NASA-CR-202447

_j_ 7_ ¸

Analytical Investigationof a

Reflux Boiler

FINAL REPORT

NASA Grant NAG9-839

October 18,1996

Lamar University

College of Engineering

Beaumont, Texas

William E. Simon, Ph.D., P.E., Principal Investigator

Fred M. Young, Ph.D., P.E., Co-Investigator

Terrence L Chambers, Ph.D., Co-Investigator

https://ntrs.nasa.gov/search.jsp?R=19960055147 2020-01-26T00:43:45+00:00Z

Ultralight Fabric Reflux Tube

Thermal Model

Imago

AbOut the Ultralight Fabric Reflux

Tube Thermal Model Report

i |

The purpose of supplying the final report in this electronic book form is to providethe user convenience and the ability to directly use the thermal model developed

since all of the calculation and plotting features of Mathcad are live. A change In any

desired parameter is accomplished by simply finding the value (or values if it is a

range variable) and changing it (them) to some other value. All values will berecalculated (provided the calculation mode remains in the automatic mode) and

plots depending on this (these) value will be redrawn.

eee

Click on the "TOC" button on the Electronic Book controls palette to go to the Table of Contents.

lille

Heat Transfer Modeling of a Single Ultralight

Fabric Reflux Tube (UFRT)

About the Format of Electronic Books

About this Electronic Book

Table of Contents

Repo_

Condensation Modelinq

Radiation Modelinq

1-5

I-1 to 16

I1-1 to 10

Model of JSC Test ArrayII1-1 to 4

Combined Model of a Sin.qle Tube IV-1 to 7

Shape Factor Between Tubes in an Array

Optimum Spacin.q of Tubes in an Array

V-1 to 3

VI-1 to 6

Thermal Conductivity of Concentric

Copper Around Nextel Fibers

VII-1 to 4

Estimation of the Free Convection Heat

Transfer Coefficient Inside the UFRTVII1-1 to 3

Thermal Modeling of a Single

Ultralight Fabric Reflux Tube

Executive Summary

A thermal model of a single Ultralight Fabric Reflux Tube (UFRT) was

constructed and tested against data for an array of such tubes tested in the

NASA-JSC facility. Modifications to the single fin model were necessary to

accommodate the change in radiation shape factors due to adjacent tubes. There

was good agreement between the test data and data generated for the same cases

by the thermal model. The thermal model was also used to generate single andlinear array data for the lunar environment (the primary difference between the testand lunar data was due to lunar gravity). The model was also used to optimize the

linear spacing of the reflux tubes in an array. The optimal spacing of the tubes wasrecommended to be about 5 tube diameters based on maximizing the heat transfer

per unit mass. The model also showed that the thermal conductivity of the Nextelfabric was the major limitation to the heat transfer. This led to a suggestion that tho

feasibility of jacketing the Nextel fiber bundles with copper strands be investigated.

This jacketing arrangement was estimated to be able to double the thermalconductivity of the fabric at a volume concentration of about 12-14%. Doubling thethermal conductivity of the fabric would double the amount of heat transferred at

the same steam saturation temperature.

The thermal model consisted of a one-dimensional, composite

convection-conduction model. The one-dimensional assumption was very good

on the shaded side of the tube that experienced condensation and was less so

when the other side experienced solar insolation of sufficient magnitude to cause

heat addition to the tube. The radiation was modeled as involving diffuse surfaces.

On the shade side radiation exchange with the lunar surface and space was

considered. On the sun side, a solar heat flux was input at the radiosity node of

the fin since it really represented the net energy exchange with the sun.

Introduction

A project was initiated at Lamar University to design and test a thermal

model of the new type heat transfer fin shown in the sketch below. The model

results were compared with data obtained in NASA's facility at JSC (Building 33,

Chamber E) and then used to predict performance on the lunar surface.

1 10117196

Section A-A

A A

Nextel 0.56 mm thick

Titanium 0.0026 in thick

CondenserSection

5.t cm

1.13 m

EvaporatorSection

Figure 1. A sketch of the Ultralight Fabric Reflux Tube (UFRT) to be modeled.

The Model

The thermal model consisted of a sequence of quasi-one-dimensional

processes: starting at the saturation temperature of the steam inside the tube the

processes are; a convection process at the tube inside wall, a conduction process

through the titanium liner, a conduction process through the containment fabric,and a radiation process from the fabric outer surface. The process is shown by

the sketch of the electric circuit analogy shown in Figure 2. below.

qToT.t ]ll="b

Rcond Rtl RNex

Ebfl n q Ebmoon

Ebspace

Figure 2. A sketch of the electric circuit analogy of the thermal model.

2 10117196

Rcond represents the condensation process inside the tube. In Appendix I It wasshown that the condensation process could be modeled as laminar film

condensation on a flat plate. Further it was shown that the magnitude of the

temperature drop across the film for the range of interest of our study was lessthan 0.1 degree Celsius. The heat transfer coefficient for the condensation

process was taken at a constant value for the JSC test and for the lunar

environment at the values shown below.

kWh JSC = 60.

2m •K

kWII lunar = 40 2

m . K

Conservative Estimates from Appendix !

Unfortunately, it was postulated in Appendix II and demonstrated in analyzing test

data in Appendix III that a simple one-dimensional model would not adequately

represent the behavior of the system. Since solar insolation impacts only one side

of the tube, the model had to be divided into two parts as shown in the sketch of

Figure 3. below.

Solar Insolation

lITtISun Side

Shade Side

Figure 3. The two parts (sides) of the thermal model.

