1. INTRODUCTION

1.1 A composite wall consist of alternative layers of fir ( 5 cm thick ) , aluminum ( 1 cm thick ), lead ( 1 cm thick ), and corkboard ( 6 cm thick ). The temperature is 60 C of the outside of the fir and 10 C on the outside of the corkboard. Plot the temperature gradient through the wall. Does the temperature profile suggest any simplifying assumptions that might be made in subsequent analysis of the wall?

Solution:

Thermal Conductivities:

kfir = 0.12 W/m.K (Table A.2, Appendix A)kalu = 237 W/m.K (Table A.1, Appendix A)kld = 35 W/m.K (Table A.1, Appendix A)kcb = 0.04 W/m.K (Table A.2, Appendix A).

Question No. 1: Plot the temperature gradient through the wall.

Answer:

Question No. 2: Does the temperature profile suggest any simplifying assumptions that might be made in subsequent analysis of the wall?

Answer:

Yes, since the thermal conductivity of aluminum and lead are very high than fir and corkboard, they are considered isothermal. Therefore consider only fir and corkboard.

fir + Tcb = 60 C 10 C = 50 K

cbfir LTk

LTkq

=

=

Lfir = 5 cm = 0.05 mLcb = 6 cm = 0.06 m

Then,

1

1. INTRODUCTION

( )( )( )

( )( )( )m

TKmWm

TKmWq cbfir

06.0/04.0

05.0/12.0

=

=

Tcb = 3.6Tfir

Then,

fir + 3.6fir = 50 Kfir = 10.87 K

( )

=

=

mKKmW

LTkq

fir 05.087.10./12.0 = 26.09 W/m2

Considering all walls:

Tfir + Talu + Tld + Tcb = 60 C 10 C = 50 K

cbldalufir LTk

LTk

LTk

LTkq

=

=

=

=

Lfir = 5 cm = 0.05 mLcb = 6 cm = 0.06 mLalu = 1 cm = 0.01 mLld = 1 cm = 0.01 m

=

alu

firfiralu

LkLk

TT

=

ld

firfirld

LkLk

TT

=

cb

firfircb

LkLk

TT

Then

2

1. INTRODUCTION

+

+

+

cbldalu

firfir

Lk

Lk

LkL

kT 1111 = 50 K

+

+

+

06.004.01

01.0351

01.02371

05.012.01firT = 50

Tfir = 10.87 K

( )

=

=

mKKmW

LTkq

fir 05.087.10./12.0 = 26.09 W/m2

There it is equal to simplified solution.

1.2 Verify Equation (1.15).

Solution:

Equation (1.15)

TT

dtdT

bodybody

For verification only

Equation (1.3)

dtdTmc

dtdUQ ==

Equation (1.16)

TTQ bodyThen

TT

dtdTmc body

TT

dtdT

body

Then

TT

dtdT

bodybody where mc is constant.

1.3 q = 5000 W/m2 in a 1 cm slab and T = 140 C on the cold side. Tabulate the temperature drop through the slab if it is made of

Silver Aluminum

3

1. INTRODUCTION

Mild steel (0.5 % carbon) Ice Spruce Insulation (85 % magnesia) Silica aerogel

Indicate which situations would be unreasonable and why.

Solution:

L = 1 cm = 0.01 m

(a) Silver Slab

SiLTkq

= = 5000 W/m2

Thermal conductivity of silver at 140 C, 99.99+ % Pure, Table A.1, Appendix Aksi = 420 W/m.K

( )

=

mTKmWq Si

01.0/420 = 5000 W/m2

TSi = 0.12 K

(b) Alumium Slab

aluLTkq

= = 5000 W/m2

Thermal conductivity of aluminum at 140 C, 99.99+ % Pure, Table A.1, App. AKalu = 237.6 W/m.K

( )

=

mTKmWq alu01.0

/6.237 = 5000 W/m2

Talu = 0.21 K

(c) Mild Steel Slab

msLTkq

= = 5000 W/m2

Thermal conductivity of mild steel at 140 C, Table A.1, Appendix AKms = 50.4 W/m.K

( )

=

mTKmWq ms01.0

/4.50 = 5000 W/m2

Tms = 0.992 K

(d) Ice Slab

4

1. INTRODUCTION

iceLTkq

= = 5000 W/m2

Thermal conductivity of ice at 140 C, Table A.1, Appendix A ice at 0 C, kice = 2.215 W/m.K Note: there is no ice at 140 C, but continue calculation at 0 C.

( )

=

mTKmWq ice01.0

/215.2 = 5000 W/m2

Tice = 22.57 K

(e) Spruce Slab

SiLTkq

= = 5000 W/m2

Thermal conductivity of spruce at 140 C, Table A.1, Appendix AKsp = 0.11 W/m.K @ 20 C (available)

( )

=

mT

KmWq Sp01.0

/11.0 = 5000 W/m2

TSp = 454.55 K

(f) Insulation (85 % Magnesia)

SiLTkq

= = 5000 W/m2

Thermal conductivity of insulation at 140 C, Table A.1, Appendix AKin = 0.074 W/m.K @ 150 C (available)

( )

=

mTKmWq in

01.0/074.0 = 5000 W/m2

TSi = 675.8 K

(g) Silica Aerogel Slab

SiLTkq

= = 5000 W/m2

Thermal conductivity of silica aerogel at 140 C, Table A.1, Appendix Aksa = 0.022 W/m.K @ 120 C

( )

=

mTKmWq sa

01.0/022.0 = 5000 W/m2

Tsa = 2,273 K

Tabulation:

5

1. INTRODUCTION

Slab Temperature Drop, KSilver 0.12

Aluminum 0.21Mild Steel (0.5 % Carbon) 0.992

Ice 22.57Spruce 454.55

Insulation (85 % Magnesia) 675.8Silica Aerogel 2273

The situation which is unreasonable here is the use of ice as slab at 140 C, since ice will melt at temperature of 0 C and above. Thats it.

1.4 Explain in words why the heat diffusion equation, eq. no. (1.13), shows that in transient conduction the temperature depends on the thermal diffusitivity, , but we can solve steady conduction problems using just k (as in Example 1.1).

