ ][ ] [ ] [ ][ ] ( )[ ] [ ] ( )[ ] ( ) 12111

1

01

1

11

1

ˆ

ˆ

ˆ

+++

+++

+++

++++

=minusminusΔ+

Δ+Δminus

minusminus

+

nnin

T

icnn

iPn

T

nin

ii

iPn

T

in

iPn

Tin

T

FdVZZTSB

dVDB

dVCDB

dVduBDBdVB

εε

σσ

σ

θ

θθ

θ

(3513)

is obtained where

[ ]( )

~C

E E E

E E E

E E E

E E

r

z

r z

=

minus minus

minus minus

minus minus

++

1 10

10

10

0 0 04 1

ν

ν ν

ν ν

νθ

(3514)


[ ] K u Fni

ni

ni

+ ++

+=θ Δ Δ11

1 (3515)


Here

[ ] [ ] [ ][ ]K B D B dVni T

nP i

+ += θ θ (3516)

and

[ ]

[ ] ( ) [ ] ( )

1 1 1

0 1 1 1

1 1 1 2

ˆ

ˆ

Ti in n n


T in n

F F B dV

B D C dV

B S T Z Z dV

θ θ

σ

σ σ ε ε

+ + +

+ + + + +

+ +

Δ = minus

+ minus + Δ + Δ


(3517)




[ ] [ ] [ ][ ]K B D B rdrdzni T

nP i

+ += θ θπ2 (3518)

[ ] [ ] [ ][ ] ( ) [ ] ( ) 2

ˆ2

2ˆ

2111

1

011

111

minusminusΔminus

Δ+Δ+minus+

minus=Δ

++

+++++

+++

rdrdzZZTSB

rdrdzCDB

rdrdzBFF

nin

T

icnnn

in

ii

iPn

T

in

Tn

in

π

εεσσπ

σπ

θθ (3519)


- 280 -







[ ]u

vN

u

v

u

v

u

v

=

1

1

2

2

8

8

(3520)


[ ] [ ]1 2 3 8

3 81 2

3 81 2

0 00 0

0 00 0

N N I N I N I N I

N NN N

N NN N

=

=

(3521)


0 1I

=

That is

1 1 2 2 3 3 8 8u N u N u N u N u= + + + +

1 1 2 2 3 3 8 8v N v N v N v N v= + + + +

hold


( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )

N

N

N

N

N

N

N

N

1

2

3

4

5

6

7

8

1 1 1 4

1 1 1 4

1 1 1 4

1 1 1 4

1 1 1 2

1 1 1 2

1 1 1 2

1 1 1 2





= + minus minus

= + + minus

= minus + minus

= minus + minus


(3522)



- 281 -


analysis

[ ]

r

z

rz

u r

v z B u

u z v rθ

εε

εεε


(3523)


[ ] [ ]B = B B B B1 2 3 8 (3524)


B i

i

i

i

i i

N

rN

zN

rN

z

N

r

=

partpart

partpart

partpart

partpart

0

0

0 (3525)



performed


r z is given by

[ ]partpartξ

partpartη

partpartξ

partpartξ

partpartη

partpartη

partpart

partpart

partpart

partpart

N

N

r z

r z

N

rN

z

N

rN

z

i

i

i

i

i

i

=

=

J (3526)

or

[ ]partpart

partpart

partpartξ

partpartη

N

rN

z

N

N

i

i

i

i

=

minusJ 1 (3527)




- 282 -

r N r

z N z

i ii

i ii

=

=

=

=

1

8

1

8


[ ]J =

= =

= =

partpartξ

partpartξ

partpartη

partpartη

Nr

Nz

Nr

Nz

ii

i

ii

i

ii

i

ii

i

1

8

1

8

1

8

1

8 (3528)


[ ] [ ]JJ

minus = =

= =

=minus

minus

1 1

8

1

8

1

8

1

8

1

det

partpartη

partpartξ

partpartη

partpartξ

Nz

Nz

Nr

Nr

ii

i

ii

i

ii

i

ii

i

(3529)

Here

[ ]det J =

minus

= = = = part

partξpartpartη

partpartη

partpartξ

Nr

Nz

Nr

Nzi

ii

ii

i

ii

i

ii

i1

8

1

8

1

8

1

8

(3530)



[ ]

[ ]

partpart

partpartη

partpartξ

partpartξ

partpartη

partpart

partpartξ

partpartη

partpartη

partpartξ

N

r

Nz

N Nz

N

N

z

Nr

N Nr

N

N

r

N

N r

i jj

j

i jj

j

i

i jj

j

i jj

j

i

i i

i ij

=

minus

=

minus

=

= =

= =

=

1

1

1

8

1

8

1

8

1

8

1

8

det

det

J

J

(3531)





[ ] [ ] [ ][ ] [ ]K B D B N r d dni T

nP i

j jj

+ +=

minusminus=

θ θπ ξ η2

1

8

1

1

1

1 det J (3533)


- 283 -

[ ] [ ]

[ ] ( ) [ ]

[ ] ( ) [ ]

81 1

1 1 11 11

81 1 0

1 1 11 11

81 1

1 1 21 11

ˆ 2 det J

ˆ2 det J

2 det J

Ti in n n j j

j


j

T in n j j

j

F F B N r d d

B D C N r d d

B S T Z Z N r d d

θ θ

θ

π σ ξ η


π ξ η

+ + +minus minus=


+ +minus minus=

Δ = minus



(3534)


f d d H H fi j i jj

m

i

n


== =

1

1

1

1

11

(3535)



applied

f d d f i jji


== =

1

1

1

1

1

2

1

2

(3536)

Here










1



Node

ξ

η

1-1

-1

1

Integrationpointa

a-a

-a


- 284 -







(4)

(1)

(2) (3) (6) (7)

(5)

(8)

r

z

Pellet

CladdingGap






- 285 -



















of pellet is fixed



Cladding

Mid-heightplane

Center


Cladding

Center


A B


- 286 -











12

13

14

15

16

17

18( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (3537)




Δv ( )1 0= (3538)

Δv ( )5 0= (3539)

hold Here ( ) Δv 1

and ( ) Δv 5









Δu( )2 0= (3540)




holds



represents





- 287 -





considered Fn+16( )





Δ Δv v( ) ( )8 8= (3541)

















F F F F F F F

F F F F F F

n nR

nR

nPC

nPP

nR

nPC

nG

nW

nR

nAX

nW

nG

+ + + + + + +

+ + + + + +

= + + + + +

+ + + + + +

1 11

12

13

14

15

16

16

17

18

18

18

18

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) (3542)


as



- 288 -

where

prime = + + ++ + + + +F F F F Fn nG

nW

nW

nG

1 16

17

18

18( ) ( ) ( ) ( ) (3544)


nR

nR

nR

1 11

12

15

18( ) ( ) ( ) ( ) (3545)


nPP

nPC

nAX

1 13

14

16

18( ) ( ) ( ) ( ) (3546)









[ ] K u Fn n n+ + +=θ Δ Δ1 1 (3547)

[ ] [ ] [ ][ ]K B D B dVn

T

nP

+ += θ θ (3548)

[ ] [ ] [ ][ ] ( ) [ ] ( )

minusminusΔminus

Δ+Δ+minus+

minus=Δ

++

+++++

+++

dVZZTSB

dVCDB

dVBFF

nnT

cnnn

inn

Pn

T

nT

nn

211

10

11

111

~ˆ

ˆ

θ

θθ εεσσ

σ

(3549)



[ ] [ ] ( ) [ ] ( )

1 1 1 1 1

01 1 1

1 1 2

1 1

ˆ

ˆ

ˆ

T

n n n n n

T P i cn n n n n n

T

n n

n n

F F F F B dV

B D C dV

B S T Z Z dV

F F

θ θ

θ

σ

σ σ ε ε

+ + + + +

+ + + + +

+ +

+ +


+ minus + Δ + Δ



(3550)

where


- 289 -

[ ] [ ] ( ) [ ] ( )

1 1 1 1

01 1 1

1 1 2

ˆ

ˆ

T

n n n n

T P i cn n n n n n

T

n n

F F F B dV

B D C dV

B S T Z Z dV

θ θ

θ

σ

σ σ ε ε

+ + + +

+ + + + +

+ +




(3551)








stiffness equation


















- 290 -



pairs









early phase of PCMI






















- 291 -












cladding







pairs



Node Couple

Pellet Gap



- 292 -


























In Eq(3542) FnPC

+13( ) and Fn


PP+14( ) and

FnAX


obtain

[ ] K u Fn n n+ + += primeθ Δ Δ1 1 (3553)


- 293 -


In this case FnPC

+13( ) and Fn


FU

VnPC n

n+

+

+

=

13 1

1

0

0

0

0

( )

(3554)

FU

VnPC n

n+

+

+

=minusminus

16 1

1

0

0

0

0

( )

(3555)

where

V Un n+ += plusmn1 1μ (3556)





i

j

k

l

m

K

u

v

u

v

U

Fn

p n

p n

c n

c n

n

n

0

0

1

1

0

0

0 0 1 0 1 0 0 0 0 0

1

1

1

1

1

1

minusplusmn

plusmn

minus

= prime

+

+

+

+

+

+

+

μ

μθ

ΔΔΔΔ

Δ

(3557)


- 294 -










equations are

K u F Uni

n n n+ + + += prime +θ( ) ( )Δ Δ1 1

11 (3558)

K u F Unj

n nj


1)(

11)( ˆ


nnk

n UFuK θ (3560)

1)(

11)( ˆ


nnl

n UFuK μθ (3561)





i

j

k

l

m

n

K

u

v

u

v

U

V

n

p n

p n

c n

c n

n

n

0 0

0 0

1 0

0 1

1 0

0 1

0 0

0 0

0 0 1 0 1 0 0 0 0 0

0 0 0 1 0 1 0 0 0 0

1

1

1

1

1

1

minusminus

minusminus

+

+

+

+

+

+

+

θ

ΔΔΔΔ

=prime

+Δ Fn 1

0

0

(3563)


- 295 -










above equations are

1)1(

11)( ˆ


n UFuK θ (3564)

K u F Vnj

n nj

n+ + + += prime +θ( ) ( )Δ Δ1 1 1 (3565)

K u F Unk

n nk

n+ + + += prime minusθ( ) ( )Δ Δ1 1 1 (3566)

1)(

11)( ˆ


nnl

n VFuK θ (3567)


minus + =+ +Δ Δu up n c n 1 1 0 (3568)

minus + =+ +Δ Δv vp n c n 1 1 0 (3569)







tt t

g gg tn

n n

n nn n+

+

+

= minusminus

+α1

1

(3571)




- 296 -





when




U p n + ge1 0 (3574)



tt t

U UU tn

n n

p n p np n n+

+

+

= minusminus

+α1

1 (3575)









Also when


+ +le1 1 (3578)


Here v p n


vc n


In contrast when


- 297 -

U p n + ge1 0 (3579)










bonded gap states









reversed











this


- 298 -






reversed























( ) ( )tZ Z

Z Z Z Zt t tn

i n j n


+ ++=


minus +α

1 11 (3580)

Here




- 299 -













That is in

[ ] F B dVn

T

n+ += 1 1σ (352)


+14( ) in the axial

direction

[ ] F F B dVnPP

n

T

n+ + += = 14

1 1( ) σ (3581)

holds




equation

( )tV

V Vt t tn

n

n nn n n+

++=

minusminus +α

11 (3582)








- 300 -


F F F F FR AX W G( ) ( ) ( ) ( ) ( )8 8 8 8 8= + + + (3583)










- 301 -



361 Creep



















t time (hour)



- 302 -




( ) ( )( )( ) ttTrTf cs





( ) ( ) ( )( ) cs






( ) ( )( )( ) ( )( )

minus=

+minus=HH

cs

Hc

TfTr

TfTr

εσσεεεσσε

(365)





( ) 1f T C Eσ ϕσ= (366)

( ) 2 r T E Cσ ϕ= (367)

3cs C Eε ϕσ= (368)

are obtained



( )c

FSεσ

part =part


2 1 3c


(3610)

( )c

H FHεε

part =part



- 303 -

2c


(3611)



( )H

GSεσ

part =part


2 1H


(3613)

( )H

H GHεε

part =part


2H


(3614)

From Eqs(369) to (3614)

( )c FFUNCε = ( )c

FSεσ

part =part

( )c

H FHεε

part =part

( )H GFUNCε =

( )H

GSεσ

part =part

( )H

H GHεε

part =part



1 1 11 1 1 1


+ + + + + + Δ = Δ = Δ + (3615)

where

1 1 3 1 1

3


n

f t F tθ θθ

ε σ θ σσ+ + + + + +

+


[ ] [ ]C t F Fnc i

n i j n

i

n

i


θ θ θ


Δ 1 1 2 (3617)


( )1

1 2

1

9

4 1

i i

i ni Hn n n

iiin nn

n Hn

f gt

ffF

gt

θ θ θ

θ θθ

θ


partε

++ + +

+ +++

+


(3618)

Ffn

i

ni2

3

2= +

+

θ

θσ (3619)

( )1 1

3

1

3

21

ii H i

n n nHn

iin

n Hn

ft g

Fg

t

θθ

θ

θ

part εpartε

σ partθpartε

+ + ++

++

+

Δ minus Δ =

minus Δ

(3620)


- 304 -




part =part

( )Hg GHε

part =part 1

c inε +Δ and













- 305 -



σpart

part

c

Hε

εpart

part

Hεσ

partpart

and

H

Hε

εpart

part


σpart

part and

Hεσ

partpart

is 1(sMPa)


be multiplied by

( )3 3 6( ) ( )36 10 36 10 ( ) 10 ( )( ) ( )

d hr d Pad s d MPa




σ ε= K n


ε true strain


( )( ) ( )1








n=00504+00001435T







- 306 -

00 05 10 15 200

200

400

600

800

1000

1200

1400300K

600K

350K

900K

800K

True

Stre

ss

(MP

a)

True Strain ()

100 s-1

1 s-1

0001 s-1




00 05 10 15 200

200

400

600

800

1000

0001s

0001s

0001s

100s

100s

900K

600K

400K

True

Stre

ss

(MP

a)

True Strain ()

352K100s








- 307 -

363 Ohta model(35)














occurs














nmid

midK

= minus310

εεσ (3623)

Plastic region 310

mnK

εσ ε minus =

(3624)


0

300

600

900

1200

1500

0 001 002 003 004True s train [frac]

True

stre

ss [M

pa]


00E+0

50E+4

10E+5

15E+5

0 001 002 003 004

True s train [frac]

dσ

dε [M

Pafr

ac]

Predict Interface

Interimregion


( )y yε σ

1 1( )ε σ

2 2( )ε σTru

e s

tress

(M

Pa)


- 308 -




Hereafter 310

εminus

m




1

1ε

minusprime =

n

y

K









2 min( 0003 175)y yε ε ε= + times (3626)



Ep2

2




m

nmid

midK

= minus311 10

εεσ

(3627)

m

nmid

midK

= minus322 10

εεσ

(3628)


midn

=

1

2

1

2

εε

σσ

(3629)

Consequently

( ) ( )( ) ( )12

12

lnln

lnln

εεσσ



- 309 -

mn

mid

mid

K

=

minus32

2

10

εε

σ

(3631)











given as follows

T≦900 K （62685）

1 2

253 4

( 27315)

( 27315) 1 exp( 1751 10 )

K L L T

L L T minus

= minus minus


1 2 3 41745 2517 526 0748L L L L= = = = (3632)


900＜T＜1000 K (62685～72685 oC)

1 2 3 4

05251 2 3 4

(

( ) 1 exp( 1751 10 )

K A T A T A TA

B T B T B TB minus

= + + +


A1＝ 3041233371times104 B1＝ 33282432 times102

A2＝－8285933869times101 B2＝ 298027653

A3＝ 7490958541times10－2 B3＝－681286681times10－3

A4＝－2242589096times10－5 B4＝ 351229783times10－6

1000 K≦T （72685oC~） 0525


61 2 3 41 10 8775 8663 0341566H H H Hminus= sdot = = = (3634)




- 310 -

















T≦700 K （42685）

1 2

05263 4

( 27315)

( 27315) 1 exp( 9699 10 )

K L L t

L L T minus

= minus minus


(3635)

L1＝1201 L2＝2508 L3＝1119 L4＝0627

700＜T≦800 K （42685～52685）

1 2 3 4

05265 6 3 4

(

( ) 1 exp( 9699 10 )

K A T A T A TA

A T A T A TA minus

= + + +


A1＝ 3656432405times104 A2＝－1361626514times102

A3＝ 1695545314times10－1 A4＝－7055897451times10－5

A5＝－3338799880times104 A6＝ 1330276514times102

800＜T＜950 K （52685～67685）



61 2 31 10 4772 9740H H Hminus= sdot = =


- 311 -


0526

2 3 4( 1 exp( 9699 10 )

non irradiated

non irradiated

K K

B T B T B TB K

minus

minusminus

=


B1＝－3617377246times104 B2＝ 1361156965times102

B3＝－1636731582times10－1 B4＝ 6386789409times10－5

950K≦T （67685oC≦T）

0526


61 2 3 41 10 4772 9740 5267783H H H Hminus= sdot = = =















T≦67315 K (400oC)

1 2 ( 27315)n C C T= minus minus (3640)

67315ltTlt1000 K （400～72685）

3 4 5( )n C T C C T= + + (3641)

C1＝0213 C2＝1808times10－4 C3＝34708518times10－2

C4＝49565251times10－4 C5＝－50245302times10－7


- 312 -

1000 K≦T （72685～）

n = C6 = 0027908 (3642)

【Note 3】


MATPRO model






the MATPRO model




irradiation

T≦730 K （45685） m＝002 (3643)

730ltT＜900 K （45685～62685）

5 6 7 8( )m A T A T A A T= + + + (3644)

900 K ≦T （62685～）


A5＝ 2063172161times101 A6＝－7704552983times10－2

A7＝ 9504843067times10－5 A8＝－3860960716times10－8

A9＝ -067times10－2 A10＝－2203times10－4




- 313 -









310

mnK

s

εσ ε minus =

(3646)


ε true strain (-)



eEεσ = (3647)



εminus

m


as





( )

11 minusprime

= minusn

E

Kn

d

df εσσ

(3650)


d

dff

n

nnn minus=+1 (3651)



solution


- 314 -


ε

σ σ0

σ1σ2 σ3

εp

nK εσ =

eEεσ =

Yσσ =cE e += εσ


ε

σ σ0

σ1σ2 σ3

εp

nK εσ =

eEεσ =

Yσσ =cE e += εσ



- 315 -

37 Skyline Method










(0) (0) (0) (0)11 1 12 2 1 1

(0) (0) (0) (0)21 1 22 2 2 2

(0) (0) (0) (0)31 1 32 2 3 3

(0) (0) (0)1 1 2 2

(1)

(2)

(3)

n n

n n

n n

n n nn n

a x a x a x b

a x a x a x b

a x a x a x b

a x a x a x





(371)






0

To eliminate 1)0(



Eq(372) we obtain

(0) (0) (0)

(0) (0) (0) (0) (0) (0)21 21 2122 12 2 2 1 2 1(0) (0) (0)

11 11 11n n n


a a a


(373)





1 xan



- 316 -

(0) (0) (0) (0)11 1 12 2 1 1

(1) (1) (1)22 2 2 2

(1) (1) (1)32 2 3 3

(1) (1) (1)2 2

(1)

(2)

(3)

( )

n n

n n

n n

n nn n n

a x a x a x b

a x a x b

a x a x b

a x a x b n



(374)

where (0) (0) (0) (0)

(1) (0) (1) (0)21 12 31 1222 22 32 32(0) (0)

11 11

(0) (0) (0) (0)(1) (0) (1) (0)21 1 31 12 2 3 3(0) (0)

11 11

a a a aa a a a

a a

a b a bb b b b

a a









=

=+

=+++

=++++

minusminus

minusminusminus

minusminus

minusminus

)1()1(

)2()2()2(

)1(2

)1(21

)1(122

)1(22

)0()0(11

)0(112

)0(121

)0(11

nnn

nnn

nnn

nnnn

nnn

nnnn

nnnnn

bxa

bxaxa

bxaxaxa

bxaxaxaxa

　　　　　　　　　　　　

　　　　　　　　

　　　　

　　　

(375)


Nkj

Nki

a

aaaa k

kk

kkj

kikk

ijk

ij

1)1(

)1()1()1()(

=+=

minus= minus


Nkiba

abb k

kkkk

kikk

ik

i 1)1()1(


minus



)1(

)1(

minus

minus

minusn

nn

nn

n a

bx (378)



- 317 -


)1(2

)1(2

)1(1

)1(

minus+

minus++

minus+

minus minusminus= k

kk

kkkkk

kkk

kk

k a

xaxabx

(379)


obtained




ib of


ib



)2(11

)2(1

)2(1)2()1(

minusminusminus

minusminus


ii

iji

iiii

iji

ij a

aaaa

)2(1)2(

11

)2(1)2()1( minus



iiii

iiii

ii

i ba

abb


)0(11

)0(1

)0(21)0(

2)1(

2 a

aaaa j

jj minus=

=

minus

minusminus

minus=

minusminus=minus=

2

1)1(

)1()1()0(

3

)1(22

)1(2

)1(32

)0(11

)0(1

)0(21)0(

3)1(22

)1(2

)1(32)1(

3)2(

3

kk

kk

kkj

kik

j

jjj

jjj

a

aaa

a

aa

a

aaa

a

aaaa

　　

so that in general

minus

=minus


1

1)1(

)1()1()0()1(

i

kk

kk

kkj

kik

iji

ij a

aaaa (3710)

holds Similarly

minus

=

minusminus

minusminus minus=

1

1

)1()1(

)1()1(

i

k

kkk

kk

kik

ii

i ba

abb (3711)

holds


nb are

obtained


- 318 -


ija and )1( minusi


ji

mj

a

aaaa

i

kk

kk

kkj

kik

iji

ij

2

21

1)1(

)1()1()0()1(

==

minus= minus

=minus


miba

abb

i

k

kkk

kk

kik

ii

i 21

1

)1()1(

)1()1( =

minus= minus

=

minusminus

minusminus (3713)



( 1) ( 1)1

( 1) (0)( 1)

1

for 2

for 2

k kiik kji

ij ij kk kk

i m j na aa a

j m i na


minus=

gt == minus gt =

　 (3714)

nmiba

abb

i

k

kkk

kk

kik

ii

i 1

1

)1(

)1(

)1()1( =

minus= minus

=

minusminus

minusminus (3715)













timestimestimes

i ｊ

Symmetry

-Li

-Lj


- 319 -



( )=

+++minusj

kkLkji

1

1 (3716)



++j

kkLk

1

1



described








minus

=


)1()1()0()1(

0

î

kk

kkj

kkiij

iij aaaa (3717)

where )1(

)1()1(ˆ minus

minusminus =

kkk

kkik

ki a

aa

mj 2=

11 minus+= jLi j

( )ji LLk max0 =

mjaaaaj

Lk

kkj

kkjjj

ijj

j

2ˆ1

)1()1()0()1( =

minus= minus

=


and

( )bi

bj

kk

kk

kkii

ii LLk

mLibabb

max

1ˆ

0

1)1()1()0()1(

0=

+=

minus= minus

=




- 320 -


mLia

bb bi

ii

iii )1(

)0()1( =

= minusminus (3720)


( 1) 1kk kx b k mminus= = (3721)

( 1) 1 2ˆ

1 1k

k k ki ii i

i m mx x a x


= minus= + minus

(3722)



kki




ija )1( minusi




minus

=


)1()1()0()1(

0

î

kk

kkj

kkiij

iij aaaa (3723)

where

)1(

)1()1(ˆ minus

minusminus =

kkk

kkik

ki a

aa (3724)

mj 2=

( )jpLi j min1 +=

( )ji LLk max0 =


part of matrix

and

pjaaaaj

Lk

kkj

kkjjj

ijj

j

2ˆ1

)1()1()0()1( =

minus= minus

=


where

( )jpLk

mi

a

aa

jk

kk

kkik

ki min

2ˆ

)1(

)1()1(

==

= minus

minusminus (3726)



- 321 -


Eq(3725)


minus

=


)1()1()0()1(

0

î

kk

kk

kkii

ii babb (3727)

where pLi b 1+=

( )bi LLk max0 =

and also

pLia

bb bi

ii

iii

i )1(

)1()1( =

= minus

minusminus (3728)

is set


minus

=


)1()1()0()1(

0

î

kk

kkj

kikij

iij aaaa (3729)

holds where

)1(

)1()1(ˆ minus

minusminus =

kkk

kikk

ki a

aa (3730)

in nmj 1+=

( )jpLi nj min1 +=

in nmi 1+=

( )jpLj ni min1 +=

and

n

j

Lk

kkj

kjkjj

ijj ppjaaaa

j

1ˆ1

)1()1()0()1( +=

minus= minus

=


where

( )jpLk

nmj

a

aa

njk

kk

kjkk

ki min

1ˆ

)1(

)1()1(

=+=

= minus

minusminus (3732)





- 322 -

n

i

kk

kk

kkii

ii ppibabb 1ˆ

1)1()1()0()1(

0

+=

minus= minus

=


where niii

iii ppi

a

bb 1)1(

)0()1( +=

= minusminus (3734)

is set


sequence

( 1) 1kk k nx b k pminus= = (3735)

( )( 1)

1 2ˆ

min 1k

k k ki ii n

i n nx x a x


= minus= minus

(3736)



References 3












- 323 -









( ) ( )K

D

TT=

minus minusminus

sdot+

+ times times

minus minus1 1

1 0 05

40 4

4641216 10 1867 104 3

ββ

exp


( ) ( ) K

DT=

minus minusminus


1 0 050 0191 1216 10 1867 104 3β

β exp



Ttemperature (degC)


Default

value

IPTHCN=1

213


IPTHCN=2





IPTHCN=3






- 324 -


4 7 01148 00035 (2475 10 8 10 )

00132exp(000188 ) (WmK)

K F B B T

F T


+ sdot sdot




IPTHCN=4 43


IPTHCN=17 215

4 01148 00040 (2475 10 000333 )

00132exp(000188 ) ( )

K F B B

F T W mK


+ sdot sdot




T(C)


IPTHCN=5 42


12 3 4 795

312

(235 10 35 10 ) (255 10 80 10 )

357 10

TD K

K

K B B T

T


minus


+ times sdot






IPTHCN=6 44

13 4 80047 173 10 (25 10 541 10 )

(WmK)






IPTHCN=7 45


2

05

10 05( 10)

