solved problems in heat transfer

7/22/2019 solved problems in heat transfer

1/51

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 thetemperature 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 assumptionsthat 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 L

Tk

L

Tkq

=

=

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

Then,

1


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1. INTRODUCTION

( )

( )( )( )

( )m

TKmW

m

TKmWq cb

fir

06.0

/04.0

05.0

/12.0 =

=

Tcb = 3.6Tfir

Then,

fir + 3.6fir = 50 K

fir= 10.87 K

( )

=

=

m

KKmW

L

Tkq

fir 05.0

87.10./12.0 = 26.09 W/m2

Considering all walls:

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

cbldalufir L

Tk

L

Tk

L

Tk

L

Tkq

=

=

=

=

Lfir = 5 cm = 0.05 m

Lcb = 6 cm = 0.06 m

Lalu = 1 cm = 0.01 m

Lld = 1 cm = 0.01 m

=

alu

fir

firalu

L

k

L

k

TT

=

ld

fir

firld

L

k

L

k

TT

=

cb

firfircb

L

k

L

k

TT

Then

2


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1. INTRODUCTION

++

+

cbldalu

fir

fir

L

k

L

k

L

kL

kT

1111 = 50 K

+

+

+

06.0

04.01

01.0

351

01.0

2371

05.012.01firT = 50

Tfir= 10.87 K

( )

=

=

m

KKmW

L

Tkq

fir 05.0

87.10./12.0 = 26.09 W/m2

There it is equal to simplified solution.

1.2 Verify Equation (1.15).

Solution:

Equation (1.15)

TTdt

dTbody

body

For verification only

Equation (1.3)

dt

dTmc

dt

dUQ ==

Equation (1.16)

TTQ bodyThen

TTdt

dTmc body

TTdt

dTbody

Then

TTdt

dT 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

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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

SiL

Tkq

= = 5000 W/m2

Thermal conductivity of silver at 140 C, 99.99+ % Pure, Table A.1, Appendix A

ksi = 420 W/m.K

( )

=

m

TKmWq Si

01.0/420 = 5000 W/m2

TSi = 0.12 K

(b) Alumium Slab

aluL

Tkq

= = 5000 W/m2

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

( )

=

m

TKmWq alu

01.0/6.237 = 5000 W/m2

Talu = 0.21 K

(c) Mild Steel Slab

msL

Tkq

= = 5000 W/m2

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

( )

=

m

TKmWq ms

01.0/4.50 = 5000 W/m2

Tms = 0.992 K

(d) Ice Slab

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1. INTRODUCTION

iceL

Tkq

= = 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.

( )

=

m

TKmWq ice

01.0/215.2 = 5000 W/m2

Tice = 22.57 K

(e) Spruce Slab

SiL

Tkq

= = 5000 W/m2

Thermal conductivity of spruce at 140 C, Table A.1, Appendix A

Ksp = 0.11 W/m.K @ 20 C (available)

( )

=

m

TKmWq

Sp

01.0/11.0 = 5000 W/m2

TSp = 454.55 K

(f) Insulation (85 % Magnesia)

SiL

Tkq

= = 5000 W/m2

Thermal conductivity of insulation at 140 C, Table A.1, Appendix A

Kin = 0.074 W/m.K @ 150 C (available)

( )

=

m

TKmWq in

01.0/074.0 = 5000 W/m2

TSi = 675.8 K

(g) Silica Aerogel Slab

SiL

Tkq

= = 5000 W/m2

Thermal conductivity of silica aerogel at 140 C, Table A.1, Appendix A

ksa = 0.022 W/m.K @ 120 C( )

=

m

TKmWq sa

01.0/022.0 = 5000 W/m2

Tsa = 2,273 K

Tabulation:

5


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1. INTRODUCTION

Slab Temperature Drop, K

Silver 0.12

Aluminum 0.21

Mild Steel (0.5 % Carbon) 0.992

Ice 22.57

Spruce 454.55Insulation (85 % Magnesia) 675.8

Silica 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)

xdt

dTcAx

dt

TTdcA

dt

dUQ

ref

net =

==

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

thermal diffusivity as the value oft

T

is not equal to zero as it I s under unsteady

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

t

T

= 0 for steady state conduction giving2

2

x

T

= 0 , sodx

dTkq = .