Solution:Equation (1.13) ( )

xdtdTcAx

dtTTd

cAdt

dUQ refnet =

==

Answer: The application of heat diffusion equation eq. no. (1.13) depends on the

thermal diffusivity as the value of tT

is not equal to zero as it I s under unsteady

state conduction. While in steady conduction depends only on k because the value of

tT

= 0 for steady state conduction giving 22

xT

= 0 , so dxdTkq = .

1.5 A 1-m rod of pure copper 1 cm2 in cross section connects a 200 C thermal reservoir with a 0 C thermal reservoir. The system has already reached steady state. What are the rates of change of entropy of (a) the first reservoir, (b) the second reservoir, (c) the rod, and (d) the whole universe, as a result of the process? Explain whether or not your answer satisfies the Second Law of Thermodynamics.

Solution:

Equation (1.9)

LTkq =

Thermal conductivity of copper at 100 C, Table A.1, Appendix A

6

1. INTRODUCTION

k = 391 W/m.KL = 1 mT = 200 C 0 C = 200 K

( )

=

mKKmWq

1200/391 = 78,200 W/m2.K

Q = qAA = 1 cm2 = 1 x 10-4 m2Q = (78,200 W/m2.K)(1 x 10-4 m2) = 7.82 W

(a) ( )KW

TQS rev

27320082.7

11 +

=

= = - 0.01654 W/K

(b) ( )KW

TQS rev

273082.7

22 +

+==

= + 0.02864 W/K

(c) =rS = 0.0 W/K (see Eq. 1.5, steady state)

(d) =+= 21 SSSUn = - 0.01654 W/K + 0.02864 W/K = + 0.0121 W/K

Since 0UnS , therefore it satisfied Second Law of Thermodynamics.1.6 Two thermal energy reservoirs at temperatures of 27 C and 43 C, respectively,

are separated by a slab of material 10 cm thick and 930 cm2 in cross-sectional area. The slab has a thermal conductivity of 0.14 W/m.K. The system is operating at steady-state conditions. What are the rates of change of entropy of (a) the higher temperature reservoir, (b) the lower temperature reservoir, (c) the slab, and (d) the whole universe as a result of this process? (e) Does your answer satisfy the Second Law of Thermodynamics?

Solution:

Equation (1.9)

LTkq =

Thermal conductivity , k = 0.14 W/m.K

A = 930 cm2 = 0.093 m2L = 10 cm = 0.10 mT = 27 C (- 43 C) = 70 KT1 = 27 + 273 = 300 KT2 = -43 + 273 = 230 K

( )

=

mKKmWq

10.070./14.0 = 98 W/m2

7

1. INTRODUCTION

Q = qA = (98 W/m2)(0.093 m2) = 9.114 W

(a) ( )KW

TQS rev

300114.9

11

=

= = - 0.03038 W/K

(b) ( )KW

TQS rev

230114.9

22

+==

= + 0.03963 W/K

(c) =rS = 0.0 W/K (see Eq. 1.5, steady state)

(d) =+= 21 SSSUn = - 0.03038 W/K + 0.03963 W/K = + 0.00925 W/K

Since 0UnS , therefore it satisfied Second Law of Thermodynamics.1.7 (a) If the thermal energy reservoirs in Problem 1.6 are suddenly replaced with

adiabatic walls, determine the final equilibrium temperature of the slab. (b) What is the entropy change for the slab for this process? (c) Does your answer satisfy the Second Law of Thermodynamics in this instance? Explain. The density of the slab is 26 lb/ft3 and the specific heat 0.65 Btu/lb-F.

Solution:

( )

= 3

33

/1/018.16/26

ftlbmkgftlb = 416.468 kg/m3

( )

=

FlbBtuKkgJFlbBtuc

./1./8.4186./65.0 = 2721.42 J/kg.K

k = 0.14 W/m.KT = 27 C (-43 C) = 70 CT1 = 27 C + 273 = 300 KT2 = - 43 C + 273 = 230 KA = 0.093 m2L = 0.10 m

(a) = 21TT TdQ

TQ

= 21TT TcVdT

TQ

( )

=

1

212 lnTTcV

TTTcV

( )

=

1

212 lnTT

TTT

8

1. INTRODUCTION

( ) ( )

=

=

300230ln

300230

ln1

2

12

TTTTT

= 263.45 K

(b) ( ) ( )

TTTcAL

TTTcV

TQS 1212 ===

( )( )( )( )( )45.263

30023010.0093.042.2721468.416 =S = - 2801 J/K

(c) This will not satisfy the Second Law of Thermodynamic since this is not a rate of entropy of production of the universe.

1.8 A copper sphere 2.5 cm in diameter has a uniform temperature of 40 C. The sphere is suspended in a slow-moving air stream at 0 C. The air stream produces a convection heat transfer coefficient of 15 W/m2.K. Radiation can be neglected. Since copper is highly conductive, temperature gradients in the sphere will smooth out rapidly, and its temperature can be taken as uniform throughout the cooling process (i.e., Bi

1. INTRODUCTION

( ) ( )

+

= TTAh

cVtTT ilnln

xi Tt

AhcVt

TTTT

=

=

ln

=

AhcVTx

xTt

i

eTTTT

=

T = 0 C + 273 = 273 K

iT = 40 C + 273 = 313 K

=

AhcVTx

3

34 rV pi=

r = (1/2)(2.5 cm) = 1.25 cm = 0.0125 m24 rA pi=

( ) hcr

rh

rc

AhcVTx 34

34

2

3

pi

pi=

==

h = 15 W/m2.K

Properties of copper, Table A.1, App. A = 8954 kg/m3cp = 384 J/kg.K = 11.57 x 10-5 m2/s2

( )( )( )( )KmW

mKkgJmkgTx ./1530125.0./3843/8954

= = 955 sec

Then:

( ) xTt

i eTTTT

=

( )

+= TeTTT xT

t

i

( ) KeT t 273273313 955 += KeT

t

27340 955 +=

95540t

eT

= C

10

1. INTRODUCTION

where t in seconds

Tabulation:Time, t, seconds Temperature, T, C

0 4010 39.620 39.240 38.460 37.680 36.9100 36.2200 32.7300 29.6400 26.8600 22800 181000 14.75000 0.310000 0.0100000 0.01000000 0.0

0.0

Plot:

11

1. INTRODUCTION

1.9 Determine the total heat transfer in Problem 1.8 as the sphere cools from 40 C to 0 C. Plot the net entropy increase resulting from the cooling process above, S vs T(K).