10 (( 75) 10)

B G B

B G B

le =ge = minus



- 325 -



23210 gAgAxAAA +++=

2210 gBgBBB ++=

A0=00254 A1=40 A2=03079 A3=122031

B0=2533x10-4 B1=8606x10-4 B2=-00154

C=10x10-12


( )( )

327384 058 10 ( 127315 )

20 ( 127315 )

T T T K

T T K

αα


= gt

IPTHCN=8 46


IPTHCN=9 214




( ) ( )

2 4

1 temperature(K)

1 396exp 6380

452 10 (mKW) 246 10 (mKW)

h T TT

A Bminus minus

=+ minus

= times = times



IPTHCN

=91 410

Latest model

( ) ( ) ( )2 4

1

Gd P

C D

A B T f Bu g Bu h T

T T

λα β


minus + (WmK)




( ) ( )

2 4

1

1 396exp 6380

453 10 (mKW) 246 10 (mKW)

h TT

A Bminus minus

=+ minus

= times = times

C=547E-9 (Wm-K3) D=229E-14 (Wm-K5)





- 326 -


( )

( )

( ) ( )

( ) ( )( )

( )

1 1 2 3 4 0

9

0 4 2

1 3265 3265

1

2

W mK

1 4715 10 16361exp W mK

00375 2165 10

109 00643 1arctan

109 00643

0019 11

3 0019 1 exp 1200 100

1 15 for s

1 1

d P P x r

d

P

P

K K K K K

T T T

K TT

KT

PK

P

λ λ

λ

ββ β β β

ββ

σσ

minus

=



minus= =+ minus

( )

( )( )3 4

4


021 1

1 exp 900 80

temperature (K)


x r

P

K KT

T

Bu Buβ minus

= = minus+ minus

= times sdot

IPTHCN=10 47


IPTHCN=13 48

KaT b

=+


Ttemperature (degC)









- 327 -


( )( )95 22 20 1 2 3 0 1 2 3

1exp (WmK)

1TD

C Dg WK


+ = + minus + + + + + + +

T = temperature (K)





W= 141MeV = 1411610-19 J

A0 = 0044819 + 0005BU+20(G - 54702) (mKW)

B0 = (24544 - 00125BU - 70(G - 54702))10-4 (mW)

A1 = 4 (-) B1 = 08 (-)

A2 = 0611 (mKW) B2 = 960310-4 (mW)

A3 = 11081 (mKW) B3 = -176810-2 (mW)

C = 5516109 (WKm) D = - 43021010 (WKm)



A =LT






IPTHCN=15 49







- 328 -


1

mat ek

K KA B T

= ++ sdot

(kWmK) for 100TD





1)

12731540 412 ( )022 ( 50)0447 000453 (50 80)0084 (80 )


Bu BuBu


2) 147315

40 174022 00003

kT KA BuB Bu

gt= + sdot= + sdot



IPTHCN=16 238


IPTHCN =18 411






ΔL


ΔL


IPTHEX=0 (default)

IPTHEX=1

34


- 329 -


ΔL




IPTHEX=2


ΔL



IPTHEX=3


ΔL



IPTHEX=4


IPTHEX=5 211


exp DL EK T K K

L kT

Δ = minus + minus



K1 = 10x10-5 K-1 K2 = 30x10-3 K3 = 40x10-2


IPTHEX=6 426

273 923T Kle le 3 6 10 2

13 3

( ) (273) 266 10 9802 10 2705 10

4391 10

L T L T T

T

minus minus minus

minus


1 6

10 2 13 3

( ) (273) (99734 10 9802 10

2705 10 4391 10 )

L T L T

T T

minus minus

minus minus


973 3120T Kle le

3 5 12 2

12 3

( ) (273) 328 10 1179 10 2429 10

1219 10

L T L T T

T

minus minus minus

minus


1 5

9 2 12 3

( ) (273) (99672 10 1179 10

2429 10 1219 10 )

L T L T

T T

minus minus

minus minus





- 330 -

(3) UO2 density





No

restrictions

213


IPLYG=2 Default







Tlt1000 K e = 08707 Tgt2050 K e = 04083


No

restrictions

34








No

restrictions

34

413


- 331 -

(8) UO2 creep


( ) ( )( )

( )( ) ( )ε

σ σσ=

+ minus

++

minus+

+ minusA A F Q RT

A D G

A Q RT

A DA F Q RT

1 2 1

32

44 5

2

56 3

exp exp exp


6 121 2

43 4

285 6

1 2 3

9728 10 324 10

877 1376 10

905 924 10

90000 132000 5200 (calmol)

A A

A A

A A

Q Q Q

minus

minus

minus

= times = times

= minus = times





Ddensity 92-98 (TD)



No restrictions

34


- 332 -


( ) ( ) ( )

( )

451 2 1 1 4 8 2 2

23 6

7 3

exp ( ) exp

( )

exp

c


A D G A D

A F Q RT

σ σε

σ

+ minus + minus= +

+ +

+ minus

(s-1)

A1=03919 A2=13100times10-19 A3=-877 A4=20391times10-25

A6=-905 A7=372264times10-35A8=00 1

1

20178848 exp 80 10 7212423

log( 199999)Q

x


1

2

20198720 exp 80 10 1115435

log( 199999)Q

x









When 05714

165474168


When 05714

165474168

Gσ gt 1 05714

165474168

Gσ = and 2σ σ=

IPCRP=2

(Default) 412







- 333 -



45 21 1 22

( )( ) exp( ) exp( )


G


(1s) ( MPa)σ =

4 17 610

1

14 33

6 5 63

2

4 13

138 10 46 10( ) 1877 10

905

259 10 863 10(1s)

905

973 10 324 106895 10

877

671 10 223 10(1s)

877

FA F

F

FA

F

ρ

ρ

ρ

ρ


minus minus

minus minusminus

minus











IPCRP=4

(Default)

414


4 303505 108577 10 exp 148 10 f





and σ=24 MPa

IPCRP=5 415


- 334 -

(9) UO2 swelling








ΔV

V T

gs

S

= times minus times

4 396 10

1645 1024


ΔV

V

gs


fissionscm3


(default )

249



ΔV

V

ss






minus3times107


minus106




V

ss


IFSWEL=1 12


- 335 -










1 1

5 51 1

2413 10 10634 10 011509

286 10 ln 20184 10 ln

VT T Bu

V

T BU T GR

minus minus

minus minus



where

V



GR grain size (μm)

IFSWEL=5 252


IFSWEL=2 213


T le 1400 028 1400 le T le 1800 ( )[ ]0 28 1 0 00575 1 400 + minusT


2200 lt T 0476


IFSWEL=3


205 (10MWdkgUO )V

V

Δ =


- 336 -


( ) ( )

( ) ( )BuBuBuBuAV

V

V

V

BuBuBuFGRCAV

V

SWSW

SW

ltminus+

Δ=

Δ

lelesdotsdotminus=

Δ

1121

111 0

whereSW

V

V

1

Δ


A1=0095 C1=0043 Bu1=57 A2=00317

SW

V

V

Δ




IFSWEL=4 251



( )BuCeV

V

V


max

ΔV


V


densification


IDENSF=0

(default value)

11




ΔV

V

max

() 10

SBU


given by CSBU

= 2 3025 (MWdt-UO2)

where exp (－23025) =01

25000


- 337 -




( )( )

ΔL

L

TD

TS= times

minusminus

22 2100

1180





BUburnup (GWdt-U)

IDENSF=1 416





( )

( ) ( )( )

ΔΔΔ Δ

ρρρ ρ

= le= + lt lt= ge

0 1

1 2

2

BU BU

m BU b BU BU BU

BU BUsn

log


(1700degCtimes24h)

BUburnup (MWdt)



burnup range

IDENSF=2 417






- 338 -



( )ΔV

V

BU

BUADST= minus

minus minus sdot

+ minus minus sdot

sdot0 01420 1 6 7943

0 00793 1 11434

exp( )

exp( )



IDENSF=4 43


Δρ = +

M

A

AD Ft

Girr irrln

10

03












AG G

Dt=

minus30

3

MA

InADt

G

th=

+

Δρ

10

3






Q 82000 calmol



IDENSF=3 418



- 339 -







bD


bD



IDENSF=30 419

420







where rcic and rpi




follows





- 340 -

(12) UO2 plasticity




( )( )

( )( )σ

εεY

p

p

T T

T T

T C

T C=


legt

66 9 0 0397 520 0 0 386

36 6 0 0144 139 5 0 06875

1200

1200

( )( )prime =


legt

HT

T

T C

T C

520 0 0 386

139 5 0 06875

1200

1200




IFY=1

(default value) 421


422

σ Y T T

T


minus times sdot

minus

minus

1176 1 1 688 8 179 10

1 293 10

4 2

7 3



IFY=0



( )( )

( )2

1 322 22

exp exp

2exp 1D

P D


RTT T

θ θθ

= + + minus minus



UO2 PuO2 Unit

K1 2967 3474 JkgK



θ 535285 571000 K


No

restrictions 34


- 341 -




sdot+minus=

m

ff

D

NNE

DK

dt

dDmax11







Efitting parameter



boundary (atomscm2)

IGRAIN = 0

(default value) 230






dD

dtk

D

B

Dm

= minus +

1 1 0 002


IGRAIN=1 231


- 342 -


( )D D k t RT20

20


IGRAIN=2 423


IGRAIN=3 424

D D k e tQ RT30

30minus = sdotminus





IGRAIN=4 213

D D e tRT40





t time (h)


IGRAIN=5 34



4




- 343 -




the followings



10 (Default)








case with IXEKR=0

IXEKR = 3



1) 012 ≦ f ≦030



where



( ) ( )

( )

1 34 12

2

0035 225 10 830 10 1

0 005

u K K

z

K T T z

z UO z

minusminus minus

minus


IPTHCN=30 426


- 344 -


( )13 4

2

exp C

KK P K K T

K T

= + +



( )( )P

D=


1 1 0 95

ββ

porosity factor




IPTHCN=31 34


IPTHCN=32 426 427



p

ppF

21

1)(

+minus=

p

pp

21

1

8610

1)(

+minussdot=α

p porosity



TK temperature (K)


( )

sdotsdot+


minusminus

312

14

103876

108852440105500093105281

K

KMOX

T

TzK

τ




2 τ


- 345 -


95

1 1 005

1 2 1 2 005

p


(WmK)

( )2951 exp( )K KK

C DA BT TT

λ = + minus +

)(0350852)( WmKxxAA +==


)(106891 9 mWKC times= 13520D K=



IPTHCN=33

428


IPTHCN=34

427

( ) 14

312

1528 000931 01055 044 2885 10( )

7638 10

KMOX

K

z TK p

T

τα

minusminus

minus



p

pp

21

1

8610

1)(




TK temperature (K)


IPTHCN=35

429

1

2 3 7 4 5 6

( )(1 ) ( )

moxfuel

C

K aK F p



(WmK)


1 1 005( )

1 2 1 2 005

pF p

p


1 2 3

4 5 6

7

404 464 184

116 10 00032

0003

a a a

a a a

a

= = == = ==

CT C deg

210 for UO

092 for MOXmoxK =

= MWdkgMOXBU



- 346 -


( )( )95 22 20 1 2 3 0 1 2 3

1exp

1TD

C Dg WK


+ = + minus + + + + + + +






IPTHCN=15 49





Condition Ref

( )95 092 ( ) 00132exp 000188 (WmK)K Q x T= +





1( ) 10789

1 05

pK p K

p

minus=+

IPTHCN=37

430


- 347 -


IPTHCN =91

410

Latest model

( ) ( ) ( )2 4

1

Gd P

C D

A B T f Bu g Bu h T

T T

λα β


minus + (WmK)




( ) ( )

2 4

1

1 396exp 6380

453 10 (mKW) 246 10 (mKW)

h TT

A Bminus minus

=+ minus

= times = times





Condition Ref

mod95 2

1 ( ) expC D

K P xT T

= + minus

( ) ( ) gad ( ) ( )

(1 09exp( 004 )) ( ) ( )


Bu g Bu h T



=Y default=00)



Bu burnup in GWdtHM

h(T)= 1 1 396expQ

T





1( ) 10789

1 05

pK p K

p

minus=+

IPTHCN =38

430


- 348 -



( ) ( )[ ]( ) ( )[ ]fBDBTQFBB

fBDBTQG

FBBc

47454

265

4332121

1exp

1exp

+minus+minus++

+minus+minus+=

σ

σε

B1=01007 B2=757times10-20 B3=333 B4=0014

B5=6469times10-25 B6=00 B7=103






When 05714

165474168


when 05714

165474168

Gσ gt

571401816547416

G=σ and 2σ σ=

IPCRP=30 34


1) fH=1 222

( )exp( )

A FQ RT C F

G

σε σ= minus +

5 13

2

263 10 893 10(1s)

877

FA

ρ






T temperature (K)



414


- 349 -

4 5 6 7 8 9 10 11 12 13 14 15 161E-10

1E-9

1E-8

1E-7

1E-6

1E-5

fH= 3 fH= 2

SKI MOX fH=1

SKI UO2

UO2Halden

MOXHalden


Fuel

cre

ep ra

te

(1s

)

10000T (1K)



μ = 0276 IPOIS=30 34


IPOIS=31 431 432




54 303505 10

8577 10 exp 973 10 fRT



IPCRP=15 415



- 350 -





IPLYG=30 34

(6) Specific heat



( )( )[ ]

CK T

T TK T

O M K E

RT

E

RTPD D=

minus+ +

minus

12

2 2 23

21 2

θ θ

θ

exp

exp

exp




ISPH=30 34





Tm melting point(K)





No restrictions


34 413


- 351 -


( ) fL

Lf

L

L

L

L

PuOUO 22

1Δ+minusΔ=Δ

ΔL


ΔL

LT T T

UO2


2

4 6 9 2 16 339735 10 84955 10 21513 10 37143 10PuO

LT T T



IPTHEX=30 34


( ) ( )xfL

Lf

L

L

L

L

PuOUOMOXsdot+sdot


22

where MOXL


2UOL


2PuOL




(279 K≦T≦923 K)

( ) 11039141070521080291097349 31321061


L

L

UO

(923 K≦T≦3120 K)

( ) 11021911042921017911096729 3122951


L

L

UO

(279 K≦T≦1693 K)

( ) ( )( )

24 6 9

2312

12232 10 75866 10 273 56948 10 273

59768 10 273PuO

LT T

LT

minus minus minus

minus




IPTHEX=31 426433


- 352 -



exp DL EK T K K

L kT

Δ = minus + minus



K1 = 90x10-6 K-1 K2 = 27x10-3 K3 = 70x10-2

IPTHEX=32 213


bD


bD



IDENSF=30 419 420


- 353 -




Ttemperature (K)

No

restrictions 213



For 27minus800degC








Ttemperature (degC)

ICATHX=0

(default value)

213


ICATHX=1

For 21minus429degC



Ttemperature (degC)


ICATHX=3 34

300ltTlt1073 K

6 3

0

444 10 124 10L

TL


6 3

0

672 10 207 10D

TD




- 354 -


Variable name








IZYG=2

Default

213

34

1) α phase 11 7







11 81 (661 10 5912 10 )K T= times + times Δ


K =088+012exp10

Φ minus







restrictions 213



- 355 -




v = 0 30















Accordingly


is obtained where




CRPEQ=0

213


- 356 -


expm p ncreep

D QA t


13111 10

0682

0550

0579

1173

A

m

p

n

Q

minus= times====

2

hours

ncm s (gt1MeV)

MPa

Kelvin

t

Tθ

φσ

sdot

CRPEQ=2 435


CRPEQ=1 436


ε ε εcthc

irrc= + (1hr)



( )

ε εthc m

HmmK=

minus11

1

where ε ε εH thc

th kc

kk

n

t( ) = = sdot= Δ

1








ceq= times minus6 64 10 25 1 23 1 34





( )d dcz

cεσ


22

( )d dzc

zcε


22



1

2



- 357 -


expm p ncreep

D QA t


1216 10

0450

0550

0579

1173

A

m

p

n

Q

minus= times====

2

hours

ncm s (gt1MeV)

MPa

Kelvin

t

Tθ

φσ

sdot

CRPEQ=3 437


expε σα = minus

2000342205 32

T ( )730 1073K T Kle le


HTCRP=1

(Default) 438


1073ltTlt1273 K

4 4

1073 107310

200 200





HTCRP=

20

21

22

439

0 10 1 exp

nG Q

AT G RT

σε = minus


(Nm-2)

( 973 1273T Kle le )








- 358 -



13


0






( ) ε partσpartσ

σ εcHf T=

(431)


ε ε ε= =HH 0 (432)

is set





ε c

ε

ε

ε

εH =0

εH =0εH =0

ε ε εH H= minus

①

②

③

④



- 359 -


( ) ε partσpartσ

σ εcHf T f=

ge 0 (433)





σ ε= K n



n

E

KKK minus+= )0020( 20

20

( )( )1





n=00504+00001435T


ICPLAS=1 11





- 360 -



prior time step

310

mnK

s

εσ ε minus =

σ true stress (Pa)



Tgt=850 K 0027908n =


cold work 2

13

7 8

0847exp( 392 ) 0153 ( 916 10 0229 )

exp373 10 2 10

RIC CWN CWN CWN

CWN

φ


times + times sdot


exponent n







900lt=T 8663exp 8755K

T = +









ICPLAS=2 34


- 361 -



prior time step

310

mnK

s

εσ ε minus =

(original)

σ true stress (Pa)



model


model




18 25

8 25

0546 554 10 26 10

0546 144 10 (26 10 )

DK CWK K

CWK K φ


KFKF



MATPRO-11 model

ICPLAS=3

00 05 10 15 200

200

400

600

800

1000

0001s

0001s

0001s

100s

100s

900K

600K

400K

True

Stre

ss

(MP

a)

True Strain ()

352K100s





- 362 -




Intermediate region

mn

midmidK

= minus310

εεσ

Plastic region 310

mnK

εσ ε minus =







T≦900 K （62685）

1 2

253 4

( 27315)

( 27315) 1 exp( 1751 10 )

K L L T

L L T minus

= minus minus


1 2 3 41745 2517 526 0748L L L L= = = =


900＜T＜1000 K (62685～72685 oC)

1 2 3 4

05251 2 3 4

(

( ) 1 exp( 1751 10 )

K A T A T A TA

B T B T B TB minus

= + + +


A1＝ 3041233371times104 B1＝ 33282432 times102

A2＝－8285933869times101 B2＝ 298027653

A3＝ 7490958541times10－2 B3＝－681286681times10－3

A4＝－2242589096times10－5 B4＝ 351229783times10－6

1000 K≦T （72685oC -）

0525


61 2 3 41 10 8775 8663 0341566H H H Hminus= sdot = = =

ICPLAS=

4 for

PWR

ICPLAS=

5 for

BWR

35


- 363 -


T≦700 K （42685）

1 2

05263 4

( 27315)

( 27315) 1 exp( 9699 10 )

K L L t

L L T minus

= minus minus


700＜T≦800 K （42685～52685）

1 2 3 4

05265 6 3 4

(

( ) 1 exp( 9699 10 )

K A T A T A TA

A T A T A TA minus

= + + +


A1＝ 3656432405times104 A2＝－1361626514times102

A3＝ 1695545314times10－1 A4＝－7055897451times10－5

A5＝－3338799880times104 A6＝ 1330276514times102

800＜T＜950 K （52685～67685）



1 2 31 10 4772 9740H H Hminus= sdot = =


05261 2 3 4( 1 exp( 9699 10 )

non irradiated

non irradiated

K K

B T B T B TB K

minus

minusminus

=


950K≦T （67685oC≦T）

0526


1 2 3 41 10 4772 9740 5267783H H H Hminus= sdot = = =


67315ltTlt1000 K (400-72685) 3 4 5( )n C T C C T= + +

C1＝0213 C2＝1808times10－4 C3＝34708518times10－2

C4＝49565251times10－4 C5＝－50245302times10－7

1000 K≦T (72685 - ) n = C6 = 0027908




B1＝－3617377246times104 B2＝ 1361156965times102

B3＝－1636731582times10－1 B4＝ 6386789409times10－5


- 364 -

A5＝

A7＝

A9＝

2063172161times101

9504843067times10－5

-067times10－2

A6＝

A8＝

A10＝

－7704552983times10－2

－3860960716times10－8

－2203times10－4


ICPLAS =6 430


prior time step

310

mnK

s

εσ ε minus =

σ true stress (Pa)




where )(TK is


750 K lt T lt 1090 K

( )

timestimes=2

66 1085000272

exp105224882T

TK







( ) ( ) 110



CWTCWf



T temperature (K)


reduction)



- 365 -



where ( )Tn is

Tlt4194 K ( ) 114050=Tn

4194 K lt T lt 10990772 K

( ) 2 3

6 2 10 3

9490 10 1165 10

1992 10 9588 10

n T T

T T

minus minus

minus minus


minus times + times


T gt 1600 K ( ) 6089531=Tn

( )φn is










300 400 500 600 700 800 900

100200

300400500

600700800

900

1000


Cla

ddin

g Y

ield

Stre

ss (

MPa

)

Temperature (K)



- 366 -


following


( )[ ][ ] [ ] [ ]ΔL


12φ

ΔL





CWcold work

ICAGRW=1

213






( )ΔL

Lt

n

0

2101 10= sdot φ ΔL

L0




ICAGRW=2


( )ΔL

Lc t

o


ΔL

Lc

o



ICAGRW=3


ICAGRW=4 445

ΔL

LA t

o

= sdot sdotφ ΔL

Lo






- 367 -





No restrictions

213





( )C C U MP= +0 φ








7





ICOPRO=1

212


ICORRO=3 (BWR)

213


3 exp









- 368 -


FCORRO


( )ds

dtT FCORRO

ds

dt= + sdot1 0 04 Δ


00









(10 026)Zr

ZrO

X

Y

ρρ

=minus

=151 or 156 or 158




T temperature (K)

IZOX=2

Default 34


restrictions





442




CNOX




00


- 369 -






No restrictions









1=Y TsolltT


No

restriction

213


- 370 -




(WcmK) ITMC=41



Not examined










12 2 15 3

1629 10 3285 10

2198 10 1629 10

T

T T

λ minus minus

minus minus



ICATHX=41


Zircaloy



Assumed to be null




- 371 -







CRPEQ =41





8 5 2

8 5 20 2

9 6 21 2

9 5 21 2

3945 10 225 10 ( 0 n cm ) (1)

7550 10 613 10 ( 15 10 n cm ) (2)

1378 10 179 10 ( 145 10 n cm ) (3)

1093 10 660 10 ( 605 10 n cm ) (4)

Y

Y

Y

Y

T t

T t

T t

T t

σ φ

σ φ

σ φ

σ φ

minus

minus

minus

minus





(Pa) (C)Y Tσ




ICPLAS=41




5 2300 1558 326 0298 956 10 (JkgK )

1558 558228p

p

T K C T T

T K C


ge = ICSPH=41 213


- 372 -






2





IGAPCN=0

(default value) 217


IGAPCN=1 213






hK

rmix

0 =+Δ δ




h PK


δ




F

ar

RF

a=

sdot

+

+1

1001429

0 3

1

2Δ



prime = minusminus

+F

x1

1

600

10001

4



- 373 -


h h hgap g s= +




( ) ( )hK

C R R g g rgg=

+ + + +1 2 1 2 Δ


He00021 Ar000029 Kr00001









hK P

a Hsm=

sdotsdot

KK K

K Km =sdot

+2 1 2

1 2

Pt Ys

ID= sdot2 a a R= 0

( )R R R= +1

2 12

22


(WcmdegC)

K

KKm

1

2

0 038

0140 05977

==

=






a0 = 05 cm12


IGAPCN=2 217


- 374 -





hP

M

K K


sdotsdot

280

1

2

1 2

1 2 1 2

12

λ λ




M2=120000psi (844kgcm2)







hK

gg=sdot052 δ


Ar00168 He0139 FP00088



h Ps = 0 6 hK

gg=


IGAPCN=3






- 375 -














Ttemperature (K)


3


( )

( )V

S ZV

Z

K Z

j S ZV

s o

s o

=+

+ prime

+

minus

part

υ part

2

2 2

1

2

21

41





IDCNST=1 228


IDCNST=2 235 ( )TD 6614exp10091 17





- 376 -


321 DDDD ++= (cm2s)


1280

11

9861

1001exp102

521

1 TD


1300

11

9861

10762exp102

440

2 TD



IDCNST=3 238


IDCNST=4 441

( )10 4

2 40

AM1 76 10 exp 7 10

BM1 ( 2 10 )v

D RT

s j V F

minus

minus





IDCNST=5 226

( )( )( )

10 4

17

22

AM1 76 10 exp 7 10

15 10 exp 10600

767 10 exp 2785

D RT

R T

R T

minus

minus

minus


+ times minus

+ times sdot minus



IDCNST=6 232

a b cD D D D= + + (m2s)







( )( )052 2 2 2








- 377 -



DkT

m n d12

1

2

122

3

8 2

1=

ππ (m2s)

mm m

m m =

+1 2

1 2


( )d d d12 1 21




Ttemperature (K)

No

restrictions 253




235U AU235=235 238U AU238=238 O AO=16


B) Density

UO2 ( )396102









- 378 -

( ) ( ) ( ) 5112 GdOGdGdOU

GdGdGd XAAXAA

XAW



85237050235950238 =times+times=UA

( ) ( ) ( )16250

511625157121685237

2515710

=times++minustimes+

=

Gd

GdGd

Gd

X

XX

X


fWW

UO

UO

Gd

OGd

sdot+

=

2

232

1

ρρ


A

XAW

GdGdO

OGd

sdot= 51


( )A

XAW GdUO

UO

minus=

12


( ) GdGdOGdUO

OGdGdO

OUUO

XAXAA

AAA

AAA

sdot+minus=

sdot+=

sdot+=

512

51

2

1

51

2



2UOW =08847

Thus ( )302310

9610

88470

208

11530950

cmg=+

=ρ




( )2 2 3314 05 07854 Pr cm cmπ = times =


- 379 -


( )( )6

10023 07854 7872

7872 10

g cm

t cmminus

times =

= times


( ) ( )( ) ( )

2

2

6

6

2378508847 07798

26985

300 10 100048871

7872 10 07798

UU UO

UO

AW W

A

MW cm dBu MWd tU

t cm

minus

minus

= sdot = sdot =

times times= =

times times


( )2

31096 095 10412 UO g cmρ = times =


( )( )6

10412 07854 8178

8178 10

g cm

t cmminus

times =

= times

88140185269

852372

2

=sdot== UOUO

UU W

A

AW

Thus

( ) ( )( ) ( )

