Ch.4

Chapter 4 Transient Heat Conduction

Chapter 4 TRANSIENT HEAT CONDUCTION

Lumped System Analysis

4-1C In heat transfer analysis, some bodies are observed to behave like a "lump" whose entire body temperature remains essentially uniform at all times during a heat transfer process. The temperature of such bodies can be taken to be a function of time only. Heat transfer analysis which utilizes this idealization is known as the lumped system analysis. It is applicable when the Biot number (the ratio of conduction resistance within the body to convection resistance at the surface of the body) is less than or equal to 0.1.

4-2C The lumped system analysis is more likely to be applicable for the body cooled naturally since the Biot number is proportional to the convection heat transfer coefficient, which is proportional to the air velocity. Therefore, the Biot number is more likely to be less than 0.1 for the case of natural convection.

4-3C The lumped system analysis is more likely to be applicable for the body allowed to cool in the air since the Biot number is proportional to the convection heat transfer coefficient, which is larger in water than it is in air because of the larger thermal conductivity of water. Therefore, the Biot number is more likely to be less than 0.1 for the case of the solid cooled in the air

4-4C The temperature drop of the potato during the second minute will be less than 4 ° since the temperature of a body approaches the temperature of the surrounding medium asymptotically, and thus it changes rapidly at the beginning, but slowly later on.

C

4-5C The temperature rise of the potato during the second minute will be less than 5 since the temperature of a body approaches the temperature of the surrounding medium asymptotically, and thus it changes rapidly at the beginning, but slowly later on.

°C

4-6C Biot number represents the ratio of conduction resistance within the body to convection resistance at the surface of the body. The Biot number is more likely to be larger for poorly conducting solids since such bodies have larger resistances against heat conduction.

4-7C The heat transfer is proportional to the surface area. Two half pieces of the roast have a much larger surface area than the single piece and thus a higher rate of heat transfer. As a result, the two half pieces will cook much faster than the single large piece.

4-8C The cylinder will cool faster than the sphere since heat transfer rate is proportional to the surface area, and the sphere has the smallest area for a given volume.

4-9C The lumped system analysis is more likely to be applicable in air than in water since the convection heat transfer coefficient and thus the Biot number is much smaller in air.

4-10C The lumped system analysis is more likely to be applicable for a golden apple than for an actual apple since the thermal conductivity is much larger and thus the Biot number is much smaller for gold.

4-1


4-11C The lumped system analysis is more likely to be applicable to slender bodies than the well-rounded bodies since the characteristic length (ratio of volume to surface area) and thus the Biot number is much smaller for slender bodies.

4-12 Relations are to be obtained for the characteristic lengths of a large plane wall of thickness 2L, a very long cylinder of radius and a sphere of radius r ro o

Analysis Relations for the characteristic lengths of a large plane wall of thickness 2L, a very long cylinder of radius and a sphere of radius are ro ro

L VA

LAA

L

L VA

r hr h

r

L VA

rr

r

c wall

c cylindero

o

o

c sphereo

o

o

,

,

,/

= = =

= = =

= = =

surface

surface

surface

22

2 2

4 34 3

2

3

2

ππ

ππ

2ro 2ro

2L

4-13 A relation for the time period for a lumped system to reach the average temperature ( ) to be obtained.

/ is

/

T Ti + ∞ 2

Analysis The relation for time period for a lumped system to reach the average temperature can be determined as

( )T Ti + ∞ 2

b0.693

b2 ln==⎯→⎯−=−

=⎯→⎯=−−

⎯→⎯=−

−+

⎯→⎯=−− −−

∞

∞−

∞

∞∞

−

∞

∞

tbt

eeTT

TTe

TT

TTT

eTTTtT btbt

i

ibt

i

i

bt

i

2ln

21

)(22)(

T∞

Ti

4-2


4-14 The temperature of a gas stream is to be measured by a thermocouple. The time it takes to register 99 percent of the initial ΔT is to be determined.

Assumptions 1 The junction is spherical in shape with a diameter of D = 0.0012 m. 2 The thermal properties of the junction are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 Radiation effects are negligible. 5 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be verified).

Properties The properties of the junction are given to be k = °35 W / m. C, , and .

ρ = 8500 kg / m3

Cp = °320 J / kg. C

Analysis The characteristic length of the junction and the Biot number are

L V

AD

DD

BihLk

c

c

= = = = =

= =°

°= <

surface2

m6

m

W / m . C m W / m. C

ππ

3

26

60 0012 0 0002

65 0 000235

0 00037 01

/ . .

( )( . )( )

. .

Since , the lumped system analysis is applicable. Then the time period for the thermocouple to read 99% of the initial temperature difference is determined from

< 0.1Bi

T t TT T

b hAC V

hC L

T t TT T

e e t

i

p p c

i

bt t

( ).

)(.

( ). ( . )

−−

=

= = =°°

=

−−

= ⎯ →⎯ = ⎯→⎯ =

∞

∞

∞

∞

− −

0 01

658500 320

01195

0 01 0 1195

ρ ρ W / m . C

( kg / m J / kg. C)(0.0002 m) s

2

3-1

s-138.5 s

Gas

h, T∞Junction

D T(t)

4-3


4-15E A number of brass balls are to be quenched in a water bath at a specified rate. The temperature of the balls after quenching and the rate at which heat needs to be removed from the water in order to keep its temperature constant are to be determined.

Assumptions 1 The balls are spherical in shape with a radius of r0 = 1 in. 2 The thermal properties of the balls are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be verified).

Properties The thermal conductivity, density, and specific heat of the brass balls are given to be k = 64.1 Btu/h.ft.°F, ρ = 532 lbm/ft3, and Cp = 0.092 Btu/lbm.°F.

Analysis (a) The characteristic length and the Biot number for the brass balls are

1.0 01820.0

)FBtu/h.ft. 1.64()ft 02778.0)(F.Btu/h.ft 42(

ft 02778.06

ft 12/26

6/

2

2

3

<=°

°==

=====

khL

Bi

DD

DAVL

c

sc π

π

Brass balls, 250°F

Water bath, 120°F

The lumped system analysis is applicable since Bi < 0.1. Then the temperature of the balls after quenching becomes

F 166 °=⎯→⎯=−−

⎯→⎯=−−

==°

°===

−−

∞

∞ )(120250120)()(

s 00858.0 h9.30ft) F)(0.02778Btu/lbm. 092.0)(lbm/ft (532

F.Btu/h.ft 42

s) 120)(s 00858.0(

1-1-3

2

-1tTetTe

TTTtT

LCh

VChA

b

bt

i

cpp

s

ρρ

(b) The total amount of heat transfer from a ball during a 2-minute period is

Btu 97.9F)166250(F)Btu/lbm. 092.0)(lbm 29.1()]([

lbm 290.16

ft) 12/2()lbm/ft 532(

6

33

3

=°−°=−=

====

tTTmCQ

DVm

ip

ππρρ

Then the rate of heat transfer from the balls to the water becomes

Btu/min 1196=×== )Btu 97.9(balls/min) 120(ballballQnQtotal &&

Therefore, heat must be removed from the water at a rate of 1196 Btu/min in order to keep its temperature constant at 120 . °F

4-4


4-16E A number of aluminum balls are to be quenched in a water bath at a specified rate. The temperature of balls after quenching and the rate at which heat needs to be removed from the water in order to keep its temperature constant are to be determined.

Assumptions 1 The balls are spherical in shape with a radius of r0 = 1 in. 2 The thermal properties of the balls are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be verified).

Properties The thermal conductivity, density, and specific heat of the aluminum balls are k = 137 Btu/h.ft.°F, ρ = 168 lbm/ft3, and Cp = 0.216 Btu/lbm.°F (Table A-3E).

Analysis (a) The characteristic length and the Biot number for the aluminum balls are

1.000852.0

)FBtu/h.ft. 137()ft 02778.0)(F.Btu/h.ft 42(

ft 02778.06

ft 12/26

6/

2

2

3

<=°

°==

=====

khL

Bi

DD

DAVL

c

cπ

π

Aluminum balls, 250°F

Water bath, 120°F

The lumped system analysis is applicable since Bi < 0.1. Then the temperature of the balls after quenching becomes

F152°=⎯→⎯=−−

⎯→⎯=−−

==°

°===

−−

∞

∞ )(120250120)()(

s 01157.0 h66.41ft) F)(0.02778Btu/lbm. 216.0)(lbm/ft (168

F.Btu/h.ft 42

s) 120)(s 01157.0(

1-1-3

2

-1tTe

tTe

TTTtT

LCh

VChA

b

bt

i

cpp

s

ρρ

(b) The total amount of heat transfer from a ball during a 2-minute period is

Btu 62.8F)152250(F)Btu/lbm. 216.0)(lbm 4072.0()]([

lbm 4072.06

ft) 12/2()lbm/ft 168(6

33

3

=°−°=−=

====

tTTmCQ

DVm

ip

ππρρ

Then the rate of heat transfer from the balls to the water becomes

Btu/min 1034=×== )Btu 62.8(balls/min) 120(ballballQnQtotal &&

Therefore, heat must be removed from the water at a rate of 1034 Btu/min in order to keep its temperature constant at 120 . °F

4-5


4-17 Milk in a thin-walled glass container is to be warmed up by placing it into a large pan filled with hot water. The warming time of the milk is to be determined.

Assumptions 1 The glass container is cylindrical in shape with a radius of r0 = 3 cm. 2 The thermal properties of the milk are taken to be the same as those of water. 3 Thermal properties of the milk are constant at room temperature. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Biot number in this case is large (much larger than 0.1). However, the lumped system analysis is still applicable since the milk is stirred constantly, so that its temperature remains uniform at all times.

Water60°C

Milk 3°C

Properties The thermal conductivity, density, and specific heat of the milk at 20°C are k = 0.607 W/m.°C, ρ = 998 kg/m3, and Cp = 4.182 kJ/kg.°C (Table A-9).

Analysis The characteristic length and Biot number for the glass of milk are

1.0> 076.2)CW/m. 607.0(

)m 0105.0)(C.W/m 120(

m 01050.0m) 03.0(2+m) m)(0.07 03.0(2

m) 07.0(m) 03.0(22

2

2

2

2

2

=°

°==

==+

==

khL

Bi

rLrLr

AVL

c

oo

o

sc ππ

πππ

π

For the reason explained above we can use the lumped system analysis to determine how long it will take for the milk to warm up to 38°C:

min 5.8s 348 ==⎯→⎯=−−

⎯→⎯=−−

=°°

===

−−

∞

∞ teeTTTtT

LCh

VChA

b

tbt

i

cpp

s

)s 002738.0(

1-3

2

-1

6036038)(

s 002738.0m) C)(0.0105J/kg. 4182)( kg/m(998

C.W/m 120ρρ

Therefore, it will take about 6 minutes to warm the milk from 3 to 38°C.

4-6


4-18 A thin-walled glass containing milk is placed into a large pan filled with hot water to warm up the milk. The warming time of the milk is to be determined.

Assumptions 1 The glass container is cylindrical in shape with a radius of r0 = 3 cm. 2 The thermal properties of the milk are taken to be the same as those of water. 3 Thermal properties of the milk are constant at room temperature. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Biot number in this case is large (much larger than 0.1). However, the lumped system analysis is still applicable since the milk is stirred constantly, so that its temperature remains uniform at all times.

Water60°C

Milk 3°C

Properties The thermal conductivity, density, and specific heat of the milk at 20°C are k = 0.607 W/m.°C, ρ = 998 kg/m3, and Cp = 4.182 kJ/kg.°C (Table A-9).

Analysis The characteristic length and Biot number for the glass of milk are

1.0> 15.4)CW/m. 607.0(

)m 0105.0)(C.W/m 240(

m 01050.0m) 03.0(2+m) m)(0.07 03.0(2

m) 07.0(m) 03.0(22

2

2

2

2

2

=°

°==

==+

==

khL

Bi

rLrLr

AVL

c

oo

o

sc ππ

πππ

π

For the reason explained above we can use the lumped system analysis to determine how long it will take for the milk to warm up to 38°C:

min 2.9s 174 ==⎯→⎯=

−−

⎯→⎯=−−

=°°

===

−−

∞

∞ teeTTTtT

LCh

VChA

b

tbt

i

cpp

s

)s 005477.0(

1-3

2

-1

6036038)(

s 005477.0m) C)(0.0105J/kg. 4182)( kg/m(998

C.W/m 240ρρ

Therefore, it will take about 3 minutes to warm the milk from 3 to 38°C.

4-7


4-19E A person shakes a can of drink in a iced water to cool it. The cooling time of the drink is to be determined.

Assumptions 1 The can containing the drink is cylindrical in shape with a radius of r0 = 1.25 in. 2 The thermal properties of the milk are taken to be the same as those of water. 3 Thermal properties of the milk are constant at room temperature. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Biot number in this case is large (much larger than 0.1). However, the lumped system analysis is still applicable since the milk is stirred constantly, so that its temperature remains uniform at all times.

Milk 3°C

Water32°F

Cola 75°F

Properties The density and specific heat of water at room temperature are ρ = 62.22 lbm/ft3, and Cp = 0.999 Btu/lbm.°F (Table A-9E).

Analysis Application of lumped system analysis in this case gives

ft 04167.0ft) 12/25.1(2+ft) ft)(5/12 12/25.1(2

ft) 12/5(ft) 12/25.1(22 2

2

2

2==

+==

πππ

πππ

oo

o

sc rLr

LrAVL

s 406=⎯→⎯=−−

⎯→⎯=−−

==°

°===

−−

∞

∞ teeTTTtT

LCh

VChA

b

tbt

i

cpp

s

)s 00322.0(

1-1-3

2

-1

32803245)(

s 00322.0 h583.11ft) F)(0.04167Btu/lbm. 999.0)(lbm/ft (62.22

F.Btu/h.ft 30ρρ

Therefore, it will take 7 minutes and 46 seconds to cool the canned drink to 45°F.

4-8


4-20 An iron whose base plate is made of an aluminum alloy is turned on. The time for the plate temperature to reach 140°C and whether it is realistic to assume the plate temperature to be uniform at all times are to be determined.

Assumptions 1 85 percent of the heat generated in the resistance wires is transferred to the plate. 2 The thermal properties of the plate are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface.

Properties The density, specific heat, and thermal diffusivity of the aluminum alloy plate are given to be ρ = 2770 kg/m3, Cp = 875 kJ/kg.°C, and α = 7.3×10-5 m2/s. The thermal conductivity of the plate can be determined from α = k/(ρCp)= 177 W/m.°C (or it can be read from Table A-3).

Analysis The mass of the iron's base plate is Air 22°C

IRON 1000 W

m V LA= = = =ρ ρ ( )( . )( . ) .2770 0 005 0 03 0 4155 kg / m m m kg3 2

Noting that only 85 percent of the heat generated is transferred to the plate, the rate of heat transfer to the iron's base plate is

& .Qin W W= × =0 85 1000 850

The temperature of the plate, and thus the rate of heat transfer from the plate, changes during the process. Using the average plate temperature, the average rate of heat loss from the plate is determined from

W21.2=C222

22140)m 03.0)(C. W/m12()( 22ave plate,loss °⎟

⎠⎞

⎜⎝⎛ −

+°=−= ∞TThAQ&

Energy balance on the plate can be expressed as

E E E Q t Q t E mC Tpin out plate in out plate plate − = → − = =Δ Δ Δ Δ& & Δ

Solving for Δt and substituting,

ΔΔ

tmC TQ Q

p=−

° − °−

plate

in out= kg J / kg. C C

(850 21.2) J / s=

& &( . )( )( )0 4155 875 140 22 51.8 s

which is the time required for the plate temperature to reach 140 . To determine whether it is realistic to assume the plate temperature to be uniform at all times, we need to calculate the Biot number,

°C

1.0<00034.0

)CW/m. 0.177()m 005.0)(C.W/m 12(

m 005.0

2=

°°

==

====

khL

Bi

LA

LAAVL

c

sc

It is realistic to assume uniform temperature for the plate since Bi < 0.1.

Discussion This problem can also be solved by obtaining the differential equation from an energy balance on the plate for a differential time interval, and solving the differential equation. It gives

⎟⎟⎠

⎞⎜⎜⎝

⎛−−+= ∞ )exp(1)( in t

mChA

hAQ

TtTp

&

Substituting the known quantities and solving for t again gives 51.8 s.

4-9


4-21 "!PROBLEM 4-21" "GIVEN" E_dot=1000 "[W]" L=0.005 "[m]" A=0.03 "[m^2]" T_infinity=22 "[C]" T_i=T_infinity h=12 "[W/m^2-C], parameter to be varied" f_heat=0.85 T_f=140 "[C], parameter to be varied" "PROPERTIES" rho=2770 "[kg/m^3]" C_p=875 "[J/kg-C]" alpha=7.3E-5 "[m^2/s]" "ANALYSIS" V=L*A m=rho*V Q_dot_in=f_heat*E_dot Q_dot_out=h*A*(T_ave-T_infinity) T_ave=1/2*(T_i+T_f) (Q_dot_in-Q_dot_out)*time=m*C_p*(T_f-T_i) "energy balance on the plate"

h [W/m2.C] time [s] 5 51 7 51.22 9 51.43

11 51.65 13 51.88 15 52.1 17 52.32 19 52.55 21 52.78 23 53.01 25 53.24

Tf [C] time [s]

30 3.428 40 7.728 50 12.05 60 16.39 70 20.74 80 25.12 90 29.51

100 33.92 110 38.35 120 42.8 130 47.28 140 51.76 150 56.27 160 60.8 170 65.35 180 69.92 190 74.51

4-10


200 79.12

5 9 13 17 21 2551

51.45

51.9

52.35

52.8

53.25

h [W/m2-C]

time

[s]

20 40 60 80 100 120 140 160 180 2000

10

20

30

40

50

60

70

80

Tf [C]

time

[s]

4-11


4-22 Ball bearings leaving the oven at a uniform temperature of 900°C are exposed to air for a while before they are dropped into the water for quenching. The time they can stand in the air before their temperature falls below 850°C is to be determined.

Assumptions 1 The bearings are spherical in shape with a radius of r0 = 0.6 cm. 2 The thermal properties of the bearings are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be verified).

Properties The thermal conductivity, density, and specific heat of the bearings are given to be k = 15.1 W/m.°C, ρ = 8085 kg/m3, and Cp = 0.480 kJ/kg.°F.

Analysis The characteristic length of the steel ball bearings and Biot number are

0.1< 0166.0

)CW/m. 1.15()m 002.0)(C.W/m 125(

m 002.06

m 012.06

6/

2

2

3

=°

°==

=====

khL

Bi

DD

DAVL

c

sc π

π

Steel balls 900°C

Air, 30°C Furnace

Therefore, the lumped system analysis is applicable. Then the allowable time is determined to be

s 3.68=⎯→⎯=

−−

⎯→⎯=−−

=°°

===

−−

∞

∞ teeTTTtT

LCh

VChA

b

tbt

i

cpp

s

)s 0161.0(

1-3

2

-1

3090030850)(

s 01610.0m) C)(0.002J/kg. 480)( kg/m8085(

C.W/m 125ρρ

The result indicates that the ball bearing can stay in the air about 4 s before being dropped into the water.

4-12


4-23 A number of carbon steel balls are to be annealed by heating them first and then allowing them to cool slowly in ambient air at a specified rate. The time of annealing and the total rate of heat transfer from the balls to the ambient air are to be determined.

Assumptions 1 The balls are spherical in shape with a radius of r0 = 4 mm. 2 The thermal properties of the balls are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be verified).

Properties The thermal conductivity, density, and specific heat of the balls are given to be k = 54 W/m.°C, ρ = 7833 kg/m3, and Cp = 0.465 kJ/kg.°C.