Solution:

T = 0 C + 273 = 273 K

24 rA pi= , 334 rV pi=

r = 0.0125 m = 8954 kg/m3cp = 384 J/kg.K = 11.57 x 10-5 m2/s2

T = 40 C 0 C = 40 K

Total Heat Transfer:Q = cVT = (8954 kg/m3)(384 J/kg.K)(4/3)(pi)(0.0125 m)3(40 K)Q = 1125 J - - - - Answer.

Plotting the net-entropy increase:

Equation (1.24)

12

1. INTRODUCTION

b

T

T b

dTTT

cVSb

b =

0

11

( )( ) ( ) bT

T b

dTTT

Sb

b =

0

110125.0343848954 3pi

=

00 lnln13.28 bbbb TT

TTTTS

= 0

0 ln13.28b

bbb

TT

TTTS

Tb0 = 40 C = 313 K

=313

ln273

31313.28 bb TTS

Tb, C Tb, K S40 313 035 308 0.062230 303 0.11725 298 0.164220 293 0.203415 288 0.234410 283 0.25695 278 0.27070 273 0.2754

Plot:

13

1. INTRODUCTION

1.10 A truncated cone 30 cm high is constructed of Portland cement. The diameter at the top is 15 cm and at the bottom is 7.5 cm. The lower surface is maintained at 6 C and the top at 40 C. The outer surface is insulated. Assume one dimensional heat transfer and calculate the rate of heat transfer in watts from top to bottom. To do this, note that the heat transfer, Q, must be the same at every cross section. Write Fouriers law locally, and integrate between this unknown Q and the known end temperatures.

Solution:

T1 = 40 C

14

1. INTRODUCTION

T2 = 6 C

dxdTkAQ =

xDD

LDD

=

121

D1 = 15 cm = 0.15 mD2 = 7.5 cm = 0.075 mL = 30 cm = 0.30 m

xDm

mmm

=

15.030.0

075.015.0

D = 0.15 m 0.25x

2

4DA pi=

dxdTDkQ

=2

4pi

( )dxdTxmkQ 225.015.0

4

=

pi

( ) dTkdxxQ

=

425.015.0 2 pi

( ) dTkdxxQm

= 425.015.03.00 2 pi

Thermal Conductivity of Portland Cement, Table A.2, Appendix A.k = 0.70 W/m.K

( ) ( )[ ] ( ) ( )6404

70.025.015.025.0

113.0

01

=

pixQ

( ) ( )( ) ( )[ ] ( ) ( )344

70.015.03.025.015.04 11

=

piQ

( ) ( ) ( )344

70.015.01

075.014

=

piQ

Q = -0.70 W Ans.

1.11. A hot water heater contains 100 kg of water at 75 C in a 20 C room. Its surface area is 1.3 m2. Select an insulating material, and specify its thickness, to keep the water from cooling more than 3 C / h . (Notice that this problem will be greatly simplified if the temperature drop in the steel casing and the temperature drop in the convective boundary layers are negligible. Can you make such assumptions? Explain.)

15

1. INTRODUCTION

Solution:

Specific heat of water at 75 C, Table A.1 , cp = 4194 J/kg.KQ = (100 kg)(4194 J/kg.K)(3 K/hr)(1 hr / 3600 s)Q = 349.5 W

A = 1.3 m2

Then:

LTkAQ =

( )( )20753.15.349

==

LkQ

Lk

= 4.89 W/m2.K

Select Magnesia, 85 % (insulation), Table A.2k = 0.071 W/m.KL = (0.071 W/m.K) / (4.89 W/m2.K) = 0.01452 m = 1.5 cm

Yes, we can make an assumption of neglecting temperature drops as above as the thermal conductivity of steel is much higher than insulation, also negligible temperature drops for thin film boundary.

1.12. What is the temperature at the left-hand wall shown in Fig. 1.17. Both walls are thin, very large in extent, highly conducting, and thermally black.

Fig. 1.17

Solution:Left: ( )LLL TThq = = 50 (100 TL)Right: ( )rrr TThq = = 20 (Tr 20)

16

1. INTRODUCTION

Equating:q = 50 (100 TL) = 20 (Tr 20)5 (100 TL) = 2 (Tr 20)100 TL = 0.4Tr 8TL = 108 - 0.4Tr o C

Then; by radiation.( )44 rL TTq = = 5.67040 x 10-8 W/m2.K4

( ) ( ) ( )[ ] ( )20202732734.01081067040.5 448 =++= rrr TTTq( ) ( ) ( )[ ] ( )20202734.03811067040.5 448 =+= rrr TTTq

By trial and error:Tr = 42 C (right hand wall)ThenTL = 108 0.4(42) = 91.2 C (left hand wall)

1.13. Develop S.I. to English conversion factors for: The thermal diffusivity, The heat flux, q The density, The Stefan-Boltzmann constant, The view factor, F1-2 The molar entropy The specific heat per unit mass, c

In each case, begin with basic dimension J, m, kg, s, C, and check your answer against Appendix B if possible.

Solution:

(1.) The thermal diffusivity,

Unit of is m2/s.The conversion factor for English units is:

( ) hs

mft 3600

3048.011 2

2

=

smhrft//750,381

2

2

= , checked with Table B.2, o.k.

(2.) The heat flux, q

Unit of q is @/m2 or J/s.m2

17

1. INTRODUCTION

The conversion factor for English units is:( )

2

23048.036000009478.01ft

mh

sJ

Btu=

2

2

2

2

//317.0

//317.01

mWfthBtu

msJfthBtu

=

= , checked with Table B.2, o.k.