6

6

300 10 100041620

8178 10 08814

MW cm dBu MWd tU

t cm

minus

minus

times times= =

times times




- 380 -

















200610 (2006)


- 381 -



(417) USNRC WASH-1236 (1972)















- 382 -





















表５SI 接頭語




リットル Ll 1L=11=1dm3=103cm3=10-3m3

トン t 1t=103 kg










カロリー cal













表し方


























































組立量




デジベル dB



名称記号












Abstract+Contents

1MAN

2M1-6

2M7_Diffusion

2M8_Rate law model

2M9--13

2M14-17_Gap_gas

3M1

3M2

3M3

3M4

3M5_2D


3M7_Skyline-Ref3

4M1_UO2

4M2_MOX

4M3_Zr

4M4-SUS

4M5-6+Ref

i






















PCMI Burn-up


ii







（2013 年 3 月 14 日受理）













iii

Contents

1Introduction 1







121 What is model 5









References 1 15














iv


































v











Model (IGASP=0) 125






2102 Swelling 132




power down 134













vi












References 2 174





















vii


















361 Creep 301


363 Ohta model 307




References 3 322


41 UO2 Pellet 323

42 MOX Pellet 343






viii



References 4 380

Acknowledgement 382


ix

目次

1序言 1









123 解析対象 7

13 全体構造 8






参考文献 1 15










x




















263 数値解法 74











xi
















2102 スエリング 132





2121 基本概念 135










xii










2171 自動制御 170



参考文献 2 174



311 基本式 178













xiii















351 要素体系 272

352 基本式 276

353 剛性方程式 278

354 境界条件 284




361 クリープ 301


363 太田モデル 307


(ICPLAS=3 6) 313


参考文献 3 322


xiv










参考文献 4 380

謝辞 382


- 1 -














rod





functions














- 2 -






























design




- 3 -

































- 4 -
















114 Usage of FEMAXI
















- 5 -

























the following steps






- 6 -















modeled
















Refs(115) and (116))



- 7 -

follows

























Pellet





Fuel rod




- 8 -




FEMAXI-7


times





times times




times times


times times

13 Whole Structure
















- 9 -








axial segment











freedom




- 10 -

Next Time Step

No

Output


Input




End of Time Step

Yes

Iteration







- 11 -

Plenum

Segment

LHR

Axial Power Profile

Zircaloy metal


Temperature



Nest ofcoaxialrings

Gas pressure













- 12 -









calculation time


















bubble growth

Nonzero element



- 13 -
























- 14 -































- 15 -








(2) Plotted output














(14)

(15)

(16)








- 16 -

(17)

(18)

(19)

(110)

(111)






















(116)





ORNL-4628 (1973)





- 17 -













rate



















- 18 -


















solution





- 19 -

Time Step Start












Next time step




- 20 -





as follows

τdzS

zQzVtHtzH

t

t primeprimeprime

+=in )(

)()()0()( in (211)



through the inlet


time t




τ time



S zd

t

t t( ) ( )

( ) ( )

( )0 0 00


+

ΔΔ

τ (212)













- 21 -

H t t H tV Q

Stj

i i

ii k

j

i( ) ( )0 0 0+ = +=Δ δ (214)


δ tL

u

L

G V

S L

WVii

i

i

i i

i i

i

= = =Δ Δ Δ

(215)

where





node j+1

node j

node j-1

node k+1

node k

segm j

segm j-1

segm k

z

zk+1

z0

δ t j

δ t j minus1

δ tk

z

zk



- 22 -

ΔL

z zi j

z z k i j

z z i ki

j j

i i

k

=

minus=

minus ne neminus =

+

+

+

1

1

1 0

2( )

( )

( )

(216)


( ) ( ) ( ) ( )( )H t H t H t H tz z

z zk k kk

k k0 0 0 1 0 0

0

1


+

(217)

















- 23 -

Inlet enthalpy

Inlet temperature

Pressure

Mass flow rate


Inlet flow

velocity







sub-segments










- 24 -









coefficient



Here

8040

e

RePr0230D

Kh =

Hea

t flu

x l

og q

D


flux point

Single phase

Nucleate

boiling

Transient

boiling

Film boiling


A

E

B

C


- 25 -

μeRe DG=





























- 26 -








SPTH

CKh 750240

sat240g

240fg

290f

50

490f

450

fP790

fw 00120 ΔΔ=

ρμσρ

(2111)




ρ density (kgm3)







temperature(Nm2)





hP

WW

N= sdottimes

012636 201 106

0 75 exp

φ (2112)





- 27 -


10

f

g

50

g

f

90

e

e

1

minus

=μμ

ρρ

x

xr

ex quality (-)

p density (kgm3)


R F

00 100

01 107

02 121

03 142

04 163

06 202

10 275

20 430

30 560

40 675

60 910

100 1210

200 2200

500 4470

1000 7600

4000 20000






FACTOR S 251

fRe F S

103 1000

104 0893

2x104 0793

3x104 0703

4x104 0629

6x104 0513

105 0375

2x105 0213

3x105 0142

4x105 0115

6x105 0093

106 0083

108 0000



- 28 -


















( ) 21gfgee

g


( ) ( ) 21gffee




ex quality(－)





(2115)

12 12g f crit

12 12g f crit

12 12crit g f crit

(low speed)

12 (highspeed)


J J J

J J J

J J J J



- 29 -




H enthalpy (Jkg)







( ) ( )( ) ( )[ ]( )

( )

( ) inf7

e

e

eee

e78

863DNB

1041292382580

05487124exp8357026640

86901571

03713563117290596114840

1098705517718exp104268117220

10237960222101543

HH

D

x

Gxxx

xPP

PqqDNBW





minus

minusminus

minus

(2117)

Here


P pressure(Nm2)







( )inf

e 03DNB 960761HH

xW DNBqq






- 30 -

( ) ( )[ ]21

gf


+minusminus=

ρρρρ

ρρσα gHq (2119)


ρ density(kgm3)





















( )DNB

satsatDNBFBDNBTB 22

expDNB T

TTTqqh

T ΔΔ

ΔminusΔminus=Δ

(2121)



DNB

FB Tq


DNBFBFBDNB

ThqT

Δ=Δ



- 31 -

( )[ ]

( ) [ ]w7

590282e

78420e

1ln2456060040g

30702w

43760g

FB Pr1045041exp89471131

RePr0810330

e

PxD

Kh

xminus

++

timesminus+

=





P pressure(Nm2)







( )

( )

=330

filmfilm

250film

film

c

PrGr738170

PrGr151443max

e

e

D

KD

K

h (2125)










ρ density(kgm3)



fwfilm

TTT

+=


- 32 -

( ) ( )

( )

ge

leΔ

gtΔminusΔ+

=

9600

1

1960

satDNB

satsatFBTB

liquid

α

α

BTq

BTThh

q (2127)

( ) ( ) satTB

BTTB ehqqBh Δminus

=Δ


sat960712 α (2128)

( ) 250

satge

fggfg3

g

1720

c

eFB 526775

Δminus

=

TD

HKDh

μρρρ

λg

(2129)

( )50

gfc

minus=

ρρσλ

g (2130)

( ) ( )( ) ( )

1 2

1 2

001368 K at 4137000 N m

001476 K at 6205500 N m

B PB

B P

minus

minus

= == = =













- 33 -



( )

( )

2503f

min 2960

minusminus

=wsatfe

fggffarla TTD

KHh

μρρρ g

(2133)

ef

ggpfff

e

gg

f

pff

f

ff

turbulent

D

VCK

D

V

K

CKh

μμρ

μμμ

ρ

310857749

02300650

minustimes=

=

(2134)

q heat flux(Wm2)




V velocity(ms)






table


- 34 -





Inlet enthalpy

Inlet temperature

Pressure


Inlet flow

velocity


Mass flow rate


Pressure


Low-flow

high-flow

NO

NO

YES

YES

NO

Water single-phase



Coolant flow rate







RETURN

RETURN

RETURN


(

RETURN

YES





(RELAP-5)

RETURN


- 35 -
















[ ]E prime Forced

convection to wall


(TwgtTB)

(TwltTB)

(TwltTB)



satTΔ

1



[D] film boiling

0

[F] Condensation

two-phase flow

xe quality

(TwgtTB)



x


Sub-cool region


- 36 -

























enters region [E]





- 37 -




is recovered










boiling transition



Dittus-Boelter

TW

ltTB


TW

gtTB




TW

gtTB




TW

gtTB



RELAP-5 MOD-1model

TW gtTB



TW

ltTB






- 38 -













using

DS

re = 4

2π (2135)








S l r= minus4 2 2π (2136)




S

Symmetrical line

l1



- 39 -





4

2 eDrS

π= (2137)










CN = 0835 + 181times10-4T (221)



(300 1478 )T Kle le













- 40 -












RCORRO times






Temperature




Meta l part

Oxide

Radius

( )Stimes PBR -1S

PBR PBR

Co

ola

nt




- 41 -





chapter again)

























dS

dt

dS += 1


- 42 -















oxide layer








- 43 -














φN = ATN + B (231)






00 02 04 06 08 10

ZrO2

1 2 3 4 5 6 7 8

Cladding

ring element number






- 44 -

















obtained by

ln(2 )N

N OXOX OX

rqT T K rπ

= minus sdot

(233)

where








- 45 -







( ) ( )[ ] ( ) ( ) ( )partpart t


Here

T temperature (K)





n-1 n n+1

kpn

Cvpn

qpn

ksn

Cvsn

qsn

hpn hsn

hpn 2 hsn 2




- 46 -





part (235)




( ) ( ) ( ) ( )

( )1

2 4 2 4

V v V v

m mn n pn pn sn sn

v pn n v sn npn


t t

T T h h h hC r C r

h

+



(236)




( ) ( ) ( ) ( )( ) ( )1 1 1 1

1 1

[ ]

2 2

V s


n npn sn


k T T k T Th hr r

h h

+ + + +minus +



(237)




hh

rh

h rh

hh

rh

hh

rh

pnV pn

npn

snV

nsn

pns

pnn

pnsns

snn

sn

= minus

= +

= minus

= +

2 4 4

1

2

1

2

(238)

D C h C hn v pn pnV

v sn snV= +


variables as

( ) ( ) ( )q r t Q r P t= (239)




- 47 -






( ) ( )( )( ) ( )

1

1

1

nm n n m m s

n n pn pn

m m sn n sn sn

V Vpn pn sn sn n

T T DT T k h

t

T T k h

Q h Q h P t

θ θ

θ θ

θ

+ + +minus

+ ++

+

minus= minus minus

Δ+ minus

+ +

(2311)



( ) ( ) ( )

( ) ( ) ( )

T T D

t

T T T Tk h

T T T Tk h

Q h Q h PP t P t

nm

nm

n nm

nm

nm

nm

pn pns

nm

nm

nm

nm

sn sns

pn pnV

sn snV

fm m

+ +minus

+minus

++

++

+

minus=

+minus

+

+ + minus +

+ ++

1 11

11

11

11

1

2 2

2 2

2

Δ

(2312)



respectively



a T b T c T dn nm

n nm

n nm

nm

minus+ +

+++ + =1

1 11

1 (2313)

where

( )

( ) ( ) ( )

ak h t

ck h t

b D a c

d a T D a c T c T

t Q h Q h PP t P t

npn pn

s

nsn sn

s

n n n n

nm

n nm

n n n nm

n nm

pn pnV

sn snV

fm m

= minus = minus

= minus minus

= minus + + + minus

+ ++

minus +

+

Δ Δ

Δ

2 2

2

1 1

1

(2314)


- 48 -



Pellet CladdingGap

n-1 n n+1 n+2

Tn-1 Tn Tn+1 Tn+2

hpn hsn+1



( ) ( ) ( )( )1

111

11

11

+

+++

+minus

++

+

minus+minusminus=Δ

primeminus

mfVpnpn

gm

nm

nspnpn

mn

mn

nm

nm

n

tPPhQ

hTThkTTt

DTT

(2315)

( ) ( ) ( )( )11

111

11

211

111

11

++

+++

++

+++

+++

++

+

minus+minusminus=Δ

primeprimeminus

mfns

snsns

mn

mng

mn

mn

nm

nm

n

tPPQ

hkTThTTt

DTT (2316)


and


( ) ( )( )

( )

11 1 1 1 1

1 1 1

1 12 1 1 1

1 1 1

m mn n n m m s

n n g p n p n

m m sn n s n s n

Vs n s n m

T T DT T h h h

t

T T k h

Q h P t

++ + + + +

+ + +

+ ++ + + +

+ + +


Δ+ minus

+

(2317)


n v s n s nV






- 49 -





( ) ( ) ( )10000

10

110

01

0+

+++

+minus=Δminus

mfVss

sss

mmsv

mm

tPPhQhkTTCt

TT (2312a)


( ) ( ) ( )1

11

1

1

++

minus+

+


mfVpnpnnn

spnpn

mn

mn

Vpnpnv

mn

mn tPPhQrhkTThC

t

TT φ (2318)


a T b T dn nm

n nm

nminus+ ++ =1

1 1 (2319)


a

k h tn

pn pns

= minusΔ

2 (2320)

b C h an v pn pnV

n= minus (2321)

( )( ) ( )

d a T C h a T

t r tQ h PP t P t

d d

n n nm

V pn pnV

n nm

nnm

nm

pn pnV

fm m

n n nm

= minus + minus

minus +

+

+


minus

++

+

1

11

1

2 2

Δ Δφ φ

φ

(2322)


Pellet center

0 1

T0 T1

hs0


n-1 n=N

Tn-1 Tn=TN

hpn

φ φn N=

Gap


- 50 -

given by

φ πnm m

nq r= 2 (2323)





(2321) are 1 11 m m





T E T Fnm

n nm

nminus+

minus+

minus= minus +11

11

1 (2324)


A T Bnm m

11

11+ ++ = φ (2325)

where




values Therefore 1m








- 51 -

500 1000 1500 20001

2

3

4

5

6

80 GWdtHM

40 GWdtHM

Ther

mal

Con

duct

ivity

(Wm

K)

Temperature (K)

NFI UO2


Halden MOX0 GWdtHM







- 52 -

p(I)＝10 - D(I) (2328)


D(i) is defined as

D(I)=Di-(3ΔLr) (2329)



by irradiation

r pellet radius




=1 00

swgVp p

V

Δ= + is used

=2 0

0 0

dens hotV Vp p

V V

Δ Δ= + + is used

=3 00 0 0

swg dens hotV V V

p pV V V


where 0

swgV

V


0

densV

V


0

hotV

V








That is





swg densV V

p pV V

Δ Δ= + +


- 53 -

























equation) (216)





- 54 -













( ) ( ) ( )( ) ( )RkK

RkK

RkIRkIRQ sdotsdot

sdotsdot+sdot= 0

11

1101 (241)

Here







( ) ( )k E DR

E Dp

= sdot +

sdot sdot0328 054

050 8

0 82

0 19

(242)

where








- 55 -




( ) ( )( ) ( ) ( )[ ]( )( ) ( )( )22

2

2 1121

112

ahhhahb

harahbrQ

bb

bb


where








( )( )pRQ

RQa

1

11= (244)






1po

po pi

T r r

pT r rdT r q dr dr









- 56 -


1 1( )

po

T r






( ) )(1

)(2

rh

Prq LHR ϕ

πsdot

minus= (247)







( )

21 1

00

1212

1 1

2

12 2 2 4

po

rT r

LHR LHRpT r r

LHR LHR LHR

rr

P P rdT r dr dr dr

r r

P P r Prdr r

λπ π

π π π

= sdot sdot = sdot

= = = minus

(248)


rP

T LHRTTp po

minus=π




( ) )1(0304

4004501

422 rr

PTT

p

LHRPO minus

sdot+=minus+=

ππλ (249)




( )( ) ( ) ( )[ ]( )( ) ( )( )22

2

1121

112)(

ahhhahb

harahbr

bb

bb




- 57 -


( )4po

TLHR

pT

PdT F rλ

π= sdot (2411)

Here

( ) [ ][ ])1)(1(2))(1(

ln)2)(()1(22)1)()(2(2

)1)(1(4

22

2222

ahhhahb

rhbhaharhabb

rarF

bb

bbbb



minusminus= ++

(2412)


( ) ( )[ ]

2

12)(

++minus+=

ba

arabr

b

ϕ (2413)

( ) )2(

)1()2(2

)1)(1(4 22

+

minus+++

minusminus

=

+

ba

rabb

ra

rF

b

(2414)


( )24 03 07( )

34

rrϕ

+= ( ) 4 203(1 ) 28(1 )

34r r

F rminus + minus

=
















- 58 -





f(x)＝arb＋C (2415)



2

bb a x af x

ba

+ minus + =+















Relative radius

of pellet


integral equation

Numerical solutions






given



pellet = 100)

00 01 02 03 04 05 06 07 08 09 10

14174114086713823113380112752111931110907196678819846482245000

141873 140997 138358 133922 127633 119412 109158 96747 82033 64848 45000

141758 140883 138246 133814 127530 119317 109075

96679 81984 64822 45000

141787 140912 138276 133844 127560 119346 109101

96701 82001 64831 45000

141758 140883 138246 133814 127530 119317 109075

96679 81984 64822 45000


- 59 -














following equation



















- 60 -

attained




Relative radius of

pellet


integral equation

Numerical solution


direction




given



00 137512 137794 137535 137686 137507 01 136408 136688 136430 136581 136403 02 133109 133381 133129 133278 133106 03 127661 127919 127677 127823 127658 04 120151 120388 120163 120302 120149 05 110726 110937 110734 110863 110725 06 99607 99784 99611 99725 99606 07 87088 87227 87090 87182 87087 08 73533 73628 73534 73599 73533

09 59357 59405 59357 59391 59357 10 45000 45000 45000 45000 45000








( ) ( ) 1 22 1 2 05

gas m cr

eff

Ph h

R HC R R g g GAP


(251)





- 61 -

Here


( ) ( )

( ) ( )λ

λ λλ λm

p po c ci

p po c ci

T T

T T=

sdot+

2 (WcmK)

hT T

T Trp c

po ci

po ci

= + minus

sdot sdotminusminus

minus1 1

1

1 4 4

ε εσ










R R Reff

22

2

2

+








1

1 1ji j

ni

gas nji

iji

x

x

λλφ

=ne

=

= minus Σ

(252)



- 62 -

Here the equation

( )( )

( )φ

λλ

ij

i

j

i

j

i

j

i j i j

i j

M

M

M

M

M M M M

M M=

+

+

+minus minus

+

1

2 1

12 41 0142

1

2

1

4

2

3

2

1

2

2

(253)







3 0668

4 0701

5 0872

5 0923

4 08462

3366 10

3421 10

40288 10 (WmK) (K)

4726 10

2091 10

He

Ar

Xe

Kr

N

k T

k T

k T k T

k T

k T

minus

minus

minus

minus

minus

= times


(254)




( )5

1 2 1 21

10n

iigas

g g g gP=

+ = Σ + sdot (255)

Here


(cm)





- 63 -
























start

t





- 64 -













conditions is given

















- 65 -

















again)













- 66 -














reaction














- 67 -






33 and 34






















history points


- 68 -






















- 69 -








is neglected



Fuel pellet

96

Cladding

978

113 Internal tube

Outer tube

Capsule outer m

antle

Heated steam

layer

Helium

layer

Circulating w

ater

Coolant w

ater

180

200

210

230

270

320



- 70 -



jl

Fixed at 40

Fuel pellet



R2 R3 R4 R5 R6 R7

Cladding

Internal tube

He layerCapsule

outer mantle

Circulating w

ater

Outer tube




( ) )(2

112220

22

1222

2222 jjjj

j

jjjj qRqR

RRHH

V

W

t

TC minus


partρ (261)

holds where

ρ density (kgm3)


T temperature (K)


V volume (m3)


R radius (m)




- 71 -

)(2

)(

112221

22

112

122

1111 jCjC

jj

jjj qRqR

RRz

TC

RR

W

t

TC minus

minus=

minus+

partpart

πpartpart

ρ (262)

is obtained


)(2

223322

23

222 jCjC

jjj qRqR

RRt

TC minus

minus=

partpart

ρ (263)


axial direction

)(2

334423

24

333 jCjC

jjj qRqR

RRt

TC minus

minus=

partpart

ρ (264)


)(

244552

425

444 jCjC

jjj qRqR

RRt

TC minus

minus=

partpart

ρ (265)


direction

)(2

)(55662

526

552

526

5555 jCjC

jj

jjj qRqR

RRz

TC

RR

W

t

TC minus

minus=

minus+

partpart

πpartpart

ρ (266)


)(2

66

7726

27

666 jCjC

jjj qRqR

RRt

TC minus

minus=

partpart

ρ (267)

is obtained

7


( )jNjjjjC BTAq 0











- 72 -


)( 677




2

32

2

2

1

2

2ln

11

R

RRR

h

h+

+prime

=

λ

(2615)




3

43

3

3

32

3

2

33

2ln

2ln

1

R

RRR

RR

RRh

++

+

=

λλ

(2616)


layer (WmK)

4

54

4

4

43

4

3

44

2ln

2ln

1

R

RRR

RR

RRh

++

+

=

λλ

(2617)


554

5

4

55 12

ln

1

hRR

RRh

prime+

+

=

λ

(2618)



6

76

6

6

5

6

2ln

11

R

RRR

h

h+

+prime

=

λ

(2619)


76

7

6

77 2

ln

1

RR

RRh

+

=

λ

(2620)


- 73 -




A) SUS316



B) Helium




( ) ( )( )

4

3

54245 10 1

1890


Pa b c

T RT

T P R


= sdot
















- 74 -


e

RePr0230D

KhW = (2621)

where μeRe DG=








21

42

41

1

21

111

TT

TThr minus

minussdotsdot

minus+=

minus

σεε

(2622)











( )

))()((2

)(

011

11

22221

22

111

111112

122

111

111

jn

jjnj

nj

njj

njj

j

nj

nj

jj

BTARTThRRR

TCTClRR

W

t

TTC


=

sdotminussdotminus

+Δminus

++

++minusminus

+

πρ

(2623)


- 75 -

))(2

2

)()(

011111

221

22

22

112

122

222

122

11111112

122

111

jn

jjnj

jjnj

nj

j

jjjnj

j

j

BTARTt

CT

RR

hR

TRR

hR

lRR

CW

t

CT

lRR

CW

minussdot+Δ

=minus

minus

minus+

minusminus

Δ+

minus

+

++minus

minus

ρ

πρ

π (2624)

For inner tube

)()(2 1

11

2221

21

33322

23

21

222

+++++


=Δminus

nj

nj

nj

nj

nj

nj

jj TThRTThRRRt

TTCρ (2625)

( ) nj

jjnj

nj

jjnj T

t

CT

RR

hRT

RR

hRhR

t

CT

RR

hR2

22132

223

33122

223

223322112

223

22 222

Δ=

minusminus

minus+

+Δ

+minus

minus +++ ρρ(2626)

For Heium gas layer

)()(2 1

21

3331

31

44423

24

31

333

+++++


=Δminus

nj

nj

nj

nj

nj

nj

jj TThRTThRRRt

TTCρ (2627)

( ) nj

jjnj

nj

jjnj T

t

CT

RR

hRT

RR

hRhR

t

CT

RR

hR3

33142

324

44132

324

334433122

324

33 222

Δ=

minusminus

minus+

+Δ

+minus

minus +++ ρρ (2628)

For outer tube

)()(2 1

31

4441

41

55524

25

41

444

+++++


=Δminus

nj

nj

nj

nj

nj

nj

jj TThRTThRRRt

TTCρ (2629)

( ) nj

jjnj

nj

jjnj T

t

CT

RR

hRT

RR

hRhR

t

CT

RR

hR4

44152

425

55142

425

445544132

425

44 222

Δ=

minusminus

minus+

+Δ

+minus

minus +++ ρρ(2630)


( )

))()((2

)(

14

1555

15

16662

526

155

115152

526

551

555

++++

++++

+


=

sdotminussdotminus

+Δminus

nj

nj

nj

nj

njj

njj

j

nj

nj

jj

TThRTThRRR

TCTClRR

W

t

TTC

πρ

(2631)

( )

nj

jjnj

nj

j

j

nj

j

jjjnj

Tt

CT

RR

hRT

lRR

CW

TRR

hRhR

lRR

CW

t

CT

RR

hR

5551

625

26

661152

526

155

152

526

556625

26

5555142

526

55

2

)(

2

)(

2

Δ=

minusminus

minus+

minus+

+minus

minusΔ

+minus

minus

+++

+

++

ρπ

πρ

(2632)


- 76 -


)()(2 1

51

6661

67772

627

61

666

++++


=Δminus

nj

nj

njj

nj

nj

jj TThRTThRRRt

TTCρ (2633)

( ) 72

627

776

66162

627

667766152

627

66 222j

nj

jjnj

jjnj T

RR

hRT

t

CT

RR

hRhR

t

CT

RR

hR

minus+

Δ=

minus+

+Δ

+minus

minus ++ ρρ (2634)



( )

))()((2

)(

110111

111

122221

22

11111

111

12

122

1111

111111

BTARTThRRR

TCTClRR

W

t

TTC

nnn

nninin

nn


=

sdotminussdotminus

+Δminus

++

+++

πρ

(2635)

where 11



( )

))()((2

)(

14

1555

15

16662

526

155

1552

526

551

555

++++

+++


=

sdotminussdotminus

+Δminus

nN

nN

nN

nN

nNN

ninin

N

nN

nN

NN

TThRTThRRR

TCTClRR

W

t

TTC

πρ

(2636)






1nNT + and 1 1

1 6 n nN NT T+ +



four axial segments


- 77 -

(2637)




obtained


- 78 -




Time Step Start














models in FEMAXI


- 79 -




is represented by

PY f q

E Nij

ij j

f A

= sdot sdotsdot

(271)

Here


( )+

minus=

50

502

i

i

ij rdrrf φπ



jLHR

pout

P

r hπ minus

















- 80 -



















matrix





N fmax

which is





the fission gas




is assumed to cease


- 81 -





















- 82 -

INTRAGRANULAR


GAS BUBBLE















- 83 -




consideration(223)

β+prime+minus

partpart+

partpart=

partpart

mbgcr

c

rr

cD

t

c 22

2

(273-1)









1 12 2






0m

mt gc b mt


(273-3)


relation



( ) 2

2

2c m c cD

t r r r


= + +

(273-5)