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 steadystate. 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)

L

Tkq=

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

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1. INTRODUCTION

k= 391 W/m.K

L = 1 m

T= 200 C 0 C = 200 K

( )

=

m

KKmWq

1

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

Q = qA

A = 1 cm2 = 1 x 10-4 m2

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

(a)( )K

W

T

QS rev

273200

82.7

1

1 +

=

= = - 0.01654 W/K

(b)( )K

W

T

QS rev

2730

82.7

2

2 ++

== = + 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 0Un

S , 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 cm 2 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) thehigher 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)

L

Tkq=

Thermal conductivity , k= 0.14 W/m.K

A = 930 cm2 = 0.093 m2

L = 10 cm = 0.10 m

T= 27 C (- 43 C) = 70 K

T1 = 27 + 273 = 300 K

T2 = -43 + 273 = 230 K

( )

=

m

KKmWq

10.0

70./14.0 = 98 W/m2

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1. INTRODUCTION

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

(a) ( )KW

T

QS rev

300

114.9

1

1

=

= = - 0.03038 W/K

(b) ( )KW

T

Q

S

rev

230

114.9

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 0Un

S , 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 satisfythe 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

ftlb

mkgftlb = 416.468 kg/m3

( )

=

FlbBtu

KkgJFlbBtuc

./1

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

k= 0.14 W/m.KT= 27 C (-43 C) = 70 C

T1 = 27 C + 273 = 300 KT2 = - 43 C + 273 = 230 K

A = 0.093 m2

L = 0.10 m

(a) = 21

T

TT

dQ

T

Q

= 21

T

TTcVdT

TQ

( )

=

1

212 lnT

TcV

T

TTcV

( )

=

1

212 lnT

T

T

TT

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1. INTRODUCTION

( ) ( )

=

=

300

230ln

300230

ln

1

2

12

T

T

TTT

= 263.45 K

(b)( ) ( )

T

TTcAL

T

TTcV

T

QS 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 ofentropy 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 willsmooth out rapidly, and its temperature can be taken as uniform throughout the

cooling process (i.e., Bi


10/51

1. INTRODUCTION

( ) ( ) +

= TT

Ah

cV

tTT ilnln

xiT

t

Ah

cV

t

TT

TT=

=

ln

=

Ah

cVTx

xT

t

i

eTT

TT

=

T = 0 C + 273 = 273 K

iT = 40 C + 273 = 313 K

=

Ah

cVTx

3

34 rV =

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

( ) hcr

rh

rc

Ah

cVTx

34

3

4

2

3

=

==

h = 15 W/m2.K

Properties of copper, Table A.1, App. A= 8954 kg/m3

cp = 384 J/kg.K

= 11.57 x 10-5 m2/s2

( )( )( )

( )KmWmKkgJmkg

Tx

./153

0125.0./3843/8954= = 955 sec

Then:

( ) xTt

ieTTTT

=

( )

+= TeTTT xT

t

i

( ) KeTt

273273313 955 +=

KeTt

27340 955 +=

95540t

eT

= C

10


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1. INTRODUCTION

where tin seconds

Tabulation:

Time, t, seconds Temperature, T, C

0 40

10 39.620 39.2

40 38.4

60 37.6

80 36.9

100 36.2

200 32.7

300 29.6

400 26.8

600 22

800 18

1000 14.75000 0.3

10000 0.0

100000 0.0

1000000 0.0 0.0

Plot:

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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 = , 334 rV =

r= 0.0125 m

= 8954 kg/m3

cp = 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)()(0.0125 m)3(40 K)

Q = 1125 J - - - - Answer.