(3.) The density

Unit of density is kg/m3The conversion factor for English units is:

( )3

33048.045359.0

11ft

mkg

lb=

3

3

//06243.01

mkgftlb

= , checked with Table B.2, o.k.

(4.) The Stefan-Boltzmann constant,

= 5.6704 x 10-8 W/m2.K4 = 5.6704 x 10-8 J/m2.s.K4The conversion factor for English units is:

( )( ) 4

4

2

2

8.136003048.00009478.01

FK

hs

ftm

JBtu

=

42

42

./../0302.01

KmWKfthrBtu

=

(5.) The view factor F1-2

The view factor is dimensionless, so there is no need for conversion factor.

(6.) The molar entropy

Unit of molar entropy, S = J/KThe conversion factor for English units is.

FK

JBtu

8.10009478.01 =

KJFBtu

//0005266.01 =

(7.) The specific heat per unit mass, c

Unit of c is J/kg.KThe conversion factor for English units is:

FK

lbkg

JBtu

8.145359.00009478.01 =

18

1. INTRODUCTION

KkgJFlbBtu

=

//00023884.01

1.14. Three infinite, parallel, black, opaque plates, transfer heat by radiation,as shown in Fig. 1.18. Find T2.

Fig. 1.18

Solution:( ) ( )43424241 TTTTq == T1 = 100 C + 273 = 373 KT3 = 0 C + 273 = 273 K

( ) ( )[ ]4442 27337321 +=TT2 = 334.1 K = 61.1 C

1.15. Four infinite, parallel black, opaque plates transfer heat by radiation, as shown in Fig. 1.19. Find T2 and T3.

Fig. 1.19

19

1. INTRODUCTION

Solution:

( ) ( ) ( )444343424241 TTTTTTq === T1 = 100 C + 273 = 373 KT4 = 0 C + 273 = 273 K

Then:4

34

14

22 TTT +=4

44

24

32 TTT +=4

44

34

2 2 TTT =

and( ) 4341444322 TTTT +=4

34

14

44

3 24 TTTT +=4

44

14

3 23 TTT +=

=4

33T (373)4 + 2 (273)4

=3T 317.45 K = 44.45 C

44

43

42 2 TTT = = 2 (317.45)4 (273)4

=2T 348.53 K = 75.53 C

1.16. Two large, black, horizontal plates are spaced a distance L from one another. The top is warm at a controllable temperature, Th, and the bottom one is cool at a

20

1. INTRODUCTION

specified temperature, Tc. A gas separates them. The gas is stationary because it is warm on top and cold on the bottom. Write the equation qrad/qcond = fn (N,

c

hT

T ), where N is dimensionless group containing , k , L, and Tc. Plot as a

function of for qrad/qcond = 1, 0.8, and 1.2 (and for other values if you wish).

Now suppose that you have a system in which L = 10 cm, Tc = 100 K, and the gas is hydrogen with an average k of 0.1 W/m.K. Further suppose that you wish to operate in such a way that the conduction and radiation heat fluxes are identical. Identify the operating point on your curve and report the value of Th that you must maintain.

Solution:

( )44 chrad TTq = ( )

LTTkq chcond

=

( )( ) ( ) ( )22

44

chchch

ch

cond

rad TTTTkL

TTTT

kL

qq

++=

=

+

+= 112

3

c

h

c

hc

cond

rad

TT

TTT

kL

qq

( )[ ] ( )[ ]1111 223 ++=++= NTkL

qq

ccond

rad

where

kLTN c

3

=

c

h

TT

=

N as a function of ;

( ) ( )11 2 ++= condradqqN

(1) 1=cond

rad

qq

( )( )111

2 ++=N

21

1. INTRODUCTION

(2) 8.0=cond

rad

qq

( )( )118.0

2 ++=N

(3) 2.1=cond

rad

qq

( )( )112.1

2 ++=N

plot of N as a function of :

For the system:L = 10 cm = 0.10 mTc = 100 Kk = 0.1 W/m.K

For qrad / qcond = 1.0

Then( )( )111 2 ++= N

Solving for N:

kLTN c

3

=

22

1. INTRODUCTION

= 5.67040 x 10-8 W/m2.K4

( )( )( )10.0

10010.0106704.5 38=N = 0.056704

Then( )( )( )11056704.01 2 ++=

( )( ) 64.1711 2 =++By trial and error:

= 2.145

Then: Th = Tc = (2.145)(100 K) = 214.5 K

1.17. A blackened copper sphere 2 cm in diameter and uniformly at 200 C is introduced into an evacuated black chamber that us maintained at 20 C.

Write a differential equation that expresses T(t) for the sphere, assuming lumped thermal capacity.

Identify a dimensionless group, analogous to the Biot number, that can be used to tell whether or not the lumped-capacity solution is valid.

Show that the lumped-capacity solution is valid. Integrate your differential equation and plot the temperature response for the

sphere.

Solution:

(1) Assuming lumped thermal capacity

dtdUQ =

( ) ( )[ ]refTTcVdrdTTA = 44( )tTT =

( ) ( )44

=

TTcV

Adt

TTd

24 rA pi=3

34 rV pi=

( ) ( ) ( ) ( )44443

2 3

344

=

=

TTcr

TTrc

rdt

TTd

pi

pi

Differential Equation, ( )tTT =( ) ( )443

=

TTcrdt

TTd

(2) Dimensionless group analogous to the Biot number

23

1. INTRODUCTION

bkLhBi =

Equivalent h ,( )

=

TTTTh

44

Biot number equivalent = ( )

( )

==

TTkTTr

krh

AkVh

bbb 33

44

(3) Showing that lumped-capacity solution is valid.