Analysis The characteristic length of the balls and the Biot number are

1.0 0018.0

)CW/m. 54()m 0013.0)(C.W/m 75(

m 0013.06

m 008.06

6/

2

2

3

<=°

°==

=====

khL

Bi

DD

DAVL

c

sc π

π

Steel balls 900°C

Air, 35°C Furnace

Therefore, the lumped system analysis is applicable. Then the time for the annealing process is determined to be

min 2.7s 163 ==⎯→⎯=

−−

⎯→⎯=−−

=°°

===

−−

∞

∞ teeTTTtT

LCh

VChA

b

bt

i

cpp

s

)ts 01584.0(

1-3

2

-1

3590035100)(

s 01584.0m) C)(0.0013J/kg. 465)( kg/m(7833

C.W/m 75ρρ

The amount of heat transfer from a single ball is

m V D

Q mC T Tp f i

= = = =

= − = ° − ° =

ρ ρπ π3

67833 0 008 0 0021

0 0021 465 900 100

( ) ( . .

[ ] ( . )( )( )

kg / m m)6

kg

kg J / kg. C C 781 J = 0.781 kJ (per ball)

33

Then the total rate of heat transfer from the balls to the ambient air becomes

W543==×== kJ/h953,1) kJ/ball781.0( balls/h)2500(ballQnQ &&

4-13


4-24 "!PROBLEM 4-24" "GIVEN" D=0.008 "[m]" "T_i=900 [C], parameter to be varied" T_f=100 "[C]" T_infinity=35 "[C]" h=75 "[W/m^2-C]" n_dot_ball=2500 "[1/h]" "PROPERTIES" rho=7833 "[kg/m^3]" k=54 "[W/m-C]" C_p=465 "[J/kg-C]" alpha=1.474E-6 "[m^2/s]" "ANALYSIS" A=pi*D^2 V=pi*D^3/6 L_c=V/A Bi=(h*L_c)/k "if Bi < 0.1, the lumped sytem analysis is applicable" b=(h*A)/(rho*C_p*V) (T_f-T_infinity)/(T_i-T_infinity)=exp(-b*time) m=rho*V Q=m*C_p*(T_i-T_f) Q_dot=n_dot_ball*Q*Convert(J/h, W)

Ti [C] time [s] Q [W] 500 127.4 271.2 550 134 305.1 600 140 339 650 145.5 372.9 700 150.6 406.9 750 155.3 440.8 800 159.6 474.7 850 163.7 508.6 900 167.6 542.5 950 171.2 576.4

1000 174.7 610.3

4-14


500 600 700 800 900 1000120

130

140

150

160

170

180

250

300

350

400

450

500

550

600

650

Ti [C]

time

[s]

Q [

W]

time

heat

4-15


4-25 An electronic device is on for 5 minutes, and off for several hours. The temperature of the device at the end of the 5-min operating period is to be determined for the cases of operation with and without a heat sink.

Assumptions 1 The device and the heat sink are isothermal. 2 The thermal properties of the device and of the sink are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface.

Properties The specific heat of the device is given to be Cp = 850 J/kg.°C. The specific heat of the aluminum sink is 903 J/kg.°C (Table A-19), but can be taken to be 850 J/kg.°C for simplicity in analysis.

Analysis (a) Approximate solution

This problem can be solved approximately by using an average temperature for the device when evaluating the heat loss. An energy balance on the device can be expressed as

E E E E Q t E t mC Tpin out generation device out generation device − + = ⎯→⎯ − + =Δ Δ Δ& & Δ

Electronic device 30 W

or, )(2generation ∞∞

∞ −=Δ⎟⎠⎞

⎜⎝⎛ −

+−Δ TTmCtT

TThAtE ps

&

Substituting the given values,

C)25)(CJ/kg. 850)(kg 02.0()s 605(C2

25)m 0005.0)(C. W/m12()s 605)(J/s 30( o22 °−°=×⎟⎠⎞

⎜⎝⎛ −

°−× TT

which gives T = 527.8°C

If the device were attached to an aluminum heat sink, the temperature of the device would be

C)25)(CJ/kg. 850(kg)02.020.0()s 605(C2

25)m0085.0)(C. W/m12()s 605)(J/s 30( 22 °−°×+=×°⎟⎠⎞

⎜⎝⎛ −

°−× TT

which gives T = 69.5°C

Note that the temperature of the electronic device drops considerably as a result of attaching it to a heat sink.

(b) Exact solution

This problem can be solved exactly by obtaining the differential equation from an energy balance on the device for a differential time interval dt. We will get

pp

s

mCE

TTmChA

dtTTd generation)(

)( &=−+

−∞

∞

It can be solved to give

⎟⎟⎠

⎞⎜⎜⎝

⎛−−+= ∞ )exp(1)( generation t

mChA

hAE

TtTp

s

s

&

Substituting the known quantities and solving for t gives 527.3°C for the first case and 69.4°C for the second case, which are practically identical to the results obtained from the approximate analysis.

4-16


Transient Heat Conduction in Large Plane Walls, Long Cylinders, and Spheres 4-26C A cylinder whose diameter is small relative to its length can be treated as an infinitely long cylinder. When the diameter and length of the cylinder are comparable, it is not proper to treat the cylinder as being infinitely long. It is also not proper to use this model when finding the temperatures near the bottom or top surfaces of a cylinder since heat transfer at those locations can be two-dimensional. 4-27C Yes. A plane wall whose one side is insulated is equivalent to a plane wall that is twice as thick and is exposed to convection from both sides. The midplane in the latter case will behave like an insulated surface because of thermal symmetry. 4-28C The solution for determination of the one-dimensional transient temperature distribution involves many variables that make the graphical representation of the results impractical. In order to reduce the number of parameters, some variables are grouped into dimensionless quantities. 4-29C The Fourier number is a measure of heat conducted through a body relative to the heat stored. Thus a large value of Fourier number indicates faster propagation of heat through body. Since Fourier number is proportional to time, doubling the time will also double the Fourier number. 4-30C This case can be handled by setting the heat transfer coefficient h to infinity since the temperature of the surrounding medium in this case becomes equivalent to the surface temperature.

∞

4-31C The maximum possible amount of heat transfer will occur when the temperature of the body reaches the temperature of the medium, and can be determined from Q mC T Tp imax ( )= −∞ .

4-32C When the Biot number is less than 0.1, the temperature of the sphere will be nearly uniform at all times. Therefore, it is more convenient to use the lumped system analysis in this case.

4-33 A student calculates the total heat transfer from a spherical copper ball. It is to be determined whether his/her result is reasonable.

Assumptions The thermal properties of the copper ball are constant at room temperature.

Properties The density and specific heat of the copper ball are ρ = 8933 kg/m3, and Cp = 0.385 kJ/kg.°C (Table A-3).

Analysis The mass of the copper ball and the maximum amount of heat transfer from the copper ball are

kJ 1064C)25200)(CkJ/kg. 385.0)(kg 79.15(][

kg 79.156

m) 15.0()kg/m 8933(

6

max

33

3

=°−°=−=

=⎥⎥⎦

⎤

⎢⎢⎣

⎡=⎟

⎟⎠

⎞⎜⎜⎝

⎛==

∞TTmCQ

DVm

ip

ππρρ

Copper ball, 200°C

Q

Discussion The student's result of 4520 kJ is not reasonable since it is greater than the maximum possible amount of heat transfer.

4-17


4-34 An egg is dropped into boiling water. The cooking time of the egg is to be determined. √

Assumptions 1 The egg is spherical in shape with a radius of r0 = 2.75 cm. 2 Heat conduction in the egg is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the egg are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The thermal conductivity and diffusivity of the eggs are given to be k = 0.6 W/m.°C and α = 0.14×10-6 m2/s.

Analysis The Biot number for this process is

Bihrk

o= =°

°=

( )( . )( . )

.1400 0 02750 6

64 2 W / m . C m W / m. C

2

Water97°C

Egg Ti = 8°C

The constants λ1 and A1 corresponding to this Biot number are, from Table 4-1,

9969.1 and 0877.3 11 == Aλ

Then the Fourier number becomes

2.0198.0)9969.1(9789770 22

1 )0877.3(1

0,0 ≈=⎯→⎯=

−−

⎯→⎯=−−

= −−

∞

∞ τθ ττλ eeATTTT

isph

Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the time required for the temperature of the center of the egg to reach 70°C is determined to be

min 17.8==×

=ατ

=−

s 1068/s)m 1014.0(

m) 0275.0)(198.0(26

22or

t

4-18


4-35 "!PROBLEM 4-35" "GIVEN" D=0.055 "[m]" T_i=8 "[C]" "T_o=70 [C], parameter to be varied" T_infinity=97 "[C]" h=1400 "[W/m^2-C]" "PROPERTIES" k=0.6 "[W/m-C]" alpha=0.14E-6 "[m^2/s]" "ANALYSIS" Bi=(h*r_o)/k r_o=D/2 "From Table 4-1 corresponding to this Bi number, we read" lambda_1=1.9969 A_1=3.0863 (T_o-T_infinity)/(T_i-T_infinity)=A_1*exp(-lambda_1^2*tau) time=(tau*r_o^2)/alpha*Convert(s, min)

To [C] time [min] 50 39.86 55 42.4 60 45.26 65 48.54 70 52.38 75 57 80 62.82 85 70.68 90 82.85 95 111.1

50 55 60 65 70 75 80 85 90 9530

40

50

60

70

80

90

100

110

120

To [C]

time

[min

]

4-19


4-36 Large brass plates are heated in an oven. The surface temperature of the plates leaving the oven is to be determined.

Assumptions 1 Heat conduction in the plate is one-dimensional since the plate is large relative to its thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the plate are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The properties of brass at room temperature are given to be k = 110 W/m.°C, α = 33.9×10-6 m2/s

Analysis The Biot number for this process is

Bi hLk

= =°

°=

(80 )( . )( )

. W / m . C m W / m. C

2 0 015110

0 0109


λ1 101039 10018= =. . and A

The Fourier number is

τ α= =

× ×= >

−tL2

6339 10 100 015

90 4 0 2( .( .

. . m / s)( min 60 s / min) m)

2

2

Plates 25°C

Furnace, 700°C

Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the temperature at the surface of the plates becomes

C 445 °=⎯→⎯=

−−

==λ=−−

=θ −τλ−

∞

∞

),(378.070025

700),(

378.0)1039.0cos()0018.1()/cos(),(

),( )4.90()1039.0(11

221

tLTtLT

eLLeATT

TtxTtL

iwall

Discussion This problem can be solved easily using the lumped system analysis since Bi < 0.1, and thus the lumped system analysis is applicable. It gives

C 448 °=°+°=−+=→=−−

=°⋅⋅×

°⋅=====

°⋅⋅×=×

°⋅==→=

−−∞∞

−

∞

∞ s) 600)(s 001644.0(

1-36

2

3626-

1-C)700-(25C700)()(

)(

s 001644.0C)s/m W10245.3m)( 015.0(

C W/m80)/()(

Cs/m W10245.3/m 1033.9

C W/m110

eeTTTtTeTTTtT

kLh

LCh

CLAhA

VChAb

skC

Ck

bti

bt

i

ppp

pp

αρρρ

αρ

ρα

which is almost identical to the result obtained above.

4-20


4-37 "!PROBLEM 4-37" "GIVEN" L=0.03/2 "[m]" T_i=25 "[C]" T_infinity=700 "[C], parameter to be varied" time=10 "[min], parameter to be varied" h=80 "[W/m^2-C]" "PROPERTIES" k=110 "[W/m-C]" alpha=33.9E-6 "[m^2/s]" "ANALYSIS" Bi=(h*L)/k "From Table 4-1, corresponding to this Bi number, we read" lambda_1=0.1039 A_1=1.0018 tau=(alpha*time*Convert(min, s))/L^2 (T_L-T_infinity)/(T_i-T_infinity)=A_1*exp(-lambda_1^2*tau)*Cos(lambda_1*L/L)

T∞ [C] TL [C] 500 321.6 525 337.2 550 352.9 575 368.5 600 384.1 625 399.7 650 415.3 675 430.9 700 446.5 725 462.1 750 477.8 775 493.4 800 509 825 524.6 850 540.2 875 555.8 900 571.4

time [min] TL [C]

2 146.7 4 244.8 6 325.5 8 391.9

10 446.5 12 491.5 14 528.5 16 558.9 18 583.9 20 604.5 22 621.4 24 635.4 26 646.8 28 656.2

4-21


30 664

500 550 600 650 700 750 800 850 900300

350

400

450

500

550

600

T∞

[C]

T L [

C]

0 5 10 15 20 25 30100

200

300

400

500

600

700

time [min]

T L [

C]

4-22


4-38 A long cylindrical shaft at 400°C is allowed to cool slowly. The center temperature and the heat transfer per unit length of the cylinder are to be determined.

Assumptions 1 Heat conduction in the shaft is one-dimensional since it is long and it has thermal symmetry about the center line. 2 The thermal properties of the shaft are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The properties of stainless steel 304 at room temperature are given to be k = 14.9 W/m.°C, ρ = 7900 kg/m3, Cp = 477 J/kg.°C, α = 3.95×10-6 m2/s

Analysis First the Biot number is calculated to be

705.0)CW/m. 9.14(

)m 175.0)(C.W/m 60( 2=

°°

==k

hrBi o

Steel shaft Ti = 400°C

Air T∞ = 150°C


λ1 110935 11558= =. . and A


1548.0m) 175.0(

s) 60/s)(20m 1095.3(2

26

2 =××

==−

Ltατ

which is very close to the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can still be used, with the understanding that the error involved will be a little more than 2 percent. Then the temperature at the center of the shaft becomes

C 390 °=⎯→⎯=

−−

===−−

= −−

∞

∞

00

)1548.0()0935.1(1

0,0

9605.0150400150

9605.0)1558.1(22

1

TT

eeATTTT

icyl

τλθ

The maximum heat can be transferred from the cylinder per meter of its length is

kJ 638,90C)150400)(CkJ/kg. 477.0)(kg 1.760(][

kg 1.760)]m 1(m) 175.0()[kg/m 7900(

max

232

=°−°=−==π=ρπ=ρ=

∞ ip

o

TTmCQLrVm

Once the constant = 0.4689 is determined from Table 4-2 corresponding to the constant J1 λ1 =1.0935, the actual heat transfer becomes

kJ 16,015==

=⎟⎠⎞

⎜⎝⎛

−−

−=λλ

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=⎟⎟⎠

⎞⎜⎜⎝

⎛

∞

∞

)kJ 638,90(177.0

177.00935.14689.0

15040015039021

)(21

1

110

max

Q

JTTTT

QQ

icyl

4-23


4-39 "!PROBLEM 4-39" "GIVEN" r_o=0.35/2 "[m]" T_i=400 "[C]" T_infinity=150 "[C]" h=60 "[W/m^2-C]" "time=20 [min], parameter to be varied" "PROPERTIES" k=14.9 "[W/m-C]" rho=7900 "[kg/m^3]" C_p=477 "[J/kg-C]" alpha=3.95E-6 "[m^2/s]" "ANALYSIS" Bi=(h*r_o)/k "From Table 4-1 corresponding to this Bi number, we read" lambda_1=1.0935 A_1=1.1558 J_1=0.4709 "From Table 4-2, corresponding to lambda_1" tau=(alpha*time*Convert(min, s))/r_o^2 (T_o-T_infinity)/(T_i-T_infinity)=A_1*exp(-lambda_1^2*tau) L=1 "[m], 1 m length of the cylinder is considered" V=pi*r_o^2*L m=rho*V Q_max=m*C_p*(T_i-T_infinity)*Convert(J, kJ) Q/Q_max=1-2*(T_o-T_infinity)/(T_i-T_infinity)*J_1/lambda_1

time [min] To [C] Q [kJ] 5 425.9 4491 10 413.4 8386 15 401.5 12105 20 390.1 15656 25 379.3 19046 30 368.9 22283 35 359 25374 40 349.6 28325 45 340.5 31142 50 331.9 33832 55 323.7 36401 60 315.8 38853

4-24


0 10 20 30 40 50 60300

320

340

360

380

400

420

440

0

5000

10000

15000

20000

25000

30000

35000

40000

time [min]

T o [

C]

Q [

kJ]

temperature

heat

4-25


4-40E Long cylindrical steel rods are heat-treated in an oven. Their centerline temperature when they leave the oven is to be determined.

Assumptions 1 Heat conduction in the rods is one-dimensional since the rods are long and they have thermal symmetry about the center line. 2 The thermal properties of the rod are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The properties of AISI stainless steel rods are given to be k = 7.74 Btu/h.ft.°F, α = 0.135 ft2/h.

Analysis The time the steel rods stays in the oven can be determined from

t = = =length

velocity ft

ft / min min = 180 s30

103

Oven, 1700°F

Steel rod, 85°F

The Biot number is

4307.0)FBtu/h.ft. 74.7(

)ft 12/2)(F.Btu/h.ft 20( 2=

°°

==k

hrBi o


λ1 10 8784 10995= =. . and A


τα

= = =t

ro2

01352 12

0 243( .( /

. ft / h)(3 / 60 h) ft)

2

2

Then the temperature at the center of the rods becomes

θ λ τ0

01

0 8784 0 24312 2

10995 0 912,( . ) ( . )( . ) .cyl

i

T TT T

A e e=−−

= = =∞

∞

− −

F228°=⎯→⎯=−−

00 912.0

1700851700

TT

4-26


4-41 Steaks are cooled by passing them through a refrigeration room. The time of cooling is to be determined.

Assumptions 1 Heat conduction in the steaks is one-dimensional since the steaks are large relative to their thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the steaks are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The properties of steaks are given to be k = 0.45 W/m.°C and α = 0.91×10-7 m2/s

Analysis The Biot number is

200.0)C W/m.45.0(

)m 01.0)(C. W/m9( 2=

°°

==k

hLBi Steaks 25°C

Refrigerated air -11°C


λ1 1 0 4328 10311= =. . and A


2.0601.5)4328.0cos()0311.1()11(25)11(2

)/cos(),(

2

21

)4328.0(

11

>=⎯→⎯=−−−−

=−−

−

−

∞

∞

τ

λ

τ

τλ

e

LLeATT

TtLT

i

Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the length of time for the steaks to be kept in the refrigerator is determined to be

min 102.6==×

== − s 6155/s)m 1091.0(

m) 01.0)(601.5(27

22

ατLt

4-27


4-42 A long cylindrical wood log is exposed to hot gases in a fireplace. The time for the ignition of the wood is to be determined.

Assumptions 1 Heat conduction in the wood is one-dimensional since it is long and it has thermal symmetry about the center line. 2 The thermal properties of the wood are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The properties of wood are given to be k = 0.17 W/m.°C, α = 1.28×10-7 m2/s


00.4)C W/m.17.0(

)m 05.0)(C. W/m6.13( 2=

°°

==k

hrBi o


4698.1 and 9081.1 11 == Aλ

Once the constant is determined from Table 4-2 corresponding to the constant

J0

λ1 =1.9081, the Fourier number is determined to be

10 cm Wood log, 10°C

Hot gases 700°C

251.0)2771.0()4698.1(50010500420

)/(),(

2

21

)9081.1(

101

=τ⎯→⎯=−−

λ=−−

τ−

τλ−

∞

∞

e

rrJeATT

TtrToo

i

o

which is above the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can be used. Then the length of time before the log ignites is

min 81.7==×

=α

τ=

−s 4904

/s)m 1028.1(m) 05.0)(251.0(

27

22ort

4-28


4-43 A rib is roasted in an oven. The heat transfer coefficient at the surface of the rib, the temperature of the outer surface of the rib and the amount of heat transfer when it is rare done are to be determined. The time it will take to roast this rib to medium level is also to be determined.

Assumptions 1 The rib is a homogeneous spherical object. 2 Heat conduction in the rib is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the rib are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The properties of the rib are given to be k = 0.45 W/m.°C, ρ = 1200 kg/m3, Cp = 4.1 kJ/kg.°C, and α = 0.91×10-7 m2/s.

Analysis (a) The radius of the roast is determined to be

Oven 163°C

Rib, 4.5°C

m 08603.04

)m 002667.0(343

34

m 002667.0kg/m 1200kg 2.3

33

33

33

===⎯→⎯=

===⎯→⎯=

πππ

ρρ

VrrV

mVVm

oo


1217.0m) 08603.0(

60)s45+3600/s)(2m 1091.0(2

27

2=

×××==

−

ortατ

which is somewhat below the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can still be used, with the understanding that the error involved will be a little more than 2 percent. Then the one-term solution can be written in the form

)1217.0(11

0,0

21

21 65.0

1635.416360 λτλθ −−

∞

∞ ==−−

⎯→⎯=−−

= eAeATTTT

isph

It is determined from Table 4-1 by trial and error that this equation is satisfied when Bi = 30, which corresponds to λ1 13 0372 19898= =. and A . . Then the heat transfer coefficient can be determined from

C. W/m156.9 2 °=°

==⎯→⎯=)m 08603.0(

)30)(C W/m.45.0(

o

o

rkBih

khr

Bi

This value seems to be larger than expected for problems of this kind. This is probably due to the Fourier number being less than 0.2.

