mtbe 3 - dp 2

PRODUCTION OF 400,000 TONNES PER YEAR OF MTBE PLANT

JUPLIN KINTI 2000337672

MOHD NAZRI ISMAIL 2001193485

NOOR HARYANI MUSTAPHA 2000132163

NORMARIAH ABDULLAH 2000337665

ROHIZAD JAMEL 2001476067

UNIVERSITI TEKNOLOGI MARA

LIST OF TABLES

TABLE PAGE

Table 1.1: Feed, Distillate and Bottom Composition 2

Table 1.2: Calculation Value of Θ 2

Table 1.3: Calculation Value of Rm 3

Table 1.4: The Value of Coefficient 4

Table 1.5: Calculation for Bubble Point Temperature 5

Table 1.6: Calculation for Dew Point Temperature 5

Table 1.7: Calculation for Bubble Point Temperature 5

Table 1.8: Viscosity of TBA, PO and Acetone ` 6

Table 1.9: Relative Volatility 6

Table 1.10: Properties of TBA, PO and Acetone 9

Table 1.11: The Sugden’s Parachor for TBA, PO and Acetone 10

Table 1.12: Calculation of Molar Volume PO 22

Table 1.13: Viscosity of TBA and PO at 356 K Temperature 23

Table 2.1: The Advantages and Disadvantages of Each

Types of Heat Exchanger 35

Table 2.2: Selection of Heat Exchanger 37

Table 2.3: Layout & Tube Size of Shell and Tube Heat Exchanger 45

Table 2.4: Summary of Chemical Engineering Design of

Heat Exchanger 55

Table 2.5 Shell-And-Tube Specification 57

Table 2.6: Material Selected for Shell-And-Tubes 59

Table 2.7: Types of Heads and Its Applications 62

Table 2.8: Standard Nozzle for Tube Size 66

Table 2.9: Standard Nozzle for Shell Side 67

Table 2.10 Steel Welding Neck Flanges 68

Table 2.11 Dimension of Selected Standard Steel Saddle 70

Table 2.12 Summary of Mechanical Engineering Design

of Heat Exchanger 70

Table 3.1: Reactor Mass Balance for Input Stream 71

Table 3.2: Reactor Mass Balance for Output Stream 72

Table 3.3: Properties of Flue Gas 80

Table 3.4: Chemical Design Specification Data 84

TABLE PAGE

Table 3.5: Mechanical Design Specification Data 96

Table 5.1: Summary of Chemical Engineering Design 150

Table 5.2: Summary of Mechanical Engineering Design 160

Table 6.1: Summary of the Piping of MTBE Plant 172

Table 7.1: Quantities Chemicals Stored/ Handled On-Site 186

Table 7.2: HAZOP Guide Words 198

Table 7.3: Typical HAZOP Process Parameter 199

Table 7.4: HAZOP Analysis of TBA Vaporizer

Table 8.1: Estimation Cost of Purchased Equipment 208

Table 8.2: Labor Cost 210

Table 8.3: Annual Cash Flow Before Tax And After Tax 214

Table 8.4: Annual IRR After Tax 215

Table 8.5: Cumulative Cash Flow After Tax 216

Table 8.6: Simple Payback Period 217

Table 8.7: Discounted Payback Period 217

Table 9.1: Data for Heat Integration 219

Table 9.2: Interval Temperatures for ΔT Min = 10°C 220

Table 9.3 Rank Order of Interval Temperature 220

Table 10.1: The Composition of Gases Discharged 224

Table 10.2: Flash Point and the Lower and Upper Flammable

Limit (LFL) and (UFL) 224

Table 10.3: Inventory of Waste Water 226

Table 10.4: Oxygen Demand of Component 227

LIST OF FIGURES

FIGURE TITLE PAGE

1.1 Trial Layout of Plate 18

1.2 Plate Specification 20

2.1 Design Procedure of Shell-and-Tube

Heat Exchanger 39

2.2 Process Operation of Shell and Tube

Heat Exchanger 41

2.3 Heat Exchanger is Separate Into 2

Heat Exchanger in Series 42

2.4 Triangular Pattern 47

2.5 Typical Standard Flange Design 68

4.1 Fixed Bed Reactor 97

6.1 Fixed Bed Reactor Control 166

6.2 Distillation Column Control 167

6.3 PFD Diagram for MTBE Process 174

7.1 Typical Plot of Plant Layout for MTBE Plant 178

7.2 TBA Vaporizer Instrumentation 203

8.1 Cumulative Cash Flow vs. Year 216

9.1 Hot and Cold Stream Composite Curves 219

9.2 Heat Cascade 221

9.3 Proposed Heat Exchanger Network 222

PRODUCTION OF 400,000 TONNES PER YEAR OF MTBE PLANT

JUPLIN KINTI 2000337672

MOHD NAZRI ISMAIL 2001193485

NOOR HARYANI MUSTAPHA 2000132163

NORMARIAH ABDULLAH 2000337665

ROHIZAD JAMEL 2001476067

A report submitted in partial fulfillment of the requirement for the award

of Bachelor of Engineering (Hons.) in Chemical Engineering

FACULTY OF CHEMICAL ENGINEERING

UNIVERSITI TEKNOLOGI MARA

SHAH ALAM

MARCH 2004

“We declared that this report is the result of our own work except for quotations and

summaries have been duly acknowledged”

(Signed) (Signed)

JUPLIN KINTI NOOR HARYANI MUSTAPHA

2000337672 2000132163

(Signed) (Signed)

MOHD NAZRI ISMAIL NORMARIAH ABDULLAH

2001193485 2000337665

(Signed)

ROHIZAD JAMEL2001476067

18thMarch 2004

“We declared that we read this report and in our point of view this report is

qualified in term of scope and quality for the purpose of awarding the

Bachelor of Engineering (Hons) in Chemical Engineering”.

Signed:………………..

Date :………………..

Supervisor

Encik Rusmi Alias

Faculty of Chemical Engineering

Universiti Teknologi MARA

Shah Alam, Selangor


Date :………………..

Supervisor

Puan Sharifah Intan Baizura Syed Ahmad Fuad



Shah Alam, Selangor

Accepted:


Date :………………..

Head of Programme

Prof. Madya Dr. Wan Shabuddin Wan Ali



Shah Alam, Selangor


Date :………………..

Coordinator

Puan Noor Fitrah Abu Bakar



Shah Alam, Selangor

CONTENTS

TITLE PAGE

CHAPTER 1 TBA DISTILLATION COLUMN

1.1 INTRODUCTION 1

1.2 CHEMICAL DESIGN OF

DISTILLATION COLUMN 2

1.2.1 The Operating Line 2

1.2.2 Calculate Bubble Point

And Dew Point 4

1.2.3 Calculate The Number

Of Stages 6

1.2.4 Pressure Drop 8

1.2.5 Calculation For Density

And Relative Molar Mass 9

1.2.6 Calculation Surface

Tension (Σ) 10

1.2.7 Column Diameter 11

1.2.8 Provisional Plate Design 13

1.2.9 Evaluation Design 13

1.2.10 Plate Efficiency ( Emv) 21

1.3 MECHANICAL DESIGN 24

1.3.1 Dead Weight of Vessel. 25

1.3.2 Weight of Plates. 26

1.3.3 Weight of Insulation. 26

1.3.4 Wind Loading. 26

1.3.5 Analysis of Stresses At Bottom 27

1.4 DESIGN OF STIFFNESS RING 29

1.5 DESIGN OF DOMED END. 30

1.6 DESIGN FOR THE

SKIRT SUPPORT. 31

CHAPTER 2 HEAT EXCHANGER DESIGN

2.1. INTRODUCTION 34

2.2 SELECTION OF EQUIPMENT 35

TITLE PAGE

2.2.1 Selection of Shell-And-Tube-Type

Of Heat Exchanger 35

2.3 BASIS DESIGN PROCEDURE

OF HEAT EXCHANGER 39

2.4 CHEMICAL DESIGN OF FLOATING

HEAD HEAT EXCHANGER 41

2.4.1 Design Specification 41

2.4.2 Properties Of Steam And TBA 42

2.4.3 Heat Load 45

2.4.4 Heat Transfer Area 45

2.4.5 Number Of Tubes 46

2.4.6 Tubes Arrangement (Pitch) 47

2.4.7 Diameter Of Shell 48

2.4.8 Tube Side Coefficient, Hi 49

2.4.9 Shell Side Coefficient, Hs 50

2.4.10 Overall Heat Transfer Coefficient, Uo 52

2.4.11 Tube Side Pressure Drop 53

2.4.12 Shell Side Pressure Drop 54

2.5 MECHANICAL DESIGN OF

HEAT EXCHANGER 56


2.6 DESIGN PRESSURE AND TEMPERATURE 57

2.7 MATERIAL OF CONSTRUCTION 58

2.8 DESIGN STRESS 59

2.9 WELDED JOINT EFFICIENCY 60

2.10 CORROSION ALLOWANCE 60

2.11 DESIGN CRITERIA 60

2.11.1 Minimum Thickness Of

Cylindrical Of The Shell 61

2.12 HEADS AND CLOSURE 62

2.1 Design of Domed

Ends-Ellipsoidal Heads 62

2.13 DESIGN LOAD 63

2.14 DESIGN OF NOZZLES 64

TITLE PAGE

2.14.1 Shell Side Nozzles 64

2.14.2 Tubes Side Nozzles 65

2.14.3 The selected tube size nozzle 66

2.14.4 Shell side nozzles 67

2.15 BOLT-FLANGED JOINTS 67

2.16 BAFFLES 68

2.17 SUPPORT DESIGN – SADDLES

SUPPORT 69

CHAPTER 3 ISOBUTYLENE REACTOR

3.1 INTRODUCTION 71

3.2 CHEMICAL DESIGN 72

3.2.1 Selection of Catalyst 743.2.2 Effective Diffusivity, De 74

3.2.3 Tube Specification 76

3.2.4 Heat Transfer Calculation 78

3.2.5 Tube Side Coefficient 78

3.2.6 Shell Side Coefficient 80

3.2.7 Overall Heat Transfer Coefficient 82




3.3.1 Design Pressure 85

3.3.2 Design Temperature 85

3.3.3 Material of Construction 85

3.3.4 Corrosion Allowance 86

3.3.5 Thickness of Cylindrical Shell 86

3.3.6 Head and Closures 86

3.3.7 Weight Load 87

3.3.8 Wind Loading 89

3.3.9 Analysis of Stresses 90

3.3.10 Elastic Stability 91

3.3.11 Vessel Support Design 92

3.3.12 General Consideration for

The Design 94

TITLE PAGE

3.3.13 Base Rings and Anchor Bolts 94

3.3.14 Pipe Size Selection for the Nozzle 96

3.3.15 Standard Flanges 96

CHAPTER 4 MTBE REACTOR

4.1 CHEMICAL ENGINEERING DESIGN

OF REACTOR 97

4.1.1Catalyst 98

4.1.2 Tube side 103

4.1.3 Shell 105

4.1.4 Condition Calculation 106

4.2 MECHANICAL DESIGN OF REACTOR 112

4.2.1 Design Consideration 112

4.2.2 The Design of Thin Walled Vessels

Under Internal Pressure 113

4.2.3 Design of Vessels Subject to

Combined Loading 116

4.2.4 Vessel Support 121

4.2.5 Base Ring and Anchor Bolt Design 124

4.2.6 Bolt Flanged Joint 127

4.2.7 Pipe Sizing 129

4.2.8 Compensation for Opening and

Branch Connections 130

CHAPTER 5 MTBE DISTILLATION COLUMN

5.1 INTRODUCTION 133


5.2.1 Determination of Key Components 134

5.2.2 Determination of Bubble Point and

Dew Point 134

5.2.3 Determination Relative Volatility, 136

5.2.4 Determination The Number of

Stages 137

5.2.5 Calculation to Determine Overall

Tray Efficiency, Eo 138

TITLE PAGE

5.2.6 Determination Of Feed Point

Location 138

5.2.7 Estimate or Gather The Physical

Properties 139

5.2.8 Determination Of Maximum And

Minimum Vapor and Liquid Flow

Factor and Flooding Velocity

For The Turn Down Ratio 140

5.2.9 Determination Of Column Diameter 141

5.2.10 Liquid Flow Arrangements 142

5.2.11 Plate Layout 142

5.2.12 Determination of Weir Length, lw 143

5.2.13 Check The Weeping Rate 143

5.2.14 Plate Pressure Drop 145

5.2.15 Downcomer Design 146

5.2.16 Check Entrainment 147


5.2.18 Number Of Hole 148

5.2.19 Column Size 149



5.3.2 Material Construction 151

5.3.3 Vessel Thickness 151

5.3.4 Heads and Closure 152

5.3.5 Column Weight 153

5.3.6 Wind Loads 154

5.3.7 Vessel Support Design

(Skirt Design) 155

5.3.8 General Consideration For Design 157


5.3.10 Feed, Top Product, Bottom Product

Piping Sizing 159

CHAPTER 6 PROCESS CONTROL AND INSTRUMENTATION


TITLE PAGE

6.2 TYPES OF CONTROL 161

6.2.1 Feedback Control 161

6.2.2 Feed forward Control 162

6.2.3 Cascade Control 162

6.3 CONTROLLING SHELL AND TUBE

HEAT EXCHANGER 163

6.4 FIXED BED REACTOR CONTROL 165

6.5 DISTILLATION COLUMN CONTROL 167

6.6 PIPING 170

6.6.1 Introduction 170



6.6.4 Fluid Velocity 171

CHAPTER 7 PLANT SAFETY

7.1 GENERAL SITE

CONSIDERATIONS 175


7.1.2 Site Layout 175

7.1.3 Plant Layout 179

7.2 PLANT SAFETY 181

7.2.1 General Overview

Of Safety 181

7.2.2 The Integration Of Safety

Procedure 182

7.2.3 Safety During Start-Up

And Shutdown 183

7.2.4 Emergency Response

Plan (ERP) 185

7.2.5 HAZOP Study 197

7.2.6 HAZOP Report 200

7.2.7 Hazard Analysis 204

CHAPTER 8 ECONOMIC EVALUATION


8.2 The specification of plant 206

TITLE PAGE

8.3 Revenue from sales 207

8.4 Cost Estimation 207

8.4.1 Capital Cost Estimation 207

8.4.2 Manufacturing Cost Estimation 209

8.4.3 Cost of Operating Labor (COL) 210

8.4.4 Cost of Utilities (CUT) 211

8.4.5 Cost of Raw Material (CRM) 212

8.5 Profitability Analysis 213

8.5.1 Before Tax and After Tax

Cash Flow 214

8.5.2 Present Worth and IRR Method 215

8.5.3 Cumulative Cash Flow After Tax 216

8.6 PAYBACK PERIOD 217

8.6.1 Simple Payback Period 217

8.6.2 Discounted Payback Period 217

8.7 CONCLUSION 217

CHAPTER 9 PROCESS INTEGRATION


9.2 PINCH TECHNOLOGY 219

9.3 THE PROBLEM TABLE METHOD 219

9.4 THE NETWORK DESIGN 221

CHAPTER 10 WASTE TREATMENT


10.2 DESCRIPTION AND PROCESS

SYNTHESIS 224

10.2.1 Air Treatment 224

10.2.2 Water Treatment 225

10.3 LAYOUT OF WASTE WATER

TREATMENT 227

10.4 MECHANICAL DESIGN WASTE

TREATMENT 228

10.5 CONCLUSION 229

PRODUCTION OF 400,000 METRIC TONNES PER YEAR OF MTBE

TBA DISTILLATION COLUMN JUPLIN KINTI

SUPERVISORS

1. EN. RUSMI BIN ALIAS

2. PN. SH. INTAN BAIZURA SYED A..FUAD DATE: 18 MARCH 2004

CHAPTER 1:

TBA DISTILLATION COLUMN

CONTENTS

TITLE PAGE

CHAPTER 1 TBA DISTILLATION COLUMN

1.3 INTRODUCTION 1

1.4 CHEMICAL DESIGN OF

DISTILLATION COLUMN 2

1.2.11 The Operating Line 2

1.2.12 Calculate Bubble Point

And Dew Point 4

1.2.13 Calculate The Number

Of Stages 6

1.2.14 Pressure Drop 8

1.2.15 Calculation For Density

And Relative Molar Mass 9

1.2.16 Calculation Surface

Tension (Σ) 10

1.2.17 Column Diameter 11

1.2.18 Provisional Plate Design 13

1.2.19 Evaluation Design 13

1.2.20 Plate Efficiency ( Emv) 21


1.3.6 Dead Weight of Vessel. 25

1.3.7 Weight of Plates. 26

1.3.8 Weight of Insulation. 26

1.3.9 Wind Loading. 26

1.3.10 Analysis of Stresses At Bottom 27

1.6 DESIGN OF STIFFNESS RING 29

1.7 DESIGN OF DOMED END. 30

1.6 DESIGN FOR THE

SKIRT SUPPORT. 31

CHAPTER 1

DESIGN DISTILLATION COLUMN

1.5 INTRODUCTION

Generally, the function of distillation column is to separate the element inside the

component by vaporization. In this method, the boiling points for each element are

important to know the top product and bottom product.

Basically the more volatile component is discharge at the top in vapour

phase and the heavy component at the bottom as a liquid. In others word, the top

product has low boiling point and the bottom product has higher boiling point

Based on our MTBE plant, multicomponent distillation are using because the

feed more than one component. There are tertiary butyl alcohol (TBA), Propylene

Oxide (PO) and Acetone. Here, distillation column is using to separate the mixture of

tertiary butyl alcohol and Propylene Oxide (PO) which the separation occurred

around the boiling point of the component. In our MTBE plant, tertiary butyl alcohol

is a bottom product that needed as a main material to synthesis MTBE

In designing the distillation column, selected materials are required based on

the characteristic of chemical properties such as the temperature, pressure and

density of component and mechanical properties such as the stress and loading,

Here, we chose stainless steel as material construction to design our distillation

column. Cost of construction and simplicity in design also the important aspect to

design our distillation column.

1.6 CHEMICAL DESIGN OF DISTILLATION COLUMN

Table: 1.1: Feed, distillate and bottom composition

component Feed

Product

Distillate Bottom

Molar

flowrate

(kmol/h)

Mole

fraction

(kmol/kmol)

Molar

flowrate

(kmol/h)

Mole

fraction

(kmol/kmol)

Molar

flowrate

(kmol/h)

Mole

fraction

(kmol/kmol)

TBA 707.9089 0.4950 - - 707.9089 0.9989

PO 707.9089 0.4950 706.9479 0.9844 0.961 0.0011

Acetone 14.3012 0.010 14.3012 0.0156 - -

Total 1430.119 1.0000 721.2491 1.0000 708.8699 1.0000

1.2.21 THE OPERATING LINE

Calculate the reflux ratio (R).

Using the Underwood (1948), equation 11.60 (Coulson & Richardson’s), vol.6 page

525.

∑ x,d ( – θ) = Rm + 1

= the relative volatility of component with respect to reference

component, usually the heavy component.

Rm = minimum reflux ratio

X,d = concentration of component in the tops at minimum reflux.

Table 1.2: Calculation value of θ

component Bottom

Top

Average

x,f xf (θ = 1.5)

xf ( – θ)

(θ = 1.45)

xf ( – θ)

TBA 1.0000 0.0000 1.0000 0.4950 0.4950 -0.9900 -1.1000

PO 4.2098 1.0255 2.6177 0.4950 1.2957 1.1593 1.1097

Acetone 0.0000 0.4540 0.4540 0.0100 0.0045 -0.0043 -0.0046

total 1.000 0.165 0.0051

θ = 1.45 acceptable because this value will approximately ∑ xf ( – θ) = 0.

Table 1.3: Calculation value of Rm

Component xd xd (xd ) / ( – θ)

TBA 0.0000 1.0000 0.0000 0.0000

PO 0.9844 2.6177 2.5768 2.2068

Acetone 0.0156 0.4540 0.0071 -0.0071

Total 1.0000 2.1997

Hence,

The minimum reflux ratio

Rm + 1 = 2.2

Rm = 1.2

Reflux ratio (R) = 1.2

The liquid and vapor streams in the column

Above the feed point:

Vapour flow rate:

Vn = D (R+1)

D = distillate molar flowrate

R = Reflux ratio

Hence,

Vn = 721.2491 (1.2 + 1)

= 1586.7480 kmol/hr

Liquid down flow:

Ln = Vn – D

= 1586.7480 - 721.249

= 865.4990 kmol/hr

Below the feed point:

Liquid flow rate:

Lm = Ln + F

F = feed molar flowrate

Hence,

Lm = 865.4990 + 1430.119

= 2295.618 kmol/hr

Vapour flow rate:

Vm = Lm – W

W = bottom molar flowrate

Hence,

Vm = 2295.618 - 708.8699

= 1586.7481 kmol/hr

The equation for the operating lines below the feed plate:

Ym = (Lm / Vm )(Xm +1) – (W / Vm) (Xw)

= (2295.618 / 1586.7481) ( Xm + 1) – (708.8699 / 1586.7481)

(0.0011)

= 1.45 ( Xm + 1) – 0.00049

The equation for the operating lines above the feed plate:

Yn = (Ln / Vn) (Xn + 1) + (D / Vn) Xd

= (865.499 / 1586.7480) (Xn + 1) + (721.2491 / 1586.7480)

(0.9844)

= 0.55 (Xn + 1) + 0.4475

1.2.22 CALCULATE BUBBLE POINT AND DEW POINT

Using Antoine equation

To get the partial pressure for each component and the value

Log P = A - B / (T+ C)

Ki = Po / P total

Table 1.4: The value of coefficient

Component A B C

TBA 16.8548 2658.29 -95.5

PO 15.327 2107.58 -64.87

HO (Acetone) 16.6513 2940.46 -35.93

Bubble point at the feed

T= 510C or 324k and the operating pressure = 760mmHg

Table 1.5: Calculation for bubble point Temperature

Component Xf = xi Pi (mmHg) Ki ∑ yi = ∑Ki xi

TBA 0.4950 194.7734 0.2563 0.1269

PO 0.4950 1373.3030 1.8070 0.8945

HO (Acetone) 0.0100 651.4247 0.8571 0.0086

1.0000 1.0299

∑ yi = ∑Ki xi value is close to 1.0, accept this temperature. Hence, the bubble point

temperature is 510C.

Dew point (Top column)

T= 350C or 308k and the operating pressure = 760mmHg.

Table 1.6: Calculation for dew point Temperature

Component Yi = xd = xi Pi (mmHg) Ki ∑ xi = ∑yi / ki

PO 0.9844 779.3566 1.0255 0.9600

HO (Acetone) 0.0156 345.0077 0.4540 0.0344

1.0000 0.9943

∑ xi = ∑yi / ki value is close to 1.0, accept this temperature. Hence, the dew point


Bubble point (Bottom column)

T= 830C or 356k and the operating pressure = 760mmHg

Table 1.7: Calculation for bubble point temperature

Component Xi = xw Pi (mmHg) Ki ∑ yi = ∑Ki xi

TBA 0.9989 772.9635 1.0171 1.0159

PO 0.0011 3254.1653 4.2818 0.0047

1.000 1.0206

∑ yi = ∑Ki xi value is close to 1.0, accept this temperature. Hence, the bubble point


1.2.23 CALCULATE THE NUMBER OF STAGES

Using the Connell”s Correlation, equation 11.67 (Coulson & Richardson’s) vol.6

page 549. This equation only considered the viscosity and volatility of each

component to determine the plate efficiency.

Eo = 51 -32.5 Log (µα)

µ = viscosity (mNs/m2)

= relative volatility for light.

Viscosity at average temperature

= (T at top + T at bottom) / 2

= (35 + 83) / 2

= 590C = 332 K

Log (viscosity) = (VISA) x { (1/T) – (1/VISB) }

Table 1.8: Viscosity of TBA, PO and Acetone.

Component VISA VISB µ

TBA 972.10 363.38 1.79

PO 377.43 213.36 0.23

HO (Acetone) 367.25 209.68 0.23

Molar average viscosity of feed, µ = 0.495 (1.79) + 0.495(0.23) + 0.01(0.23)

= 1.0022 mNs/m2.

Relative volatility for light key

α = K light key / K heavy key

Here, the light key is PO and the heavy key is TBA

Table 1.9: Relative volatility

PO Top column Bottom column

ki 1.0255 4.2818

α 1.0255 4.2098

Average relative volatility = (1.0255 + 4.2098)/ 2

= 2.618

Hence, the plate efficiency

Eo = 51 -32.5 Log (µα)

= 51 -32.5 log (1.0022 x 2.618)

= 0.37

Minimum number of stages:

Fenske equation, equation 11.58 (Coulson & Richardson’s) vol.6 page 524

Nm = Log [x LK / x HK] d [x HK / x LK] b

Log α LK

= log [0.9844] [0.9989/0.0011]

Log (2.618)

= 7 stages

The actual stages

= 2Nm – 1

Eo

= 2(7)-1

0.37

= 35 stages

Determination of Feed Point Location

In order to find the feed point location, estimation can be made by using the Fenske

equation to calculate the number of stages in the rectifying and stripping section

separately, but this requires an estimate of the feed point temperature. As an

alternative approach, here I use the empirical equation given by Kirkbride (1944) as

a matter for the same objective.

Log [Nr/ Ns] = 0.206 log [ ( B/D) ( Xf, HK / Xf, LK) (Xb, HK / Xb, HK)2 ]

Where,

Nr = number of stages above the feed, include the condenser

Ns = number of stages below the feed, include the reboiler

B = molar flow bottom product

D = molar flow top product

Xf, HK = concentration of the heavy key in the feed

Xf, LK = concentration of the light key in the feed

Xd,HK = concentration of the heavy key in the top product

Xb, LK = concentration of the light key in the bottom product

Hence,

Xb, HK =1/ B = 1/ 708.8699

= 0.00014

Xb, HK =1/ D = 1/ 721.2491

= 0.00013

So,

Log [Nr/Ns]= 0.206log[708.87/ 721.25) (0.495/ 1.2957) (0.00014/0.00013)2 ]

Log [Nr/ Ns] = 0.206 log (0.4355)

[Nr/ Ns] = 0.0744

Nr = 0.0744Ns

From previous calculation, number of stages, excluding the reboiler = 35

Nr + Ns = 35

Ns = 35 – Nr = 35 – 0.0744Ns

1.0744Ns = 35

Ns = 35 / 1.0744

= 32.58

Hence, we take the location of feed point is at stages 33.

1.2.24 PRESSURE DROP

Pt = (ρl) x g x ht x number of real stages (N)

Assume pressure drop 200 mm liquid per plate

( ρl ) at bottom column =0.9989 (787) + 0.0011(829)

= 787.1 kg/m3

g = 9.81m2/s

ht = 200 x 10-3 m

N = 35

Hence, pt = 787.1 x 9.81 x 200 x 10-3 x 35

= 54050.157 N/m2 or Pa

Top pressure 1 bar = 100 x 103 pa

Estimated bottom pressure = 100 x 103 + 54050.157

= 154050.157 Pa

= 1.54 bar

1.2.25 CALCULATION FOR DENSITY AND RELATIVE MOLAR MASS

Table 1.10: Properties of TBA, PO and Acetone

Component Molecular

weight

Feed

Mol

fraction

Top

product

Mol

fraction

Bot.

product

Mol

fraction

Liquid

density

TBA 74 0.4950 - 0.9989 787.0

PO 58 0.4950 0.9844 0.0011 829.0

HO(acetone) 58 0.0100 0.0156 - 790.0

Calculation for relative molar mass

Feed = 0.495 (74) + 0.495 (58) + 0.010 (58)

= 66.0 kg/kmol

Top product = 0.9844 (58) + 0.0156 (58)

= 58 kg/kmol

Bottom product = 0.9989 (74) + 0.0011(58)

= 74.0 kg/kmol

Calculation for density ( ρ)

Bottom product:

Liquid density (ρL) = 0.9989 (787) + 0.0011(829)

= 787.1 kg/m3

Vapour density (ρv) = (74 / 22.4) (273 / 356) (1.54 / 1.00)

= 3.90 kg/m3

Top product:

Liquid density (ρL) = 0.9844 (829) + 0.0156 (790)

= 828.4 kg/m3

Vapour density (ρv) = (58 / 22.4) (273 / 308) (1.0 / 1.0)

= 2.30 kg/m3

1.2.26 CALCULATION SURFACE TENSION (Σ)

Using Sugden (1924), equation 8.23 (Coulson & Richardson’s) vol.6 page 334

σ = [ {Pch (ρL – ρv)} / M ]4 x 10-12

Where,

σ = surface tension, MJ/ m2

Pch = Sugden’s parachor

ρL = liquid density

ρv = Vapour density

M = relative molecular weight

For mixture

σm = σ1x1 + σ2 x2

Table 1.11: The Sugden’s Parachor for TBA, PO and Acetone

Component Formula Pch

contribution

Top product

Mol fraction

Bot. product

Mol fraction

TBA C4H10O 184.4 - 0.9989

PO CH3(CHCH2)O 137 0.9844 0.0011

HO(acetone) CH3COCH3 137 0.0156 -

Pch at top = 0.9844 (137) + 0.0156(137)

= 137

Pch at bottom = 0.9989 (184.4) + 0.0011 (137)

= 184.3

Hence,

Surface tension at top Column,

σm = {[137 (828.4 - 2.30)] / 58} 4 x 10-12

= 14.5 dyne /cm or 14.5 x 10-3 N/m

Surface tension at bottom column,

σm = {[184.3 (787.1 -3.90)] / 74} 4 x 10-12

= 14.5 dyne / cm or 14.5 x 10-3 N/m

1.2.27 COLUMN DIAMETER

F LV = Lm,n { ρv / ρL)0.5

Vm,n

Where, FLV is a liquid flow factor.

FLV (Top) = 0.55 {2.30 / 828.4)0.5

= 0.029

Where, 0.55 is the distillate operating line.

FLV (Bottom) = 1.45 {3.90 / 787.1)0.5

= 0.102

Where, 1.45 is the bottom operating line.

Assume initially 0.9m of tray spacing, to know that flooding occurred or not.

From figure 11.27 (Coulson & Richardson’s, Vol. 6 page 567) we get the value of K i.

(Please refer APPENDIX A-1)

Hence,

Bottom Ki = 1.04 x 10-1

Top Ki = 1.05 x 10-1

Take hole active area 10 %,

Bottom Ki = 1.04 x 10-1 x 1.0 = 0.104

Top Ki = 1.05 x 10-1 x 1.0 = 0.105

Correction for surface tension,

= K1 x (σ / 0.02)2

Which, liquid surface tension is 0.02 N/m

Bottom Ki = (14.5 x 10-3 / 0.02)0.2 x 0.104 = 0.098

Top Ki = (14.5 x 10-3 / 0.02)0.2 x 0.105 = 0.099

Flooding velocity,

Uf = Ki √ (ρL – ρv) / ρv

Bottom Uf = 0.098 √ (787.1-3.9) / 3.9 = 1.39 m/s

Top Uf = 0.099 √ (828.4 – 2.3) / 2.3 = 1.87 m/s

The flooding percentage was assumed to be 90%, this is based on flooding velocity

for design, a value of 70 % to 90 %.

Bottom Uf = 1.39 m/s x 0.9 = 1.25 m/s

Top Uf = 1.8 m/s x 0.9 = 1.68 m/s

Maximum volumetric flow rate,

Bottom = { Vm ( RMM)} / ρv

RMM = Relative molecular mass

Bottom = 1586.7481 x 74 = 8.36 m3/s

3.90 x 3600

Top = { Vn ( RMM)} / ρv

= 1586.7480 x 58 = 11.12 m3/s

2.30 x 3600

Net area required = Volumetric flow rate / flooding velocity

Bottom = 8.36 / 1.25 = 6.69 m2

Top = 11.12 / 1.68 = 6.62 m2

As first trial, take downcomer area as 10% of total column cross sectional area

Bottom = 6.69 / 0.90 = 7.43 m2

Top = 6.62 / 0.90 = 7.36 m2

Column Diameter

Hence, Column cross sectional area, A = (diameter / 2)2

Bottom = √ (7.43 x 4) / = 3.08 m

Top = √ (7.36 x 4) / = 3.06 m

For the design take the whole diameter as 3.00 m.

1.2.28 PROVISIONAL PLATE DESIGN

Column diameter Dc = 3.00 m

Column area Ac = 7.07 m2

Downcomer area Ad = 0.10 x 7.07 (at 10%) = 0.707 m2

Net area An = Ac – Ad = 7.07 – 0.707 = 6.36 m2

Active area Aa = Ac – 2Ad

= 7.07 – 2(0.707) = 5.66 m2

Hole area Ah = 0.566 m2 (take 10% of Aa as first trial)

Weir length

(Ad / Ac) x 100% = 0.707/ 7.07 x 100% = 10%

(From figure 11.31, Coulson & Richardson’s, vol. 6 page 572),


Iw = 0.73 x Dc

= 0.75 x 3.00 = 2.19m

Take weir height, hw = 50mm

Hole diameter, dh = 5mm

Plate thickness = 5mm

1.2.29 EVALUATION DESIGN

Check weeping (enough vapour to prevent liquid flow through hole).

Max. Liquid flowrate = 2295.618 kmol/hr x 74 kg/kmol

3600 s

= 47.19 kg/s

Minimum Liquid rate = 0.7 x 47.19 kg/s (70% turn down ratio)

= 33.03 kg/s.

Weir Liquid crest

how = 750 (Lw / (ρL x Iw) 2/3

Where, Iw = weir length

Lw = liquid flow rate, kg/s

ρL = Liquid density

Maximum, how = 750 {47.19 / (787.1 x 2.19)}2/3

= 68.13 mm liquid

Minimum, how = 750 {33.03 / (787.1 x 2.19)}2/3

= 53.17 mm liquid

At minimum rate, clear liquid depth,

how + hw = 53.71 + 50 = 103.71 mm liquid.

From figure 11.30, Coulson & Richardson’s, vol. 6, page 571)


When, how + hw = 103.71 mm liquid

k2 = 31.0

Weep point

The purpose to calculate this weep point is to know the lower limit of the operating

range occurs when liquid leakage through the plate holes becomes excessive.

Minimum vapour velocity through the holes based on the holes area.

Uh (min) = k2 - 0.9(25.4 – dh)

(ρv)1/2

= 31- 0.9 (25.4 – 5)

(3.9)1/2

= 6.40 m/s

Actual minimum Vapour velocity,

= minimum vapour rate / Ah

= (0.7 x 8.36) / 0.566

= 10.34 m/s

So, minimum operating rate will above weep point.

Plate pressure drop

Dry plate drop ( hd)

Maximum vapour velocity through holes

Uh (Max.) = volumetric flow rate / hole area (Ah)

= 8.36 / 0.566 = 14.77 m/s

From figure 11.34, (Coulson & Richardson’s), Vol 6 page 576.