Plotting the net-entropy increase:

Equation (1.24)

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1. INTRODUCTION

b

T

Tb

dTTT

cVS

b

b

= 0

11

( )( ) ( )b

T

T b

dTTT

S

b

b

= 0

110125.0

3

43848954

3

=

0

0 lnln13.28b

b

b

b TTTT

TTS

=

0

0 ln13.28b

bbb

T

T

T

TTS

Tb0 = 40 C = 313 K

=

313ln

273

31313.28

bbTT

S

Tb, C Tb, K S40 313 0

35 308 0.0622

30 303 0.117

25 298 0.1642

20 293 0.2034

15 288 0.2344

10 283 0.2569

5 278 0.2707

0 273 0.2754

Plot:

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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. Todo 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


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1. INTRODUCTION

T2 = 6 C

dx

dTkAQ =

x

DD

L

DD =

121

D1 = 15 cm = 0.15 mD2 = 7.5 cm = 0.075 m

L = 30 cm = 0.30 m

x

Dm

m

mm =

15.030.0

075.015.0

D = 0.15 m 0.25x

2

4DA

=

dx

dTDkQ

=

2

4

( )dx

dTxmkQ

2

25.015.04

=

( ) dTkdxxQ = 4

25.015.02

( ) dTkdxxQm

= 425.015.0

3.0

0

2

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

( ) ( )[ ] ( ) ( )6404

70.025.015.025.0

11

3.0

0

1 =

xQ

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

70.015.03.025.015.0411

=

Q

( ) ( ) ( )344

70.015.0

1

075.0

14

=

Q

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 surfacearea 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 inthe convective boundary layers are negligible. Can you make such assumptions?

Explain.)

15


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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:

L

TkAQ=

( )( )20753.15.349 ==L

kQ

L

k= 4.89 W/m2.K

Select Magnesia, 85 % (insulation), Table A.2k = 0.071 W/m.K

L = (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 forthin film boundary.

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

Fig. 1.17

Solution:

Left: ( )LLLTThq = = 50 (100 TL)

Right: ( )rrr

TThq = = 20 (Tr 20)

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1. INTRODUCTION

Equating:

q = 50 (100 TL) = 20 (Tr 20)

5 (100 TL) = 2 (Tr 20)100 TL = 0.4Tr 8

TL = 108 - 0.4Tro C

Then; by radiation.

( )44rL

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)

Then

TL = 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 dimensionJ, m, kg,s, C, and check your answer againstAppendix B if possible.

Solution:

(1.) The thermal diffusivity,

Unit of is m2/s.

The conversion factor for English units is:

( ) hs

m

ft 3600

3048.0

11

2

2

=

sm

hrft

/

/750,381

2

2

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

(2.) The heat flux, q

Unit ofq is @/m2 or J/s.m2

17


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1. INTRODUCTION


( )2

23048.036000009478.0

1ft

m

h

s

J

Btu=

2

2

2

2

/

/317.0

/

/317.01

mW

fthBtu

msJ

fthBtu =

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

(3.) The density

Unit of densityis kg/m3


( )3

33048.0

45359.0

11

ft

m

kg

lb=

3

3

/

/06243.01

mkg

ftlb= , 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.K4


( )

( ) 44

2

2

8.1

36003048.00009478.01

F

K

h

s

ft

m

J

Btu=

42

42

./

../0302.01

KmW

KfthrBtu=

(5.) The view factorF1-2

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.