(b) The temperature at the surface of the rib is

C 159.5 °=⎯→⎯=

−−

==−−

= −−

∞

∞

),(0222.01635.4

163),(0372.3

)rad 0372.3sin()9898.1(/

)/sin(),(),( )1217.0()0372.3(

1

11

221

trTtrT

err

rreA

TTTtrT

tr

oo

oo

oo

i

ospho λ

λθ τλ

4-29


(c) The maximum possible heat transfer is

kJ2080C)5.4163)(C kJ/kg.1.4)( kg2.3()(max =°−°=−= ∞ ip TTmCQ

Then the actual amount of heat transfer becomes

Q

QQ Q

o sphmax

,

max

sin( ) cos( )( . ) sin( . ) ( . ) cos( . )

( . ).

. ( . )(

= −−

= −−

=

= = =

1 3 1 3 0 65 3 0372 3 0372 3 03723 0372

0 783

0 783 0 783 2080

1 1 1

13 3θ

λ λ λλ

kJ) 1629 kJ

(d) The cooking time for medium-done rib is determined to be

hr 3≅==

×==

=⎯→⎯=−−

⎯→⎯=−−

=

−

−−

∞

∞

min 181s 866,10/s)m 1091.0(m) 08603.0)(1336.0(

1336.0)9898.1(1635.416371

27

22

)0372.3(1

0,0

221

ατ

τθ ττλ

o

isph

rt

eeATTTT

This result is close to the listed value of 3 hours and 20 minutes. The difference between the two results is due to the Fourier number being less than 0.2 and thus the error in the one-term approximation.

Discussion The temperature of the outer parts of the rib is greater than that of the inner parts of the rib after it is taken out of the oven. Therefore, there will be a heat transfer from outer parts of the rib to the inner parts as a result of this temperature difference. The recommendation is logical.

4-44 A rib is roasted in an oven. The heat transfer coefficient at the surface of the rib, the temperature of the outer surface of the rib and the amount of heat transfer when it is well-done are to be determined. The time it will take to roast this rib to medium level is also to be determined.

Assumptions 1 The rib is a homogeneous spherical object. 2 Heat conduction in the rib is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the rib are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The properties of the rib are given to be k = 0.45 W/m.°C, ρ = 1200 kg/m3, Cp = 4.1 kJ/kg.°C, and α = 0.91×10-7 m2/s

Analysis (a) The radius of the rib is determined to be

Oven 163°C

Rib, 4.5°C

m 08603.04

)m 00267.0(343

34

m 00267.0kg/m 1200kg 2.3

33

33

33

===⎯→⎯=

===⎯→⎯=

πππ

ρρ

VrrV

mVVm

oo


1881.0m) 08603.0(

60)s15+3600/s)(4m 1091.0(2

27

2=

×××==

−

ortατ

which is somewhat below the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can still be used, with the understanding that the error involved will be a little more than 2 percent. Then the one-term solution formulation can be written in the form

4-30


)1881.0(11

0,0

21

21 543.0

1635.416377 λτλθ −−

∞

∞ ==−−

⎯→⎯=−−

= eAeATTTT

isph

It is determined from Table 4-1 by trial and error that this equation is satisfied when Bi = 4.3, which corresponds to 7402.1and 4900.2 11 == Aλ . Then the heat transfer coefficient can be determined from.

C. W/m22.5 2 °=°

==⎯→⎯=)m 08603.0(

)3.4)(C W/m.45.0(

o

o

rkBih

khr

Bi

(b) The temperature at the surface of the rib is

C 142.1 °=⎯→⎯=

−−

==−−

= −−

∞

∞

),(132.01635.4

163),(49.2

)49.2sin()7402.1(

/)/sin(),(

),( )1881.0()49.2(

1

11

221

trTtrT

err

rreA

TTTtrT

tr

oo

oo

oo

i

ospho λ

λθ τλ

(c) The maximum possible heat transfer is

kJ2080C)5.4163)(C kJ/kg.1.4)( kg2.3()(max =°−°=−= ∞ ip TTmCQ


kJ 1512===

=−

−=−

−=

kJ)2080)(727.0(727.0

727.0)49.2(

)49.2cos()49.2()49.2sin()543.0(31

)cos()sin(31

max

331

111,

max

QQQ

Qspho λ

λλλθ

(d) The cooking time for medium-done rib is determined to be

hr 4===

×=

ατ

=

=τ⎯→⎯=−−

⎯→⎯=−−

=θ

−

τ−τλ−

∞

∞

min 0.240s 403,14/s)m 1091.0(m) 08603.0)(177.0(

177.0)7402.1(1635.416371

27

22

)49.2(1

0,0

221

o

isph

rt

eeATTTT

This result is close to the listed value of 4 hours and 15 minutes. The difference between the two results is probably due to the Fourier number being less than 0.2 and thus the error in the one-term approximation.

Discussion The temperature of the outer parts of the rib is greater than that of the inner parts of the rib after it is taken out of the oven. Therefore, there will be a heat transfer from outer parts of the rib to the inner parts as a result of this temperature difference. The recommendation is logical.

4-31


4-45 An egg is dropped into boiling water. The cooking time of the egg is to be determined.

Assumptions 1 The egg is spherical in shape with a radius of r0 = 2.75 cm. 2 Heat conduction in the egg is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the egg are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The thermal conductivity and diffusivity of the eggs can be approximated by those of water at room temperature to be k = 0.607 W/m.°C, α = k Cp/ ρ = 0.146×10-6 m2/s (Table A-9).


Water100°C

Egg Ti = 8°C

2.36)CW/m. 607.0(

)m 0275.0)(C.W/m 800( 2=

°°

==k

hrBi o


9925.1and 0533.3 11 == Aλ

Then the Fourier number and the time period become

1633.0)9925.1(100810060 22

1 )0533.3(1

0,0 =⎯→⎯=

−−

⎯→⎯=−−

= −−

∞


isph

which is somewhat below the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can still be used, with the understanding that the error involved will be a little more than 2 percent. Then the length of time for the egg to be kept in boiling water is determined to be

min 14.1==×

== − s 846/s)m 10146.0(

m) 0275.0)(1633.0(26

22

ατ or

t

4-32


4-46 An egg is cooked in boiling water. The cooking time of the egg is to be determined for a location at 1610-m elevation.

Assumptions 1 The egg is spherical in shape with a radius of r0 = 2.75 cm. 2 Heat conduction in the egg is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the egg and heat transfer coefficient are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The thermal conductivity and diffusivity of the eggs can be approximated by those of water at room temperature to be k = 0.607 W/m.°C, α = k Cp/ ρ = 0.146×10-6 m2/s (Table A-9).


Water94.4°C

Egg Ti = 8°C

2.36)CW/m. 607.0(

)m 0275.0)(C.W/m 800( 2=

°°

==k

hrBi o


9925.1and 0533.3 11 == Aλ


1727.0)9925.1(4.9484.9460 22

1 )0533.3(1

0,0 =⎯→⎯=

−−

⎯→⎯=−−

= −−

∞


isph

which is somewhat below the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can still be used, with the understanding that the error involved will be a little more than 2 percent. Then the length of time for the egg to be kept in boiling water is determined to be

min 14.9==×

== − s 895/s)m 10146.0(

m) 0275.0)(1727.0(26

22

ατ or

t

4-33


4-47 A hot dog is dropped into boiling water, and temperature measurements are taken at certain time intervals. The thermal diffusivity and thermal conductivity of the hot dog and the convection heat transfer coefficient are to be determined.

Assumptions 1 Heat conduction in the hot dog is one-dimensional since it is long and it has thermal symmetry about the center line. 2 The thermal properties of the hot dog are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The properties of hot dog available are given to be ρ = 980 kg/m3 and Cp = 3900 J/kg.°C.

Analysis (a) From Fig. 4-14b we have

15.01

1

17.094599488

==

⎪⎪⎭

⎪⎪⎬

⎫

==

=−−

=−−

∞

∞

o

o

o

o

o

hrk

Birr

rr

TTTT

Water94°C

Hot dog The Fourier number is determined from Fig. 4-14a to be

20.047.0

94209459

15.01

2==

⎪⎪⎭

⎪⎪⎬

⎫

=−−

=−−

==

∞

∞ o

i

o

o

rt

TTTT

hrk

Bi ατ

The thermal diffusivity of the hot dog is determined to be

/sm 102.017 27−×===α⎯→⎯=α

s 120m) 011.0)(2.0(2.0

20.022

2 tr

rt o

o

(b) The thermal conductivity of the hot dog is determined from

CW/m. 0.771 °=°×=αρ= − C)J/kg. )(3900kg/m /s)(980m 10017.2( 327pCk

(c) From part (a) we have 15.01==

ohrk

Bi. Then,

m 0.00165m) 011.0)(15.0(15.0 0 === rhk

Therefore, the heat transfer coefficient is

C.W/m 467 2 °=°

=⎯→⎯=m 0.00165

C W/m.771.000165.0 hhk

4-34


4-48 Using the data and the answers given in Prob. 4-43, the center and the surface temperatures of the hot dog 4 min after the start of the cooking and the amount of heat transferred to the hot dog are to be determined.

Assumptions 1 Heat conduction in the hot dog is one-dimensional since it is long and it has thermal symmetry about the center line. 2 The thermal properties of the hot dog are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The properties of hot dog and the convection heat transfer coefficient are given or obtained in P4-47 to be k = 0.771 W/m.°C, ρ = 980 kg/m3, Cp = 3900 J/kg.°C, α = 2.017×10-7 m2/s, and h = 467 W/m2.°C.


Water94°C

Hot dog

66.6)C W/m.771.0(

)m 011.0)(C. W/m467( 2=

°°

==k

hrBi o


5357.1and 0785.2 11 == Aλ


2.04001.0m) 011.0(

s/min) 60min /s)(4m 10017.2(2

27

2 >=××

==−

Ltατ

Then the temperature at the center of the hot dog is determined to be

C 73.8 °=⎯→⎯=

−−

===−−

= −−

∞

∞

00

)4001.0()0785.2(1

0,

2727.0942094

2727.0)5357.1(22

1

TT

eeATTTT

icylo

τλθ

From Table 4-2 we read =0.2194 corresponding to the constant J0 λ1 =2.0785. Then the temperature at the surface of the hot dog becomes

C 89.6 °=⎯→⎯=

−−

===−− −−

∞

∞

),(05982.09420

94),(

05982.0)2194.0()5357.1()/(),( )4001.0()0785.2(

10122

1

trTtrT

errJeATT

TtrT

oo

ooi

o λτλ

The maximum possible amount of heat transfer is

[ ]

J 13,440C)2094)(CJ/kg. 3900)(kg 04657.0()(kg 04657.0m) 125.0(m) 011.0()kg/m 980(

max

2.32

=°−°=−=====

∞TTmCQLrVm

ip

o πρπρ

From Table 4-2 we read = 0.5760 corresponding to the constant J1 λ1 =2.0785. Then the actual heat transfer becomes

kJ 11,409==⎯→⎯=−=λλ

θ−=⎟⎟⎠

⎞⎜⎜⎝

⎛)kJ 440,13(8489.08489.0

0785.25760.0)2727.0(21

)(21

1

11,

maxQ

JQ

Qcylo

cyl

4-35


4-49E Whole chickens are to be cooled in the racks of a large refrigerator. Heat transfer coefficient that will enable to meet temperature constraints of the chickens while keeping the refrigeration time to a minimum is to be determined.

Assumptions 1 The chicken is a homogeneous spherical object. 2 Heat conduction in the chicken is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the chicken are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The properties of the chicken are given to be k = 0.26 Btu/h.ft.°F, ρ = 74.9 lbm/ft3, Cp = 0.98 Btu/lbm.°F, and α = 0.0035 ft2/h.

Analysis The radius of the chicken is determined to be

RefrigeratorT∞ = 5°F

Chicken Ti = 72°F

ft 2517.04

)ft 06676.0(343

34

ft 06676.0lbm/ft 9.74lbm 5

33

33

33

===⎯→⎯=

===⎯→⎯=

πππ

ρρ

VrrV

mVVm

oo

From Fig. 4-15b we have

75.11

1

75.0545535

==

⎪⎪⎭

⎪⎪⎬

⎫

==

=−−

=−−

∞

∞

o

o

o

o

o

hrk

Birr

rx

TTTT

Then the heat transfer coefficients becomes

F.Btu/h.ft 0.590 2 °=°

==ft) 2517.0(75.1

)FBtu/h.ft. 26.0(75.1 orkh

4-36


4-50 A person puts apples into the freezer to cool them quickly. The center and surface temperatures of the apples, and the amount of heat transfer from each apple in 1 h are to be determined.

Assumptions 1 The apples are spherical in shape with a diameter of 9 cm. 2 Heat conduction in the apples is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the apples are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The properties of the apples are given to be k = 0.418 W/m.°C, ρ = 840 kg/m3, Cp = 3.81 kJ/kg.°C, and α = 1.3×10-7 m2/s.


Apple Ti = 20°C

Air T∞ = -15°C 861.0

)C W/m.418.0()m 045.0)(C. W/m8( 2=

°°

==k

hrBi o


2390.1 and 476.1 11 == Aλ


τα

= =× ×

= >−t

r02

713 10 3600 s0 045

0 231 0 2( .( .

. . m / s)(1 h / h) m)

2

2

Then the temperature at the center of the apples becomes

C11.2°=⎯→⎯==−−−−

⎯→⎯=−−

= −−

∞

∞0

)231.0()476.1(01

0, 749.0)239.1(

)15(20)15( 22

1 TeT

eATTTT

ispho

τλθ

The temperature at the surface of the apples is

C2.7°=⎯→⎯=

−−−−

===−−

= −−

∞

∞

),(505.0)15(20

)15(),(

505.0476.1

)rad 476.1sin()239.1(

/)/sin(),(

),( )231.0()476.1(

1

11

221

trTtrT

err

rreA

TTTtrT

tr

oo

oo

oo

i

ospho λ

λθ τλ

The maximum possible heat transfer is

[ ] kJ 76.42C)15(20)CkJ/kg. 81.3)(kg 3206.0()(

kg 3206.0m) 045.0(34)kg/m 840(

34

max

3.33

=°−−°=−=

=⎥⎦⎤

⎢⎣⎡ π=πρ=ρ=

∞TTmCQ

rVm

ip

o


kJ 17.2===

=−

−=λ

λλ−λθ−=

kJ) 76.42)(402.0(402.0

402.0)476.1(

)rad 476.1cos()476.1()rad 476.1sin()749.0(31

)cos()sin(31

max

331

111,

max

QQ

QQ

spho

4-37


4-51 "!PROBLEM 4-51" "GIVEN" T_infinity=-15 "[C]" "T_i=20 [C], parameter to be varied" h=8 "[W/m^2-C]" r_o=0.09/2 "[m]" time=1*3600 "[s]" "PROPERTIES" k=0.513 "[W/m-C]" rho=840 "[kg/m^3]" C_p=3.6 "[kJ/kg-C]" alpha=1.3E-7 "[m^2/s]" "ANALYSIS" Bi=(h*r_o)/k "From Table 4-1 corresponding to this Bi number, we read" lambda_1=1.3525 A_1=1.1978 tau=(alpha*time)/r_o^2 (T_o-T_infinity)/(T_i-T_infinity)=A_1*exp(-lambda_1^2*tau) (T_r-T_infinity)/(T_i-T_infinity)=A_1*exp(-lambda_1^2*tau)*Sin(lambda_1*r_o/r_o)/(lambda_1*r_o/r_o) V=4/3*pi*r_o^3 m=rho*V Q_max=m*C_p*(T_i-T_infinity) Q/Q_max=1-3*(T_o-T_infinity)/(T_i-T_infinity)*(Sin(lambda_1)-lambda_1*Cos(lambda_1))/lambda_1^3

Ti [C] To [C] Tr [C] Q [kJ] 2 -1.658 -5.369 6.861 4 -0.08803 -4.236 7.668 6 1.482 -3.103 8.476 8 3.051 -1.97 9.283 10 4.621 -0.8371 10.09 12 6.191 0.296 10.9 14 7.76 1.429 11.7 16 9.33 2.562 12.51 18 10.9 3.695 13.32 20 12.47 4.828 14.13 22 14.04 5.961 14.93 24 15.61 7.094 15.74 26 17.18 8.227 16.55 28 18.75 9.36 17.35 30 20.32 10.49 18.16

4-38


0 5 10 15 20 25 30-10

-5

0

5

10

15

20

25

Ti [C]

T o [

C]

T0

Tr

0 5 10 15 20 25 306

8

10

12

14

16

18

20

Ti [C]

Q [

kJ]

4-39


4-52 An orange is exposed to very cold ambient air. It is to be determined whether the orange will freeze in 4 h in subfreezing temperatures.

Assumptions 1 The orange is spherical in shape with a diameter of 8 cm. 2 Heat conduction in the orange is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the orange are constant, and are those of water. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The properties of the orange are approximated by those of water at the average temperature of about 5°C, k = 0.571 W/m.°C and (Table A-9). α ρ= = × = × −k Cp/ . / ( ) .0 571 1000 4205 0136 10 6 m / s2


Orange Ti = 15°C

Air T∞ = -15°C Bi

hrk

o= =°

°= ≈

( )( . )( . )

. .15 0 040 571

1051 10 W / m . C m W / m. C

2


λ1 1 15708 12732= =. . and A


2.0224.1m) 04.0(

s/h) 6003 h/s)(4m 10136.0(2

26

2 >=××

==−

Ltατ

Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the temperature at the surface of the oranges becomes

C 5.2 - °=⎯→⎯=−−−−

===−−

= −−

∞

∞

),(0396.0)6(15

)6(),(

0396.05708.1

) rad5708.1sin()2732.1(

/)/sin(),(

),( )224.1()5708.1(

1

11

221

trTtrT

err

rreA

TTTtrT

tr

oo

oo

oo

i

ospho λ

λθ τλ

which is less than 0°C. Therefore, the oranges will freeze.

4-40


4-53 A hot baked potato is taken out of the oven and wrapped so that no heat is lost from it. The time the potato is baked in the oven and the final equilibrium temperature of the potato after it is wrapped are to be determined.

Assumptions 1 The potato is spherical in shape with a diameter of 8 cm. 2 Heat conduction in the potato is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the potato are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The properties of the potato are given to be k = 0.6 W/m.°C, ρ = 1100 kg/m3, Cp = 3.9 kJ/kg.°C, and α = 1.4×10-7 m2/s.

Oven T∞ = 170°C

Potato T0 = 70°C

Analysis (a) The Biot number is

Bihrk

o= =°

°=

( )( . )( . )

.25 0 040 6

167 W / m . C m W / m. C

2


λ1 118777 14113= =. . and A


2.0203.0)4113.1(69.01702517070 22

1 )8777.1(1

0,0 >=⎯→⎯==

−−

⎯→⎯=−−

= −−

∞


isph

The baking time of the potatoes is determined to be

t ro= =×

= =−τα

2

70 203 0 0414 10

2320( . )( .( .

m) m / s)

s2

2 38.7 min

(b) The maximum amount of heat transfer is

kJ 166.76C)25170)(CkJ/kg. 900.3)(kg 295.0()(

kg 295.0m) 04.0(34)kg/m 1100(

34

max

3.33

=°−°=−=

=⎥⎦⎤

⎢⎣⎡ π=πρ=ρ=

∞ ip

o

TTmCQ

rVm


kJ 87.5===

=−

−=−

−=

kJ)76.166)(525.0(525.0

525.0)8777.1(

)8777.1cos()8777.1()8777.1sin()69.0(31

)cos()sin(31

max

331

111,

max

QQQ

Qspho λ

λλλθ

The final equilibrium temperature of the potato after it is wrapped is

C101°=°

+°=+=⎯→⎯−=)CkJ/kg. 9.3)(kg 295.0(

kJ 87.5C25)(p

ieqvieqvp mCQTTTTmCQ

4-41


4-54 The center temperature of potatoes is to be lowered to 6°C during cooling. The cooling time and if any part of the potatoes will suffer chilling injury during this cooling process are to be determined. Assumptions 1 The potatoes are spherical in shape with a radius of r0 = 3 cm. 2 Heat conduction in the potato is one-dimensional in the radial direction because of the symmetry about the midpoint. 3 The thermal properties of the potato are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal conductivity and thermal diffusivity of potatoes are given to be k = 0.50 W/m⋅°C and α = 0.13×10-6 m2/s. Analysis First we find the Biot number:

Bi W / m . C) m0.5 W / m C

2= =

°°

=hrk

0 19 0 03 114( ( . ).