For the thickness / hole diameter = 1

Ah / Ap = Ah / Aa = 0.566 / 5.66

= 0.1


So, (Orifice coefficient) Co = 0.84

hd = 51(Uh / Co)2 (ρv/ ρL)

= 51(14.77/0.84)2(3.9/ 787.1)

= 78.13 mm liquid

Residual head ( hr) = (12.5 x 103) / ρL

= (12.5 x 103) / 787.1

= 15.88 mm liquid

Pressure drop per plate ( ht) = hd + (hw+how) + hr

= 78.13+ (50+68.13) + 15.88

= 212.14 mm liquid

Note: 200 mm liquid was assumed to calculate the bottom pressure. The calculation

could be repeated but the small change in physical properties will have effect on the

plate design. Hence, 212.14 mm liquid per plate is considered acceptable.

Down comer liquid back–up

Downcomer pressure loss.

Take hap = hw - 10mm

= 50 - 100 = 40 mm

Where, hap is the height of the bottom edge of the apron above the plate.

Area under apron, Aap = hap x Iw

= 40 x 10-3 x 2.19 m2

= 0.0876 m2

Where, Aap is the clearance area under downcomer.

As this less than Ad = 0.707 m2

hdc = 166 ( Lwd / ρL Am)2

Where, hdc = head loss in downcomer, mm

Lwd = liquid flow-rate in downcomer, kg/s

Am = either the downcomer area Ad / the

clearance area under the downcomer

Aap, whichever is the smaller.

Hdc = 166 (47.19 / 787.1 x 0.0876)2

= 77.76 mm.

Back up in the downcomer, (hb),

Hbc = hw + how + hf + hdc

= 50 + 68.13 + 212.14 + 77.76

= 408.03 mm liquid @ 0.408 m.

Check,

0.408 ‹ 1/2 (plate spacing + weir height)

‹ 1/2 (0.9 + 0.05) m

‹ 0.475 m

So, tray spacing is acceptable, (to avoid flooding).

Residence time, t

Sufficient residence time must be allowed in the downcomer for the entrained

vapour to disengage from the liquid stream, to prevent heavily “aerated “liquid being

carried under the downcomer. A time at least 3 seconds is recommended.

Check residence time:

tr = (Ad x hbc x ρL ) / Lwd

= (0.707 x 0.408 x 787.1) / 47.19

= 4.81 s.

tr is greater than 3.0 which recommended so tr here is satisfactory.

Check Entrainment.

Actual velocity, Uv = Uf( max.) / An(net area)

= 8.36m3 /s / 6.36m2

= 1.31 m/s

% Flooding, = Un actual velocity (based on net area)

Uf

= (1.31 / 1.39) x 100

= 94%

Flv (bottom) = 0.102

From figure 11.29 (Coulson & Richardson’s), Vol.6 page 570


θ

Ψ value = 0.071

Ψ value is below 0.1, so the column diameter, which is proposed earlier, is

acceptable.

Trial Layout

Use cartridge type construction.

Allow 50 mm unperforated

50 mm wide calming zone.

50 mm

Iw=2.19

Dc = 3.0m

50 mm

Figure 1.1: Trial layout of plate

Perforated area:

From figure 11.32 (Coulson & Richardson’s), Vol.6 page 573.

At Iw / Dc = 2.19 / 3.00

= 0.73

θc = 960

Angle subtended at plate edge by unperforated strips

= 180 – 98

= 840

Mean length, unperforated edge strips

= (3.00 – 50 x 10-3) x x 84/180

= 4.32 m

Area of unperforated edge strips,

= 50 x 10-3 x 4.32m

= 0.216 m2

Area of Claming zone,

= 2 x 50 x 10-3 (3.00 - 50 x 10-3) sin (98/2)

= 0.22 m2

Total area available for perforation, Ap:

Ap = Active area - (area of unperforated edge + area of calming)

= 5.66 - (0.216 + 0.220)

= 5.22 m2

Ah / Ap,

= 0.566 / 5.22 = 0.11

From figure 11.33, (Coulson & Richardson’s) vol.6 page 574.


Ip / dh,

= 2.85

Satisfactory, range normally within 2.5 – 4.0

Number of holes:

Area of one hole = 1.964 x 10-5 (with diameter 5mm)

Number of holes = 0.566 / 1.964 x 10-5

= 28819

Plate Specification

50 mm

Iw=2.19 m Dc =3.0m

50 mm

Figure 1.2: Plate specification.

Plate no. = 1 (from bottom column)

Plate ID = 3.00

Hole size diameter = 5 mm

Hole pitch = 12.5 Δ

Active holes = 28,819

Turn down ratio = 70% at max.Liquid

Plate Material = stainless

Downcomer material = stainless

Plate spacing = 0.9 m

Plate thickness = 5 mm

Plate pressure drop = 212 mm liquid.

1.2.30 PLATE EFFICIENCY ( EMV)

Using Van Winkle correlation, equation 11.69 (Coulson & Richardson’s) vol.6 page

551

Emv = 0.07 Dg 0.14 Sc 0.25 Re 0.08

Where,

Emv = Plate efficiency

Dg = surface tension number = σL / (µL µV)

µV = superficial vapour velocity

σL = liquid surface tension

µL = liquid Viscosity

Sc = liquid schimdt number= µL / (ρL DLK)

ρL = liquid density

DLK = Liquid Diffusivity, Light Key Component

Re = Reynol’s number = (hw uv ρv) / (µL FA)

Hw = weir height

Ρv = Vapour density

FA = fractional area = area of holes

Total column cross sectional area

To calculate liquid diffusivity (for light key component)

Using equation develop by Wilke and Chang (1955), to predict the liquid diffusivity.

DL = 1.173 x 10 -13 ( ØM) 0.5 T

µVm 0.6

DL = liquid diffusivity

Ø = an association factor for solvent

(Take Ø as 1.0 for unassociated solvent)

M = Molecular weight of solvent

µ = viscosity of solvent

T = temperature, K

Vm = molar volume of solute at its boiling point, m3 / kmol

Here, solvent is TBA.

To estimate Viscosity of solvent (TBA)

Estimation based at temperature 830C or 356 K (bottom column temperature)


= 972.1 x { 1/356 – 1/363.8)

= 0.0586

Viscosity = 1.14 mNs/m2.

Actual viscosity,

= 1.14 x 0.9874

= 1.1256 mNs/m2.

Molecular weight of TBA at bottom column = 74.0 kg / kmol.

Note: reference for viscosity calculation from physical property data, C&R,vol 6

To estimate the molar (Vm) of the solute (PO) at its boiling point

This can be estimated from the group contributions given in table 8.6 (C & R),Vol. 6

page 333.

Propylene oxide, (PO),

Formula : C3H6O

Table 1.12: Calculation of molar volume PO

Atom Volume Number of Volume x Number of

C 0.0148 3 0.0444

H 0.0037 6 0.0222

O 0.0074 1 0.0074

total 0.0740

Therefore, actual molar volume ( Vm) of PO = 0.0011 x 0.0740

= 8.14 x 10-5 m3 / kmol

Liquid diffusivity, DL

DL = 1.173 x 10 -13 x (1.0 x 74)0.5 x 356

1.1256 x (8.14 x 10-5)0.6

= 9.07 x 10 -8 m2 /s

To calculate liquid viscosity (at bottom )


Using the T = 356 K (Bottom Temperature)

Table 1.13: Viscosity of TBA and PO at 356 K temperature

Component VISA VISB µ

TBA 972.10 363.38 1.145

PO 377.43 213.36 0.195

Liquid viscosity mixture (µ L)

µL = 0.9989 (1.145) + 0.0011 (0.145)

= 1.144 mNs/m2

Superficial vapour velocity ( Uv)

Uv = vapour volumetric flow rate


= 8.36 m3/s / 7.07 m2

= 1.18 m/s

FA (Fractional area),

= hole area


= 0.566 m2 / 7.36 m2

= 0.0769

Liquid surface tension σL = 14.6 x 10 -3 N/m

Dg = surface tension number

= σL / (µL µV)

= 14.5 x 10 -3

1.18 x 1.144 x 10 -3

= 10.74

Sc (Liquid Schmidt number) = µL / (ρL DLK)

= 1.144 x 10 -3 / (787.1 x 9.07 x 10-8 )

= 16.02

Reynolds’s number = (hw uv ρv) / (µL FA)

50 x 10-3 x 1.18 x 3.9

0.0769 x 1.144 x 10 -3

= 2615.56

Therefore, plate efficiency,

Emv = 0.07 Dg 0.14 Sc 0.25 Re 0.08

= 0.07(10.74)0.14 (16.02)0.25 (2615.56) 0.08

= 0.37 x 100

= 37%

1.3 MECHANICAL DESIGN

Column design specification

Total column height = tray spacing x no. of stages

= 0.9 x 35

= 31.5 m

Allow, 2 m for clearance height

= 31.5 + 2

= 33.5 m

Internal Diameter, Dc = 3.00 m

Operating pressure,

Top column = 1.0 bar

Bottom column = 1.39 bar

Take column operating at = 1.39 bar

Material of column = stainless steel

Operating temperature = 350C to 83 0C

Tray type = sieve tray (35 trays)

Material of type = stainless steel

Insulation column = mineral wool 75 mm thick

Design stress (σ des) = 175 N/mm2

Take design pressure as 10% above operating pressure,

= 1.39 x 1.1

= 1.529 bar 0r 0.1529 N/mm2

Minimum thickness required for pressure loading (t),

t = (ΔP x Dc) / (2 σ des – ΔP)

= 0.1529 x 3.00x 103

2 (175) – 0.1529

= 8.57 mm.

A much thicker wall will be needed at the column base to withstand the wind and

dead weight load. As a trial, divide the column into two sections with the thickness

increasing by 2 mm per section. Try 8 and 12 mm with mean thickness 10 mm.

1.3.11 Dead weight of vessel.

Wv = 240 x Cv x Dm x ( Hv + 0.8 Dm ) t x 10-3 kN

Where,

Wv = total weight of shell, excluding internal fitting such as

plates

Cv = a factor to account for the weight of nozzles, man ways and

internal supports. (In this case for distillation column take Cv as 1.15).

Dm = mean diameter of vessel ( Dc + t x 10-3 ) m

Hv = height or length between tangent lines, m

t = wall thickness, m

Note: (Equation above applies strictly to vessel with uniform thickness, but it can

be used to get rough estimation of the weight of the vessel.)

First try take,

T = 10 mm or 10 x 10-3 m.

Cv = 1.15

Dm = (3.00 + 10 x 10-3 ) = 3.01m

Hv = 33.5 m

Wcv = 240 x 1.15 x 3.01 (33.5 + 0.8(33.5)) x 10 x 10-3

= 500.9 kN

1.3.12 Weight of plates.

Plate area = ( / 4) x 32 = 7.07m2

Weight of plate = 1.2 x 7.07 = 8.48 kN.

(Where, 1.2 is factor for contacting plates, steel including typical liquid loading in

kN/m2 )

For 35 plates = 35 x 8.48

= 296. kN

1.3.13 Weight of insulation.

Mineral wool density = 130 kg/m3.

Approximate volume of insulation = x 3.0 x 33.5(75 x 10-2)

= 23.68 m3.

Weight = 23.68 x 130 x 9.81

= 30199.1 N or 30.20 kN.

Double this value to allow for fitting = 60.4 kN.

Total weight ( Wv ),

Shell = 500.9 kN

Plates = 296.8 kN

Insulation = 60.40 kN

Total = 858.1 kN

1.3.14 Wind Loading.

Dynamic wind pressure = ½ x Cd x a x Uw2

For smooth cylinder = 0.05 Uw

2

Design for 160 km/hr = 0.05 x (160)2

= 1280 N/m2.

Mean diameter, including insulation = 3.0 + 3.0 (10+75) x10-3.

= 3.26 m

Loading per unit length, Fw = 1280 x 3.26

= 4172.8 N/m.

Bending moment at bottom tangent line, Mx :

Mx = ∫ ox Fwxdx

Where x = Hv = 33.5 m (column height).

Mx = Fw (x2 / 2)

= 4172.8 {(33.5) 2 / 2}

= 2,341,462.4 N/m.

1.3.15 Analysis of stresses at bottom.

At bottom tangent line:

Pressure stresses:

L = PD / 4t

Where P = operating pressure = 0.1529 N/mm2

D = column diameter = 3.00 m

T = thickness = 12 mm.

L = 0.1529 x 3x103 / (4 x 12)

= 9.56 N/mm2

h = PD / 2t

= 0.1529 x 3x103 / (2 x 12)

= 19.11 N/mm2

Dead weight stress

w = Wv ( Di + t ) t

= 858.1

x ( 3x103 + 12) 12

= - 7.56 N/mm2

(- ve sign because compressive stress)

Bending stress

b = M / Iv ( Di / 2 + t )

Where Iv is the second moment of area,

Iv = / 64 (Do4 - Di

4 )

Do = 3000 + 2 x 12 = 3024 mm

Di = 3000 mm

Iv = / 64 x (30244-30004) =1.29 x 1011mm4

b = 2,341,462.4 x 103 x (3000/ 2 + 12)

1.29 x 1011

= 27.44 N/mm2.

The resultant longitudinal stress is

z = L + w b

w is compressive stress and therefore –ve sign.

z (upwind) = 9.56 - 7.56 + 27.44

= 29.44 N/mm2

z (downwind) = 9.56 - 7.56 - 27.44

= - 25.44 N/mm2

As there is no torsional shear stress, the principle stresses will be z and h. The

radial stress r is negligible.

Check failure against greatest difference of the principle stresses:

i.e 19.11 - (- 25.44) = 44.55 N/mm2

Design stress = 175 N/mm2. ( stainless steel ).

Failure is well below the design stress.

Check elastic stability (buckling).

Critical buckling stress, c :

c = 2 x 104 ( t / Do) N/mm2

= 2 x 104 ( 12 / 3024)

= 79.37 N/mm2.

When the vessel is not under pressure (where the maximum stress occur)

= w + b

= 27.44 + 7.56 = 35.0 N/mm2.

The maximum stress is well below the critical buckling stress. Hence, design is

satisfactory.

1.8 DESIGN OF STIFFNESS RING

Take Rings = 75 mm wide.

Rings = 10 mm deep.

Plate spacing = 0.9 m.

Take design pressure as 1 bar external or 105 N/m2.

The load each ring :

Fr = PcLs

Where, Pc = External pressure

Ls = Spacing between the rings

So, the load per unit length on the ring

Fr = 105 N/m2. x 0.9 m

= 0.9 x 105 N/m.

Taking E (young’s modulus) for the steel at temperature 830C (column maximum

operating temperature) as 150.000 N/mm2 or 1.5 x 1011 N/m2 and using a factor of

safety of 6, the second moment of area of the ring to ovoid buckling is given:

PcLs = 24 E Ir

Dr3 x factor of safety

Where, Ir = second moment of area of the ring cross-section

Dr = diameter of the ring (approximately equal to the shell outside

diameter) = 3.00 m.

0.9 x 105 N/m = 24 x 1.5 x 0.9 x 1011 Ir

33 x 6

Ir = 4.5 x 10-7 m4

For the rectangular section, the second moment of area is given by:

I = breath x depth3

12

So, Ir for the support rings = 10 x (75) x 3 10-12

12

= 3.5 x 10-7 m4

And the support rings is adequate size to be considered as a stiffening

Ring,

L’ = 0.9 / 3 = 0.3Do

Where L’ = plate spacing

Do = internal diameter

Do = 3000 / 10 = 300

t

Where t = column shell mean thickness (10 mm).

From figure 13.16, (Coulson & Richardson’s), vol.6 pg. 825,


Kc = 101

From equation 13.52, (Coulson & Richardson’s), vol.6 pg. 751,

Pc = Kc x E x (t / Do)3.

= 101 x 1.5 x 1011 x (10 / 3000)3.

= 5.6 x 105 N/m2.

This is above the maximum design pressure of 1.0 x 105 N/m2. So, design of the

support rings to support the plate is satisfied.

1.9 DESIGN OF DOMED END.

Taken an ‘standard’ ellipsoidal head, ratio major: minor axes = 2 : 1.

This type of head is chosen because it would be the most economical.

Material of construction is stainless steel.

e = Pi Di

2 J f – 0.2 Pi

Where, e = minimum thickness of the plate required

Pi = internal pressure, 0.1529 N/mm2

Di = internal diameter, 3.00 m

f = design stress , 175 N/mm2

J = Joint factor ( for ellipsoidal head J= 1 )

Therefore, minimum thickness required,

e = 0.1529 x 3.0 x 1000

2 x 1x 175 - 0.2 (0.1529)

= 1.31 mm

Add 2 mm for corrosion allowance = 3.31 mm. say 4.0 mm.

So, thickness for the Domed End with ellipsoidal head is 4 mm.

1.6 DESIGN FOR THE SKIRT SUPPORT.

Design for a straight cylindrical skirt, ( s - 900 )

Material of construction stainless steel

Design stress = 175 N/mm2

Young’s modulus = 150,000 N/mm2

The maximum dead weight load on the skirt will occur when the vessel is full with

TBA.

Approximate weight = / 4 x 3.02 x 33.5 x 787.1 x 9.81

= 1828420.65 N = 11828.42 kN.

Weight of vessel from previous calculation = 858.1 kN.

Total weight = 1828.42 + 858.1

= 2686.52 kN

Wind loading from previous calculation = 4.17 kN

Take skirt support height as 1 m,

Bending moment at base skirt=4.17 x (Column height+skirt support height)2

2

= 4.17 (33.5 + 1)2 / 2

= 2481.67 kNm.

As a first trial, take skirt thickness as the same that of the bottom section of the

vessel, 12 mm. The skirt thickness must be sufficient to withstand the dead weight

loads and bending moments imposed on it by the vessel; it will not be under the

pressure vessel.

The resultant stresses in the skirt will be :

s (tensile) = bs - ws

and s (compressive) = bs + ws

where bs = bending stress in the skirt

ws = dead weight stress in the skirt

bs = 4 Ms

( Ds + ts ) ts Ds

Where, Ms = maximum bending moment, evaluated at the

base of the skirt (due to the wind, seismic and

Eccentric loads)

Ds = inside diameter of the skirt, at the base

ts = skirt thickness

bs = 4 x 2481.67 x 103 x 103

(3000 + 12) 3000 x 12

= 29.14 N/mm2

ws (test) = W

( Ds + ts ) ts

= 1828.42 x 103

(3000 + 12) 12

= 16.10 N/mm2

ws (operating) = 858.1 x 103

(3000 + 12) 12

= 7.56 N/mm2

Maximum s (compressive) = 29.14 + 16.10

= 45.24 N/mm2

Maximum s (tensile) = 29.14 - 7.56

= 21.58 N/mm2

Take joint factor as 0.85,

Criteria for design:

s (maximum, tensile) < fs J sin

21.58 < 0.85 x sin 90o

21.58 < 148.7

s (maximum, compressive) < 0.12 E ( ts / Ds ) sin

45.24 < 0.125 x 150000 x (12/3000) sin 90

45.24 < 75

Both criteria are satisfied. Add 2 mm for corrosion allowance.

Therefore for the design thickness = 14 mm.

REFERENCES

R. K. Sinnott. 2000. Chemical Engineering Design. Volume 6. Third Edition. Great

Britain. Butterworth Heinmann.

H. Perry and W.Green, 1998. Perry’s Chemical Engineer’s Handbook, Seventh

Edition, United State.

Douglas, James M., 1988,Conceptual Design of Chemical Processes, Singapore,

McGraw-Hill Book Co.

Elvers B.,1989, Ullman’s Encyclopedia of Industrial Chemistry, Volume 13,

Germany, VCH Verlagsgesellschuft.

Scott, Doug and Crawley Frank. 1992, Process Plant design and Operation,

Warwickshire, UK, Institution of Chemical Engineers.

Fogler, H.Scott, 1999, Elements of Chemical Reaction Engineering, Third Edition,

Upper Saddle River, New Jersey, Prentice Hall, Inc.

Ludwig, E. Ernest, 1964, Applied Process Design for Chemical and Petrochemical

Plants, Vol. 1, Houston, Gulf Publishing Company.

LIST OF NOMENCLATURE

Dimensions in M, L, T

Aa Active area of plate L2

Aap Clearance area under apron L2

Ac Total column cross sectional – area L2

Ad Downcomer cross - sectional area L2

Ah Total hole area L2

An Net area available for vapour – liqud disengagement L2

Ap Perforated area L2

Co Orifice coefficient -

D Mols of distillate per unit time MT-1

Dc Column diameter L

dh Hole diameter L

Emv Plate efficiency -

g Gravitational acceleration -

hap Apron clearance LT-2

hb Height of liquid back – up in down comer L

hbc Down comer back – up in term of clear liquid head L

hd Dry plate pressure drop, head of liquid L

hdc Head loss in down comer L

how Height of liquid crest over down comer weir L

hr Plate residual pressure drop L

ht Total plate pressure drop L

hw Weir height L

K1 Constant -

Lm Molar flow rate of liquid per unit area ML-2T-1

Lw Liquid flow rate L2T-1

Lwd Liquid mass flow rate MT-1

Ip Pitch of holes (distance between centre) L

Iw Weir length L

Nm Minimum number of stages -

NT Theoretical number of stages -

pt Total plate pressure drop ML-1T-2

Po Partial pressure ML-1T-2

q Heat to vaporize one mol of feed divided by molar latent heat -

R Universal gas constant L2T2-1

R Reflux ratio -

Rm Minimum reflux ratio -

Ua Vapour velocity based on active area LT-1

Uf Vapour velocity through holes LT-1

Uv Superficial velocity (based on total cross sectional area) LT-1

V Vapour flow rate per unit time MT-1

Vw Vapour mass flow rate MT-1

xi Mole fraction of component I -

xd Mole fraction of component in distillate -

yi Mole fraction of component I -

L Liquid viscosity -

Viscosity of solvent ML-1T-1

L Liquid density ML-1T-1

v Vapour density ML-3

Surface tension MT-2

Dm Mean diameter L

E Young’ Modulus ML-1T-2

Hv Height between tangent L

Pi Internal pressure ML-1T-2

Mx Bending moment at base of the skirt ML-1T-2

Ms Bending moment at point x from free end column ML2T-2

t Thickness of plate (shell) L

ts Skirt thickness L

J Joint factor -

b Bending stress ML-1T-2

w Dead weight stress ML-1T-2

cw Compressive stress ML-1T-2

ws Stress in skirt due to weight of vessel ML-1T-2


HEAT EXCHANGER DESIGN NOOR HARYANI BINTI MUSTAPHA

SUPERVISORS


2. PN. SH. INTAN BAIZURA SYED A.FUAD DATE: 18 MARCH 2004

CHAPTER 2:

HEAT EXCHANGER DESIGN

CONTENTS

TITLE PAGE

CHAPTER 2 HEAT EXCHANGER DESIGN

2.1. INTRODUCTION 34

2.2 SELECTION OF EQUIPMENT 35

2.2.1 Selection of Shell-And-Tube-Type

Of Heat Exchanger 35

2.3 BASIS DESIGN PROCEDURE

OF HEAT EXCHANGER 39

2.4 CHEMICAL DESIGN OF FLOATING

HEAD HEAT EXCHANGER 41


2.4.2 Properties Of Steam And TBA 42

2.4.3 Heat Load 45

2.4.4 Heat Transfer Area 45

2.4.5 Number Of Tubes 46

2.4.6 Tubes Arrangement (Pitch) 47

2.4.7 Diameter Of Shell 48

2.4.8 Tube Side Coefficient, Hi 49

2.4.9 Shell Side Coefficient, Hs 50

2.4.10 Overall Heat Transfer Coefficient, Uo 52



2.5 MECHANICAL DESIGN OF

HEAT EXCHANGER 56


2.6 DESIGN PRESSURE AND TEMPERATURE 57

2.7 MATERIAL OF CONSTRUCTION 58

2.8 DESIGN STRESS 59

2.9 WELDED JOINT EFFICIENCY 60

2.10 CORROSION ALLOWANCE 60

2.11 DESIGN CRITERIA 60

TITLE PAGE

2.11.1 Minimum Thickness Of

Cylindrical Of The Shell 61

2.12 HEADS AND CLOSURE 62

2.1 Design of Domed

Ends-Ellipsoidal Heads 62

2.13 DESIGN LOAD 63

2.14 DESIGN OF NOZZLES 64

2.14.1 Shell Side Nozzles 64

2.14.2 Tubes Side Nozzles 65

2.14.3 The selected tube size nozzle 66

2.14.4 Shell side nozzles 67

2.15 BOLT-FLANGED JOINTS 67

2.16 BAFFLES 68

2.17 SUPPORT DESIGN – SADDLES

SUPPORT 69

CHAPTER 2

CHEMICAL AND MECHANICAL DESIGN OF HEAT EXCHANGER

2.1. INTRODUCTION

Heat exchanger is a device that provides the flow of thermal energy between 2

or more fluids at different temperature. Heat exchanger are used in a wide variety of

application which include power production; process, chemical and etc. In

production of MTBE, heat exchanger is one of the important equipment to design, in

spite of distillation column, reactor and separator. The purpose of this equipment is

to increase or decrease the mixture to the desired temperature that is from

temperature 82oC to temperature 316oC. Type of heat exchanger that has been

chosen is the shell-and-tube heat exchanger.

The shell and tube heat exchanger is the most common of the various types of

unfired heat transfer equipment used in industry. Although it is not especially

compact, it is robust and its shapes make it well suited to pressure operation. Shell-

and-tube heat exchanger gives a lot of advantage, which may include;

a) Good mechanical layout; a good shape for pressure operations.

b) It can provide a large transfer area in a small space.

c) It also can be constructed from a wide range of materials.

d) It can clean easily.

e) It used well-established fabrication technique and design procedure.

Shell-and-tube heat exchangers are built of round tubes mounted in long

cylindrical shell with the tubes axis parallel to that of the shell. One fluids stream flow

through the tube while the other flows on the shell side, across or along the tubes. A

number of shell-and-tube flow arrangements are used in shell-and-tube heat

exchanger depending on heat duty, pressure drop, pressure level, fouling

manufacturing technique and cost, and cleaning problems. Shell-and-tube heat

exchanger is design on a custom basis for any capacity and operating condition and

this is contrary to many heat exchanger types.

2.2 SELECTION OF EQUIPMENT

Basically there are three types of heat exchanger used in industries, which are

a) Shell-and-tube heat exchanger

b) Plate heat exchanger

c) Spiral heat exchanger

Among the three types of exchanger, the one that have been chosen is shell-

and-tubes heat exchanger because it is the most widely used and can be designed

for virtually application. Besides, it also relatively cheaper than other heat exchanger

with a sufficient in its applications.

2.2.1 Selection of Shell-and-Tube-Type of Heat Exchanger

There are various types of heat exchanger used in industries; each one of it can

give its own advantages and disadvantages. Table 2.1 shows the advantages and

disadvantages of each type of heat exchanger:

Table 2.1: The Advantages and Disadvantages of Each Types of Heat Exchanger

Construction Advantages Disadvantages

Non-

removable

Bundle,

Fixed Tube

Sheet

Less costly

Give maximum

heat transfer

surface per given

size of shell and

tubes

Shell side can be cleaned only

by chemical means.

Table 2.1: The Advantages and Disadvantages Of Each Types Of Heat Exchanger

(Continue)


Provides multi-

tube pass

arrangement

Removable Bundle,

Packed Floating Tube

Sheet

Shell side can be

mechanically

cleaned

Bundle can be

easily replaced or

repair

Shell side fluids

limited to non

volatile

Tube side

arrangement

limited to one or

Less costly than

pull, internal

floating head

types

Maximum surface

per given shell

and tube size

2 passes

Tubes expand as

a group, not

individually, so

sudden shock’s

should be

avoided

Limits design

pressure and

temperature

Removable Bundle,

Internal Clamp ring,

Types Floating head

cover.

Good for

handling

flammable or

toxic fluids

High surface per

given shell and

tubes size

Provides multi

tubes pass

arrangement

More costly than

fixed tube sheet

or U tube heat

exchanger design

Shell cover,

clamp ring and

floating head

cover must be

removed prior to

removing the

bundle. Results

Table 2.1: The Advantages and Disadvantages of Each Types of Heat Exchanger

(Continue)


in higher

maintenance cost

Removable Bundle U

tube

Less costly than

floating head or

packed floating

tube sheet design

Provides multi

tube pass

arrangement

High surface area

Tube side only

can be cleaned

by chemical

means

Individual tube

replacement is

not practical

Cannot made

Capable of

withstanding

thermal shock

single tube pass

Draining tube

side difficult in

vertical position

TEMA (Tubular Exchanger Manufacturer Association) give classification of heat

exchanger. Table 2.2 give the types of heat exchanger that have been chosen and

reasons why it being selected.

Table 2.2: Selection of Heat Exchanger

Type Reasons of Selection

Front End

Stationary

Head Types

Type A

Channel and

removable

cover

Good for frequent cleaning of

tubes

Shell Types Type E

One pass

shell

The most commonly used in

industries

More cheaper and simple

Table 2.2: Selection of Heat Exchanger (Continue)

Type Reasons of Selection

Rear Ends Head Types Type T

Pull through

floating head

Give a smaller

number of tubes

Reduced

maintenance time

because the

bundle can be

withdrawn from

the shell without

removing shell or

floating-head

covers

2.3 BASIS DESIGN PROCEDURE OF HEAT EXCHANGER

An algorithm for the design of shell-and tube exchanger is shown in figure 2.1.

Figure 2.1: Design Procedure of Shell-and-Tube Heat Exchanger

Step 1

Specification

Define duty

Make energy balance if needed

to calculate unspecified flow

rates of temperature

Step 10

Decide baffle spacing and

estimate shell-side heat transfer

coefficient

Calculate overall heat transfer

coefficient including fouling

factors, Uo, calc

NO

Step 11

asso

assocalco

U

UU

,

., %300

Step 2

Collect physical properties

Step 3

Assume value of overall

coefficient Uo, ass

Step 4

Decide number of shell and

tubes passes. Calculate ,

correction factor, F, and

Step 5

Determine heat transfer area

required:

Step 6

Calculate number of tube

Estimate tube and shell-side

pressure drop

Set Uo, ass = Uo, cal

Estimate cost of heat

exchanger

Step 12

Step 8

Pressure drop

within

specification?Step 13

YES

NO

YES

(Sources: Coulson & Richardson, Vol. 6, 2002)

2.4 CHEMICAL DESIGN OF FLOATING HEAD HEAT EXCHANGER

2.4.1 DESIGN SPECIFICATION

Fresh feed TBA to the plant is 52385.26 kg per hour. The stream

specification at the vaporization stage can be obtained from the material and energy

balance. At this stage, the first heat exchanger is used to heat up the TBA stream

from 82oC to 316oC before entering the first reactor. The process operation is shown

above;

Calculate shell diameter

Step 9

Estimate tube side heat transfer

coefficient

Can design be

optimized to

reduce cost?YES

NO

Accept design

Decide type, tube size, material layout.

Assign fluids to shell or tube sideStep 7

Step 14

Figure 2.2: Process Operation Of Shell And Tube Heat Exchanger

But because the temperature used to heat up the TBA is 316oC, then, 2 units of heat

exchanger are needed to get the desired temperature. The process operation will

becomes as follows;

Figure 2.3: Heat Exchanger Is Separate Into 2 Heat Exchanger In Series

2.4.2 PROPERTIES OF STEAM AND TBA

Assumption

1) Heat losses are negligible

2) The rate of each fluid flow is constant.

3) The specific heat of each flux flow is constant

4) All steam have been condensed

E-101

Tin = 82oC

Tout = 316oC

E-101

Tin = 82oC

Tout = 316oCtout = 250oC

tin = 110oC

E-101

Flow rate of steam= 2500 kg/hr

= 0.69 kg/s

Properties of Steam at Temperature = 250 o C & 110 o C

Temperature

Item Units T=250oC T=110oC

Flow rate kg/s 2500 2500

Specific Heat, cp kJ/kg.oC 1.9898 2.06

Dynamic viscosity, kg/m.s 1.776E-05 1.271E-05

Thermal conductivity, k W/m.oC 0.0355 0.0246

Density of fluids, kg/m3 0.4245 0.5863

Latent heat kJ/kg 1716 2230

Properties of Steam at Mean Temperature Tm= 180 o C

Item Units Temperature (180oC)

Flow rate kg/s 0.6944

Specific Heat, cp kJ/kg.oC 1.98

Dynamic viscosity, kg/m.s 1.525E-05

Thermal conductivity, k W/m.K 0.0299

Density of fluids, kg/m3 0.4902

Latent heat kJ/kg 2015

(Source: Incropera Dewitt, 2002) (Refer APPENDIX 13)

Properties of TBA

Temperature (oC)

Item Units 82 119

Flow rate kg/h 52385.26 52385.26

Specific Heat kJ/kg.oC 3.50 2.44

Dynamic viscosity kg/m.s 1.5500E-03 7.06E-06

Thermal conductivity, k W/m.K 0.11 0.1018

Density of fluids kg/m3 705.00 650.00

(Source: R.W.Gallant, Vol.1, 1992)

Mean Temperature

Mean temperature different :

= (2.1)

where;

= True temperature different

Ft = temperature correction factor

= Logarithmic mean temperature

=

(2.2)

where;

t1 = inlet shell-side fluid temperature

t2 = outlet shell-side temperature

T1 = inlet tube-side temperature

T2 = outlet tube-side temperature

= 66.8oC

Using Figure 12.19 (Coulson & Richardson, Vol. 6) the temperature correction

factor can be obtained

(2.3)

= 2.7

(2.4)

= 0.22

From figure 12.19 (Coulson & Richardson, Vol. 6) (Refer APPENDIX B-1)

Ft = 0.92

Substitute value above into equation 1.1

= 0.92 (66.8)

= 61.5oC

2.4.3 HEAT LOAD

Shell side Q = w. Cp (t2 - t1) (2.5)

= (52385.2586/3600) x (3.5) x (119-82)

= 1.88 kW

Tube side Q = w. Cp (t2 - t1)

= (2500/3600) x (1.98) x (250-110)

= 192.5 kW

2.4.4 HEAT TRANSFER AREA

Types of heat exchanger = Shell and tube with floating head

Passes = 1 shell pass & 2 tube passes

The tube layout and tube size of shell and tube heat exchanger with pull-through

floating head are shown above:

Table 2.3: Layout & Tube Size Of Shell And Tube Heat Exchanger

Unit Dimension

Tube Length, L m 4.8

Outer Diameter, OD mm 20

Inside Diameter, ID mm 16

Pitch, Pt mm 25

Birmingham wire gage

(BWG)

- 14

Birmingham wire gage (BWG) with value 14 is chosen because it can give

moderate flow area and wall thickness to withstand significant pressure drop.