- 84 -


1 0mt

gc b mmb


Here 1

1

1 ii

i

mtb

m bα minus

minus




( )i

i

c c mg

αα

= ++

(273-8)

holds


( ) ( ) ( )2

2

2i

i

c m c m c mD

t g r r r


+ + += + + +

(273-9)



Ψ = +c m (273-10)


2

2

2D

t r r r


Ψ Ψ Ψprime= + +

(273-11)

Here

i

i

DD

g

αα

prime =+

(273-12)











- 85 -


276 below






( )2152 oN R Zα πλ= + (274-1)

Here








4g DRNπ= (274-2)



( )2

0303b F R Zπλprime = + (274-3)

Here







Namelist input BFCT







- 86 -








( )PR

V m B m kT0

2+

minus =γ (274-4)

( )


(274-5) where





3πR cm3)




T temperature (K)



surface tension

2

gas extP PR

γ= + (274-6)





( )

3 24

3 2ext

ext

P RRm

B P R kTR

γπγ

+= sdot+ +

(274-7)


- 87 -

is obtained







c m= minusΨ

( )prime + =b g m N g Ψ (274-10)

is obtained


( )( )2 0303 4 4F R Z DRN m DRπλ π π+ + = Ψ (274-11)


( )( ) ( )3

2

0

( 2 )303 4 3

2ext

ext

R ZR P R

F DRN DB P R kTR

γπλ πγ

+++ sdot = Ψ

+ + (274-12)


respect to R

( )( ) ( )

32

0 2

608 ( 2 )303 3

2ext

exto

DR R P RF R Z D

B P R kTRR Z

α γπλγλ

+ + + sdot = Ψ + ++ (274-13)







2m

R which satisfies 2




- 88 -










3N Rπ minus sdot




22 ( ) ( )F R Z f R Nπ λsdot + (275-1)




follows

0

00

10 ( )

( )( )

k

R R

f R RR R

R

le= ge

(275-2)




3 20

42 1 2 ( ) ( )

3

dNF N R F R Z f R N


(275-3)






Then


- 89 -

3 20

3 20

42 2 ( ) ( )

3

2 2

4( ) ( )

3

dNF F R R Z f R N

dt

F FKN

K R R Z f R

α πα πλ

α

πα πλ

= minus + + sdot

= minus

= + +

(275-4)






m m N= (275-6)





4g DRNπ= (275-7)






Ψ = +c m (275-8)


20303 ( )b F R Zπλprime = + (275-9)

2

04 303 ( )

( )

dmDRNc F R Z m

dtgc b m g g b m






32 4

3

RP m B m kT

R


(275-12)


- 90 -



2gas extP P

R

γ= + (275-13)

holds





4 2v v P

H gas

dn D RP

dt kT R

π γσ = + minus

(275-14)

gas

v v

m kTP

n=

Ω (275-15)

3

4

3

v vV n m B

Rπ

= Ω +

= (275-16)




4 2

041

085 140 10 (Jm )

P SV SV

SV T

γ γ γγ minus

= sdot =

= minus times

GSV (0ltTlt2850C)





3Ω =34454E-10 m




Ttemperature (K)





- 91 -






1i i vv v

dnn n t


1

11


dn D RP

dt kT R

π γσminus

minusminus

= + minus

(275-19)




dt=





variables




method

1

1

1

422

22

i ii

i iv


R RD P Pdn

PR Rdt kT

π γσ

minus

+

minus

+ +

= + minus +

(275-20)



dt= is explained


4 20

iiv vH gasi

dn D RP

dt kT R

π γσ = + minus =

(275-21)


- 92 -


( ) 2 0iH gasP Rσ γ+ minus = (275-22)


gasv

AP

n= (275-23)

3

vR nα β= + (275-24)


are cubed

33 3( ) 8H gasP Rσ γ+ = (275-25)



(275-26)


-------------------------------------------------------------------------------------------------







Number density 0 0

00762 Bu 104( ) ( )

98N Bu T N T

minus sdot += (cm-3) (276-1)




(Cut-off)



- 93 -

1 3

0

3

4

m MR

Nπρ prime

=

(276-3)


GROU=40 (default)



04 ( ) ( ) ( )g g g








8000

273159

ln0005

mgTBu

= +minus

(276-5)















- 94 -

expressed as

0 3

3 ( )( )

4g

g

m r t MN r t

Rπ ρprime

= (276-7)


2) R Rρge 0

N

iniR

R

ρ

ρ ρ =















(277-1)


( ) ( ) ( ) ( )Ψ Ψ Ψt r t r t rj jj

asymp = φ




ˆ 2Dt

rtR (277-2)





- 95 -

equation

( )R t r dviV Ψ φ = 0 (277-3)




2ˆ

ˆ 0iVD dv

tβ φ


2ˆ

ˆ 0i iV Vdv D dv

tβ φ φ



( ) ( ) ( )

( ) ( ) ˆˆ

ˆˆˆ2



nabla V ii

V iV iV i

dvDsdD

dvdvDdvD

φφ

φφφ

part

(277-5)



( ) ( )ˆ

ˆ 0i iV Vdv D dv

tβ φ φ



jΨ is the





ie iφ ＝jφ


that


t

partφ β φ φ φ


(277-7)



- 96 -



t


partΨ







where

Veji j i

j iji Ve

i Ve i

E dV

A dVr r

H dV

φ φ


φ

=

=

=

(277-10)



( ) ( )

( )

( ) ( )ξξξφ

ξξφ

ξξξφ

+=

minus=

minusminus=

12

1

1

12

1

3

22

1

(277-11)











each node


- 97 -

( ) ( )1

11




( ) ( )3

11




( ) ( ) ( ) ( )21 2 3

1 1( 1 ) 1 ( 1 )


(277-12)

and

( ) 11Ψ minus = Ψ ( ) 20Ψ = Ψ ( ) 31Ψ = Ψ (277-13)

holds






boundary layer




- 98 -

( )

minus

Δ+

Δminus

Δ+Δ=

==

1

1

2

2

4

4

ξξφφπ

φφπφφ

dRRR

drrdVE

kij

RR

RR ijVe ijji

K

K (277-14)

( )

2

1 2

1

4

4

k

Ve k

R Rj ji iji R R

j ik

A dV r drr r r r

R R dR


+Δ

minusΔ

minus

= =

= + ΔΔ

(277-15)

( )1 22

14 4

k

Ve k

R R


+Δ



( )( )1

1 1 0n nj j n n

ji j j ji ij

E D A Ht

θ θ β+


Here

( )( )

D D Dn n

n n


prime = minus +

+

+

1

1

1

1

θ θβ θ β θβ

(277-18)






j+Ψ for









- 99 -

( ) 1 1 11n n


E D A E D A Ht t



1ij ji jiW E D A

kθ= +

Δ (277-20)

( ) 11 n

i ji ji j ij

Q E D A Ht





[ ] W Qij jn

iΨ + =1 (277-22)




















boundary (1)

A B


- 100 -




























tdVβΔ (atoms)



total+Ψ (atoms)


boundary (2)


- 101 -






( )

( ) ( )

32

1

3

1

ˆ4

ˆ

1

e

k ktotal V

e

k kj j

j

r r dr

r r

k n n

π

φ

=

=

Ψ = Ψ Ψ = Ψ

= +

(279-2)







3

11 nntotal

n

a

a++

sdot Ψ

(279-3)


( )31

4


(279-4)

consequently 3


n

an tdV

aβ+ ++


(279-5)

is obtained









- 102 -



ofga

3

143

igrain

g

Naπ

= (2710-1)

holds











fn ie





max 1 2 max 114n n






fN )



3

1 nnf

n

an

a+

sdot

(2710-5)


- 103 -



3


n

an n n n

a+ + ++

= sdot + Δ

(2710-6)







(2710-7)

where 1rls n





(2710-9)



3

1 nnf

n

an

a+

sdot

(2710-10)



3


n

an n n n

a+ + ++

= sdot + Δ minus

(2710-11)






- 104 -




3


n

an n n n

a+ + ++

Δ = sdot + Δ le

(2711-3)

3 3


n n

a an n n n n n n

a a+ + + + ++ +


(2711-4)



1

12

1

nn rr

n

nN

aπ

++

+

ΔΔ = (2711-5)







is given by

1

nkr

k

n=

Δ (2711-7)



1

nkr

kntotal

nFGR

FPG=

Δ=

(2711-8)


- 105 -





3

1 1ng

n

af

a+

= minus

(2712-1)

Here





an+1

an



da

dtK

a

N N

af f

m

= minus+

1 1 max

(μmh) (2712-2)









- 106 -



dt dt= times






Boundary

layer

Node







11

2

2n

n nn

a Rr r

a R+

+minus Δ=

minus Δ (2712-3)

where









- 107 -


given by

Ψn+10(r) = Ψn(r) (0 le r le an) (2712-4)


where









( )

11 max 1

21

1 max 1

min 4

min

n nf fn n

f fn

n n nf f f

n nN N

a

N N N

π

++ +

+

+ +

+ Δ=

= + Δ

(atomscm2) (27-12-5)


Δ+sdot=

Δ+sdot

=

+++

+

+

++

+

1max11

1max2

1

1

3

1

1

min

4

min

nf

nf

nf

n

n

nf

n

nf

nf

n

n

nf

NNNa

a

Na

nna

a

Nπ

(2712-6)

holds where24

nfn

fn

nN

aπ= and

11

214

nfn

fn

nN

aπ

++

+

ΔΔ =






- 108 -



3

11

12

1

3

2 11

21

11

4

4

4

n nnf f

nnf

n

n nnn f f

n

n

n nnf f

n

an n

aN

a

aa N n

a

a

aN N

a

π

π

π

++

+

+

++

+

++

sdot + Δ

=

sdot + Δ

=

= sdot + Δ

(2712-7)

holds where 24

nfn

fn

nN

aπ= and

11

214

nfn

fn

nN

aπ

++

+

ΔΔ =





1nfN +






( )3

2 2 10114 4n n n nn


aa N a N

aπ π ++

+


(2712-8)




n+1 0 give

10 12

14

ng totaln nn

f fn n

f aN N

a aπ+ +

+

Ψ= + (2712-9)

where 2

14

ng total

n

f

aπ +





- 109 -


1

1 12

1

4

n

ng totaln i n in

f fr an n

f aN N D t tdV

a a rβ

π+

+ +

=+





( ) ( )1 1 11

2 2i n n i

f f

bN N ADDF


Δ (2712-11)







1( 2 )nna r+



1nr a

D tr

+=


(2712-12)

is determined



updated


1 1 1

1

n i n if f

n if

N N

Nε

+ + minus

+





( ) 1

1 12

1

1

4


f fn n

f tdvaN N

a a

βπ

++ +

+

+ Ψ minus Ψ + Δ= + (2712-13)



- 110 -



1

4





considered












( ) ( )1 11 ADDF

2 2n nf f

bN N



Here Nf



(atomscm2)






- 111 -




~β β β= + prime (2713-3)



















We have

sinb fr r θ= (2714-1)





- 112 -









1

1 gas

FP B

kT

=

+

(2714-3)


34( )



2

gas extf

P Pr

(2714-5)




on the boundary

Grain boundary


- 113 -

12

2

( sin )

4 ( ) 2

3 sin

ff

fext

f

mN

r

r fP F

kT r

π θ

θ γθ

=

= +

(2714-6)





( )max

2

4 2

3 sinf f

f b extf

r fN f P F

kT r

θ γθ

= +

( )2atomscm (2714-8)













thermal stress








- 114 -









500 1000 1500 200010-22

10-21

10-20

10-19

10-18

10-17

10-16

Diff

usio

n C

onst

ant

(m2 s

)

Temperature (K)




- 115 -

4 5 6 7 8 9 10 11 12 13 14 15 161E-22

1E-21

1E-20

1E-19

1E-18

1E-17

1E-16

1E-15

Diff

usio

n co

nsta

nt (

m2 s

)

10000T (1K)



(FGDIF0-1)

FGDIF FGDIF010000

9 exp 7

a

a a aT




(default)




about 1250K


indispensable



- 116 -

5 6 7 8 9 100

2

4

6

8

10

FGDIF0=5 a=6

FGDIF0=10 a=4

FGDIF0=10 a=8

FGD

IF M

agni

fyin

g fa

ctor

10000T (1K)




factor FGDIF


- 117 -


























inside of the grain



- 118 -




1 densN d= (μm2) d=183x183 (μm2) (282-2)






34( )


(282-3)

where





2

2 2 2 22

sin 4 sin4

fb f bb g bb

g

rf r N a N

aπ θ π θ= sdot = (282-4)

Here by setting

2 2

2

sinfc

rF

s

π θ= (282-5)


( )2 2

2 2 2 2 22 2

sin 4 sin sin4 4


g g

r rf N a N r N


2 2sin

bdens

f

fN

rπ θ=

sdot (282-8)

hold




- 119 -







dt


1

nf

f kk

drr t

dt=

= Δ (282-9)

2

22

2FRAD

4 ( )kT

Vf bd gb

b ext ff f

dr s DP P k

dt r f r

δ γθ


(282-10)


plenum gas pressure



When 0fdr










2 2

4

sing b

f

a fK

r θsdot

=


boundary)


- 120 -

When 0fdr





22

8(1 )

( 1)(3 ) 2 lnfks

φφ φ φ


(282-11)

2 2 2

2

sin sinf f br r f

s s

θ θφ

π = = =

(282-12)

3

z rextP θσ σ σ+ += (282-13)



boundary


2 2

2 sinf

b

rs

f

π θ= (282-15)







2 2 2 2 22


HN r r H

rφ θ θ θ= = =


constant value


- 121 -




That is 2densf

HN




HN

r= = times (Refer





RINITfr ge holds




bb

Gm


and m change








1new


MGB

NGB (282-16)



- 122 -




calculation


nf

f kk

drr t

dt=



0fdr


2b ext

f

P Pr

γ= + holds

In other words


342

( ) 3

f

f

rP f m b m kT

r

πγ θ






2 (100 1)100 100 100 100f f f f f f f f

f f f f f f







1

f if i f i i

drr r t

dt+minus = Δ (c)


Δ=i

jji tt

1








- 123 -


2


V N P Sn

kTη= sdot (282-17)

22669

MTης







24bb g densN a Nπ=








4

0 12 32 210 (m )

16 3c

aV

L











- 124 -
















invi

bb grain

Gm

N N=

sdot (282-25)


element


2 2 2

2 2 2

sin

sin 1h f b c b

ef c

s r P F P

s r F

σ π θ σσπ θ


(282-26)

h ext gapP Pσ = + (282-27)



2 2 2 2

2 2 2

sin sin

sinf b f

c bf

r f rF f

s r

π θ π θπ θ

= = = (282-28)

is obtained


- 125 -




number of atoms














In other words 2

2 2

4

singr b

f

a fK

r θsdot

= (291-2)



2

2

6816 b gr

f

f aK

r

sdot= (291-3)




- 126 -










Eqs(279-5) and (2710-3) as


3

1

1


n

an tdV

aβminus

minus


(291-6)





That is 2

gas extf

P Pr

γ= + (291-7)


34( )

3f

gas f

rP f n B n kT

πθ


(291-8)



2

26816

nf f

b gr

r nn

f a

sdot=

sdot (291-9)


- 127 -


3 2 2

2 2

42( )

3 6816 6816

n nf f f f f

ext ff b gr b gr

r r n r nP f B kT

r f a f a

πγ θ sdot sdot



nf

b gr

nQ

f a=

sdot

42( )

3f

ext ff

rP f QB QkT

r

πγ θ


From Eq(291-11)

( ) ( ) 0223

4

3

4 2 =minus





4

3




a

acbbrf 2






fr =05 μm






- 128 -

292 Bubble swelling




2

6816 b gr

f

f aK

r

sdot= (291-3)




34( )

3f

bb f

rV f

πθ= sdot (292-1)


34

3gr

gr

aV

π= (292-2)


0 2

Bubble

bb

gr

V V K

V V

Δ sdot=

(292-3)




In this context 0

001436Bubble

V

V

Δ =







- 129 -

3intra

0

4

3

BubbleV

R NV

π Δ = sdot

(292-4)




0

4

3

Bubble

bb

grTotal

V V KR N

V Vπ

Δ sdot= sdot +

(292-5)

and 0

Bubble

Total

Vp

V

ΔΔ =

(292-6)


hold


















- 130 -











2b ext gapb

P P Pr

γle + + (2101-1)

03c bF f= ge = FCON (2101-2)

FCON=03 (default)















assumed


- 131 -




bubbles 1 is

2 21 sinS M rπ θ= sdot (2101-3)



1 1r M r= sdot (2101-4)







1

2γ

r

to2

2γ




3 3

1 2

32 1

4 4

3 3r M r

r M r

π πtimes =

= sdot (2101-5)






expressed by K


- 132 -

2

2 2

4

singr b

f

a fK

r θsdot

= (2101-7)


2

2 3

1

bb

r ff

r M

sdot =

(2101-8)





have a radius

2 1 1 sinr r M r θ= = sdot (2101-9)




indicates

32

1

1 rM

M r

=

(2101-11)





is also identical







- 133 -




fN remains




dependent on extP





















r zST




- 134 -





stress and 2

0extPr


08r

γminus times


















internal pressure







- 135 -





Eq(2714-8)







recovery




2121 Basic concept



















- 136 -

































- 137 -






( )W Bu= minus2 19 48 8 (2122)

















( ) ( )1 1 1

1 12

1

1

4


f fn n

f tdv R M R MaN N

a a

βπ

+ + ++ +

+

+ Ψ minus Ψ + Δ + minus= + (2125)






- 138 -





R F Pr r r= (2126)




















as described below






- 139 -

50 60 70 80 90 100 110 120000

001

002

003

004

005

Frac

tion

of ri

m a

rea

Local burnup (GWdt)

Cunningham model




times (2128)

holds

















using Eq(2122)


- 140 -



























- 141 -




1tanXv

effB Buα δ

π

























- 142 -



temperature history





11 0exp ( )n




11 0exp ( )n n n










rod

00 01 02 03 04 050

10

20

30

40

50

60

70

80

90

Burn

up (G

WdtU

)

Pellet Radius (cm)




- 143 -



value

















[ ]1

FPOR 10 FSOLID


= minus= minus

(2123-6)


Here







- 144 -



3

3

23


131 (gmol)

6022 10 (atomsmol)

ρtimestimes

times sdot (2123-8)

is used






FPOR As a result









GS0a

0 50 100 150 20000

05

10


Xe

conc

(w

t)

Local burnup (GWdt)




- 145 -





















defines as

OPR=ATHMR (2123-11)







(2123-12)



- 146 -

20 40 60 80 100 120 1400

5

10

15

20

Min

Por

osity

(

)

Local burnup (GWdt)

Max











in Fig2124

2

2

PROMAX 0005 ( 40) 25

PROMIN 00013 ( 40) 25




Bu

Bu

p

(2123-15)



RMPRS=2 mode


- 147 -



















1

095rim

rim Bx

pλ λ minus = sdot

(2123-19)








( )Bx T Bxλ λ= (2123-20)



- 148 -

































- 149 -



of pellet is

100( ) ( )TDF p T Buλ λ= sdot (2131)





1) IPRO=0 0p p= (2132)

2) IPRO=1 00

swgVp p

V

Δ= + (2133)

3) IPRO=2 00 0

dens hotV Vp p

V V

Δ Δ= + + (2134)

4) IPRO=3 00 0 0

swg dens hotV V V

p pV V V

Δ Δ Δ= + + + (2135)


0

swgV

V


0

densV

V

Δ densification

0

hotV

V

Δ hot press


swgV

V


IFSWEL






- 150 -

Billaux equation

251

095rim

rim Bx

pλ λ minus = sdot

(2123-19)


conductivity




swsV

V






10Df =

and when FDENSF=1 0

10 swsD

Vf

V

Δ= minus (2138)


FEMAXI model










defined as rimS



- 151 -




00025ss

solid

VS SWSLD

V

Δ = = sdot


SWSLD=10 (Default)


2) 00025ss

solid

VS SWSLD

V

Δ = = sdot


SWSLD=10 (Default)











growth model


model



GBFIS=0


- 152 -


migration GBFIS=1









- 153 -











A1 0

Equili-

brium

model

0 2

3 4 5



0 default


1


A2 1



0 default


1


B1 2

Rate-

Law

model

0 2

3 4 5



0 default


1


B2 1



0 default


1



- 154 -



Time Step Start














- 155 -









Xenon




rod










T temperature (K)

V volume (m3)


Subscripts are




- 156 -





TpTj+1Tj

2lj 2lj+1



N n V

C n VP

RC

V

T

i j i j j

i j j j i jj

j

=

= = (2141)





( )G j S Dn n

lj jj

j1

12 1 1rarr =minus

(2142)


( )G j S Dn n

lj jj

j1 1 1

12 1 1 1

1

1 rarr + =minus

+ ++

+ (2143)


( ) ( )G j G j1 1 1rarr = rarr +

holds and

S Dn n

lS D

n n

lj jj

jj j

j

j

12 1 11 1

12 1 1 1

1

minus=

minus+ +

+

+ (2144)


nS D l n S D l n


j j j j j j1

121 1 1 1

121 1

1 112 12

1

=+

++ + + +

+ + + (2145)



- 157 -

( ) ( )

( )G j j G j

S D S D


j j j j j jj j

1 1

121 1

12

1 112 12

11 1 1

1rarr + = rarr

=+

minus+ +

+ + ++

(2146)

Here by setting

FS D S D


j j j j j j +

+ +

+ + +

=minus1

121 1

12

1 112 12

1

(2147)


(2148)


(2149)

is obtained



(21410)


n n n n n nj j j j j j= + = ++ + +1 2 1 1 1 2 1

holds











nP

R Tn

P

R Tjj

jj

= =++

1 11

1

(21413)




- 158 -



Here

nP

R

C

Tn

P

R

C

Tjj

jj

11

1 11

1 = =+

+ (21416)

CN N

V V

n V n V

V Vj j

j j

j j j j

j j1

1 1 1

1

1 1 1 1

1

=++

=++

+

+

+ +

+

(21417)

nP

R

C

Tn

P

R

C

Tjj

jj

22

2 12

1 = =+

+ (21418)

CN N

V V

n V n V

V Vj j

j j

j j j j

j j2

2 2 1

1

2 2 1 1

1

=++

=++

+

+

+ +

+

(21419)

C C1 2 1+ = (21420)








Vn

tJ Jj

jR S

partpart

11 1

= minus (21423)




- 159 -

( ) ( )[ ]

( ) ( )[ ]n n

t VF n n n n

F n n n n

jk

jk

jj j j j j j

j j j j j j

11

1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1

11 2 2 1 1

1 2 2 1 1

+

minus minus minus

minus + +



θ θ θ θ θ

θ θ θ θ θ (21424)

Here




θ 1 and θ 2







( )n n nj jk

jk

12

2 1 2 111

θ θ θ= minus + + (21425)


( )[ ( )

( )]

n n

t

F

Vn

F F

Vn

F

Vn

VF n n

n n F

jk

jk

j j

jj

k j j j j

jj

k

j j

jj

k

jj j j

kj

k

j j j j

11

1 2 11 1

1 2 1 1

11

2 11 1

11 2 1 1 1

1 1 1 1

1

1

1 1

1

1

1 1

1

1 1 1

11

1

+minus

minus+ + minus +

minus+

+minus minus

minus +

minus= minus

+

+ + minus minus


Δθ θ

θθ

θ

θ

θ

θ θ

θ

θ

θ θθ

θ θ θ ( )[ ( ) ( )]2 1 1 1 1 1 11 1n n n nj

kj


θ θ

(21426)


A n B n C n Dj jk

j jk

j jk

j1 11

11

1 11

minus+ +

+++ + sdot = (21427)

Here

AF

Vtj

j i

j

= minus minusθ θ

θ2 1

1

1

Δ (21428)

B

F F

Vtj

j i j j

j

= + minus+minus +

12 1 1

1 1

1

θ θ θ

θ Δ (21429)


- 160 -

CF

Vtj

j j

j

= minus +θ θ

θ2 1

1

1

Δ (21430)

( ) ( )( )

( ) ( )

DV

F t nV

F F t n

VF t n

VF n n F n n

jj

j j jk

jj j j j j

k

jj j j

k

jj j j j j j j j



minus minus minus +

+ + + + minus minus

11 1

11

11

1

1

1

1

1 1

1

1

1

1 1 1 1 1

1 2 1 1 1 1 2 1

1 2 1 1 1 1 1 1 1 1 1 1

θθ

θθ θ

θθ

θθ θ θ θ θ

θ θ

θ

Δ Δ

Δ ( )θ1 Δt

(21431)



( ) ( )[ ]V

n

tJ

F n n n n

S11 1

1

1 2 1 1 1 2 1 1 1 2

partpart

= minus


(21432)


( )

( ) ( )

n n

t

F

Vn

F

Vn

F

Vn

F

Vn

F

Vn n

k kk k k

k

1 11

1 1 2 1 2

11 1

1 2 1 2

11 2

1 2 1 2

11 1

2 1 2

11 2

1 2

11 1 1 2

1

1

1

1

1

1

1

1

1

1

1 1

1

1

++ +minus


+minus

+ minus

Δθ θ θ

θ

θ

θ

θ

θ

θ

θθ

θ

θ

θθ θ

(21433)


( ) ( ) ( )

BF

Vt C

F

Vt

DF

Vt n

F

Vtn

F

Vn n tk k

12 1 2

11

2 1 2

1

12 1 2

11 1

2 1 2

11 2

1 2

11 1 1 2

1

11 1

1

1

1

1

1

1

1

1

1

1

1 1

= + = minus

= minusminus

+

minus+ minus

θ θ

θ θ

θ

θ

θ

θ

θ

θ

θ

θ

θ

θθ θ

Δ Δ

Δ Δ Δ (21434)

and we obtain

B n C n Dk k1 1 1

11 2

11sdot + sdot =+ +

(21435)



( ) ( )[ ]V

n

tJ

F n n n n

nn

R

n n n n n n

partpart

11

1 1 1 1 1 1 1

= minus


(21436)


A

F

Vt B

F

Vtn

n n

nn

n n

n


θ

θ

θ2 1 2 1

1

1

1

11 Δ Δ


- 161 -

( ) ( )

( ) tnnV

F

ntV

Fnt

V

FD

nnn

nn

knn

n

nnkn

n

nnn

Δminusminus

Δ

minusminus+Δ

minus=

minusminus

minusminus

minus

11

1

1

1

1

1

1

1111

112

1112 1

11

θθθ

θ

θ

θ

θ

θ θθ

(21437)

we obtain

A n B n Dn nk

n nk


1 11

11

(21438)