Dimensionless group must be

1. INTRODUCTION

( )( )( ) ( )

( )( )222222222

222244 211

axaxaaxax

axaxax ++

=

+=

( ) ( )22222244 21

211

axaaxaax +

=

( ) ( )( ) ( )222244 2

1)(22

11axaaxaxa

axaxaax +

+

+=

( ) ( ) ( )222244 21

21

21

211

axaaxaaxaaax +

+

=

( )222344 2111

411

axaaxaxaax +

+

=

+

+

=

12

111411

24

344

axa

axaxaax

+

+

=

12

111

411

23

344

axa

aaxaxaax

Then,

44 ax dx = CaxArcaax axa + + tan21ln41 33Applying:

+

+

=

TTArcTTArcTTT TTTT TTTTT dT iiiTTi tantan21lnln41 3344Substitute values:

( ) ( )

+

+

=

293473tan293tan2934 2293473 293473ln293293ln2934 1 3344 ArcTArcTTTT dTTTi( ) cr

tTArcTT

TTdT

T

Ti

348062.3293

tan2293293ln

29341

344

=

+

+

=

( )

( )( )( )( )01.03848954

106704.5348062.3293

tan2293293ln

29341 8

3

tTArcTT

=

+

+

( ) tTArc

TT 0004978.048062.3

293tan2

293293ln

29341

3 =

+

+

Tabulation:

25

1. INTRODUCTION

T, C T, K T, s200 473 0182 455 93.5164 437 206.8146 419 346.1128 401 520.9110 383 745.892 365 104674 347 1468.556 329 2119.838 311 3340.6

Plot :

26

1. INTRODUCTION

1.18. As part of space experiment, a small instrumentation packaged is released from a space vehicle. It can be approximated as a solid aluminum sphere, 4 cm in diameter. The sphere is initially at 30 C and it contains a pressurized hydrogen component that will condense and malfunction at 30 K. If we take the surrounding space to be at 0 K, how long may we expect the implementation package to function properly? Is it legitimate to use the lumped-capacity method in solving the problem? (Hint: See the directions for Problem 1.17).

Solution:

Properties of aluminum, Table A.1 = 2707 kg/m3cp = 905 J/kg.Kkb = 237.2 W/m.K @ 30 C

From Prob. 1.17, usingiT = 30 C + 273 = 303 K

T = 0 KT = 30 K = 5.6704 x 10-8 W/m2.Kr = (1/2)(4 cm) = 2 cm = 0.02 m

Check for the legitimacy of lumped-capacity method.

( )( )

TTkTTr

b3

44=

( )( )( )( )( )03032.2373

030302.0106704.5 448

= 0.000044

1. INTRODUCTION

1.19. Consider heat calculation through the wall as shown in Fig. 1.20. Calculate q and the temperature of the right-hand side of the wall.

Fig. 1.20

Solution:( ) ( )

=

= TThL

TTkq 221

1T = 200 C

T = 0 Ck = 2 W/m2.KL = 0.5 mh = 3 W/m2.K

( )( ) ( )( )0350.0

20022

2=

= TTq

22 34800 TT =T2 = 114.286 C

q = (3)(114.286 0) = 343 W/m2.

1.20. Throughout Chapter 1 we have assumed that the steady temperature distribution in a plane uniform wall is linear. To prove this, simplify the heat diffusion equation to the form appropriate for steady flow. Then integrate it twice and eliminate the two constant using the known outside temperatures Tleft and Tright at x = 0 and x = wall thickness, L.

28

1. INTRODUCTION

Solution:

Eq. 1.14, one dimensional heat diffusion equation,

tT

tkTc

xT

=

12

2

Use 22

xT

= 0 for steady flow.

1CdxdT

=

21 CxCT +=at T = Tleft, x = 0Tleft = 0 + C2C2 = Tleft

At T = Tright, x = LTright = C1L + Tleft

LTT

C leftright

=1

Then, ( )left

leftright TxL

TTT +

=

LTT

xTT leftrightleft

=

, therefore linear.

1.21 The thermal conductivity in a particular plane wall depends as follows on the wall temperature: k = A + BT, where A and B are constants. The temperatures are T1 and T2 on either side of the wall, and its thickness is L. Develop an expression for q.

Solution:

dxdTkq =

( )dxdTBTAq +=

( )dTBTAqdx +=

29

1. INTRODUCTION

( ) +=L TT dTBTAdxq 0 212

1

2

21

T

T

BTATqL

+=

( ) ( )

+=

21

2212 2

1 TTBTTAqL

( ) ( )L

TTBTTAq

+

=

21

2212 2

1

1.22 Find k for the wall shown in Fig. 1.21. Of what might it be made?

Figure 1.21.

Solution:L = 0.08 m( ) ( )leftrightleft TThL

TTkq =

=

( ) ( )( )2010020008.0

020=

k

k = 64 W/m.K

30

1. INTRODUCTION

From Table A.1, @ 10 C, k = 64 W/m.KThis might be Steel, AISI 1010, k = 64.6 W/m.K

1.23 What are Ti, Tj, and Tr in the wall shown in Fig. 1.22?

Fig. 1.22.

Solution:L1 = 2 cm = 0.02 mk1 = 2 W/m.CL2 = 6 cm = 0.06 mk2 = 1 W/m.CL3 = 4 cm = 0.04 mk3 = 5 W/m.CL4 = 4 cm = 0.04 mk4 = 4 W/m.C

( ) ( ) ( ) ( )4

4

3

3

2

2

1

12525100

LTTk

LTk

LTk

LTkq rjjii

=

=

=

=

( ) ( )2

2

1

1 25100L

TkL

Tk ii =

( ) ( ) ( )( )06.0

25102.0

1002 =

ii TT

256600 = ii TT

iT =89.29 C( ) ( )

3

3

1

125100

LTk

LTk ji

=

( )( ) ( )( )04.0

25502.0

29.891002 jT=

jT = 16.43 C

31

1. INTRODUCTION

( ) ( )4

4

1

1 100L

TTkL

Tk rji =

( )( ) ( )( )04.043.164

02.029.891002 rT

=

( )( ) ( )( )04.043.164

02.029.891002 rT

=

rT = 5.72 C

1.24 An aluminum can of beer or soda pop is removed from the refrigerator and set on the table. If h is 13.5 W/m2.K, estimate when the beverage will be at 15 C. Ignore thermal radiation. State all of your other assumptions.