F

K

J

Btu

8.1

0009478.01 =

KJ

FBtu

/

/

0005266.01=

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

Unit ofc is J/kg.K


F

K

lb

kg

J

Btu

8.1

45359.00009478.01 =

18


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1. INTRODUCTION

KkgJ

FlbBtu

=

/

/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:

( ) ( )4

3

4

2

4

2

4

1 TTTTq == T1 = 100 C + 273 = 373 K

T3 = 0 C + 273 = 273 K

( ) ( )[ ]4442

2733732

1+=T

T2 = 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


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1. INTRODUCTION

Solution:

( ) ( ) ( )44

4

3

4

3

4

2

4

2

4

1TTTTTTq ===

T1 = 100 C + 273 = 373 KT4 = 0 C + 273 = 273 K

Then:4

3

4

1

4

22 TTT +=

4

4

4

2

4

32 TTT +=4

4

4

3

4

22 TTT =

and

( ) 43

4

1

4

4

4

322 TTTT +=

4

3

4

1

4

4

4

324 TTTT +=

4

4

4

1

4

323 TTT +=

=43

3T (373)4 + 2 (273)4

=3T 317.45 K = 44.45 C

4

4

4

3

4

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

=2

T 348.53 K = 75.53 C

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

20


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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

h

TT ), whereNis dimensionless group containing , k,L, and Tc. Plot as a

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

Now suppose that you have a system in whichL = 10 cm, Tc = 100 K, and the gas

is hydrogen with an average kof 0.1 W/m.K. Further suppose that you wish tooperate in such a way that the conduction and radiation heat fluxes are identical.

Identify the operating point on your curve and report the value ofTh that you must

maintain.

Solution:

( )44chradTTq =

( )LTTkq ch

cond=

( )( )

( ) ( )2244

chch

ch

ch

cond

rad TTTTk

L

TT

TT

k

L

q

q++=

=

+

+= 11

2

3

c

h

c

h

c

cond

rad

T

T

T

TT

k

L

q

q

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

L

q

qc

cond

rad

where

k

LTN c

3

=

c

h

T

T=

Nas a function of ;

( )( )112

++= condrad qqN

(1) 1=cond

rad

q

q

( )( )111

2 ++=N

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1. INTRODUCTION

(2) 8.0=cond

rad

q

q

( )( )118.0

2 ++=N

(3) 2.1=cond

rad

q

q

( )( )112.1

2 ++=N

plot of N as a function of:

For the system:

L = 10 cm = 0.10 mTc = 100 K

k= 0.1 W/m.K

Forqrad / qcond= 1.0

Then

( )( )111 2 ++= N

Solving forN:

k

LTN c

3

=

22


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1. INTRODUCTION

= 5.67040 x 10-8 W/m2.K4

( )( )( )10.0

10010.0106704.538

=N = 0.056704

Then

( ) ( )( )11056704.012 ++=

( )( ) 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

dt

dUQ =

( ) ( )[ ]ref

TTcVdr

dTTA =

44

( )tTT =( ) ( )44

=

TT

cV

A

dt

TTd

24 rA =

3

3

4rV =

( ) ( ) ( ) ( )4

44

4

3

23

3

4

4 =

= TTcrTT

rc

r

dt

TTd

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

=

TT

crdt

TTd

(2) Dimensionless group analogous to the Biot number

23


24/51

1. INTRODUCTION

bk

LhBi =

Equivalent h ,

(

= TTTT

h

4

4

Biot number equivalent =(( )

==TTk

TTr

k

rh

Ak

Vh

bbb33

44

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

Dimensionless group must be


25/51

1. INTRODUCTION

( )( )( ) ( )

( )( )222222222

222244

2

11

axaxa

axax

axaxax ++

=+

=

( ) ( )22222244 21

2

11

axaaxaax +

=

( ) ( )( ) ( )222244 21

)(2211

axaaxaxaaxax

aax +

+ +=

( ) ( ) ( )222244 21

2

1

2

1

2

11

axaaxaaxaaax +

+

=

( )222344 2111

4

11

axaaxaxaax +

+

=

+

+

=

12

111

4

112

4

344

a

xa

axaxaax

+

+

=

12

1

11

4

112

3

344

a

xa

a

axaxaax

Then,

44 ax dx = CaxArcaax axa + +

tan2

1ln

4

133

Applying:

+

+

=

TTArcTTArcTTT TTTT TTTTT dT iiiT

Ti

tantan2

1lnln

4

13344

Substitute values:

( ) ( )

+

+

= 293473tan293tan2934 2293473 293473ln293293ln2934 1 3344 ArcTArcTTTT dT

T

Ti

( ) crtT

ArcT

T

TT

dTT

Ti

348062.3

293tan2

293

293ln

2934

1344

=

+

+

=

( )( )

( ) ( )( )01.03848954106704.5348062.3

293tan2

293293ln

29341

8

3tTArc

TT

=

+

+

( )t

TArc

T

T0004978.048062.3

293tan2

293

293ln

2934

13

=

+

+

Tabulation:

25


26/51

1. INTRODUCTION

T, C T, K T, s

200 473 0

182 455 93.5

164 437 206.8

146 419 346.1

128 401 520.9110 383 745.8

92 365 1046

74 347 1468.5

56 329 2119.8

38 311 3340.6

Plot :

26


27/51

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 indiameter. 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 tofunction 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/m3

cp = 905 J/kg.K

kb = 237.2 W/m.K @ 30 C

From Prob. 1.17, usingi

T = 30 C + 273 = 303 KT = 0 K

T= 30 K

= 5.6704 x 10-8 W/m2.K

r= (1/2)(4 cm) = 2 cm = 0.02 m

Check for the legitimacy of lumped-capacity method.

(

( )

TTk

TTr

b3

44

=( )( )( )

( )( )03032.2373030302.0106704.5 448

= 0.000044


28/51

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:

( )( )=

= TTh

L

TTkq

2

21

1T = 200 CT = 0 C

k= 2 W/m2.K

L = 0.5 m

h = 3 W/m2.K

( )( )( )( )03

50.0

20022

2 == TTq

2234800 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 distributionin 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 atx= 0 andx = wall thickness,L.

28


29/51

1. INTRODUCTION

Solution:

Eq. 1.14, one dimensional heat diffusion equation,

t

T

tk

Tc

x

T

=

12

2

Use2

2

x

T

= 0 for steady flow.

1C

dx

dT=

21CxCT +=

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

At T= Tright,x =L

Tright = C1L + Tleft

L

TTC

leftright

=1

Then,

left

leftrightTx

L

TTT +

=

L

TT

x

TTleftrightleft

=

, therefore linear.

1.21 The thermal conductivity in a particular plane wall depends as follows on the walltemperature: k= A + BT, where A and B are constants. The temperatures are T1and T2 on either side of the wall, and its thickness is L. Develop an expression for

q.

Solution:

dx

dTkq =

( )dx

dTBTAq +=

( )dTBTAqdx +=

29


30/51

1. INTRODUCTION

( ) +=L T

T

dTBTAdxq

0

2

1

2

1

2

2

1T

T

BTATqL

+=

( ) ( )

+=2

1

2

212 2

1

TTBTTAqL

( ) ( )

L

TTBTTA

q

+=

2

1

2

2122

1

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

Figure 1.21.

Solution:L = 0.08 m

( )left

rightleftTTh

L

TTkq =

=

( )( ) ( )20100200

08.0

020=

k

k= 64 W/m.K

30


31/51

1. INTRODUCTION

From Table A.1, @ 10 C, k= 64 W/m.K

This 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 m

k1 = 2 W/m.CL2 = 6 cm = 0.06 m

k2 = 1 W/m.C

L3 = 4 cm = 0.04 m

k3= 5 W/m.C

L4 = 4 cm = 0.04 m

k4 = 4 W/m.C

( ) ( )

4

4

3

3

2

2

1

12525100

L

TTk

L

Tk

L

Tk

L

Tkq

rjjii

=

=

=

=

( ) ( )