3 10117/96

Some combinations of the radiant boundary conditions cause the direction of heattransfer to reverse. Such reversal changes the internal convective process fromcondensation to free convection. In Appendix VIII the free convection heat transfer

coefficients are calculated to be:

watt

h .ISC = 57. 2m , K

Conservative Estimates from Appendix VIII

watth lunar = 31. 2

m . K

Comparison with Experiment

In Appendix III the thermal model was used to generate data for the cases

of the NASA-JSC test.

Conclusions

1. The thermal model contained in this report closely reproduces the test data

generated in the NASA-JSC test facility.

2. From a heat transfer perspective, the optimal tube length to diameter ratio does

not appear to have been reached. It appears that the optimal value would be

limited by the thickness of the laminar condensate layer at the bottom of the tube

which could be increased above the values calculated.

3. A simple model of the tube, heat transfer fluid system, and support structure forthe fins indicates that the optimal pitch of a linear tube array should be about 5

tube diameters. This result is based on a mass of heat transfer fluid, fluid lines and

support structure which would be about three times that of the tube itself per meterof pitch spacing. The parameter optimized, heat transfer per Kg of mass, is verysensitive to smaller values of pitch to diameter ratio but much less sensitive to

larger values (see Appendix VI).

4. As expected, the thermal conductivity of the fabric containment limits the heat

4 10/17198

transfer from the tube. A method of introducing a small volume fraction (about12-14%) of copper fibers into the Nextel fabric is suggested in Appendix Vii whichit is estimated would double the effective thermal conductivity of the fabric and

thereby double the heat transfer from the tube at the same saturation temperatUre

of the steam inside the tubes.

i

End of Section

Table of Contents

Next Section

5 10117196

APPENDIX I

Thermal Modeling of the Condensation ProcessInside a Single Ultralight Fabric Reflux Tube (UFRT)

I-1 I0117196

Project Objective: To formulate and test a thermal model of the new type heattransfer fin shown in the sketch below. The model is to be tested against data

obtained in NASA's facility at JSC and then used to predict performance on the

Lunar surface.

Section A-A

-- Nextel 0.56 mm thick

-- Titanium 0.0026 in thick

[A A

CondellserSection

1.13 m

5.1 cm

EvaporatorSection

kI! ex

kW-0.00012.

m.K

k ti 8.4 kW Ti-2AI-2Mnm K

3Vol water 20 cm

The first step will be to examine the condensation process in the condenser

to determine the complexity of the model needed. We will begin by making

simple assumptions of laminar filmwise condensation on a vertical flat platewith no interface shear between the vapor and liquid film. We must test to

see whether each of these assumptions is valid.

To begin, we will take saturated steam at 20 Ib/in2 abs condensing on theinside of a vertical 1 in diameter tube 1.13 m tall. The tube wall temperature

is constant and varied from 227.8 to 227.95 oF. The heat transfer

coefficient, the heat transfer and the condensate flow will be calculated for

each wall temperature. Take Tsar = 227.96 °F and hfg = 960.1 Btu/Ib.

In

j 0.. 15 {j 227.8 , j..01 g 9.8. 2,gPC

I-2 10117196

Twj (460 , tj).R Tsal (460 _ 227.96) R

Tw. _ Tsat

Tfi _ .I -- AT i Tsar Tw.2, K • ,I

h f_

L

d

: 960.1.BTU

Ib

5.1-cm

N READ(sz_h2o) Number of data points in water properties

i __

Th i

kh i

Chp i

Pf..!

P!o

krl

P vii

P vi

0..(N I)

READ(T h2o) ph i

READ( k_h2o ) GrPr i

REA I)( mu_h2o ) ph i REAl)( ro_h2o )

READ( GrPr_h2o )

READ( CP_h2o )

'Th ,lth ,T kglinle,'l, ( f.i ). m. sec

Th,ph,T ').li,,let'p ( fJ kg3' ' I!1

wall

li,,,erp(Th,kh,T fi )' m-K

T f. 400.I

0.5542 t- - . ( 0.490250

100 p v lj kg

14.7 3m

0.5542 )

h.i0.943.

I

P fid" (T sat Twt)

Eqn. 9-10 in Holman

q.I hj'Tt d' L (T sat Tw.i )Eqn. 9-15 in Holman

l- 3 1011"/190

500

Condensation lleat Transfer vs. Del(Temp)

I I I I I I

L.

'!.1

:::[:

-i(}o

300

200

I00

I I I I I I I I0.(} I (1.02 0.03 0.04 (I.05 0.06 0.07 0.08 0.09

Delta Temperature - dclzrccs K

I

41tfkf, iL'(Tsat T wi)]4.i ..... : ....... 7

Eqn. 9-6 in Holman

w.i

P fi'(P fj- PVj )'g {_j) 3'/i'd

3. p f,i

I - 4 10117196

i

.,u+

..J

E2

r'n

O.O45

0.ll4

0.035

q).03

q).025

0.02

Condensate Layer ThicknessI I I I I I I I

l I I I I I I I0.(I I 0.02 0.1}3 11.0-1 0.t|5 0.06 0.()7 0.08 0.1)9

Delia Temper;llure - degrees K

LJ

i

t,.

O.q|8

ll.06

II.I)-I

11.4)2

Conndensalc Mass FIowrale

I I I I I ! I I

I I I I I I I I0.111 0.02 0.113 l).04 11.I)5 0.06 I).07 0.(18 0.09

Delia Temperature - degrees K

4. w.iRe : ............

[i +. d. li f..!