Solution:

Properties of aluminum, Table A.1, App. A = 2707 kg/m3cp = 905 J/kg.Kk = 237 W/m.K = 9.61 x 10-5 m2/s

Assume size of can is 50 mm diameter x 100 mm heightiT = 0 C, and room at T = 20 C

Time constant,

+

==

DLDh

LDc

AhcVT

pipi

pi

2

42

2

( )LDhcDLT

22 +=

D = 0.05 m L = 0.10 mh = 13.5 W/m2.K

( )( )( )( )( ) ( )( ) =+= 10.025.05.132

10.005.09052707T 648.1 ns

Eq. 1.22.T

t

i

eTTTT

=

at T = 15 C

32

1. INTRODUCTION

1.648

2002015 te=

t = 898.5 s = 15 minutes

1.25. One large, black wall at 27 C faces another whose surface is 127 C. The gap between the two walls is evacuated. If the second wall is 0.1 m thick and has a thermal conductivity of 17.5 W.m.K, what is its temperature on the back side? (Assume steady state).

Solution:

T3 = temperature on the back side.

( ) ( )L

TTkTTq 23414

2

==

L = 0.1 mT1 = 27 C + 273 = 300 KT2 = 127 C + 273 = 400 K = 5.6704 x 10-8 W/m2.Kk = 17.5 W/m.K

( )( ) ( ) ( )10.0

4005.17300400106704.5 3448 == Tq

T3 = 405.67 K = 132.67 C.

1.26. A 1-cm diameter, 1 % carbon steel sphere, initially at 200 C, is cooled by natural convection, with air at 20 C. In this case, h is not independent of temperature.

33

1. INTRODUCTION

Instead, h =3.51(t C)1/4 W/m2.K. Plot Tsphere as a function of t. Verify the lumped-capacity assumption.

Solution:

Properties of 1% carbon steel, Table A.1 = 7801 kg/m3cp = 473 J/kg.Kk = 42 W/m.K = 1.17 x 10-5 m2/s

Verify the lumped-capacity assumption:

kLhBi

t = 200 C 20 C = 180 Ch =3.51(180)1/4 W/m2.K = 12.86 W/m2.K

3434

2

3

rr

r

AVL ===

pi

pi

r = (1/2)(1 cm) = 0.005 mL = 0.005 m / 3 = 0.001667 m

( )( )42

001667.086.12Bi = 0.00051

1. INTRODUCTION

( ) ( )cV

AtTTT

Ti 51.34 4

1

=

( )cV

AtTTT

Ti 51.3

11 1

45

45

=

++

( ) ( ) cVAt

TTTT i 8775.011

4/14/1 =

rVA 3

=

Then,

( ) ( ) ( ) crtt

rcTTTT i 6325.238775.011

4/14/1 =

=

Substitute value,

( ) ( ) ( )( )( )005.047378016325.2

202001

201

4/14/1

tT

=

( ) 273012.0000143.0201

4/1 +=

tT

( )273012.0000143.0

120 4/1+

=

tT

( )1909

699320 4/1+

=

tT

Ct

T 201909

69934

+

+=

Tabulation:t,s T, C0 200

100 166.8200 140.9300 120.4400 104.7500 91600 80.4800 64.41000 53.41200 45.61400 401600 35.81800 32.62000 30.2

Plot:

35

1. INTRODUCTION

1.27. A 3-cm diameter, black spherical heater is kept at 1100 C. It radiates through an evacuated space to a surrounding spherical shell of Nichrome V. The shell has a 9 cm inside diameter and is 0.3 cm thick. It is black on the inside and is held at 25 C on the outside. Find (a) the temperature of the inner wall of the shell and (b) the heat transfer, Q. (Treat the shell as a plane wall.)

Solution:

Properties of Nichrome V, Table A.1, Appendix A. = 8,410 kg/m3cp = 466 J/kg.Kk = 10 W/m.K = 0.26 x 10-5 m2/s

Radiation ( )42411 TTAQrad =T1 = 1100 C = 1373 KT3 = 25 C + 273 = 298 K = 5.6704 x 10-8 W/m2.C

Conduction( )L

TTkAQcond 232

=

211 4 rA pi= , r1 = (1/2)(3 cm) = 1.5 cm = 0.015 m

( ) 21 015.04pi=A m2

36

1. INTRODUCTION

222 4 rA pi= , r2 = (1/2)(9 cm) = 4.5 cm = 0.045 m

( ) 21 045.04pi=A m2L = 0.3 cm = 0.003 m

Thencondrad QQ =

( ) ( )L

TTkATTA 322424

11

=

( )( )( ) ( )[ ] ( )( )( ) ( )003.0

298045.04101373015.04106704.5 22

42

428 =

TT pipi

( ) ( )29810290632.51373 211424 = TTBy trial and error method.

(a) Inner Wall Temperature = T2 = 304.7 K = 31.7 C(b) Heat Transfer, Q

( ) ( ) ( ) ( ) ( ) ( )[ ] WTTAQ 4.5687.3041373015.04106704.5 442842411 === pi1.28. The sun radiates 650 W/m2 on the surface of a particular lake. At what rate (in

mm/hr) would the lake evaporate if all of this energy went to evaporating water? Discuss as many other ways you can think of that this energy can be distributed (hfg for water is 2,257,000 J/kg). Do you suppose much of the 650 W/m2 goes to evaporation?

Solution:

q = 650 W/m2 = 2,340,000 J/hr.m2

Evaporation rate = kgJmhrJ

/000,257,2./000,340,2 2

= 1.036774 kg/hr.m2

Density of water = 1000 kg/m3

Evaporation rate =

mmm

mkgmhrkg

11000

/1000./036774.1

3

2

=1.036774 mm/hr

There are other ways that this energy can be distributed such as cloud barrier, heating up of the lake up to evaporation, haze or atmosphere.

Yes, much of the 650 W/m2 goes to evaporation especially on a clear day.