The hot fluids used is TBA and cold fluids is steam, then from Figure 12.1 assume

the Overall Coefficient,

Uo, ass = 400 W/m2.oC (Refer APPENDIX B-2)

Start with one shell pass and 2 tubes pass

Heat transfer area in tubes:

Area needed, Ao

(2.6)

Ao = 8.38E+06/(900*147.9)

= 62.94 m2

2.4.5 NUMBER OF TUBES

Standard pipe are taken from Table 12.3 (Coulson & Richardson, Vol. 6)

Inside diameter, di = 16 mm

Outside diameter, do = 20 mm

Length of the tubes is assumed as

Length, L = 4.8 m

For a given surface area, the use of longer tubes will reduce the shell

diameter, which will result lower cost of exchanger.

Using a floating head exchanger for efficiency and ease of cleaning

Area of one tube: neglecting thickness of tubes sheets Area,

2.7)

a = 3.142 X 0.02 X 4.8

= 0.3016 m2

Number Of tubes,

Nt = Ao/a

Nt = 245.67

say = 246

So for 2 passes, tube per pass = 246/2

= 123

2.4.6 TUBES ARRANGEMENT (PITCH)

Bundle and Shell diameter: the tube in a heat exchanger arranged in triangular

patterns because it give a low-pressure drop.

The tubes pitch (distance between tubes center) = 1.25

Pt = 1.25 x do (2.8)

= (1.25) x (0.02)

= 0.03 m

Figure 2.4: Triangular pattern

2.4.7 DIAMETER OF SHELL

From Table 14.2 for 2 tubes per pass (Refer APPENDIX B-3)

K1 = 0.249

n1 = 2.207

Bundle diameter

Db = do X (Nt/Kt)1/n1 (2.9)

= 0.02*(246/0.249) 1/2.207

Db = 0.4547 m @ 454.67 mm

For a pull-through floating head exchanger, the typical shell clearance from figure

12.10 is 90 mm (Refer APPENDIX B-4)

Bundle diameter clearance = 90 mm

Shell diameter, Ds

Ds = 0.4547+0.09

= 0.5447 m @ 544.67 mm

Flow

Pt

2.4.8 TUBE SIDE COEFFICIENT, hi

Parameter Units Value

Mean temperature of the

tube t mean

oC = 180

Cross sectional area m2 = 3.142 x d2

= 2.0109E-04

Tube per pass tube/pass = 246/2

= 123

Total flow rate area m2 = (Cross sectional area) x (tube/pass)

= 2.0109E-04 x 123

= 0.0247

Steam mass velocity, Gt kg/m2.s =(Steam flow rate)/(total flow rate area)

= (2500/3600)/0.0247

= 28.1142

Steam linear velocity, ui m/s = (Gt) / (steam density)

= 28.1142 x 0.4902

= 57.3526

Reynolds Number, Re= (2.10)

= (0.4902) x (57.3526) x (0.016)/1.525E-05

= 2.4131x104

Prandtl number, Pr= (2.11)

= (1.98 x 1.525E-05)/0.0299

= 1.0099

Heat transfer factor (Coulson & Richardson, Vol.6)

Ratio of L/di = 300

Reynolds Number, Re = 2.9497E+04 (Refer APPENDIX B-5)

jh = 3.70E-03

Tube side coefficient, hi can be calculated using equation below

hi = (2.12)

= (3.7E-03 x 0.0299 x 2.9497E+04 x 1.0099)/0.016

= 204.6164 W/m2.oC

2.4.9 SHELL SIDE COEFFICIENT, hs

Parameter Formula Value Units

Take baffle spacing

as 1/5 from the shell

diameter, Baffle

spacing,

IB = Ds/5

= 0.5447/5= 0.1089

m

Tube pitch, Pt Pt = 1.25 x do

= (1.25) x (0.02)

= 0.025 m

Flow area, As =

= (0.025-0.02) x (0.5447) x 0.1089

=0.0119 m2

Mass velocity, Gs = Ws / As

= (52385.2586/3600)/0.0119

=1226.2667 kg/m2s

Shell side velocity, us

= (2.13)

= 1226.2667/705

= 1.7394 m/s

Shell side equivalent diameter for triangular pitch arrangement

de = (2.14)

= (1.10/0.02) x (0.0252 – 0.917 x 0.022)

= 0.0142 m

Calculate Reynolds number, Re

Re = (2.15)

= (1226.2667 x 0.0142)/1.55 E-03

= 22368.0704

Prandtl number, Pr

Pr = (2.16)

= (3.5 x 1.55E-03)/0.109

= 21.9377

Choose baffle cut of 25%

REASON: Because generally a baffle at this percentages will be the

optimum, giving good heat transfer rate, without excessive drop

From figure 12.20 (Coulson & Richardson, Vol.6), we can obtained

jf = 4.00E-03 (Refer APPENDIX B-6)

Assume that the viscosity correction is negligible

hs = (2.17)

= 0.109 x4.00E-03 x 22368.0704 x 21.93771/3

= 1817.06

2.4.10 OVERALL HEAT TRANSFER COEFFICIENT, Uo

From table 12.6, the conductivity of metals at temperature

Material of construction = Stainless Steel

Reason = The material, steam in the tube more

corrosive than TBA in the shell. Its also can

corrode the tube wall

Thermal conductivity of

the tube Wall, kw = 16 W/moC

(Refer APPENDIX B-7)

From Table 12.2 (Coulson & Richardson, Vol.6) take dirt Coefficient as


hid = 10000 W/m2oC

hod = 5000 W/m2oC

(2.18)

= 0.00711

Uo = 1/0.00711

= 140.6185 W/m2oC

2.4.11 TUBE SIDE PRESSURE DROP

Reynolds number, Re

=

= (0.4902) x (57.3526) x (0.016)/1.525E-05

= 2.4131x104



jf = 5.50E-01

Neglecting the viscosity correction term

(2.19)

= 2 8(5.5E-01)(4.8/0.016)+2.5 x ((0.4902 x 57.34262)/2)

= 2132432.078 N/m2

= 2132.4321 kPa

= 21.0455 Psi

= 0.0002 Bar

The pressure drop in the tubes is 2132.4321 kPa

2.4.12 SHELL SIDE PRESSURE DROP

Reynolds number, Re

Re = (2.15)

= (1226.2667 x 0.0142)/1.55 E-03

= 22368.0704



jf = 4.00E-03

Shell side pressure drop can be calculated using equation below

(2.20)

= 8(4.00E-03) x (0.5447/0.0142) x (4.8/0.1089) x ((705x1.7394)/2)

= 33158.6806 N/m2

= 33.1587 kPa

= 0.3273 psi

= 0.3316 Bar

The pressure drop in the shell is 33.1587 kPa

Summaries of Chemical Engineering Design Of Heat Exchanger is shown in Table

2.4 above:

Table 2.4: Summary Of Chemical Engineering Design Of Heat Exchanger

Parameter Value

Process Condition :

Heat load, Q

Overall coefficient, Uo

1.88 kW

400 W/m2K

Tube Side: Saturated Steam

Inlet temperature, t1

Outlet temperature, t2

Flow rate, W

Tube inside diameter, ID

Tube outside diameter, OD

BWG

Length, L

Pitch, Pt

Number of tubes, Nt

Passes

Area per pass

Heat transfer coefficient, hi

Pressure drop, ∆Pt

150oC

250 oC

25000 kg/hr

16 mm

20 mm

12

4.8

25 mm

246

2

301.6 mm2

204.6146 W/m2.oC

2132.4321 kPa

Shell Side: TBA

Inlet temperature, T1

Outlet temperature, T2

Flow rate, W

Shell side diameter, Ds

Pass

Baffles spacing with 25% cut, IB

82 oC

316 oC

52385.2586 kg/h

544.6667 mm

1

108.9 mm

Equivalent triangular pitch arrangement, de

Heat transfer coefficient, hs

Pressure drop, ∆Ps

14.2 mm

1858.9766 W/m2.oC

33.4587 kPa

2.5 MECHANICAL DESIGN OF HEAT EXCHANGER

British Standard, BS 3274, covers mechanical design features, fabrication,

and material of construction and testing of shell and tube heat exchanger. The

standard of the American Tubular Heat Exchanger, the TEMA standards, are also

used. The standard give the preferred shell and tube dimensions: the design and

manufacturing tolerances; corrosion allowances; and the recommended design

stresses for material of construction.

The details in designing heat exchanger may include the following;

1. Design Pressure And Temperature

2. Material Of Construction

3. Design Stress

4. Welded Joint Efficiency

5. Corrosion Allowance

6. Minimum Thickness Of Cylindrical Of The Shell

7. Longitudinal Stress

8. Circumferential Stress

9. Minimum Thickness Of Head And Closure

10. Minimum Thickness Of The Channel Cover

11. Design Load

12. Nozzle Design

13. Standard Flange

14. Baffles

15. Saddles Support

2.5.1 DESIGN SPECIFICATION

Table 2.5: Shell-and-tube Specification

Shell Side Tube Side

Types of fluids Tert-Butanol

(TBA)

Types of fluids Steam

Inlet

Temperature, oC

82 Inlet Temperature, oC 250

Outlet

temperature, oC

119 Outlet temperature, oC 110

Internal Diameter,

mm

454.6667 Number of Tubes, Nt 246

Baffles space,

mm

108.9 Length, m 4.8

Pass 1 OD, mm 20

BWG 14

Pitch, mm 25

Passes 2

Flow area per tubes, a

(mm2)

301.6

2.6 DESIGN PRESSURE AND TEMPERATURE

A vessel must be designed to withstand the maximum pressure to which it is

likely to be subjected in operation. Under internal pressure, the pressure is normally

taken as the pressure at which the relieve valve is set; normally 5 to 10 percent

above the normal working pressure. The purpose is to avoid spurious operation

during minor process upsets. The increasing in temperature will decreased the

strength of the metals, where the maximum allowable stress will depends on the

material of the temperature.

By taking a safety factor of 10%


Operating Pressure, bar 11 1

Design Pressure, bar 12.1 1.1

Operating Temperature oC 119 250

Design Temperature, oC 130.9 275

2.7 MATERIAL OF CONSTRUCTION

Selection of a suitable material must be taken into consideration, firstly is the

suitability of the material for fabrication as well as the compatibility of the material

with the process environment.

The most economical material selected for both chemical and mechanical

requirements should be selected; this will be the material that gives lowest cost over

the working life of the plant, allowing for maintenance and replacement.

The applied material selected must suitable for the various specific operation

conditions. A few factors that need to be considered:

Corrosion resistance

Operating conditions

Economic feasibility

Suitability for fabrication (welding)

Process safety.

Table 2.6: Material Selected for Shell-and-tubes

Shell Tubes

Material selected Carbon steel Stainless steel

Advantages Commonly used as a

engineering material

Impregnated with

chemically resistant resins

and used for specialized

equipment especially heat

exchanger

It has high conductivity and

good resistance to most

chemicals (except

oxidizing acid more than

30%)

Since the fluid properties in

the shell are not corrosive,

then the carbon steel have

been chosen

It is more economical and

easy to get compare to

stainless steel

It also easily fabricated

and have high strength

Widely used in

industries

Suitable for

corrosive material

2.8 DESIGN STRESS (NOMINAL DESIGN STRENGTH)

For design purpose it is necessary to decide a value for the maximum

allowable stress that have been decided and accepted in the material of

construction. This is determined by applying a suitable “design stress factor” (factor

of safety) to the maximum stress that the material should be expected to withstand

without failure under standard test conditions.

The typical design stress factor for pressure component:

Material used Design Stress (N/mm2)

Shell: Carbon steel 115 at 119 oC

Tubes: Brass 95 at 250oC

2.9 WELDED JOINT EFFICIENCY

The strength of a welded joint will depend on the types of joint and the

quality of the welding. Normally, the large size of the vessel (large d) is made from

the large plate. The plate will form using the machine and its joint will be welding.

The possible lower strength of a welded joint compared with the virgin plate is

usually allowed for in design by multiplying the allowable design stress for the

material by a “welded joint factor” J. taking the factor as 1.0 implies that the joint is

equally as strong as the virgin plate; this is achieve by radio graphing the complete

weld length and cutting out and remaking any defects.

2.10 CORROSION ALLOWANCE

The “corrosion allowance” is the additional thickness of metal added to allow

for material lost by corrosion and erosion, or scaling. For carbon and low-alloy

steels, where severe corrosion is not expected, a minimum allowance of 2.0 mm

should be used. Most design codes and standards specify a minimum allowance of

1.0 mm.

2.11 DESIGN CRITERIA

Minimum Practical Wall Thickness

A minimum wall thickness is necessary because to ensure that any vessel is

sufficiently rigid to withstand its own weight, and any incidental loads.

Shell inside diameter, di = 544.67 mm @ 0.5447 m

2.11.1 Minimum Thickness of cylindrical of the shell

For a cylindrical of a shell, the minimum thickness required to resist internal

pressure can be determined

From equation

(2.21)

Where;

e = minimum thickness

Pi = design pressure

= 1.21 N/mm2

di = shell diameter

di = 454.6667 [email protected]

f = design stress (From Table 13.2, Coulson &

Richardson, Vol. 6)

f = 85

J = welded joint efficiency

= 1

e = (1.21 x 0.5447)/(2(85)-1.21)

= 3.9 mm @ 0.0039 m

Adding corrosion allowance = 2 mm

e = 3.9 + 2

= 5.9 mm

Round up the number of the thickness

e = 6 mm

So the minimum thickness of the cylindrical shell after adding corrosion allowance is

5.9 mm

2.12HEADS AND CLOSURE

Heads closes the end of a cylindrical heat exchanger. The typical used of heads are

as follows:

Table 2.7: Types of Heads and its applications

Types of heads The applications

1) Flat plates & formed flat heads Used as covers for many man

ways, and as the channel

covers of heat exchanger

Limited to low-pressure and

small-diameter vessel

1) Hemispherical heads The strongest shape

Higher cost

Used for high pressure

1) Ellipsoidal heads Most economical for operation

above 15 bar

1) Torispherical Most common for operation up

to 15 bar

2.12.1 Design of domed ends-ellipsoidal heads

These types of heads have been chosen because it gives economical evaluation

compared to other heads and since it save cost the minimum thickness of the heads

is calculated

The minimum thickness of torispherical head can be calculated by equation

(2.22)

Where



Di = shell diameter

f = design stress

= 70.00 N/mm2

e = (1.21x 544.6667)/2(1)+0.2(1.21)

= 4.7 mm

Adding corrosion allowance

= 4.7 + 2

e = 6.7 mm

The minimum thickness of domed ends is 6.7 mm

2.13DESIGN LOAD

The major sources of dead weight loads are as follows

1) The vessel shell and tubes

2) The fluids to fill the vessel (TBA)

3) The fluid to fill the tubes (steam)

4) The insulator

2.13.1 Dead weight of vessel

A) Weight of the shell

(2.23)

Wv = 1580.56 N

B) Weight of tubes

(2.24)

Wt = 34190.24 N

C) Weight of insulation

Approximate volume of insulation

V = (3.142 x d x L) x t (2.25)

= 0.41m3

Weight

Wi = V g (2.26)

= 523.79 N

Total weight of heat exchanger

WT = Wv + Wt + Wi

= 36294.59 N

= 36.29 kN

2.14DESIGN OF NOZZLES

There are four opening or known as nozzles in one heat exchanger for steam

inlet and outlet and also for TBA inlet and outlet. Designing tube side and shell side

nozzles are based on TEMA heat exchanger standard.

2.14.1 Shell Side Nozzles

Pipe size for TBA at inlet

Material used Mineral Wool insulation

Density, kg/m3 130

Length, m 4.8

Thickness of insulator, mm 50

Shell diameter, m 0.54

Material of construction = Carbon steel

Density TBA inlet = 705.00 kg/m3

Flow rate of TBA, GTBA = 14.55 kg/s

Diameter pipe for TBA inlet, DTBA

(2.27)

DTBA = 107.00 mm

Pipe size for TBA at outlet

Material of construction = Carbon steel

Density TBA inlet = 705.00 kg/m3

Flow rate of TBA, GTBA = 14.55 kg/s

Diameter pipe for TBA inlet, DTBA

(2.28)

DTBA = 107.00 mm

2.14.2 Tubes Side Nozzles

Pipe size for inlet steam

Material of construction = Aluminium Brass

Density Steam inlet = 0.33 kg/m3

Flow rate of Steam, Gsteam = 0.69 kg/s

Diameter of the pipe for steam inlet

(2.29)

D steam, in = 365.21 mm

Pipe size for outlet steam

Diameter of the pipe for steam outlet

Material of construction = Aluminium Brass

Density Steam inlet = 0.37 kg/m3

Flow rate of Steam, Gsteam = 0.69 kg/s

Diameter of the pipe steam outlet

(2.30)

Dsteam,out= 348.85 mm

2.14.3 The selected tube size nozzle

Fluid: Steam

By taking D = 107 mm (4.21”)

Table 2.8: Standard Nozzle for Tube size

Nominal

pipe size, in

Outside

diameter, in

Schedule

No.

Wall

thickness, in

Inside

diameter, in

4 4.5

(114.3 mm)

40ST 0.237

(6.02mm)

4.026

(102.26mm)

2.14.4 Shell side nozzles

Fluid: TBA

By taking D = 365.21 mm (14.38’), the selected shell size nozzle

Table 2.9: Standard Nozzles for Shell Side

Nominal

pipe size, in

Outside

diameter, in

Schedule

No.

Wall

thickness, in

Inside

diameter, in

14 14

(355.6 mm)

ST 0.375

(9.53 mm)

13.250

(336.55mm)

(Source: Robert H. Perry, 1998)


2.15 BOLT-FLANGED JOINTS

A flanged joint are used for connecting pipes and instruments to vessels, for

manhole covers and for removable vessel heads when ease of access is required.

Flanged joints are also used to connect pipes to other equipment, such as pump

and valves. There are several types of flanges used for various applications

(Coulson & Richardson, Vol.6, 1999).

For the design, standard flanges are specified. The standards are adapted

from the British standard (BS 4504) with nominal pressure of 6 bars. Type of flange

chosen is full neck welding neck flange.

Table 2.10: Steel Welding Neck Flanges

Nominal

pipe

Pipe,

o.d

d1

Flanged Raised

face

Bolting Drilling Neck

D b h1 d4 f No d2 k d3 h2 r

100 114.3 210 16 45 148 3 M16 4 18 170 130 10 8

350 355.6 490 22 62 415 4 M20 12 22 445 385 15 12

(All units in mm) (Refer APPENDIX B-11)

Figure 2.3: Typical standard Flange Design

2.16 Baffles

Have two fuction:

To support the tubes for structural rigidity, preventing tube vibration and

sagging

To divert the flow across the bundle to obtain a higher heat transfer

coefficient

Types = Transverse baffle

d4

k

D

de

d3

d1

Baffles thickness = 5.5 mm

Diameter of tubes holes in baffles, Dh

Dh outer diameter of tube = 5.5

= 20 + 5.5

= 25.5 mm

= 0.03 m

Baffles spacing, IB

Ds = 544.67 mm

= 0.5447 m

= Ds/5 mm

IB = 0.11 m

Baffles cut = 25%

No. of Baffles = L/Ds (2.31)

= 8.81

Round up to = 9 Baffles

2.17SUPPORT DESIGN – SADDLES SUPPORT

The methods used to support a vessel will depend on

Design temperature and temperature

Internal and external fitting and attachment

Size and shape

Weight of the vessel

Vessel location and arrangement.

Heat exchanger is mounted with two-saddle support, which must be designed to

carry the weight of the vessel and contents and any super imposed loads such as

wind load.

Table 2.11: Dimension of Selected Standard Steel Saddle.

Vessel Maximum Dimensions (m) mm

Diameter, m Weight, kN V Y C E J G t2 t1 Bolt diam. Bolt holes

0.6 35 0.48 0.15 0.55 0.24 0.19 0.095 6 5 20 25


Table 2.12 above show the summaries of calculation on mechanical design of shell

and tube heat exchanger

Table 2.12: Summary Of Mechanical Engineering Design Of Heat Exchanger

Parameter Value


Design pressure, Pi 12.1 bar 1.1 bar

Design temperature, TD 275 oC 130.9oC

Material of construction Carbon Steel Stainless Steel

Corrosion allowance 2 mm 2 mm

Thickness 6 mm 4 mm

Nozzle diameter 107 mm 365.21 mm

Type of flange Welding neck

Head and closer

Domed and type

Thickness

Ellipsoidal

6.7 mm

Insulation Thickness 31.24 mm

Support type Saddle

Total weight 36294.59 N

REFERENCES

R.K Sinnott.. Chemical Engineering Design. Vol.6. Butterworth Heinemann 1999.

E. AD Saunders, Heat Exchanger Selection, Design and Construction. Longman

Scientific & Technical1988.

J.P Gupta. Working with Heat Exchanger. Hemisphere Publishing Corporation.ical

Industry 1990.

Green W. Don & Perry Robert H. Perry’s Chemical Engineers’ Handbook.

Seventh Edition Kansas. McGraw Hill, 1997.

Sadik Kakac, Hong Tan Liu. Heat Exchanger, Selection, Rating and Thermal

Design. CR C Press.

Dr.Brian Spulding,J.Tab Orela, Heat exchanger Theory and Design Handbook,

McGraw-Hill, 1990

R.W. Gallent and Jay M. Railey, Physical Properties of Hydrocarbon Volume 2

and Volume 1, Gulf Publishing Company, 1992

Yunus A. Cengel, Michael A. Boles, Thermodynamics An Engineering Approach,

Third Edition, McGraw-Hill, 1998

www.yahoo.com

www.google.com

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http://www.yahoo.com/


ISOBUTYLENE REACTOR ROHIZAD BIN JAMEL

SUPERVISORS



CHAPTER 3:

ISOBUTYLENE REACTOR

CONTENTS

TITLE PAGE

CHAPTER 3 ISOBUTYLENE REACTOR

3.1 INTRODUCTION 71


3.2.1 Selection of Catalyst 743.2.2 Effective Diffusivity, De 74

3.2.3 Tube Specification 76

3.2.4 Heat Transfer Calculation 78

3.2.5 Tube Side Coefficient 78

3.2.6 Shell Side Coefficient 80

3.2.7 Overall Heat Transfer Coefficient 82





3.3.2 Design Temperature 85


3.3.4 Corrosion Allowance 86

3.3.5 Thickness of Cylindrical Shell 86

3.3.6 Head and Closures 86

3.3.7 Weight Load 87

3.3.8 Wind Loading 89

3.3.9 Analysis of Stresses 90

3.3.10 Elastic Stability 91

3.3.11 Vessel Support Design 92

3.3.12 General Consideration for

The Design 94

CHAPTER 3

ISOBUTYLENE REACTOR

3.1 INTRODUCTION

The high purity isobutylene can be economically produced by dehydrating tertiary

butyl alcohol (TBA). Vapor phase TBA dehydration have been developed in an

isothermal fixed bed reactor using a silica alumina cracking catalyst. Increased

demand for oxygenates in reformulated gasoline has focused the attention of the

petrochemical industry on isobutylene etherification. Potential sources of isobutylene

are dehydrogenation of isobutene and dehydration of TBA. TBA is a major

byproduct of the process for propylene oxide manufacture from propylene and

tertiary hydroperoxide. TBA can be converted into Methyl Tertiary Butyl Ether

(MTBE) which is environmentally accepted blending component for reformulated

gasoline by a two-step process. In the two-step process, TBA is first dehydrated to a

high purity isobutylene and water and then isobutylene is reacted with methanol to

produce MTBE. Furthermore, it was decided that the heat of reaction to be supplied

by flue gas and that in order to achieve 98% conversion, the temperature and

pressure in the reactor is kept maintain at 316ºC and 20 psig (3 bars) respectively.

Table 3.1: Reactor mass balance for input stream

Components Mass flowrate

(kg/h)

Mass

fraction

(w/w)

Molar flowrate

(kmol/h)

Mole fraction

(kmol/kmol)

TBA 52385.2586 0.9989 707.9089 0.9986

Propylene oxide 55.7380 0.0011 0.9610 0.0014

Total (Gm) 52440.9966 1.0000 708.8700 1.0000

Table 3.2: Reactor mass balance for output stream

Components Mass flowrate

(kg/h)

Mass

fraction

(w/w)

Molar flowrate

(kmol/h)

Mole fraction

(kmol/kmol)

TBA 1047.7069 0.0200 14.1582 0.0200

Propylene oxide 55.7380 0.0011 0.9610 0.0011

Isobutylene 38850.0392 0.7408 693.7507 0.7408

Water 12487.5126 0.2381 693.7507 0.2381

Total (Gm) 52440.9966 1.0000 1402.6206 1.0000

3.2 CHEMICAL DESIGN

Chemical design is carried out to determine the dimensions of the reactor as shown

in following sections. The reactor volume, combination of shell and tubes

dimensions, heat transfer coefficients and pressure drop are determined.

Dehydration of TBA is characterized by endothermic reaction and from the energy

balance the heat required for the reaction is 2.379x107 kJ/h. For isothermal fixed

bed reactor and first order reaction, the reaction rate constant for TBA dehydration is

given by equation below, (Journal of Hydrocarbon Processing, please refer to

APPENDIX C1):

Reaction involved:

C4H10O C4H8 + H2O

TBA Isobutylene Water

(3.1)

K = Reaction rate constant

X = Fraction of TBA converted

LHSV = Ratio of feed rate and the amount of catalyst

X = 0.98 (TBA conversion)

= 6.844

At 20 psig (3 bars), the equation relating reaction rate constant and temperature is

given by:

ln K = 21.7483 – 17992/T (3.2)

T = Temperature in degree Rankine, 316ºC = 1060ºR

ln K = 21.7483 – 17992/1060

K = 118.48 h

(3.3)

Where,

V = Volume of catalyst

v = Feed volume flowrate

Volume flowrate of TBA = mass flowrate of TBA x specific gravity of TBA

= 52385.2586 kg/h x 1.2706x10-3 m3/kg

= 66.5632 m3/h

Volume of catalyst, Vc (3.4)

= 66.5632 m3/h x 0.0578 h

= 3.8474 m3

Volume of reactor, VR = Vc/ (1 – ε) (3.5)

= 3.8474 / (1-0.4)

= 6.4123 m3

3.2.1 Selection of Catalyst

A catalyst is a substance that increases a rate of reaction by participating chemically

in intermediate stages of reaction and is liberated near the end in a chemically

unchanged form. Over a period of time, however, permanent changes in the catalyst

such as deactivation may occur. Many catalysts have specific actions in that they

influence only one reaction or group of definite reactions. The catalyst that used for

the fixed bed reactor is silica alumina cracking catalyst. The properties of catalyst

are shown below (Perry’s Handbook, 1997).

Surface area SA: 0.35x106 m2/kg

Diameter of particle, dp: 3 mm

Pore diameter, θd: 6.1 nm

Porosity of particle, θp: 0.56

Particle density, ρp: 1062 kg/m3

Specific surface area, Sg:1840 m2/m3

Tortuosity, : 2

Voidage, ε: 0.4

Bulk density, ρB = (1-ε)(ρp) (3.6)

= (1- 0.4)(1062 kg/m3)

= 637.2 kg/m3

Weight of catalyst, Wc = (Vc)(ρB) (3.7)

= (3.8474 m3)(637.2 kg/m3)

= 2451.56 kg

3.2.2 Effective Diffusivity, De

The resistance to diffusion in a catalyst pore is due to collisions with other molecules

and with the walls of the pore. The corresponding diffusivities are called bulk

diffusivity and Knudsen diffusivity DK. The actual diffusivity in common porous

catalysts usually is intermediate between bulk and Knudsen. Moreover, it depends

on the pore size distribution and on the true length path. The effective diffusivity, D e

is given by the equation below:

(3.8)

Where,

θ= particle porosity

= Tortuosity of the pores

D = Dk + DB (3.9)

Where,

Dk = Knudsen Diffusivity

DB = Bulk Diffusivity

Neglecting DB term, hence D = Dk, Knudsen diffusivity (m2/s) in a straight cylindrical

pore can be expressed (based on the kinetic theory of gases) as:

(3.10)

Where,

rp = the pores radius = dp/2 = 6.1 nm/2 =3.05 nm

T = temperature in K = 589 K

mt = the mean molecular weight of tube side material

= 54.445 kg/kmol

= 9.73x10-7 m2/s

Therefore,

= 2.72x10-7 m2/s

3.2.3 Tube Specification

In order to decide the tube length and tube diameter used in this design, the

following criteria is followed:

Square pitch arrangement is chosen for ease of cleaning (Kern, 1965).

From table 12.3 (Colson & Richardson’s, Chemical Engineering), we take standard

tube of:

Inside diameter, di = 44.5 mm

Outside diameter, do = 50.8 mm

Length of tube, L = 1.22 m

Area of the tube can be calculated using equation below.

Assumed thickness of the tube is negligible.

Cross sectional area of tube,

As = (3.11)

=

= 2.0268x10-3 m2

Number of tube, Nt = (3.12)

=

= 1556 tubes

Residence time, = (3.13)

=

= 208.08 s

= 3.47 minutes

Flow through each tube, Vt = Vf / Nt (3.14)

=

= 3.0402x10-3 m3/s

Superficial velocity, uc = (3.15)

=

= 1.51 m/s

Tube side mass flowrate per unit area,

(3.16)

Fixed bed reactor is designed as the shell and tube heat exchanger which catalysts

are in the tube.

Approximate tube bundle diameter, Db

Db = (3.17)

Where,

k1 and n1 are constants,

for square pitch, pt = 1.25do

for 1 pass, k1 = 0.215

n1= 2.207

Therefore, Db =

= 2846 mm

Square pitch is chosen, pt = 1.25do (3.18)

hence, pt = 1.25(50.8) mm

= 63.5 mm

Allow 50 mm for shell-inside diameter to bundle diameter

Therefore, approximate shell diameter, Ds = (2846 + 50) mm

= 2896 mm

3.2.4 Heat Transfer Calculation

From energy balance, heat required for the process is 2.379x103 h.

Assuming overall heat transfer calculation, Uo = 0.06kW/m2K

Heat transfer area available, A

A = NtπL(do + di) / 2 (3.19)

= 1556π x 1.22 x (0.0505 + 0.0445) / 2 m2

= 283.2780 m2

By using Q = UoAΔT for isothermal condition.

ΔT = (3.20)

=

= 389 K

3.2.5 Tube Side Coefficient

The temperature profile in the bed is constant which is not likely to happen. For the

worst condition, the temperature profile in the bed is parabolic, and to ensure that

the design is in the safe region, therefore the resistance in the bed should be

considered.

The tube side coefficient is split into two parts to account for the resistance in the

region very near the wall and for the resistance in the rest of the packed bed.

Wall coefficient, hw,

Rec = (3.21)

=

= 412

Pr = (3.22)

=

= 0.89

= 1.6(Re)0.51(Pr)0.33 (3.23)

(3.24)

=

= 469 W/m2K

Bed coefficient, hbed,

(3.25)

Where ke = effective thermal conductivity of bed

r = radius of the inside tube

(3.26)

Hence ke = k(5 + 0.1RePr)

ke = 0.0424(5+0.1(412)(0.89)) W/m2K

ke = 1.77 W/m2K

318 W/m2K

Tube side coefficient, hi,

(3.27)

=

190 W/m2K

Correcting this coefficient to the heat transfer area corresponding to the centre of

the tube wall.

Tube side heat transfer coefficient, hiw

(3.28)

= 177 W/m2K

3.2.6 Shell Side Coefficient

The source of heating medium is flue gas. It is assumed that the flue gas is available

at 760 K, therefore temperature of flue gas at outlet is 371 K.

Table 3.3: Properties of flue gas

Property Value

Heat capacity, Cps

Viscosity, µs

Thermal conductivity, ks

Density, ρs

1.195 kJ/kgK

0.0255x10-3 Ns/m2

0.0287 W/m2K

0.78 kg/m3

Mass flowrate of flue gas required, ms = (3.29)

=

= 14.22 kg/s

Take baffle spacing as 1/5 from the shell diameter.

Baffle spacing, Ib = Ds/5 (3.30)

= 2896/5 mm

= 579.2 mm

Tube pitch, pt = 1.25do (3.31)

= 1.25(0.0508 m)

= 0.0635 m

Cross flow area, Ac = (pt – do)IbDs/pt (3.32)

= (0.0635 – 0.0508) m (0.5792 m) (2.896 m)/0.0635m

= 0.3355 m2

Mass velocity, Gs = m/Ac (3.33)

= 14.22 kg/s / 0.3355 m2

= 42.3845 kg/m2s

Reynolds number, (3.34)

From figure 12.31 (Coulson & Richardson, 1999, please refer to APPENDIX C2),

heat transfer factor for cross flow tube bank, jh = 3.7x10-3

Prandtl number, (3.35)

=

= 1.06

(3.36)

neglecting term,

(3.37)

Correcting to the tube wall centre heat transfer area.

Shell side coefficient,

(3.38)

=

= 192 W/m2K

3.2.7 Overall Heat Transfer Coefficient

Scale factor, Organic vapor, his = 5000 W/m2K

Flue gas, hos = (2000 – 5000) W/m2K

Take mean value = 3500 W/m2K

Tube type is stainless steel and it’s thermal conductivity at 600ºC, kw = 36 W/m2K

Hence,

Overall heat transfer coefficient:

(3.39)

3.2.8 Tube Side Pressure Drop

Hougan and Watson equation will be used to calculate tube side pressure drop

(3.40)

Where,

Z = length of tube (1.22 m)

V = S(1-e)

= 1840(1-0.4)

= 1102

Where f = 2.6(Re”)-0.13 for 150>Re”>10 and f = 1.23(Re”)-015 for 300>Re”>150

Re” = (3.41)

= 276

Therefore, f = 1.23(276)-0.15

= 0.53

= 10637 N/m2

= 10.6 kPa

3.2.9 Shell Side Pressure Drop

From figure 12.36 (Coulson & Richardson`s, 1999, please refer to APPENDIX C3)

at Re = 84436, for 1.25 Δpitch, jf = 4.8x10-2

Shell side flue gas velocity,

(3.42)

=

= 47.4 m/s

(3.43)

Neglecting viscosity term,

(3.44)

= 2095 N/m2

= 2.1 kPa

Table 3.4: Chemical Design Specification Data

Parameter Value

Volume of catalyst 3.8474 m3

Weight of catalyst 2451.56 kg

Volume of reactor 6.4123 m3

Residence time 3.47 minutes

Length of vessel 8 m

Tube OD 50.8 mm

Tube ID 44.5 mm

Number of tube 1556 mm

Tube pitch 63.5 mm

Tube bundle diameter 2846 mm

Shell diameter 2896 mm

Tube side heat transfer coefficient 190 W/m2K

Shell side heat transfer coefficient 192 W/m2K

Tube side pressure drop 10.6 kPa

Shell side pressure drop 2.1 kPa

.


The mechanical design is a function of the equipment, operating pressure and

temperature, material of construction and equipment dimensions. This mechanical

design for the fixed bed reactor is carried out based on the approach to Sinnot,

(Coulson & Richardson’s,1999).