Eqn. 9-13 in Holman

I - 5 I0117196

t

p.,

E0

an

(|.O45

o.q)-I

0.035

0.03

0.025

q).(J2 o

Condensate Layer Thickness

1 I I I I I I I

I I I __1 I I I IILl| i 0.02 0.03 (I,(M l).05 0.l)6 0.07 0.08 0.09

Delia Temperature - deErt'es K

¢.)

i

'o

LL

0.08

0.06

0.(|4 -

0.02 -

0 0

Condensate Mass Flowrate

I I I I I I I I

I I I . I I I I II).(I I ll.02 (I.II3 (l.04 0.05 l).06 0.07 0.08 0.09

Dclla Tenlperature - dc_rees K

4. Wj

Re f.i g.d._ f..I

Eqn. 9-13 in Holman

I - 5 10/17/96

E

7-

15

Condensate Reynohl's Number

I I I I

I I I0

II 0.02 0.(1-1 1l.I)6

Dcll:l Temperature - (Icl_rccs K

I0.08 I_. I

Ileal Transfer Coefficient

120 i i i i

rl<

' !00

8(I0

6(I

n,)

I I I I4O

(I 0.02 0.04 0.(16 0.118 O.I

Delia Temperature - degrees K

From which we can see that assuming an h = 60 KwlmZoK would be

conservative.

condkW

60. -2

m • K

I-6 I0117196

Therefore, we can examine our assumptions. The filmwise condensation

process is laminar as shown by Reynold's Numbers in the range of 2 to 14

in the figure above. The figure for condensate layer thickness at the

bottom of the tube gives a maximum of:

5max :" 3.5.10 .=11

at the maximum heat dissipation that we are interested in of 150 Watts.

The ratio of diameter to condensate thickness is then:

ratiomax

(|

ratio = 1.378- 10 .3

which validates our vertical flat plate assumption.

We will calculate and plot the upward vapor velocity to determine

whether our no interface shear assumption is approximately correct.

We know that for steady state the upward mass flowrate of vapor must

be equal to the downward flow of condensate.

x i d 2._i )2 wj

A v. -- - w vapor =w condensate V v. - "

.j 4 .I p v.i. A v.t

i

i::,L_

'o

>

>

0.04

0.03

0.02

0.01

Upward Vapor VelocityI 1 I I

I I I I0 0.02 0.04 0.06 0.08

Delta Temperature - degrees K

0.1

I-7 10117196

which validates our assumption of no interface shear.

We know that the initial water charge is 20 cm 3. Let us check to see themaximum volume of water contained in the film to be sure that theevaporator section does not go dry. The volume of water contained in thefilm is given by:

.i+Volume= 7t.d 5 dx

Vohlme= _ d

"LI

4-B.k.x. (T sat Tw+

_ltrg+[Pr(Pr, Pv)l

dx

Voluine = _ d

I 5

4. p.k.(Tsa! T w ) /_(i,)_

Volume i : _ d.

I5

4_t fkf(Tsatj.i ' " Tw').; t4.......<L,4,oo_....

but at

ft 3v 20.089.

g Ib

Illvapor

+t.d 2 L + 5.I.cin

4 vg

T sat :: 5110. K

invapor

ft 3v -: 3.788.-

g Ib

= 4.566.10 4 . kg

I - 8 10117196

I!1vapor

7c.(!2 L _ 5.1 cm

4 vg

nlvapor

= 2.421.10 :} .kg

"o

i

E¢,

_u

E-6>

2.5

2

1.5

Volume of Water in Film

I I I Ii

I 0 0.02 0.04 0.()6 0.(18 (I. I

Della Temperature - de_rees K

So about 10% of the 20 cm 3 charge is contained in the film at the maximum

anticipated heat load of 150 Watts, however, the mass of water vapor becomes

the same order of magnitude for a saturation temperature of 500oK.

Lunar Environment

Now we need to repeat the calculations for the lunar environment.

9.8 mg . . __

6 2sec

Again, we will take saturated steam at 20 Ib/in2 abs condensing on theinside of a vertical 1 in diameter tube 1.13 m tall. The tube wall temperature

is constant and varied from 227.8 to 227.95 oF. The heat transfer

coefficient, the heat transfer and the condensate flow will be calculated for

each wall temperature. Take Tsat = 227.98 °F and hfg = 960.1 Btullb.

I - 9 10117196

j 0..15 t,i : 227.8 _ j.0!

Twj I460 , t i).R Tsat (460 _ 227.96).R hfg

I" , 1" sat

T r.I wj _ AT. :T T L :2 K .; sa! w.td

960. I

1.13m

I-in

BTU

Ib

5.1 cm

N = READ( sz_h2o ) Number of data points in water properties

i 0.,(N t)

Th i

kh i

READ( T_h2o ) ph i REAl)( mu_h2o ) ph i -- READ( ro_h2o )

READ(k_h2o) GrPr i - READ(GrPr_h2o)

Chp i

"!i

Of..I

REAl)( CP_h2o )

i) kglinlerp{ Th,ph,T f. m.sec

linterp4'Th ph Tf.il. kg' s • 3

m

kr.I

PVI..I

P vj

linterl)iTh,kh T fi, ) .....watt• . m. K

T f. 40O

0.5542 + -- .! .( 0.490250

100. p vii kg

14.7 3m

0.5542)

hj: 0.943

!