37

1. INTRODUCTION

1.29. It is proposed to make picnic cups, 0.005 m thick, of a new plastic for which k = ko(1 + aT2), where T is expressed in C, ko = 15 W/m.K, and a = 10-4 C-2. We are concerned with thermal behavior in the extreme case in which T = 100 C in the cup and 0 C outside. Plot T against position in the cup wall and find the heat loss, q.

Solution:

dxdTkq =

( )dTaTkqdx o 21+=( )+= 2

1

21T

T

o dTaTkxq

[ ] 21

331 T

To aTTkxq +=

( ) ( )[ ]3131132312 aTTaTTkxq o ++=( ) ( )3131132312 aTTaTTk xq o ++=

( ) ( )okxqaTTaTT +=+ 31311

323

12

Solving for q if,T1 = 100 CT2 = 0 Cx = 0.005 m

( ) ( )3231231311 aTTaTTk xq o ++=

( ) ( ) ( )( )[ ] ( ) ( )( )[ ]34313431 01001001010015.0 005.0 ++=qq = 4000 W

Plotting:Use T1 = 100 C, a = 10-4 C-2, q = 4000 W, ko = 0.15 W/m.K( ) ( )

okxqaTTaTT +=+ 31311

323

12

( ) ( )[ ]q

aTTaTTkx o3

231

23

131

1 ++=

( ) ( ) ( )( ) ( )[ ]4000

1001010015.0 3231234

31 aTTx ++=

( )[ ]4000

1015.020 324

31

2 TTx+

=

38

1. INTRODUCTION

Tabulation:T2, C x, m100 080 0.0013660 0.0024840 0.0034220 0.004240 0.00500

Heat loss , q = 4000 W

1.30. A disc-shaped wafer of diamond 1 lb is the target of a very high intensity laser. The disc is 5 mm in diameter and 1 mm deep. The flat side is pulsed intermittently with 1010 W/m2 of energy for one microsecond. It is then cooled by natural convection from that same side until the next pulse. If h = 10 W/m2.K and T=30 C, plot Tdisc as a function of time for pulses that are 50 s apart and 100 s apart. (Note that you must determine the temperature the disc reaches before it is pulsed each time.)

Solution:

Properties of Diamond, Table A.2 = 3250 kg/m3cp = 510 J/kg.Kkb = 1350 W/m.K = 8.1 x 10-4 m2/s

39

1. INTRODUCTION

L = 1 mm = 0.001 m

bkLhBi =

h = 10 W/m2.K

T =30 C

( )( )1350

001.010=Bi = 0.0000074

1. INTRODUCTION

t = 50 s75.165

50

3033.9030

=

eT

T = 74.62 C

Second 50 s, Ti = 60.33 C + 74.62 C = 134.95 Ct = 25 s

75.16525

3095.13430

=

eT

T = 120.26 C

t = 50 s75.165

50

3095.13430

=

eT

T = 107.62 C

Third 50 s, Ti = 60.33 C + 107.62 C = 167.95 Ct = 25 s

75.16525

3095.16730

=

eT

T = 148.64 C

t = 50 s75.165

50

3095.16730

=

eT

T = 132.03 CAnd so on. Tabulation:

t, s Tdisc, C1st 50 s 0 90.33

25 81.8850 74.62

2nd 50 s 50 134.9575 120.26100 107.62

3rd 50 s 100 167.95125 148.64150 132.03

41

1. INTRODUCTION

Plot:

For 100 s pulse apart

First 100 s, Ti = 90.33 Ct = 50 s

75.16550

3033.9030

=

eT

T = 74.62 C

t = 100 s

42

1. INTRODUCTION

75.165100

3033.9030

=

eT

T = 63.00 C

Second 100 s, Ti = 60.33 C + 63.00 C = 123.33 Ct = 50 s

75.16550

3033.12330

=

eT

T = 99.03 C

t = 100 s75.165

100

3033.12330

=

eT

T = 81.05 C

Third 100 s, Ti = 60.33 C + 81.05 C = 141.38 Ct = 50 s

75.16550

3038.14130

=

eT

T = 148.64 C

t = 100 s75.165

100

3038.14130

=

eT

T = 112.38 C

And so on. Tabulation:

t, s Tdisc, C1st 100 s 0 90.3.3

50 74.62100 63.00

2nd 100 s 100 123.33150 99.03200 81.05

3rd 100 s 200 141.38250 112.38300 90.93

43

1. INTRODUCTION

Plot:

1.31 A 150 W light bulb is roughly a 0.006 m diameter sphere. Its steady surface temperature in room air is 90 C, and h on the outside is 7 W/m2.K. What fraction of the heat transfer from the bulb is by radiation directly from the filament through the glass? (Stae any additional assumptions.)

Solution:

Assume black body radiation.

( ) ( )44 abab TTATTAhQ += ( ) ( )44 abab TTTThAQ +=

44

1. INTRODUCTION

bT = 90 + 273 = 363 Kh = 7 W/m2.K. = 5.6704 x 10-8 W/m2.K4.Q = 150 W but change to 0.150 W as light bulb is very small. It may be a typographical error.

22 4 rDA pipi ==D = 0.006 mThen:

( ) ( )( ) ( )( )4482 363106704.53637006.0150.0 aa TT += pi( ) ( ) ( ) ( )448 363106704.536371326 aa TT +=

By Trial and Error Method:Ta = 270.5 K = -2.5 C

Fraction = ( )

( ) ( )4444

abab

ab

TTTThTT

AQ

+

=

=

( )( )( )( ) ( )( )448

448

5.270363106704.55.27036375.270363106704.5

+

Fraction = 0.5126

1.32 How much entropy does the light bulb in Problem 1.31 produce?

Solution:

( )

=

=

3631

5.270115.011

baUn TT

QS = 0.0001413 W/K

1.33 Air at 20 C flows over one side of a thin metal sheet ( h = 10.6 W/m2.K). Methanol at 87 C flows over the other side ( h = 141 W/m2.K). The metal functions as an electrical resistance heater, releasing 1000 W/m2. Calculate (a) the heater temperature, (b) the heat transfer from the methanol to the heater, and (c) the heat transfer from the heat of the air.