.

From Table 4-1 we read, for a sphere, λ1 = 1.635 and A1 = 1.302. Substituting these values into the one-term solution gives

θ τλ τ τ0 1

63512 26 2

25 21302 0 753=

−−

= →−−

= →∞

∞

− −T TT T

A e eo

i =. .(1. )

Potato Ti = 25°C

Air 2°C

4 m/s

which is greater than 0.2 and thus the one-term solution is applicable. Then the cooling time becomes

h 1.45==×

==⎯→⎯= s 5213s/m 100.13

m) 03.0)(753.0( 26-

220

20 α

τατr

tr

t

The lowest temperature during cooling will occur on the surface (r/r0 = 1), and is determined to be

001

0010

001

0010

0

01

011 /

)/sin(=

/)/sin()(

/

)/sin()( 21

rrrr

TTTT

rrrr

TTTrT

rrrr

eATTTrT

iii λλ

λλ

θλλτλ

∞

∞

∞

∞−

∞

∞

−−

=−−

→=−−

Substituting, C4.44=)( 1.635

rad)635.1sin(225

26225

2)(0

0 °⎯→⎯⎟⎠⎞

⎜⎝⎛

−−

=−−

rTrT

which is above the temperature range of 3 to 4 °C for chilling injury for potatoes. Therefore, no part of the potatoes will experience chilling injury during this cooling process. Alternative solution We could also solve this problem using transient temperature charts as follows:

15a)-4 (Fig. 75.0t

=174.0

22526

877.0m)C)(0.03.W/m(19

CW/m.50.0Bi1

2

o2

o

=

⎪⎪

⎭

⎪⎪

⎬

⎫

=−−

=−−

===

∞

∞ o

i

o

o

rTTTThrk

ατ

Therefore, tr

ss= =

×= ≅−

τα

02 2

60 75 0 03

013 105192

( . )( . ). /m 2 1.44 h

The surface temperature is determined from

10 877

10 60

0

0

BiFig. 4 -15b)

= =

=

⎫

⎬⎪⎪

⎭⎪⎪

−−

=∞

∞

khr

rr

T r TT T

.( )

. (

which gives T T T Tsurface o= + − = + − =∞ ∞0 6 2 0 6 6 2 4 4. ( ) . ( ) . º C The slight difference between the two results is due to the reading error of the charts.

4-42


4-55E The center temperature of oranges is to be lowered to 40°F during cooling. The cooling time and if any part of the oranges will freeze during this cooling process are to be determined. Assumptions 1 The oranges are spherical in shape with a radius of r0 =1.25 in = 0.1042 ft. 2 Heat conduction in the orange is one-dimensional in the radial direction because of the symmetry about the midpoint. 3 The thermal properties of the orange are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal conductivity and thermal diffusivity of oranges are given to be k = 0.26 Btu/h⋅ft⋅°F and α = 1.4×10-6 ft2/s. Analysis First we find the Biot number:

Bi Btu / h.ft . F) ft0.26 Btu / h.ft. C

2= =

°°

=hrk

0 4 6 125 12 1843( . ( . / ) .

From Table 4-1 we read, for a sphere, λ1 = 1.9569 and A1 = 1.447. Substituting these values into the one-term solution gives

θ τλ τ τ0 1

956912 240 25

78 251447 0 426=

−−

= →−−

= →∞

∞

− −T TT T

A e eo

i =. .(1. )

Orange D = 2.5 in

85% water Ti = 78°F

Air 25°F 1 ft/s


τ α τα

= → = =×

= =t

rt r

02

02 0 426 125 12 3302 ft)

1.4 10 ft s s

2

-6 2( . )( . /

/55.0 min


T r T

T TA e r r

r rT r T

T Tr r

r rT TT T

r rr ri i i

( ) sin( / )/

( ) sin( / )/

sin( / )/

−−

= →−

−=

−−

∞

∞

− ∞

∞

∞

∞1

1 0

0

00

1 0 0

0 0

0 1 0

0 012λ τ 0λ

λθ λ

λλ

λ1 1 =

1

Substituting, F32.1=)( 1.9569

rad)9569.1sin(25782540

257825)(

00 °⎯→⎯⎟

⎠⎞

⎜⎝⎛

−−

=−−

rTrT

which is above the freezing temperature of 31 °C for oranges . Therefore, no part of the oranges will freeze during this cooling process. Alternative solution We could also solve this problem using transient temperature charts as follows:

15a)-4(Fig. 43.0283.0

25782540

543.0ft) F)(1.25/12.ºBtu/h.ft (4.6

FBtu/h.ft.º0.261

2

2=

α=τ

⎪⎪⎭

⎪⎪⎬

⎫

=−−

=−−

===

∞

∞ o

i

o

o

rt

TTTT

rhk

Bi

Therefore, min 55.5s 3333/sft101.4

5/12ft)(0.43)(1.226

22

==×

== −ατ ort

The lowest temperature during cooling will occur on the surface (r/r0 =1) of the oranges is determined to be

15b)4(Fig. 0.45)(

1

543.01

−=−−

⎪⎪⎭

⎪⎪⎬

⎫

=

==

∞

∞

TTTrT

rr

rhk

Bi

o

o

o

which gives T T T Tsurface o= + − = + − =∞ ∞0 45 25 0 45 40 25. ( ) . ( ) 31.8º F

The slight difference between the two results is due to the reading error of the charts.

4-43


4-56 The center temperature of a beef carcass is to be lowered to 4°C during cooling. The cooling time and if any part of the carcass will suffer freezing injury during this cooling process are to be determined. Assumptions 1 The beef carcass can be approximated as a cylinder with insulated top and base surfaces having a radius of r0 = 12 cm and a height of H = 1.4 m. 2 Heat conduction in the carcass is one-dimensional in the radial direction because of the symmetry about the centerline. 3 The thermal properties of the carcass are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal conductivity and thermal diffusivity of carcass are given to be k = 0.47 W/m⋅°C and α = 0.13×10-6 m2/s. Analysis First we find the Biot number:

Bi W / m . C) m0.47 W / m C

2= =

°°

=hrk

0 22 012 562( ( . ).

.

From Table 4-1 we read, for a cylinder, λ1 = 2.027 and A1 = 1.517. Substituting these values into the one-term solution gives

θ τλ τ τ0 1

2 02712 24 6

37 61517 0 456=

−−

= →− −− −

= →∞

∞

− −T TT T

A e eo

i =( )

( ). .( . )

Beef 37°C

Air -6°C

1.8 m/s


h 14.0==×

==→= s 558,50s/m 100.13

m) 12.0)(456.0( 26-

220

20 α

τατr

tr

t


T r T

T TA e J r r T r T

T TJ r r T T

T TJ r r

i i i

( ) ( / ) ( ) ( / ) ( /−−

= → )−−

=−−

∞

∞

− ∞

∞

∞

∞1 0 1 0

00 0 1 0

00 1 0 01

2λ τ λ θ λ = λ

Substituting, C-3.9=)( 0485.02084.02326.0)()6(37)6(4

)6(37)6()(

0100 °⎯→⎯=×=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−−−

=−−−−

rTJrT

λ

which is below the freezing temperature of -1.7 °C. Therefore, the outer part of the beef carcass will freeze during this cooling process. Alternative solution We could also solve this problem using transient temperature charts as follows:

)14a4(Fig. 0.423.0

)6(37)6(4

178.0m) C)(0.12W/m².º (22

CW/m.º0.471

2−=

α=τ

⎪⎪⎭

⎪⎪⎬

⎫

=−−−−

=−−

===

∞

∞ o

i

o

o

rt

TTTT

rhk

Bi

Therefore, h12.3s44,308/sm100.13

m) (0.4)(0.1226

22

≅=×

== −ατ ort


14b)4(Fig. 17.0)(

1

178.01

−=−−

⎪⎪⎭

⎪⎪⎬

⎫

=

==

∞

∞

TTTrT

rr

rhk

Bi

o

o

o

which gives T T T Tsurface o= + − C= − + − − = −∞ ∞017 6 017 4 6 4 3. ( ) . [ ( )] . º

The difference between the two results is due to the reading error of the charts.

4-44


4-57 The center temperature of meat slabs is to be lowered to -18°C during cooling. The cooling time and the surface temperature of the slabs at the end of the cooling process are to be determined. Assumptions 1 The meat slabs can be approximated as very large plane walls of half-thickness L = 11.5 cm. 2 Heat conduction in the meat slabs is one-dimensional because of the symmetry about the centerplane. 3 The thermal properties of the meat slabs are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). 6 The phase change effects are not considered, and thus the actual cooling time will be much longer than the value determined. Properties The thermal conductivity and thermal diffusivity of meat slabs are given to be k = 0.47 W/m⋅°C and α = 0.13×10-6 m2/s. These properties will be used for both fresh and frozen meat. Analysis First we find the Biot number: Air

-30°C 1.4 m/s

Meat 7°C

Bi W / m . C) m0.47 W / m C

2= =

°°

=hrk

0 20 0115 4 89( ( . ).

.

From Table 4-1 we read, for a plane wall, λ1 = 1.308 and A1=1.239. Substituting these values into the one-term solution gives

θ τλ τ τ0 1

30812 218 30

7 301239 0 783=

−−

= →− − −

− −= →∞

∞

− −T TT T

A e eo

i =( )

( ). .(1. )


τ α τα

= → = =×

= =t

Lt L

2

2 0 783 0115 79 650 m)0.13 10 m s

s2

-6 2( . )( .

/, 22.1 h

The lowest temperature during cooling will occur on the surface (x/L = 1), and is determined to be

T x T

T TA e x L

T L TT T

L LT TT Ti i

( )cos( / )

( )cos( / ) cos( )

−−

= →−

−=

−−

∞

∞

− ∞

∞

∞

∞1 1 0 1

01

12λ τ λ θ λ =

iλ

Substituting,

C26.9°−=⎯→⎯=×=⎟⎟⎠

⎞⎜⎜⎝

⎛−−−−−

=−−−−

)( 08425.02598.03243.0)cos()30(7

)30(18)30(7

)30()(1 LT

LTλ

which is close the temperature of the refrigerated air. Alternative solution We could also solve this problem using transient temperature charts as follows:

13a)4(Fig. 75.0324.0

)30(7)30(18

0.204m) C)(0.115W/m².º (20

CW/m.º0.471

2−=

α=τ

⎪⎪⎭

⎪⎪⎬

⎫

=−−−−−

=−−

===

∞

∞ Lt

TTTThLk

Bi

i

o

Therefore, tro= =

×= ≅−

τα

276 300(0.75)(0.115 m)

0.13 10 m / ss 21.2 h

2

6 2 ,


13b)4(Fig. 22.0)(

1

204.01

−=−−

⎪⎪⎭

⎪⎪⎬

⎫

=

==

∞

∞

TTTxT

Lx

hLk

Bio

which gives T T T Tsurface o= + − = − + − − − = −∞ ∞0 22 30 0 22 18 30. ( ) . [ ( )] 27.4º C

The slight difference between the two results is due to the reading error of the charts.

4-45


4-58E The center temperature of meat slabs is to be lowered to 36°F during 12-h of cooling. The average heat transfer coefficient during this cooling process is to be determined. Assumptions 1 The meat slabs can be approximated as very large plane walls of half-thickness L = 3-in. 2 Heat conduction in the meat slabs is one-dimensional because of symmetry about the centerplane. 3 The thermal properties of the meat slabs are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The thermal conductivity and thermal diffusivity of meat slabs are given to be k = 0.26 Btu/h⋅ft⋅°F and α=1.4×10-6 ft2/s.

Air 23°F

Meat 50°F

Analysis The average heat transfer coefficient during this cooling process is determined from the transient temperature charts for a flat plate as follows:

)13a4(Fig. 7.01

481.023502336

968.0ft)² (3/12

s) 3600ft²/s)(1210(1.4²

6

−=

⎪⎪⎭

⎪⎪⎬

⎫

=−−

=−−

=××

=α

=τ

∞

∞

−

BiTTTT

Lt

i

o

Therefore,

FºBtu/h.ft². 1.5===ft (3/12)F)(1/0.7)Btu/h.ft.º(0.26

LkBih

Discussion We could avoid the uncertainty associated with the reading of the charts and obtain a more accurate result by using the one-term solution relation for an infinite plane wall, but it would require a trial and error approach since the Bi number is not known.

4-46


4-59 Chickens are to be chilled by holding them in agitated brine for 2.5 h. The center and surface temperatures of the chickens are to be determined, and if any part of the chickens will freeze during this cooling process is to be assessed.

Assumptions 1 The chickens are spherical in shape. 2 Heat conduction in the chickens is one-dimensional in the radial direction because of symmetry about the midpoint. 3 The thermal properties of the chickens are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). 6 The phase change effects are not considered, and thus the actual the temperatures will be much higher than the values determined since a considerable part of the cooling process will occur during phase change (freezing of chicken).

Properties The thermal conductivity, thermal diffusivity, and density of chickens are given to be k = 0.45 W/m⋅°C, α = 0.13×10-6 m2/s, and ρ = 950 kg/ m3. These properties will be used for both fresh and frozen chicken.

Analysis We first find the volume and equivalent radius of the chickens:

cm³1789m³)g/(0.95g/c1700/ === ρmV

Brine -10°C

Chicken Ti = 15°C

m 0753.0cm 53.7cm³ 178943

43 3/13/1

==⎟⎠⎞

⎜⎝⎛

π=⎟

⎠⎞

⎜⎝⎛

π= Vro

Then the Biot and Fourier numbers become

0.2063

m)² (0.0753s) 3600m²/s)(2.510(0.13

6.73C.W/m 0.45

)m 0753.0(C).W/m 440(Bi

6

20

20

=××

==

=°

°==

−

rtk

hr

ατ

Note that τ = >0 207 0 2. . , and thus the one-term solution is applicable. From Table 4-1 we read, for a sphere, λ1 = 3.094 and A1 = 1.998. Substituting these values into the one-term solution gives

C3.1°−⎯→⎯=−−−−

→=−−

= −−

∞

∞ = 0.277=998.1)10(15)10(

0)2063.0()094.3(0

1022

1 TeT

eATTTT

i

o τλθ


T r T

T TA e r r

r rT r T

T Tr r

r rT TT T

r rr ri i i

( ) sin( / )/

( ) sin( / )/

sin( / )/

−−

= →−

−=

−−

∞

∞

− ∞

∞

∞

∞1

1 0

0

00

1 0 0

0 0

0 1 0

0 012λ τ 0λ

λθ λ

λλ

λ1 1 =

1

Substituting, C9.9°−=→=−−−−

)( 3.094

rad)094.3sin(0.277 )10(15

)10()(0

0 rTrT

The entire chicken will freeze during this process since the freezing point of chicken is -2.8°C, and even the center temperature of chicken is below this value.

Discussion We could also solve this problem using transient temperature charts, but the data in this case falls at a point on the chart which is very difficult to read:

15)4(Fig. ?? 30.0....15.00.0136

)C)(0.0753mW/m².º(440CW/m.º0.451

0.206m)² (0.0753

s) 3600m²/s)(2.510(0.13 6

20 −=

−−

⎪⎪

⎭

⎪⎪

⎬

⎫

===

=××

=α

=τ

∞

∞

−

TTTT

rhk

Bi

rt

i

o

o

4-47


Transient Heat Conduction in Semi-Infinite Solids

4-60C A semi-infinite medium is an idealized body which has a single exposed plane surface and extends to infinity in all directions. The earth and thick walls can be considered to be semi-infinite media.

4-61C A thick plane wall can be treated as a semi-infinite medium if all we are interested in is the variation of temperature in a region near one of the surfaces for a time period during which the temperature in the mid section of the wall does not experience any change.

4-62C The total amount of heat transfer from a semi-infinite solid up to a specified time t0 can be determined by integration from

∫ ∞−=ot

dtTtTAhQ0

]),0([

where the surface temperature T(0,t) is obtained from Eq. 4-22 by substituting x = 0.

4-63 The water pipes are buried in the ground to prevent freezing. The minimum burial depth at a particular location is to be determined.

Assumptions 1 The temperature in the soil is affected by the thermal conditions at one surface only, and thus the soil can be considered to be a semi-infinite medium with a specified surface temperature. 2 The thermal properties of the soil are constant.

Properties The thermal properties of the soil are given to be k = 0.35 W/m.°C and α = 0.15×10-6 m2/s.

Analysis The length of time the snow pack stays on the ground is

t = = ×(60 days)(24 hr / days)(3600 s / hr) 5.184 10 s6 Ts =-8°C

Soil Ti = 8°C

The surface is kept at -18°C at all times. The depth at which freezing at 0°C occurs can be determined from the analytical solution,

⎟⎠⎞

⎜⎝⎛=⎯→⎯

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

××=

−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛

α=

−−

−

7636.1444.0

)s 10184.5)(/sm 1015.0(288

80

),(

626

xerfc

xerfc

txerfc

TTTtxT

is

i

Water pipe

Then from Table 4-3 we get m 0.934=⎯→⎯= xx 5297.07636.1

Discussion The solution could also be determined using the chart, but it would be subject to reading error.

4-48


4-64 An area is subjected to cold air for a 10-h period. The soil temperatures at distances 0, 10, 20, and 50 cm from the earth’s surface are to be determined.

Assumptions 1 The temperature in the soil is affected by the thermal conditions at one surface only, and thus the soil can be considered to be a semi-infinite medium with a specified surface temperature. 2 The thermal properties of the soil are constant.


Analysis The one-dimensional transient temperature distribution in the ground can be determined from

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟⎠

⎞⎜⎜⎝

⎛+−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−

∞ kth

txerfc

kth

khx

txerfc

TTTtxT

i

i αα

αα 2

exp2

),(2

2

Soil Ti =10°C

Winds T∞ =-10°C

where

11387.33

7.33C W/m.0.9

)s 360010)(s/m 10(1.6C). W/m40(

22

2

2

2-52

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

=°

××°=

kth

kth

kth

αα

α

Then we conclude that the last term in the temperature distribution relation above must be zero regardless of x despite the exponential term tending to infinity since (1) 4for 0)( >→ ξξerfc (see Table 4-3) and (2) the term has to remain less than 1 to have physically meaningful solutions. That is,

03.332

1138exp2

exp2

2≅

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞

⎜⎝⎛ +=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟⎠

⎞⎜⎜⎝

⎛+

txerfc

khx

kth

txerfc

kth

khx

αα

αα

Therefore, the temperature distribution relation simplifies to

⎟⎟⎠

⎞⎜⎜⎝

⎛−+=→⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−

∞∞ t

xerfcTTTtxTt

xerfcTT

TtxTii

i

i

αα 2)(),(

2

),(

Then the temperatures at 0, 10, 20, and 50 cm depth from the ground surface become

x = 0: C10°−==×−+=−+=⎟⎟⎠

⎞⎜⎜⎝

⎛−+= ∞∞∞∞ TTTTerfcTTT

terfcTTTT iiiiii 1)()0()(

20)()h 10,0(α

x = 0.1m:

C8.5°−=×−=−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

××−−+=

−

9257.02010)066.0(2010

)s/h 3600h 10)(/sm 106.1(2

m 1.0)1010(10)h 10,m 1.0(25

erfc

erfcT

x = 0.2 m:

C7.0°−=×−=−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

××−−+=

−

8519.02010)132.0(2010

)s/h 3600h 10)(/sm 106.1(2

m 2.0)1010(10)h 10,m 2.0(25

erfc

erfcT

x = 0.5 m:

C2.8°−=×−=−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

××−−+=

−

6418.02010)329.0(2010

)s/h 3600h 10)(/sm 106.1(2

m 5.0)1010(10)h 10,m 5.0(25

erfc

erfcT

4-49


4-65 "!PROBLEM 4-65" "GIVEN" T_i=10 "[C]" T_infinity=-10 "[C]" h=40 "[W/m^2-C]" time=10*3600 "[s]" "x=0.1 [m], parameter to be varied" "PROPERTIES" k=0.9 "[W/m-C]" alpha=1.6E-5 "[m^2/s]" "ANALYSIS" (T_x-T_i)/(T_infinity-T_i)=erfc(x/(2*sqrt(alpha*time)))-exp((h*x)/k+(h^2*alpha*time)/k^2)*erfc(x/(2*sqrt(alpha*time))+(h*sqrt(alpha*time)/k))

x [m] Tx [C] 0 -9.666

0.05 -8.923 0.1 -8.183

0.15 -7.447 0.2 -6.716

0.25 -5.993 0.3 -5.277

0.35 -4.572 0.4 -3.878

0.45 -3.197 0.5 -2.529

0.55 -1.877 0.6 -1.24

0.65 -0.6207 0.7 -0.01894

0.75 0.5643 0.8 1.128

0.85 1.672 0.9 2.196

0.95 2.7 1 3.183

4-50


0 0.2 0.4 0.6 0.8 1-10

-8

-6

-4

-2

0

2

4

x [m]

T x [

C]

4-51


4-66 The walls of a furnace made of concrete are exposed to hot gases at the inner surfaces. The time it will take for the temperature of the outer surface of the furnace to change is to be determined.