3.3.1 Design Pressure

A reactor must be designed to withstand the maximum pressure to which it is likely

to be subjected for operation. The operating pressure for this reactor is10 bar. For

safety reason, the design pressure of this reactor is taken as 10% above the

operating pressure

Design pressure, Pi = (10 – 1) x 1.1

= 9.9 bar

= 0.99 N/mm2

3.3.2 Design Temperature

The design temperature at which the design stress is valuated should be taken as

the maximum working temperature of the material. For this reactor the design

temperature is 500˚C.

3.3.3 Material of Construction

The material used is stainless steel (18Cr/8Ni, 304). This material is good for creep

resistance, intergranular cracking and last longer. For this material, the design

stress at 500ºC (Coulson & Richardson’s,1999, please refer to APPENDIX C4).

Design stress, f = 90 N/mm2

Tensile strength = 510 N/mm2

3.3.4 Corrosion Allowance

Corrosion and erosion or scaling will cause material lost, so an additional thickness

of material is needed and it is corrosion allowable. The recommended corrosion

allowance is 2 mm.

3.3.5 Thickness of Cylindrical Shell

A minimum thickness is required to ensure that the vessel is sufficiently rigid to

withstand its own weight and any incidental load.

(3.45)

Where,



Ds = Diameter of shell

f = Friction factor

e =

= 16.20

Add corrosion allowance = 2 mm

e = (16.20 + 2) mm

= 18.20 mm, take thickness as 19 mm

3.3.6 Head and Closures

The ends of a cylindrical vessel are closed by heads of various shapes. There are

three types of commonly used domed head:

1) Hemispherical head

2) Ellipsoidal head

3) Torispherical head

Torispherical head had been choosing for this reactor. The selection of head

depends on the cost and the thickness required for the head. The design equation

and chart for the various types of domed heads are given in the codes and standard

BS 5500 used in this design.

Take, crown radius, Rc = Di = 2.896 m

Knuckle radius, Rk = 6%Rc = 0.174 m

A head for this size would be form by pressing: no joints, so J = 1.0

Cs = (3.46)

=

= 1.771

Therefore, minimum thickness:

e = (3.47)

=

= 16 mm

Add corrosion allowance = 2 mm

e = (16 + 2) mm

= 18 mm

3.3.7 Weight Load

Dead weight of vessel, Wv

For a steel vessel,

(3.48)

Where,

Dm = mean diameter, m (Di + t)

Cv = a factor, take 1.15


t = wall thickness

Wv = (240)(1.15)(2.915)(1.22+0.8(2.915))19

= 50991.77 N

= 51 kN

Weight of tubes, Wt

Wt N t do2 di

2 Lmg(3.49)

= 1556π(0.05082-0.04452)(1.22)(3000)(9.81)

= 105376.31 N

= 105 kN

Weight of insulation

Material used = mineral wool insulation

Insulation thickness = 50mm = 0.05m

Density = 130kg/m3

Approximate volume of insulation

V Hv (r r1)2 r2 (3.50)

= π(1.22)[(0.38 + 0.05)2 – (0.38)2]

= 0.16 m3

Wi Vg (3.51)

= (0.16)(130)(9.81)

= 197.96 N

= 0.2 kN

Weight of catalyst, Wc

Wc = (mc)(g) (3.52)

Where; mc is weight of catalyst

Wc = (2451.56 kg) (9.81m/s2)

= 24049.8 N

= 24 kN

Total weight of vessel

Wt = Wv + Wt +Wi + Wc (3.53)

= 51 kN + 105 kN + 0.2 kN + 24 kN

= 156.2 kN

3.3.8 Wind Loading

A vessel must be designed to withstand the highest wind speed that is likely to be

encountered at the site throughout the life span of the plant. A wind speed of 160

km/h is used for the preliminary design studies.

Pw = 0.05Uw2 (3.54)

= 0.05(160)2

= 1280 N/m2

Loading per unit length of reactor, Fw

Fw = PwDeff (3.55)

Where,

Deff = Effective reactor diameter

= Diameter shell + 2(tshell + tinsulation)

= 2896 + 2(19 + 50)

= 3034 mm

= 3.034 m

Therefore,

Fw = (1280)(3.034)

= 2973.32 N/m

Bending Moment

Mx = Fw (X)2/2 (3.56)

Where,

X = Distance measure from the free end

= 5 m

Therefore,

Mx = 2973.32(5)2/2

= 37166.5 Nm

= 37.17 kNm

3.3.9 Analysis of Stresses

Longitudinal pressure stress,

(3.57)

= (0.99)(2896)/4(19)

= 37.72 N/mm2

Circumferential pressure stress,

(3.58)

= (0.99)(2896)/2(19)

= 75.45 N/mm2

Dead weight stress,

(3.59)

= 0.8977 N/mm2

Bending stress,

(3.60)

Where,

M = total bending moment

Iv = (3.61)

Iv = second moment of area

Where,

Di = 2896 mm

Do= (2896 + 2(19))

= 2934 mm

so,

Iv

= 1.848x1011mm4

Therefore,

= 0.30 N/mm2

The resulted longitudinal stress, σz is,

σz(upwind) = σL – σw + σb (3.62)

= 37.72 –0.8977 + 0.30

= 37.12 N/mm2

σz(downwind) = σL – σw - σb (3.63)

= 37.72 – 0.8977 + 0.30

= 36.52 N/mm2

3.3.10 Elastic Stability

Critical bulking stress

(3.64)

=

= 129.52 N/mm2

Maximum compressive stress will occurs when the vessel not under pressure, σmax

= σw + σb (3.65)

= 0.8977+ 0.30

=1.1977 N/mm2

This is below critical bulking stress, so acceptable.


The method used to support a vessel will depend on the size, shape and weight of

the vessel, design pressure and temperature and vessel location and arrangement.

Since reactor is vertical vessel, skirt support is used in this design. A skirt support

consists of a cylinder or conical shell welded to the base of the vessel.

Type of support :Straight cylindrical skirt

θs :90°

Material construction :Carbon steel

Design stress, fs :135 N/mm2

Skirt height :1.0 m

Young modulus :200, 000 N/mm2

Approximate weight,

Wapprox = (3.66)

= π/4(2.896)2(1.44)(1000)(9.81)

= 93050.45 N

= 93.1 kN

Weight of vessel = 180.2 kN

Total weight = 93.1 kN + 180.2 kN

= 273.3 kN

Wind load,

Fw = 2973.32 N/m

= 2.97 kN/m

Bending moment at skirt base,

Ms = (3.67)

=

= 8.84 kNm

As a first trial, take skirt thickness as same as the thickness of the bottom section of

the vessel, ts = 19 mm.

Bending stress in skirt,

(3.68)

Where,

Ms = maximum bending moment (at the base of the skirt)


Ds = inside diameter of the skirt base

= 2.896 m

Therefore,

= 7.07 N/mm2

Dead weight stress in the skirt,

(3.69)

Therefore,

= 5.35 N/mm2

= 0.90 N/mm2

Thus, the resulting stress in the skirt, σs:

Maximum σs (compressive) = σbs - σws(test) (3.70)

= 7.07 + 5.35

= 12.42 N/mm2

Maximum σs (tensile) = σbs + σws(operating) (3.71)

= 7.07 – 0.90

= 6.17 N/mm2

3.3.12 General Consideration for the Design

Take the joint factor as J as 0.85,

σs (tensile) < fs J sin θs

σs (compressive) < 0.125 E

Where,

fs = maximum allowable design stress for the skirt material (135 N/mm2)

J = weld joint factor

θs = base angle of a conical skirt

E = modulus Young (200, 000N/mm2)

Therefore,

σs (tensile) < 135 x 0.85 sin 90

6.17 N/mm2 < 113.08 N/mm2

σs (compressive) < (0.125)(200, 000)

12.42 N/mm2 < 161.53 N/mm2

Both criteria are satisfied, add 2 mm for corrosion, give design thickness of 21 mm.

3.3.13 Base Rings and Anchor Bolts

Assume pith circle diameter = 3.0 m

Circumference of bolt circle = 4000π

Bolt stress design, fb = 125 N/mm2

Recommended spacing between bolts = 600 mm

Minimum number bolt required, Nb =

= 20.9

Closest multiple of 4 = 12 bolts

Bending moment at base skirt, Ms = 80.8 kNm

Total weight of vessel, W = 273.3 kN

Area of bolt,

Ab = (3.72)

=

= 580.25 mm2

Bolt root diameter,

(3.73)

= 27.18 mm

Total compressive load on the base ring per unit length,

(3.74)

=

= 37.88 kN/m

Assuming that a pressure of 5 N/mm2 is one of the concrete foundation pad, fc so

minimum width of the base ring,

(3.75)

= 7.58 mm

3.3.14 Pipe Size Selection for the Nozzle

Material of construction = Stainless steel

Density of TBA = 787 kg/m3

Flowrate of TBA = 14.55 kg/s

Diameter pipe for TBA

DTBA = 260 G0.52 ρ-0.37 (3.76)

= 260 (14.55)0.52 (787)-0.37

= 88.75 mm

3.3.15 Standard Flanges

Flanges used in this design are chosen from the standard flanges. Here standard

flanges are adapted from the British standard (BS 4504), nominal pressure of 6 bar.

(Please refer to APPENDIX C5).

Table 3.5: Mechanical Design Specification Data

Parameter Value

Design pressure 9.9 bar

Design temperature 500ºC

Material of construction Stainless steel (18Cr/8Ni, 304)

Design stress 90 N/mm2

Tensile stress 510 N/mm2

Tube thickness 3 mm

Shell thickness 19 mm

Torispherical head thickness 18 mm

Manhole 800 mm [BS470:1984]

Longitudinal pressure stress 37.22 N/mm2

Circumferential pressure stress 75.45 N/mm2

Longitudinal stress (upwind) 37.12 N/mm2

Longitudinal stress (downwind) 36.52 N/mm2

Critical buckling stress 129.52 N/mm2

Maximum compressive stress 1.1977 N/mm2

Type of skirt support Straight cylindrical skirt

Material construction of skirt support Carbon steel

REFERENCES

Coulson, J M and Richardson, J F, 1998, Coulson & Richardson’s Chemical

Engineering, Vol. 1 : “Fluid Flow, Heat Transfer and Mass Transfer”, Oxford,

Pergamon,

Douglas, James M., 1988, Conceptual Design of Chemical Processes, Singapore,

McGraw-Hill Book Co.

Elvers B. et. al., 1989, Ullman’s Encyclopedia of Industrial Chemistry, Volume 13,

Germany, VCH Verlagsgesellschuft.

Felder, Richard M. and Rousseau, Ronald W., 1986, Elementary Principles of

Chemical Processes, Second Edition, United States, John Wiley & Sons, Inc.

Fogler, H.Scott, 1999, Elements of Chemical Reaction Engineering, Third Edition,

Upper Saddle River, New Jersey, Prentice Hall, Inc.

Ludwig, E. Ernest, 1964, Applied Process Design for Chemical and Petrochemical

Plants, Vol. 1, Houston, Gulf Publishing Coompany.

Perry R.H., Green D.W., 1997, Perry’s Chemical Engineer’s Handbook, 7th edition,

USA, McGraw-Hill.

Richardson, J F and Peacock, D G, 1994, Coulson & Richardson’s Chemical

Engineering, Vol. 3 : “Chemical & Biochemical Reactors & Process Control”, Oxford,

Pergamon.

Scott, Doug and Crawley, Frank, 1992, Process Plant Design and Operation,

Warwickshire, UK, Institution of Chemical Engineers.

Sinnott, R.K, 1999, Coulson & Richardson’s Chemical Engineering, Vol. 6 :

“Chemical Engineering Design”, Oxford, Butterworth-Heinemann.

Smith R., 1995, Chemical Process Design, USA, McGraw-Hill.

Hydrocarbon Processing, 1992, Vol. 71(February).

LIST OF NOMENCLATURES

Ab - By-pass area m2

Ac - Cross flow area m2

Ah - Heat transfer area m2

As - Tube cross-sectional area m2

Cps - Flue gas heat capacity kJ/kgK

Db - Bundle diameter m

De - Effective diffusivity m2/s

Dk - Knudsen Diffusivity m2/s

Ds - Shell inside diameter m

di - Inside tube diameter m

do - Outside tube diameter m

dp - Diameter of particle m

e - Thickness m

f - Design stress N/m2

Gm - Vapor mixture mass flowrate kg/h

Gs - Shell side mass flowrate per unit area kg/m2s

GT - Tube side mass flowrate per unit area kg/m2s

G - Gravitational acceleration m/s2

hi - Tube side heat transfer coefficient W/m2K

hs - Shell side heat transfer coefficient W/m2K

hw - Wall heat transfer coefficient W/m2K

hbed - Bed heat transfer coefficient W/m2K

jh - Heat transfer factor -

jf - Friction factor -

k - Reaction rate constant s

L - Tube length m

Ib - Baffle spacing m

Mt - Molecular weight of tube side material kg/kmol

ms - Flue gas flowrate kg/s

Nt - Number of tube -

Pi - Design pressure N/m2

pt - Tube pitch m

ΔPs - Shell side pressure drop kPa

ΔPt - Tube side pressure drop kPa

Q - Heat transfer required W

Rc - Crown radius m

Rk - Knuckle radius m

S - Surface area m2

ts - Shell thickness m

tt - Tube thickness m

ΔT - Temperature different °C

Uc - Superficial velocity m/s

V - Volume of catalyst m3

v - Feed volume flowrate m3/h

x - Fraction of TBA converted -

ρp - Particle density kg/m3

ρs - Flue gas density kg/m3

µs - Flue gas viscosity Ns/m2

LIST OF FORMULA

(3.1)

ln K = 21.7483 – 17992/T (3.2)

(3.3)

Vc (3.4)

VR = Vc/ (1 – ε) (3.5)

ρB = (1-ε)(ρp) (3.6)

Wc = (Vc)(ρB) (3.7)

(3.8)

D = Dk + DB (3.9)

(3.10)

As = (3.11)

Nt = (3.12)

= (3.13)

Vt = Vf / Nt (3.14)

uc = (3.15)

(3.16)

Db = (3.17)

pt = 1.25do (3.18)

A = NtπL(do + di) / 2 (3.19)

ΔT = (3.20)

Rec = (3.21)

Pr = (3.22)

= 1.6(Re)0.51(Pr)0.33 (3.23)

(3.24)

(3.25)

(3.26)

(3.27)

(3.28)

ms = (3.29)

Ib = Ds/5 (3.30)

pt = 1.25do (3.31)

Ac = (pt – do)IbDs/pt (3.32)

Gs = m/Ac (3.33)

(3.34)

(3.35)

(3.36)

(3.37)

(3.38)

(3.39)

(3.40)

Re” = (3.41)

(3.42)

(3.43)

(3.44)

(3.45)

Cs= (3.46)

e = (3.47)

(3.48)

Wt N t do2 di

2 Lmg(3.49)

V Hv (r r1)2 r2 (3.50)

Wi Vg (3.51)

Wc = (mc)(g) (3.52)

Wt = Wv + Wt +Wi + Wc (3.53)

Pw = 0.05Uw2 (3.54)

Fw = PwDeff (3.55)

Mx = Fw (X)2/2 (3.56)

(3.57)

(3.58)

(3.59)

(3.60)

Iv = (3.61)

σz(upwind) = σL – σw + σb (3.62)

σz(downwind) = σL – σw - σb (3.63)

(3.64)

σmax = σw + σb (3.65)

Wapprox = (3.66)

Ms = (3.67)

(3.68)

(3.69)

σs (compressive) = σbs - σws(test) (3.70)

σs (tensile) = σbs + σws(operating) (3.71)

Ab = (3.72)

(3.73)

(3.74)

(3.75)

DTBA = 260 G0.52 ρ-0.37 (3.76)


MTBE REACTOR NORMARIAH BINTI ABDULLAH

SUPERVISORS



CHAPTER 4:

MTBE REACTOR

CONTENTS

TITLE PAGE

CHAPTER 4 MTBE REACTOR

4.1 CHEMICAL ENGINEERING DESIGN

OF REACTOR 97

4.1.1Catalyst 98

4.1.2 Tube side 103

4.1.3 Shell 105

4.1.4 Condition Calculation 106

4.2 MECHANICAL DESIGN OF REACTOR 112

4.2.1 Design Consideration 112

4.2.2 The Design of Thin Walled Vessels

Under Internal Pressure 113

4.2.3 Design of Vessels Subject to

Combined Loading 116

4.2.4 Vessel Support 121

4.2.5 Base Ring and Anchor Bolt Design 124

4.2.6 Bolt Flanged Joint 127


4.2.8 Compensation for Opening and

Branch Connections 130

CHAPTER 4

MTBE REACTOR

4.1 CHEMICAL ENGINEERING DESIGN OF REACTOR

The major equipment in the MTBE process plant is reactor, where the

conversion of reactant to products takes place. The reactor use is fixed bed reactor

which operates isothermally. The fixed bed reactor consists of a number of tubes

packed with catalyst particles and operated at vertical position. The condition of the

reactor is plug flow reactor where the reactants flow through the tube without back

mixing with concentration changing down the tube as a result of the reaction.

The amount of catalyst required can be determined by calculating the rate of

reaction. The catalyst effectiveness factor must be taken into account in overall rate

constant calculation because the intra- particle diffusion has a great effect on the

rate of reaction.

Feed

Coolingwater inlet

CoolingWater Outlet

Product

Figure 4.1: Fixed Bed Reactor

4.1.1 Catalyst

The catalyst used for this process is sulfonic ion resin which is having 16000

hours operating life, longer life compared to other catalyst.

Catalyst properties:

Diameter of catalyst (dp) = 0.04 mm

Bulk density (b) = 700 kg/ m3

Surface area (Sa) = 100 m2/ g

Total voidage (b) = 0.54

Void fraction (p) = 0.32

Pore volume (V) = 4.7 x 10-7 m3/ g

Specific surface area (O) = 0.020 m2/ g

4.1.1.1 Particle solid density

From Perry’s:

p =

1

b

=

= 1029 kg/ m3

4.1.1.2 Pore radius of catalyst

Brunauer -Emmet-Teller (BET) showed that the pore radius is related to the

specific surface area, O perunit mass to the pore volume, V by the equation below: (

G.H Osborn, 1961)

r = 2.7

= 6.35 x 10 -5 m

4.1.1.3 Effective Diffusivity

Knudsen diffusion

Dk =

= 4.68 cm2/ s

4.1.1.4 Thiele Modulus

From Levenspiel, Thiele Modulus for sphere is given by: (Octave Levenspiel,

1999)

= 1.1

4.1.1.5 Effectiveness factor,

From Perry’s:

=

= 0.93

4.1.1.6 Reaction Rate

For a reversible first order reaction on exothermic conditions, the rate of

reaction for the suphonic cation exchange resin catalyst is:

CH3OH + CH3C(CH3)CH2 (CH3)3CH2COCH3

A B C

-rB = k1CB – k2Cc A1 = 6.5 x 105

R = 8.314 J/ mol.K A2 = 1.36 x 108

Tin = 65oC = 338 K E1 = 4.74 x 104

Tout = 200oC = 473 K E2 = 7.04 x 104

Pin = 2 bar

Pout = 10 bar

K1 = A1 e(-E1/ RT)

= 3.07 x 10-2 hr-1

K2 = A1 e(-E2/ RT)

= 2.285 hr-1

By using ideal gas law for isobutylene :

CBO =

= 71.17 mol/m3

Component Density (kg/m3) Volume flowrate (m3/hr)

C4H8 600 64.75

CH3OH 791.5 28.04

H2O 998.2 0.23

PO 312 0.18

Input flowrate = 61335.1630 kg/hr

Volume of mixture = 93.20 m3/hr

Density of mixture =

=

= 658.10 kg/m3

CCO =

F = mole flowrate of the feed

V = volume of the feed

CCO =

= 15.03 mol/m3

M =

= 0.211

-rB = k1CB – k2Cc

= k1(CBO – CBOXB) – k2(MCBO + CBOXB)

= -196.77

4.1.1.7 Weight of catalyst

W = weight of catalyst needed

F = molar flowrate of the feed

X = conversion of reactant

=

= 7.11 m3

= 7.11 m3 x 700 kg/ m3

= 4977 kg

4.1.1.6 Pressure Drop

For gas-liquid reaction at high pressure, the change in pressure may effects

the global rate significantly. Also, the pressure drop is needed for designing

pumping equipment which usually estimates the economic structure of a reactor

system. For packed bed, the pressure drop may be estimated from the Ergun

equation as below:

Where:

f = friction factor

u = superficial linear velocity

= density of fluid

L = depth of the bed

d’p = effective particle diameter = 5 x 10-4 m

i. Superficial linear velocity

L = 4.88 m

D = 2.0 m

U =

=

= 3.377 x 10-3 m/s

ii. Reynolds number

Re =

= 658.10 (3.377 x 10 -3 ) (5x10 -4 )

1.89 x 10-3

= 0.59

iii. Friction factor

f =

f =

f = 346.76

iv. Pressure drop

= 346.76 (3.377 x 10 -3 ) 2 (658.10)

5 x 10-4

= 5204.90 N/m3

4.1.1.9 Height of the bed

The preferred lengths of the tubes length are 6 ft, 8 ft, 12 ft, 16 ft, 20 ft and

24 ft ( Coulson and Richardson,1999). The height of the bed is taken as 16 ft

(4.88m). The height of the bed is selected to suit the criteria that the optimum value

of pressure drop is between 5 to 15% of the total pressure.

Pressure drop (- = 5204.90 N/m3x 4.88m

= 25399.95 N/m2

4.1.1.10 Volume of catalyst bed

Vp =

=

= 4.84 m3

4.1.2 Tube Side

4.1.2.1 Total cross section of the tube

= volume of catalyst bed Height of the bed

= 4.84 m 3 4.88 m

= 0.99 m2

4.1.2.2 Tube diameter (O.D)

The standard dimensions for steel tube diameter is in the range of 16 mm to

50 mm. The smaller diameters (16 mm to 25 mm) are preferred for most duties.

Larger tubes are easier to clean by mechanical methods and would be selected for

heavily fouling fluids. Therefore the tube diameter of 50 mm is choose as they will

give more compact and therefore cheaper.

4.1.2.3 Wall thickness

A 2.0 mm of wall thickness is the standard wall thickness for 50 mm tube

diameter that are given in BS 3274 used in this reactor.

4.1.2.4 Inside diameter

DI = 50 mm – 2(2.0mm)

= 0.046 m

4.1.2.5 Total number of tube

cross section of one tube =

=

= 1.66 x 10-3 m2

nt = Total cross section of tube

cross section of one tube

=

= 596 tubes

4.1.2.6 Tube arrangements

The tubes are usually arranged in an equilateral triangular, square or rotated

square pattern. Since this process required high heat transfer to maintain isothermal

condition in the reactor, the triangular arrangement is recommended. (Please refer

APPENDIX D10)

4.1.2.7 Tube pitch

The recommended tube pitch (distance between tube centre) is 1.25 times

the tube outside diameter.

Pt = 1.25 x O.D

= 0.0625 m

4.1.2.8 Tube side passes

Since the inlet flow rate is very high, this exchanger is build with one tube

passes.

4.1.2.9 Bundle diameter

The bundle diameter will depend not only on the number of tubes but also

the number of the tubes passes. For triangular patterns;

where :

Nt = number of tubes

Db = bundle diameter

Do = tube outside diameter

= 1.68 m

4.1.2.10 Holding time

Vtube = x0.0462 x 4.88

= 8.11 x 10-3 m3

Voutlet = 83.77 m3/hr

th =

= 0.35 s

4.1.3 Shell

4.1.3.1 Shell types

A single shell pass type is used

4.1.3.2 Shell diameter

The shell diameter must be selected to give as close as fit to the tube bundle

as in practical to reduce bypassing round the outside of the bundle. The clearance

required between the outermost tubes in the bundle and the shell inside diameter

will depend on the type of exchanger and manufacturing tolerances. The split ring

floating head type is used in this reactor.

From figure 12.10 (Coulson & Richardson vol.6);(Please refer APPENDIX

D7)

Ds – Db = 100 mm

= 100 mm + 1680 mm

= 1.780 m

4.1.3.3 Baffles

Baffles are used in the shell to increase the fluid velocity and to improve the

rate of the heat transfer. 25% baffles cut is used for this shell.

4.1.3.4 Baffle spacing

The baffle spacing used range from 0.2 to 1.0 shell diameters. The optimum

spacing usually between 0.3 to 0.5 times the shell diameter.

Bs = 0.4 x 1.780 m

= 0.712 m

4.1.3.5 Number of baffles

Nb = 6 baffles

4.1.3.6 Cross flow area

As =

=

= 0.2535 m2

4.1.3.7 Volume of reactor

VR = x bed height

= 12.15 m3

4.1.4 Condition Calculation

4.1.4.1 Tube side

Feed = 61335.1630 kg/hr

Outside diameter = 0.050 m

Inside diameter = 0.046 m

Pitch = 0.0625 m

Length = 4.88 m

Number of tubes = 596

Passes = 1

Cross section of one tube = 0.00166 m2

Total cross section = 0.99 m2

4.1.4.1.1 Heat transfer coefficient in tube side

Gmax =

Where;

Gmax = maximum mass flowrate

M = total mass flowrate

Amin = total minimum free flow area

Gmax = 61335.1630 kg/hr

0.99 m2

= 61954.7101 kg/hr

= 17.21 kg/s

Re =

Where;

D = outside diameter

= average viscosity

Re =

= 455.29

L/D = 4.88 m / 0.046m

= 106.09

From figure 12.23 (Coulson & Richardson vol.6);

Heat transfer factor, jh = 8 x 10-3 (Please refer APPENDIX D1)

= w

Neglect

w

Pr = = 84.45

= 106.09

=

= 75.267 W/m2.K

4.1.4.1.2 Correction for tube heat transfer coefficient

The heat transfer coefficient that calculated is based on the inside diameter.

In order to obtain heat transfer coefficient that based on outside diameter, correction

is;

=

= 69.247 W/m2.K

4.1.4.1.3 Tube side pressure drop

From figure 12.24 for Re=3154,

Jf = 1.8 x10-1 (Please refer APPENDIX D2)

∆ P = 8jf

= 33.981 N/m2

4.1.4.2 Shell side

Flow area = 0.2535 m2

Inside diameter = 1.780 m

Baffle spacing = 0.712 m

4.1.4.2.1 Shell side heat transfer coefficient

Shell side mass velocity, Gs

= Ws

As

Ws = fluid flowrate on the shell side

Gs = 61335.1630 kg/hr 0.2535 m2

= 67.209 kg/m2.s

4.1.4.2.2 Shell side equivalent diameter

For an equilateral triangular pitch arrangements,

De =

= 0.037 m

4.1.4.2.3 Reynolds Number

Re =

=

= 2045

4.1.4.2.4 Heat transfer factor

From figure 12.29 (Coulson & Richardson vol. 6)

Jh = 1.5 x 10-1 (Please refer APPENDIX D3)

4.1.4.2.5 Heat transfer coefficient

Neglect

=

= 5145.90 W/m2.K

4.1.4.2.6 Overall heat transfer coefficient

The overall heat transfer coefficient can be determined from Fourier

equation. By neglecting the wall effect, the equation is;

= 74.182 W/m2.K

4.1.4.2.7 Friction factor

From figure 12.30 (Coulson & Richardson vol.6)

Jf = 7.0 x 10-2 (Please refer APPENDIX D4)

4.1.4.2.8 Pressure drop

∆ Ps = 8jf

Neglect

∆ Ps = 8 (7.0 x 10-2)

∆ Ps = 632.139 N/m2

4.1.4.2.9 Total heat transfer area

A = DoLNt

= (4.88)(596)(0.05)

= 456.922 m2

4.1.4.2.10 Design overall coefficient

Let dirt factor,Rd = 0.001

=

Ud = 69.059 W/m2.K

Overall heat transfer,

Q = UdA∆Tm

Log Mean Temperature Different (LMTD)

∆Tm =

= 716.16 K

TLMTD = To - T L

ln(To / TL)

By trial and error, outlet temperature of cooling water = 155 oC

4.1.4.2.11 Reactors cooling system

Cooling water is flow outside the reactor tubes where the reaction took place. This is

to maintain a constant operating temperature and to prevent any excessive heating

happen.

Mass of cooling water enter,

mfCp(T1 –T2) = mcCp(t2 –t1) = Q

mc = 39.673 kg/s

4.2 MECHANICAL DESIGN OF REACTOR

4.2.1 Design Consideration

4.2.1.1 Design pressure

For vessels under internal pressure, the design pressure is normally taken

as the pressure at which the relief device is set. This will normally be 5 to 10 percent

above the normal working pressure, to avoid spurious operation during minor

process upsets. The design pressure is taken as 10% above the operating pressure.

PD = (PI -Po ) x 1.1

= (10-1 ) x 1.1

= 9.9 bar

4.2.1.2 Design temperature

The maximum allowable design stress is depended on the temperature of

material because the strength of metals decreases with increasing temperature. The

design temperature at which the design stress is evaluated is taken as the maximum

working temperature of the material, that is T = 200 oC

4.2.1.3 Material

A suitable material must take into account the suitability of material for

fabrication as well as the compatibility of the material with the process environment

since the maximum working temperature at this reactor is

200 oC because it will oxidize rapidly at high temperature. Stainless steel is

recommended in construction of vessel tubes and shell.

4.2.1.4 Design stress (nominal design strength)

A maximum allowable stress that can be accepted in the material of

construction is necessary to decide for design purpose in which the material could

be expected to withstand without failure under standard test conditions. By using

stainless, the design stress is given as 115 N/mm2 (Please refer APPENDIX D5)

4.2.1.5 Welded joint efficiency

The strength of a welded joint will depend on the type of joint and the quality

of the welding. For reactor, the joint factor is taken as 1.0 which implies that the joint

is equally as strong as the virgin plate. This highest category, requires 100% non

destructive testing welds.

4.2.1.6 Corrosion allowance

The corrosion allowance is the additional thickness of metal added to allow

for material lost by corrosion and erosion. For carbon and low- alloy steels, where

severe corrosion is not expected, a minimum corrosion allowance of 2.0 mm is used

since the influent and effluent gas of the reactor is not corrosive.

4.2.2 The Design of Thin Walled Vessels Under Internal Pressure

4.2.2.1 Cylinders shell minimum practical wall thickness

A minimum wall thickness is required to ensure that any vessel is sufficiently

rigid to withstand its own weight, and any incidental loads. For a cylindrical shell the

minimum thickness required to resist internal pressure can be determined from

equation below:

where:

e = minimum wall thickness, m

Pi = internal pressure, N/mm2

f = design stress, N/mm2

J = joint efficiency

Di = internal diameter of shell, mm

e = 1 N/mm 2 (1780mm)

2 (115 N/mm2)-1N/mm2

= 7.773 mm

By adding corrosion allowance of 2 mm,

9.773 mm

4.2.2.2 Heads and closure

The ends of a cylindrical vessel are closed by heads of various shapes. The

commonly types used are:

i. Domed heads

a. Hemispherical heads

b. Ellipsoidal heads

c. Torispherical heads

ii. Flat heads

Design equations and charts for the various types of domed heads are given

in the codes and standards and values for design constant Cp and the nominal plate

diameter De of flat end closures are given in the design codes and standards for

various arrangements of flat end closures. The selection of head depends on the

thickness required for the head which contributed to cost.

a) Torispherical heads.

The minimum thickness of head can be calculated from equation

below:

where :

Cs = stress concentration factor for torispherical

heads

Rc = crown radius = shell outside diameter

Rk = Knuckle radius

Where Rc / Rk should not be less than 0.06 and to avoid buckling;

crown radius Rc should not be greater than diameter of the cylinder

section.

For formed head (no joints in the head), the joint factor J is taken as

1.0.

Rc = 1.78 m

Rk = 0.06 Rc

= 0.1068 m

= 1.771 m

Add corrosion allowance of 2 mm

1.773 m

b) Ellipsoidal heads with major and minor axis ratio of 2:1. The minimum

thickness required can be determined by equation below:

e = 1 N/mm 2 (1780mm)

2 (115 N/mm2)-0.2(1N/mm2)

= 7.746 mm

Add corrosion allowance of 2 mm

e = 9.746 mm

c) Flat heads

The minimum thickness required is given by equation below:

Where Cp = a design constant, dependent on the edge

constraint

De = nominal plate diameter

f = design stress

For bolted cover with a full face gasket (to avoid leakage) take Cp = 0.4 and De equal to the bolt circle diameter, take as approximately 1.7 m

= 0.063 m

Add corrosion allowance of 2 mm;

= 65 mm

This shows the inefficiency of flat head. It would be better to use a

flanged domed head.

4.2.3 Design of Vessels Subject To Combined Loading

Pressure vessels are subjected to other loads in addition to pressure and

must be designed to withstand the worst combination of loading without failure. A

trial thickness must be assumed (based on that calculated for pressure alone) and

the resultant stress from all loads to ensure that the maximum allowable stress

intensity is not exceeded at any point.

The main sources of load to consider are;

a. Pressure

b. Dead weight of vessel and contents

c. Wind

d. Earthquake

e. External loads imposed by piping and attached equipment.

4.2.3.1 Stresses Analysis

4.2.3.1.1 Stresses resulting from internal pressure

The longitudinal and circumferential stresses due to pressure are given by:

= (1 N/mm 2 )(1780 mm)

4(9.773)

= 45.534 N/mm2

= (1 N/mm 2 )(1780 mm)

2(9.773)

= 91.067 N/mm2

4.2.3.1.2 Dead weight stress

The major sources of dead weight loads are:

a. The vessel shell

b. The vessel fittings: manways, nozzles

c. Internal fittings; ladders,platforms,

piping.

4.2.3.1.2.1 Weight of cylindrical vessel

The approximate weight of a cylindrical vessel with domed ends, and uniform

wall thickness, can be estimated from the following equation:

For a stainless steel vessel, the equation reduces to:

where;

Wv = total weight of the shell, excluding internal

fittings

Cv = a factor to account for the weight of nozzles

manways, internal supports,etc, Cv is taken as

1.08 for vessel with only a few internal fittings.

Hv = Height between tangent lines

t = wall thickness

g = gravitational acceleration, 9.81m/s2

m = density of vessel material, kg/m3

Dm = mean diameter of vessel

= ( Di + t x10-3)

= (1.780 + 9.773 x 10-3)

= 1.790 m

Wv =240 (1.080)(1.790)(4.88 + 0.8(1.790))9.773

= 28.62 kN

4.2.3.1.2.2 Weight of tubes

From Perry’s (Robert H. Perry,1997), the mass per length of steel tube is

equal to 1.905 kg/m.

The weight of one tube = 1.905 x 4.88

= 9.2968 kg

Total weight of tubes = 596 x 9.2968

= 54.353 kN

4.2.3.1.2.3 Weight of insulation

For high operating temperature, mineral wool is normally used as insulator.