• 3] 4IJ.fid(Tsa I - Twl' )

q.i h.i. _- d. I,. (T sa t T w,i )

I - 10 10117198

l-4

F-_

%)

31)1)

25q)

21)1)

150

Iq)O

5O

Condensation lleat Transfer vs. Del(Temp)

I I I I I I I I

I I I I I I I I0.01 11.02 0.03 11.1)4 0.1)5 11.06 0.07 0.08 0.09

Delta Temperature - dc_rees K

4PfikfL(T,i sat T w.I)

p,:,l]

w.i

Pfi(_fJ _vi,_ _.i)331L f.

,I

I - 11 10117196

i

,j

O

Condensate Layer Thickness0.()7 I I I I I I I I

(I.06

().05

(I.04

(I.(I.]I I I I I I I I

(|.ll I 0.02 0.03 0.04 0.()5 0.06 0.07 0.08 0.09

Della Temperature - degree_ K

i

'D

o

0.05

0.0-1

0.03

(1.1)2

o.(I I

Condensate Mass Flowrate

I I I I I I I t 1

I I I I I I I I0 0.0 ! 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

Dellal Temperature - (lellrees K

. W.

I I 12 1011"D96

4 W,i

Re fi -

EZ

0

10

6-

4-

2

Condensate Reynohl's NumberI I I [

I I I Il).02 0.04 O.ll6 0.(18 (). I

Della Temper-'lture - de_rees K

7O

rl<

E60

5()@

lleat Transfer Coefficient

I I I

I I I I30 l) 0.02 0.04 0.06 0.08 O.I

Della Temperature - dellree,q K

From which we can see that assuming an h = 40 Kw/m2oK would be

conservative. I - 13 10/17/90

conservative.kW

II cond 40.-2

m •K

Therefore, we can examine our assumptions. The filmwise condensation

process is laminar as shown by Reynold's Numbers in the range of 2 to 10

in the figure above. The figure for condensate layer thickness at the

bottom of the tube gives a maximum of:

54.2.I0 .m

nlax

at the maximum heat dissipation that we are interested in of 150 Watts.

The ratio of diameter to condensate thickness is then:

nlflXratio .....

dratio -- 1.654.10

which validates our vertical flat plate assumption.

We will calculate and plot the upward vapor velocity to determinewhether our no interface shear assumption is approximately correct.

We know that for steady state the upward mass flowrate of vapor must

be equal to the downward flow of condensate.

;t.(d- 2._i )2

A v i 4 ' w vapor =w condensate V

w.i

v.i p v I. A v.i

u

i

'u

>

>

0.025

0.02

(I.01

0.01

0.005

0

Upward Vapor VelocityI I I 1

I I I I0 0.0 2 0.04 0.06 0.08 O.I

I - 14 10117198

Della Temperature - dcl_rees K

which validates our assumption of no interface shear.

We know that the initial water charge is 20 cm 3. Let us check to see the

maximum volume of water contained in the film to be sure that the

evaporator section does not go dry. The volume of water contained in the

film is given by:

Volllllle = . "i __. d. (_ (Ix

Vohlme- 7£. d-

LI

4ll k. x. (T sat Twl 14........... dx

I5

I 4|tk(Tsal Tw) ]4Volume=_. d. ' ' ( L)4

Volume.i

I5

I 411 f(k fi"/T sat T wi} (L) 4 1003_" d" -: " - ................ " -

fi3v : 20.089 ....

g Ib

_. (!2

mvapor

L i 5.1 cm

1-15Ill

vapor= 4.566- 10 4 kg

10117196

m vapor = 4.566 10 kgm vapor - 4g

but at T sat500 K V

g

fl 3

3.788 ....Ib

m vapor

d27[ L t 5.1cm

4 vg

nlvapor

= 2.421.10 3 . kg

;

G

>

4.5

4

3.5

3

2.5

Volume of Water in Film

I I

I I I I2 0 0.02 0.04 0.06 0.08 0. I

Delta Temperature - degrees K

So about 25% of the 20 cm 3 charge is contained in the film at the maximum

anticipated heat load of 150 Watts, however, the mass of water vapor increases

the total to about 35% at a saturation temperature of 500oK.

............ =6 IIH, IIII I IIII II IIIIIIt ii ill

End of Section

Table of Contents

I - 16 10117196

APPENDIX II

Radiation Modeling

II - 1 10/17/96

Radiation Heat Transfer

If solar insolation is present, we might note that the insolation is incident ononly half the fin area as shown in the sketch below.

Solar Insolation

Sun Side

Shade Side

The area of the fin will be composed of equal areas on the sun side and

on the shade side.

Radiant energy exchange will be assumed to be diffuse. The electric

circuit analogy for the radiant energy exchange between the fin, the

lunar surface, the sun and black space can be represented by the

partial sketch below.

Ebsun

1 " I:_lur I

'3buwl _ _u,_

}Im

Ebfin " Et_oo,

, - _l _ _ '/V_J.... /_-_:-

I " l:m_rt

"li I":m©°n Am°°n

'I

Ebspace

The reason we have presented only a partial sketch is that we will show

that it is all that is necessary to determine the net radiant energy

exchange with the fin. In each of the surface resistances there is an area

term in the denominator. All other t ir will be expressed with the alr_p_)/f96

term in the denominator. All other terms will be expressed with the area of

the fin in the denominator. Since Aspac e, Amoon, and Asu n are much much

larger than Afin, their associated resistances can be taken as zero. Inaddition, we know the net heat transfer with the sun to be the solar

insolation value. Hence, the modified electric circuit analogy would look

like that below.