B CA B C

A B C

A B CA B C

A B

n

n

n

n

n

n

n n n

n n n

n n

k

k

k

nk

nk

nk

1 1

2 2 2

3 3 3

2 2 2

1 1 1

1 11

1 21

1 31

1 21

1 11

11



minus minus minus

minus minus minus

+

+

+

minus+

minus+

+

=


minus

minus

DDD

DDD

n

n

n

1

2

3

2

1

(21439)




DkT

m n d12

1

2

122

3

8 2

1=

ππ

(21440)

where

mm m

m m =

+1 2

1 2







N Nj iji

= (21441)


- 162 -



jj

N N= (21442)

Upperpl enum

Segment

5

Segment


4

Counter-diffusion

Segment3

Segment2

Segment1

Lower

pl enum

Previous step

Current step

He Xe He Xe He Xe



Since

P N RV

T jj

= sdot sdot

1 (21443)



prime =

NP

R

V

Tjj

(21444)


- 163 -








prime = minus


il l jl

(21445)



















- 164 -


V r lhj

pi Zj= π 2 (2151)





equations

V r lp po= π 2 2Δ (2152)

V r u lcrack porel

p= 2π (2153)

hold where





( )V V V r l r u ll

lj

dish cham po porel

pZj

pint = + + + sdotπ π2 2 2Δ (2154)












Parameter AY

pellet



- 165 -




length ~


( )~l lZ pj

thj

denj

ssj

Zj= + + +1 ε ε ε (2155)





~lZ Cj is

( )~ l lZ Cj

c thj

c irrj

Zj= + +1 ε ε (2156)





( )2

1

NAXIj j


V r l l l lπ=

Δ = minus + minus

(2158)



~V V Vpl pl pl= + Δ (2159)







2 3chamfer

Dp aV abπ = minus

(21511)


- 166 -


2 2( 3 )6dish

dV d r

π= + (21512)



a

b

Dp

Dp

r

d


- 167 -


2 21tan ( )


















θ

Din

Center hole

Dp


- 168 -




int

1

Tgas j j jM



n RP

V V V V V

T T T T T=

sdot=

+ + + +

(2161)

where









Vgapj





jci

j+







surface temperature





- 169 -






























- 170 -






















strain






- 171 -



EFCOEF1 sdotsdot

leΔcE

tε

σ (2171)












Eq(21423) as

( )

Δtn n V

J Jji j j j

R S

=minus

minus1 1

1 1

(2172)













- 172 -


using ( )J R j1

( )ΔtN

Jj

R j

2

1

= max

(2173)






ΔtN

jj

3 = max

β (2174)


( )Δ Δ Δ Δt t t tn21 11

21 1


( )Δ Δ Δ Δt t t tn22 12

22 2= min (2176)

( )Δ Δ Δ Δt t t tn23 13

23 3= min (2177)

Δt2 is given by

( )Δ Δ Δ Δt t t t2 21 22 23= min (2178)


code









- 173 -


as

22

3 005 FMULTR

tD


(2179)





( )2

64

min

10i

rt

K

Δ Δ = sdot

(21710)
















- 174 -

References 2















Company Inc(1972)



EPRI-NP 2789 (1982)





- 175 -




CRFD-1075 (1962)

(218) Wilson MP Jr GA-1355 (1960)




















- 176 -


























- 177 -




fuel JNM 321 pp 249-255(2003)
















- 178 -















311 Basic equations

















- 179 -

10r rr rz R



part part part

10zrz z rz Z



210r z r



0rr


part

rr

u

rε part=

part1r uu

r rθ

θεθ

part= +part

zz

u

zε part=

part

[ ]1( ) pl cr sw


[ ]1( ) pl cr sw


[ ]1( ) pl cr sw


0r zθε ε ε+ + =








for shear strain




- 180 -







follows


assumed



assumed









( ) tt





Here ( )nt

f σαpartpart=




f

partpart




- 181 -


f

t

f σσpartpart=

partpart













Then

[ ] B uε = (313)










T

FW Fν= (314)


[ ] TT

INW Bν σ= (315)





- 182 -

[ ]

[ ]

TT T

V

TT

V

F B dV

B dV

ν ν σ

ν σ

=

=

(316)



[ ] T

VF B dVσ= (317)













ie at time tn+1

[ ] ++ = dVBF nT

n 11 σ (318)

holds where






T

n

T

n+ += + 1 1σ σΔ (319)


n n

T

nΔσ σ+ += minus 1 1 (3110)


quantity



- 183 -



[ ] 1e

n nDσ ε+ +Δ = Δ (3111)










Δε ne


[ ] ( )[ ] [ ]D D Dn n n+ += minus + =

θ θ θ θ1

1

21 and



[ ] Δ Δε σθne

n nC+ + +=1 1 (3112)


[ ] [ ] [ ] B D dV F B dVT

ne

n

T

nΔε σ+ += minus 1 1 (3113)



10




nnnT σεε 1

011 (3115)




nT

nnT σεε 0

111 (3116)



- 184 -

[ ] [ ][ ] [ ] [ ] [ ]

1

01 1

T

n

T T

n n n

B D B dV u

F B D dV B dVε σ

+

+ +

Δ

= + Δ minus

(3117)




[ ] [ ] [ ] [ ] [ ]

T


B D B dV u F


Δ = Δ

+ Δ + Δ + Δ + Δ + Δ + Δ + Δ + Δ

(3118)


[ ] [ ][ ]

[ ] [ ] [ ]

11

01 1

1 1

MT

i i i i ni

M MT T i

n i i n i i n ii i

B D B V u

F B D V B Vε σ

+=

+ += =

Δ Δ

= + Δ Δ minus Δ

(3119)




be obtained




f

t

f σσpartpart=

partpart














- 185 -









( )σε F=Δ (3122)

















[ ] [ ] [ ]εσ DDD == (3123)




- 186 -






















01σΔ

εΔ

σ

εε

ε

σ

σ

)(a

)(b

1



- 187 -

















section














- 188 -

cn

Pn

HPn

swn

denn

crkn

reln

thn

enn

111

1111111

+++

+++++++

Δ+Δ+Δ+



(3124)






Δε nden


Δε nsw






crk+1

Δε nHP+1 Δε n

P+1 and Δε n




reln


n 1+Δε




[ ] e Cε σΔ = Δ (3125)

[ ] Δ Δε σθne

n nC+ + +=1 1 (3126)

where [ ] ( )[ ] [ ]

=+minus= ++ 2









- 189 -

1 1 1 1 1 1 1

1 1 1


HP P cn n n


ε ε ε+ + + + + + +

+ + +





swn

denn

reln

thnn 1111

01 +++++ Δ+Δ+Δ+Δ=Δ εεεεε (3128)





cHn

PHn

crknnn

en 111


where








[ ] 11 ++ Δ=Δ nn uBε (3130)



[ ] [ ] 01110


PHn











- 190 -







nonlinear equation



ψ( x)=0 (3133)



following manner

1+ix = ix + Δ 1+ix

Here

i

ii

dx

dxx

minus=Δ + ψψ )(

1




0)()( 122 =Δminus= +PHnf εσσψ (3135)

)(xψ

idx

d

ψ

ix1+ix

)( ixψ

１



- 191 -

0)()( 133 =Δminus= +cHnf εσσψ (3136)

also

)( 11 σσψ fdd = (3137)

)( 22 σσψ fdd = (3138)

)( 33 σσψ fdd = (3139)














[ ] ( )

1 11 1

0 1 1 11 1 1 1 1 0

i i in n n


C d B u

C

θ

θ

σ

ε ε ε ε σ σ

+ ++ + +

+ + ++ + + + + +

minus Δ

+ Δ + Δ + Δ + Δ + minus =

(3140)

Here d ni

ni

niσ σ σ+

+++

+= minus11

11

1


[ ] [ ] V

T

ni

V

T

ni

nB dV B d dV F + ++

++ =σ σ1 11

1 (3141)



11

Δεncrk i+

+1

1 1

1+

+Δ iPHnε 1

1+

+Δ icHnε and Δun

i++

11

Here the



0)()( 1111 =Δminus= +

+icrk

nf εσσψ (3142)

0)()( 1122 =Δminus= +

+iPH

nf εσσψ (3143)

0)()( 1133 =Δminus= +

+icH

nf εσσψ (3144)


- 192 -


unknown values 11

++

indσ Δε n

crk i+

+1

1 1

1+

+Δ iPHnε 1

1+

+Δ icHnε and Δun

i++

11



















- 193 -


iPHn

1ˆ +Δε and icH

n

1ˆ +Δε








[ ] in

in

in FuK 1

11 +

+++ Δ=Δθ



++

indε and 1

1++

indσ are


++Δ i


++

inε and

11

++







- 194 -




Time Step Start














- 195 -







(1) Geometry







U P PER PL ENUM

S E GMENT( M)

S E GMENT( M- 1 )

S E GMENT( 2 )

S E GMENT( 1 )

L O WER PL ENUM

Pellet

Cladding

Clad Pellet


- 196 -











Uz2

Lz9Lz3

Lz2Lz1


r12

Uz3Uz1

Uz10

r1 r2 r3 r10 r11

Lz10

r13 r14 r15 r16

Lz11Lz12

Lz13Lz14

Uz11Uz12

Uz13 Uz14Uz9

Mz1Mz2

Mz3

Mz9Mz10


00 02 04 06 08 10

ZrO2

1 2 3 4 5 6 7 8

Cladding

ring element number






- 197 -






1 10L Lz z - 1 10


U Uz z - as shown

below

( )( )

( )( )U

CUUUU

LC

LLLL

Up

UUU

Lp

LLL

zzzzz

zzzzz

zzzz

zzzz

====

====

====

====

14131211

14131211

1021

1021

(321)













for its importance

ri ri+1

ZiU

ZiL




- 198 -











oxide elements





















- 199 -



Bedworth ratio













Uiz and




111

1+

++

+

minusminus

+minusminus

= iii

ii

ii

i urr

rru

rr

rru (323)

( )L

M Lii iM L

i i

z z


minus

or ( )M

U Mii iU M

i i

z z


minus (324)




- 200 -

[ ] r

z

ur

u B ur

z

θ

partpartε

ε εε partν

part

= = equiv

(325)


( ) ( ) 111

1

111

11

+++

+

+++

minusminus

+minusminus

=

minus+

minusminus=

iii

ii

ii

i

iii

iii

urrr

rru

rrr

rr

r

u

urr

urrr

u

partpart

(326)

Here by putting 1

2i ir r


( ) ( )

1 11

11 1

1 1

11 1

2 2

2 21 1

i i i ii i

i ii i i i

i i i i

i ii i i i

r r r rr ru

u ur r r rr r r r r

u ur r r r

+ ++

++ +

+ +

++ +


= ++ +

(327)

M Li iM Li iz z z

partν ν νpart

minus=minus


partν ν νpart

minus=minus

(328)


1 1

1

1 1

1 10 0

1 10 0

1 10 0

ii i i i

r i

i

ii i i i L

i

z iM

M L M L ii i i i

ur r r r

u

r r r r

z z z z

θ

ε

εν

εν

+ +

+

+ +


minus minus minus

Or

1 1

1

1 1

1 10 0

1 10 0

1 10 0

ii i i i

r i

i

ii i i i M

i

z iU

U M U M ii i i i

ur r r r

u

r r r r

z z z z

θ

ε

εν

εν

+ +

+

+ +


minus minus minus

(329)



- 201 -

[ ]

minusminusminus

++

minusminusminus

=++

++

Li

Mi

Li

Mi

iiii

iiii

zzzz

rrrr

rrrr

B

1100

0011

0011

11

11

or [ ]

minusminusminus

++

minusminusminus

=++

++

Mi

Ui

Mi

Ui

iiii

iiii

zzzz

rrrr

rrrr

B

1100

0011

0011

11

11

(3210)




( )( )Li


221π

or ( )( )Mi


221π (3211)



[ ] Δ Δσ ε= D e (3212)



[ ] Δ Δε σe C= (3213)


[ ]


=1

1

11

νννννν

EC (3214)


- 202 -






[ ] F B dVn

T

n+ += 1 1σ (3215)


Here






T

n

T

n+ += + 1 1σ σΔ (3216)


n n

T

nΔσ σ+ += minus 1 1 (3217)



[ ] [ ] [ ] B D dV F B dVT

ne

n

T

nΔε σ+ += minus 1 1 (3218)




en εεε (3219)



nnnT σεε 1

011 (3220)




nT

nnT σεε 0

111 (3221)


- 203 -


[ ] [ ][ ]

[ ] [ ] [ ] 1

01 1

T

n

T T

n n n

B D B dV u

F B D dV B dVε σ

+

+ +

Δ

= + Δ minus

(3222)



[ ] [ ] [ ]

[ ] [ ] [ ]

11

01 1

1 1

MT

i i i i ni

M MT T i

n i i n i i n ii i

B D B V u

F B D V B Vε σ

+=

+ += =

Δ Δ

= + Δ Δ minus Δ

(3223)

















321 (2))






- 204 -




































MiZ

ir

UiZ

LiZ

1ir +

centerline

Outer surface


M3Ζ

M7Ζ

Chamfer space

Dish space

U1Ζ U

2Ζ U3Ζ

U4Ζ

U5Ζ U

6Ζ U7Ζ U

8Ζ U9Ζ U

10Ζ

M10Ζ

M1Ζ

M4Ζ

M5Ζ

M2Ζ

L1Ζ L

2Ζ L3Ζ L

4Ζ L5Ζ L

6Ζ L7Ζ L

8Ζ L9Ζ L

10Ζ

M6Ζ

M8Ζ

M9Ζ



- 205 -








Miz and










U U U Ui pz z z z= = = = =

(Refer to Eq(321))









element








- 206 -














space elements








z

zz l

SEk = (3224)









- 207 -

















zs zz

lk k

l

Δ= sdot (3226)




( )zzs z zd z

z

lk k k k FAC

l






[ ] enn D 11 ++ = εσ (3228)




[ ] 111 +++ = nnen C σε (3229)


- 208 -


324 Pellet cracking












StrongInteraction

W eakInteraction

NoInteraction

Ec

E

ε eminusε rel







- 209 -

value



























- 210 -

( )

minus=lt


=lt

=

reli

ei

ei

reliccrel

i

ei

eic

i

E

EEE

E

E

εε

εεεε

ε

　

　　

　　

0

0

(3230)






introduced as

ri

eii εεε += (3231)



iei εε minus= 0=r

iε That is the


iε ) and when reli



range 0ltltminus ei




Elastic region

Elastic strain

Relocation strain

-εrel

εrel

σ

εi

minusσR


- 211 -

In the range relii




function




(3232)


bad

dE i

i

ii +== ε

εσ

2 (3233)



02 =++ cba rel

ireli εε (3234)

From Eq(3233)




reli

ccreli

c EEc

EEb

EEa ε

ε 4

3

2

4

+minus=

+=

minusminus= (3237)


reli

ci

cirel

i

ci

EEEEEE εεεε

σ4

3

242 +

minus+

+minus

minus= (3238)


ci

EE εσ4

3+minus=



recovers completely









- 212 -




[ ]~C

E E E

E E E

E E E

r

z

=

minus minus

minus minus

minus minus

1

1

1

ν ν

ν ν

ν νθ

(3239)







( )

0

00

0

0

0

c i i


i

reli i i

E

E E E E

E

ε ε

ε ε ε ε εε

ε ε ε


(3240)






en nC+ + +=1 1 (3241)



ncrk i

n niC+ +

++ +

++ =1 11

11 ~

(3242)



- 213 -

[ ] [ ] Δ Δε σ

ν ν

ν ν

ν ν

ν ν

ν ν

ν ν

θ θ

θ

ncrk i

n n ni

r

z

C C

E E E

E E E

E E E

E E E

E E E

E E E

++

+ + ++= minus

=

minus minus

minus minus

minus minus

minus

minus minus

minus minus

minus minus

11

11

1

1

1

1

1

1

~

=

minus

minus

minus

+

minus

++

++

++

+

+

+

Δ

Δ

Δ

Δ

Δ

Δ

σσσ

σ

σ

σ

θ

θθ

r ni

z ni

ni

rr ni

zz ni

ni

rE E

E E

E E

E E

11

11

11

1

1

1

1 1

1 1

1 1

1 1

minus

minus

++

++

++

d

E Ed

E Ed

r ni

zz ni

ni

σ

σ

σθ

θ

11

11

11

1 1

1 1

(3243)


+1




+ +Δ Δ and 1

1i

nθσ ++Δ and one






Eq(3140) gives

[ ] ( ) ( )

1 1 01 1 1

11 1

1 11 1 1 0

i i in n n n

i i i in n n n n


C d B u

C C d

C

θ

θ θ

θ

σ ε

σ σ σ

ε ε σ σ

+ ++ + + +

++ + + +

+ ++ + + +

minus Δ + Δ

+ minus + minus

+ Δ + Δ + minus =


[ ] [ ] [ ] ( )~ ~ C d B u Cni

ni

ni

n nP i

nc i

ni

ni

n+ ++

++

+ ++

++


11

10

11

11

1 0Δ Δ Δ Δ (3244)







- 214 -



















( ) ( ) ( ) ( )

( )

2 22 2

12 2

36

2

3

r z z r zr

r z

h H F GF G H θ θ

θ


α σ σ σ


+ + +

(3245)





have


- 215 -

( ) ( ) ( ) ( )2222 32



( ) ( ) ( ) ( )22222 362


(3247)


( ) ( ) ( )05

2 2 22 1



( ) ( ) ( )05

2 2 2 22 1 3



2 2 22( ) ( ) ( )

3pl pl pl pl

r zd d d dθε ε ε ε = + + (3250)


( )2 2 2 2 22 1( ) ( ) ( ) ( ) ( )





1

npl pl

ii

dε ε=

= (3252)






- 216 -







(IFEMRD=1)

( ) ( ) ( ) ( )

( )

2 22

12 2

3

2

3

r z z r

r z

h H F GF G H θ θ

θ

σ σ σ σ σ σ

α σ σ σ σ


+ + + equiv

(3253)

holds where






( )2

23 27

3r z

r zθ

θσ σ σα σ σ σ α + + + + =






( )( )( )

( )

( )

( )

22

322

2 23

2 22

3

r zr z

r ST r z

z rz ST r z r z

ST r zr z

r z

θθ

θ

θθ θ

θ θ θθ




(3254)

Here the term 3r z



- 217 -


2

32

03

2

3

r z

r ST

z rz ST

STr z

θ

θ

θθ

σ σ σ


σ σ σ σ σ

minus minus


(3255)









2

273


is added inside the



1 1P Pn n

n θ

σε εσ+ +

+



( )3 22

2 3r z

r zr

θθ



holds Accordingly

1 3 05 3 05 3

11 3 05 3

1 3n

n n Symθ

θ θ


α+

+ +


+

(3258)


( ) 1

11 3

1

1

P P

PP P Pnn n

n P P

v v

v v

v vθ θ

θ

α εε σσ

++ +

+


minus minus

(3259)

where v P = minus+

0 5 3

1 3

αα

(3260)


- 218 -




( ) ( )( )

( )

Δ Δ Δ Δ

Δ

Δ

ε ε ε ε

α εσ

σ σ σ

α εσ

σ σ σ

θ

θ

θ

h nP

nP r

nP z

nP

nP

Pr z

nP

r z

v

+ + + +

+

+

= + +

=+

minus + +

= + +

1 1 1 1

1

1

1 31 2

9

(3261)


+1 due to


( )Δ Δε α εσ

σ σ σθh nc n

c

r z ++= + +119

(3262)



( )( )

Δ Δ Δ

Δ Δ

ε ε εα

σε ε σ σ σ θ

h n h nP

h nc

nP

nc

r z

+ + +

+ +

= +

= + + +

1 1 1

1 1

9 (3263)



A) BETAX









11

1

BETAX ( )

0 ( )

nn

i

n

D DD D

D D

D D

α

α

++

+

minus= ltminus

= ge (3264)


- 219 -









value=0













strain increment


[ ] [ ] [ ]

[ ] [ ]

T

TH rel swl den

T

crk c p HOT

B D B dV u F

B D dVε ε ε ε

ε ε ε ε

Δ = Δ

Δ + Δ + Δ + Δ + + Δ + Δ + Δ + Δ

(3265)



- 220 -

[ ] [ ]

[ ] [ ]

T

TH rel swl den

T

crk c p HOT

B D dV F

B D dV

ε

ε ε ε ε

ε ε ε ε

Δ = Δ

Δ + Δ + Δ + Δ + + Δ + Δ + Δ + Δ

(3266)











e TH rel swl den

crk c p HOT

ε ε ε ε ε ε

ε ε ε ε

Δ = Δ + Δ + Δ + Δ + Δ

+ Δ + Δ + Δ + Δ (3267)


[ ] [ ] T eB D dV FεΔ = Δ (3268)


[ ] TF B dVσΔ = Δ (3269)


[ ] [ ] [ ] T TeB D dV B dVε σΔ = Δ (3270)




increments

0FΔ = (3271)

holds and

0eεΔ = (3272)

holds



- 221 -





[ ] eDσ εΔ = Δ (3273)


















en n n

crknc

np


1 1 1 (3277)





- 222 -


Δε nc







- 223 -



( ) ε σ ε φc Hf T F= (331)

Here




T temperature (K)




( ) ε σ ε φH Hf T F= (332)





( ) ε β σ εc H= (333)





n nc

n nHt+ + + + += =1 1

(334)

where

( )

( )σ θ σ θ σ

ε θ ε θεθ

θ

n n n

nH

nH

nH

+ +

+ +

= minus +

= minus +

1

1

1

1

(here 0≦ θ ≦1)






- 224 -












ε ε σσ σ

σc c d

d

f=

= prime3

2 (335)




FEMAXI calculation














- 225 -











is pursued





















- 226 -



























calculation






- 227 -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -



( ) ( )( )

( ) ( )



σ σ θ σ

σ σ θ σ

θ θ θ

θ θ

θ

θ

nc i

ni

ni

nH i

nH i

ni

ni

nH i

nH i

ni

ni

ni

ni

d d

f d d

d

d

++

+ ++

+ ++

+ ++

+ ++

+ ++

+ ++

= + +

=+ +

+

times prime +

111

11

11

11

11

11

3

2 (336)




( ) 1 1 11 1c i i i H i H i



indσ +

+ and 11

H indε +


obeys the flow rule

( )( ) ( ) ( )

1 11 1 1 1

111

1 11

3

2

3

2


n n ni in n

i in ni

n

f d dd

d

f

θ θθ θ

θ

θ θθ


σ σ θ σ

σσ

+ ++ + + ++ +

+ + +++ +

+ ++ ++

+

+ +prime= times +

+

prime=

(338)

holds


unknown 11

indσ +

+ and 11

H indε +

+ as follows

( ) ( )

( )

( )( )

( ) 11

11

11

11

111

2

3

2

3

2

3

2

3

2

3

2

3

+++

++

++++

+

+++++

+

++++

+++

+

++

+++

+

prime

part

part+

primepartpart+

primepartpartprime+

prime


iHn

in

i

n

H

H

in

in

in

Hini

n

in

in

in

Hini

n

in

in

Hini

n

in

Hini

n

in

ini

n

df

df

df

dfff

εθσε

εσσ

σθσεσσσ

σθσσ

σεσσ

σθσεσσσ

σεσσ

σσ

θθθ

θθθ

θθθθ

θθθ

θθθ

θθθ

(339)

Then


- 228 -

( )

1 112

1 11 1

11

3 3

2 2

3 3

2 2

3

2


nn n

i i ii i i i

n n n ni in n nn n

ii H i

n ni Hnn

f f d

ff d d

fd


θ θθ θ θθ θ

θθθ




+ ++ + + + + +

++ +

+ ++ + + +

+ + ++ +

++ +

++

prime prime= minus

prime prime+ +

prime+


1 11

1 11 1

3

2

3 3

2 2


n nn n


n nn n

f fd

f fd d

θθ θ

θ θθ θ

θθ

θ θθ θ



+ +++ + +

+ ++ +

+ +++ + +

+ ++ +

prime= + minus


(3310)


σ

= prime3

2

( ) [ ]

ε ε

σpartpartσ σ

σ σ θ σ

σpartσpartσ

θ σσ

σ partpartε

θ ε

θ θ

θ θ

θ

θθ

θ

θ θ θθ

θ

nc i

nc i

ni

n

i

ni

ni i j n

i

ni

ni

ni

n

i

ni

ni n

iH

n

i

nH

f fd

fd

fd

++

+

+ +

+

++ +

+

+

+ +++

++

++

=

+

minus

prime prime

+ prime

+ prime

1

2 11

11

1

9

4

3

2

3

2i+1

(3311)

is obtained



type









+ + by


- 229 -



( ) ( )( )

1 1 11 1

1 11 1 1

H i i H in n n n


t g

t g d d

θ θ

θ θ

ε σ ε


+ + ++ + + +

+ ++ + + + +

Δ = Δ

= Δ + + (3312)


1 11 1 1

11 1 1

11 1

H i H i H in n n

i ii i

n n n nn n

iH i

n nHn

d

gt g t d

gt d

θθ θ

θ

ε ε ε


partθ εpartε

+ ++ + +

++ + + +

+ +

++ +

+

Δ = Δ +

part = Δ + Δ part

+ Δ

(3313)

From Eq(3313) 11

H indε +

+ is expressed as

( ) 1

1 1 1 1 11

11

i ii H i i

n n n n nH i n n

n i

n Hn

gt g t d

dg

t

θθ θ

θ


partε

++ + + + +

+ + ++

++


(3314)


H indε +



H indε +


++

indσ and 1

1H i

ndε ++ in the



1 1 11 1 2 1 3 1

1

iic i c i i i i

n n i j n n nnn

F d F d Fθ θ θ


+ + ++ + + + ++

+


is obtained where

( )

1

1 2

1

9

4 1

i i

i ni Hn n n

iiin nn

n Hn

f gt

ffF

gt

θ θ θ

θ θθ

θ


partε

++ + +

+ +++

+


Ffn

i

ni2

3

2= +

+

θ

θσ (3316)

( )1 1

3

1

3

21

ii H i

n n nHn

iin

n Hn

ft g

Fg

t

θθ

θ

θ

part εpartε

σ partθpartε

+ + ++

++

+

Δ minus Δ =

minus Δ


- 230 -
















2 ( ) nH f T tε σ= sdot (3318)