Density of mineral wool = 130 kg/m3

The thickness of insulator = 75 mm

Approximate volume of insulation; = DiHvt

Vi = (1.780)(4.88)(75 x 10-3)

= 2.047 m3

Weight of insulator = Vig

= 130 (2.04)(9.81)

= 2.61 kN

4.2.3.1.2.4 Weight of catalyst

Weight of catalyst, Wc = 48.824 kN

4.2.3.1.2.5 Total weight

Total weight, WT = Wv + Wt + Wi + Wc

= 134.408 kN

4.2.3.1.2.6 Calculation of dead weight stress

The dead weight stress can be calculated by equation below:

= 134.408 (1780 + 9.773)9.773

= 2.446 N/mm2

4.2.3.1.3 Bending stress

4.2.3.1.3.1 Wind loads

A vessel installed in the open must be designed to withstand the weight

bending stress caused by wind loading. The wind loading is a function of the wind

velocity, air density and the shape of structure. A wind speed of 160 km/hr is used

for preliminary design.

For a cylindrical column, the following semi-empirical equation can be used to

estimate the wind pressure.

where ; Pw = wind pressure, N/m2

Uw = wind speed, km/hr

= 0.07 (160)

= 1792 N/m2

The loading per unit length of the column can be obtained from the wind pressure by

multiplying by the effective column diameter

= 3494.40 N/mm2

where Fw = Loading per unit length

= Pwx(mean diameter including insulation)

Deff = Effecting column diameter, the outside

diameter plus allowance for the thermal

insulation.

= 1.780 + 2(9.773+75)x10-3

= 1.950 m

4.2.3.1.3.2 Bending moment

For a uniformly loaded cantilever, the bending moment at any plane is given

by:

= 3494.40 (4.88) 2

2

= 41608.52 Nm

where X = distance measured from the free end (Hv)

W = Fw = load per unit length (N/m)

4.2.3.1.3.3 Calculation for bending stress

The bending stresses will be compressive or tensile, depending on location,

and are given by;

= 0.106 N/mm2

where Iv = second moment of area of the vessel

about the place of bending

= 3.522 x 10 11 mm4

Outside diameter of vessel

=

= 1780 + 2(9.7730

= 1799.55 mm

4.2.3.1.3.4 Principle stresses

The resultant longitudinal stress is :

is compressive and therefore negative

(upwind) = 45.534 – 2.446 +0.106 =43.194 N/mm2

(downwind)= 45.534 -2.446 -0.106 =42.982N/mm2

As there is no torsional shear stress, the principal stresses will be z and h.

h = 43.406 N/mm2

The greatest principal stresses that acted on the vessel is 0.212N/mm2 which is well

below the maximum allowable design stress.

4.2.3.1.3.5 Check Elastic Stability (Buckling)

A vessel design must be checked to ensure that the maximum value of

resultant axial stress (compressive) does not exceed the critical value at which

buckling will occur. For steel cylindrical vessels, the critical buckling stress is given

by:

= 2 x 10-4 (9.773/1780)

= 109.81 N/mm2

The maximum compressive stress will occur when the vessel is not under pressure.

Maximum compressive stress =

= 2.552 N/mm2

which is well below the critical buckling stress and maximum allowable design

stress.

4.2.4 Vessel Support

The method used to support a vessel will depend on size, shape and weight

of the vessel, the design temperature and pressure, the vessel location and

arrangement: the internal and external fittings and attachments.

Since the design reactor is a vertical vessel, a skirt support is recommended

as it does not impose concentrated loads on the vessel shell. Supports will impose

localized loads on the vessel wall, and the design must be checked to ensure that

the resulting stress concentrations are below the maximum allowable design stress.

4.2.4.1 Skirt supports

A skirt support consists of a cylindrical or conical shell welded to the base of

vessel. A flange at the bottom of the skirt transmits the load to the foundations.

Openings must be provided in the skirt for access and for any connecting pipes.

4.2.4.2 Skirt thickness

The skirt thickness must be sufficient to withstand the dead weight loads and

bending moments imposed on it by the vessel; it will not be under the vessel

pressure.

4.2.4.3 Structure of skirt

The skirt is not required to withstand the pressure in the vessel and in the

condition of the fluid, then the selection of material is not limited to steels permitted

by the pressure vessel codes. A straight cylindrical skirt of plain carbon steel with

design stress 105 N/mm2 is used.

4.2.4.4 Height of the skirt

The height of the skirt is taken as I m

4.2.4.5 Stresses analysis on skirt

The resultant stresses in the skirt will be:

s(tensile) = bs - ws

s(compressive) = bs + ws

where bs = bending stress in the skirt

=

ws = the dead weight stress in the skirt

=

where Ms = maximum bending moment

W = total weight of the vessel and contents

Ds = inside diameter of skirt


The skirt thickness should be such that under the worst combination of wind and

dead weight loading the following design criteria not exceeded

where fs = maximum allowable design stress for the skirt material,

normally taken at ambient temperature.

J = weld joint factor, if applicable

s= base angle of a conical skirt (80o to 90o)

4.2.4.6 Calculation of bending stress at the base of the skirt

Wind loading, Fw = 3494.40 N/m

Bending moment at base of skirt,

Ms = Fw x ½ (Hv + Hs)2

= 3494.40 x ½ (4.88 + 1)2

= 60408.39 Nm

Where Hv = Height of vessel

Hs = Height of skirt

4.2.4.7 Calculation of bending stress in the skirt

Weight of vessel = 134.408 kN

bs =

bs = 4(60408.39 x 10 3 ) (1780 + 18)18(1780)

= 1.335 N/mm2

ws =

ws = 134.408 x10 3 (1780 + 18)18

= 1.322 N/mm2

Maximum = 1.335 + 1.322 = 2.657 N/mm2

Maximum = 1.335 – 1.322 = 0.013 N/mm2

Take joint factor J as 0.85 because type of joint is double welded butt and requires

less non-destructive testing but places some limitations on the materials which can

be used and the maximum plate thickness, and Young’s Modulus 200000N/mm2.

Criteria for design:

0.85(105) sin 90o

89.25N/mm2

0.125 (200000)(18/1780)sin 90o

252.81 N/mm2

Both criteria are satisfied, add 2 mm for corrosion, gives a design thickness of 20

mm.

4.2.5 Base Ring and Anchor Bolt Design

The loads carried by the skirt are transmitted to the foundation slab by the

skirt base ring (bearing plate). The moment produced by the wind and other lateral

loads will tend to overturn the vessel: this will be opposed by the couple set up by

the weight of the vessel and the tensile load in the anchor bolts.

Since reactor is considered as small vessels, the simplest type rolled angle rings is

recommended.

Scheiman’s method can be used for preliminary design.

4.2.5.1 Calculation for area of bolt

The anchor bolts are assumed to share the overturning load equally, and the

bolt area required is given by:

where Ab = area of one bolt at the root of the thread,mm2

Nb = number of bolts

fb = maximum allowable bolts stress, N/mm2 :

typical design value 125 N/mm2 (18,000 psi)

Ms = bending (overturning) moment at the base, Nm

W = weight of the vessel, N

Db = bolt circle diameter, m

Scheiman gives several guide rule for selecting the anchor bolts.

a. Bolts smaller than 25 mm (1 in) diameter should not be used.

b. Minimum number of bolts 8

c. Use multiples of 4 bolts.

d. Bolt pitch should not be less than 600 mm (2 ft)

Let the pitch circle diameter = 0.49 m

Circumference of bolt circle = 1540 mm

Number of bolts required, at minimum recommended bolt spacing

= 1540 600= 2.57

Since the minimum number of bolts is 8, therefore 8 bolts are used.

Take bolts design stress =125 N/mm2

Ms = 60408.39 Nm

Take W= operating value = 134.408 kN

= 359 mm2

From BS 4190 : 1967, M24 bolts with root area of 353 can be used.

Bolt root diameter = (353 x 4/ )1/2

= 21.20 mm

4.2.5.2 Calculation for minimum thickness of base ring

The base ring must be sufficiently wide to distribute the load to the

foundation. The total compressive load on the base ring is given by:

= 48304.81 N/m

where Fb = the compressive load on the base ring, Newtons per

linear metre

Ds = skirt diameter, m

The minimum width of the base ring is given by:

where Lb = base ring width, mm

fc = the maximum allowable bearing pressure on the

concrete foundation pad, which will depend on the mix

used, and will typically range from 3.5 to 7 N/mm2 (500 to

1000 psi)

Taking bearing pressure as 5 N/mm2

= (48304.81/5) x (1/103)

= 9.66 mm

Actual width required = Lr + ts + 50 mm

= 64 + 18 + 50

= 132 mm

From M24 (BS 4190 : 1967), Lr = 64

Actual bearing pressure on concrete foundation:

48304.81 / 132 x 103

= 0.366 N/mm2

The minimum thickness is given by:

where tb = Base ring thickness, mm

Lr = The distance from the edge of the skirt to the outer edge

of the ring, mm

f’c = Actual bearing pressure on base , N/mm2

fr = Allowable design stress in the ring material, typically

140 N/mm2

= 64 ((3 x 0.366)/140)1/2

= 5.67 mm

4.2.6 Bolt Flanged Joint

Flanged joints are used for connecting pipes and instruments to vessels, for

manhole covers, and for removable vessel heads when ease of access is required.

Flanges may also be used on the vessel body, when it is necessary to divide the

vessel into sections for transport or maintenance.

4.2.6.1 Selection of Flange

Since the operating temperature of the reactor is to be considered high,

welding–neck flanges are recommended which are suitable for extreme service

conditions such as high temperature. They will normally be specified for the

connections and nozzles on process vessels and process equipment. They have a

long tapered hub between the flange ring and the welded joint. This hub provides a

more gradual transition from the flange ring thickness to the pipe wall thickness,

thereby decreasing the discontinuity stresses and consequently increasing the

strength of the tube flange.

4.2.6.2 Selection of Gaskets

Gaskets are used to make a leak-tight joint between two surfaces. It is

impractical to machine flanges to the degree of surface finish that would be required

to make a satisfactory seal under pressure without a gasket. Gaskets are made from

‘semi-plastic’ materials; which will deform and flow under load to fill the surface

irregularities between the flange faces, yet retain sufficient elasticity to take up the

changes in the flange alignment that occur under load. An Iron or soft steel is

recommended for this vessel since they are normally used for higher temperature.

4.2.6.3 Flange Faces

The raised face, narrow faced which is probably the most commonly used

types of flange are used for all the flanges.

4.2.6.4 Flange Design

The bolts hold the flange faces together, resisting the forces due to internal

pressure and gasket sealing pressure. As these forces offset, the flange is subject to

a bending moment. A flange assembly must be sized so as to have sufficient

strength and rigidity to resist this bending moment.

The total moment acting on the flange is given by:

where = gasket reaction (pressure force) =

= pressure force on the flange face =

= total pressure force =

= pressure force on the area inside the flange

=

= mean diameter of the gasket

= inside diameter of the flange

= effective gasket pressure width

= effective gasket sealing width

The minimum required bolt load under the operating condition is given by:

The moment Matm is given by:

where Wmz is the bolt load required to seat the gasket, given by:

where y is the gasket seating pressure (stress)

The flange stresses are given by:

Longitudinal hub stress,hb = F1M

Radial flange stress, rd = F2M

Tangential flange stress,hb = F3M – F4rd

Where M is taken as Mop or Matm, whichever is the greater.F1 and F4 are the

flange type and dimensions, are obtained from equations and graphs given

in BS5500.

The design criteria of flange are:

where is the maximum allowable design stress for the flange material at

the operating conditions.

4.2.7 Pipe Sizing

The pipe diameter can be obtained from the following equation below:

Carbon steel pipe;

doptimum = 293 G0.53 -0.37

Stainless steel pipe;

doptimum = 260 G0.52 -0.37

where d = optimum diameter of the pipe, mm

G = flow rate of fluid in the pipe, kg/s

= density of fluid, kg/m3

Equation below can be used to calculate the thickness where the pipe diameter is

considerably large.

where Di = optimum diameter of pipe.

4.2.7.1 Calculation of Pipe Diameter

i. Feed Stream

Flow rate = 17.04 kg/s

Density of the stream = 658.10 kg/m3

Stainless steel is recommended for the construction of the pipe.

For stainless steel,

doptimum= 260(17.04)0.52(658.10)-0.37

= 104.06 mm

ii. Inlet and Outlet Stream for Cooling Water

Stainless steel is recommended for the construction of this pipe

For stainless steel pipe,

doptimum = 260 (39.673)052(998.20)-0.37

= 136.92 mm

4.2.8 Compensation for Openings and Branch Connections

The presence of openings and branches weakens the shell and give rise to

stress concentrations. Sufficient reinforcement must be provided to compensate for

the weakening effect of the opening.

The “equal area method” is chosen because it is the simplest method used

for calculating the amount of reinforcement required and experience has proved it to

be satisfactory for a wide range of application.

I. Feed stream

From Perry’s Handbook,

For d = 104.06 mm

Nominal pipe size = 127.00 mm

Outside diameter = 141.30 mm

Nominal wall thickness = 19.05 mm

Minimum thickness of branch , e1 = PiDi

(2f-Pi)

= 0.454 mm

The nominal pipe wall thickness is above minimum thickness of branch,

so no reinforcement of the branch is required.

II. Cooling Water Stream

For d = 136.92 mm

Nominal pipe size = 127.00 mm

Nominal wall thickness = 2.767

Minimum thickness of branch, e1 = 0.60 mm

The nominal pipe wall thickness is above the minimum thickness of

branch, so no reinforcement of the branch is required.

4.2.8.1 Manholes

The maximum length of manhole is dependent on the manhole diameter.

The length is perpendicular distance fro the face of the opening including lining or

any projection of the branch within the vessel.

Type of branch connection: flush nozzle

Inside diameter = 598.50 mm

Nominal size = 600 mm

Outside diameter = 609.60 mm

Nominal wall thickness = 5.54 mm

4.2.8.1.1 Compensation for manholes

Actual thickness

ta = (do – di) / 2

= 5.55 mm

Minimum thickness

e1 = PiDi

(2f-Pi)

= 2.614 mm

Distance, N = 2.5 ta

= 13.875 mm

Length, S = di /2

= 299.25 mm

Area removed, X = edi

2 = (7.773)(598.5)

2

= 2326.07 mm2

Compensation area, Y = Nta – Ne1 + Stc

= 40.737 +299.25tc

tc is the thickness for compensation

Area X = Area Y

2326.07 = 40.737 +299.25tc

tc = 7.637 mm

4.2.1.8.2 Flat end closure for manholes

Flat plates are used to blank off flange connections, and as covers for

manholes and inspection parts. Flat end closures are blind flanges, bolted cover

with a full face gasket,

The thickness required will be depend on the degree of constraint at the

plate periphery. The minimum thickness required is given by:

e = CpDe(Pi/f)1/2

where Cp = design constant = 0.4

De = bolt circle diameter = 490 mm

f = design stress, 115 N/mm2

Minimum thickness for flat end closures, e = 18.277 mm

Add 2 mm for corrosion allowance, e = 20.277 mm

REFERENCES

Coulson and Richardson. 1999. Chemical Engineering Volume 6.

Butterworth Heinemann.

Coulson and Richardson. 1971,Chemical Engineering Volume 3.

Pergammon Press.

Massimo Morbidelli. 2001. Catalyst Design- Optimal Distribution of Catalyst

in Pellets reactors and membranes. Cambridge University Press.

130.

M.J Slater. 1992. Ion Exchange Advances-Proceedings of IEX’s. Elsevier

Science Publisher Limited.

G.H. Osborn. 1961. Synthetic Ion Exchange. London Chapman and Hall

Limited. Page 1-17.

Robert C. Reid. The Properties of Gases and Liquid. Fourth Edition.

McGraw Hill Inc..Page 433.

James M. Douglas. 1998. Conceptual Design of Chemical Process.

McGraw Hill International Editions. Page 329.

Michael Streat. 1988. Ion Exchange for Industry. Ellis Herwood Limited.

Page 585.

M. Necati Ozisik. 1985. Heat Transfer-A Basic Approach. McGraw Hill Book

Company. Page 385-397

Octave Levenspiel. 1999. Chemical Reaction Engineering. John Wiley and

Sons. Page 367-509.

Robert H. Perry. 1997. Perry’s Chemical Engineer’s Handbook. 7th Edition.

McGraw Hill.


Dimension

A Total Heat Transfer Area for Tubes L-2

Ab Area of One Bolt At The Root of the Thread

As Cross section Area of Shell L-2

At Total Cross Section Area of Tubes L-2

Bs Baffles Spacing L

Cs Stress Concentration Factor for Torispherical

Head

Cv Account Factor

d Particle diameter

Db Bolt Circle Diameter L

Db Bundle diameter L2

De Effective diffusivity L2T-1

De Effective Column Diameter L

De’ Equivalent Diameter of Shell L2

Dk Knudsen Diffusivity L2T-1

Di Inner Diameter of Tube L

Di Inner Diameter L

Dm Vessel Mean Diameter L

Do Outer Diameter of Tube L

Ds Shell Diameter L

Ds Skirt Diameter L

dc Diameter of Catalyst L

E Young’s Modulus ML-1T-2

e Shell Thickness L

eh Domed Head Thickness L

Fb Compressive Load on the Base Ring MT-2

Fw Wind Loading MT-2

fb Maximum Allowable Bolt Stress ML-1T-2

fc Maximum Allowable Bearing Pressure on

Concrete Foundation ML-1T-2

fs Maximum Allowable Design Stress for the

Skirt Material ML-1T-2

f’c Actual Bearing Stress on Base ML-1T-2

g Gravitational Acceleration LT-2

Hs Skirt Height L

Hv Height Between Two Tangent Lines of A Vessel L

Hi Tube Side Heat Transfer Coefficient MT-3-1

Hio Corrected Tube Side Coefficient MT-3-1

Hs Shell Side Heat Transfer Coefficient MT-3-1

Iv Second Moment of Area L4

J Welded Joint Efficiency

jf Shell Side Friction Factor

jHS Shell Side Heat Transfer Factor

jHt Tube Side Heat Transfer Factor

K1,K2 Velocity constant of Reaction T-1

Kf Thermal Conductivity of Fluid In Tubes MLT-3-1

L Length of Tube L

Lb Base Ring Width L

M Mass flowrate

Mx Bending Moment ML2T-2

Nb Number of Bolts

Nc Number of Crosses

nt Total number of Tubes

PD Design Pressure ML-1T-2

Pi Internal Pressure ML-1T-2

Pw Wind Pressure ML-1T-2

P Pressure Drop In the Tube Side ML-1T-1

Pr Prandt Number

Ps Pressure Drop In Shell Side ML-1T-1

Pt Tube Pitch L

Rc Crown Radius L

Re Reynold number

r Pore radius

Sg Total Surface area of Catalyst M-1L2

T Operating temperature

t Wall Thickness L

ts Skirt Thickness L

TLMTD Log Mean Temperature

Ud Design Overall Heat Transfer Coefficient MT-3-1

V Volume flowrate M-2L3

Vg Void Volume of Catalyst M-2L3

W Weight of Catalyst M

Wc Weight of Catalyst MLT-2

W Total Weight MLT-2

Wt Weight of Tubes MLT-2

Wi weight of Insulation MLT-2

Wv Weight of Vessel MLT-2

X Total Conversion of Reactant

b Voidage of Catalyst

p Internal void fraction

Fluid density ML-3

p Particle Solid Density ML-3

Thiele Modulus

Effectiveness Factor

D Design Stress ML-1T-2

L Longitudinal Stress ML-1T-2

b Bending Stress ML-1T-2

h Circumferential Stress ML-1T-2

m Maximum Compressive Stress ML-1T-2

w Dead Weight Stress ML-1T-2

z Resultant Longitudinal Stress ML-1T-2

r Radial Stress ML-1T-2

Fluid viscosity at the bulk fluid temperature ML-1T-2

w Fluid viscosity at the wall ML-1T-2


MTBE DISTILLATION COLUMN MOHD. NAZRI BIN ISMAIL

SUPERVISORS



CHAPTER 5:

MTBE (C5H12O) DISTILLATION

COLUMN

CONTENTS

TITLE PAGE

CHAPTER 5 MTBE DISTILLATION COLUMN



5.5.1 Determination of Key Components 134

5.5.2 Determination of Bubble Point and

Dew Point 134

5.5.3 Determination Relative Volatility, 136

5.5.4 Determination The Number of

Stages 137

5.5.5 Calculation to Determine Overall

Tray Efficiency, Eo 138

TITLE PAGE

5.5.6 Determination Of Feed Point

Location 138

5.5.7 Estimate or Gather The Physical

Properties 139

5.5.8 Determination Of Maximum And

Minimum Vapor and Liquid Flow

Factor and Flooding Velocity

For The Turn Down Ratio 140

5.5.9 Determination Of Column Diameter 141

5.5.10 Liquid Flow Arrangements 142


5.5.12 Determination of Weir Length, lw 143

5.5.13 Check The Weeping Rate 143

5.5.14 Plate Pressure Drop 145

5.5.15 Downcomer Design 146

5.5.16 Check Entrainment 147


5.5.18 Number Of Hole 148

5.5.19 Column Size 149



5.6.2 Material Construction 151

5.6.3 Vessel Thickness 151

5.6.4 Heads and Closure 152

5.6.5 Column Weight 153

5.6.6 Wind Loads 154


(Skirt Design) 155

5.6.8 General Consideration For Design 157


5.6.10 Feed, Top Product, Bottom Product

Piping Sizing 159

CHAPTER 5

MTBE (C5H12O) DISTILLATION COLUMN

5.1 Introduction

Basically, the function of distillation is to separate by vaporization, a liquid

mixture of miscible and volatile substances into individual components or

some into groups of components. It also known as a method used to

separate the components of a liquid solution, which depends upon the

distribution of these various components between a vapor and a liquid

phase. All components are present in both phases. The vapor phase is

created from the liquid phase by vaporization at the boiling point.

In our project design, the MTBE Distillation Column is been

selected as a part of equipment design for objective to separate MTBE

composition from Methanol (CH3OH). This distillation column is important

for the MTBE plant production for recycle back methanol from the MTBE

Distillation Column. Besides that, the equipment design of the MTBE

Distillation Column is also consider the multicomponent distillation method

which means that for this distillation method it consist the feed component

with more than one component. Therefore, the determination of the

minimum number of stages of this MTBE Distillation Column, Nm by using

the Frenske Equation: Overall minimum total trays with total condenser

(Reference: Applied Process Design, Volume 2: Third Edition) has been

selected as a methodology for this equipment design.

The characteristics in chosen types of distillation column are

requirement of separation objective satisfied with this distillation column,

the cost of construction and simplicity in design. The design of a distillation

column can be divided into the following steps:

1. Specify the degree of separation required; set product specifications.

2. Select the operating conditions: batch or continuous; operating

pressure.

3. Select the type of contacting device: plates or packing.

4. Determine the stage and reflux requirements: the number of equilibrium

stages.

5. Size of the column: diameter, number of real stages.

6. Design the column internals: plates, distributors, packing support.

7. Mechanical design: vessel and internal fittings.

5.2 CHEMICAL DESIGN

5.2.1 Determination of Key Components

Firstly, we must determine the key components which involving in this

distillation method. There are 2 main key components in the distillation

method:

1. Heavy Key Component, KHK

2. Light Key Component, KLK

Therefore, the determination of the key components are :

1. Heavy Key Component, KHK = MTBE (C5H12O)

2. Light Key Component, KLK = Methanol (CH3OH)

5.2.2 Determination of Bubble Point and Dew Point

The vapor pressure can be calculated from Antoine equations:

ln P* = A -

Where P* = vapor pressure (mm Hg)

A, B and C = The Antoine coefficients (All these value are referred

Appendix D, Coulson and Richardson, Volume 6, 1999)

T = Temperature, K

The designing of an evaporation or condensation process, the

most important that we must know that the conditions which the transition

from liquid to vapor and from vapor to liquid takes place. This principle is

also considered in designing other processes such as distillation,

absorption and stripping which requires information on the conditions at

phase transitions occur and on the compositions of the resulting phases.

The bubble – point temperature, Tbp is the temperature at which

the first vapor bubble forms when the liquid is heated slowly at constant

pressure. Meanwhile the dew – point temperature, Tdp is the temperature at

which the first liquid droplet forms when a gas or vapor is cooled slowly at a

constant pressure.

For bubble – point conditions. By using Raoult’s law for an ideal

liquid solution and contains species such as A, B, C. With known mole

fraction each component, xA, xB, xC…. Let assume that the vapor is ideal

(follows the ideal gas equation of state) and since the vapor is in

equilibrium with liquid, therefore the partial pressures of the components

are given by Raoult’s law,

pi = xipi* (Tbp) (5.1)

Where, pi* = The vapor pressure of component i at bubble – point

temperature

i = Components

Therefore, the sum of the partial pressures must be the total

system pressure, P;

P = xA pA*(Tbp) + xB pB*(Tbp) +………. (5.2)

This bubble point temperature may be calculated by trial and error

as the value of (Tbp) that satisfies this equation. Once (Tbp) is known, the

composition of the vapor phase can be easily be determined by evaluating

the partial pressures each component from Equation 1 and determining

each vapor – phase mole fractions as

yi = pi / P (5.3)

Equation 5.2 can be used to determine such a pressure for an

ideal liquid solution at a specific temperature and mole fractions in the

vapor in equilibrium with the liquid can be determined as

yi = pi / P = xipi* (T) / Pbp (5.4)

For dew – point conditions. This calculation is using the similar

method from bubble – point temperature estimation. There are suppose a

gas phase contains the condensable components A, B, C … and a

noncondensable components at fixed temperature. By assuming applying

Raoult’s law, the liquid – phase mole fractions may be calculated as;

x i =

(5.5)

Where, i = components, A, B, C…excluding noncondensable components

= The mole fraction of component i in the gas

The mole fractions of the liquid components (those that are

condensable) at the dew point of the gas mixture must sum to 1.

xA + xB +xC +……….. = 1 (5.6)

From Equation 5.4,

+ +…….=1 (5.7)

The value of Tdp can be found by trial and error once expressions

for ) have been substituted. The dew – point pressure can be

determined from Equation 5.6 with Tdp replaced by system temperature, T.

From the calculation which are included at Table 5.2 (APPENDIX

E), The Appendix given, the value of the boiling point temperature, Tbp and

the dew point temperature, Tdp:

1). At Feed Stream, Tbp = 400.6 K

2). At Distillate (Top) Stream, Tdp = 335.3 K

3). At Bottom Stream, Tbp = 408.2 K

5.2.3 Determination Relative Volatility,

The determination of relative volatility, of the components can be

determined as the ratio between K values of light key component to heavy

key component:

= (5.8)

Where = Light key component, Methanol (CH3OH)

= Heavy key component, MTBE (C5H12O)

For both K values, they are determined as shown in formula below:

K = (5.9)

Where, = Mole fraction (liquid) component

= Mole fraction (vapor) component

Determination of relative volatility, can be referred to Table 5.4

from the Appendix and the values of K can be referred to Table 5.3

(APPENDIX E)

5.2.4 Determination The Number of Stages

The determination of the minimum number of stages of this MTBE

Distillation Column, Nm by using the Frenske Equation :Overall minimum

total trays with total condenser (Reference : Coulson and Richardson,

Volume 6, 1999)

Nm = (5.10)

Nm =

Nm =

Nm = 2.1285

In order to the number of theoretical stages by using the Gilliland

Correlations. Normally after using the Fenske’s Equation, the value of Nmin

is given by the equation below to get the number of stages, NT,

NT = 2Nm (5.11)

NT = 2(2.1285)

= 4.2570

5 stages

5.2.5 Calculation to Determine Overall Tray Efficiency, Eo

By using O’Connell Correlations equation,

Eo = (5.12)

From calculation,

1. For feed, overall viscosity, = 0.355419cP

2. For Distillate, overall viscosity, = 0.1294 cP

3. For Bottom, overall viscosity, = 0.1998 cP

Average viscosity between distillate and bottom

viscosity, =

= 0.1646 cP

From the Excel calculation, the average volatility, or a between distillate

and bottom is 315.1148.

Therefore, Eo = =

Eo = 0.1863

Eo = 18.63 %

Finally, to determine the real number stages of this MTBE

Distillation Column.

Number of real stages = (5.13)

No of stages =

= 21.4707

22 stages

5.2.6 Determination Of Feed Point Location

In order to find the feed point location, estimation can be made by using the

Fenske equation to calculate the number of stages in the rectifying and

stripping section separately, but this requires an estimate of the feed point

temperature. As an alternative approach, here I use the empirical equation

given by Kirkbride (1944) as a matter for the same objective.

(5.14)

Where,

Nr = number of stages above the feed, include the condenser

Ns = number of stages below the feed, include the reboiler

B = molar flow bottom product

D = molar flow top product

Xf, HK = concentration of the heavy key in the feed

Xf, LK = concentration of the light key in the feed

Xd,HK = concentration of the heavy key in the top product

Xb, LK = concentration of the light key in the bottom product

From previous calculation, number of stages, excluding the reboiler = 21

Nr + Ns = 21

Ns = 21 – Nr = 21 – 0.0782Ns

Ns = 19.4769

Ns 20

5.2.7 Estimate or Gather The Physical Properties

The properties consider in this design are liquid flow rate, LW, vapor flow

rate, VW, liquid surface tension, σ, liquid density, ρl and vapor density, ρv.

This physical properties evaluated at the system temperature by using

HYSIS generated data or estimate manually from mass and energy

balance data. The useful properties data is given as below:

Feed;

Liquid flow rate, LW = 55802.0640 kg/hr = 15.5006 kg/s

= 0.013245 N/m

Liquid density, ρl

=

= 0.6673g/ml

= 667.3 kg/m3

Distillate;

Vapor flow rate, VW = 5531.1456 kg/hr = 1.5364 kg/s

Vapor density, ρv (5.15)

RMM = Relative molecular mass

*Most data evaluate at system temperature and pressure

5.2.8 Determination Of Maximum And Minimum Vapor And Liquid

Flow Factor And Flooding Velocity For The Turn Down Ratio

Liquid-vapour flow factor were determine by using below equation

FLV = (5.16)

FLV

Assumption were made for initial tray spacing based on value of FLV by

referring to figure 11.27 from Coulson Richardson Chemical engineering

volume page 567. The data were used to determine the constant, K1 for

estimation of flooding velocity.

So, Assumption initially 0.5m of tray spacing, the value k1 = 0.080

(constant) and correction factor are used as equation below:

K1 = k1 (5.17)

= 0.080

= 0.0736

and flooding velocity, Uf determine by equation 5.18.

(5.18)

Uf =

The flooding percentage was assumed to be 85%, this is based on flooding

velocity for design, a value of 80 to 85 %. Therefore, Uv were found by

using below,

UV = 0.80 (Uf) (5.19)

= 0.80 (0.6284 m/s)

= 0.5027 m/s

0.50 m/s

5.2.9 Determination Of Column Diameter

Based on flooding (distillate) consideration by using equation 5.20,

Dc = (5.20)

Dc =

= 0.6582m

0.7 m @ 700mm

5.2.10 Liquid Flow Arrangements

Before deciding liquid flow arrangement, maximum volumetric flowrate

were determined by using,

VL = (5.21)

VL =

= 1.7166 m3/s

Based on value of volumetric flow rate and column diameter, DC. Figure

11.28 from Coulson Richardson Chemical engineering volume 6, page

568. Therefore, types of liquid flow found as single pass.

5.2.11 Plate Layout

The value of downcomer area, active area, hole area, hole size, and weir

height were determined based on above value calculated, trial plate layout

column area determine by using below,

Column area, AC = (5.22)

AC =

= 2.0000 m2

Where Um = Velocity at below plate,

Down comer area were found by assume 20% of column area and using

below,

Down comer Area, Ad = 0.2 AC (5.23)

Ad = 0.2(2.0000m2)

= 0.4 m2

Net area and active area were determined by using equation 5.24 and

equation 5.25,

Net Area, An = Ac - Ad (5.24)

= 2.000 – 0.4000 m2

= 1.6000 m2

Active area, Aa = Ac - 2Ad (5.25)

= 2.0000 - 2(0.4000)

= 1.2000 m2

Hole Area, AH are determine with trial value of 10% active area by equation

5.27,

Hole Area, AH = 0.10(Aa) (5.26)

= 0.10(1.2000)

= 0.1200 m2

Weir Length, lw calculated by referring figure 11.31 from Coulson

Richardson Chemical engineering volume 6, page 572 which determined

based value the ratio of Ad/Ac to get the ratio of lw/ Dc

5.2.12 Determination of Weir Length, lw

The weir height determine from standard from as below

Weir Height = 50 mm (Standard)

Hole diameter = 5 mm (Standard)

Plate Thickness = 5 mm (Standard)

From figure 11.31,Coulson Richardson Chemical engineering volume 6,

page 572,

When Ad/Ac x 100% = 18%, lw/ Dc = 0.85

From calculation,

DC =0.7000 m

lw/ DC = 0.85

lw = 0.85DC

lw = 0.5950 m

5.2.13 Check The Weeping Rate

By using Francis equation from Coulson Richardson Chemical engineering

volume 6, page 571 to determine the height over the weir ;

how = (5.27)

=

= 86.3114 mm liquid

Where,

lw = Weir length

how = weir crest

Lw = liquid flow rate

So at minimum liquid flow rate determine by adding weir height hw and weir

crest, how. After that, the constant, K2 where find based on the value and

referring to figure 11.30 from Coulson Richardson Chemical engineering

volume 6, page 571;

hw + how (mm) = 50 mm + 86.3114 mm

hw + how (mm) = 136.3114 mm (since this value is too big, I use hw + how =

100 mm )

hw + how (mm) 100mm

K2 = 31

Minimum vapor velocity Uh, were determine by equation 5.29

Uh = (5.28)

=

= 4.2064 m/s

4.2 m/s

And actual minimum vapor velocity is ratio of Minimum vapor rate / Ah were

determine and comparing to value of weep point. Satisfaction value must

above weep point. Therefore, the calculation as below;

Actual minimum vapor velocity = Minimum vapor rate / Ah (5.29)

=

=

= 5.8637 m/s

6 m/s

and minimum vapor velocity is above weep point velocity.