"I - _fiR

I-'_r t Afl=,

Jfin

Solar Insolation

Ebfin I Ebm°°n

II,

Ebspace

1

Afi n F rin moon

1

Afi" F fin.space

For which we can sum the heat flows into the node Jfin and set them to zero.

E bfin ,I fin

I _fin )fin A fin

E bspace 3 fin

,- I )'A fin F fin space

E bmoon ,I fin

....... I ..... \ ' q sun

)A fin F fin moon

=0

which applies on the sun side and a similar equation would apply on the

shade side.

II - 3 10/17/96

E bfin2 J fin2

I _ fin

fin A fin

E bspace J fin2 E bmoon - J fin2

E =_T 4 andIn these equations we know that Ebspace=0, Ebfin=(_Tfin 4, bmoon moon

Sfin=0.89. The configuration factors may be determined in the following

manner: consider the sketch below,

fin --

dA -

space '

seesspace

I/

sees lunar surface

lunar surface

From the figure above it is obvious that FdA.moon and FdA.spac e are both 0.5

for any position on the surface of the fin (assumes lunar surface is an

infinite flat plate). Therefore Ffin_moon and Ffin.spac e are 0.5 for either thewhole fin area or half the fin area (divided into sun side and shade side).

The equations above then reduce to on .the sun side taking qsun to be the

energy incident on the fin per unit area times the perpendicular area:

4J fin =_ fin oT fill t (! 6. T 2oon4 q StillCfin ) ......... _ ........

A fin

II -4 10/17/96

and on the shade side:

/1110011

Jfin2-_fit,._.Tfi,,2 , (I I:fi,,) .... 2 -

where Afi n is half the total fin area, Tfi n is the outside temperature of the

Nextel exposed to the sun, and Tfin 2 is the Nextel temperature on the

shade side.

We can now examine the simple but important case of what minimum heat

transfer is necessary to keep the fin from freezing-up during the lunar night.

For this time we will take Tmoon = 88 OK, Tsa t = 273°K and no solar insolation.

d : I.ill 1, 1.13-m

_fin

dO

0.89 T sat 273. K T moo,1 88. K

d , 2..00056.111 _ 2..0026. in do =0"027"nl

We will iterate for the outside Nextel surface temperature such that the heat

loss by radiation is equal to the heat convected and conducted through the

cylindrical fin wall. We will take only half the fin area so that we will beconsistent later on when we must consider sun side and shade side.

x.d o-LA fin

2A fin = 0.047.m 2

kW kW kWk ti _ 8.4 .... k = 0.00012 ..... h cond = 60 ....

m. K nex m. K In 2. K

The iteration process is shown in the sketch below.

q

Tsar _ To

Rcond Rtl RNex

Ebfi n q Ebrnoon

II - 5 10/17/96

R coralI

h coral' g'd !,R ti

In (d , .0026.2. in /d ........ /\

7t.k li.L

R Nex

d o

I,,(_ , .002612 in

x.k Lngx

There is a more compact way to do the calculations below, however,

this method has been used to try to illustrate the calculation

methodology.

AssumeT 271.694 K

0

qlR

T sat T o

cond_ Rli , RNex

5.669 I0 8 watt

2 K-Im .

Jfin : aliaif'To 4 ' (I _ Grin}.

o. T 04 - J fin

\_ fin 'A fin/

q 1 = 12.935.watt

wattJ fin = 275.114 ........2

m

q 2 = 12.935.watt

Therefore it takes q tot _ 2q I q tot = 25.871 .watt

to keep the fin from freezing during the lunar night.

Increasing the saturation temperature to 300oK. T sat

0 q 0 T 273ki 0 4 T skj 11,i S0

II - 6

300. K

q lot

q 1o war!

10117196

Assume T 298.104 KO

ql

T sat T o

Rcond + Rti t RNexq I = 18.779.watt

O 5.669 10 8 watt

2 K 4Ill .

J fin

q2

tincT ° 1

o T 04 - J fin

'i fi. 1I_ fin "A fill/

(I t Inoon4

o.T\

2J fin = 398.631.

q 2 = 18.778.watt

q tot 2. q I

at T sat = 300. K qtot =37"558"watt Tsl : 300 qt

watt

2In

q tot

walt

Increase T sa{ 350. K

Assume T 346.532. KO

ql

T sat T o

Rcond _ Rti _ RNexq 1 = 34.349. watt

.I fin /oT _ oon 4)

wattJ fin = 727.749. 2

In

q2

o T 04 - J fin

t_fin A fin/

q 2 = 34.348. watt

q tot - 2q I

II - 7 10/17/96

at T sat =350. K qtot =68"698"watt Ts2 _ 350 q t2

q tot

watt

Increase "Fsat : 400. K

Assume T : 394.189. Ko

ql

T sat T o

Rcond + Rti _ RNexq I = 57.556. watt

.I fin E fin. _To 4 , (I _;fin)

wattJ fin = 727.749- -

2m

_.T 04 J fin

q2 ...............I _ fin

_"fin A fin

q 2 = 57.558-watt

q tot 2. q !

at T sat = 400. K q tot -- I 15.11 I .watt T : 400s 3 q t3

q tot

watt

Increase T sat : 500 K

Assume T o 486.507. K

T sat T o

ql ...................Rcond + Rti * RNex

q 1 = 133.643 .watt

II - 8 10/17/96

J fin oon)_fin.o-To'l _ (I _-_fin) ................ J fin = 2.827.103 • w.:__al!2

m

q2

o. T o 4 J fill

I _ fin

fin A fin

q 2 = 133.644 .watt

q tot : 2q 1

at Tsar = 500.Kq tot

q tot = 267.286.walt s4 = 500 q t4 watt

i

'D

300

250

2(}1}

150

100

Lunar Night Fin q vs Sat. Temperature

5O -

°2so

I I I I

I

I I I I300 350 400 450 ,00

Saturation Temperature Inside Fin - K

II - 9 10117196

End of Section

Table of Contents

II - 10 10117196

APPENDIX III

Thermal Model of the JSC Test

III - 1 10/t7/96

JSC Test Model

'I • l-;flol '...... +.