12 ( ) n



[ ] ( )1 1

2( ) ( )n


minus

= = (3320)






H

f



( )1 2

9

4

i inii

n nn

f fF θ

θ θθ

partpartσ σσ

+

+ ++

= minus


- 231 -

2

3

2

inin

fF θ

θσ+

+

=

(3321)

3 0F =

are obtained


indσ +

+


gives 1 1 1


+ + + + + + Δ = Δ = Δ + (3322)

where

1 1 3 1 1

3


n

f t F tθ θθ

ε σ θ σσ+ + + + + +

+


1 1 2

iic i

n n i j nn

C t F Fθ θ θ



(3323)




c inε +


[ ] ( )

1 1 0 1 11 1 1 1 1

1 0


i in n n

C d B u

C

θ

θ

σ ε ε ε

σ σ

+ + + ++ + + + + +

+ +


+ minus =

(3244)


( ) [ ] ( )

1 1 0 1 1 1 1 1 1

1 0


i in n n

C C d B u

C

θ θ

θ

σ ε ε ε

σ σ

+ + ++ + + + + + +

+ +


+ minus =

(3324)

[ ]( ( )

)

1 11 1 1

0 11 1 1


c i P in n n


ε ε ε

+ ++ + + + +

++ + +

= Δ minus minus


（3325)

where

[ ] [ ] [ ]( ) 1111

~ˆ minus

+++ += icn

in

in CCD


nC 1+ to creep


indσ +

+ In this equation

11

c inε +




- 232 -









( ) ε partσpartσ

σ εcHf T=

(3326)


ˆ 0HHε ε ε= = (3327)

is set





ε c ε

ε

ε

εH =0

εH =0 εH =0

ε ε εH H= minus

①

②

③

④




( ) ε partσpartσ

σ εcHf T f=

ge 0 (3328)

time


- 233 -




sign is performed




13

IPUGH IPUGH2


1


( ) ( )h K TPσ ε= (3329)

where


T temperature












( ) ( ) ( ) ( )

( )

2 22

12 2

3

2

3

r z z r

r z

h H F GF G H θ θ

θ


α σ σ σ


+ + +

(3245A)

where


- 234 -










nP

nP

n

h+ +

+

=

1 1 (3330)










iteration



( ) ( )h K T Tn ni

nP

nP i

n nσ σ ε ε+ = + +++

++

+Δ Δ Δ11

11

1 (3331)



in nh σ σ +



that ( ) 11 1P P i



inσ +

+Δ and 11

P inε +

+Δ so


P inε +


inσ +



- 235 -


nP i

nP i

n

ih

++

++

+

=

11

11 (3332)


11

ΔεnP i+

+1


( ) 11

111 +

+

++

+

++

+

Δ

+Δ

+=Δ

+ n

n

iPn

i

nPn

Pn

in

i

nn T

T

KKTK

hh

θθθ partpartε

εpartpartεσ

partσpartσ (3333)

By substituting

nin

in


++++

111

11 (3334)


( ) ( ) ( )

Δ

Δ

ε partpartσ

σ σ σ ε

partpart

partpartσ

σ

θ θ

θ θ

nP i

ni

n

i

ni

n n nP

n

nn

n

i

ni

H

hh K T

K

TT

hd

++

+ ++

++

+++

=prime

minus + minus

minus

+

11

1

1 11

1

(3335)


P inε +

+Δ by using 11

inσ +

+Δ instead of 11

inσ +

+Δ

Here

prime =

+

+

HK

ni

P n

i

θθ

partpartε

(3336)


indσ +

+



( ) ( )

( )

[ ] [ ] [ ] ( )


+

minus

=

+minusΔ

+Δprime

+

++++++

+++

+

++

++

+++

i

n

iPn

icnnn

in

in

in

in

i

n

nin

i

n

nP

nnnn

iPn

in

hCuBD

h

TKhTT

KH

θθθ

θ

θ

θ


partσpart

σσpartσpartσ

εσpartpartε

11

1

011

11

1

11

11

~ˆ

(3337)


++1

1 we obtain


- 236 -

[ ]( ( ) )1 0 1 1 1 1

11

ˆ

ˆ


P i nn i i

i in n

n n

hD B u C

h hH D

θ θθ

θ θθ θ



++ + + + + +

+ ++

+ ++ +


( ) ( ) ( )1 1

ˆ

ii Pn n n n n n

n ni i

i in n

n n

h KT h K T

T

h hH D

θ θ

θ θθ θ



+ ++ +

+ ++ +


(3338)

Here 1

1indσ +



P inε +



[ ] [ ]( [ ] ( ) )

[ ]

( ) ( ) ( )

[ ]

i

n

i

n

in

i

n

in

nP

nnnn

nin

i

n

i

n

in

i

n

in

icnnn

in

in

in

in

i

n

i

n

iPn

iPH

h

hD

hH

TKhTT

Kh

hD

hH

CuBDh

h

1

11

1

011

11

1

11

1

ˆ

ˆ

~ˆ

+

++

++

++

++

++

++

+++++++

+

+

++

+

+prime

minus+Δ

minusminus

+

+prime


=

Δ=Δ

partσpart

partσpart

partσpart

εσpartpartσσ

partσpart

partσpart

partσpart

εεσσpartσpart

partσpartεε

θθ

θθ

θθ

θθ

θθ

θθθ

(3339)


P inε +

+Δ the following


[ ]( ( ) )

[ ]( ( )

1 1 0 1 1 1 1 1

1 01 1 1 1

ˆ

ˆ ˆ

ˆ


i ii in n


i in n

n n

d D B u C

h hD D

B u Ch h

H D

θ θ

θ θθθ

θ

θ θθ θ

σ σ σ ε ε


+ ++ + + + + + +

+ +++++ + + + +

+ ++ +



)i

( ) ( ) ( )1 1

ˆ

ˆ


n nn

iii i

n nn n

h h KD T h K T

T

h hH D

θθ θθ

θ θθ θ



+ + ++ ++

+ ++ +


(3340)


- 237 -


[ ][ ]( [ ] ( ) )

211

1

011

11

11

ˆ

ZZTS

CuBDd

nin

icnnn

in

in

in

ipn

in

minusminusΔ+


++

+++++++

++

θ







[ ] [ ][ ] [ ]

[ ]

D D

Dh h

D

Hh

Dh

nP i

ni

ni

n

i

n

i

ni

ni

n

i

ni

n

i+ +

++ +

+

++

++

= minus

prime +

θ θ

θθ θ

θ

θθ

θθ

partpartσ

partpartσ

partpartσ

partpartσ

(3342)

[ ]

[ ]S

Dh K

T

Hh

Dh

ni

ni

n

i

n

ni

n

i

ni

n

i+

++ +

++

++

=

prime +

θ

θθ θ

θθ

θθ

partpartσ

partpart

partpartσ

partpartσ

(3343)

[ ] ( )

[ ]Z

Dh h

Hh

Dh

ni

n

i

n

i

ni

n

ni

n

i

ni

n

i1

1

=

minus

prime +

++ +

+

++

++

θθ θ

θθ

θθ

partpartσ

partpartσ

σ σ

partpartσ

partpartσ

(3344)

[ ] ( ) ( )( )

[ ]Z

Dh h

h K T

Hh

Dh

ni

n

i

n

i

n nP

n

ni

n

i

ni

n

i21 =

minus

prime +

++ +

++

++

θθ θ

θθ

θθ

partpartσ

partpartσ

σ ε

partpartσ

partpartσ

(3345)


+1








- 238 -





1 1 1 1 1 1e th P c irr

















1 1X Xn n nC θε σ+ + + Δ = Δ (3347)






1 1 1 1 1 1e th P c irr




- 239 -

[ ] ( ) 1 1 1 1e P c th irr



n n n nC C C Cθ θ θ θ+ + + + = + + (3350)

1

n nD Cθ θ

minus

+ + = (3351)


[ ] ( )1 1 1 1th irr


where

[ ] 1 1

T

n nF B dVσ+ +Δ = Δ (3353)


[ ] [ ] ( )( )1 1 1 1


and finally

[ ] [ ] [ ] ( )1 1 1 1





1

n nD Cθ θ

minus

+ + = and 1nF +Δ


1 1en n nC θε σ+ + + Δ = Δ (3356)




expressed

( )

1 1 1 1

1 1 1 1 1

1 1 1 1

1 1

( )

( ) ( ) ( )

( ) ( ) ( )

( )

c Hn n n n


HH H H


H cn n n n

f T t

f T t f T t f T t

df df df T t T T t

d d d

f T t C

ε σ ε

σ ε σ ε σ ε


σ ε σ

+ + + +

+ + + + +

+ + + +

+ +

Δ = Δ

= Δ + Δ minus Δ

= Δ + + sdot Δ Δ = Δ + Δ

(3357)


- 240 -


T temperature











expanded form as

[ ] [ ] [ ] ( )

1

1 1 1 0 1

T

n n

T th irr cn n n n n

B D B dV u

F B D dV

θ

θ ε ε ε

+ +

+ + + + +

Δ =

Δ + Δ + Δ + Δ

(3359)




( ) ( )h K TPσ ε= (3360)


T temperature


h yield function







- 241 -



Setting


and

( )1 1 1P Pn n n n n nP

n n n

h K Kh K T T

Tθ θ θ


+ + +

+ Δ = + Δ + Δ (3362)

holds


1 1 1P

n n nPn n n

h K KT

Tθ θ θ


+ + +

Δ = Δ + Δ (3363)

Here setting nP n

KH θ

θ

partpartε+

+

prime =


nP

nP

n

h+ +

+

=

1 1 Eq(3363)

becomes

1

1 1 1

n

Pn n n n

n n

h h KH T

Tθ

θθ θ


+

minus

+ + + ++ +

primeΔ = Δ + Δ

(3364)


1 1 1

1

1

1 0 1

1

n n

n

n

Pn n n

n nn

nn n

nn n

P Pn n n

h h K hT

H T

h h KT

hT

H H

C

θ θ

θ

θ

θ θθ

θ θ

θ θ

θ



partσ

σ ε

+ +

+

+

+ + ++ ++

++ +

++ +

+ + +



= Δ minus Δ

(3365)


form

[ ] [ ] [ ] ( )

1

1 1 1 0 1 0 1

T

n n


B D B dV u

F B D dV

θ

θ ε ε ε ε

+ +

+ + + + + +

Δ

= Δ + Δ + Δ + Δ + Δ

(3366)





- 242 -







( ) 1

0 1

n

Pn n n n

nPn

n

KT K T h

T h

Hθ

θ

θ


partσ+

++

++


is used



indσ +

+ 11

inu +


[ ] ( ) ( )

1 1 0 1 1 1 1 1

1 1 2


in n

d D B u C

S T Z Z

θ θ

θ

σ σ σ ε ε+ ++ + + + + + +

+ +


+ Δ minus minus

(3368)


strain amounts



[ ] K u Fni

ni

ni

+ ++

+=θ Δ Δ11

1 (3369)



equations

[ ] [ ] [ ][ ]K B D B rdrdzni T

nP i

+ += θ θπ2 (3370)


- 243 -

[ ] ( )

[ ] ( ) [ ]

1 1

0 1 1

1 1 2

1 1

ˆ ˆ2

2

2

Ti P i i in n n n n

c in n

T in n

T in n

F B D C

rdrdz

B S T Z Z rdrdz

F B rdrdz

θ θ

θ

π σ σ

ε ε

π

π σ

+ + + +

+ +

+ +

+ +

Δ = minus

+ Δ + Δ


+ minus

(3371)




11




- 244 -




[ ] K u Fni

ni

ni

+ ++

+=θ Δ Δ11

1 (3369)

[ ] ( )

[ ] ( ) [ ]

1 1

0 1 1

1 1 2

1 1

ˆ ˆ2

2

2

Ti P i i in n n n n

c in n

T in n

T in n

F B D C

rdrdz

B S T Z Z rdrdz

F B rdrdz

θ θ

θ

π σ σ

ε ε

π

π σ

+ + + +

+ +

+ +

+ +

Δ = minus

+ Δ + Δ


+ minus

(3371)



[ ] K u Fn n n+ + +=θ Δ Δ1 1 (341)


[ ] B u F Bee n n et nΔ + + += +1 1 1 (342)

Here [ ] [ ] B K

B F F

ee n

et n n n

=

= minus+

+ + +

θ

1 1 1Δ


explained next





shown in Fig 341


v2U v1

U

① ② ③ ④ ⑤


v2L v1

L


- 245 -




+

+++

++

+

++

+

+++

+

+++

++++

++

++++

U

U

L

L

v

v

vv

u

u

u

u

u

u

u

BB

BB

BBB

BBB

B

BB

B

B

BB

BB

B

BB

BB

BBB

B

B

BB

BB

B

B

BBBBBB

BBBBBBBBB

BBB

BBBBBB

BBB

BBB

BBBBBB

BBBBBB

BBB

2

1

2

1

7

6

5

4

3

2

1

544

444

534

434

344

244

144

334

234

134

524

514

424

414

324

314

224

214

124

114

543

443

533

433

343

243

143

542

532

541

442

531

432

441

431

342

341

242

241

142

141

333

233

133

333

332

233

231

132

131

523

522

521

513

423

512

511

422

421

413

412

411

323

322

321

213

223

312

311

222

221

213

123

212

211

122

121

113

112

111

0

0

00

0

0

0

0

0

0

0

0

0

0

000

0

0

0

0

0

0

0

00000

000000

00000

000000

000000

00000

00000

000000

+

++

+++

+

++

+

minus

=

54

44

34

24

14

53

43

33

23

13

52

51

42

41

32

31

22

21

12

11

0

0

00

0

0

0

0

0

etet

etetet

etet

etetet

et

etet

et

et

etet

etet

et

pcb

pca

BB

BBB

BB

BBB

B

BB

B

B

BB

BB

B

F

F (343)




times

+++++++++

++

++++

+++++++++

times

544

444

542

541

442

441

543

443

344

244

144

342

341

242

241

142

141

343

243

143

524

523

514

424

513

423

414

413

22477

313

314

224

313

223

214

124

213

123

114

113

534

434

532

531

432

431

533

433

334

234

134

332

331

232

231

132

131

333

233

133

000000

0000

00

00

00

00

00

00

00

000000

00000

BBBBBBBB

BBBBBBBBBBBB

BB

BBBB

BB

BBB

BBBB

BBBB

BB

BBBBBBBB

BBBBBBBBBBBB


- 246 -

+++

+

++

+++

+

minus=


54

44

34

24

14

52

51

42

41

32

31

22

21

12

11

53

43

33

23

13

2

1

7

6

5

4

3

2

1

2

1

0

0

0

0

0

0

0

0

0

etet

etetet

et

etet

et

et

etet

etet

et

etet

etetet

pcb

pca

U

U

L

L

BB

BBB

B

BB

B

B

BB

BB

B

BB

BBB

F

F

v

vu

uu

u

u

u

u

v

v

(344)






set as

L

jU

j

Lj

Uj

vv

vv

112

111

+

+

=

= (345)


+

=

ΔΔΔ

ΔΔ=ΔΔ=Δ

Δ

ΔΔ=ΔΔ=Δ

Δ

ΔΔΔ

++ 11

32

31

37

31

3222

3111

27

21

2212

2111

17

11

12

11

3

2

1

netn

U

U

LU

LU

LU

LU

L

L

ee

ee

ee

BF

v

v

u

u

vv

vv

u

u

vv

vv

u

u

v

v

B

B

B

(346)


- 247 -


Eqs(341) and (342) as

[ ] FuBee Δ=Δ (347)








z

rrca

PELLET CLADDINGF3

lz

F3

Fpca Fpcb

F2

F2

rcb



F3 axial force






- 248 -




A) Upper plenum




surface (I) is







( ) ( )onUszUsUs


where



Upper plenum spring

Ppcb

Ppca lsz

rca

rcb

(I)

(II)

PELLET CLADDING

lsz

(I)

(II)

Ppca－Ppcb


- 249 -




plenum







where











( )( )

ΔΔ

minus

minusminus

minusminusminus

=

ΔΔminus

+

+U

N

sz

UsUs

UsUs

nnUszUsUs

nnUszUsUs

z

z

v

v

kk

kk

TTlk

TTlk

F

F

1

1

1

1

1

αα

(3412)




[ ] K u Fni

ni

ni

+ ++

+=θ Δ Δ11

1 (3413)


- 250 -

[ ] ( )

[ ] ( ) [ ]

1 1

0 1 1

1 1 2

1 1

ˆ ˆ2

2

2

Ti P i i in n n n n

c in n

T in n

T in n

F B D C

rdrdz

B S T Z Z rdrdz

F B rdrdz

θ θ

θ

π σ σ

ε ε

π

π σ

+ + + +

+ +

+ +

+ +

Δ = minus

+ Δ + Δ


+ minus

(3414)






temperature





















- 251 -

++

++

minus=

ΔΔΔΔΔ

++++++

+++

54

44

53

43

52

51

42

41

2

7

6

5

544

444

543

443

542

541

442

441

534

434

533

433

532

531

432

431

524

523

522

521

514

424

513

423

512

511

422

421

414

413

412

411

0

0

0

0

etet

etet

et

etet

et

pcz

pcb

pca

sz

UN

BB

BB

B

BB

B

F

F

F

v

v

u

u

u

BBBBBBBB

BBBBBBBB

BBBB

BBBBBBBB

BBBB

(3415)

where



the top segment)


(3415)

ΔΔΔΔΔΔ

+++minus++++

minus

+++

sz

UN

UN

UsUs

UsUs

v

v

v

u

u

u

kBBBBkBBBB

BBBBBBBB

kk

BBBB

BBBBBBBB

BBBB

2

1

7

6

5

544

444

543

443

542

541

442

441

534

434

533

433

532

531

432

431

524

523

522

521

514

424

513

423

512

511

422

421

414

413

412

411

0

0000

00

0

00

( )

( )

minus+++

minusminus

+

+

minus=

+

+

nnUszUsUsetet

etet

nnUszUsUs

et

etet

et

pcz

pcb

pca

TTlkBB

BB

TTlk

B

BB

B

F

F

F

15

44

4

53

43

1

52

51

42

41

0

0

0

α

α (3416)

is obtained



B) Lower plenum


boundary


- 252 -







LszLsLs


where






( ) ( )nnLszLsLs



( )

( )

ΔΔ

minus

minusminus

minusminusminus

=

ΔΔminus

+

+L

Lsz

LsLs

LsLs

nnLszLsLs

nnLszLsLs

z

z

v

v

kk

kk

TTlk

TTlk

F

F

11

1

1

1

1

αα

(3419)

holds





Lower plenum spring

Ppcb

Ppca lsz

rca

rcb

(I)

(II)

PELLET CLADDING

lsz


(II)



- 253 -

ΔΔΔΔΔΔ

+++minus++++

minus

+++

Lsz

L

L

LsLs

LsLs

v

v

v

u

u

u

kBBBBkBBBB

BBBBBBBB

kk

BBBB

BBBBBBBB

BBBB

12

11

7

6

5

544

444

543

443

542

541

442

441

534

434

533

433

532

531

432

431

524

523

522

521

514

424

513

423

512

511

422

421

414

413

412

411

0

0000

00

0

00

( )

( )

minusminus++

minus

+

+

minus

minus=

+

+

nnLszLsLsetet

etet

nnLszLsLs

et

etet

et

pcz

pcb

pca

TTlkBB

BB

TTlk

B

BB

B

F

F

F

15

44

4

53

43

1

52

51

42

41

0

0

0

α

α (3420)



( )

+minus

++

minus=

ΔΔΔΔΔ

++

++

+5

34

3

1

52

51

42

41

12

11

7

6

5

533

433

532

531

432

431

523

522

521

513

423

512

511

422

421

413

412

411

0

0

0

0

0000

00

0

00

etet

nnLszLsLs

et

etet

et

pcb

pca

L

L

Ls

BB

TTlk

B

BB

B

F

F

v

v

u

u

u

BBBBBB

k

BBB

BBBBBB

BBB

α (3421)





+

=

Δ

+++

111

1

netnn

UPee

nee

ee

LPee

BFu

B

B

B

B

(3422)


eeB is the





- 254 -




























equal)


zero


- 255 -


















bonded state when









( )

22

2μ

ππμ sdot

minus=

lr

rr

i

io (3423)





UV μge

UV μlt

ir

orμ


- 256 -








jrjz FF μgt













segments









- 257 -

Pellet stacksegment

Initial state


A B



deformed state B


























described in A











- 258 -














statesrdquo







and cladding













Cladding

Pellet



- 259 -































( )3 2

2 2AMU2

2j j

io i

F F

rlr r ππ= sdot

minus (3429)


( )2 2

AMU22

o i

i

r r

rl

πμ

πminus

= sdot (3430)


- 260 -

cf

pf+cf

cf

cf

cf

3

2

1

cf

cf

cf

cf


Plenum space










and cladding






















- 261 -










11 12j ju uΔ = Δ (3431)

jj FF 23 μ= (3432)



Δ=

ΔΔ

ΔΔ

minus

minus

minus

00001100

0

0

0

0

1

1

0

0

2

12

11

F

F

u

u

B

j

Ujc

Ujp

j

j

ee

νν

μμ

(3433)




Row l1 =

Δ=+ΔL

lljjll FFuK

1211 11

(3434)

Row l2 =

Δ=minusΔL

lljjll FFuK

1212 22

(3435)

Row l3 =

Δ=+ΔL

llj

UjPll FFvK

12 33

μ (3436)

Row l4 =

Δ=minusΔL

llj

UjCll FFvK

12 34

μ (3437)


- 262 -














vkF lz sdot= (3439)













treatment applies


- 263 -


Segment No

Contact state




Plenum － 0 0 uF













conditions
















results


- 264 -











11 12j ju uΔ = Δ (3431) U

jCU

jP vv Δ=Δ (3440)



segment j

U


of segment j


segment j


Δ=

Δ=Δ

Δ=Δ

F

u

B

jCU

jP

jj

ee

1211

νν

μ

(3441)



UjPv Δ=

UjCv Δ a common


- 265 -










the total matrix U

jCU

jP vv Δ=Δ (3440)


of segment j U


of segment j


Δ=

Δ=Δ

F

B

jCU

jP

ee

νν

(3442)






- 266 -














cladding
















11 12j ju uΔ = Δ (3431)

jj FF 23 μ= (3444)



- 267 -

Δ=

ΔΔ

ΔΔ

minus

minus

minus

00001100

0

0

0

0

1

1

0

0

2

12

11

F

F

u

u

B

j

Ujc

Ujp

j

j

ee

νν

μμ

(3445)


























- 268 -







cladding





external force















po j ci ju uΔ = Δ (=Eq(3431)) (3446)

UjC

UjP vv Δ=Δ (3440)

Here




j of the cladding



- 269 -







(3447)




Row l1 =

Δ=+ΔL

lljlll FFuK

12 11

(3448)

Row l2 =

Δ=minusΔL

lljlll FFuK

12 22

(3449)

Row l3 =

Δ=+ΔL

lljlll FFuK

13 33

(3450)

Row l4 =

Δ=minusΔL

lljlll FFuK

13 44

(3451)



UjP vv (3453)








2

3

0 0

1

1

0

0

1

1

0 0 1 1 0 0 0 0

0 1 1 0 0 0

0 0

p j

c j

po j

ci jee

Up

Uc

j

j

u

uBF

Fv

Fv

F

F



- 270 -






axial direction





















- 271 -

Time Step Start


















- 272 -



















Fuel Rod

Gap

Cladding

R

Pellet


1 segment


- 273 -


Upper pellet

Centerline of rod

1

1

1

Gap



Oxide layer



















- 274 -

00 02 04 06 08 10

R(ridge)M(mid)



Cladding

1 2 3 4

B(body)

2n-1 2n

n




analysis



follows


outer oxide layer











- 275 -










from each other










1-D analysis

2-D analysis





- 276 -

ε

ε

ε

ε

γpartpart

partpart

θ

=

=

+

r

z

rz

du

dr

dv

dz

u

r

v

r

u

z

(351）





analysis

352 Basic equations



of its importance


work as

[ ] B dV FT

n n + +minus =σ 1 1 0 (352)

Here







given by

[ ] σ εn neD+ +=1 1Δ (353)

where


Δε ne



- 277 -

[ ] ( )[ ] [ ]D D Dn n n+ += minus + =

θ θ θ θ1

1

21




[ ] Δ Δε σθne

n nC+ + +=1 1 (354)


[ ]CE

=

minus minus

minus minus

minus minus

+

1

1 0

1 0

1 0

0 0 0 2 1

ν ν

ν ν

ν ν

ν( )

(355)




n n ncrk

nc

np


1 1 1 (356)

where




Δε nc




strain on the block


[ ] ( )Δε n n nB u u+ += minus1 1 (357)

where



[ ] ( ) [ ] ( ) C B u un n n n n n np

nc


1 1 0Δ Δ Δ (358)

Here

[ ] [ ] [ ]

[ ] [ ]

1

1

1(1 )

2n n n

n n

C C C

C D

θ θ θ θ+ +

minus

= minus + =

=


- 278 -





( ) ( ) ( ) ( ) ( ) 052 22 2 232 3











[ ] ( ) [ ] ( ) C d B u du un ni

ni

n ni

ni

n n np

nc

+ + ++

+ ++


1 11

10

1 1 0Δ Δ Δ

(359)

Here

d ni

ni

niσ σ σ+

+++

+= minus11

11

1 (3510)

du u uni

ni

ni

++

++

+= minus11

11

1 (3511)


as


ni T

ni

nσ σ+ ++

+ + =1 11

1 (3512)





That is


- 279 -

[ ] [ ] [ ][ ] [ ] [ ][ ] ( )[ ] [ ] ( )[ ] ( ) 12111

1

01

1

11

1

ˆ

ˆ

ˆ

+++

+++

+++

++++

=minusminusΔ+

Δ+Δminus

minusminus

+

nnin

T

icnn

iPn

T

nin

ii

iPn

T

in

iPn

Tin

T

FdVZZTSB

dVDB

dVCDB

dVduBDBdVB

εε

σσ

σ

θ

θθ

θ

(3513)

is obtained where

[ ]( )

~C

E E E

E E E

E E E

E E

r

z

r z

=

minus minus

minus minus

minus minus

++

1 10

10

10

0 0 04 1

ν

ν ν

ν ν

νθ

(3514)


[ ] K u Fni

ni

ni

+ ++

+=θ Δ Δ11

1 (3515)


Here

[ ] [ ] [ ][ ]K B D B dVni T

nP i

+ += θ θ (3516)

and

[ ]