Assumptions 1 The temperature in the wall is affected by the thermal conditions at inner surfaces only and the convection heat transfer coefficient inside is given to be very large. Therefore, the wall can be considered to be a semi-infinite medium with a specified surface temperature of 1800°F. 2 The thermal properties of the concrete wall are constant.

Properties The thermal properties of the concrete are given to be k = 0.64 Btu/h.ft.°F and α = 0.023 ft2/h.

Analysis The one-dimensional transient temperature distribution in the wall for that time period can be determined from

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−−

txerfc

TTTtxT

is

i

α2

),(

But,

)85.2(00006.0 00006.0701800701.70),(

erfcTT

TtxT

is

i =→=−−

=−−

(Table 4-3)

1800°F 70°F

L =1.5 ft

Q&

Wall

Therefore,

min 181==×

=×

=⎯→⎯= h 01.3)/hft 023.0()85.2(4

ft) 5.1()85.2(4

85.22 22

2

2

2

ααxt

tx

4-52


4-67 A thick wood slab is exposed to hot gases for a period of 5 minutes. It is to be determined whether the wood will ignite.

Assumptions 1 The wood slab is treated as a semi-infinite medium subjected to convection at the exposed surface. 2 The thermal properties of the wood slab are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface.

Properties The thermal properties of the wood are k = 0.17 W/m.°C and α = 1.28×10-7 m2/s.

Analysis The one-dimensional transient temperature distribution in the wood can be determined from

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟⎠

⎞⎜⎜⎝

⎛+−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−

∞ kth

txerfc

kth

khx

txerfc

TTTtxT

i

i αα

αα 2

exp2

),(2

2

where

L=0.3 m

Wood Slab

Ti = 25°C Hot

gases T∞ = 550°C

x0

628.1276.1

276.1C W/m.0.17

)s 605)(s/m 10(1.28C). W/m35(

22

2

2

2-72

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

=°

××°=

kth

kth

kth

αα

α

Noting that x = 0 at the surface and using Table 4-3 for erfc values,

630.0

)0727.0)(0937.5(1

)276.10()628.10exp()0(25550

25),(

=−=

++−=−− erfcerfctxT

Solving for T(x, t) gives

C356°=),( txT

which is less than the ignition temperature of 450°C. Therefore, the wood will not ignite.

4-53


4-68 The outer surfaces of a large cast iron container filled with ice are exposed to hot water. The time before the ice starts melting and the rate of heat transfer to the ice are to be determined.

Assumptions 1 The temperature in the container walls is affected by the thermal conditions at outer surfaces only and the convection heat transfer coefficient outside inside is given to be very large. Therefore, the wall can be considered to be a semi-infinite medium with a specified surface temperature. 2 The thermal properties of the wall are constant.

Properties The thermal properties of the cast iron are given to be k = 52 W/m.°C and α = 1.70×10-5 m2/s.

Analysis The one-dimensional transient temperature distribution in the wall for that time period can be determined from

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−−

txerfc

TTTtxT

is

i

α2

),(

Ice, 0°C

Ice chest Hot water

60°C But,

)225.2(00167.0 00167.006001.0),(

erfcTT

TtxT

is

i =→=−−

=−−

(Table 4-3)

Therefore,

s 7.4=×

=×

=⎯→⎯= − )/sm 107.1()225.2(4m) 05.0(

)225.2(4225.2

2 252

2

2

2

ααxt

tx

The rate of heat transfer to the ice when steady operation conditions are reached can be determined by applying the thermal resistance network concept as

W28,990=°−

=−

=

°=++=++=

°≅×∞

==

°=×°

==

°=×°

==

C/W0.00207C)060(

C/W00207.0000040.000167.0

C/W0)m 2)(1.2(

11

C/W00040.0)m 2C)(1.2 W/m.(52

m 05.0

C/W00167.0)m 2C)(1.2. W/m(250

11

o12

2,1,

20,

2

22,

total

convwallconvtotal

oconv

wall

iiconv

RTT

Q

RRRRAh

R

kALR

AhR

&

Rconv, i Rwall Rconv ,o

T1 T2

4-54


Transient Heat Conduction in Multidimensional Systems

4-69C The product solution enables us to determine the dimensionless temperature of two- or three-dimensional heat transfer problems as the product of dimensionless temperatures of one-dimensional heat transfer problems. The dimensionless temperature for a two-dimensional problem is determined by determining the dimensionless temperatures in both directions, and taking their product.

4-70C The dimensionless temperature for a three-dimensional heat transfer is determined by determining the dimensionless temperatures of one-dimensional geometries whose intersection is the three dimensional geometry, and taking their product.

4-71C This short cylinder is physically formed by the intersection of a long cylinder and a plane wall. The dimensionless temperatures at the center of plane wall and at the center of the cylinder are determined first. Their product yields the dimensionless temperature at the center of the short cylinder.

4-72C The heat transfer in this short cylinder is one-dimensional since there is no heat transfer in the axial direction. The temperature will vary in the radial direction only.

4-55


4-73 A short cylinder is allowed to cool in atmospheric air. The temperatures at the centers of the cylinder and the top surface as well as the total heat transfer from the cylinder for 15 min of cooling are to be determined.

Assumptions 1 Heat conduction in the short cylinder is two-dimensional, and thus the temperature varies in both the axial x- and the radial r- directions. 2 The thermal properties of the cylinder are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The thermal properties of brass are given to be ,ρ = 8530 kg / m3 CkJ/kg 389.0 °⋅=pC ,

, and C W/m110 °⋅=k α = × −3 39 10 5. m / s2 .

Analysis This short cylinder can physically be formed by the intersection of a long cylinder of radius D/2 = 4 cm and a plane wall of thickness 2L = 15 cm. We measure x from the midplane.

(a) The Biot number is calculated for the plane wall to be

D0 = 8 cm

L = 15 cm

z

Brass cylinder Ti = 150°C

r

Air T∞ = 20°C

02727.0)C W/m.110(

)m 075.0)(C. W/m40( 2=

°°

==k

hLBi


0050.1 and 164.0 11 == Aλ


τ α= =

× ×= >

−tL2

5339 100 075

5424 0 2( .( .

. . m / s)(15 min 60 s / min) m)

2

2

Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the dimensionless temperature at the center of the plane wall is determined from

869.0)0050.1( )424.5()164.0(1

0,

221 ===

−−

=θ −τλ−

∞

∞ eeATTTT

iwallo

We repeat the same calculations for the long cylinder,

01455.0)C W/m.110(

)m 04.0)(C. W/m40( 20 =

°°

==k

hrBi

λ1 101704 10038= =. . and A

τα

= =× ×

= >−t

ro2

5339 100 04

19 069 0 2( .( .

. . m / s)(15 60 s) m)

2

2

θ λ τo cyl

i

T TT T

A e e,( . ) ( . )( . ) .=

−−

= = =∞

∞

− −01

0 1704 19 06912 2

10038 0577

Then the center temperature of the short cylinder becomes

C85.1°=⎯→⎯=−−

=×=θ×θ=⎥⎦

⎤⎢⎣

⎡−

−

∞

∞

),0,0(501.020150

20),0,0(

501.0577.0869.0),0,0(

,,

tTtT

TTTtT

cylowallo

cylindershorti

4-56


(b) The center of the top surface of the cylinder is still at the center of the long cylinder (r = 0), but at the outer surface of the plane wall (x = L). Therefore, we first need to determine the dimensionless temperature at the surface of the wall.

857.0)164.0cos()0050.1()/cos(),(

),( )424.5()164.0(11

221 ===

−−

= −−

∞

∞ eLLeATT

TtxTtL

iwall λθ τλ

Then the center temperature of the top surface of the cylinder becomes

C84.2°=⎯→⎯=−−

=×=×=⎥⎦

⎤⎢⎣

⎡−

−

∞

∞

),0,(494.020150

20),0,(

494.0577.0857.0),(),0,(

,

tLTtLT

tLTT

TtLTcylowall

cylindershorti

θθ

(c) We first need to determine the maximum heat can be transferred from the cylinder

[ ]

kJ 325C)20150)(CkJ/kg. 389.0)(kg 43.6()(kg 43.6m) 15.0(m) 04.0()kg/m 8530(

max

2.32

=°−°=−==π=ρπ=ρ=

∞TTmCQLrVm

ip

o

Then we determine the dimensionless heat transfer ratios for both geometries as

135.0164.0

)164.0sin()869.0(1)sin(

11

1,

max=−=

λλ

θ−=⎟⎟⎠

⎞⎜⎜⎝

⎛wallo

wallQQ

427.01704.00846.0)577.0(21

)(21

1

11,

max=−=−=⎟⎟

⎠

⎞⎜⎜⎝

⎛λλ

θJ

QQ

cylocyl

The heat transfer ratio for the short cylinder is

504.0)135.01)(427.0(135.01maxmaxmaxmax

=−+=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛

wallplane

cylinderlong

wallplane

cylindershort Q

QQ

QQ

QQ

Q

Then the total heat transfer from the short cylinder during the first 15 minutes of cooling becomes

kJ 164=== kJ) 325)(504.0(503.0 maxQQ

4-57


4-74 "!PROBLEM 4-74" "GIVEN" D=0.08 "[m]" r_o=D/2 height=0.15 "[m]" L=height/2 T_i=150 "[C]" T_infinity=20 "[C]" h=40 "[W/m^2-C]" "time=15 [min], parameter to be varied" "PROPERTIES" k=110 "[W/m-C]" rho=8530 "[kg/m^3]" C_p=0.389 "[kJ/kg-C]" alpha=3.39E-5 "[m^2/s]" "ANALYSIS" "(a)" "This short cylinder can physically be formed by the intersection of a long cylinder of radius r_o and a plane wall of thickness 2L" "For plane wall" Bi_w=(h*L)/k "From Table 4-1 corresponding to this Bi number, we read" lambda_1_w=0.2282 "w stands for wall" A_1_w=1.0060 tau_w=(alpha*time*Convert(min, s))/L^2 theta_o_w=A_1_w*exp(-lambda_1_w^2*tau_w) "theta_o_w=(T_o_w-T_infinity)/(T_i-T_infinity)" "For long cylinder" Bi_c=(h*r_o)/k "c stands for cylinder" "From Table 4-1 corresponding to this Bi number, we read" lambda_1_c=0.1704 A_1_c=1.0038 tau_c=(alpha*time*Convert(min, s))/r_o^2 theta_o_c=A_1_c*exp(-lambda_1_c^2*tau_c) "theta_o_c=(T_o_c-T_infinity)/(T_i-T_infinity)" (T_o_o-T_infinity)/(T_i-T_infinity)=theta_o_w*theta_o_c "center temperature of short cylinder" "(b)" theta_L_w=A_1_w*exp(-lambda_1_w^2*tau_w)*Cos(lambda_1_w*L/L) "theta_L_w=(T_L_w-T_infinity)/(T_i-T_infinity)" (T_L_o-T_infinity)/(T_i-T_infinity)=theta_L_w*theta_o_c "center temperature of the top surface" "(c)" V=pi*r_o^2*(2*L) m=rho*V Q_max=m*C_p*(T_i-T_infinity) Q_w=1-theta_o_w*Sin(lambda_1_w)/lambda_1_w "Q_w=(Q/Q_max)_w" Q_c=1-2*theta_o_c*J_1/lambda_1_c "Q_c=(Q/Q_max)_c" J_1=0.0846 "From Table 4-2, at lambda_1_c" Q/Q_max=Q_w+Q_c*(1-Q_w) "total heat transfer"

4-58


time [min] To,o [C] TL,o [C] Q [kJ]

5 119.3 116.8 80.58 10 95.18 93.23 140.1 15 76.89 75.42 185.1 20 63.05 61.94 219.2 25 52.58 51.74 245 30 44.66 44.02 264.5 35 38.66 38.18 279.3 40 34.12 33.75 290.5 45 30.69 30.41 298.9 50 28.09 27.88 305.3 55 26.12 25.96 310.2 60 24.63 24.51 313.8

0 10 20 30 40 50 6020

40

60

80

100

120

50

100

150

200

250

300

350

time [min]

T o,o

[C

]

Q [

kJ]

temperature

heat

4-59


0 10 20 30 40 50 6020

40

60

80

100

120

time [min]

T L,o

[C

]

4-60


4-75 A semi-infinite aluminum cylinder is cooled by water. The temperature at the center of the cylinder 10 cm from the end surface is to be determined. Assumptions 1 Heat conduction in the semi-infinite cylinder is two-dimensional, and thus the temperature varies in both the axial x- and the radial r- directions. 2 The thermal properties of the cylinder are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of aluminum are given to be k = 237 W/m.°C and α = 9.71×10-5m2/s. Analysis This semi-infinite cylinder can physically be formed by the intersection of a long cylinder of radius ro = D/2 = 7.5 cm and a semi-infinite medium. The dimensionless temperature 5 cm from the surface of a semi-infinite medium is first determined from

8951.01049.01),(

1049.0)7308.0)(0468.1(8699.0)2433.0()0458.0exp()1158.0(

237)608)(1071.9()140(

)608)(1071.9(2

05.0

)237()608)(1071.9()140(

237)05.0)(140(exp

)608)(1071.9(2

05.0

2exp

),(

inf

5

5

2

52

5

2

2

=−=−−

=θ

=−=−=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ ××+

×××

⎟⎟⎠

⎞⎜⎜⎝

⎛ ××+−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

××=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ α+

α⎟⎟⎠

⎞⎜⎜⎝

⎛ α+−⎟⎟

⎠

⎞⎜⎜⎝

⎛

α=

−−

∞

∞−

−

−

−

−

∞

TTTtxT

erfcerfc

erfc

erfc

kth

txerfc

kth

khx

txerfc

TTTtxT

isemi

i

i

The Biot number is calculated for the long cylinder to be

0443.0)C W/m.237(

)m 075.0)(C. W/m140( 2=

°°

==k

hrBi o

The constants 11 and Aλ corresponding to this Biot number are, from Table 4-1, λ1 = 0.2948 and A1 = 1.0110 The Fourier number is

τα

= =× ×

= >−t

ro2

59 71 100 075

8 286 0 2( .( .

. . m / s)(8 60 s) m)

2

2 D0 = 15 cm

z

Semi-infinite cylinder

Ti = 150°C r

Water T∞ = 10°C

Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the dimensionless temperature at the center of the plane wall is determined from

4921.0)0110.1( )286.8()2948.0(1

0,

221 ===

−−

=θ −τλ−

∞

∞ eeATTTT

icylo

The center temperature of the semi-infinite cylinder then becomes

C71.7°=⎯→⎯=⎥⎦

⎤⎢⎣

⎡−−

=×=θ×θ=⎥⎦

⎤⎢⎣

⎡−

−

−

−−∞

∞

),0,(4405.010150

10),0,(

4405.04921.08951.0),(),0,(

cylinderinfinitesemi

,infsemi

cylinderinfinitesemi

txTtxT

txTT

TtxTcylo

i

4-61


4-76E A hot dog is dropped into boiling water. The center temperature of the hot dog is do be determined by treating hot dog as a finite cylinder and also as an infinitely long cylinder. Assumptions 1 When treating hot dog as a finite cylinder, heat conduction in the hot dog is two-dimensional, and thus the temperature varies in both the axial x- and the radial r- directions. When treating hot dog as an infinitely long cylinder, heat conduction is one-dimensional in the radial r- direction. 2 The thermal properties of the hot dog are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The thermal properties of the hot dog are given to be k = 0.44 Btu/h.ft.°F, ρ = 61.2 lbm/ft3 Cp = 0.93 Btu/lbm.°F, and α = 0.0077 ft2/h.

Analysis (a) This hot dog can physically be formed by the intersection of a long cylinder of radius ro = D/2 = (0.4/12) ft and a plane wall of thickness 2L = (5/12) ft. The distance x is measured from the midplane.

After 5 minutes

First the Biot number is calculated for the plane wall to be

Bi hLk

= =°

°=

( . /( .

.120 2 5 120 44

56 8 Btu / h.ft . F)( ft Btu / h. ft. F)

2 ) Water212°F

Hot dog x

r

The constants λ1 and A1

A

corresponding to this Biot number are, from Table 4-1,

= 2728.1and 5421.1 11 =λ


τ α= = = <

tL2

0 00772 5 12

0 015 0 2( .( . /

. ft / h)(5 / 60 h) ft)

2

2 . (Be cautious!)

Then the dimensionless temperature at the center of the plane wall is determined from

1)2728.1( )015.0()5421.1(1

0,

221 ≅==

−−

= −−

∞

∞ eeATTTT

iwallo

τλθ


Bihrk

o= =°

°=

( . /( .

.120 0 4 120 44


2 )

λ1 12 1589 15618= =. . and A

τα

= = = >t

ro2

0 00770 4 12

0 578 0 2( .( . /

. . ft / h)(5 / 60 h) ft)

2

2

θ λ τo cyl

i

T TT T

A e e,( . ) ( . )( . ) .=

−−

= = =∞

∞

− −01

2 1589 0 57812 2

15618 0106


F194°=⎯→⎯=−

−

=×=θ×θ=⎥⎦

⎤⎢⎣

⎡−

−

∞

∞

),0,0(106.021240

212),0,0(

106.0106.01),0,0(

,,

cylindershort

tTtT

TTTtT

cylowalloi

After 10 minutes

4-62


τ α= = = <

tL2

0 00772 5 12

0 03 0 2( .( . /

. ft / h)(10 / 60 h) ft)

2

2 . (Be cautious!)

1)2728.1( )03.0()5421.1(1

0,

221 ≅==

−−

= −−

∞

∞ eeATTTT

iwallo

τλθ

τα

= = = >t

ro2

0 00770 4 12

1156 0 2( .( . /

. . ft / h)(10 / 60 h) ft)

2

2

θ λ τo cyl

i

T TT T

A e e,( . ) ( . )( . ) .=

−−

= = =∞

∞

− −01

2 1589 115612 2

15618 0 007

F211°=⎯→⎯=−

−

=×=×=⎥⎦

⎤⎢⎣

⎡−

−

∞

∞

),0,0(007.021240

212),0,0(

007.0007.01),0,0(

,,

cylindershort

tTtT

TTTtT

cylowalloi

θθ

After 15 minutes

τ α= = = <

tL2

0 00772 5 12

0 045 0 2( .( . /

. ft / h)(15 / 60 h) ft)

2

2 . (Be cautious!)