r.fl n A +m

Jlln

Simulated Solar Insolation

,>-_+A:,+_'pq--,_

,|+

/ E hchamh/

Simulated Lunar

Heat Flux

1

k 0.00012. --nex

kW

m K

wallk ti _ 8.4+

m. K

fin _ 0.89

d : d , 2..00056. m , 2..0026. ino

x.d o.L

A fin ........2

R cond

I

Lhcond. X.d +

2

R ti

Reqv Rcond + Rti t RNex

L 1.13. m 5.I.cm

d I.in

d = 0.027.mo

A fin = 0.045.m 2

in( d _ .00262in)d

x.k ti.L

Reqv2

h cond: ........ R

h fc

h cond

kW60.

2m .K

o _ 5.669.10_l watt

2 K4m •

walthr. : 57 ......

IlL 2m.K

do '

R Nex ........................z. k nex' L

cond _ Rli _- RNex

III-2 10117198

Therefore the energy balance at the radiosity node is:

E bfin = J fin

(I.[ _fin I ass

\e fin A fin/

+ qsl +_

E bchamb + J fill...................... _0

( A tin-(F+ filn chamb )

I 4 q ss q sl 4....... 7_fin_Tosun t ..... , - - - _ F fin chamb oTchamb

(i-_ n,,) An,, An,, -

Jfin= -- I _ + ! , afin F fin ¢hamb

l/I +"n,,/

or without solar:

J fin =

I 4 q sl 4+--.efin._Tosh _ --- _ F fin chamb _Tchamb

(1- _fin) Aria -

',7......... , ........... 1++(1 + _ _ +fin Ffin chamb 1

fin) - /

Then an energy balance setting the energy transfer through the composite Wall

equal to the energy radiated from the outer surface gives:

T chamb 120. K F fin chamb I

qsl 16120 ' wal---tA fin qss 1370 .watt.... d.L2 2

m m

Z(Ts,T

(I -- afin) E fin. oTo + A fin ,- Ffin_chamb. O.Tchamb

(I + e fin) + fin - F fin_chamb

_('r)Ts To _To o

R I _, lit - 3 10117196

$ O Reqv I - _fin

r fin A tilt

Fs(To)

1 4 _ q sl q ss _ .,. 4

7 .... _.Efin.o. To A .... _ A ........ _ F fin chamb6tchambI - _ fin finIC n,) - _...................

I( i- -t-finiefin- ITfin chamb

Zs(Ts,T o )

T s T O o"To 4- Fs(To)

R eqv I - _ fin

fin A fin

T TS 0

Zs2(Ts To) :.........' R eqv2

oTo 4 F s(To)

\_ fin. A fin/

T sat 373. K T on 300. K

Tosh root(Z(Tsat Ton)Ton) T osh = 433.294. K

End of Section

Table of Contents

III - 4 101t7196

APPENDIX IV

Modeling of a Single Fin on the Lunar Surface

p

IV - 1 10/17/96

Single Fin Model for the Lunar Environment

Solar Insolation

Ebfin ' T / Ei)m°°r_

1 - Efin I

_:[_,,Att,_ Ebspace

J fin

1

Afi n F fin ntom_

1

Ar_nF ri,,.sp,,o

Test Series 6: Single Fin - Lunar night just prior to sunrise

kWk 0.00015 !, 1.13 Ill 5.1 cm

IIPX m, K

W _1t I

kli 8.4 m K d I.in

r, fin

do

0.89

d _ 2 .00056m _ 2 .0026 in do =: 0.027.m

_(I o.I. 2A fin A fin :: 0.045. m

2

I-- R tiR cond I,

h condTtd2

d , .0026 2. inIn i i

d

7t k lil,

Rcqv Rcond t Rli ' R Ncx Rcqv2

h cond

h fc

h COlld

kW40.

2m K

5.669 I 0x wall

2 K-IIll

wailh fc 31. 2

m K

R Nex

,' tl4 O

In=d , .00262 in

_.k ,!,IlPX

• Rcond ' Rii ' RNex

T 88. KIIIOOll

4E hun _ T moon

IV - 2 10/17/96

Z, Toe TsaleJT sate T

Reqv

oe

4efin,(t, Toc , ,' !

I _"finF

I:fin A fill

/ E

r bm le fin !q

2

j o .. 20 T s. ( 273 , I0 ,i)K.i

T 3oo KOll

T oZi ,-,,otiZ(T ,,, T on ion, T s.i ,,

q I..; R

T Ts. o2.J .t 2

eqv

Single Fin, Just Prior to Sunrise

500

45O

t

_. -8OO"-1

351)F_

300

25O

l,unar Night q front FinI I I 1

I I I I50 Iq)o 150 200

Tsar

Toutsurf

q - W;ttts

25(1

IV - 3 10/17/96

Test Series 1 - 5

To model the different test series, set the model input

parameters as shown - results plotted below.