2

2

1

125100

L

Tk

L

Tkii

=

( )( ) ( )( )

06.0

251

02.0

1002 =

iiTT

256600 =ii

TT

iT

=89.29 C( )

3

3

1

125100

L

Tk

L

Tk ji =

( ) ( ) ( )

04.0

255

02.0

29.891002 jT=

jT = 16.43 C

31


32/51

1. INTRODUCTION

( )

4

4

1

1100

L

TTk

L

Tk rji =

( )( ) ( )( )04.0

43.164

02.0

29.891002rT

=

( )( ) ( )( )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/m3

cp = 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 height

iT = 0 C, and room at T = 20 C

Time constant,

+

==DL

Dh

LDc

Ah

cVT

2

42

2

( )LDhcDL

T22 +

=

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.

Tt

i

eTT

TT

=

at T= 15 C

32


33/51

1. INTRODUCTION

1.648

200

2015 te

=

t= 898.5 s = 15 minutes

1.25. One large, black wall at 27 C faces another whose surface is 127 C. The gapbetween 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 23

4

1

4

2

==

L = 0.1 mT1= 27 C + 273 = 300 K

T2= 127 C + 273 = 400 K

= 5.6704 x 10-8 W/m2.K

k= 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 naturalconvection, with air at 20 C. In this case, h is not independent of temperature.

33


34/51

1. INTRODUCTION

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

capacity assumption.

Solution:

Properties of 1% carbon steel, Table A.1= 7801 kg/m3

cp = 473 J/kg.K

k= 42 W/m.K

= 1.17 x 10-5 m2/s

Verify the lumped-capacity assumption:

k

LhBi

t= 200 C 20 C = 180 C

h =3.51(180)1/4 W/m2.K = 12.86 W/m2.K

34

3

4

2

3

r

r

r

A

VL ===

r= (1/2)(1 cm) = 0.005 m

L = 0.005 m / 3 = 0.001667 m( )( )

42

001667.086.12Bi = 0.00051


35/51

1. INTRODUCTION

( ) ( )cV

AtTT

T

Ti

51.34 4

1 =

( )cV

AtTT

T

Ti

51.3

1

1 14

5

4

5

=

+

+

( ) ( ) cV

At

TTTTi

8775.0114/14/1

=

rV

A 3=

Then,

( ) ( )( )

cr

tt

rcTTTTi

6325.238775.0114/14/1

=

=

Substitute value,

( ) ( ) ( )( )( )005.047378016325.2

20200

1

20

14/14/1

t

T =

( )273012.0000143.0

20

14/1

+=

tT

( )273012.0000143.0

120

4/1

+=

tT

( )1909

699320

4/1

+=

tT

Ct

T 201909

69934

+

+

=

Tabulation:

t,s T, C

0 200

100 166.8

200 140.9

300 120.4

400 104.7

500 91

600 80.4

800 64.4

1000 53.41200 45.6

1400 40

1600 35.8

1800 32.6

2000 30.2

Plot:

35


36/51

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 Con 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/m3

cp = 466 J/kg.Kk= 10 W/m.K

= 0.26 x 10-5 m2/s

Radiation

( )42

4

11TTAQ

rad=

T1= 1100 C = 1373 K

T3= 25 C + 273 = 298 K

= 5.6704 x 10-8 W/m2

.C

Conduction

( )L

TTkAQ

cond

232

=

2

114 rA = , r1 = (1/2)(3 cm) = 1.5 cm = 0.015 m

( ) 21

015.04=A m2

36


37/51

1. INTRODUCTION

2

224 rA = , r2 = (1/2)(9 cm) = 4.5 cm = 0.045 m

( ) 21

045.04=A m2

L = 0.3 cm = 0.003 m

Then

condradQQ =

( ) ( )L

TTkATTA 322

4

2

4

11

=

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

298045.04101373015.04106704.5 2

2

4

2

428 = T

T

( ) ( )29810290632.513732

114

2

4 = TT

By 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 442842

4

11===

1.28. 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 =kgJ

mhrJ

/000,257,2

./000,340,22

= 1.036774 kg/hr.m2

Density of water= 1000 kg/m3

Evaporation rate =

m

mm

mkg

mhrkg

1

1000

/1000

./036774.13

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


38/51

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 thecup and 0 C outside. Plot Tagainst position in the cup wall and find the heat loss,

q.