Solution:

(a) q = 1000 W/m2

( ) ( )CThCThq hh 8720 21 +=( )( ) ( )( )87141206.10 + hh TT = 1000Th = 88.9 C

45

1. INTRODUCTION

(b) ( ) ( )( )879.88141872 == hm Thq

mq = 267.9 W

(c) ( ) ( )( )209.886.10201 == ha Thq

aq = 730.3 W

1.34 A planar black heater is simultaneously cooled by 20 C air ( h =14.6 W/m2.K) and by radiation to a parallel black wall at 80 C. What is the temperature of the heater if it delivers 9000 W/m2 ?

Solution:

( ) ( ) ( )[ ]44 2738027320 +++= TThq = 9000 W/m2( )( ) ( ) ( ) ( )[ ]448 353273106704.5206.14 ++= TTq = 9000 W/m2

By Trial and error method.

T = 294.3 C

1.35 An 8-oz. can of beer is taken from a 3 C refrigerator and placed in a 25 C room. The 6.3 cm diameter by 9 cm high can is placed on an insulated surface ( h =7.3 W/m2.K). How long will it take to reach 12 C? Ignore thermal radiation and discuss your other assumption.

Solution:

Tt

i

eTTTT

=

Assume aluminum material for the can of beer,Properties of aluminum, Table A.1, Appendix A = 2707 kg/m3cp = 905 J/kg.Kk = 237 W/m.K

Then,

T = 25 CiT = 3 C

T = 12 C

46

1. INTRODUCTION

Time constant:

AhcVT =

LDV 24

=

pi

D = 6.3 cm = 0.063 m

( ) ( )09.0063.04

2

=

piV = 2.8055 x 10-4 m3

( ) ( ) ( ) ( )( )09.0063.02063.04

24

22 pipi

pipi

+

=+

= DLDA = 0.02405 m2

( )( )( )( )( )02405.03.7

108055.29052707 4=T = 3915 s

3915

2532512 te

=

t = 2314 sec = 38.6 min

1.36 A resistance heater in the form a thin sheet runs parallel with 3 cm slabs of cast iron on either side of an evacuated cavity. The heater, which releases 8000 W/m2, and the cast iron are very nearly black. The outside surfaces of the cast iron slabs are kept at 10 C. Determine the heater temperature and the inside slab temperatures.

Solution:

q = 8000 W/m2

47

1. INTRODUCTION

Properties of cast iron, Table A.1, Appendix A = 7272 kg/m3cp = 420 J/kg.Kkb = 52 W/m.K

L = 3 cm = 0.03 m

Inside slab temperature,( )

LTCkq = 10 = 8000 W/m2

( ) ( )03.0

1052 Tq =

T = 14.62 C

Heater temperature,( )44 TTq h = = 8000 W/m2( ) ( ) ( )[ ]448 27362.14273106704.5 ++= hTq = 8000 W/m2

Th = 347.2 C

1.37 A black wall at 1200 C radiated to the left side of a parallel slab of type 316 stainless steel, 5 mm thick. The right side of the slab is to be cooled convectively and is not to exceed 0 C. Suggest a convective proceed that will achieve this.

Solution:

1.38 A cooler keeps one side of a 2 cm layer of ice at 10 C. The other side is exposed to air at 15 C. What is h just on the edge of melting? Must h be raised or lowered if melting is to progress?

48

1. INTRODUCTION

Solution:

Melting point of ice = 0 CThermal Conductivity of ice at 0 C = 2.215 W/m.K

( ) ( )3212 TThLTTkq ==

( ) ( )2312 TThLTTk

=

( ) ( )2312 TThLTTk

=

T1 = -10 CT2 = 0 CT3 = 15 CL = 2 cm = 0.02 mThen,( ) ( )( ) ( )015

02.0100215.2

=

h

h = 73.83 W/m2.KIf the melting is to progress the thickness will reduce and h must be raised.

1.39 At what minimum temperature does a black heater deliver its maximum monochromatic emissive power in the visible range? Compare your result with Fig. 10.2.

Solution:

Figure 1.15 or Wiens Law, Eq. (1.29)( )

max= eT = 2898 m.K

Minimum visible range, = 0.4545 m

Then:(0.4545 m)(Tmin) = 2898 m.K

Tmin = 6376 K

From Fig. 10.2 , T = 5900 K

1.40 The local heat transfer coefficient during the laminar flow of fluid over a flat plate of length L is equal to F / x1/2, where F is a function of fluid properties and the flow velocity. How does h compares with h (x = L). (x is the distance from the leading edge of the plate.)

49

1. INTRODUCTION

1.41 An object is initially at a temperature above that of its surroundings. We have seen that many kinds of convection processes will bring the object into equilibrium with its surroundings. Describe the characteristics of a process that will do so with the least net increase of the entropy of the universe.

Solution:

bbT

boTb

dTTT

cVS

=

11

Determine bT for least net increase of the entropy of the universe.

bTT11

= 0

=TTb

=

bo

bo

TT

TTTcVS ln

=

TT

TTTcVS bobo ln

The characteristic of the process is unsteady state conduction having Biot number increasing from less than one to more than one when reaching equilibrium at =TTb .

1.42 A 250 C cylindrical copper billet, 4 cm in diameter and 8 cm long is cooled in air at 25 C. The heat transfer coefficient is 5 W/m2.K Can this be treated as lump-capacity cooling? What is the temperature of the billet after 10 minutes?

Solution:

Check Biot Number

Properties of copper, Table A.1, App. A = 8954 kg/m3cp = 384 J/kg.Kkb = 398 W/m.K

Time constant:

AhcVT =

LDV 24

=

pi

D = 4 cm = 0.04 m, L = 8 cm = 0.08 m

( ) ( )08.004.04

2

=

piV = 1.0053 x 10-4 m3

50

1. INTRODUCTION

( ) ( ) ( ) ( )( )08.004.0204.04

24

22 pipi

pipi

+

=+

= DLDA = 0.012566 m2

( )( )( )( )( )012566.05

100053.13848954 4=T = 5501 s

Biot Number

mLkLhBib

02.0=

=

( )( )398

02.05=Bi = 0.00025