Test Series Tmoon Theta1 388 90

2 352 45

3 288 60

4 238 45

5 88 0

90T 388. K 0

moon 180

BTUq' 442

Ill" ft 2

4

E I)m r_ T moonq sun q'.d I,. cos(0)

Ej, bm q sun

Eq :2 A fin

Z I ("r oe,T sate)

T sale T

Reqv

oe

Z 2iT oc'T sale )T sale T

R eqv2

oe

I oe. It, fin.c_.T , ( I

I _fin )r. fin A fin ,,

_'fin° Toe 4 , I I

('e fin A fin,

r,fin I Eq

T o2n 300. K

IV - 4 10/17196

j 0 10 Tsal. 273 K _ j, 10 K.I

fr osun. root i Z I'f T o2n,T sat..I .I

•T o2n

Toshade. roo! Z(To2n Tsar. ,To2n.I .I

T osunfc..I

tool Z2!To2n,Tsal. i,To2nl.I

q sun.j R

T TS;i{. Ogllll.

.I .I

eqv

q sunfc..I

T sat. T oshade.J .I

Reqv

T Silt. T Olllll..I .I

Reqv2

q it(kk)

q Iol..I

I

-- q lunld i q slialleld _ )

(! SuilfCkk + q shndekk)

q It(J)

if Tsat :-Tkk

otherwise

OSll !!1( k

IV - 5 10117196

L_i

38O

360

34O

320

300

28O

260

Single Fin on the Lunar Surface

Fin q vs. Temperature

I I I I _

////

I l I II 0 20 31) 40 50 611

q - Waits

Ts;i t

T sun side

T shade side

q _unrc.i

Free Convection and Condensation Heat Transfer

2O

I 0

(|

i(!

I I I I

ZOIo I I I I20 30 40 50 6O

q loll

IV - 6 10/17/96

End of Section

Table of Contents

IV - 7 10/17/96

APPENDIX V

The Shape Factor Between AdjacentTubes in a Linear Array

V - 1 10/17/96

To determine the shape factor between adjacent tubes in a linear arraywe will treat them as infinitely long. Then the shape factor is simply thedifference in the sum of the crossed strings minus the uncrossed stringsdivided by the base length. The sketch below shows two tubes ofdiameter, d, in such an array separated by a pitch distance p.

I +__ P

"0

T

Then F1. 2 is:

4(s

F I 2 = ....

q) 4, p2

(!7[.

2

_+_

but s= / p! (124

S(s +d

or

q

(I

P

2, df

S

d

l( i'P I

,]_,a/

2

= ;1sin ((|

\ p

and q = asin/(d

d _,p

therefore

F I 2 = ) ,+...,,,p •d ..... d

V - 2 10/17196

F c¢( pd ) /( iI I)dpd)2 ! , _lsin(,pd

0.. 20Plij 4 , j !; 12j F cc t I)11i)

©

L_

r,_,

0.14

0.12

(I.I

0.0X

0.06

0.11-1

O.O2

Shape F;iclor Bclween Two Infinile Cyl.I I t

I I

11 ._ I II I _ 20 25

pilch/dilimeler

..................... ..._-/--7_L_..:...;7;7-----_:;77........... 7.7_ __;;L_-.:_--_-_-7_;_7_7...----_T7_-___7__7__77:7..................................

End of Section

Table of Contents

V - 3 10117/96

APPENDIX Vl

Pitch Optimization of Linear Tube Arrays

VI - 1 10/17/96

J fi=_

Solar Insolation

Ebfin t Ebm°°rl,,, .y "

1

Afin F ri,, moon

1

Afin Ffin.gpace

kI|('X

kti

r. fin

do

kW0.00012 - -

m.K

wall8.4

m K

0.89

d , 2 .00056 m ,

!, ! m 5.1 ¢m

d I. in

2..0026 in do

:: 0.027. m

h cond 60.kW

2m •K

o 5.669 I 08 wall

2 K 4m •

walth f¢ 31 2

m K

A fin

r coild

I

h cond x. d2

Rli

A flit = 0.04.m 2

I,IId t .0026 2 in)

_.k li, I,R Nex

In ( d d o ).0026 2 in

_k .!,ilex

Reqv R cond _ R li R Nex R

eqv2hcond Rcond ' Rti ' RNex

h fc

VI - 2 10/17/96

E bfin J fin

i: ! _:fin_l

I e fin A fin/

0 J fin

I 'iA fin'Ff s

J(tT II1 .l fin

/ I 1

" fill f m,

q sun A fin =0

.I fin =

4e fin crT oe

I r fin

i l;'fmi'°"T-I

III

r fin i

, 2. Ff m!r.fin - r

q SIIII

but FF f s= f m

T m 80. K q sun

C fille l-

i [ Ill

(;_ TOP ' li"f m ) e ftt-T

B'FIIq' 442.

hr. fl2

q'. cos(O)

4 4oe ' _ F f III'T I!i P (I Sllll

45

180

Z I(Toe, T sale, F f III )

T sale T oe

Reqv

G(Toe,F f m)

(eft 2. Ff m)

I

e fA fin

Zi T oe, Tsale,Ff m)T sale T

Reqv

efo.T 4 m 4oe * °FfmT

(ef, 2Ff m)

e f A fit)

VI - 3 10117196

Z 2( T Me'T sate, F f m!'F sale T

R eqv2

o.T oe 4

Me

|?G(T oe, f m

!ef+ 2. Ff m

,)erA fin

T o211 300. K

j o 40 Pd.i I _ .i-.5 3" sat 373. K

F co( pd )

f

,/( pd )2

r

1, asin l Ipd

pd

TMSIIil.