Solution:

dx

dTkq =

( )dTaTkqdxo

21+=

( )+=2

1

2

1

T

T

odTaTkxq

[ ] 21

3

3

1T

ToaTTkxq +=

( ) ( )[ ]313

1

1

3

23

1

2aTTaTTkxq

o

++=

( ) ( )313

1

1

3

23

1

2aTTaTT

k

xq

o

++=

( ) ( )o

k

xqaTTaTT

+=+ 3

13

1

1

3

23

1

2

Solving forq if,

T1 = 100 C

T2 = 0 C

x = 0.005 m

( ) ( )323

1

2

3

13

1

1aTTaTT

k

xq

o

++=

( )( ) ( )( )[ ] ( ) ( )( )[ ]3431

34

31 010010010100

15.0

005.0 ++=q

q = 4000 W

Plotting:

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

( ) ( )o

k

xqaTTaTT

+=+ 3

13

1

1

3

23

1

2

( ) ( )

q

aTTaTTkx o

3

23

1

2

3

13

1

1++=

( ) ( ) ( )( (4000

1001010015.03

23

1

2

34

3

1 aTTx

++=

( )4000

1015.0203

2

4

3

1

2TT

x+=

38


39/51

1. INTRODUCTION

Tabulation:

T2, C x, m100 0

80 0.0013660 0.00248

40 0.00342

20 0.00424

0 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/m3

cp = 510 J/kg.K

kb = 1350 W/m.K

= 8.1 x 10-4 m2/s

39


40/51

1. INTRODUCTION

L = 1 mm = 0.001 m

bk

LhBi =

h = 10 W/m2.K

T =30 C

( )( )1350

001.010=Bi = 0.0000074


41/51

1. INTRODUCTION

t= 50 s

75.165

50

3033.90

30 =

e

T

T= 74.62 C

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

t= 25 s

75.165

25

3095.134

30 =

e

T

T= 120.26 C

t= 50 s

75.165

50

3095.13430

= eT

T= 107.62 C

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

t= 25 s

75.165

25

3095.167

30 =

e

T

T= 148.64 C

t= 50 s

75.165

50

3095.167

30 =

e

T

T= 132.03 CAnd so on.

Tabulation:

t, s Tdisc, C

1st 50 s 0 90.33

25 81.8850 74.62

2nd 50 s 50 134.95

75 120.26

100 107.62

3rd 50 s 100 167.95

125 148.64

150 132.03

41


42/51

1. INTRODUCTION

Plot:

For 100 s pulse apart

First 100 s, Ti = 90.33 C

t= 50 s75.165

50

3033.90

30 =

e

T

T= 74.62 C

t= 100 s

42


43/51

1. INTRODUCTION

75.165

100

3033.90

30 =

e

T

T= 63.00 C

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

75.165

50

3033.123

30 =

eT

T= 99.03 C

t= 100 s

75.165

100

3033.123

30 =

eT

T= 81.05 C

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

t= 50 s

75.165

50

3038.141

30 =

eT

T= 148.64 C

t= 100 s

75.165100

3038.141

30 =

eT

T= 112.38 C

And so on.Tabulation:

t, s Tdisc, C

1st 100 s 0 90.3.3

50 74.62

100 63.002nd 100 s 100 123.33

150 99.03

200 81.05

3rd 100 s 200 141.38

250 112.38

300 90.93

43


44/51

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 fractionof 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.