5.2.14 Plate Pressure Drop

Plate pressure Drop was calculated by:

Maximum vapor velocity through hole,

Uh = (5.30)

=

= 8.3767 m/s

8.4 m/s

By referring to figure 11.34 from Coulson Richardson Chemical Engineering

volume 6, page 576

Ah/Ap = 0.1

Ah/Ap x 100% = 0.1 x 100% = 10

Plate thickness/Hole diameter = 0.005m/0.005m = 1

Orifice coefficient, Cd = 0.84

Then pressure Drop through dry plate

hd =51 (5.31)

= 51

= 69.0109 mm

70 mm

Residual pressure drop

hr = (5.32)

=

= 18.7322 mm liquid

Total pressure drop,

ht = hd+(hw + how) + hr (5.33)

= 70 + 100 + 18.7322

= 188.7 mm liquid

190 mm liqiuid (H2O)

5.2.15 Downcomer Design

Downcomer pressure drop

hap = hw –10 (5.34)

= 50 – 10 mm

= 40 mm

Area under downcomer

Aap = hap x lw (5.35)

= 0.04 m x 0.5950m

= 0.0238 m2 (less than Ad = 0.4 m2)

Since Aap is less than Ad so we use Aap to calculate head loss in downcomer

hdc

hdc = (5.36)

=

= 0.33821mm liquid

Where

Lwd = liquid flow rate in downcomer

Am = Ad (downcomer) area or Aap (Area under downcomer) either is

smaller

,hdc = head loss in downcomer

Downcomer backup, hb

hb = (hw +how) + ht + hdc (5.37)

= 100 + 190 + 0.3382

= 290.3382 mm liquid

hb @ ½ (lt + hw)

Resident time, tr were determined by equation 6.44,

tr = (5.38)

=

= 108.08660 second

108 second

The value is relevant and recommended to proceed for another design.

Therefore, plate layout details, calming zones, unperforated area,

and check hole pitch will be decide as below section.

5.2.16 Check Entrainment

Entrainment were checked by determined actual flooding percentage, Uv by

using equation 6.40 and equation 6.41,

Actual % of flooding, Uv

Uv = (5.39)

=

= 0.5026 m/s

Then determine the percentage of flooding,

% of flooding = (5.40)

=

= 79.9809%

80%

After that, fractional entrainment was getting based on this percentage and

FLV = 1.1736, by referring figure 11.29 from Coulson Richardson Chemical

engineering volume 6, page 570. So:

Fractional entrainment Ψ = 0.08

5.2.17 Plate Layout

Angle subtended by unperforated strip were determine from figure 11.32

from Coulson Richardson Chemical engineering volume 6, page 573. An

example likes below:

For, lw/Dc = (5.41)

= 0.8500

θc = 130o

So angle = 180 o – 130 o (5.42)

= 50o

.Mean Length unperforated edge strip = (Dc – 0.05) (5.43)

Mean Length unperforated edge strip = (0.7000 -0.05)

= 0.8727 m

Area of unperforated edge strip = 0.005m x 0.8727m

= 0.0044 m2

Mean length of claming zone = (Dc – 0.05) sin (5.44)

= (0.7000 – 0.05) sin

= 0.5891 m

Area of claming zone = 2(0.05)(Mean length of claming zone) (5.45)

= 2(0.05)(0.5891)

= 0.0589 m2

Total Area available for perforation, Ap

Ap = Aa – (Area of unperforated + area of claming zone) (5.46)

= 1.200 – (0.044 + 0.0589)

= 1.0971 m2

Determine ratio (5.47)

=

= 0.1094

and referring to figure 11.32 from Coulson Richardson Chemical

engineering volume 6, page 574.

= 2.8 satisfactory, within 2.5 to 4.0 (5.48)

5.2.18 Number Of Hole

Area of hole

AH = (5.49)

=

= 0.00001964 m2

= 1.9635 x10-5 m2

Number of Hole = (5.50)

=

= 6111.5498unit

≈ 6112 units

5.2.19 Column Size

Column Diameter = 0.7000 m

≈ 700 mm

Column Height;

= (No stage –1) (tray spacing) + (tray spacing x 2) +

(No stage-1)(thickness of Plate) (5.51)

= (22 –1)(0.4)+(0.4)(2) + (22-1)(0.05)

= 8.4 m + 0.8 m +1.05 m

= 10.25 m = 10250 mm

Table 5.5: Summary of Chemical Engineering Design

5.1 Item Value Unit

Column diameter 0.70 m

No of plate 22

Plate spacing 0.4 m

Plate thickness 0.005 m

Total column height 10.25 m

Plate pressure drop 0.07 m

Plate material SS 304

Downcomer area 0.4 m2

Downcomer material SS 304

Column area 2.0000 m2

Net area 1.6000 m2

Active area 1.2000 m2

Hole area 0.1200 m2

Number of Hole 6112 Units

Weir Length 0.5950 m

Weir height (standard) 0.005 m

Resident time 108 s


5.3.1 Design Pressure

In mechanical design, there are two parameters such as temperature and

pressure are important properties in order evaluate the thickness and the

stress of material. Therefore, the safety factor is added as precaution and

determined by certain consideration such as corrosion factor, location and

process characteristic.

The operating pressure is 802.5548 kPa or 8.025548 atm or

8.025548 bars and the safety factor is 10% above operating pressure. The

design pressure calculated as below equation.

Design Pressure, Pi = (Operating P –1) x 1.1 (5.52)

= (8.03 –1) x 1.1

= 7.733 bar

= 0.7733 N/mm2

Operating Temp, T = 135.21oC

Design Temp. , T = Operating T (ºC) x 1.1 (5.53)

= 135.21ºC x 1.1

= 148.731ºC

5.3.2 Material Construction

The material used is stainless steel (18Cr/8Ni, 304). For this material, the

design stress at 150 ºC is obtained from table 13.2, page 809 Chemical

Engineering Vol. 6.

Design stress, f = 130 N/mm2

= 1.30 x 108 N/m2

Diameter vessel, Di = 0.7000 m

Tensile strength, = 510 N/mm2

= 5.1 x 108 N/m2

5.3.3 Vessel Thickness

The thickness of column and other design are calculated based on

equation below;

e = (5.54)

Where, Pi = Design pressure

Di = Column diameter,

,f = joint factor

e =

= 2.0882 mm

= 2.0882 mm + 2 mm (corrosion allowance)

= 4.0882 mm

For vessel diameter less than 1 m, a minimum thickness required is 5mm,

these values include a corrosion allowance of 2 mm.

As a first trial, divide the column into five sections (courses), with

the thickness increasing by 2 mm per section. Try 7, 9, 11, 13 and 15 mm

to determine the thickness average.

Therefore, average thickness = 7+9+11+13+15 mm / 5 = 11 mm (5.55)

5.3.4 Heads and Closure

Torispherical head had been choose because of operating pressure below

10 bars, and suitable for liquid vapor phase process in inconsistent high

pressure. The calculations as below with take

Crown radius, Rc = Di = 0.7000m

Knuckle radius, Rk = 6% Rc = 0.0420m

A head of this size would be form by pressing: no joints, so J = 1.0

Cs = (5.56)

=

= 1.7706

Where

Crown radius, Rc = Di

Knuckle radius, Rk = 6% Rc

Therefore, minimum thickness: e =

(5.57)

=

= 3.6692mm

For welding purposes, the thickness of head were taken as same as

thickness of the vessel, = 5 mm. Its, matching to joint factor were taken as

1.

5.3.5 Column Weight

1. Dead Weight of Vessel, Wv

For a steel vessel, the equation 6.66 are used

Wv = 240 Cv Dm (Hv + 0.8Dm) t (5.58)

Where, Dm = mean diameter, m

= (Di + t) (5.59)

Cv = a factor, take 1.15 for distillation


t = wall thickness, m

To get a rough estimate of the weight of this vessel is by using the average

thickness. 11 mm.

Dm = 0.7000 + 0.011

= 0.7110 m

Hv = 10.25 m

So,

Wv = 240 (1.15) (0.7110) (10.25 + 0.8(0.7110)) 11

= 23353.4 N

= 23.3534 Kn

2. Weight of Plates, Wp

From Nelson Guide, page 833 Chemical Engineering Volume 6; take

contacting plates, 1.2 kN/m2. The total of weight of plate determine by

multiply the value with number of plate design.

Weight of plate = Ac x 1.2 (5.60)

= 0.3848 x 1.2

= 0.4618 kN

Weight of 22 plates, Wp = 0.4618 x 22 (5.61)

= 10.1596 kN

10.2 kN

3. Weight of Insulation, Wi

The fiberglass was choosing as insulation material. By referring to Coulson

Richardson Chemical engineering volume 6, page 833,

Density, of fiber glass = 100 kg/m3

Thickness = 50 mm = 0.05 m

Volume of insulation,VI = x Dm x Hv x thickness of insulation (5.62)

= (0.7000) (10.25) (0.05)

= 1.1270m3

Weight of insulation, WI = Volume of insulation x x g (5.63)

= 1.1270 x 100 x 9.81

= 1105.587 N

= 1.1056 kN

Double this value to allow fittings, so weight of insulation will be = 2.2112

kN

4. Total Weight

Double this value to allow fittings. The total weight is the summation of

dead weight of vessel, weight of insulation, weight of plates,

Total weight = Wv + Wp + WI (5.64)

= 23.3534 + 10.2 + 2.2112

= 35.7646 kN

36 kN

5.3.6 Wind Loads

This factor also is be considered and calculated based on location and

weather surrounding. Since our plant, situated at Teluk Kalong Industrial

Park, near the Kemaman Port, therefore it has higher wind speed. This is

because it is located to the near sea.

Win speed, Uw = 160 km/hr

For a smooth cylindrical column stack, the following semi-empirical

equation can be used to estimate wind pressure using below equation

Pw = 0.05Uw2 (5.65)

= 0.05(160) 2

= 1280 N/m2

Loading per Unit Length of column, Fw

Fw =Pw Deff] (5.66)

Where, Deff = Effective column diameter

= Diameter + 2(tshell + tinsulation) (5.67)

= 0.7000 + 2(0.011 + 0.03)

= 0.98 m

Fw = 1280 x 0.98

= 1254.4N/m

Bending Moment

Mx = (5.68)

Where, X = Distance measure from the free end

= 10.25 m

Therefore,

Mx =

= 65895.2 Nm

5.3.7 Vessel Support Design (Skirt Design)

Type of support: Straight cylindrical skirt

s : 90º

Material construction: Carbon steel

Design stress, fs : 135 N/mm2 at ambient temperature, 20ºC

Skirt height: 2.5 m

Young modulus: 200, 000 N/mm2

At this condition of ambient temperature, the maximum dead weight load on

the skirt will occur when the vessel is full of the mixture.

Approximate weight, Wapprox = x DI 2 x Hv x L x g (5.69)

= (0.72)(10.25)(667.3)(9.81)

= 25822.5995N

26 kN

Weight of vessel + insulation + plates = 36kN

Therefore, total weight = Wv + Wp + WI + Wapprox (5.70)

= 62 kN

Wind load, Fw = 1516.42 N/m

= 1.51642 kN/m

Bending moment at skirt base, Ms = Fw (5.71)

= 1.2544

= 101.9593kNm

102kNm

As a first trial, take skirt thickness as same as the thickness of the bottom

section of the vessel, ts = 11 mm

Bending stresses in skirt, bs = (5.72)

Where, Ms = maximum bending moment (at the base of

the skirt)


Ds = inside diameter of the skirt base

= 0.7 m = 700 mm

Therefore, bs =

= 23721.9270 kN/m2

= 23.7219 N/mm2

Dead weight stress in the skirt, ws = (5.73)

Where, W = Total weight of the vessel and content

= 62 kN

Therefore, ws(test) =

= 1.0582 N/mm2

wbs,(operating) =

= 1.4652 N/mm2

Thus, the resulting stress in the skirt, s:

Maximum s (compressive) = ws (test) + wbs (5.74)

= 1.0582 +1.4652 N/mm2

= 2.5234N/mm2

Maximum s (tensile) = bs - wbs (operating) (5.75)

= 23.7219 – 1.4652 N/mm2

= 22.2567 N/mm2

5.3.8 General Consideration For Design

Take the joint factor J as 0.85,

s (tensile) < fs J sin s (5.76)

s (compressive) < 0.125 E (5.77)

Where , fs = maximum allowable design stress for the skirt material

= 135 N/mm2

J = weld joint factor

s = base angle of a conical skirt

E = modulus young

= 200, 000 N/mm2

Therefore, s (tensile) < 135 x 0.85 sin 90

22.2567 N/mm2 < 114.75 N/mm2

s (compressive) < (0.125)(200,000)

2.5234N/mm2 < 392.8571 N/mm2

Both criteria are satisfied, add 2 mm for corrosion, and give design

thickness = 13mm

5.3.9 Base Rings and Anchor Bolts

Assume pitch circle diameter,Db = 0.9 m

Circumference of bolt circle,Db = 900

= 2827 mm

According to Scheiman (Coulson Richardson Chemical engineering volume

6, page 848),

Bolt stress design, fb = 125 N/ mm2

Recommended spacing between bolts = 600 mm

Minimum number bolt required, Nb = (5.78)

= 4.7124

8

Bending moment at base skirt, Ms= 102 kN/m

Total weight of vessel, W = 36 kN

Area of bolt, Ab = (5.79)

=

= 417mm2

Bolt root diameter, d = (5.80)

= 23.04 mm

23 mm

Total compressive load on the base ring per unit length,

Fb = (5.81)

=

= 173066N/m

By assuming that a pressure of 5 N/mm2 is one of the concrete foundation

pad, fc

Minimum width of the base ring, Lb = (5.82)

=

= 34.61 mm

With this minimum width, can get actual width

Use M56 bolts (BS 4190:1967) root area = 2030 mm2, figure 13.30, page

849, Chemical Engineering Volume 6, 1996.

Actual width required = Lr + ts + 50 (5.83)

= 150 + 11 + 50

= 211 mm

Actual bearing pressure on concrete foundation

f'c = (5.84)

=

= 0.8202 N/mm2

Actual minimum base thickness, tb = Lr (5.85)

Where , fc= actual bearing pressure on base, N/mm2

fr= allowable design stress in the ring material, typically 140 N/mm2

Therefore, tb = 150

= 19.8860 mm

20 mm

5.3.10 Feed, Top Product, Bottom Product Piping Sizing

By assuming that the flow of the pipe is turbulent flow, therefore to

determine optimum duct diameter is

Optimum duct diameter, dopt,t = 260G0.52-0.37 (For Stainless Steel) (5.86)

Where, G = flow rate, kg/s

= Density, kg/ m3

Nozzle thickness, t = (5.87)

Where Ps = Operating pressure, N/mm2

= Design stress at working temperature, N/mm2

Optimum duct diameter, dopt,t= 226G0.52-0.37

Where, G = flow rate = 61333.2096 kg/hr

= 17.0370 kg/s

= density = 698.0384 kg/ m3

Therefore, dopt = 260 (17.0370)0.52 (698.0384)-0.37

= 100.7076 mm

125 mm

Nozzle thickness, t =

Where Ps = Operating pressure = 8.02 N/mm2

= Design stress at working temperature = 135 N/mm2

Therefore, t =

= 0.3702 mm

So, thickness of nozzle = corrosion allowance + 0.3702 mm

= 2 + 0.3702 mm

= 2.3702 mm

3 mm

For the top vapor output calculated as below,

Optimum duct diameter, dopt,t = 260G0.52-0.37

Where, G = flowrate = 5531.1456 kg/hr

= 1.5364 kg/s

= density = 9.0296 m3

Therefore, dopt = 260(1.5364)0.52 (9.0296)-0.37

= 143.9986 mm

= 150 mm

For the bottom liquid output calculated as below,

Optimum duct diameter, dopt,t = 260G0.52-0.37

Where, G = flow rate = 55802.0640 kg/hr

=15.5006 kg/s

= density = 667.3 kg/ m3

Therefore, dopt = 260 (15.5006)0.52 (667.3)-0.37

= 97.4890 mm 100 mm

TABLE 5.6: SUMMARY OF MECHANICAL ENGINEERING DESIGN

Column types Pressure vessel

Column material Stainless steel (SS 304)

Design temperature 135.2 0C

Operating Pressure 8.02 bar

Design Pressure 7.733 bar (10% of safety factor)

Design Stress 135 N/mm2

Skirt Height 2500 mm

Total column height 10250 mm

Column head Torispherical head

Column diameter 700 mm

Insulation material Fiberglass

Insulation thickness 50 mm

No of manhole 2

Manhole diameter 500 mm [BS 470: 1984]

REFERENCES

J. M. Coulson, J. F. Richardson, Chemical Engineering, Volume Two,

Third Edition, The Pergamon Press, 1977.

R. K Sinnot, Coulson & Richardson’s Chemical Engineering,

Chemical Engineering Design, Volume Six, Butterworth

Heinemann, 1999.

Robert H. Perry, Don W. green, Perry’s Chemical Engineer’s

Handbook, Seventh Edition, McGraw-Hill, 1998.

James, M. Douglas, Conceptual Design of Chemical Processes,

McGraw-Hill Book Company, 1988.

Martyn S. Ray and David, W. Johnston, Chemical Engineering,

Design Project: A Case Study Approach, Gordon and Breach

Science Publishers, 1989.

Carl R. Branan, Rules of Thumb for Chemical Engineers, Gulf

Publishing Company, 1994.

Billet, R., Distillation Engineering, Heydon Publishing, 1979.

King, C. J., Separation Processes, Second Edition, McGraw-Hill,

1992.

Kister, H. Z., Distillation Design, McGraw-Hill, 1992.

Lockett, M. J., Distillation Tray Fundamentals, Cambridge University

Press, 1986.

Normans, W. S., Absorption, Distillation and Cooling Towers,

Longmans, 1961.

Oliver, E. D., Diffusional Separation Procesess, John-Wiley, 1966.

Robinson, C.S., and Gilliland, E.R., Elements of Fractional

Distillation, McGraw-Hill, 1950.

Smith, R., Chemical Process Design, McGraw-Hill, 1995.

Van Winkle, M., Distillation, McGraw-Hill, 1967.

Micheal J. Barber, Handbook of Hose, Pipes, Couplings and

Fittings, First Edition, The Trade & Technical Press Limited,

1985.

Louis Gary Lamit, Piping Systems: Drafting and Design, Prentice-

Hall, Inc., 1981.

David H. F. Liu, Bela. G. Liptak, Wastewater Treatment, Lewis

Publishers, 2000.


Dimensions in M, L, T

Aa Active area of plate L2

Aap Clearance area under apron L2

Ac Total column cross sectional – area L2

Ad Downcomer cross - sectional area L2

Ah Total hole area L2

An Net area available for vapour – liqud disengagement L2

Ap Perforated area L2

Co Orifice coefficient -

D Mols of distillate per unit time MT-1

Dc Column diameter L

dh Hole diameter L

Emv Plate efficiency -

g Gravitational acceleration -

hap Apron clearance LT-2

hb Height of liquid back – up in down comer L

hbc Down comer back – up in term of clear liquid head L

hd Dry plate pressure drop, head of liquid L

hdc Head loss in down comer L

how Height of liquid crest over down comer weir L

hr Plate residual pressure drop L

ht Total plate pressure drop L

hw Weir height L

K1 Constant -

Lm Molar flow rate of liquid per unit area ML-2T-1

Lw Liquid flow rate L2T-1

Lwd Liquid mass flow rate MT-1

Ip Pitch of holes (distance between centre) L

Iw Weir length L

Nm Minimum number of stages -

NT Theoretical number of stages -

pt Total plate pressure drop ML-1T-2

Po Partial pressure ML-1T-2

q Heat to vaporize one mol of feed divided by molar latent heat -

R Universal gas constant L2T2-1

R Reflux ratio -

Rm Minimum reflux ratio -

Ua Vapour velocity based on active area LT-1

Uf Vapour velocity through holes LT-1

Uv Superficial velocity (based on total cross sectional area) LT-1

V Vapour flow rate per unit time MT-1

Vw Vapour mass flow rate MT-1

xi Mole fraction of component I -

xd Mole fraction of component in distillate -

yi Mole fraction of component I -

L Liquid viscosity -

Viscosity of solvent ML-1T-1

L Liquid density ML-1T-1

v Vapour density ML-3

Surface tension MT-2

Dm Mean diameter L

E Young’ Modulus ML-1T-2

Hv Height between tangent L

Pi Internal pressure ML-1T-2

Mx Bending moment at base of the skirt ML-1T-2

Ms Bending moment at point x from free end column ML2T-2

t Thickness of plate (shell) L

ts Skirt thickness L

J Joint factor -

b Bending stress ML-1T-2

w Dead weight stress ML-1T-2

cw Compressive stress ML-1T-2

ws Stress in skirt due to weight of vessel ML-1T-2


PROCESS CONTROL AND

INSTRUMENTATION NORMARIAH BINTI ABDULLAH

NOOR HARYANI BINTI

MUSTAPHA

SUPERVISORS



CHAPTER 6:

PROCESS CONTROL AND

INSTRUMENTATION

CONTENTS

TITLE PAGE

CHAPTER 6 PROCESS CONTROL AND INSTRUMENTATION


TITLE PAGE

6.2 TYPES OF CONTROL 161

6.2.1 Feedback Control 161

6.2.2 Feed forward Control 162

6.2.3 Cascade Control 162

6.3 CONTROLLING SHELL AND TUBE

HEAT EXCHANGER 163

6.4 FIXED BED REACTOR CONTROL 165

6.5 DISTILLATION COLUMN CONTROL 167

6.6 PIPING 170




6.6.4 Fluid Velocity 171

CHAPTER 6

CONTROL AND INSTRUMENTATION

6.1 INTRODUCTION

Process control is important in chemical plant that operated under

known and specified conditions. It is important in safety and environmental,

in operability so that plant can achieve desired reaction and in economics

to meet market requirement of purity product.

A chemical plant might be thought of as a collection of tanks in

which materials are heated, cooled and reacted, and of pipes through

which they flow. Such a system will not, in general, naturally maintain itself

in a state such that precisely the temperature required by a reaction is

achieved, a pressure in excess of the safe limits of all vessels be avoided,

or a flow rate just sufficient to achieve the economically optimum product

composition arise.

6.2 TYPES OF CONTROL

There is several control approach that has been applied in the

production plant, the basic concepts of these control are stated below.

6.2.1 Feedback Control

The feedback control system function is to bring the measured

quantity to its required value or set point. The feedback control system uses

direct measurements of the controlled variables to adjust the values of the

manipulated variables. The main advantage of the feedback control is the

corrective actions occur as soon as the controlled variable deviates from

the set point regardless of its source and the type of disturbances. Minimal

knowledge of the process is sufficient to set up this type of control. It is also

both versatile and robust which means that if the process condition

changes, re-tuning will still give a satisfactory result. However, this type of

control also has certain disadvantages, which are, there is no corrective

action taken until after a deviation in the controlled variable occurs. In

addition, it does not provide a predictive control action to compensate for

the effects of known or measurable disturbances. If the process encounters

large and frequent disturbance, the action of the controller will be such that

the process will operate continually in a transient state and never attain the

desired steady state.

6.2..2 Feedforward Control

The basic idea of the feedforward control is to measure the

important load variables and take the corrective actions before they upset

the process. However there are disadvantages of this control technique as

the load disturbances must be measured online and in many applications

this is not feasible. For this technique to be effective, we need to have

some basic knowledge about the process to construct a process model.

Ideal feedforward control theoretically is capable of achieving perfect

control but in reality it may not be physically realizable. There are times

when the combination of both feedback and feedforward control strategies

are required such as in the level control.

6.2.3 Cascade Control

The cascade control uses a secondary measurement point and a

secondary feedback controller in order to improve its dynamic response to

the load changes. The secondary measurement point is located so that it

recognizes the upset conditions sooner than the controlled variable. The

cascade control system utilizes multiple feedback loops in a process. It has

two distinguish features. The first feature is that the output signal of the

master controller serves as the set point of the slave controller. The second

feature is that it consists of two nested feedback control loops with the

secondary loop (slave controller) located inside the primary loop (master

controller). The advantages of cascade control are; 1) the control will

eliminate the effect of disturbance entering the secondary loop, 2) the

control will reduce the response time of the element inn the secondary loop,

which in turn will affect the primary loop, 3) the control will make the closed-

loop less sensitive to model error.

6.3 CONTROLLING SHELL AND TUBE HEAT EXCHANGER

6.3.1 INTRODUCTION

The principle of operation for heat exchanger is simple enough: two

fluid of different temperature are brought into close contact and prevented

from mixing by a physical barrier. But, actually shell and tube heat

exchanger are among the most confusing pieces of equipment for the

process control engineers.

The temperature of the 2 fluids will tend to equalize by arranging counter

current flow it is possible for the outlet temperature of each fluids to

approach the inlet temperature of the other. The heat contents are simply

exchanged from one fluid to the other and vice versa and no energy is

added or removed.

The heat exchanger it self is not constant and its characteristic

change with time. The most common changes are a reduction of heat

transfer rate because of the surface fouling.

The heat exchanger must be controlled to make it operate at the

particular rate required by the process at every moment in time.

6.3.2 CONTROL OBJECTIVE

The purpose of controlling in heat exchanger is to control the temperature

from the outlet stream. Precisely, it is important that the temperature

involved in the process must be maintained because it will influence the

temperature in the next stages or equipment and it must in a certain

temperature so that the process can operate smoothly.

6.3.3 HOW TO MAINTAIN THE TEMPERATURE

a) Thermocouple is used to determine any changes in temperature,

especially in the outlet the transmitter, which will convert the signal to

the electrical signal, detects stream and then the changes.

b) Temperature controller is used to interpret the electrical signal and will

send an output to the transducer. The function of the transducer is to

convert the electrical signal into air pressure. These are because the air

pressure is used to open and closed the valve in order to maintain the

temperature.

c) Here, the controller will correct the increased in temperature in the

outlet stream by using the valve of the inlet stream. When this happen

the heat transfer will reduced in spite of reduce in temperature.

6.3.4 CONTROL SYSTEM FOR HEAT EXCHANGER

6.1.4.1 FEEDBACK CONTROL OF HEAT EXCHANGER

Feedback Controller

T set

F

T1 = Temperature of inlet fluid T = Transducer

T2 = Temperature of outlet fluid Tset = Temperature set

TT = Temperature Transmitter

F = Fluid flowrate

6.3.4.3 TYPES OF CONTROLLING

The controller used is feedback control, which have lots of

advantages. These types of controller can detect any changes

of temperature in the outlet stream and corrective action then

occurred as soon as the controlled variables deviates from the

set point, regardless of the source and types of disturbance.

Besides, it also required minimum knowledge about the

process to be controlled.

1.1.1.1. Manipulated variable A steam flow rate, F

1.1.1.2. Process variable/control variable Temperature of the outlet liquid (Tout), T2

1.1.1.3. Load/disturbance Temperature and flow rate of liquid in temperature outlet.

Tout = f (Ti,Fi)

6.4 FIXED BED REACTOR CONTROL

6.4.1 INTRODUCTION

Heat Exchanger

TT

T

Steam

Condensate

T2

F

T1

FLiquid in

Condensate

The schemes used for reactor, if a reliable on line analyzer is

available, and the reactor dynamics are suitable, the product composition

can be monitored continuously and the reactor conditions are feed flows

controlled automatically to maintain the desired product composition and

yield. Reactor temperature will normally be controlled by regulating the flow

of the heating or cooling medium. Pressure is usually held constant.

Material balance control will be necessary to maintain the correct flow of

reactants to the reactor and the flow of product and un reacted materials

from the reactor.

6.4.2 CONTROL OBJECTIVE

i. To overcome the temperature increase of inlet cooling water

that may cause unsatisfactory performance by adjusting the

control valve on inlet stream.

ii. To maintain the temperature of exit liquid at the desired

value.

TC

INLET FEED

COOLING WATER OUT

COOLING WATER IN OUTLET PRODUCT

REACTOR

Figure 6.1: Fixed Bed Reactor control

6.4.3 CASCADE CONTROL

The cooling water is passed through the reactor jacket to regulate the

reactor temperature. The reactor temperature is affected by changes in

disturbances variables such as reactant feed temperature or composition.

The control strategy to handle such disturbances is by adjusting a control

valve on the cooling water inlet stream. By adding cascade control on the

feedback controller will overcome the increase of the inlet cooling water

temperature that may cause unsatisfactory performance. Cascade control

measures the jacket temperature, compares to it set point, and uses the

resulting error as the input to a controller for the cooling water makeup,

thus maintaining the heat removal rate from the reactor at constant level.

The controller set point and both measurements are used to adjust a single

manipulated variable, the cooling water makeup.

6.4.4 ADVANTAGE OF CASCADE CONTROL

i. The output signal of the master controller serves as the set

point for the slave controller.

ii. The two feedback control loops are nested, with the

secondary control loop( for the slave controller) located

inside the primary control loop (for the master controller)

6.5 DISTILLATION COLUMN CONTROL

Figure 6.2: Distillation Column control

6.5.1 INTRODUCTION

The final, overall objective of any process control application should

always be to maximize the profitability of the process under control. This is

normally achieved via a rationalization of the value added by the process

with the energy that is consumed by the process. In the distillation column,

increasing the internal vapor and liquid flows nearly always increase the

separation of key component and therefore, increase either the product

yield or its value. However, the increase in internal flow rates in only

achieved at the expense of additional energy consumption in both the

condenser and reboiler. With most reactors and many other unit operations,

this principle often manifests itself with respect to the heating or cooling

requirement, or the recycle rate. An effective control application adjusts the

process operation towards an optimum where the incremental value added

is just less than the incremental cost of the energy and raw materials.

6.5.2 DEGREE OF FREEDOM ANALYSIS

A simple two product distillation column with a single feed and a

total condenser has five degrees of freedom. These correspond to control

valves that vary the following quantities:

The distillate product draw rate (D):

The bottoms product draw rate (B):

The reboiler duty (QR or V to donate the internal vapour rate):

The reflux rate (R):

The condenser duty(Qc):

The condenser and reboiler duty usually cannot be manipulated directly but

the designation, QR and QC, are used to represent the group of variables

which could not be used to adjust the duty in each case, for example the

control valve which is designated to regulate the condenser duty might

actually manipulate the coolant flow rate (either directly or indirectly by

regulating the bypass rate) the active surface area of the condenser or the

rate at which vapour is withdrawn from the from the column. Similarly, the

method of regulating the reboiler duty could be the heating medium flow

rate, the reboiler exchange area or the process flow through the reboiler.

The column pressure, the reboiler sump level and the reflux

accumulation level (i.e. the column vapour and liquid inventory) must all be

stabilized for the column to operate in a steady state. The column pressure

is almost always controlled via the condenser duty (Luyben, 1990) and tight

control is usually achievable with a simple SISO (single-input, single-

output) control loop (Dale E. Seborg,1989). The liquid inventory can usually

be controlled by two simple SISO controllers provided either the distillate

rate or the reflux accumulator level and either the bottoms rate or the

reboiler duty is use to control the reboiler sump level.

Therefore, two degrees of freedom remain for the control of the

process objectives. If neither of these variables is used within a control loop

(i.e. the process operator manipulates the control valve directly), the

column is said to be operated in open-loop or manual. If only one of these

variables is manipulated automatically to control a measured property, a

one-point or single composition control scheme is deemed to be in used. In

this case, the remaining degree of freedom is usually fixed at a constant

value or manipulated only occasionally to reflect capacity constraints (e.g.

maximum reboiler duty or flooding). Finally, both available degrees of

freedom can be utilized within control loops. This is known as two-point or

dual composition control.

6.5.3 OPEN-LOOP CONTROL

The most basic distillation control system consider only the column

inventory and relies on the process operator to counteract disturbances to

the process by adjusting (when required) on the manipulated variables

which are not being used for inventory control. The effectiveness of this

approach depends on the variable pairings (i.e. the control configuration).

It is convenient to adopt a nomenclature to concisely describe the

variable pairings or control configuration. The most widely accepted method

of describing control configuration employs two letter designations that

correspond to the variables which are not used for inventory control.

6.5.4 ONE-POINT CONTROL

One-point control schemes have been the backbone of industrial

distillation control for many years, although the advent of multivariable

predictive controllers (e.g. Dynamic Matrix Control, DMC) has recently seen

a shift towards more complex strategies. However, one-point control is still

widely practiced and has some inherent advantages compared with open

loop and two-point control.

One-point control is relatively easy to implement, is not subject to

interactions between opposing composition control loops and provides a

form of effective constraint management. Distillation columns are almost

always illconditioned due to the presence of high gain variables (e.g.

internal flows which change the energy balance). If a two-point control

scheme is applied, the illconditioning can restrict the attainable closed-loop

performance and, in extreme cases, create instability due to excessive

interaction.

6.6 PIPING

6.6.1 INTRODUCTION

The installation cost of piping systems varies widely with the materials of

construction and the complexity of the system. The economics also depend

on the pipe size and fabrication techniques employed. Therefore, it is

important to choose pipe sizes which give a minimum total cost for pumping

and fixed changes.

6.6.2 MATERIAL OF CONSTRUCTION

There are several considerations that need to be evaluated when

selecting the piping material such as corrosiveness, brittle failure and the

ability of thermal insulation. For this process, almost all the components are

not corrosive. So, carbon steel is the most suitable for piping system

because of its low cost of material. However for the reactor with cooling

water, stainless steel is used because of its corrosiveness.

6.6.3 PIPE SIZING

An approximate estimation of the economic pipe diameter can be

obtained by the following equation, (Coulson and Richardson, 1999)

For carbon steel pipe

doptimum = 293 G0.53 -0.37

For stainless steel

doptimum = 260 G0.52 -0.37

where d = optimum diameter of the pipe, mm

G = flow rate of fluid in the pipe, kg/s

= density of fluid, kg/m3

The minimum pipe wall thickness can be determined by : (Coulson and

Richardson, 1999)

where t = minimum wall thickness, mm

Pi = Internal pressure of the pipe, N/mm2

f = maximum allowable stress, N/mm2

D = Pipe outer diameter, mm

6.6.4 FLUID VELOCITY

The fluid velocity of each stream is given by the equation below;

(Robert H. Perry, 1997)

where U = Fluid velocity, m/s

D = Selected internal diameter, m

The fluid velocity should be kept below erosion is likely to occur. For gases

and vapor, the velocity must not exceed the critical velocity.