.I

0.5 F eei pdi)

'Zll M2n, sat, fm.iroot T T F •T M211t

J

T oshade..I

rMOI Z!ITo2n ,T sat, F f,,,.i ) ,T M2n)

T osunfcj ,'oot(Z2(To2n Tsat Ffm.i ),TM2',)

q NIIll..I

1" sat T Msun..I

--4--

Reqv

q shade..I

T sat T oshade..!

Reqv

Vl - 4 10117196

T sat T osun.i

q sunfcj --R eqv2

q it (kk) q sunkk q sh,ldeld_ ) if T snl T

iq sttnfckk i q sh:i(lek k :, olherwise

osttnkk

q lot I..I

q tt(.i)

131.7. gin _ l)di.d. 131.7, gm• !!1

A mass fraction of 1, i.e. each

meter of spacing adds a

mass equal to the reflux tubemass.

q IoI2i±

131.7. gm

q ll(.i)

Pdi. d. 131.7. gin. 2I!1

(! tot3.i! 31.7. _m

q tt(i)

Pdi d 131.7 gm 3Ill

Vl - 5 10/17196

For a Saturation Temperature of 373oK, LunarTemperature of 80oK, and a Solar Angle of 45°

300

200 -

_h

100i

E

0

lq)O 0

lira! Transfcn'/Mas,s vs Pitrh/Diamclcr

I I I I

I I I I

5 10 15

PilchlDiam('ler - l,inear Sl)acin E

mass frac=l

ma,_s fra¢=2

I!i _l,gS frfl(---3

2O 25

This figure indicates that a resonable design pitch/diameter ratio would be

about 5. A higher saturation temperature would perhaps push this ratio downto 4 to 4.5 as would more massive structure and piping arrangements. Even

in these cases, the use of a ratio of 5 would not be costly since the function is

not as sensitive to the right of the peak as it is to the left.

End of Section

Table of Contents

VI - 6 10/17/96

APPENDIX VII

Equivalent Thermal Conductivity of Concentric Layers

of Copper and Nextel

VII - 1 10/17/96

The model below shows copper wires covering Nextel threads in a compositefabric weave. The purpose of the fine copper wires is to increase the thermalconductivity of the fabric while minimizing the loss of strength provided by theNextel fibers. The copper layer is shown around the Nextel fibers so that theweave of the fabric allows the heat to be conducted along the copper wires or

strands and thereby carries the heat from the inside surface to the cross layersof fibers and from these cross layers to the outside surface as shown in the

figure below. Copper is used instead of gold to avoid an undesireable decreasein outside surface emmisivity.

:Z AI" ,r"t= , ,AT .... L2AI"I<Ncx w • + k cu w.. _-,2.x.l_ . I_'kr.Nvi .

t _Z --|wl _'c°` t

,t

Ikk =

c

2w

Nex , kcu. t+xt

(w z

kc

-- _ I +

k Ncx

2k cu /' I =

I ] .x

k Nex ' w

,t=k Nex + k• I ]2

£11 w ,]

VII - 2 10117/96

k Nex

kCtl

kW11.00012. I .0026 in w .05 ill

m. K

IOvV0.386 --

m. K

k cu _ I 2 k ¢u

ke(xx) I t t xxk Nex ' w k Nex

= 3.217- 103

i 0 .211 x. 0 , i .0075 2

('1W,

krat i : ke_:x i)

= 2.7O4. I 0

7_

.:A

>

2.5

2

1.5

I

I I I 1 I I I

II 0.(12 11.04 tt.I)6 11.08 (I. I O. ! 2 I). ! 4

Volume Fraction - Cnppt'r

O. 16

VII - 3 10117196

End of Section

Table of Contents

VII - 4 10117196

VII - 1 10/17/96

iii!i! iiii!i!iiiiiii __ii!iiiiii_iiiiiiiiiii_iiiiiiil_

VII - 2 10117196

VII - 3 10/17/96

VII - 4 10/17/96

APPENDIX VIII

Estimation of the Free Convection Heat Transfer

Coefficient Inside the Vertical Tube

VIII - 1 10/17/96

Estimation of the free convection heat transfer coefficient between the

inside tube wall and the saturated steam from the evaporator. This

condition exists when the radiant boundary conditions produce an

external surface temperature above the saturated steam temperature.

The heat transfer coefficient will be taken as that for a vertical flat plate

even though the tube will not behave in this fashion. This will give a

larger value which in our case will produce a higher heat transfer to the

tube and thereby give conservative results.

Properties of steam at 573oK.

6p 19.7 10

p 37.4 kg3

m

I13

573 K

waltk .064 I.

m K

Pa sec Pr 1.443

Gr

5 K L

g 13AT i,3

1.13 m gin

9.82

Gr : 4.447.10 It therefore, turbulent

Nil

h

I

0.15 (Gr. Pr) 3

16

9 27

16

0.492 iPr ,

kNu. -

Lh = 56.695"

watt

2m . K

Nu : 999.452

h = 9.984.BTU

hr. ft 2. R

VIII - 2 10/17/96

Use h = 57 watt/m2°K for the JSC test and:

gg

6

Gr g [3AT !"3 Gr = 7.412.10 i°

p/

Nu Nu -- 550.02

!

0.15.(Gr Pr) 3

16

, )9 127I _ ( 0.492' 16Pr j

kIi Np h = 31.2.

I,

still turbulent

watt BTIIh = 5.495.

2m K hr-ll 2. R

h = 31 watt/m2°K for the lunar surface.

End of Section

Table of Contents

Next Section

VIII - 3 10/17/96