[ ] ( ) [ ] ( )

1 1 1

0 1 1 1

1 1 1 2

ˆ

ˆ

Ti in n n


T in n

F F B dV

B D C dV

B S T Z Z dV

θ θ

σ

σ σ ε ε

+ + +

+ + + + +

+ +

Δ = minus

+ minus + Δ + Δ


(3517)




[ ] [ ] [ ][ ]K B D B rdrdzni T

nP i

+ += θ θπ2 (3518)

[ ] [ ] [ ][ ] ( ) [ ] ( ) 2

ˆ2

2ˆ

2111

1

011

111

minusminusΔminus

Δ+Δ+minus+

minus=Δ

++

+++++

+++

rdrdzZZTSB

rdrdzCDB

rdrdzBFF

nin

T

icnnn

in

ii

iPn

T

in

Tn

in

π

εεσσπ

σπ

θθ (3519)


- 280 -







[ ]u

vN

u

v

u

v

u

v

=

1

1

2

2

8

8

(3520)


[ ] [ ]1 2 3 8

3 81 2

3 81 2

0 00 0

0 00 0

N N I N I N I N I

N NN N

N NN N

=

=

(3521)


0 1I

=

That is

1 1 2 2 3 3 8 8u N u N u N u N u= + + + +

1 1 2 2 3 3 8 8v N v N v N v N v= + + + +

hold


( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )

N

N

N

N

N

N

N

N

1

2

3

4

5

6

7

8

1 1 1 4

1 1 1 4

1 1 1 4

1 1 1 4

1 1 1 2

1 1 1 2

1 1 1 2

1 1 1 2





= + minus minus

= + + minus

= minus + minus

= minus + minus


(3522)



- 281 -


analysis

[ ]

r

z

rz

u r

v z B u

u z v rθ

εε

εεε


(3523)


[ ] [ ]B = B B B B1 2 3 8 (3524)


B i

i

i

i

i i

N

rN

zN

rN

z

N

r

=

partpart

partpart

partpart

partpart

0

0

0 (3525)



performed


r z is given by

[ ]partpartξ

partpartη

partpartξ

partpartξ

partpartη

partpartη

partpart

partpart

partpart

partpart

N

N

r z

r z

N

rN

z

N

rN

z

i

i

i

i

i

i

=

=

J (3526)

or

[ ]partpart

partpart

partpartξ

partpartη

N

rN

z

N

N

i

i

i

i

=

minusJ 1 (3527)




- 282 -

r N r

z N z

i ii

i ii

=

=

=

=

1

8

1

8


[ ]J =

= =

= =

partpartξ

partpartξ

partpartη

partpartη

Nr

Nz

Nr

Nz

ii

i

ii

i

ii

i

ii

i

1

8

1

8

1

8

1

8 (3528)


[ ] [ ]JJ

minus = =

= =

=minus

minus

1 1

8

1

8

1

8

1

8

1

det

partpartη

partpartξ

partpartη

partpartξ

Nz

Nz

Nr

Nr

ii

i

ii

i

ii

i

ii

i

(3529)

Here

[ ]det J =

minus

= = = = part

partξpartpartη

partpartη

partpartξ

Nr

Nz

Nr

Nzi

ii

ii

i

ii

i

ii

i1

8

1

8

1

8

1

8

(3530)



[ ]

[ ]

partpart

partpartη

partpartξ

partpartξ

partpartη

partpart

partpartξ

partpartη

partpartη

partpartξ

N

r

Nz

N Nz

N

N

z

Nr

N Nr

N

N

r

N

N r

i jj

j

i jj

j

i

i jj

j

i jj

j

i

i i

i ij

=

minus

=

minus

=

= =

= =

=

1

1

1

8

1

8

1

8

1

8

1

8

det

det

J

J

(3531)





[ ] [ ] [ ][ ] [ ]K B D B N r d dni T

nP i

j jj

+ +=

minusminus=

θ θπ ξ η2

1

8

1

1

1

1 det J (3533)


- 283 -

[ ] [ ]

[ ] ( ) [ ]

[ ] ( ) [ ]

81 1

1 1 11 11

81 1 0

1 1 11 11

81 1

1 1 21 11

ˆ 2 det J

ˆ2 det J

2 det J

Ti in n n j j

j


j

T in n j j

j

F F B N r d d

B D C N r d d

B S T Z Z N r d d

θ θ

θ

π σ ξ η


π ξ η

+ + +minus minus=


+ +minus minus=

Δ = minus



(3534)


f d d H H fi j i jj

m

i

n


== =

1

1

1

1

11

(3535)



applied

f d d f i jji


== =

1

1

1

1

1

2

1

2

(3536)

Here










1



Node

ξ

η

1-1

-1

1

Integrationpointa

a-a

-a


- 284 -







(4)

(1)

(2) (3) (6) (7)

(5)

(8)

r

z

Pellet

CladdingGap






- 285 -



















of pellet is fixed



Cladding

Mid-heightplane

Center


Cladding

Center


A B


- 286 -











12

13

14

15

16

17

18( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (3537)




Δv ( )1 0= (3538)

Δv ( )5 0= (3539)

hold Here ( ) Δv 1

and ( ) Δv 5









Δu( )2 0= (3540)




holds



represents





- 287 -





considered Fn+16( )





Δ Δv v( ) ( )8 8= (3541)

















F F F F F F F

F F F F F F

n nR

nR

nPC

nPP

nR

nPC

nG

nW

nR

nAX

nW

nG

+ + + + + + +

+ + + + + +

= + + + + +

+ + + + + +

1 11

12

13

14

15

16

16

17

18

18

18

18

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) (3542)


as



- 288 -

where

prime = + + ++ + + + +F F F F Fn nG

nW

nW

nG

1 16

17

18

18( ) ( ) ( ) ( ) (3544)


nR

nR

nR

1 11

12

15

18( ) ( ) ( ) ( ) (3545)


nPP

nPC

nAX

1 13

14

16

18( ) ( ) ( ) ( ) (3546)









[ ] K u Fn n n+ + +=θ Δ Δ1 1 (3547)

[ ] [ ] [ ][ ]K B D B dVn

T

nP

+ += θ θ (3548)

[ ] [ ] [ ][ ] ( ) [ ] ( )

minusminusΔminus

Δ+Δ+minus+

minus=Δ

++

+++++

+++

dVZZTSB

dVCDB

dVBFF

nnT

cnnn

inn

Pn

T

nT

nn

211

10

11

111

~ˆ

ˆ

θ

θθ εεσσ

σ

(3549)



[ ] [ ] ( ) [ ] ( )

1 1 1 1 1

01 1 1

1 1 2

1 1

ˆ

ˆ

ˆ

T

n n n n n

T P i cn n n n n n

T

n n

n n

F F F F B dV

B D C dV

B S T Z Z dV

F F

θ θ

θ

σ

σ σ ε ε

+ + + + +

+ + + + +

+ +

+ +


+ minus + Δ + Δ



(3550)

where


- 289 -

[ ] [ ] ( ) [ ] ( )

1 1 1 1

01 1 1

1 1 2

ˆ

ˆ

T

n n n n

T P i cn n n n n n

T

n n

F F F B dV

B D C dV

B S T Z Z dV

θ θ

θ

σ

σ σ ε ε

+ + + +

+ + + + +

+ +




(3551)








stiffness equation


















- 290 -



pairs









early phase of PCMI






















- 291 -












cladding







pairs



Node Couple

Pellet Gap



- 292 -


























In Eq(3542) FnPC

+13( ) and Fn


PP+14( ) and

FnAX


obtain

[ ] K u Fn n n+ + += primeθ Δ Δ1 1 (3553)


- 293 -


In this case FnPC

+13( ) and Fn


FU

VnPC n

n+

+

+

=

13 1

1

0

0

0

0

( )

(3554)

FU

VnPC n

n+

+

+

=minusminus

16 1

1

0

0

0

0

( )

(3555)

where

V Un n+ += plusmn1 1μ (3556)





i

j

k

l

m

K

u

v

u

v

U

Fn

p n

p n

c n

c n

n

n

0

0

1

1

0

0

0 0 1 0 1 0 0 0 0 0

1

1

1

1

1

1

minusplusmn

plusmn

minus

= prime

+

+

+

+

+

+

+

μ

μθ

ΔΔΔΔ

Δ

(3557)


- 294 -










equations are

K u F Uni

n n n+ + + += prime +θ( ) ( )Δ Δ1 1

11 (3558)

K u F Unj

n nj


1)(

11)( ˆ


nnk

n UFuK θ (3560)

1)(

11)( ˆ


nnl

n UFuK μθ (3561)





i

j

k

l

m

n

K

u

v

u

v

U

V

n

p n

p n

c n

c n

n

n

0 0

0 0

1 0

0 1

1 0

0 1

0 0

0 0

0 0 1 0 1 0 0 0 0 0

0 0 0 1 0 1 0 0 0 0

1

1

1

1

1

1

minusminus

minusminus

+

+

+

+

+

+

+

θ

ΔΔΔΔ

=prime

+Δ Fn 1

0

0

(3563)


- 295 -










above equations are

1)1(

11)( ˆ


n UFuK θ (3564)

K u F Vnj

n nj

n+ + + += prime +θ( ) ( )Δ Δ1 1 1 (3565)

K u F Unk

n nk

n+ + + += prime minusθ( ) ( )Δ Δ1 1 1 (3566)

1)(

11)( ˆ


nnl

n VFuK θ (3567)


minus + =+ +Δ Δu up n c n 1 1 0 (3568)

minus + =+ +Δ Δv vp n c n 1 1 0 (3569)







tt t

g gg tn

n n

n nn n+

+

+

= minusminus

+α1

1

(3571)




- 296 -





when




U p n + ge1 0 (3574)



tt t

U UU tn

n n

p n p np n n+

+

+

= minusminus

+α1

1 (3575)









Also when


+ +le1 1 (3578)


Here v p n


vc n


In contrast when


- 297 -

U p n + ge1 0 (3579)










bonded gap states









reversed











this


- 298 -






reversed























( ) ( )tZ Z

Z Z Z Zt t tn

i n j n


+ ++=


minus +α

1 11 (3580)

Here




- 299 -













That is in

[ ] F B dVn

T

n+ += 1 1σ (352)


+14( ) in the axial

direction

[ ] F F B dVnPP

n

T

n+ + += = 14

1 1( ) σ (3581)

holds




equation

( )tV

V Vt t tn

n

n nn n n+

++=

minusminus +α

11 (3582)








- 300 -


F F F F FR AX W G( ) ( ) ( ) ( ) ( )8 8 8 8 8= + + + (3583)










- 301 -



361 Creep



















t time (hour)



- 302 -




( ) ( )( )( ) ttTrTf cs





( ) ( ) ( )( ) cs






( ) ( )( )( ) ( )( )

minus=

+minus=HH

cs

Hc

TfTr

TfTr

εσσεεεσσε

(365)





( ) 1f T C Eσ ϕσ= (366)

( ) 2 r T E Cσ ϕ= (367)

3cs C Eε ϕσ= (368)

are obtained



( )c

FSεσ

part =part


2 1 3c


(3610)

( )c

H FHεε

part =part



- 303 -

2c


(3611)



( )H

GSεσ

part =part


2 1H


(3613)

( )H

H GHεε

part =part


2H


(3614)

From Eqs(369) to (3614)

( )c FFUNCε = ( )c

FSεσ

part =part

( )c

H FHεε

part =part

( )H GFUNCε =

( )H

GSεσ

part =part

( )H

H GHεε

part =part



1 1 11 1 1 1


+ + + + + + Δ = Δ = Δ + (3615)

where

1 1 3 1 1

3


n

f t F tθ θθ

ε σ θ σσ+ + + + + +

+


[ ] [ ]C t F Fnc i

n i j n

i

n

i


θ θ θ


Δ 1 1 2 (3617)


( )1

1 2

1

9

4 1

i i

i ni Hn n n

iiin nn

n Hn

f gt

ffF

gt

θ θ θ

θ θθ

θ


partε

++ + +

+ +++

+


(3618)

Ffn

i

ni2

3

2= +

+

θ

θσ (3619)

( )1 1

3

1

3

21

ii H i

n n nHn

iin

n Hn

ft g

Fg

t

θθ

θ

θ

part εpartε

σ partθpartε

+ + ++

++

+

Δ minus Δ =

minus Δ

(3620)


- 304 -




part =part

( )Hg GHε

part =part 1

c inε +Δ and













- 305 -



σpart

part

c

Hε

εpart

part

Hεσ

partpart

and

H

Hε

εpart

part


σpart

part and

Hεσ

partpart

is 1(sMPa)


be multiplied by

( )3 3 6( ) ( )36 10 36 10 ( ) 10 ( )( ) ( )

d hr d Pad s d MPa




σ ε= K n


ε true strain


( )( ) ( )1








n=00504+00001435T







- 306 -

00 05 10 15 200

200

400

600

800

1000

1200

1400300K

600K

350K

900K

800K

True

Stre

ss

(MP

a)

True Strain ()

100 s-1

1 s-1

0001 s-1




00 05 10 15 200

200

400

600

800

1000

0001s

0001s

0001s

100s

100s

900K

600K

400K

True

Stre

ss

(MP

a)

True Strain ()

352K100s








- 307 -

363 Ohta model(35)














occurs














nmid

midK

= minus310

εεσ (3623)

Plastic region 310

mnK

εσ ε minus =

(3624)


0

300

600

900

1200

1500

0 001 002 003 004True s train [frac]

True

stre

ss [M

pa]


00E+0

50E+4

10E+5

15E+5

0 001 002 003 004

True s train [frac]

dσ

dε [M

Pafr

ac]

Predict Interface

Interimregion


( )y yε σ

1 1( )ε σ

2 2( )ε σTru

e s

tress

(M

Pa)


- 308 -




Hereafter 310

εminus

m




1

1ε

minusprime =

n

y

K









2 min( 0003 175)y yε ε ε= + times (3626)



Ep2

2




m

nmid

midK

= minus311 10

εεσ

(3627)

m

nmid

midK

= minus322 10

εεσ

(3628)


midn

=

1

2

1

2

εε

σσ

(3629)

Consequently

( ) ( )( ) ( )12

12

lnln

lnln

εεσσ



- 309 -

mn

mid

mid

K

=

minus32

2

10

εε

σ

(3631)











given as follows

T≦900 K （62685）

1 2

253 4

( 27315)

( 27315) 1 exp( 1751 10 )

K L L T

L L T minus

= minus minus


1 2 3 41745 2517 526 0748L L L L= = = = (3632)


900＜T＜1000 K (62685～72685 oC)

1 2 3 4

05251 2 3 4

(

( ) 1 exp( 1751 10 )

K A T A T A TA

B T B T B TB minus

= + + +


A1＝ 3041233371times104 B1＝ 33282432 times102

A2＝－8285933869times101 B2＝ 298027653

A3＝ 7490958541times10－2 B3＝－681286681times10－3

A4＝－2242589096times10－5 B4＝ 351229783times10－6

1000 K≦T （72685oC~） 0525


61 2 3 41 10 8775 8663 0341566H H H Hminus= sdot = = = (3634)




- 310 -

















T≦700 K （42685）

1 2

05263 4

( 27315)

( 27315) 1 exp( 9699 10 )

K L L t

L L T minus

= minus minus


(3635)

L1＝1201 L2＝2508 L3＝1119 L4＝0627

700＜T≦800 K （42685～52685）

1 2 3 4

05265 6 3 4

(

( ) 1 exp( 9699 10 )

K A T A T A TA

A T A T A TA minus

= + + +


A1＝ 3656432405times104 A2＝－1361626514times102

A3＝ 1695545314times10－1 A4＝－7055897451times10－5

A5＝－3338799880times104 A6＝ 1330276514times102

800＜T＜950 K （52685～67685）



61 2 31 10 4772 9740H H Hminus= sdot = =


- 311 -


0526

2 3 4( 1 exp( 9699 10 )

non irradiated

non irradiated

K K

B T B T B TB K

minus

minusminus

=


B1＝－3617377246times104 B2＝ 1361156965times102

B3＝－1636731582times10－1 B4＝ 6386789409times10－5

950K≦T （67685oC≦T）

0526


61 2 3 41 10 4772 9740 5267783H H H Hminus= sdot = = =















T≦67315 K (400oC)

1 2 ( 27315)n C C T= minus minus (3640)

67315ltTlt1000 K （400～72685）

3 4 5( )n C T C C T= + + (3641)

C1＝0213 C2＝1808times10－4 C3＝34708518times10－2

C4＝49565251times10－4 C5＝－50245302times10－7


- 312 -

1000 K≦T （72685～）

n = C6 = 0027908 (3642)

【Note 3】


MATPRO model






the MATPRO model




irradiation

T≦730 K （45685） m＝002 (3643)

730ltT＜900 K （45685～62685）

5 6 7 8( )m A T A T A A T= + + + (3644)

900 K ≦T （62685～）


A5＝ 2063172161times101 A6＝－7704552983times10－2

A7＝ 9504843067times10－5 A8＝－3860960716times10－8

A9＝ -067times10－2 A10＝－2203times10－4




- 313 -









310

mnK

s

εσ ε minus =

(3646)


ε true strain (-)



eEεσ = (3647)



εminus

m


as





( )

11 minusprime

= minusn

E

Kn

d

df εσσ

(3650)


d

dff

n

nnn minus=+1 (3651)



solution


- 314 -


ε

σ σ0

σ1σ2 σ3

εp

nK εσ =

eEεσ =

Yσσ =cE e += εσ


ε

σ σ0

σ1σ2 σ3

εp

nK εσ =

eEεσ =

Yσσ =cE e += εσ



- 315 -

37 Skyline Method










(0) (0) (0) (0)11 1 12 2 1 1

(0) (0) (0) (0)21 1 22 2 2 2

(0) (0) (0) (0)31 1 32 2 3 3

(0) (0) (0)1 1 2 2

(1)

(2)

(3)

n n

n n

n n

n n nn n

a x a x a x b

a x a x a x b

a x a x a x b

a x a x a x





(371)






0

To eliminate 1)0(



Eq(372) we obtain

(0) (0) (0)

(0) (0) (0) (0) (0) (0)21 21 2122 12 2 2 1 2 1(0) (0) (0)

11 11 11n n n


a a a


(373)





1 xan



- 316 -

(0) (0) (0) (0)11 1 12 2 1 1

(1) (1) (1)22 2 2 2

(1) (1) (1)32 2 3 3

(1) (1) (1)2 2

(1)

(2)

(3)

( )

n n

n n

n n

n nn n n

a x a x a x b

a x a x b

a x a x b

a x a x b n



(374)

where (0) (0) (0) (0)

(1) (0) (1) (0)21 12 31 1222 22 32 32(0) (0)

11 11

(0) (0) (0) (0)(1) (0) (1) (0)21 1 31 12 2 3 3(0) (0)

11 11

a a a aa a a a

a a

a b a bb b b b

a a









=

=+

=+++

=++++

minusminus

minusminusminus

minusminus

minusminus

)1()1(

)2()2()2(

)1(2

)1(21

)1(122

)1(22

)0()0(11

)0(112

)0(121

)0(11

nnn

nnn

nnn

nnnn

nnn

nnnn

nnnnn

bxa

bxaxa

bxaxaxa

bxaxaxaxa

　　　　　　　　　　　　

　　　　　　　　

　　　　

　　　

(375)


Nkj

Nki

a

aaaa k

kk

kkj

kikk

ijk

ij

1)1(

)1()1()1()(

=+=

minus= minus


Nkiba

abb k

kkkk

kikk

ik

i 1)1()1(


minus



)1(

)1(

minus

minus

minusn

nn

nn

n a

bx (378)



- 317 -


)1(2

)1(2

)1(1

)1(

minus+

minus++

minus+

minus minusminus= k

kk

kkkkk

kkk

kk

k a

xaxabx

(379)


obtained




ib of


ib



)2(11

)2(1

)2(1)2()1(

minusminusminus

minusminus


ii

iji

iiii

iji

ij a

aaaa

)2(1)2(

11

)2(1)2()1( minus



iiii

iiii

ii

i ba

abb


)0(11

)0(1

)0(21)0(

2)1(

2 a

aaaa j

jj minus=

=

minus

minusminus

minus=

minusminus=minus=

2

1)1(

)1()1()0(

3

)1(22

)1(2

)1(32

)0(11

)0(1

)0(21)0(

3)1(22

)1(2

)1(32)1(

3)2(

3

kk

kk

kkj

kik

j

jjj

jjj

a

aaa

a

aa

a

aaa

a

aaaa

　　

so that in general

minus

=minus


1

1)1(

)1()1()0()1(

i

kk

kk

kkj

kik

iji

ij a

aaaa (3710)

holds Similarly

minus

=

minusminus

minusminus minus=

1

1

)1()1(

)1()1(

i

k

kkk

kk

kik

ii

i ba

abb (3711)

holds


nb are

obtained


- 318 -


ija and )1( minusi


ji

mj

a

aaaa

i

kk

kk

kkj

kik

iji

ij

2

21

1)1(

)1()1()0()1(

==

minus= minus

=minus


miba

abb

i

k

kkk

kk

kik

ii

i 21

1

)1()1(

)1()1( =

minus= minus

=

minusminus

minusminus (3713)



( 1) ( 1)1

( 1) (0)( 1)

1

for 2

for 2

k kiik kji

ij ij kk kk

i m j na aa a

j m i na


minus=

gt == minus gt =

　 (3714)

nmiba

abb

i

k

kkk

kk

kik

ii

i 1

1

)1(

)1(

)1()1( =

minus= minus

=

minusminus

minusminus (3715)













timestimestimes

i ｊ

Symmetry

-Li

-Lj


- 319 -



( )=

+++minusj

kkLkji

1

1 (3716)



++j

kkLk

1

1



described








minus

=


)1()1()0()1(

0

î

kk

kkj

kkiij

iij aaaa (3717)

where )1(

)1()1(ˆ minus

minusminus =

kkk

kkik

ki a

aa

mj 2=

11 minus+= jLi j

( )ji LLk max0 =

mjaaaaj

Lk

kkj

kkjjj

ijj

j

2ˆ1

)1()1()0()1( =

minus= minus

=


and

( )bi

bj

kk

kk

kkii

ii LLk

mLibabb

max

1ˆ

0

1)1()1()0()1(

0=

+=

minus= minus

=




- 320 -


mLia

bb bi

ii

iii )1(

)0()1( =

= minusminus (3720)


( 1) 1kk kx b k mminus= = (3721)

( 1) 1 2ˆ

1 1k

k k ki ii i

i m mx x a x


= minus= + minus

(3722)



kki




ija )1( minusi




minus

=


)1()1()0()1(

0

î

kk

kkj

kkiij

iij aaaa (3723)

where

)1(

)1()1(ˆ minus

minusminus =

kkk

kkik

ki a

aa (3724)

mj 2=

( )jpLi j min1 +=

( )ji LLk max0 =


part of matrix

and

pjaaaaj

Lk

kkj

kkjjj

ijj

j

2ˆ1

)1()1()0()1( =

minus= minus

=


where

( )jpLk

mi

a

aa

jk

kk

kkik

ki min

2ˆ

)1(

)1()1(

==

= minus

minusminus (3726)



- 321 -


Eq(3725)


minus

=


)1()1()0()1(

0

î

kk

kk

kkii

ii babb (3727)

where pLi b 1+=

( )bi LLk max0 =

and also

pLia

bb bi

ii

iii

i )1(

)1()1( =

= minus

minusminus (3728)

is set


minus

=


)1()1()0()1(

0

î

kk

kkj

kikij

iij aaaa (3729)

holds where

)1(

)1()1(ˆ minus

minusminus =

kkk

kikk

ki a

aa (3730)

in nmj 1+=

( )jpLi nj min1 +=

in nmi 1+=

( )jpLj ni min1 +=

and

n

j

Lk

kkj

kjkjj

ijj ppjaaaa

j

1ˆ1

)1()1()0()1( +=

minus= minus

=


where

( )jpLk

nmj

a

aa

njk

kk

kjkk

ki min

1ˆ

)1(

)1()1(

=+=

= minus

minusminus (3732)





- 322 -

n

i

kk

kk

kkii

ii ppibabb 1ˆ

1)1()1()0()1(

0

+=

minus= minus

=


where niii

iii ppi

a

bb 1)1(

)0()1( +=

= minusminus (3734)

is set


sequence

( 1) 1kk k nx b k pminus= = (3735)

( )( 1)

1 2ˆ

min 1k

k k ki ii n

i n nx x a x


= minus= minus

(3736)



References 3












- 323 -









( ) ( )K

D

TT=

minus minusminus

sdot+

+ times times

minus minus1 1

1 0 05

40 4

4641216 10 1867 104 3

ββ

exp


( ) ( ) K

DT=

minus minusminus


1 0 050 0191 1216 10 1867 104 3β

β exp



Ttemperature (degC)


Default

value

IPTHCN=1

213


IPTHCN=2





IPTHCN=3






- 324 -


4 7 01148 00035 (2475 10 8 10 )

00132exp(000188 ) (WmK)

K F B B T

F T


+ sdot sdot




IPTHCN=4 43


IPTHCN=17 215

4 01148 00040 (2475 10 000333 )

00132exp(000188 ) ( )

K F B B

F T W mK


+ sdot sdot




T(C)


IPTHCN=5 42


12 3 4 795

312

(235 10 35 10 ) (255 10 80 10 )

357 10

TD K

K

K B B T

T


minus


+ times sdot






IPTHCN=6 44

13 4 80047 173 10 (25 10 541 10 )

(WmK)






IPTHCN=7 45


2

05

10 05( 10)

10 (( 75) 10)

B G B

B G B

le =ge = minus



- 325 -



23210 gAgAxAAA +++=

2210 gBgBBB ++=

A0=00254 A1=40 A2=03079 A3=122031

B0=2533x10-4 B1=8606x10-4 B2=-00154

C=10x10-12


( )( )

327384 058 10 ( 127315 )

20 ( 127315 )

T T T K

T T K

αα


= gt

IPTHCN=8 46


IPTHCN=9 214




( ) ( )

2 4

1 temperature(K)

1 396exp 6380

452 10 (mKW) 246 10 (mKW)

h T TT

A Bminus minus

=+ minus

= times = times



IPTHCN

=91 410

Latest model

( ) ( ) ( )2 4

1

Gd P

C D

A B T f Bu g Bu h T

T T

λα β


minus + (WmK)




( ) ( )

2 4

1

1 396exp 6380

453 10 (mKW) 246 10 (mKW)

h TT

A Bminus minus

=+ minus

= times = times

C=547E-9 (Wm-K3) D=229E-14 (Wm-K5)