1)2728.1( )045.0()5421.1(1

0,

221 ≅==

−−

= −−

∞

∞ eeATTTT

iwallo

τλθ

τα

= = = >t

ro2

0 00770 4 12

1734 0 2( .( . /

. . ft / h)(15 / 60 h) ft)

2

2

θ λ τo cyl

i

T TT T

A e e,( . ) ( . )( . ) .=

−−

= = =∞

∞

− −01

2 1589 1 73412 2

15618 0 0005

F 212 °=⎯→⎯=−

−

=×=×=⎥⎦

⎤⎢⎣

⎡−

−

∞

∞

),0,0(0005.021240

212),0,0(

0005.00005.01),0,0(

,,

cylindershort

tTtT

TTTtT

cylowalloi

θθ

(b) Treating the hot dog as an infinitely long cylinder will not change the results obtained in the part (a) since dimensionless temperatures for the plane wall is 1 for all cases.

4-63


4-77E A hot dog is dropped into boiling water. The center temperature of the hot dog is do be determined by treating hot dog as a finite cylinder and an infinitely long cylinder.

Assumptions 1 When treating hot dog as a finite cylinder, heat conduction in the hot dog is two-dimensional, and thus the temperature varies in both the axial x- and the radial r- directions. When treating hot dog as an infinitely long cylinder, heat conduction is one-dimensional in the radial r- direction. 2 The thermal properties of the hot dog are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The thermal properties of the hot dog are given to be k = 0.44 Btu/h.ft.°F, ρ = 61.2 lbm/ft3 Cp = 0.93 Btu/lbm.°F, and α = 0.0077 ft2/h.

Analysis (a) This hot dog can physically be formed by the intersection of a long cylinder of radius ro = D/2 = (0.4/12) ft and a plane wall of thickness 2L = (5/12) ft. The distance x is measured from the midplane.

After 5 minutes

First the Biot number is calculated for the plane wall to be

Bi hLk

= =°

°=

( . /( .

.120 2 5 120 44


2 ) Water202°F

Hot dog x

r

The constants λ1 and A1

A

corresponding to this Biot number are, from Table 4-1,

= 2728.1and 5421.1 11 =λ


τ α= = = <

tL2

0 00772 5 12

0 015 0 2( .( . /

. ft / h)(5 / 60 h) ft)

2

2 . (Be cautious!)

Then the dimensionless temperature at the center of the plane wall is determined from

1)2728.1( )015.0()5421.1(1

0,

221 ≅==

−−

= −−

∞

∞ eeATTTT

iwallo

τλθ


Bihrk

o= =°

°=

( . /( .

.120 0 4 120 44

9 1 Btu / h.ft . F)( ft Btu / h.ft. F)

2 )

λ1 12 1589 15618= =. . and A

τα

= = = >t

ro2

0 00770 4 12

0 578 0 2( .( . /

. . ft / h)(5 / 60 h) ft)

2

2

θ λ τo cyl

i

T TT T

A e e,( . ) ( . )( . ) .=

−−

= = =∞

∞

− −01

2 1589 0 57812 2

15618 0106


F185°=⎯→⎯=−−

=×=×=⎥⎦

⎤⎢⎣

⎡−

−

∞

∞

),0,0(106.020240

202),0,0(

106.0106.01),0,0(,,

cylindershort

tTtT

TTTtT

cylowalloi

θθ

After 10 minutes

4-64


τ α= = = <

tL2

0 00772 5 12

0 03 0 2( .( . /

. ft / h)(10 / 60 h) ft)

2

2 . (Be cautious!)

1)2728.1( )03.0()5421.1(1

0,

221 ≅==

−−

= −−

∞

∞ eeATTTT

iwallo

τλθ

τα

= = = >t

ro2

0 00770 4 12

1156 0 2( .( . /

. . ft / h)(10 / 60 h) ft)

2

2

θ λ τo cyl

i

T TT T

A e e,( . ) ( . )( . ) .=

−−

= = =∞

∞

− −01

2 1589 115612 2

15618 0 007

F201°=⎯→⎯=−

−

=×=×=⎥⎦

⎤⎢⎣

⎡−

−

∞

∞

),0,0(007.020240

202),0,0(

007.0007.01),0,0(

,,

cylindershort

tTtT

TTTtT

cylowalloi

θθ

After 15 minutes

τ α= = = <

tL2

0 00772 5 12

0 045 0 2( .( . /

. ft / h)(15 / 60 h) ft)

2

2 . (Be cautious!)

1)2728.1( )045.0()5421.1(1

0,

221 ≅==

−−

= −−

∞

∞ eeATTTT

iwallo

τλθ

τα

= = = >t

ro2

0 00770 4 12

1734 0 2( .( . /

. . ft / h)(15 / 60 h) ft)

2

2

θ λ τo cyl

i

T TT T

A e e,( . ) ( . )( . ) .=

−−

= = =∞

∞

− −01

2 1589 1 73412 2

15618 0 0005

F 202 °=⎯→⎯=−

−

=×=×=⎥⎦

⎤⎢⎣

⎡−

−

∞

∞

),0,0(0005.020240

202),0,0(

0005.00005.01),0,0(

,,

cylindershort

tTtT

TTTtT

cylowalloi

θθ

(b) Treating the hot dog as an infinitely long cylinder will not change the results obtained in the part (a) since dimensionless temperatures for the plane wall is 1 for all cases.

4-65


4-78 A rectangular ice block is placed on a table. The time the ice block starts melting is to be determined.

Assumptions 1 Heat conduction in the ice block is two-dimensional, and thus the temperature varies in both x- and y- directions. 2 The thermal properties of the ice block are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The thermal properties of the ice are given to be k = 2.22 W/m.°C and α = 0.124×10-7 m2/s.

Analysis This rectangular ice block can be treated as a short rectangular block that can physically be formed by the intersection of two infinite plane wall of thickness 2L = 4 cm and an infinite plane wall of thickness 2L = 10 cm. We measure x from the bottom surface of the block since this surface represents the adiabatic center surface of the plane wall of thickness 2L = 10 cm. Since the melting starts at the corner of the top surface, we need to determine the time required to melt ice block which will happen when the temperature drops below at this location. The Biot numbers and the corresponding constants are first determined to be

0°C

Ice block -20°C

Air 18°C

1081.0)C W/m.22.2(

)m 02.0)(C. W/m12( 21

wall,1 =°

°==

khL

Bi ⎯ →⎯ = =λ1 10 3208 10173. . and A

2703.0)C W/m.22.2(

)m 05.0)(C. W/m12( 23

wall,3 =°

°==

khL

Bi ⎯ →⎯ = =λ1 10 4951 10408. . and A

The ice will start melting at the corners because of the maximum exposed surface area there. Noting that and assuming that τ > 0.2 in all dimensions so that the one-term approximate solution for

transient heat conduction is applicable, the product solution method can be written for this problem as τ α= t L/ 2

hours 30.04==⎯→⎯

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡ ×−×

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡ ×−=

⎥⎦⎤

⎢⎣⎡ λ⎥⎦

⎤⎢⎣⎡ λ=

−−−

θθ=θ

−

−

τλ−τλ−

s 135,108

)4951.0cos()05.0(

)10124.0()4951.0(exp)0408.1(

)3208.0cos()02.0(

)10124.0()3208.0(exp)0173.1(4737.0

)/cos()/cos(1820

180

),(),(),,,(

2

72

2

2

72

3311

2

1111

wall,232

wall,11321

21

21

t

t

t

LLeALLeA

tLtLtLLL block

Therefore, the ice will start melting in about 30 hours.

Discussion Note that

2.0536.0m) 05.0(

s/h) 35/s)(1108,1m 10124.0(2

27

2 >=×

==−

Ltατ

and thus the assumption of τ > 0.2 for the applicability of the one-term approximate solution is verified.

4-66


4-79 "!PROBLEM 4-79" "GIVEN" 2*L_1=0.04 "[m]" L_2=L_1 2*L_3=0.10 "[m]" "T_i=-20 [C], parameter to be varied" T_infinity=18 "[C]" h=12 "[W/m^2-C]" T_L1_L2_L3=0 "[C]" "PROPERTIES" k=2.22 "[W/m-C]" alpha=0.124E-7 "[m^2/s]" "ANALYSIS" "This block can physically be formed by the intersection of two infinite plane wall of thickness 2L=4 cm and an infinite plane wall of thickness 2L=10 cm" "For the two plane walls" Bi_w1=(h*L_1)/k "From Table 4-1 corresponding to this Bi number, we read" lambda_1_w1=0.3208 "w stands for wall" A_1_w1=1.0173 time*Convert(min, s)=tau_w1*L_1^2/alpha "For the third plane wall" Bi_w3=(h*L_3)/k "From Table 4-1 corresponding to this Bi number, we read" lambda_1_w3=0.4951 A_1_w3=1.0408 time*Convert(min, s)=tau_w3*L_3^2/alpha theta_L_w1=A_1_w1*exp(-lambda_1_w1^2*tau_w1)*Cos(lambda_1_w1*L_1/L_1) "theta_L_w1=(T_L_w1-T_infinity)/(T_i-T_infinity)" theta_L_w3=A_1_w3*exp(-lambda_1_w3^2*tau_w3)*Cos(lambda_1_w3*L_3/L_3) "theta_L_w3=(T_L_w3-T_infinity)/(T_i-T_infinity)" (T_L1_L2_L3-T_infinity)/(T_i-T_infinity)=theta_L_w1^2*theta_L_w3 "corner temperature"

Ti [C] time [min] -26 1614 -24 1512 -22 1405 -20 1292 -18 1173 -16 1048 -14 914.9 -12 773.3 -10 621.9 -8 459.4 -6 283.7 -4 92.84

4-67


-30 -25 -20 -15 -10 -5 00

200

400

600

800

1000

1200

1400

1600

1800

Ti [C]

time

[min

]

4-68


4-80 A cylindrical ice block is placed on a table. The initial temperature of the ice block to avoid melting for 2 h is to be determined.

Assumptions 1 Heat conduction in the ice block is two-dimensional, and thus the temperature varies in both x- and r- directions. 2 Heat transfer from the base of the ice block to the table is negligible. 3 The thermal properties of the ice block are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Air

T∞ = 20°C

x

r

(ro, L)

Insulation

Ice block Ti

Properties The thermal properties of the ice are given to be k = 2.22 W/m.°C and α = 0.124×10-7 m2/s.

Analysis This cylindrical ice block can be treated as a short cylinder that can physically be formed by the intersection of a long cylinder of diameter D = 2 cm and an infinite plane wall of thickness 2L = 4 cm. We measure x from the bottom surface of the block since this surface represents the adiabatic center surface of the plane wall of thickness 2L = 4 cm. The melting starts at the outer surfaces of the top surface when the temperature drops below 0 at this location. The Biot numbers, the corresponding constants, and the Fourier numbers are

°C

1171.0)C W/m.22.2(

)m 02.0)(C. W/m13( 2

wall =°

°==

khLBi ⎯→⎯ = =λ1 10 3318 10187. . and A

05856.0)C W/m.22.2(

)m 01.0)(C. W/m13( 2

cyl =°

°==

khr

Bi o ⎯→⎯ = =λ1 10 3407 10146. . and A

τ αwall

2

2 m / s)(2 h / h)

m)= =

× ×= >

−tL2

70124 10 3600 s0 02

0 2232 0 2( .( .

. .

τα

cyl

2

2 m / s)(2 h / h)

m)= =

× ×= >

−tro

2

70124 10 3600 s0 01

0 8928 0 2( .( .

. .

Note that τ > 0.2 in all dimensions and thus the one-term approximate solution for transient

heat conduction is applicable. The product solution for this problem can be written as

[ ][ )9707.0()0146.1( )3318.0cos()0187.1(20200

)/()/cos(20200

),(),(),,(

)8928.0()3407.0()2232.0()3318.0(

10111

cylwallblock

22

21

21

−−

−−

=−−

⎥⎦⎤

⎢⎣⎡⎥⎦⎤

⎢⎣⎡=

−−

=

eeT

rrJeALLeAT

trtLtrL

i

ooi

oo

λλ

]

θθθ

τλτλ

which gives Ti = − °4 C

Therefore, the ice will not start melting for at least 2 hours if its initial temperature is -4°C or below.

4-69


4-81 A cubic block and a cylindrical block are exposed to hot gases on all of their surfaces. The center temperatures of each geometry in 10, 20, and 60 min are to be determined.

Assumptions 1 Heat conduction in the cubic block is three-dimensional, and thus the temperature varies in all x-, y, and z- directions. 2 Heat conduction in the cylindrical block is two-dimensional, and thus the temperature varies in both axial x- and radial r- directions. 3 The thermal properties of the granite are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The thermal properties of the granite are given to be k = 2.5 W/m.°C and α = 1.15×10-6 m2/s.

Analysis:

Cubic block: This cubic block can physically be formed by the intersection of three infinite plane walls of thickness 2L = 5 cm.

After 10 minutes: The Biot number, the corresponding constants, and the Fourier number are

400.0)C W/m.5.2(

)m 025.0)(C. W/m40( 2=

°°

==k

hLBi ⎯→⎯ = =λ1 105932 10580. . and A

τ α= =

× ×= >

−tL2

6115 100 025

1104 0 2( .( .

. . m / s)(10 min 60 s / min) m)

2

2

To determine the center temperature, the product solution can be written as

[ ]

{ }C323°=

==−

−

⎟⎠⎞⎜

⎝⎛=

−−

θ=θ

−

τλ−

∞

∞

),0,0,0(

369.0)0580.1(50020

500),0,0,0(

),0,0,0(),0(),0,0,0(

3)104.1()5932.0(

3

1

3wallblock

2

21

tT

etT

eATT

TtTtt

i

After 20 minutes

τ α= =

× ×= >

−tL2

6115 100 025

2 208 0 2( .( .

. . m / s)(20 min 60 s / min) m)

2

2

{ } C445°=⎯→⎯==−

− − ),0,0,0(115.0)0580.1(50020

500),0,0,0( 3)208.2()5932.0( 2tTetT

Ti = 20°C

Hot gases 500°C

5 cm × 5 cm × 5 cm

Ti = 20°C

5 cm × 5 cm

After 60 minutes

τ α= =

× ×= >

−tL2

6115 100 025

6 624 0 2( .( .

. . m / s)(60 min 60 s / min) m)

2

2

{ } C500°=⎯→⎯==−

− − ),0,0,0(00109.0)0580.1(50020

500),0,0,0( 3)624.6()5932.0( 2tTe

tT

Note that τ > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction is applicable.

4-70


Cylinder: This cylindrical block can physically be formed by the intersection of a long cylinder of radius ro = D/2 = 2.5 cm and a plane wall of thickness 2L = 5 cm.

After 10 minutes: The Biot number and the corresponding constants for the long cylinder are

400.0)C W/m.5.2(

)m 025.0)(C. W/m40( 2=

°°

==k

hrBi o ⎯ →⎯ = =λ1 10 8516 10931. . and A

To determine the center temperature, the product solution can be written as

[ ][ ]

{ }{ } C331°=⎯→⎯==−

−

⎟⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛=

−−

=

−−

−−

∞

∞

),0,0(352.0)0931.1()0580.1(50020

500),0,0(

),0,0(

),0(),0(),0,0(

)104.1()8516.0()104.1()5932.0(

11

22

21

21

tTeetT

eAeATT

TtT

ttt

cylwalli

cylwallblock

τλτλ

θθθ

After 20 minutes

{ }{ } C449°=⎯→⎯==−

− −− ),0,0(107.0)0931.1()0580.1(50020

500),0,0( )208.2()8516.0()208.2()5932.0( 22tTeetT

After 60 minutes

{ }{ } C500°=⎯→⎯==−

− −− ),0,0(00092.0)0931.1()0580.1(50020

500),0,0( )624.6()8516.0()624.6()5932.0( 22tTee

tT


4-71


4-82 A cubic block and a cylindrical block are exposed to hot gases on all of their surfaces. The center temperatures of each geometry in 10, 20, and 60 min are to be determined.

Assumptions 1 Heat conduction in the cubic block is three-dimensional, and thus the temperature varies in all x-, y, and z- directions. 2 Heat conduction in the cylindrical block is two-dimensional, and thus the temperature varies in both axial x- and radial r- directions. 3 The thermal properties of the granite are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The thermal properties of the granite are k = 2.5 W/m.°C and α = 1.15×10-6 m2/s.

Analysis:

Cubic block: This cubic block can physically be formed by the intersection of three infinite plane wall of thickness 2L = 5 cm. Two infinite plane walls are exposed to the hot gases with a heat transfer coefficient of h = 40 W / °m . C2 and one with h = °80 W / m . C2 .

After 10 minutes: The Biot number and the corresponding constants for h = °40 W / m . C2 are

400.0)C W/m.5.2(

)m 025.0)(C. W/m40( 2=

°°

==k

hLBi ⎯ →⎯ = =λ1 10 5932 10580. . and A

The Biot number and the corresponding constants for h = °80 W / m . C2 are

800.0)C W/m.5.2(

)m 025.0)(C. W/m80( 2=

°°

==k

hLBi

⎯ →⎯ = =λ1 10 7910 11016. . and A


τ α= =

× ×= >

−tL2

6115 100 025

1104 0 2( .( .

. . m / s)(10 min 60 s / min) m)

2

2

To determine the center temperature, the product solution method can be written as

[ ] [ ]

{ } { }

C364°=

==−

−

⎟⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛=

−−

θθ=θ

−−

τλ−τλ−

∞

∞

),0,0,0(

284.0)1016.1()0580.1(50020

500),0,0,0(

),0,0,0(),0(),0(),0,0,0(

)104.1()7910.0(2)104.1()5932.0(

1

2

1

wall2

wallblock

22

21

21

tT

eetT

eAeATT

TtTttt

i

Ti = 20°C

Hot gases 500°C

5 cm × 5 cm × 5 cm

Ti = 20°C

5 cm × 5 cm

After 20 minutes

τ α= =

× ×= >

−tL2

6115 100 025

2 208 0 2( .( .

. . m / s)(20 min 60 s / min) m)

2

2

{ } { }

C469°=⎯→⎯

==−

− −−

),0,0,0(

0654.0)1016.1()0580.1(50020

500),0,0,0( )208.2()7910.0(2)208.2()5932.0( 22

tT

eetT

After 60 minutes

4-72


τ α= =

× ×= >

−tL2

6115 100 025

6 624 0 2( .( .

. . m / s)(60 min 60 s / min) m)

2

2

{ } { }

C500°=⎯→⎯

==−

− −−

),0,0,0(

000186.0)1016.1()0580.1(50020

500),0,0,0( )624.6()7910.0(2)624.6()5932.0( 22

tT

eetT


Cylinder: This cylindrical block can physically be formed by the intersection of a long cylinder of radius ro = D/2 = 2.5 cm exposed to the hot gases with a heat transfer coefficient of h = °40 W / m . C2 and a

plane wall of thickness 2L = 5 cm exposed to the hot gases with h = °80 W / m . C2 .