George R. Kent shows the formulas which can be used to estimate

maximum fluid velocity: (Robert H. Perry, 1997)

Liquids, U =

Gases, U =

Table 6.1: Summary of the piping for MTBE plant

Stream Pi(bar) T(K)Mwmix

(kg/kmol) (N/mm2) (kg/m3) G (kg/h) G

(kg/s)d

(opt),mm t(mm) u (m/s) D (mm)

S1 1.000 324.000 66.990 165.000 461.990 19525.130 26.190 159.135 0.4822 2.8502438 254.8322

S2 1.000 309.000 58.000 165.000 315.750 41490.910 11.620 120.060 0.3638 3.2506812 267.4558

S3 1.500 356.000 73.980 145.000 714.010 61016.040 14.570 99.857 0.3443 2.6056159 254.1877

S4 11.000 589.000 73.980 85.000 721.930 41655.890 14.570 99.450 0.5850 2.5981534 326.9544

S5 15.000 477.400 45.300 105.000 664.690 41655.890 14.570 102.537 0.4883 2.6545600 376.1654

S6 15.000 371.000 56.000 145.000 599.290 579.890 10.810 91.225 0.3146 2.7597723 298.2495

S7 10.000 473.000 22.330 115.000 969.360 41076.000 3.760 44.091 0.1917 2.5404945 533.3017

S8 5.000 308.000 56.000 135.000 599.290 41076.000 10.810 91.225 0.3379 2.7597723 271.7491

S9 2.000 338.000 47.170 145.000 658.300 414.910 17.040 111.634 0.3849 2.6446178 310.1788

S10 8.020 473.000 84.136 115.000 732.190 19360.150 17.040 107.325 0.4666 2.5724736 274.7422

S11 4.080 400.000 46.057 115.000 667.180 605.940 1.550 31.934 0.1388 2.9006508 341.4829

S12 7.850 335.000 87.950 125.000 740.670 18754.210 15.480 101.664 0.4067 2.5746525 226.1472

S13 8.050 408.000 45.150 115.000 474.970 18755.210 1.610 36.934 0.1606 3.1637755 348.3276

S14 5.000 340.000 54.030 125.000 612.840 18756.210 1.000 26.238 0.1050 3.0178778 290.6757

S15 2.000 340.000 30.610 125.000 808.360 18757.210 0.610 18.315 0.0733 2.8643195 386.1842

S16 1.000 335.000 31.860 125.000 792.520 18758.210 0.560 17.647 0.0706 2.8889613 375.7390

S17 1.000 330.000 18.000 115.000 998.550 18759.210 0.050 4.613 0.0201 2.9965184 496.1437

S18 1.000 340.000 31.860 145.000 793.160 18760.210 5.670 58.796 0.2027 2.6329025 378.5326

S19 1.000 340.000 31.860 125.000 793.120 18761.210 6.230 61.749 0.2470 2.6230361 378.5326

PC

FC

LC

FC

LC

TT

T

T SET

FB CONTROLLER

T TRAP

TC

COOLINGWATER IN

COOLINGWATER OUT

PC

FC

LC

LC

FC

I/P FC

FTFC

FEED

SYMBOL ;

CONTROL VALVE

CENTRIFUGAL PUMP

PC PRESSURE CONTROLLER

FC FLOW CONTROLLER

TC TEMPERATURE CONTROLLER

LC LEVEL CONTROLLER

TTTEMPERATURE TRANSMITTER

DESIGNED BY;

NORMARIAH ABDULLAH

ISOBUTYLENE

TBA

TBA

ISOBUTYLENE

INERT

PROPYLENE OXIDE

DISTILLATION COLUMN

HEAT EXCHANGER

REACTOR

DISTILLATION COLUMN

COOLING WATER IN

CONDENSATE

S1

S2

S3

S4

S5

S6

S7

S8

Figure 6.3 : PFD Diagram for MTBE process

FC

ISOBUTYLENE

METHANOL

T SET

FEEDBACK CONTROLLER

FT

FCI/P

TT

TTFC

PC

FC

LC

LC

FC

TT

PC

LC

FC

LC

FC

COOLING WATER OUT

ISOBUTYLENE

MTBE

MTBE

METHANOL + WATER

MTBE REACTOR

DISTILLATION COLUMN

WASH WATER COLUMN

METHANOL RECOVERY

PURGE

HEAT EXCHANGER

METHANOL RECYCLE

TT

COOLING WATER

S8

S11

S12

S13

S14

S15

S17S19 S9

S10

S18

S16

Figure 6.3 : PFD Diagram for MTBE process

REFERENCES

Coulson and Richardson. 1999. Chemical Engineering Volume 6.

Butterworth Heinemann.

Coulson and Richardson. 1971,Chemical Engineering Volume 3.

Pergammon Press.

Robert H. Perry. 1997. Perry’s Chemical Engineer’s Handbook.

7th Edition. McGraw Hill.

Dale E. Seborg. 1989. Process Dynamics and Control.

John Wiley and Sons.

William L. Luyben. 1990. Process Modelling, Simulation and Control for

Chemical Engineers. McGraw Hill.


T Temperature

T Transducer

TC Temperature controller

F Fluid flowrate

Mw Molecular weight

Design stress

Density

G Mass flowrate

D Optimum diameter

t Thickness

u Velocity

D Outer diameter


PLANT SAFETY NOOR HARYANI BINTI

MUSTAPHA

MOHD NAZRI ISMAIL

SUPERVISORS



CHAPTER 7:

PLANT SAFETY

CONTENTS

TITLE PAGE

CHAPTER 7 PLANT SAFETY

7.1 GENERAL SITE

CONSIDERATIONS 175


7.1.3 Site Layout 175

7.1.3 Plant Layout 179

7.2 PLANT SAFETY 181

7.2.1 General Overview

Of Safety 181

7.2.8 The Integration Of Safety

Procedure 182

7.2.9 Safety During Start-Up

And Shutdown 183

7.2.10 Emergency Response

Plan (ERP) 185

7.2.11 HAZOP Study 197

7.2.12 HAZOP Report 200

7.2.13 Hazard Analysis 204

CHAPTER: 7

PLANT SAFETY

7.1 GENERAL SITE CONSIDERATIONS

7.1.1 INTRODUCTION

In determining plant layout designers should consider all the factors that

have been outlined. Plant layout actually is often a compromise between a

numbers of factors, which will discuss later in this chapter. The most

important factors of plant layout as far as safety aspects are concerned are:

1. Prevent, limit and/or mitigate escalation of adjacent events (domino

effects- fire explosion and toxic release)

2. Ensure safety within on-site occupied buildings

3. Control access of unauthorized personnel

4. Facilitate access for emergency service

To avoid aggregation and trapping of flammable/toxic vapors which could

lead to hazardous event, building should be designed so that all parts of the

buildings were well ventilated by natural or forced ventilation.

Basically, evacuation routes should not be blocked by poor plant

layout, and personnel with more general site responsibilities should usually

be housed in buildings sited in a non-hazard area near the main entrance.

7.1.4 SITE LAYOUT

A suitable site must be found for a new project and the site and equipment

layout planned. Ancillary building must have the provision and service

needed for plant operation. All of these should lie out in order to give the

most economical flow of materials and personnel around the site.

Considerations have to be take into account especially for the hazardous

processes where the site must be located at a safe distance from other

buildings.

The proper layout of industrial facilities is an important factor in the

prevention of catastrophic fires. In spite of ancillary buildings and services

required on the site, others additional main processing units also must take

into consideration, which may include:

1. Storage for raw materials and products: tank farms and warehouse

2. Maintenance workshop

3. Stores for maintenance and operating supplies

4. Laboratories for process control

5. Fire station and other emergency services

6. Utilities; steam boilers, compressed air, power generation,

refrigeration, and transformer station.

7. Effluent disposal plant

8. Offices for general administration

9. Cafeteria and other amenity buildings, such as medical center

10. Car parks and bicycle sheds

Normally, during roughing out the preliminary site layout, the first and

foremost to be sited and arranged is the process unit, in order to give the

smooth flow of material starting from the raw material until the final product

storage.

The process unit normally being spaced at least 30 m apart or may

be greater especially for hazardous process (MTBE & TBA). Then the

location of the principal ancillary building should be decided. The

arrangement of the ancillary building should minimize the time spending by

the personnel to traveling between the buildings.

The best location for the administration offices, cafeteria, mosque

and laboratory (occupied buildings) is should be far away from the

potentially hazardous process since there are many people staying there.

The distance between occupied building and plant building will be governed

by the need to reduce the danger of explosion, fire and toxicity. In all cases,

occupied buildings should be sited downwind of hazardous plant areas,

where consideration should be given to site of occupied buildings outside

the main fence.

While for the controller room also must located at a safer distance

from the hazardous processes but normally it is located adjacent to the

process unit. Layout for plant roads, pipe alleys and drains also must be

taken into consideration in locating the main process units. Each building

for construction, operation and maintenance needs the easy access road,

so that the process can be done smoothly. Site location for utilities building

and process units must give the most economical run of pipes to and from

the process center.

Medical center also must be located, in case there are some

workers or personnel injured or infected by chemicals or radiation. The

medical center also must be located far from the hazardous process, so

that it may not affect the patients. Besides, future expansions needed to be

considered also in order the company to expand their business/production

in the future.

Finally is the storage area to keep the products and raw materials

where these areas should be placed between loading and unloading

facilities and the process units they serve. But for hazardous storage tank

the site should be located at least 70 m from the site boundary. Flammable

storage should be sited in the open air so that minor leaks or thermal out-

breathing can be dissipated by natural ventilation.

(Source: Coulson & Richardson, Vol. 6, 2002)

A typical plot plan is in Figure 7.1.

20

N

N

1

23

4

5

6 7

8

9

10

11

12

13

14 1

5

16

17

18

19

21

Figure 7.1: Typical Plot of Plant Layout for MTBE Plant

ROADROAD

EW

7.1.3 PLANT LAYOUT

How well the plant and equipment specified on the process flow sheet is laid out

may influence the economic construction and efficiency of the process operation units.

There are 7 principal factors that to be considered in a plant layout which include:

a) Economic consideration; construction and operating costs

The cost of construction can be minimize by adopting layout that gives

the shortest run of connecting pipe between equipment and the lest

amount of structural steel work.

The need to keep distance for transfer of materials between

plant/storage units to a minimum to reduce costs and risks.

b) The process requirement

Interaction with existing or planned facilities on site, such as existing

roadways, drainage and utilities routings.

Interaction with other on site plants.

c) Operation

Convenience of equipment that needs a frequent attention from the

operator should be located near to the control room.

Sufficient working space and headroom must be provided to allow easy

access to equipment convenience.

d) Maintenance

Heat exchange need to be sited so that the tube bundles can be easily

withdrawn for cleaning and tube replacement.

Equipment that requires dismantling for maintenance, such as

compressor and large pumps should be placed under cover.

e) Safety

At least 2 escape roots for operators must be provided from each level in

process building.

Site boundaries and people living in the local neighborhoods need to be

locating as far as possible from hazardous materials facilities.

The need to prevent confinement where release of flammable

substances may occur.

The need to provide access for emergency services

f) Plant expansion

Equipment should be located so that in the future expansion of the

process it can be conveniently tied in.

Some space should be left to on pipe alley for future needs and service

pipes over sized to allow for future requirements.

g) Modular construction

The modules may include the equipment structural steel, piping and

instrumentation.

Advantages of modular construction

1. Improved quality control

2. Reduced construction cost

3. Less need for skilled labor on site

4. Less need for skilled personnel on overseas site

Disadvantages

1. Higher design cost

2. More structural steel work

3. More flanged connection

4. Possible problems, with assembly on site

(Source: Coulson & Richardson, Vol. 6, 2002)

7.2 PLANT SAFETY

7.2.1 GENERAL OVERVIEW OF SAFETY

Process safety can be applied in any process industry, but the term and the

approaches have been particularly widely used in the process industries, where its

usually means the same as loss prevention. Safety can be defined as; concerned with

adverse reactions to prolonged exposure to dangerous but less intense hazard. Safety

is also good business; the good management practices needed to ensure safe

operation would also ensure efficient operation.

The engineers should aware of hazard associated with the chemical used and

the process conditions and ensure that through the sound engineering practice

application, the risk can be reduced to acceptable levels. When a physician makes an

error, only one victim suffer, but when an engineer makes mistakes an error in design of

a product, many persons may suffer.

For many years, since chemical plant was opened, accident tragedies happened

in chemical industry plant, which have injured thousand of people and many more were

killed due to the leaks of safety considerations and no considerations in the factor of

safety (FOS) and in engineering designed. But fortunately nowadays, almost all

chemical industries have given full attention on considering the safety in their plant,

start from the research, development on new process, construction manufacture and

distribution and lastly are the use of the product.

Anyway, there are few steps that may steps in order to prevent losses in

chemical plant industries design, which included;

Identification and assessment of the hazard

Control of the hazard; by containment of flammable and toxic material

Control of the process; automatic controlling system alarm system and interlock

system.

Limitation of loss; pressure relief, plant layout, provision of fire fighting

equipment.

Occupational Safety & Health Act (OSHA) 1994 is an organization in Malaysia that

has listed some legal and administrative forced to promote, stimulate the high quality

standards and encourage safety at work place. The aims of OSHA 1994 only can be

achieved by;

Encouraging employers and employees in their effort to reduce the number of

occupational safety and health hazards at the workplace.

By setting mandatory occupational safety and health standard.

By providing for research in the field of occupational safety and health, including

the psychological factors involved.

By exploring ways to discover latent disease, establishing causal connections

between disease and work in environmental condition.

7.2.14 THE INTEGRATION OF SAFETY PROCEDURE

Employers, users and distributors of hazardous chemical should be given an

appropriate MSDS for each of the component and it is should be readily

accessible to all employees.

Operations, maintenance, R&D, engineering design, contraction, contractor and

other personnel who are involved in the plant process and design project need

to understand the safety and health hazard of the chemicals and process for the

protection of all personnel and the public nearby community (Kemaman).

Operators and maintenance personnel need training in order to posses the

knowledge and proficiency to perform there assigned job functions.

The management and employees should communicate effectively and be open

to constructive ideas from all the parties that will result in mutually desirable

improvements.

The interface between employee and the process equipment should be

compatible. In process control, the alarm and information systems displays

should be user friendly.

It is important during the process and plant operation to study the near-misses

incident because a serious consequence could have occurred but did not.

Process and plant personnel need to identify and mitigate the causes of the

incident and near misses of similar operation that have been previously

constructed. The main objective is to learn from past experience and thereby

avoid repeating the mistake.

Maintenance personnel should be having good mechanical ability and a well-

rounded background in their assigned work areas. They also should have more

in depth safety training than other employees because the hazard condition in

their workplace changes from day to day they should know how to use many

specialty tools, machines and personnel protective equipment (PPE).

Maintenance crew should be provided with a proper training to create

awareness of job hazard and control. The maintenance employees need to

know the properties and hazard control of all irritating, toxic or corrosive material

present in work place.

By doing a lots of maintenance activities through out the plan, the personnel

become familiar with all of the processes of normal plant operation and the

related equipment in the plant. With this exposure, this will allow maintenance

employees the ability to identify, analyze and correct many unsafe conditions in

the plant and its process units.

7.2.15 SAFETY DURING START-UP AND SHUTDOWN

Operating facilities need to go through a start up period before normal operations been

reached. The start-up and shutdown of the plant must proceed safely and easily, yet be

flexible enough to be carried out in several ways. The operating limits of the plant must

not exceed and dangerous mixtures must not be formed.

During start-up and normal operating phase of the new facility, procedures should

be in placed to maintain the integrity of process equipment, where hazardous materials

are involved. The probability for a successful start-up is greatly enhances through

operator training and process design that anticipates start-up problems before they

actually occurred. Some errors that could occur during start-up of the plant may include:

1. Wrong routing, involving failure to ensure that correct valves are closed.

2. Setting of wrong valves for operating parameters (e.g.; jacket temperature in the

reactor and reflux in the distillation column).

3. Drain valves are left open resulting in loss of material and possibly endangering

the lives of workers.

4. Valves left closed resulting in over pressure in the vessel.

5. Failure to complete purging cycle before admission of fuel air mixture.

6. Backflow of material because of the flow from high pressure to low-pressure

system.

Safety during shutdown

Basically shutdown procedure is important on occasion for the maintenance,

emergency situations and to do an adjustment in product inventories. During the

shutdown phase, deviations outside of normal operating range can also be expected.

The operating personnel and plant designer have to make sure that the operating

problem is connected with plant shutdown. A successful start-up of operating facility

may results a smooth shutdown procedure in one plant. If a shutdown procedure is not

well done or not done in well concieved, then it will cause;

a) Plugged lines

b) Damage to instrumentation

c) Stress and corrosion of construction materials

d) Poor documentation procedures followed,

e) And other safety hazard may results.

7.2.16 EMERGENCY RESPONSE PLAN (ERP)

Emergency planning is an important component of any workplace Occupational,

Safety and Health Management System. In large Chemical Work Emergency Response

Plans are likely to separate for on-site and off-site emergency, but they must be

consistent with each other i.e. they must be related to the same assessed emergency

conditions.

Basically, emergency plan should cover fires, natural disaster (flood,

earthquakes, etc.) and other such incident. But in our country, Malaysia natural disaster

like hurricane, earthquakes and tornado will never happens, so the emergency

response neither plans for this types of emergency will nor take into considerations.

Furthermore, emergency planning can actually minimize losses and top

management is the ones who responsible with the plan, which is usually, lie on the

guidance of the safety, health and environmental protection sector of the organization.

Planning should focus on protecting the health and safety of employee and the public,

as well as property and the environment and on restoring normal operation after an

accident. Content of the Emergency Response Plan (ERP) that should be included are

as follows;

a) Purpose of the plan

b) Nature and quantities of chemicals on-site

c) Description of potential emergencies

d) Allocate responsibility

e) Communications

f) Backup resources

g) Test emergency procedures

h) Notification of authorities

i) Notification of neighbors

j) Evacuation

k) Incident investigation

l) Media interest

(Refer APPENDIX G)

A) PURPOSE OF THE PLAN

A few reasons on why planning for emergency;

a) Emergency will happen; it’s only a question of time

b) Proper implementation of an appropriate ERP can;

~ Minimize cost

~ protect people, properties and environment

c) Minimize losses caused by emergency required

~understanding on the responsibility

~trained experience people

~accepted accountability

~designated authority

~planned procedure

d) Too late to plan when emergency occurred

e) Lack of preparing can turn an emergency to disaster.

B) NATURE AND QUANTITIES OF CHEMICAL ON-SITE

Table 7.1: Quantities Of Chemicals Stored/Handled On-Site

BIL TYPES OF COMPONENT QUANTITIES STORED/

HANDLED (kg/h)

1. MTBE 55555.56

2. TBA 94.79

3. ISOBUYLENE 3396.9

4. PO 55.738

5. ACETONE 829.46

6. DI-ISOBUTYLENE 27.87

Material safety data sheet (MSDS) for each of the component are shown in

APPENDIX G

C) DESCRIPTION OF POTENTIAL EMERGENCIES

The potential emergency, which could occur in MTBE plant, may include fire,

flooding (plumbing failure), explosion, natural gas leak, and boiler plant failure. The

entire potential emergencies need some evacuations procedure either to the workers,

engineers’ manager or foreman so that any disaster won’t happen.

1. FIRE

a) If see fire or smoke, immediately push/pull the nearest fire alarm station

to warn occupant.

b) Call the Kemaman Fire Department at 9-994, given the location and

description of the fire.

c) If the fire small, then use the fire extinguisher, but do not put yourself at

undue risk while fighting the fire

d) If the alarm sounds, turn off any electrical equipment that been operating

and evacuation the building immediately.

- Close all doors to help preventing the fire from spreading, exit via stairwells

e) Go to the assigned assembly area and keep away from the building. Do

not enter the building until authorized by the police.

f) Call 994 (HAZMAT) to give location and extent of fire and notify the

management to report the fire.

– State if there are any circumstance, such as dangerous chemical

g) Do not use elevators, when fire occurred, it safe if use the stairways.

2. FLOODING/PLUMBING FAILURE

a) Do not use electrical equipment in the area of the flooding, if necessary

evacuate the area. Keep unauthorized personnel from entering the area.

REMEMBER! THERE IS AN EXTREME DANGER OF ELECTRIC SHOCK!

b) Workers will not enter a flooded area until the power has been shut off.

c) All electrical equipment used for emergency purpose will be connected to a

ground fault circuit interrupter (GFCI).

3. EXPLOSION

a) Pull the fire alarm and call 9-991

b) Get out of the building as quickly and calmly as possible. Do not try to extinguish

fire and go to the assigned assembly area. Stay well away from the building.

4. NATURAL GAS LEAK

a) If you smell natural or hear blowing or hissing noise, turn off possible ignition

source.

b) Do not switch on light or any electrical equipment

c) Activate fire alarm if you believe there is a potential damage to building

occupants.

d) Once outside move to a clear area that is at least 500 ft away from the affected

building.

5. BOILER PLANT FAILURE (BOILER RUPTURE/ PIPE RUPTURE, TANK

RUPTURE & STEAM LINE

a) Evacuate the area and keep people from entering the area

b) If the heat too great and the building valve cannot be turned off, shut off the

main isolator valves for the area, but under covered by the boiler plant operator.

c) In certain situations, the entire plant system may need to be shut down where it

is made by Boiler operator if there is a catastrophic problem

d) In the event of a potential steam explosion inside the boiler plant immediately hit

the ‘kill switch’ by the door and leave the facility. Notify supervision.

D) RISK ASSESSMENTS

a) In industrial operations employees, materials and equipment come together in

the work environment to produce product. Productivity is best when the

operation at the facility runs smoothly, thereby allowing time and resources to be

used efficiently and effectively.

b) Risk assessment in safety, health and environmental protection often involves

exposure to potential chemical hazards. The health risk begin with identify of the

chemical and their exposure levels for the various employees activities and

behavior pattern.

MTBE

MTBE may cause minor eye irritation, moderate skin irritant; over-exposure may

produce anesthetic or narcotic effects.

Prolonged over-exposure may cause coughing, shortness of breath, dizziness

and intoxication.

TBA

TBA is extremely flammable liquid. Can cause severe eye irritant, skin irritant,

inhalation and ingestion hazard.

Prolonged over-exposure may also cause coughing, shortness of breath,

dizziness and intoxication.

METHANOL

Methanol also a flammable liquid that may cause skin irritation and also can

cause central nervous system depression.

It may absorb through the skin and also can cause kidney damage

More worse is that it may be fatal or blindness if swallowed it.

Methanol also can cause severe eye irritation and possible injury in spites of

causing respiratory and digestive tract irritation.

ACETONE

Inhalation of acetone vapors irritates the respiratory tract, cause coughing,

dizziness, dullness and headache.

Higher concentration can produce central nervous system depression, narcosis

and unconsciousness.

It also irritating due to default action on skin.

Prolonged or repeated skin contact may produce severe irritation and dermatitis.

PROPYLENE OXIDE

Ingestion of PO can cause gastrointestinal irritation and diarrhea.

PO splashed in eye can cause severe burning, tearing, etc.

It also can cause severe skin irritation and blistering.

Excessive inhalation of vapor can cause nasal and respiratory irritation, central

nervous system such as dizziness, weakness, fatigue and etc.

ISOBUTENE

It can cause irritation, nausea, headache, symptoms of drunkenness, coma, etc.

in short-term exposure.

Major health hazard is central nervous system depression and difficulty to

breath.

(Refer APPENDIX G)

E) ALLOCATE RESPONSIBILITY

On-site Plans

The emergency plan details how the accident is dealt with, the name of the person

responsible for on-site safety and the names of the persons authorized to take action

under the plan

a) Incident Controller

to proceed the scene of the incident and take control

assesses the emergency and decides if the major emergency is to be activated

direct all operations within the affected area with priorities to secure the safety of

personnel, minimize damage to plant property and environment and to minimize

loss of material

action is taken to shut down

means of controlling release and spillages must be considered

established appropriate communications an all activities of the site main

controller are taken until the arrival of the designated person

Closing the emergency after consulting the site main controller and the

emergency services.

b) The Site Main Controller (SMC)

The SMC goes to emergency control centre and takes overall control of the

emergency in the work

All outside emergency services are called in upon declaration of emergency

Key personnel are called in

Emergency response is then coordinated in communication with the emergency

services

The emergency duties end on assertion of the emergency

c) Spokesperson

A senior manager be appointed as the sole authoritative source of information

F) COMMMUNICATIONS

An effective round-the-clock communications system is essential. Commercial

telephone or walkie-talkie will be the primary means of communication between the

locations. The primary means of communications is by radio between fixed and mobile

location. The following communications device should be available

1. telephones

2. fax

3. voice mail

4. cell phones

5. e-mail

6. 2-ways radio

G) EVACUATION

a) Exit the plant as calmly and quickly as possible using the nearest safety exit. DO

NOT USE ELEVATOR

b) Alert all persons in the area. Turn off all ignition sources if possible.

c) Wear personal protective equipment (PPE) such as coals, shoes, and take a

wet towel to place over the face in case of smoke or fire.

d) Proceed to the assembly area outside the plant are and wait for the further

instruction. Stay away from the plant as well far as possible.

e) Evacuation plans should be posted on every floor or every placed in the plant

area either the control room, workshop, or store. The plans are showing the

location and updated by the safety officer.

Do not re-enter the building until and unless the safety officer or fire

personnel have been determined that it is safe.

H) INCIDENT INVESTIGATION

a) Major accident might be defined as one having the potential to kill three or

more people or damage a specific area of the environment or cause property

damage and loss in excess of a particular sum.

b) Most major accident in the process industries will involved a large accidental

release of chemicals or the energy from their reaction, in such a away as to

cause appreciable damage.

c) The major accident may give adverse effect either to humans, or

environment.

The loss of public and employee support and confidence

The terminations of operating and siting permits

Kill the people and cause tremendous property damages.

d) The main objective of this investigation is that to learnt from the past

experience and thereby avoid from repeating the same mistake.

e) The incident investigation may include;

Study the near misses Process and plant operation must study the near misses because a

serious incident may occur but did not.

The person who responsible is the process and plant personnel, they

need to identify and mitigate the causes of the accident near misses

of similar operations.

Management of risk A prime consideration and precaution to the location of hazardous

process, materials and product, when a new operating facility, major

plant expansion or significant process modification is being planned.

The management of risk should involve the health and safety of the

employees and community, the protection of the environment and the

protection of the plant property.

The hazard analysis include; maximum release of flammable,

explosive, reactive and toxic materials, interruption of business

activities, and the determination of exposure to employees and

community

Reviewed near misses Whenever possible the near misses that can be major incident must

be reviewed if they could relate to the process and plant.

This information can be applied to improvement of the process and

plant when another occurrences of near misses (repeated errors)

An effective compliance audits.

A compliance audit is used to verify compliance with regulatory

standard and management practice. An effective one usually

includes a review of the relevant documentation and process safety

information, an inspection of the operating facility and interviews with

selected plant personnel.

In the process and plant design phase the results of a compliance

audit at a similar facility would greatly aid and guide design engineers

about situations that could adversely affect the performance of the

operating facility.

I) MEDIA INTEREST

a) Effective media such as radio, TV, Internet and press communication is

essential both in implementing ERP and dealing with the community interest

during and after any significant incident. The news media have played major

roles in making people fearful of chronic risk to chemical exposure.

b) But these media also one of the safety tools in chemical process especially

when any accident happen either fire, explosion or release of hazardous

chemical, where it will inform the community about the incident and told them to

take safety precaution or stayed at home if hazardous chemical release. These

media also will inform the community when the plant is back to normal.

J) NOTIFICATION OF AUTHORITIES

a) If an incident happen either minor or major, it is very important to contact the

emergency services such as fire department, police department ambulance

(red cross)

b) For our MTBE plant, the nearest police department is in Kemaman. If any

fire occurred, there is a fire station in the plant itself. But if major fires

occurred then supervisor should call fire department in Kemaman, because

it involved dangerous situation. (Call 994). Police also should be contact so

that they can control the situation in the incident place.

c) If the incident such as explosion, chemical release or electrical shocks then

the safety officer should call 999 or ambulance to bring the victims to the

hospitals.

d) The manager of the plant should call the accident investigator team to

discover the real situation of the disaster by determining what are the main

causes, if there any leakage or any significant chemical or radiation release.

K) NOTIFICATION OF NEIGHBOURS

a) It is very important to our plant to get to know the neighbor, so that if

anything happens they know something happening to our company

b) If the neighbor especially the community saw smoke, fire or suffer from the

toxic release then they will informed us because of our good relationship.

They also will call police or fire department or ambulance in case there is

incident happen.

c) So we have to have the list of contact number, so that if we have problem

with our plant we can inform the community or other neighbor about it.

L) BACKUP RESOURCES

Backup resources such as personal protective gear, monitoring equipment,

absorbent litter, fire fighting equipment, access to earth moving machinery and

waste containment should be available to deal with emergency.

The desirable technique to solve safety hazard in engineering and management

is the use of personal protective equipment (PPE). The personal protective

equipment that mostly used in industries may include:

a) Protection of the head – hard hats, safety hats, and safety helmets.

b) Protection of hearing

Types of hearing protection; earplugs, hearing bands and earmuff.

Employees are expose to a steady noise level more than 85 dBA

c) Protection of the face and eyes

Safety eyewear/glasses

Goggles

d) Protection from falls

Employees who are working at the elevated conditions must be

qualified and trained properly used the equipment.

Debris net designed to protect employee or the other people from

falling tools, foreign object and construction debris.

e) Protection of respiratory system

There are 2 main types of respirators to protect against hazard:

1) Air-supplying respirators

Self-contained breathing apparatus (SCBA)

2) Air-purifying respirators

Chemical cartridge respirators

Gas masks

Disposable dust, mist and fume respirators

Powered air purifying respirators.

f) Protection of the foot

- Electrical hazard shoes- minimize hazard contact with

electrical; current

- Chemical protection footwear (from dirt, mud, water to

hazardous contaminates, corrosive splashes.

- Disposable and disposable, rubber boots and shoes.

g) Protection of the hand. Arms, and body

- Gloves, coveralls, aprons, coats.

h) Safety showers and eyewash fountains.

For personal protection from fire and corrosive chemicals.

i) Another types of safety backup is the fire fighting facility;

The fire hose – placed strategically throughout the facility.

Fire detection – alarm alert personnel to escape and take

action to control the fire

Types of fire extinguisher that normally provided;

CO2 extinguisher

Foam extinguisher

Dry chemical fire extinguisher

(Refer APPENDIX G)

M) TEST EMERGENCY PROCEDURES

The emergency response plan (ERP) should be periodically tested to ensure

that the organization is prepared, and response procedures work in an adequate

and timely manner.

The emergency coordinator is responsible for overall coordination of training of

assigned emergency services and volunteer personnel.

Personnel assigned to provide the following emergency services then will

receive initial training and refresher training annually;

1. Notification and warning

2. Communications

3. Evacuation assembly center operations

4. Emergency operation

5. Law enforcement

6. Fire and resume.

7.2.17 HAZOPS STUDY

a) The hazard and operability, commonly referred to as HAZOP Study, is a

procedure for systematic, critical, examination of operability of a process.

When applied to a process design or an operating plant, it indicates

potential hazards that may arise from deviations from the intended

design conditions.

b) It provides a better understanding about plant operation and should lead

to improved plant efficiency.

c) Only an expert and experience analysis team that is familiar with HAZOP

analysis should do HAZOP analysis.

d) HAZOP analysis uses guidewords that shown in Table 7.2 and applied in

the HAZOP process parameter as shown in Table 7.3;

Table 7.2: HAZOP Guide Words

Guide words Meaning

OSHA-required

No

Less

More

Part of

As well as

Reverse

Other than

Other possible guide words

Yes

Same as

Forward

Begin

End

Reached

Never, none

Quantitative decrease, low, too short

Quantitative increase, high too long Qualitative decrease, too little

Qualitative increase, contaminates, too much

Opposite of forward or intent

Complete substitution, another

Always

Constant

Opposite of reverse

Start

Completion

Achieved

Table 7.3: Typical HAZOP Process Parameter

Typical HAZOP Process Parameter

Pressure

Temperature

Flow

Level

Time

Composition

pH

Reaction

Heating

Cooling

Mixing

Addition

Data

Information

Separation

Viscosity

Voltage

Frequency

Speed

Density

Solubility

(Source: A. Charles, 1998)

e) Consider that the steam supply line and associated control

instrumentation (Refer figure 7.2). The engineer’s intention is that steam

shall be supplied at a pressure and flow rate to match the required TBA

demand and Table 7.4 shows the HAZOP analysis of TBA vaporizer.

7.2.18 HAZOP REPORT

The HAZOP report for heat transfer equipment and reactor as shown in Table 7.4

and Table 7.5

Table 7.4: The HAZOP Report for Reactor (R-102)

Deviation Guide

Word

Cause Consequence Action

Vessel: Reactor

Intention: Reaction of isobutylene with Methanol to form MTBE

FLOW NO ~Valve failure

~Compressor

failure

~Pump failure

~Reduce the

reaction rate

~Shut down

compressor 1 & start

up compressor 2

~Fit low flow alarm

~Maintenance of

pump, valve &

compressor

LESS ~Partial failure

pump

~Failure of

compressor

~Valve partially

closed

~Fall in

reaction rate

~Shut down pump

~Fit low level alarm

~Maintenance of

pump & compressor

MORE ~Failure of valve

& pumps

~Failure of

compressor

~Failure of

~Reduce the

reaction rate

~Increase the

reactor

temperature

~Fit temperature

alarm

~Automatic pump

shutdown

~Maintenance on

Table 7.4: The HAZOP Report for Reactor (R-102)-(Continue)

Deviation Guide

Word


control system pumps, compressor

& valve

REVERSE ~Pumps

and

compresso

r fails

~Failure of

valve

~Reduce in

reaction rate

~High reactor

temperature

~Fit level alarm

~Fit non-return

valve (NRV)

~Automatic

pump shut

down

~Automatic

compressor

shut down.

LEVEL LOW ~Failure of level

control

~Leakage

~Electric failure

~Valve fails to

open

~Valve closed

~Failure of

pump

~Low level

~Fall in

reaction rate

~Fit level control

~Shut pump 1 and

start up pump 2

~Automatic

generator start-up

~Fit level alarm

~Maintenance on

pump

HIGH ~Failure of level

control

~Valve fully

open

~Failure of

pump

~Miss operation

~Valve fails to

closed

~Level

exceeded

~Hazardous

explosion

~High

concentration of

methanol and

low

concentration of

isobutylene

~Fit high level alarm

~Automatic pump

shut-down

~Maintenance on

pump and valve

Table 7.4: The HAZOP Report for Reactor (R-102)-(Continue)

Deviation Guide

Word


PRESSURE HIGH ~Failure of

pressure control

~Failure of high

pressure alarm

~Explosion on

reactor

~No reaction

occurred

~Excessive reactant

in reactor

~Fit high

pressure

alarm

LOW ~Failure of

pressure control

~Failure of low

pressure alarm

~No reaction

occurred

~Damage to reactor

~Excessive reactant

in reactor

~Fit low

pressure

alarm

TEMPERATURE LOW ~Failure of

temperature

control

~Failure of low

temperature

alarm

~No reaction

occurred

~Fall in reaction

rate

~Fit low

temperature

alarm

HIGH ~Failure of

temperature

controller

~High

temperature

~controller fails

~Fails of heat

transfer

equipment

Failure of pump

and valve

~Reactor running at

high temperature

~Damages to

reactant and

catalyst

~Falls in reaction

rate

Reactor explosion

~More

supplies of

cooling water

~Fit high

temperature

alarm

~Pump and

compressor

shut down

~Maintenance

on heat

transfers

equipment.

Figure 7.2, shows the operability of vaporizer while Table 7.5 shows the HAZOP report

of the vaporizer operation.