( ) ( )44ababTTATTAhQ +=

( ) ( )44abab

TTTThA

Q+=

44


45/51

1. INTRODUCTION

bT = 90 + 273 = 363 K

h = 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 ==D = 0.006 m

Then:

( )( )( ) ( )( )448

2363106704.53637

006.0

150.0aa

TT +=

( ) ( ) ( ) ( )448 363106704.536371326aa

TT +=

By Trial and Error Method:

Ta = 270.5 K = -2.5 C

Fraction = ( )( ) ( )44

44

abab

ab

TTTTh

TT

A

Q

+

=

=

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

448

5.270363106704.55.2703637

5.270363106704.5

+

Fraction = 0.5126

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

Solution:

( )

=

=

3631

5.270115.011

ba

UnTT

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/m 2. Calculate (a) theheater 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

( ) ( )CThCThqhh

872021+=

( )( ) ( )( )87141206.10 +hhTT = 1000

Th = 88.9 C

45


46/51

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:

T

t

i

eTT

TT

=

Assume aluminum material for the can of beer,Properties of aluminum, Table A.1, Appendix A

= 2707 kg/m3

cp = 905 J/kg.Kk= 237 W/m.K

Then,T = 25 C

iT = 3 C

T = 12 C

46


47/51

1. INTRODUCTION

Time constant:

Ah

cVT

=

LDV2

4=

D = 6.3 cm = 0.063 m

( ) ( )09.0063.04

2

=

V = 2.8055 x 10-4 m3

( ) ( ) ( ) ( )( )09.0063.02063.04

24

22

+

=+

= DLDA = 0.02405 m2

( )( )( )( )( )02405.03.7

108055.29052707 4=T = 3915 s

3915

253

2512 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 castiron on either side of an evacuated cavity. The heater, which releases 8000 W/m 2,

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 slabtemperatures.

Solution:

q = 8000 W/m2

47


48/51

1. INTRODUCTION

Properties of cast iron, Table A.1, Appendix A

= 7272 kg/m3

cp = 420 J/kg.Kkb = 52 W/m.K

L = 3 cm = 0.03 m

Inside slab temperature,( )

L

TCkq

=

10= 8000 W/m2

( )( )

03.0

1052

Tq

=

T= 14.62 C

Heater temperature,

( )

44

TTqh =

= 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 316stainless 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 orlowered if melting is to progress?

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1. INTRODUCTION

Solution:

Melting point of ice = 0 C

Thermal Conductivity of ice at 0 C = 2.215 W/m.K

( ) ( )3212 TTh

LTTkq ==

( )( )

23

12 TThL

TTk=

( )( )

23

12 TThL

TTk=

T1 = -10 C

T2 = 0 CT3 = 15 C

L = 2 cm = 0.02 m

Then,( ) ( )( )( )015

02.0

100215.2=

h

h = 73.83 W/m2.K

If 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=

e

T = 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 plateof 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.)

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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 intoequilibrium 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:

b

bT

boT

b

dTTT

cVS

=

11

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

bTT

11

= 0

=TTb

=

bo

bo

T

T

T

TT

cVS ln

=

T

T

T

TTcVS 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-capacitycooling? What is the temperature of the billet after 10 minutes?

Solution:

Check Biot Number

Properties of copper, Table A.1, App. A

= 8954 kg/m3

cp = 384 J/kg.K

kb = 398 W/m.K

Time constant:

Ah

cVT

=

LDV2

4=

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

( ) ( )08.004.04

2

=

V = 1.0053 x 10-4 m3

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1. INTRODUCTION

( ) ( ) ( ) ( )( )08.004.0204.04

24

22

+

=+

= DLDA = 0.012566 m2

( )( ) ( )( )( )012566.05

100053.13848954 4=T = 5501 s

Biot Number

mL

k

LhBi

b

02.0=

=

( )( )

398

02.05=Bi = 0.00025

solved problems in heat transfer

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