After 10 minutes: The Biot number and the corresponding constants for the long cylinder are

400.0)C W/m.5.2(

)m 025.0)(C. W/m40( 2=

°°

==k

hrBi o ⎯ →⎯ = =λ1 10 8516 10931. . and A

To determine the center temperature, the product solution method can be written as

[ ][ ]

{ }{ }

C370°=

==−

−

⎟⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛=

−−

=

−−

−−

∞

∞

),0,0(

271.0)0931.1()1016.1(50020

500),0,0(

),0,0(

),0(),0(),0,0(

)104.1()8516.0()104.1()7910.0(

1wall

1

wallblock

22

21

21

tT

eetT

eAeATT

TtT

ttt

cyli

cyl

τλτλ

θθθ

After 20 minutes

{ }{ } C471°=⎯→⎯==−

− −− ),0,0(06094.0)0931.1()1016.1(50020

500),0,0( )208.2()8516.0()208.2()7910.0( 22tTee

tT

After 60 minutes

{ }{ } C500°=⎯→⎯==−

− −− ),0,0(0001568.0)0931.1()1016.1(50020

500),0,0( )624.6()8516.0()624.6()7910.0( 22tTeetT


4-73


4-83 A cylindrical aluminum block is heated in a furnace. The length of time the block should be kept in the furnace and the amount of heat transfer to the block are to be determined. Assumptions 1 Heat conduction in the cylindrical block is two-dimensional, and thus the temperature varies in both axial x- and radial r- directions. 2 The thermal properties of the aluminum are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (it will be verified). Properties The thermal properties of the aluminum block are given to be k = 236 W/m.°C, ρ = 2702 kg/m3, Cp = 0.896 kJ/kg.°C, and α = 9.75×10-5 m2/s. Analysis This cylindrical aluminum block can physically be formed by the intersection of an infinite plane wall of thickness 2L = 20 cm, and a long cylinder of radius ro = D/2 = 7.5 cm. The Biot numbers and the corresponding constants are first determined to be

0339.0)C W/m.236(

)m 1.0)(C. W/m80( 2=

°°

==k

hLBi 0056.1and 1811.0 11 ==⎯→⎯ Aλ

0254.0C W/m.236

)m 075.0)(C. W/m80( 20 =

°°

==k

hrBi 0063.1and 2217.0 11 ==⎯→⎯ Aλ

Noting that and assuming τ > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction is applicable, the product solution for this problem can be written as

τ α= t L/ 2

7627.0)075.0(

)1075.9()2217.0(exp)0063.1()1.0(

)1075.9()1811.0(exp)0056.1(1200201200300

),0(),0(),0,0(

2

52

2

52

cyl1

wall1cylwallblock

21

21

=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡ ×−

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡ ×−=

−−

⎟⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛=θθ=θ

−−

τλ−τλ−

tt

eAeAttt

Solving for the time t gives t = 241 s = 4.0 min. We note that

2.035.2m) 1.0(

s) /s)(241m 1075.9(2

25

2 >=×

==−

Ltατ

and thus the assumption of τ > 0.2 for the applicability of the one-term approximate solution is verified. The maximum amount of heat transfer is

[ ]

kJ 100,10C)120020)(CkJ/kg. 896.0)(kg 550.9()(kg 550.9m) 2.0(m) 075.0()kg/m 2702(

max

2.320

=°−°=−=====

∞TTmCQLrVm

ip

πρπρ


2415.01811.0

)1811.0sin()7627.0(1)sin(

11

1,

max=−=

λλ

θ−=⎟⎟⎠

⎞⎜⎜⎝

⎛wallo

wallQQ

2425.02217.01101.0)7627.0(21

)(21

1

11,

max=−=

λλ

θ−=⎟⎟⎠

⎞⎜⎜⎝

⎛ JQ

Qcylo

cyl


4254.0)2415.01)(2425.0(2415.01maxmaxmaxmax

=−+=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛

wallplane

cylinderlong

wallplane

cylindershort Q

QQ

QQ

QQ

Q

Lz

Cylinder Ti = 20°C

r0

Furnace T∞ = 1200°C

L

Then the total heat transfer from the short cylinder as it is cooled from 300°C at the center to 20°C becomes kJ 4297=== kJ) 100,10)(4254.0(4236.0 maxQQ

which is identical to the heat transfer to the cylinder as the cylinder at 20°C is heated to 300°C at the center.

4-74


4-84 A cylindrical aluminum block is heated in a furnace. The length of time the block should be kept in the furnace and the amount of heat transferred to the block are to be determined. Assumptions 1 Heat conduction in the cylindrical block is two-dimensional, and thus the temperature varies in both axial x- and radial r- directions. 2 Heat transfer from the bottom surface of the block is negligible. 3 The thermal properties of the aluminum are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of the aluminum block are given to be k = 236 W/m.°C, ρ = 2702 kg/m3, Cp = 0.896 kJ/kg.°C, and α = 9.75×10-5 m2/s. Analysis This cylindrical aluminum block can physically be formed by the intersection of an infinite plane wall of thickness 2L = 40 cm and a long cylinder of radius r0 = D/2 = 7.5 cm. Note that the height of the short cylinder represents the half thickness of the infinite plane wall where the bottom surface of the short cylinder is adiabatic. The Biot numbers and corresponding constants are first determined to be

0678.0)C W/m.236(

)m 2.0)(C. W/m80( 2=

°°

==k

hLBi 0110.1and 2568.0 11 ==⎯→⎯ Aλ

0254.0)C W/m.236(

)m 075.0)(C. W/m80( 20 =

°°

==k

hrBi 0063.1and 2217.0 11 ==⎯→⎯ Aλ


τ α= t L/ 2

7627.0)075.0(

)1075.9()2217.0(exp)0063.1()2.0(

)1075.9()2568.0(exp)0110.1(1200201200300

),0(),0(),0,0(

2

52

2

52

cyl1

wall1cylwallblock

21

21

=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡ ×−

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡ ×−=

−−

⎟⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛=θθ=θ

−−

τλ−τλ−

tt

eAeAttt

Solving for the time t gives t = 285 s = 4.7 min. We note that

2.069.0m) 2.0(

s) /s)(285m 1075.9(2

25

2 >=×

==−

Ltατ

and thus the assumption of τ > 0.2 for the applicability of the one-term approximate solution is verified. The maximum amount of heat transfer is

[ ]

kJ 100,10C)120020)(CkJ/kg. 896.0)(kg 55.9()(kg 55.9m) 2.0(m) 075.0()kg/m 2702(

max

2.3

=°−°=−=====

∞TTmCQLrVm

ip

o πρπρ

Furnace T∞ = 1200°C


2457.02568.0

)2568.0sin()7627.0(1)sin(

11

1,

max=−=

λλ

θ−=⎟⎟⎠

⎞⎜⎜⎝

⎛wallo

wallQQ

2425.02217.01101.0)7627.0(21

)(21

1

11,

max=−=

λλ

θ−=⎟⎟⎠

⎞⎜⎜⎝

⎛ JQ

Qcylo

cyl


4286.0)2457.01)(2425.0(2457.01wallplanemax

cylinderlongmax

wallplanemax

cylindershortmax

=−+=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛Q

QQ

QQ

QQ

Q

Lz

Then the total heat transfer from the short cylinder as it is cooled from 300°C at the center to 20°C becomes kJ 4239=== kJ) 100,10)(4286.0(4255.0 maxQQ which is identical to the heat transfer to the cylinder as the cylinder at 20°C is heated to 300°C at the center.

Cylinder Ti = 20°C

r0

L

4-75


4-85 "!PROBLEM 4-85" "GIVEN" 2*L=0.20 "[m]" 2*r_o=0.15 "[m]" T_i=20 "[C]" T_infinity=1200 "[C]" "T_o_o=300 [C], parameter to be varied" h=80 "[W/m^2-C]" "PROPERTIES" k=236 "[W/m-C]" rho=2702 "[kg/m^3]" C_p=0.896 "[kJ/kg-C]" alpha=9.75E-5 "[m^2/s]" "ANALYSIS" "This short cylinder can physically be formed by the intersection of a long cylinder of radius r_o and a plane wall of thickness 2L" "For plane wall" Bi_w=(h*L)/k "From Table 4-1 corresponding to this Bi number, we read" lambda_1_w=0.1439 "w stands for wall" A_1_w=1.0035 tau_w=(alpha*time)/L^2 theta_o_w=A_1_w*exp(-lambda_1_w^2*tau_w) "theta_o_w=(T_o_w-T_infinity)/(T_i-T_infinity)" "For long cylinder" Bi_c=(h*r_o)/k "c stands for cylinder" "From Table 4-1 corresponding to this Bi number, we read" lambda_1_c=0.1762 A_1_c=1.0040 tau_c=(alpha*time)/r_o^2 theta_o_c=A_1_c*exp(-lambda_1_c^2*tau_c) "theta_o_c=(T_o_c-T_infinity)/(T_i-T_infinity)" (T_o_o-T_infinity)/(T_i-T_infinity)=theta_o_w*theta_o_c "center temperature of cylinder" V=pi*r_o^2*(2*L) m=rho*V Q_max=m*C_p*(T_infinity-T_i) Q_w=1-theta_o_w*Sin(lambda_1_w)/lambda_1_w "Q_w=(Q/Q_max)_w" Q_c=1-2*theta_o_c*J_1/lambda_1_c "Q_c=(Q/Q_max)_c" J_1=0.0876 "From Table 4-2, at lambda_1_c" Q/Q_max=Q_w+Q_c*(1-Q_w) "total heat transfer"

4-76


To,o [C] time [s] Q [kJ]

50 44.91 346.3 100 105 770.2 150 167.8 1194 200 233.8 1618 250 303.1 2042 300 376.1 2466 350 453.4 2890 400 535.3 3314 450 622.5 3738 500 715.7 4162 550 815.9 4586 600 924 5010 650 1042 5433 700 1170 5857 750 1313 6281 800 1472 6705 850 1652 7129 900 1861 7553 950 2107 7977 1000 2409 8401

0 200 400 600 800 10000

500

1000

1500

2000

2500

0

1000

2000

3000

4000

5000

6000

7000

8000

9000

To,o

time

Q [k

J]

time

heat

4-77


Special Topic: Refrigeration and Freezing of Foods

4-86C The common kinds of microorganisms are bacteria, yeasts, molds, and viruses. The undesirable changes caused by microorganisms are off-flavors and colors, slime production, changes in the texture and appearances, and the spoilage of foods.

4-87C Microorganisms are the prime cause for the spoilage of foods. Refrigeration prevents or delays the spoilage of foods by reducing the rate of growth of microorganisms. Freezing extends the storage life of foods for months by preventing the growths of microorganisms.

4-88C The environmental factors that affect of the growth rate of microorganisms are the temperature, the relative humidity, the oxygen level of the environment, and air motion.

4-89C Cooking kills the microorganisms in foods, and thus prevents spoilage of foods. It is important to raise the internal temperature of a roast in an oven above 70ºC since most microorganisms, including some that cause diseases, may survive temperatures below 70ºC.

4-90C The contamination of foods with microorganisms can be prevented or minimized by (1) preventing contamination by following strict sanitation practices such as washing hands and using fine filters in ventilation systems, (2) inhibiting growth by altering the environmental conditions, and (3) destroying the organisms by heat treatment or chemicals.

The growth of microorganisms in foods can be retarded by keeping the temperature below 4ºC and relative humidity below 60 percent. Microorganisms can be destroyed by heat treatment, chemicals, ultraviolet light, and solar radiation.

4-91C (a) High air motion retards the growth of microorganisms in foods by keeping the food surfaces dry, and creating an undesirable environment for the microorganisms. (b) Low relative humidity (dry) environments also retard the growth of microorganisms by depriving them of water that they need to grow. Moist air supplies the microorganisms with the water they need, and thus encourages their growth. Relative humidities below 60 percent prevent the growth rate of most microorganisms on food surfaces.

4-92C Cooling the carcass with refrigerated air is at -10ºC would certainly reduce the cooling time, but this proposal should be rejected since it will cause the outer parts of the carcasses to freeze, which is undesirable. Also, the refrigeration unit will consume more power to reduce the temperature to -10ºC, and thus it will have a lower efficiency.

4-78


4-93C The freezing time could be decreased by (a) lowering the temperature of the refrigerated air, (b) increasing the velocity of air, (c) increasing the capacity of the refrigeration system, and (d) decreasing the size of the meat boxes.

4-94C The rate of freezing can affect color, tenderness, and drip. Rapid freezing increases tenderness and reduces the tissue damage and the amount of drip after thawing.

4-95C This claim is reasonable since the lower the storage temperature, the longer the storage life of beef. This is because some water remains unfrozen even at subfreezing temperatures, and the lower the temperature, the smaller the unfrozen water content of the beef.

4-96C A refrigerated shipping dock is a refrigerated space where the orders are assembled and shipped out. Such docks save valuable storage space from being used for shipping purpose, and provide a more acceptable working environment for the employees. The refrigerated shipping docks are usually maintained at 1.5ºC, and therefore the air that flows into the freezer during shipping is already cooled to about 1.5ºC. This reduces the refrigeration load of the cold storage rooms.

4-97C (a) The heat transfer coefficient during immersion cooling is much higher, and thus the cooling time during immersion chilling is much lower than that during forced air chilling. (b) The cool air chilling can cause a moisture loss of 1 to 2 percent while water immersion chilling can actually cause moisture absorption of 4 to 15 percent. (c) The chilled water circulated during immersion cooling encourages microbial growth, and thus immersion chilling is associated with more microbial growth. The problem can be minimized by adding chloride to the water.

4-98C The proper storage temperature of frozen poultry is about -18ºC or below. The primary freezing methods of poultry are the air blast tunnel freezing, cold plates, immersion freezing, and cryogenic cooling.

4-99C The factors, which affect the quality of frozen, fish are the condition of the fish before freezing, the freezing method, and the temperature and humidity during storage and transportation, and the length of storage time.

4-79


4-100 The chilling room of a meat plant with a capacity of 350 beef carcasses is considered. The cooling load, the air flow rate, and the heat transfer area of the evaporator are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Specific heats of beef carcass and air are constant.

Properties The density and specific heat of air at 0°C are given to be 1.28 kg/m3 and 1.0 kJ/kg⋅°C. The specific heat of beef carcass is given to be 3.14 kJ/kg⋅°C.

Analysis (a) The amount of beef mass that needs to be cooled per unit time is

kg/s2.27s) 3600s)/(12h kg/carcas280(350

time)oolingcooled)/(cmassbeef(Total=××=

=beefm&

The product refrigeration load can be viewed as the energy that needs to be removed from the beef carcass as it is cooled from 35 to 16ºC at a rate of 2.27 kg/s, and is determined to be

Beef 35°C

280 kg

Lights, 2 kW

Fans, 22 kW

11kW

kW135C16)ºC)(35kJ/kg.ºkg/s)(3.14 (2.27

)(=−=

Δ= beefpbeef TCmQ &&

Then the total refrigeration load of the chilling room becomes

kW 170=+++=+++= 11222135roomchillingtotal, gainheatlightsfanbeef QQQQQ &&&&&

(b) Heat is transferred to air at the rate determined above, and the temperature of air rises from -2.2ºC to 0.5ºC as a result. Therefore, the mass flow rate of air is

kg/s63.0C]2.2)(C)[0.5kJ/kg. (1.0

kW170)(

=°−−°

=Δ

=airp

airair TC

Qm

&&

Then the volume flow rate of air becomes

m³/s 49.2==ρ

=kg/m³ 1.28kg/s 63.0

air

airair

mV

&&

4-80


4-101 Turkeys are to be frozen by submerging them into brine at -29°C. The time it will take to reduce the temperature of turkey breast at a depth of 3.8 cm to -18°C and the amount of heat transfer per turkey are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The thermal properties of turkeys are constant.

Brine -29°C

Turkey Ti = 1°C

Properties It is given that the specific heats of turkey are 2.98 and 1.65 kJ/kg.°C above and below the freezing point of -2.8°C, respectively, and the latent heat of fusion of turkey is 214 kJ/kg.

Analysis The time required to freeze the turkeys from 1°C to -18ºC with brine at -29ºC can be determined directly from Fig. 4-45 to be

t ≅180 min. ≅ 3 hours

(a) Assuming the entire water content of turkey is frozen, the amount of heat that needs to be removed from the turkey as it is cooled from 1°C to -18°C is

Cooling to -2.8ºC: kJ 79.3C](-2.8)-C)[1kJ/kg kg)(2.98 7()( freshfreshcooling, =°°⋅=Δ= TCmQ

Freezing at -2.8ºC: kJ 1498kJ/kg) kg)(214 7(latentfreezing === hmQ

Cooling -18ºC: kJ 175.6C18)](2.8C)[kJ/kg.kg)(1.65(7)( frozenfrozencooling, =°−−−°=Δ= TCmQ

Therefore, the total amount of heat removal per turkey is

kJ1753≅++=++= 6.17514983.79frozencooling,freezingfreshcooling,total QQQQ

(b) Assuming only 90 percent of the water content of turkey is frozen, the amount of heat that needs to be removed from the turkey as it is cooled from 1°C to -18°C is

Cooling to -2.8ºC: kJ 79.3C](-2.98)-C)[1kJ/kg kg)(2.98 7()( freshfreshcooling, =°°⋅=Δ= TCmQ

Freezing at -2.8ºC: Q mhfreezing latent (7 0.9 kg)(214 kJ / kg) 1,348kJ= = × =

Cooling -18ºC: Q mC Tcooling,frozen frozen (7 0.9 kg)(1.65kJ / kg.º C)[ 2.8 ( 18)]º C 158kJ= = × − − − =( )Δ

kJ 7.31]C18)º(-2.8)[CkJ/kg.º 2.98)(kg0.17()( freshunfrozencooling, =−−×=Δ= TCmQ

Therefore, the total amount of heat removal per turkey is

kJ 1617=+++=++= 7.3115813483.79unfrozen&frozencooling,freezingfreshcooling,total QQQQ

4-81


4-102E Chickens are to be frozen by refrigerated air. The cooling time of the chicken is to be determined for the cases of cooling air being at –40°F and -80°F. Assumptions 1 Steady operating conditions exist. 2 The thermal properties of chickens are constant. Analysis The time required to reduce the inner surface temperature of the chickens from 32ºF to 25ºF with refrigerated air at -40ºF is determined from Fig. 4-44 to be t ≅ 2.3 hours

Air -40°C

Chicken 6 lbm

32°F

If the air temperature were -80ºF, the freezing time would be t ≅ 1.4 hours Therefore, the time required to cool the chickens to 25°F is reduced considerably when the refrigerated air temperature is decreased.

4-103 Chickens are to be cooled by chilled water in an immersion chiller. The rate of heat removal from the chicken and the mass flow rate of water are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The thermal properties of chickens are constant.

Properties The specific heat of chicken are given to be 3.54 kJ/kg.°C. The specific heat of water is 4.18 kJ/kg.°C (Table A-9).

Immersion chilling, 0.5°C

3°C 15°C

210 kJ/min

Analysis (a) Chickens are dropped into the chiller at a rate of 500 per hour. Therefore, chickens can be considered to flow steadily through the chiller at a mass flow rate of

&mchicken (500 chicken / h)(2.2 kg / chicken) 1100 kg / h = 0.3056 kg / s= =

Then the rate of heat removal from the chickens as they are cooled from 15°C to 3ºC at this rate becomes

kW13.0 C3)ºC)(15kJ/kg.ºkg/s)(3.54(0.3056)( chickenchicken =−=Δ= TCmQ p&&

(b) The chiller gains heat from the surroundings as a rate of 210 kJ/min = 3.5 kJ/s. Then the total rate of heat gain by the water is

& & & . . .Q Q Qwater chicken heat gain kW= + = + =130 35 165

Noting that the temperature rise of water is not to exceed 2ºC as it flows through the chiller, the mass flow rate of water must be at least

kg/s 1.97==Δ

=C)C)(2ºkJ/kg.º (4.18

kW16.5)( water

waterwater TC

Qm

p

&&

If the mass flow rate of water is less than this value, then the temperature rise of water will have to be more than 2°C.

4-82


4-104 The center temperature of meat slabs is to be lowered by chilled air to below 5°C while the surface temperature remains above -1°C to avoid freezing. The average heat transfer coefficient during this cooling process is to be determined.

Assumptions 1 The meat slabs can be approximated as very large plane walls of half-thickness L = 5-cm. 2 Heat conduction in the meat slabs is one-dimensional because of symmetry about the centerplane. 3 The thermal properties of the meat slab are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The thermal properties of the beef slabs are given to be ρ = 1090 kg/m3, C 3.54 kJ/kg.°C, k

= 0.47 W/m.°C, and α = 0.13×10p =

-6 m2/s.

Analysis The lowest temperature in the steak will occur at the surfaces and the highest temperature at the center at a given time since the inner part of the steak will be last place to be cooled. In the limiting case, the surface temperature at x = L = 5 cm from the center will be -1°C while the mid plane temperature is 5°C in an environment at -12°C. Then from Fig. 4-13b we obtain

95.0Bi1

65.0)12(5)12(1),(

1=cm 5cm 5

==

⎪⎪⎭

⎪⎪⎬

⎫

=−−−−−

=−−

=

∞

∞ hLk

TTTtLT

Lx

o

Air

-12°C

Meat 15°C which gives

C. W/m9.9 2 °=⎟⎠⎞

⎜⎝⎛°

==0.95

1m 0.05

C W/m.47.0BiLkh

Therefore, the convection heat transfer coefficient should be kept below this value to satisfy the constraints on the temperature of the steak during refrigeration. We can also meet the constraints by using a lower heat transfer coefficient, but doing so would extend the refrigeration time unnecessarily.