Figure 7.2: TBA vaporizer Instrumentation

PC

LC1

LA2

LA1

LC2

Steam

CRV3

NRV

Trap

CRV2

CRV1

TBA Feed

Vapor reactor

Table 7.4: HAZOP Analysis of TBA Vaporizer

Item Deviation Cause Consequence Action

Steam to

vaporizer

TBA to

Vaporizer

No flow

More flow

Reverse

flow

Less flow

Blockage, Valve

Failure, Failure of

steam supply

CV1 sticking, LC1

fails

Pumps fails,

vaporizer press

higher than

delivery

Partial failure pump

Loss of TBA flow

Vaporizer floods,

liquids to reactor

Flow of vapor in

storage tank

Level falls in

vaporizer

Fit low level alarm (LA1)

Fit high level alarm

(LA2) with automatic

pump shut-down

LA1 alarm, fit non-return

valve (NRV)

Fit low level alarm (LA1)

(Source: Coulson & Richardson, Vol.6, 2002)

7.2.19 HAZARD ANALYSIS

An operability study will identify potential hazards, but gives no guidance on the

likelihood on an incident occurring or loss suffer.

Incident usually occurred through the coincident failure of two or more items;

failure of equipment, control system and instrumentation and miss-operation.

The sequence of events that leads to a hazardous incident can be shown as a

fault tree (logic tree) as shown in the figure 7.2, which shows the set of

circumstances that would result in the flooding of the TBA vaporizer shown in

figure 7.3. The AND symbol is used where coincident inputs are necessary

before the system fails and OR symbols, where failure of any input by itself

would cause failure of the system. (Source: A.

Charles, 1998)

Figure 7.3: Simple Fault Chart

REFERENCES

Failure of Steam Trap

Failure of flow valve

Failure of level control

Failure of high-level S / D system

Flooding of

vaporizer

liquid TBA

to reactor

AND

OROR

Green W. Don & Perry Robert H. Perry’s Chemical Engineers’ Handbook. Seventh

Edition Kansas. McGraw Hill, 1997.

Charles A.Wentz, Safety, Health, And Environmental Protection, McGraw Hill, 1998

R.K Sinnott. Coulson & Richardson, Chemical Engineering Design. Vol.6. Butterworth

Heinemann 1999.

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ECONOMIC EVALUATION MOHD NAZRI ISMAIL

JUPLIN KINTI

SUPERVISORS



CHAPTER 8:

ECONOMIC EVALUATION

CONTENTS

TITLE PAGE

CHAPTER 8 ECONOMIC EVALUATION


8.9 The specification of plant 206

TITLE PAGE

8.10 Revenue from sales 207

8.11 Cost Estimation 207

8.11.1 Capital Cost Estimation 207

8.11.2 Manufacturing Cost Estimation 209

8.11.3 Cost of Operating Labor (COL) 210

8.11.4 Cost of Utilities (CUT) 211

8.11.5 Cost of Raw Material (CRM) 212

8.12 Profitability Analysis 213

8.12.1 Before Tax and After Tax

Cash Flow 214

8.12.2 Present Worth and IRR Method 215

8.12.3 Cumulative Cash Flow After Tax 216

8.13 PAYBACK PERIOD 217

8.13.1 Simple Payback Period 217

8.13.2 Discounted Payback Period 217

8.14 CONCLUSION 217

CHAPTER 8

ECONOMIC EVALUATION

8.1 INTRODUCTION

In this chapter, the costing of equipment which has been designed will be estimated

and the feasibility of MTBE production will be evaluated by profitability analysis to make

sure the project is economically attractive. There are some general assumptions to this

chapter;

i. The plant life span is fifteen years.

ii. The currency exchange rate of US dollar to Ringgit Malaysia is fixed at

3.8 as fixed by Malaysian Government.

iii. The price of raw materials, catalyst and product is fixed for the whole

period of operation.

8.2 THE SPECIFICATION OF PLANT

For the specification of plant, the following data throughout the lifetime of the project are

as:

Cost of raw material : Tert-Butanol – RM 1035/ metric ton

(Reference: Appendix H1)

: Methanol – RM 832/metric ton


Price of product : MTBE – RM 1235/ metric ton


Price of by-product : Propylene Oxide – RM 1411/ metric ton


8.3 REVENUE FROM SALES

Price of MTBE = RM 1235 / metric on

The capacity of MTBE = 400,000 ton / year

Sales income =

= RM 494,000,000.00 / year

Capacity of Propylene Oxide = 295,625 ton / year

Sales income =

= RM 417,8126,875.00 / year

Total sales income = RM 911,126,875.00 / year

8.4 COST ESTIMATION

8.4.1 Capital Cost Estimation

CTC = CFC + CWC + CL (8.1)

Where,

CTC = total capital cost

CFC = fixed capital cost

CWC = working capital cost

CL = cost of land & other non-depreciable costs

FP = Pressure factor to account for high pressure

FM = Material factor to account for material of constructions

CP = Purchase cost for base condition

FBM = Bare module cost factor

CBM = Bare module equipment cost for base condition

C°BM = Bare module equipment cost for actual condition

Table 8.1: Estimation Cost of Purchased Equipment

Equipment

Size(m2)/

Diameter

(m)/Power(kW)

Material of

Construction

Operating

Pressure

(bar)

FP/Fq FM FBM FOBM Cp($) CBM ($) Co

BM ($)

R - 504 3m SS 3 1.4 4 1 5.6 40000 160000 224000

R - 508 456.922m2 SS 9.9 1.1 1 1 - 250000 250000 -

E-506 62.94 m2 CS/Cu 11 1 1.25 1.25 1.2563 10500 13125 13191.15

E-507 62.94 m2 CS/Cu 11 1 1.25 1.25 1.2563 10500 13125 13191.15

C - 503 500kW SS - - - 1.5 3.5 200000 300000 700000

P-100A/B 65kW SS - 1 2.4 3.31 5.424 18000 119160 195264

DC - 501 3m SS 1.5 1.1 2 5.5 6 150,000 825000 900,000

35 Trays - SS 1.5 1 - 1.2 2 950 33250 66500

DC - 509 3m SS 1.5 1.1 2 5.5 6 150,000 825000 900,000

35 Trays - SS 1.5 1 1.2 2 950 33250 66500

DC - 505 0.7m SS 8.02 1.4 4 4 12 22000 88,000 264,000

22 Trays - SS 8.02 2 - 1.2 2 298 7867.2 13112

DC – 511 3 SS 1.5 1.1 2 5.5 6 150,000 825000 900,000

35 Trays - SS 1.5 1 1.2 2 950 33250 66500

V-100 3 SS 1.5 1.1 2 5.5 6 150,000 825000 900,000

Total 4351027 5222258

Total Module Cost, CTM = ∑C°TM (Reference: Appendix H5) (8.2)

= 1.18 (∑C°BM)

= 1.18 (5222258)

= 6162265 x 3.8

= RM 23416607

Grass Root Cost, CGR = CTM + 0.35 (∑ CBM) (Reference: Appendix H5)(8.3)

= 6162265 + 0.35 (4351027)

= 7685124 x 3.8

= RM 29203471

Since, Grass Root Cost (CGR) is:

CGR = CFC + CL (8.4)

When,

CWC = 15% of fixed capital cost (CFC) (8.5)

(Coulson & Richardson,1999)

So,

CGR = 1.15 CFC

i. CFC = (Coulson & Richardson,1999) (8.6)

=

= RM 25394323

ii. CL =

= RM 24,281,136

iii. CWC = 15% fixed capital cost (CFC)

= 0.15(25394323)

= RM 3809148

Total capital cost (CTC) = CFC + CWC + CL

= RM 25394323 + RM 3809148 + RM 24,281,136

= RM 53,484,607

8.4.2 Manufacturing Cost Estimation

The equation below is used to evaluate the cost of manufacture:

(8.7)

COM = 0.304FCI + 2.73COL + 1.23(CUT + CWT + CRM)

The cost of manufacturing (COM) can be determined when the following costs are

known or can be estimated:

1. Fixed Capital Investment (FCI): (CTM or CGR)

2. Cost of Operating Labor (COL)

3. Cost of Utilities (CUT)

4. Cost of Waste Treatment (CWT)

Cost of manufacture (COM) = Direct Manufacturing Cost (DMC) +

Fixed Manufacturing Cost (FMC) +

General Expenses (GE)

5. Cost of Raw Material (CRM)

8.4.3 Cost of Operating Labor (COL)

Table 8.2: Labor Cost

Equipment typeNo of

equipmentOperators per shift per

equipmentOperator per

shiftHeat exchangers 1 0.1 0.1Heater 1 0.5 0.5Reactor 2 0.5 1.0Vessels 1 0.0 0.0Pumps 1 0.0 0.0Compressor 1 0.15 0.15Towers 4 0.35 1.4Waste Treatment 1 2.0 2.0Total 5.15

Since, a single operators works on the average 49 weeks (3 weeks time off for vacation

and sick leave) a year, five 8-hour shifts a week.

1 operator =

=

Operating shift per Year =

=

So, the number operator needed =

= 3.7 operators

Thus,

Operating Labor = 3.7operators x 5.15 operator per shift

= 19.1 operator

= 20 operator

A mechanical engineers maximum wages per year (MIDA Jan, 2003) RM 54,000.00

Thus,

Labor Cost (2003) = 20 x RM 54,000.00

= RM 1,080,000.00

8.4.4 Cost of Utilities (CUT)

Yearly costs = flowrate x costs x period x stream factor

Since, assuming the plants operating days per year = 300 days

So,

Stream factor (SF) = no. of day’s plant operates per year

no. of days per year

=

= 0.82

1. Heater (E-100)

Duty =

Thus,

Yearly cost = (Q) (C steam) (t)

=

= RM 91157.77

2. Pump

Power (shaft) = 2.34x105kJ/h = 65 kW

Effeciency of drives, ξdr = 91.3% (Reference: Appendix H6)

Electric Power, Pr =

=

= 71.19kW

Yearly cost = (Reference: Appendix G7)

= RM 25218.35

3. Compressor

Power (shaft) = 1.8x107kJ/h = 5000kW

Effeciency of drives, ξdr = 96% (Reference: Appendix H6 & H7)

Electric Power, Pr =

=

= 5208.33 kW

Yearly cost =

= RM 1, 844, 998.82

Total of utilities costs = RM (91157.76 + 25218.35 + 1, 844, 998.82)

= RM 1, 961, 374.93/yr

8.4.5 Cost of Raw Material (CRM)

1. TBA = 94273.44 kg/h (Price = RM1035/Metric Ton = RM1.035/kg)

=

= RM 576,071,053.4/yr

2. Methanol = 22429.40 kg/h Price RM 0.8816/kg

=

= RM 116,744,273.4/yr

3. Copper catalyst (solid) = 1,677 tonne/yr

Price RM 2.00/kg (Reference: Appendix H8)

= 1,677 tonne x 1000 kg x 2 x RM 2

tonne yr kg

= RM 6708000/yr

Total cost of raw material cost (CRM) = RM 699523327.8

The estimation of total manufacturing cost:

COM = 0.304FCI + 2.73COL + 1.23(CUT + CWT + CRM)(Reference:

Appendix H9)

Since, the cost of Waste Treatment (CWT) = RM 516600

Thus,

COM = 0.304 (53484607) + 2.73 (1080000) + 1.23 (1961374 + 516600 +699523328)

COM = RM 883260746/yr

8.5 PROFITABILITY ANALYSIS

The feasibility of MTBE production in Malaysia is evaluated by profitability analysis.

The profitability of the project will be the largest factor that makes a project

economically attractive. To this stage, almost all the design and cost information

required for the profitability analysis were obtained. Based on the information available,

the best methods assessing the profitability of alternatives are based on projections of

the cash flows during the project file.

8.5.1 Before Tax and After Tax Cash Flow

Table 8.3: Annual Before Tax(CFBT) and After Tax Cash Flow (CFAT)

Year Gross Income

ExpensesInvestment &

Salvage Value

CFBT DepreciationTaxable Income (38%)

Taxes CFAT

0 -25394323 -25394323 -25394323 -3809148 -3809148 -3809148 -24,281,136 -24,281,136 -24,281,1361 911,126,875 -883260746 27866129 3628848.757 24237280.24 9210166.492 18655962.512 911,126,875 -883260746 27866129 6219069.703 21647059.3 8225882.533 19640246.473 911,126,875 -883260746 27866129 4441467.093 23424661.91 8901371.525 18964757.484 911,126,875 -883260746 27866129 3171750.943 24694378.06 9383863.662 18482265.345 911,126,875 -883260746 27866129 2267713.044 25598415.96 9727398.063 18138730.946 911,126,875 -883260746 27866129 2265173.612 25600955.39 9728363.048 18137765.957 911,126,875 -883260746 27866129 2267713.044 25598415.96 9727398.063 18138730.948 911,126,875 -883260746 27866129 1132586.806 26733542.19 10158746.03 17707382.979 911,126,875 -883260746 27866129 0 27866129 10589129.02 17276999.98

10 911,126,875 -883260746 27866129 0 27866129 10589129.02 17276999.9811 911,126,875 -883260746 27866129 0 27866129 10589129.02 17276999.9812 911,126,875 -883260746 27866129 0 27866129 10589129.02 17276999.9813 911,126,875 -883260746 27866129 0 27866129 10589129.02 17276999.9814 911,126,875 -883260746 27866129 0 27866129 10589129.02 17276999.9815 911,126,875 -883260746 27866129 0 27866129 10589129.02 17276999.9815 2539432.3 28090284 2539432.3 964984.274 29055268.27

Estimated salvage value = 10%CFC(Coulson & Richardson, 1999)(8.8)

= 0.1 x RM 25394323

= RM 25394323.3

8.5.2 Present Worth and IRR Method

In theory, the Minimum Attractive Rate of Return (MARR) is choosen higher than the

rate expected from the bank or some safe investment that involved minimal

investment risk. The MARR for after taxes is selected at 15%. (Analysis and Design

of Chemical Processes)

Table 8.4: Annual IRR After Tax

Year CFAT 30% PW 40% PW 0 -25394323 -25394323 -25394323 -3809148 -3809148 -3809148 -24,281,136 -24,281,136 -24,281,1361 18655962.51 0.76923 14350726.04 0.71129 13269799.572 19640246.47 0.59172 11621526.64 0.5102 10020453.753 18964757.48 0.45517 8632188.66 0.36443 6911326.5674 18482265.34 0.35013 6471195.563 0.26031 4811118.495 18138730.94 0.26933 4885304.403 0.18593 3372534.2436 18137765.95 0.20718 3757782.35 0.13281 2408876.6967 18138730.94 0.15937 2890769.549 0.09486 1720640.0178 17707382.97 0.12259 2170748.078 0.06776 1199852.279 17276999.98 0.0943 1629221.098 0.0484 836206.79910 17276999.98 0.07254 1253273.579 0.03457 597265.889311 17276999.98 0.0558 964056.5989 0.02469 426569.129512 17276999.98 0.04292 741528.8391 0.01764 304766.279613 17276999.98 0.03302 570486.5393 0.0126 217690.199714 17276999.98 0.0254 438835.7995 0.009 155492.999815 17276999.98 0.01954 337592.5796 0.00643 111091.109915 29055268.27 0.01954 567739.9421 0.00643 186825.375

7798369.258 -6934097.616

After interpolation it is found that the value of IRR is equal to 34.65% and therefore,

since it is bigger than the value of MARR (15%) this project is acceptable

8.5.3 Cumulative Cash Flow After Tax

Table 8.5: Cumulative Cash Flow After Tax (CFBT)

Year CFBTCumulative cash flow

0 -53,484,607 -53,484,6071 27866129 -25,618,4782 27866129 2,247,6513 27866129 30,113,7804 27866129 57,979,909

5 27866129 85,846,0386 27866129 113,712,1677 27866129 141,578,2968 27866129 169,444,4259 27866129 197,310,554

10 27866129 225,176,68311 27866129 253,042,81212 27866129 280,908,94113 27866129 308,775,07014 27866129 336,641,19915 28090284 392,597,612

Cumulative Cash Flow (RM) vs Year

-100,000,000

-50,000,000

0

50,000,000

100,000,000

150,000,000

200,000,000

250,000,000

300,000,000

350,000,000

400,000,000

450,000,000

1 3 5 7 9 11 13 15

Year

Cu

mu

lati

ve C

ash

Flo

w

Figure 8.1: Cumulative Cash Flow vs Year

8.6 Payback Period

8.6.1 Simple Payback Period

Table 8.6: Simple Payback Period

Year Cash FlowCumulative Cash Flow

0 -53484607 -534846071 27866129 -256184782 27866129 2247651

From Table 8.6 it is found that the simple payback period is in the second year.

8.6.2 Discounted Payback Period

Table 8.7: Discounted Payback Period

Year Cash FlowCumulative Cash Flow

0 -53484607 -53484607 1 27866129 -25618478 -29461249.72 27866129 -1595120.7 -1834388.8053 27866129 26031740.2 29936501.22

From Table 8.7 it is found that the discounted payback period is in the third year of

operation.

8.7 Conclusion

Based on this chapter, the economic evaluation plant are made through

study in all aspect including feasibility study, process synthesis and flow sheeting

and designed of major equipment. From the cash flow analysis, the payback period

is about 3 years. By looking to the IRR value (34.65%) which is bigger than the

MARR therefore it can be concluded that this project is profitable and acceptable.

Furthermore, it should be stated that the present work is primarily illustrated based

on the method of engineering economic analysis of chemical processes.

REFERENCES

J. M. Coulson, J. F. Richardson, Chemical Engineering, Volume Two, Third

Edition, The Pergamon Press, 1977.

R. K Sinnot, Coulson & Richardson’s Chemical Engineering, Chemical

Engineering Design, Volume Six, Butterworth Heinemann, 1999.

Robert H. Perry, Don W. green, Perry’s Chemical Engineer’s Handbook,

Seventh Edition, McGraw-Hill, 1998.

James, M. Douglas, Conceptual Design of Chemical Processes, McGraw-Hill

Book Company, 1988.

Martyn S. Ray and David, W. Johnston, Chemical Engineering, Design

Project: A Case Study Approach, Gordon and Breach Science

Publishers, 1989.

Carl R. Branan, Rules of Thumb for Chemical Engineers, Gulf Publishing

Company, 1994.

Smith, R., Chemical Process Design, McGraw-Hill, 1995.


C°BM = Bare module equipment cost for actual condition

CL = cost of land & other non-depreciable costs

CP = Purchase cost for base condition

CBM = Bare module equipment cost for base condition

CGR = Grass Root Cost

CFC = Fixed capital cost

COL = Cost of Operating Labor

CTC = Total capital cost

CTM = Total Module Cost

CUT = Cost of Utilities

CWC = Working Capital Cost

CWT = Cost of Waste Treatment

COM = Cost of Manufacturing

CFAT = Cash Flow After Tax

CFBT = Cash low Before Tax

FM = Material factor to account for material of constructions

FP = Pressure factor to account for high pressure

FBM = Bare module cost factor

FCI = Fixed Capital Investment

PW = Present Worth

MARR = Minimum Attractive Rate of Return


PROCESS INTEGRATION ROHIZAD JAMIL

SUPERVISORS



CHAPTER 9:

PROCESS INTEGRATION

CONTENTS

TITLE PAGE

CHAPTER 9 PROCESS INTEGRATION


9.2 PINCH TECHNOLOGY 219

9.3 THE PROBLEM TABLE METHOD 219

9.4 THE NETWORK DESIGN 221

\CHAPTER 9

PROCESS INTEGRATION

9.1 INTRODUCTION

Pinch technology was introduced by Linnhoff and Vredeveld to represent a new set

of thermodynamics based methods to minimized energy level in design of heat

exchanger networks. Process integration can lead to a substantial reduction in

energy requirements and increase the efficiency of a plant. One of the generally

useful techniques is pinch technology. The term derives from the fact that in a plot of

the system temperature versus the heat transferred, a pinch occurs at the minimum

temperature difference between the hot and cold stream, refer to Figure 9.1. It has

been shown that the pinch represents a distinct thermodynamic break in the system.

Temperature, oC

Pinch ∆Tmin

Enthalpy, kW

Figure 9.1: Hot and cold stream composite curves (not according to the value in this

case)

.

Hot Stream

Cold Stream

9.2 PINCH TECHNOLOGY

In this problem, the hot stream, stream 11-12 which requires cooling and cold

stream, stream 3, 4 has to be heated. Each starts at from a source temperature Ts,

and is to be treated to a target temperature Tt. The heat capacity of each stream is

shown as CP where given by:

CP = mCp (9.1)

Where m = mass flow rate, kg/s

Cp = average specific heat capacity between Ts and Tt, kW kg-1 oC-1

The heat load is the total heat required to be exchange as the objective is to reduce

the power consumption.

Table 9.1: Data for heat integration

Stream

number

Type Heat Capacity, CP

KW/oC

Ts

oC

Tt

oC

Heat Load

KW

3-4 Cold 30.5 55 282 -6923.5

11-12 Hot 106.55 400 330 7458.5

9.3 THE PROBLEM TABLE METHOD

The actual stream temperatures Tact need to be converted into stream interval

temperatures, Tint. The use of interval temperatures rather than the actual

temperatures allows the minimum temperature difference to be taken into account.

Hot stream Tint = Tact - (9.2)

= 400 – 10/2

= 395 oC

Cold stream Tint = Tact + (9.3)

= 55 + 10/2

= 60ºC

Table 9.2: Interval temperatures for oC

Stream

number

Actual Temperature Interval Temperature

Ts,oC Tt,oC Ts,oC Tt,oC

3-4 55 282 60 287

11-12 400 330 395 325

The bracketed temperature indicates the duplicated temperature. All Tint are ranked

in order of magnitude and carry out a heat balance for the streams falling within

each temperature interval.

(9.4)

where = net heat required in that interval

=sum of heat capacities of the cold stream

= sum of heat capacities of the hot stream

= Interval temperature difference

Table 9.3: Rank order of interval temperatures

RankInterval( )

oC

-

kW/oC kW

395

325 70 -106.55 -7458.5

287 38 0 0

60 227 -76.5 -17263.35

Cascading the heat from one interval to the next implies that the temperature

difference is such that the heat can be transferred between the hot and cold

streams. The pinch occurs where the heat flow in the cascade is zero. This is

because the rule of heat integration says that for minimum utility requirements no

heat flows across the pinch. The supply of external heat only occurs above the

pinch, and external cooling only below the pinch.

Interval temperature

395 oC

325 oC

287 oC

60ºC

0 kW

-7458.5

0

-17263.35

0 kW

7458.5 kW

7458.5 kW

24721.85 kW

Figure 9.2: Heat cascade

From the Figure 9.2 above, the pinch occurs at interval temperature = 395C

9.4 THE NETWORK DESIGN

For the case which CPhot CPcold, the heat transfer can only occurs below the pinch.

Stream 3-4 will received the full load amount of heat required to bring up the

temperature to the Tt.

(9.5)

= 30.5 (282 - 55)

= 6923.5 kW

The load of stream 11-12 is (9.6)

= 106.55 ( 400 -330)

= 7458.5 kW

Heat being transferred, = received by stream 3-4

= 6923.5 kW

Heat being cooled by chiller, = 7458.5 – 6923.5

= 535 kW

400C 330 oC

535 kW

282 oC 55oC

6923.5 kW

Figure 9.3: Proposed heat exchanger network

The network shown in Figure 9.3 was designed to give the maximum heat recovery

and minimum energy consumption, hence increase the efficiency. Before process

integration, the process requires 7458.5 kW for cooling and 6923.5 kW for heating,

which will total up to 14382 kW. However, after heat exchanger was designed, the

process only requires 535 kW for cooling.

11-12

3-4

A

A

REFERENCES

Sinnott, R.K, 1999, Coulson & Richardson’s Chemical Engineering, Vol. 6 :

“Chemical Engineering Design”, Oxford, Butterworth-Heinemann.


CP - Steam heat capacity kW/°C

ΣCPc - Sum of heat capacities of cold stream kW/°C

ΣCPh - Sum of heat capacities of hot stream kW/°C

ΔH - Change in enthalpy kW

ΔHex - Heat transfer in exchanger kW

Tact - Actual stream temperature °C

Tint - Interval temperature °C


WASTE TREATMENT JUPLIN KINTI

NORMARIAH BINTI ABDULLAH

SUPERVISORS



CHAPTER 10:

WASTE TREATMENT

CONTENTS

TITLE PAGE

CHAPTER 10 WASTE TREATMENT


10.7 DESCRIPTION AND PROCESS

SYNTHESIS 224

10.2.3 Air Treatment 224

10.2.4 Water Treatment 225

10.8 LAYOUT OF WASTE WATER

TREATMENT 227

10.9 MECHANICAL DESIGN WASTE

TREATMENT 228

10.10 CONCLUSION 229

CHAPTER 10

WASTE TREATMENT

10.11 INTRODUCTION

Generally, MTBE plant produces waste into air and water during operation.

This waste will be treating before discharge to the environment. The

purpose to treat this waste is to make sure that component have not been

hazardous to the animal, plants and human health. The entire waste

component must be treated to fulfill the requirement of Environmental

Quality Act (1974) before discharge to the environment

In aspect air pollution, air discharge from MTBE plant contains the

Volatile Organic Compound (VOC) such as Acetone. These materials are

been hazardous in high concentration in the air.

The location of MTBE plant in Teluk Kalong Industrial Park that is

located at Terengganu Darul Iman and nearby the sources of water supply.

Hence, the wastewater treatment is under the effluent parameter in

standard B, Environmental Quality (Sewage and Industrial Effluents)

Regulation 1979.

So, the characteristic of waste water must achieve the required in standard

B before discharge to the river.

In our MTBE plant, most of the product discharge as a waste such

as Methanol, TBA and Isobutylene are recycle or reuse again to minimize

the waste discharge and reduce the pollution occurred to the environment

and directly reduce the cost to treat the waste.

10.12 DESCRIPTION AND PROCESS SYNTHESIS

10.2.5 Air Treatment

Generally, the waste that released into air in our MTBE plant is Volatile

Organics Component (VOC). There is acetone that produces in the first

distillation column. The concentration of acetone must be compare with the

Air Quality, Industrial Emission Standards for Organic Substances (1995)

which the limitation value is 2400 mg / m3. (Please refer to APPENDIX J-1)

This waste is flammable gases and has been low of flash point.

Based on this characteristic, direct combustion with flare is using to dispose

this waste with high temperature to make sure the complete combustion

occurred.

Table 10.1: The composition of gases discharged

Types of gas Mole fraction (kmol/kmol) Molar flowrate (kmol/h)

Acetone 0.0198 14.3012

In this case, we must consider the flammable limits of each

component to make sure that our waste is not lean to burn or explode. The

lower flammable limit (LFL) and upper flammable limit (UFL) are important

to detect the volume of gases mixture with air or oxygen for burning.

(Please refer to APPENDIX J-2)

Table 10.2: The Flash Point and the lower flammable limit (LFL) and

upper flammable limit (UFL).

Properties Acetone

Flash point (o C) -20

Auto-ignition (o C) 465

(LFL) % in ambient air 2.5

(UFL) % in ambient air 12.8

Acetone passed through at the stack flare and burned with high

temperature 10900C to make sure these waste burned in complete

combustion. This combustion will produce CO2 and dust into air. The

concentration of CO2 must be compare with level of emissions

Environmental Quality (Clean Air) Regulation 1978.

10.2.6 Water Treatment

In our MTBE plant, we discharged three types of waste water. There are

Methanol, Dimethyl and Propylene Oxide. This component must be treated

before discharge into river because it will be toxicity in human health, skin

and eyes irritation. For the long term, this component can reduce the

productive of animals and plant. The waste water treatment divided into

three parts. There are primary water treatment, secondary water treatment

and sludge water treatment.

Primary Treatment

In this treatment, the waste water passed through the settling tank to

remove oil and grease from waste water by disperses process and

sedimentation will occur by reducing the velocity of flowrate. Aeration tank

are using to keep the waste in suspension. In suspension, the waste having

a higher specific gravity then the liquid tend to settle while the lower specific

gravity will tend to rise. The precipitation component will pass through the

sludge digester.

Secondary Treatment

In secondary treatment, the waste is subjected to biological decomposition

by bacteria. Here, activated sludge tank are using which oxygen will supply

from air into tank and the temperature to be maintain about 370C.The

process biological occurs when the microorganism oxidize the organic

material to be stable compound

The sludge will discharge into sludge digester and remain passing

through the secondary settling using Granular medium filtration. Here, the

flocs settles out in this clarifying tank and pass through the digester tank.

Sludge Treatment

Generally, raw sludge that produces is untreated non-stabilized sludge and

it tends to acidify digestion and produce odor. Hence, we need sludge

digester to eliminate nuisances and reduce health related threat. Aerobic

digestion are apply with no external food supplied into tank because the

bacteria can metabolic their own protoplasm and oxidation the sludge be

stable and not harmful.

Dewatering sludge will reducing the amount of water in the sludge

so that it can be handled and disposed of as a solid rather than a liquid.

Here we use sand drying beds. After that sludge will pass through the

storage tank and disposed by combustion using incineration.

Inventory of Waste Water

Table 10.3: Inventory of waste water

Component Mass fraction (w/w) Mass flow rate (kg/h)

Methanol 0.0001 5.5739

Dimethyl 0.0005 27.8697

Propylene oxide 0.0010 55.7395

Oxygen required.

a. Methanol

CH3OH + 3/2 O2 CO2 + 2H20

Hence, 1 mol methanol required 3/2 mol of oxygen.

b. Dimethyl

C7H16 + 11 02 7 CO2 + 8 H20

Hence, 1 mol Dimethyl required 11 mol of oxygen.

c. Propylene oxide

CH3 (CHCH2) 0 + 4 O2 3 C02 + 3 H2O

Hence, 1 mol propylene oxide required 4 mol of oxygen.

Table 10.4: Oxygen Demand of component

Component Oxygen demand (g O2)

Methanol 3/2 x (16 x 2) = 48

Dimethyle 11 x (16 x 2) = 352

Propylene oxide 4 x (16 x 2) = 128

Average density = ∑ mass fraction of component x density of component

= {(0.0001x 791) + (0.0005 x 674) + (0.001 x 829)} kg / m3

= 1.25 kg / m3

5.1 Total oxygen demand = (48 + 352 + 128) g O2

= 528 g O2

Generally, the value of total oxygen related with Biological oxygen demand

(BOD) and chemical oxygen demand (COD). This total oxygen demand is

high and must be reduced before discharged into river. Based on the

Environmental Quality (Sewage and Industrial Effluents) Regulation 1979

the BOD discharge must below 50 mg / l and COD 100 mg / l in standard B.


10.13 LAYOUT OF WASTE WATER TREATMENT

Grit camber

Generally, Grid chamber as a function to remove inorganic like’s sand in

the waste water. In this treatment, grid chamber provided to remove the oil

and grease in the settling tank.

Primary settling tank

Here, two existing primary tank will be installed for high efficiency of settling

will be occurred. Generally, more than one tank will install which one of the

tank as a function stand by tank. The rotating arm skims the floatable

materials (oils and greases) from the water surface. Another rotating arm at

the bottom of the tank sweeps sludge into a collecting chamber.

Aeration Tank

Oxygen will supply from the air into aeration tank and consumed by the

microorganism. The process biological occurs when the microorganism

oxidize the organics material to be a stable compound. This process will

provide a higher degree of treatment.

Pump

The waste water from the aeration tank will pass through to the secondary

settling tank by the centrifugal pump. This pump is needed to raise and

distribute the waste water to settling tank.

Secondary settling tank

Here, more than one tank will install and the other tanks as a stand by tank.

Waste water flows into secondary settling tanks or secondary clarifiers

where bacterial cells form clumps called floc. The floc settles out in this

clarifying tank and is piped over to the sludge holding tank for disposal.

Sludge storage tank

These tanks are covered to keep out precipitation. Here, the biological

treatment still occurred inside the sludge tank and oxidize the sludge to be

stable and non-hazardous. Sludge thickening is accomplished in a tank

equipped with slowly rotating that breaks the bridge between sludge,

thereby increasing settling and compaction.


10.14 MECHANICAL DESIGN WASTE TREATMENT

Stack Flame Design

Stack flame is the equipment to disposed gases that produce in the plant.

The combustion of gases will occurred at the top of stack with continuous

flare at high temperature so that the gases will burn in complete

combustion to prevent any hazardous gases released.

The stack design must be high and designed plus 2.5 with higher

structure in the plant. The purpose is to make sure the plume to upward

level. In our MTBE plant the higher structure is 33.5 meter (Distillation

Column). Based on the EPA guidelines, the flare designed has the

operating temperature between 1370 K and 1920 K (1090 0C and 1650 0C).


h = (33.5 + 2.5) m

= 36 m.

Settling Tank Primary and Secondary Design

Flow rate average = Total waste discharged x average density

89 .18 kg / h x 1.25 kg /m3

= 69.13 m3 / hr.

The detention time = 1 hr

Depth Of tank = 5 m

The volume of tank = flow rate average x time detention

= 69.13 m3 / hr x 1 hr

= 69.13 m3.

The diameter of tank = (volume tank/ x depth)

(69.13/ x 5)

= 2.10 m

10.15 CONCLUSION

In our MTBE plant the waste water treatment discharge must be followed

the standard B in Environmental Quality (Sewage and Industrial Effluents)

Regulation 1979 which the value of BOD not exceed 50 mg/L and COD 100

mg / L. The waste water will continuously treat until achieved this

requirement.

Besides that, certain product such as TBA and isobutylene are not

disposed but reuse again to the plant operation because these components

are raw material to produce MTBE. This directly reduce cost production of

MTBE and pollution into environmental.

The other methods to minimize the waste in our MTBE plant is

packaging the certain product such as Propylene Oxide for commercialize

to get profit in the market.

REFERENCES

Christ. 1999. Production-Integrated Environmental Protection and Waste

Management in the chemical industry.. Germany. Wliley –VCH

C. Stern, W. Boubel. 1984. Fundamentals of Air Pollution. Second Edition.

Academic Press

Environmental Quality Act 1974 ( Act 127). 1999. Laws of Malaysia. Kuala

Lumpur . International Law Book services

European Chemical Bureau, European Commission. 2002. MTBE,

Summary Risk Assessment Report. Finland.

Hammer. 2004. Water and Waste Technology . Fifth Edition. New Jersey.

Pearson Education

H. Perry and W.Green, 1998. Perry’s Chemical Engineer’s Handbook,

Seventh Edition, United State.

R. Brunner, 1994. Hazardous waste incineration. Second Edition,

Singapore, McGraw-hill.

Regina D. 2002. An Overview of MTBE Remediation Process, Chemical

perspective. Inquiry reports MTBE. 7 August.

R. K. Sinnott. 2000. Chemical Engineering Design. Volume 6. Third Edition.

Great Britain. Butterworth Heinmann.


BOD biochemical oxygen demand

h height of stack

H effective height

LFL lower flammable limit

UFL upper flammable limit

VOC Volatile Organics Component

mtbe 3 - dp 2

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