- 326 -


( )

( )

( ) ( )

( ) ( )( )

( )

1 1 2 3 4 0

9

0 4 2

1 3265 3265

1

2

W mK

1 4715 10 16361exp W mK

00375 2165 10

109 00643 1arctan

109 00643

0019 11

3 0019 1 exp 1200 100

1 15 for s

1 1

d P P x r

d

P

P

K K K K K

T T T

K TT

KT

PK

P

λ λ

λ

ββ β β β

ββ

σσ

minus

=



minus= =+ minus

( )

( )( )3 4

4


021 1

1 exp 900 80

temperature (K)


x r

P

K KT

T

Bu Buβ minus

= = minus+ minus

= times sdot

IPTHCN=10 47


IPTHCN=13 48

KaT b

=+


Ttemperature (degC)









- 327 -


( )( )95 22 20 1 2 3 0 1 2 3

1exp (WmK)

1TD

C Dg WK


+ = + minus + + + + + + +

T = temperature (K)





W= 141MeV = 1411610-19 J

A0 = 0044819 + 0005BU+20(G - 54702) (mKW)

B0 = (24544 - 00125BU - 70(G - 54702))10-4 (mW)

A1 = 4 (-) B1 = 08 (-)

A2 = 0611 (mKW) B2 = 960310-4 (mW)

A3 = 11081 (mKW) B3 = -176810-2 (mW)

C = 5516109 (WKm) D = - 43021010 (WKm)



A =LT






IPTHCN=15 49







- 328 -


1

mat ek

K KA B T

= ++ sdot

(kWmK) for 100TD





1)

12731540 412 ( )022 ( 50)0447 000453 (50 80)0084 (80 )


Bu BuBu


2) 147315

40 174022 00003

kT KA BuB Bu

gt= + sdot= + sdot



IPTHCN=16 238


IPTHCN =18 411






ΔL


ΔL


IPTHEX=0 (default)

IPTHEX=1

34


- 329 -


ΔL




IPTHEX=2


ΔL



IPTHEX=3


ΔL



IPTHEX=4


IPTHEX=5 211


exp DL EK T K K

L kT

Δ = minus + minus



K1 = 10x10-5 K-1 K2 = 30x10-3 K3 = 40x10-2


IPTHEX=6 426

273 923T Kle le 3 6 10 2

13 3

( ) (273) 266 10 9802 10 2705 10

4391 10

L T L T T

T

minus minus minus

minus


1 6

10 2 13 3

( ) (273) (99734 10 9802 10

2705 10 4391 10 )

L T L T

T T

minus minus

minus minus


973 3120T Kle le

3 5 12 2

12 3

( ) (273) 328 10 1179 10 2429 10

1219 10

L T L T T

T

minus minus minus

minus


1 5

9 2 12 3

( ) (273) (99672 10 1179 10

2429 10 1219 10 )

L T L T

T T

minus minus

minus minus





- 330 -

(3) UO2 density





No

restrictions

213


IPLYG=2 Default







Tlt1000 K e = 08707 Tgt2050 K e = 04083


No

restrictions

34








No

restrictions

34

413


- 331 -

(8) UO2 creep


( ) ( )( )

( )( ) ( )ε

σ σσ=

+ minus

++

minus+

+ minusA A F Q RT

A D G

A Q RT

A DA F Q RT

1 2 1

32

44 5

2

56 3

exp exp exp


6 121 2

43 4

285 6

1 2 3

9728 10 324 10

877 1376 10

905 924 10

90000 132000 5200 (calmol)

A A

A A

A A

Q Q Q

minus

minus

minus

= times = times

= minus = times





Ddensity 92-98 (TD)



No restrictions

34


- 332 -


( ) ( ) ( )

( )

451 2 1 1 4 8 2 2

23 6

7 3

exp ( ) exp

( )

exp

c


A D G A D

A F Q RT

σ σε

σ

+ minus + minus= +

+ +

+ minus

(s-1)

A1=03919 A2=13100times10-19 A3=-877 A4=20391times10-25

A6=-905 A7=372264times10-35A8=00 1

1

20178848 exp 80 10 7212423

log( 199999)Q

x


1

2

20198720 exp 80 10 1115435

log( 199999)Q

x









When 05714

165474168


When 05714

165474168

Gσ gt 1 05714

165474168

Gσ = and 2σ σ=

IPCRP=2

(Default) 412







- 333 -



45 21 1 22

( )( ) exp( ) exp( )


G


(1s) ( MPa)σ =

4 17 610

1

14 33

6 5 63

2

4 13

138 10 46 10( ) 1877 10

905

259 10 863 10(1s)

905

973 10 324 106895 10

877

671 10 223 10(1s)

877

FA F

F

FA

F

ρ

ρ

ρ

ρ


minus minus

minus minusminus

minus











IPCRP=4

(Default)

414


4 303505 108577 10 exp 148 10 f





and σ=24 MPa

IPCRP=5 415


- 334 -

(9) UO2 swelling








ΔV

V T

gs

S

= times minus times

4 396 10

1645 1024


ΔV

V

gs


fissionscm3


(default )

249



ΔV

V

ss






minus3times107


minus106




V

ss


IFSWEL=1 12


- 335 -










1 1

5 51 1

2413 10 10634 10 011509

286 10 ln 20184 10 ln

VT T Bu

V

T BU T GR

minus minus

minus minus



where

V



GR grain size (μm)

IFSWEL=5 252


IFSWEL=2 213


T le 1400 028 1400 le T le 1800 ( )[ ]0 28 1 0 00575 1 400 + minusT


2200 lt T 0476


IFSWEL=3


205 (10MWdkgUO )V

V

Δ =


- 336 -


( ) ( )

( ) ( )BuBuBuBuAV

V

V

V

BuBuBuFGRCAV

V

SWSW

SW

ltminus+

Δ=

Δ

lelesdotsdotminus=

Δ

1121

111 0

whereSW

V

V

1

Δ


A1=0095 C1=0043 Bu1=57 A2=00317

SW

V

V

Δ




IFSWEL=4 251



( )BuCeV

V

V


max

ΔV


V


densification


IDENSF=0

(default value)

11




ΔV

V

max

() 10

SBU


given by CSBU

= 2 3025 (MWdt-UO2)

where exp (－23025) =01

25000


- 337 -




( )( )

ΔL

L

TD

TS= times

minusminus

22 2100

1180





BUburnup (GWdt-U)

IDENSF=1 416





( )

( ) ( )( )

ΔΔΔ Δ

ρρρ ρ

= le= + lt lt= ge

0 1

1 2

2

BU BU

m BU b BU BU BU

BU BUsn

log


(1700degCtimes24h)

BUburnup (MWdt)



burnup range

IDENSF=2 417






- 338 -



( )ΔV

V

BU

BUADST= minus

minus minus sdot

+ minus minus sdot

sdot0 01420 1 6 7943

0 00793 1 11434

exp( )

exp( )



IDENSF=4 43


Δρ = +

M

A

AD Ft

Girr irrln

10

03












AG G

Dt=

minus30

3

MA

InADt

G

th=

+

Δρ

10

3






Q 82000 calmol



IDENSF=3 418



- 339 -







bD


bD



IDENSF=30 419

420







where rcic and rpi




follows





- 340 -

(12) UO2 plasticity




( )( )

( )( )σ

εεY

p

p

T T

T T

T C

T C=


legt

66 9 0 0397 520 0 0 386

36 6 0 0144 139 5 0 06875

1200

1200

( )( )prime =


legt

HT

T

T C

T C

520 0 0 386

139 5 0 06875

1200

1200




IFY=1

(default value) 421


422

σ Y T T

T


minus times sdot

minus

minus

1176 1 1 688 8 179 10

1 293 10

4 2

7 3



IFY=0



( )( )

( )2

1 322 22

exp exp

2exp 1D

P D


RTT T

θ θθ

= + + minus minus



UO2 PuO2 Unit

K1 2967 3474 JkgK



θ 535285 571000 K


No

restrictions 34


- 341 -




sdot+minus=

m

ff

D

NNE

DK

dt

dDmax11







Efitting parameter



boundary (atomscm2)

IGRAIN = 0

(default value) 230






dD

dtk

D

B

Dm

= minus +

1 1 0 002


IGRAIN=1 231


- 342 -


( )D D k t RT20

20


IGRAIN=2 423


IGRAIN=3 424

D D k e tQ RT30

30minus = sdotminus





IGRAIN=4 213

D D e tRT40





t time (h)


IGRAIN=5 34



4




- 343 -




the followings



10 (Default)








case with IXEKR=0

IXEKR = 3



1) 012 ≦ f ≦030



where



( ) ( )

( )

1 34 12

2

0035 225 10 830 10 1

0 005

u K K

z

K T T z

z UO z

minusminus minus

minus


IPTHCN=30 426


- 344 -


( )13 4

2

exp C

KK P K K T

K T

= + +



( )( )P

D=


1 1 0 95

ββ

porosity factor




IPTHCN=31 34


IPTHCN=32 426 427



p

ppF

21

1)(

+minus=

p

pp

21

1

8610

1)(

+minussdot=α

p porosity



TK temperature (K)


( )

sdotsdot+


minusminus

312

14

103876

108852440105500093105281

K

KMOX

T

TzK

τ




2 τ


- 345 -


95

1 1 005

1 2 1 2 005

p


(WmK)

( )2951 exp( )K KK

C DA BT TT

λ = + minus +

)(0350852)( WmKxxAA +==


)(106891 9 mWKC times= 13520D K=



IPTHCN=33

428


IPTHCN=34

427

( ) 14

312

1528 000931 01055 044 2885 10( )

7638 10

KMOX

K

z TK p

T

τα

minusminus

minus



p

pp

21

1

8610

1)(




TK temperature (K)


IPTHCN=35

429

1

2 3 7 4 5 6

( )(1 ) ( )

moxfuel

C

K aK F p



(WmK)


1 1 005( )

1 2 1 2 005

pF p

p


1 2 3

4 5 6

7

404 464 184

116 10 00032

0003

a a a

a a a

a

= = == = ==

CT C deg

210 for UO

092 for MOXmoxK =

= MWdkgMOXBU



- 346 -


( )( )95 22 20 1 2 3 0 1 2 3

1exp

1TD

C Dg WK


+ = + minus + + + + + + +






IPTHCN=15 49





Condition Ref

( )95 092 ( ) 00132exp 000188 (WmK)K Q x T= +





1( ) 10789

1 05

pK p K

p

minus=+

IPTHCN=37

430


- 347 -


IPTHCN =91

410

Latest model

( ) ( ) ( )2 4

1

Gd P

C D

A B T f Bu g Bu h T

T T

λα β


minus + (WmK)




( ) ( )

2 4

1

1 396exp 6380

453 10 (mKW) 246 10 (mKW)

h TT

A Bminus minus

=+ minus

= times = times





Condition Ref

mod95 2

1 ( ) expC D

K P xT T

= + minus

( ) ( ) gad ( ) ( )

(1 09exp( 004 )) ( ) ( )


Bu g Bu h T



=Y default=00)



Bu burnup in GWdtHM

h(T)= 1 1 396expQ

T





1( ) 10789

1 05

pK p K

p

minus=+

IPTHCN =38

430


- 348 -



( ) ( )[ ]( ) ( )[ ]fBDBTQFBB

fBDBTQG

FBBc

47454

265

4332121

1exp

1exp

+minus+minus++

+minus+minus+=

σ

σε

B1=01007 B2=757times10-20 B3=333 B4=0014

B5=6469times10-25 B6=00 B7=103






When 05714

165474168


when 05714

165474168

Gσ gt

571401816547416

G=σ and 2σ σ=

IPCRP=30 34


1) fH=1 222

( )exp( )

A FQ RT C F

G

σε σ= minus +

5 13

2

263 10 893 10(1s)

877

FA

ρ






T temperature (K)



414


- 349 -

4 5 6 7 8 9 10 11 12 13 14 15 161E-10

1E-9

1E-8

1E-7

1E-6

1E-5

fH= 3 fH= 2

SKI MOX fH=1

SKI UO2

UO2Halden

MOXHalden


Fuel

cre

ep ra

te

(1s

)

10000T (1K)



μ = 0276 IPOIS=30 34


IPOIS=31 431 432




54 303505 10

8577 10 exp 973 10 fRT



IPCRP=15 415



- 350 -





IPLYG=30 34

(6) Specific heat



( )( )[ ]

CK T

T TK T

O M K E

RT

E

RTPD D=

minus+ +

minus

12

2 2 23

21 2

θ θ

θ

exp

exp

exp




ISPH=30 34





Tm melting point(K)





No restrictions


34 413


- 351 -


( ) fL

Lf

L

L

L

L

PuOUO 22

1Δ+minusΔ=Δ

ΔL


ΔL

LT T T

UO2


2

4 6 9 2 16 339735 10 84955 10 21513 10 37143 10PuO

LT T T



IPTHEX=30 34


( ) ( )xfL

Lf

L

L

L

L

PuOUOMOXsdot+sdot


22

where MOXL


2UOL


2PuOL




(279 K≦T≦923 K)

( ) 11039141070521080291097349 31321061


L

L

UO

(923 K≦T≦3120 K)

( ) 11021911042921017911096729 3122951


L

L

UO

(279 K≦T≦1693 K)

( ) ( )( )

24 6 9

2312

12232 10 75866 10 273 56948 10 273

59768 10 273PuO

LT T

LT

minus minus minus

minus




IPTHEX=31 426433


- 352 -



exp DL EK T K K

L kT

Δ = minus + minus



K1 = 90x10-6 K-1 K2 = 27x10-3 K3 = 70x10-2

IPTHEX=32 213


bD


bD



IDENSF=30 419 420


- 353 -




Ttemperature (K)

No

restrictions 213



For 27minus800degC








Ttemperature (degC)

ICATHX=0

(default value)

213


ICATHX=1

For 21minus429degC



Ttemperature (degC)


ICATHX=3 34

300ltTlt1073 K

6 3

0

444 10 124 10L

TL


6 3

0

672 10 207 10D

TD




- 354 -


Variable name








IZYG=2

Default

213

34

1) α phase 11 7







11 81 (661 10 5912 10 )K T= times + times Δ


K =088+012exp10

Φ minus







restrictions 213



- 355 -




v = 0 30















Accordingly


is obtained where




CRPEQ=0

213


- 356 -


expm p ncreep

D QA t


13111 10

0682

0550

0579

1173

A

m

p

n

Q

minus= times====

2

hours

ncm s (gt1MeV)

MPa

Kelvin

t

Tθ

φσ

sdot

CRPEQ=2 435


CRPEQ=1 436


ε ε εcthc

irrc= + (1hr)



( )

ε εthc m

HmmK=

minus11

1

where ε ε εH thc

th kc

kk

n

t( ) = = sdot= Δ

1








ceq= times minus6 64 10 25 1 23 1 34





( )d dcz

cεσ


22

( )d dzc

zcε


22



1

2



- 357 -


expm p ncreep

D QA t


1216 10

0450

0550

0579

1173

A

m

p

n

Q

minus= times====

2

hours

ncm s (gt1MeV)

MPa

Kelvin

t

Tθ

φσ

sdot

CRPEQ=3 437


expε σα = minus

2000342205 32

T ( )730 1073K T Kle le


HTCRP=1

(Default) 438


1073ltTlt1273 K

4 4

1073 107310

200 200





HTCRP=

20

21

22

439

0 10 1 exp

nG Q

AT G RT

σε = minus


(Nm-2)

( 973 1273T Kle le )








- 358 -



13


0






( ) ε partσpartσ

σ εcHf T=

(431)


ε ε ε= =HH 0 (432)

is set





ε c

ε

ε

ε

εH =0

εH =0εH =0

ε ε εH H= minus

①

②

③

④



- 359 -


( ) ε partσpartσ

σ εcHf T f=

ge 0 (433)





σ ε= K n



n

E

KKK minus+= )0020( 20

20

( )( )1





n=00504+00001435T


ICPLAS=1 11





- 360 -



prior time step

310

mnK

s

εσ ε minus =

σ true stress (Pa)



Tgt=850 K 0027908n =


cold work 2

13

7 8

0847exp( 392 ) 0153 ( 916 10 0229 )

exp373 10 2 10

RIC CWN CWN CWN

CWN

φ


times + times sdot


exponent n







900lt=T 8663exp 8755K

T = +









ICPLAS=2 34


- 361 -



prior time step

310

mnK

s

εσ ε minus =

(original)

σ true stress (Pa)



model


model




18 25

8 25

0546 554 10 26 10

0546 144 10 (26 10 )

DK CWK K

CWK K φ


KFKF



MATPRO-11 model

ICPLAS=3

00 05 10 15 200

200

400

600

800

1000

0001s

0001s

0001s

100s

100s

900K

600K

400K

True

Stre

ss

(MP

a)

True Strain ()

352K100s





- 362 -




Intermediate region

mn

midmidK

= minus310

εεσ

Plastic region 310

mnK

εσ ε minus =







T≦900 K （62685）

1 2

253 4

( 27315)

( 27315) 1 exp( 1751 10 )

K L L T

L L T minus

= minus minus


1 2 3 41745 2517 526 0748L L L L= = = =


900＜T＜1000 K (62685～72685 oC)

1 2 3 4

05251 2 3 4

(

( ) 1 exp( 1751 10 )

K A T A T A TA

B T B T B TB minus

= + + +


A1＝ 3041233371times104 B1＝ 33282432 times102

A2＝－8285933869times101 B2＝ 298027653

A3＝ 7490958541times10－2 B3＝－681286681times10－3

A4＝－2242589096times10－5 B4＝ 351229783times10－6

1000 K≦T （72685oC -）

0525


61 2 3 41 10 8775 8663 0341566H H H Hminus= sdot = = =

ICPLAS=

4 for

PWR

ICPLAS=

5 for

BWR

35


- 363 -


T≦700 K （42685）

1 2

05263 4

( 27315)

( 27315) 1 exp( 9699 10 )

K L L t

L L T minus

= minus minus


700＜T≦800 K （42685～52685）

1 2 3 4

05265 6 3 4

(

( ) 1 exp( 9699 10 )

K A T A T A TA

A T A T A TA minus

= + + +


A1＝ 3656432405times104 A2＝－1361626514times102

A3＝ 1695545314times10－1 A4＝－7055897451times10－5

A5＝－3338799880times104 A6＝ 1330276514times102

800＜T＜950 K （52685～67685）



1 2 31 10 4772 9740H H Hminus= sdot = =


05261 2 3 4( 1 exp( 9699 10 )

non irradiated

non irradiated

K K

B T B T B TB K

minus

minusminus

=


950K≦T （67685oC≦T）

0526


1 2 3 41 10 4772 9740 5267783H H H Hminus= sdot = = =


67315ltTlt1000 K (400-72685) 3 4 5( )n C T C C T= + +

C1＝0213 C2＝1808times10－4 C3＝34708518times10－2

C4＝49565251times10－4 C5＝－50245302times10－7

1000 K≦T (72685 - ) n = C6 = 0027908




B1＝－3617377246times104 B2＝ 1361156965times102

B3＝－1636731582times10－1 B4＝ 6386789409times10－5


- 364 -

A5＝

A7＝

A9＝

2063172161times101

9504843067times10－5

-067times10－2

A6＝

A8＝

A10＝

－7704552983times10－2

－3860960716times10－8

－2203times10－4


ICPLAS =6 430


prior time step

310

mnK

s

εσ ε minus =

σ true stress (Pa)




where )(TK is


750 K lt T lt 1090 K

( )

timestimes=2

66 1085000272

exp105224882T

TK







( ) ( ) 110



CWTCWf



T temperature (K)


reduction)



- 365 -



where ( )Tn is

Tlt4194 K ( ) 114050=Tn

4194 K lt T lt 10990772 K

( ) 2 3

6 2 10 3

9490 10 1165 10

1992 10 9588 10

n T T

T T

minus minus

minus minus


minus times + times


T gt 1600 K ( ) 6089531=Tn

( )φn is










300 400 500 600 700 800 900

100200

300400500

600700800

900

1000


Cla

ddin

g Y

ield

Stre

ss (

MPa

)

Temperature (K)



- 366 -


following


( )[ ][ ] [ ] [ ]ΔL


12φ

ΔL





CWcold work

ICAGRW=1

213






( )ΔL

Lt

n

0

2101 10= sdot φ ΔL

L0




ICAGRW=2


( )ΔL

Lc t

o


ΔL

Lc

o



ICAGRW=3


ICAGRW=4 445

ΔL

LA t

o

= sdot sdotφ ΔL

Lo






- 367 -





No restrictions

213





( )C C U MP= +0 φ








7





ICOPRO=1

212


ICORRO=3 (BWR)

213


3 exp









- 368 -


FCORRO


( )ds

dtT FCORRO

ds

dt= + sdot1 0 04 Δ


00









(10 026)Zr

ZrO

X

Y

ρρ

=minus

=151 or 156 or 158




T temperature (K)

IZOX=2

Default 34


restrictions





442




CNOX




00


- 369 -






No restrictions









1=Y TsolltT


No

restriction

213


- 370 -




(WcmK) ITMC=41



Not examined










12 2 15 3

1629 10 3285 10

2198 10 1629 10

T

T T

λ minus minus

minus minus



ICATHX=41


Zircaloy



Assumed to be null




- 371 -







CRPEQ =41





8 5 2

8 5 20 2

9 6 21 2

9 5 21 2

3945 10 225 10 ( 0 n cm ) (1)

7550 10 613 10 ( 15 10 n cm ) (2)

1378 10 179 10 ( 145 10 n cm ) (3)

1093 10 660 10 ( 605 10 n cm ) (4)

Y

Y

Y

Y

T t

T t

T t

T t

σ φ

σ φ

σ φ

σ φ

minus

minus

minus

minus





(Pa) (C)Y Tσ




ICPLAS=41




5 2300 1558 326 0298 956 10 (JkgK )

1558 558228p

p

T K C T T

T K C


ge = ICSPH=41 213


- 372 -






2





IGAPCN=0

(default value) 217


IGAPCN=1 213






hK

rmix

0 =+Δ δ




h PK


δ




F

ar

RF

a=

sdot

+

+1

1001429

0 3

1

2Δ



prime = minusminus

+F

x1

1

600

10001

4



- 373 -


h h hgap g s= +




( ) ( )hK

C R R g g rgg=

+ + + +1 2 1 2 Δ


He00021 Ar000029 Kr00001









hK P

a Hsm=

sdotsdot

KK K

K Km =sdot

+2 1 2

1 2

Pt Ys

ID= sdot2 a a R= 0

( )R R R= +1

2 12

22


(WcmdegC)

K

KKm

1

2

0 038

0140 05977

==

=






a0 = 05 cm12


IGAPCN=2 217


- 374 -





hP

M

K K


sdotsdot

280

1

2

1 2

1 2 1 2

12

λ λ




M2=120000psi (844kgcm2)







hK

gg=sdot052 δ


Ar00168 He0139 FP00088



h Ps = 0 6 hK

gg=


IGAPCN=3






- 375 -














Ttemperature (K)


3


( )

( )V

S ZV

Z

K Z

j S ZV

s o

s o

=+

+ prime

+

minus

part

υ part

2

2 2

1

2

21

41





IDCNST=1 228


IDCNST=2 235 ( )TD 6614exp10091 17





- 376 -


321 DDDD ++= (cm2s)


1280

11

9861

1001exp102

521

1 TD


1300

11

9861

10762exp102

440

2 TD



IDCNST=3 238


IDCNST=4 441

( )10 4

2 40

AM1 76 10 exp 7 10

BM1 ( 2 10 )v

D RT

s j V F

minus

minus





IDCNST=5 226

( )( )( )

10 4

17

22

AM1 76 10 exp 7 10

15 10 exp 10600

767 10 exp 2785

D RT

R T

R T

minus

minus

minus


+ times minus

+ times sdot minus



IDCNST=6 232

a b cD D D D= + + (m2s)







( )( )052 2 2 2








- 377 -



DkT

m n d12

1

2

122

3

8 2

1=

ππ (m2s)

mm m

m m =

+1 2

1 2


( )d d d12 1 21




Ttemperature (K)

No

restrictions 253




235U AU235=235 238U AU238=238 O AO=16


B) Density

UO2 ( )396102









- 378 -

( ) ( ) ( ) 5112 GdOGdGdOU

GdGdGd XAAXAA

XAW



85237050235950238 =times+times=UA

( ) ( ) ( )16250

511625157121685237

2515710

=times++minustimes+

=

Gd

GdGd

Gd

X

XX

X


fWW

UO

UO

Gd

OGd

sdot+

=

2

232

1

ρρ


A

XAW

GdGdO

OGd

sdot= 51


( )A

XAW GdUO

UO

minus=

12


( ) GdGdOGdUO

OGdGdO

OUUO

XAXAA

AAA

AAA

sdot+minus=

sdot+=

sdot+=

512

51

2

1

51

2



2UOW =08847

Thus ( )302310

9610

88470

208

11530950

cmg=+

=ρ




( )2 2 3314 05 07854 Pr cm cmπ = times =


- 379 -


( )( )6

10023 07854 7872

7872 10

g cm

t cmminus

times =

= times


( ) ( )( ) ( )

2

2

6

6

2378508847 07798

26985

300 10 100048871

7872 10 07798

UU UO

UO

AW W

A

MW cm dBu MWd tU

t cm

minus

minus

= sdot = sdot =

times times= =

times times


( )2

31096 095 10412 UO g cmρ = times =


( )( )6

10412 07854 8178

8178 10

g cm

t cmminus

times =

= times

88140185269

852372

2

=sdot== UOUO

UU W

A

AW

Thus

( ) ( )( ) ( )

6

6

300 10 100041620

8178 10 08814

MW cm dBu MWd tU

t cm

minus

minus

times times= =

times times




- 380 -

















200610 (2006)


- 381 -



(417) USNRC WASH-1236 (1972)















- 382 -





















表５SI 接頭語




リットル Ll 1L=11=1dm3=103cm3=10-3m3

トン t 1t=103 kg










カロリー cal













表し方


























































組立量




デジベル dB



名称記号












Abstract+Contents

1MAN

2M1-6

2M7_Diffusion

2M8_Rate law model

2M9--13

2M14-17_Gap_gas

3M1

3M2

3M3

3M4

3M5_2D


3M7_Skyline-Ref3

4M1_UO2

4M2_MOX

4M3_Zr

4M4-SUS

4M5-6+Ref

light water reactor fuel analysis code femaxi-7

Documents