Discussion We could avoid the uncertainty associated with the reading of the charts and obtain a more accurate result by using the one-term solution relation for an infinite plane wall, but it would require a trial and error approach since the Bi number is not known.

4-83


Review Problems

4-105 Two large steel plates are stuck together because of the freezing of the water between the two plates. Hot air is blown over the exposed surface of the plate on the top to melt the ice. The length of time the hot air should be blown is to be determined.

Assumptions 1 Heat conduction in the plates is one-dimensional since the plate is large relative to its thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the steel plates are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The thermal properties of steel plates are given to be k = 43 W/m.°C and α = 1.17×10-5 m2/s

Analysis The characteristic length of the plates and the Biot number are

1.0 019.0

)CW/m. 43()m 02.0)(C.W/m 40(

m 02.0

2<=

°°

==

===

khL

Bi

LAVL

c

sc

Since , the lumped system analysis is applicable. Therefore, Bi < 0.1

min 8.0s 482 ==⎯→⎯=

−−−

⎯→⎯=−−

=°×°

===

−−

∞

∞ teeTTTtT

LCh

VChA

b

bt

i

cpp

s

)ts 000544.0(

1-36

2

-1

5015500)(

s 000544.0m) (0.02)C.J/m 10675.3(

C.W/m 40ρρ

Steel platesTi = -15°C

Hot gases T∞ = 50°C

where C.J/m 10675.3/sm 1017.1

C W/m.43 3625

°×=×

°=

α=ρ

−

kC p

Alternative solution: This problem can also be solved using the transient chart Fig. 4-13a,

2.015769.0

5015500

6.52019.011

2>==

⎪⎪⎭

⎪⎪⎬

⎫

=−−

−=

−−

==

∞

∞ oi

o rt

TTTTBi ατ

Then,

s 513=×

=α

τ=

− /s)m 1017.1(m) 02.0)(15(

25

22or

t

The difference is due to the reading error of the chart.

4-84


4-106 A curing kiln is heated by injecting steam into it and raising its inner surface temperature to a specified value. It is to be determined whether the temperature at the outer surfaces of the kiln changes during the curing period.

Assumptions 1 The temperature in the wall is affected by the thermal conditions at inner surfaces only and the convection heat transfer coefficient inside is very large. Therefore, the wall can be considered to be a semi-infinite medium with a specified surface temperature of 45°C. 2 The thermal properties of the concrete wall are constant.

Properties The thermal properties of the concrete wall are given to be k = 0.9 W/m.°C and α = 0.23×10-5 m2/s.

Analysis We determine the temperature at a depth of x = 0.3 m in 3 h using the analytical solution,

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−−

txerfc

TTTtxT

is

i

α2

),(

2°C42°C

30 cm

Kiln wall

x0

Substituting,

C 9.1 °=

==

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

××=

−−

−

),(

1782.0)952.0(

)s/h 3600h 3)(/sm 1023.0(2

m 3.0542

2),(25

txT

erfc

erfctxT

which is greater than the initial temperature of 2°C. Therefore, heat will propagate through the 0.3 m thick wall in 3 h, and thus it may be desirable to insulate the outer surface of the wall to save energy.

4-107 The water pipes are buried in the ground to prevent freezing. The minimum burial depth at a particular location is to be determined.

Assumptions 1 The temperature in the soil is affected by the thermal conditions at one surface only, and thus the soil can be considered to be a semi-infinite medium with a specified surface temperature of -10°C. 2 The thermal properties of the soil are constant.

Soil Ti = 15°C

Water pipe

Ts =-10°C

x


Analysis The depth at which the temperature drops to 0°C in 75 days is determined using the analytical solution,

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−−

txerfc

TTTtxT

is

i

α2

),(

Substituting,

m 7.05=⎯→⎯

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

×××=

−−−

−

x

xerfc

)s/h 3600h/day 24day 75)(/sm 104.1(21510150

25

Therefore, the pipes must be buried at a depth of at least 7.05 m.

4-85


4-108 A hot dog is to be cooked by dropping it into boiling water. The time of cooking is to be determined.

Assumptions 1 Heat conduction in the hot dog is two-dimensional, and thus the temperature varies in both the axial x- and the radial r- directions. 2 The thermal properties of the hot dog are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The thermal properties of the hot dog are given to be k = 0.76 W/m.°C, ρ = 980 kg/m3, Cp = 3.9 kJ/kg.°C, and α = 2×10-7 m2/s.

Analysis This hot dog can physically be formed by the intersection of an infinite plane wall of thickness 2L = 12 cm, and a long cylinder of radius ro = D/2 = 1 cm. The Biot numbers and corresponding constants are first determined to be

37.47)C W/m.76.0(

)m 06.0)(C. W/m600( 2=

°°

==k

hLBi 2726.1 and 5381.1 11 ==⎯→⎯ Aλ

895.7)C W/m.76.0(

)m 01.0)(C. W/m600( 2=

°°

==k

hrBi o 5515.1 and 1251.2 11 ==⎯→⎯ Aλ


τ α= t L/ 2

2105.0)01.0(

)102()1251.2(exp)5515.1(

)06.0()102()5381.1(exp)2726.1(

100510080

),0(),0(),0,0(

2

72

2

72

112

12

1

=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡ ×−×

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡ ×−=

−−

⎟⎠⎞⎜

⎝⎛⎟⎠⎞⎜

⎝⎛==

−

−

−−

t

t

eAeAttt cylwallblockτλτλθθθ

Water 100°C

2 cm Hot dog Ti = 5°C

which gives

min 4.1 s 244 = =t

Therefore, it will take about 4.1 min for the hot dog to cook. Note that

2.049.0m) 01.0(

s) /s)(244m 102(2

27

2>=

×==

−

ocyl

rtατ

and thus the assumption τ > 0.2 for the applicability of the one-term approximate solution is verified.

Discussion This problem could also be solved by treating the hot dog as an infinite cylinder since heat transfer through the end surfaces will have little effect on the mid section temperature because of the large distance.

4-86


4-109 A long roll of large 1-Mn manganese steel plate is to be quenched in an oil bath at a specified rate. The temperature of the sheet metal after quenching and the rate at which heat needs to be removed from the oil in order to keep its temperature constant are to be determined.

Assumptions 1 The thermal properties of the balls are constant. 2 The heat transfer coefficient is constant and uniform over the entire surface. 3 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be checked).

Properties The properties of the steel plate are given to be k = 60.5 W/m.°C, ρ = 7854 kg/m3, and Cp = 434 J/kg.°C (Table A-3).

Analysis The characteristic length of the steel plate and the Biot number are

1.0 036.0

CW/m. 5.60)m 0025.0)(C.W/m 860(

m 0025.0

2<=

°°

==

===

khL

Bi

LAVL

c

sc

Steel plate 10 m/min

Oil bath 45°C

Since , the lumped system analysis is applicable. Therefore, Bi < 0.1

s 30min 5.0m/min 10

m 5velocitylengthtime

s 10092.0m) C)(0.0025J/kg. 434)( kg/m(7854

C.W/m 860 1-3

2

====

=°°

===cpp

s

LCh

VChA

bρρ

Then the temperature of the sheet metal when it leaves the oil bath is determined to be

C82.53°=⎯→⎯=−−

⎯→⎯=−− −−

∞

∞ )(4582045)()( s) 30)(s 10092.0( -1

tTetTeTTTtT bt

i

The mass flow rate of the sheet metal through the oil bath is

kg/min 4.785m/min) 10(m) 005.0(m) 2)(kg/m 7854( 3 ==ρ=ρ= wtVVm &&

Then the rate of heat transfer from the sheet metal to the oil bath and thus the rate at which heat needs to be removed from the oil in order to keep its temperature constant at 45°C becomes

kW 213.2 =J/min 10279.1C)4553.82)(CJ/kg. 434)(kg/min 4.785(])([ 7×=°−°=−= ∞TtTCmQ p&&

4-87


4-110E A stuffed turkey is cooked in an oven. The average heat transfer coefficient at the surface of the turkey, the temperature of the skin of the turkey in the oven and the total amount of heat transferred to the turkey in the oven are to be determined. Assumptions 1 The turkey is a homogeneous spherical object. 2 Heat conduction in the turkey is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the turkey are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions are applicable (this assumption will be verified). Properties The properties of the turkey are given to be k = 0.26 Btu/h.ft.°F, ρ = 75 lbm/ft3, Cp = 0.98 Btu/lbm.°F, and α = 0.0035 ft2/h. Analysis (a) Assuming the turkey to be spherical in shape, its radius is determined to be

Oven T∞ = 325°F

Turkey Ti = 40°F

ft 3545.04

)ft 1867.0(343

34

ft 1867.0lbm/ft 75

lbm 14

33

33

33

===⎯→⎯=

===⎯→⎯=

πππ

ρρ

VrrV

mVVm

oo

The Fourier number is 1392.0ft) 3545.0(

h) /h)(5ft 105.3(2

23

2=

×==

−

ortατ

which is close to 0.2 but a little below it. Therefore, assuming the one-term approximate solution for transient heat conduction to be applicable, the one-term solution formulation at one-third the radius from the center of the turkey can be expressed as

1

1)14.0(1

1

11

333.0)333.0sin(

491.032540325185

/)/sin(),(

),(

21

21

λλ

λλ

θ

λ

τλ

−

−

∞

∞

==−−

=−−

=

eA

rrrr

eATT

TtxTtx

o

o

isph

By trial and error, it is determined from Table 4-1 that the equation above is satisfied when Bi = 20 corresponding to λ1 12 9857 19781= =. and A . . Then the heat transfer coefficient can be determined from

F.Btu/h.ft 14.7 2 °=°

==⎯→⎯=)ft 3545.0(

)20)(FBtu/h.ft. 26.0(

o

o

rkBih

khr

Bi

(b) The temperature at the surface of the turkey is

F 317 °=⎯→⎯

===−− −−

),(

02953.09857.2

)9857.2sin()9781.1(

/)/sin(

32540325),( )14.0()9857.2(

1

11

221

trT

err

rreA

trT

o

oo

ooo

λλτλ

(c) The maximum possible heat transfer is Btu 3910=F)40325)(FBtu/lbm. 98.0)(lbm 14()(max °−°=−= ∞ ip TTmCQ


Btu 3238===

=−

−=λ

λλ−λθ−=

Btu) 3910)(828.0(828.0

828.0)9857.2(

)9857.2cos()9857.2()9857.2sin()491.0(31

)cos()sin(31

max

331

111,

max

QQ

QQ

spho

Discussion The temperature of the outer parts of the turkey will be greater than that of the inner parts when the turkey is taken out of the oven. Then heat will continue to be transferred from the outer parts of the turkey to the inner as a result of temperature difference. Therefore, after 5 minutes, the thermometer reading will probably be more than 185 . °F

4-88


4-111 The trunks of some dry oak trees are exposed to hot gases. The time for the ignition of the trunks is to be determined. Assumptions 1 Heat conduction in the trunks is one-dimensional since it is long and it has thermal symmetry about the center line. 2 The thermal properties of the trunks are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of the trunks are given to be k = 0.17 W/m.°C and α = 1.28×10-7 m2/s. Analysis We treat the trunks of the trees as an infinite cylinder since heat transfer is primarily in the radial direction. Then the Biot number becomes

24.38)C W/m.17.0(

)m 1.0)(C. W/m65( 2=

°°

==k

hrBi o

The constants λ1 and A1 corresponding to this Biot number are, from Table 4-1, 5989.1 and 3420.2 11 == Aλ


184.0m) 1.0(

s/h) 6003h /s)(4m 1028.1(2

27

20

=××

==−

rtατ

which is slightly below 0.2 but close to it. Therefore, assuming the one-term approximate solution for transient heat conduction to be applicable, the temperature at the surface of the trees in 4 h becomes

D = 0.2 m

Tree Ti = 30°C Hot

gases T∞ = 520°C

C410 >),(01935.0)0332.0()5989.1(

52030520),(

)/(),(

),(

)184.0()3420.2(

101

2

21

°°=⎯→⎯==−−

λ=−−

=θ

−

τλ−

∞

∞

C 511trTetrT

rrJeATT

TtrTtr

oo

oi

ocylo

Therefore, the trees will ignite. (Note: is read from Table 4-2). J0

4-112 A spherical watermelon that is cut into two equal parts is put into a freezer. The time it will take for the center of the exposed cut surface to cool from 25 to 3°C is to be determined. Assumptions 1 The temperature of the exposed surfaces of the watermelon is affected by the convection heat transfer at those surfaces only. Therefore, the watermelon can be considered to be a semi-infinite medium 2 The thermal properties of the watermelon are constant. Properties The thermal properties of the water is closely approximated by those of water at room temperature, k = 0.607 W/m.°C and α = k Cp/ ρ = 0.146×10-6 m2/s (Table A-9).

Analysis We use the transient chart in Fig. 4-23 in this case for convenience (instead of the analytic solution),

10

2

595.0)12(25

)12(31),(

1=

⎪⎪⎭

⎪⎪⎬

⎫

==

=−−−−

−=−−

−∞

∞

kth

tx

TTTtxT

i α

αξ

Therefore, min 46.7==×°

°== s 2804

/s)m 10146.0()C. W/m30(C) W/m.607.0()1(

26-22

2

2

22

αhkt

Freezer T∞ = -12°C

WatermelonTi = 25°C

4-89


4-113 A cylindrical rod is dropped into boiling water. The thermal diffusivity and the thermal conductivity of the rod are to be determined.

Assumptions 1 Heat conduction in the rod is one-dimensional since the rod is sufficiently long, and thus temperature varies in the radial direction only. 2 The thermal properties of the rod are constant.

Properties The thermal properties of the rod available are given to be ρ = 3700 kg/m3 and Cp = 920 J/kg.°C.

Analysis From Fig. 4-14b we have

Water 100°C

2 cm Rod Ti = 25°C

25.01

1

28.01007510093

==

⎪⎪⎭

⎪⎪⎬

⎫

==

=−−

=−−

∞

∞

o

o

o

o

o

hrk

Birr

rx

TTTT

From Fig. 4-14a we have

40.033.0

1002510075

25.01

2==

⎪⎪⎭

⎪⎪⎬

⎫

=−−

=−−

==

∞

∞ o

i

o

o

rt

TTTT

hrk

Bi ατ

Then the thermal diffusivity and the thermal conductivity of the material become

C W/m.0.756

/sm 102.22 27

°=°×==⎯→⎯=

×=×

==

−

−

C)J/kg. )(920kg/m /s)(3700m 1022.2(

s/min 60min 3m) 01.0)(40.0(40.0

327

22

pp

o

CkCk

tr

αρα

α

α

4-90


4-114 The time it will take for the diameter of a raindrop to reduce to a certain value as it falls through ambient air is to be determined.

Assumptions 1 The water temperature remains constant. 2 The thermal properties of the water are constant.

Properties The density and heat of vaporization of the water are ρ = 1000 kg/m3 and hfg = 2490 kJ/kg (Table A-9).

Analysis The initial and final masses of the raindrop are

kg 0000141.0m) 0015.0(

34)kg/m 1000(

34

kg 0000654.0m) 0025.0(34)kg/m 1000(

34

333

333

====

====

ππρρ

ππρρ

fff

iii

rVm

rVm

whose difference is

kg 0000513.00000141.00000654.0 =−=−= fi mmm

The amount of heat transfer required to cause this much evaporation is

kJ 1278.0kJ/kg) kg)(2490 0000513.0( ==Q

The average heat transfer surface area and the rate of heat transfer are

J/s 0.2777=C)518()m 10341.5(C).W/m 400()(

m 10341.52

m) (0.0015+m) 0025.0[(42

)(4

252

252222

°−×°=−=

×==+

=

−∞

−

TThAQ

rrA

is

fis

&

ππ

Air T∞ = 18°C

Raindrop 5°C

Then the time required for the raindrop to experience this reduction in size becomes

min 7.7====Δ⎯→⎯Δ

= s 460J/s 0.2777J 8.127

QQt

tQQ

&&

4-91


4-115E A plate, a long cylinder, and a sphere are exposed to cool air. The center temperature of each geometry is to be determined.

Assumptions 1 Heat conduction in each geometry is one-dimensional. 2 The thermal properties of the bodies are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The properties of bronze are given to be k = 15 Btu/h.ft.°F and α = 0.333 ft2/h.

Analysis After 5 minutes

Plate: First the Biot number is calculated to be 2ro

01944.0)FBtu/h.ft. 15(

)ft 12/5.0)(F.Btu/h.ft 7( 2=

°°

==k

hLBi

2roThe constants λ1 and A1 corresponding to this Biot number are, from Table 4-1,

0033.1 and 1410.0 11 == Aλ

2L


2.098.15ft) 12/5.0(

min/h) min/60 /h)(5ft 333.0(2

2

2>===

Ltατ

Then the center temperature of the plate becomes

F312°=⎯→⎯==−−

⎯→⎯=−−

= −−

∞

∞0

)98.15()1410.0(01

0,0 730.0)0033.1(

7540075 22

1 TeT

eATTTT

iwall

τλθ

Cylinder:

Bi = 0 02. 0050.1 and 1995.0 1119 ==⎯⎯⎯ →⎯ − ATable λ

F248°=⎯→⎯==−−

⎯→⎯=−−

= −−

∞

∞0

)98.15()1995.0(01

0,0 532.0)0050.1(

7540075 22

1 TeT

eATTTT

icyl

τλθ

Sphere:

Bi = 0 02. 0060.1 and 2445.0 1119 ==⎯⎯⎯ →⎯ − ATable λ

F201°=⎯→⎯==−−

⎯→⎯=−−

= −−

∞

∞0

)98.15()2445.0(01

0,0 387.0)0060.1(

7540075 22

1 TeT

eATTTT

isph

τλθ

After 10 minutes

2.097.31ft) 12/5.0(

min/h) min/60 /h)(10ft 333.0(2

2

2>===

Ltατ

Plate:

F248°=⎯→⎯==−−

⎯→⎯=−−

= −−

∞

∞0

)97.31()1410.0(01

0,0 531.0)0033.1(

7540075 22

1 TeT

eATTTT

iwall

τλθ

4-92


Cylinder:

F167°=⎯→⎯==−−

⎯→⎯=−−

= −−

∞

∞0

)97.31()1995.0(01

0,0 282.0)0050.1(

7540075 22

1 TeT

eATTTT

icyl

τλθ

Sphere:

F123°=⎯→⎯==−−

⎯→⎯=−−

= −−

∞

∞0

)97.31()2445.0(01

0,0 149.0)0060.1(

7540075 22

1 TeT

eATTTT

isph

τλθ

After 30 minutes

τ α= = = >

tL2

0 33305 12

959 0 2( .( . /

. . ft / h)(30 min / 60 min / h) ft)

2

2

Plate:

F123°=⎯→⎯==−−

⎯→⎯=−−

= −−

∞

∞0

)9.95()1410.0(01

0,0 149.0)0033.1(

7540075 22

1 TeT

eATTTT

iwall

τλθ

Cylinder:

F82°=⎯→⎯==−−

⎯→⎯=−−

= −−

∞

∞0

)9.95()1995.0(01

0,0 0221.0)0050.1(

7540075 22

1 TeT

eATTTT

icyl

τλθ

Sphere:

F76°=⎯→⎯==−−

⎯→⎯=−−

= −−

∞

∞0

)9.95()2445.0(01

0,0 00326.0)0060.1(

7540075 22

1 TeT

eATTTT

isph

τλθ

The sphere has the largest surface area through which heat is transferred per unit volume, and thus the highest rate of heat transfer. Consequently, the center temperature of the sphere is always the lowest.

4-93


4-116E A plate, a long cylinder, and a sphere are exposed to cool air. The center temperature of each geometry is to be determined. √

Assumptions 1 Heat conduction in each geometry is one-dimensional. 2 The thermal properties of the geometries are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The properties of cast iron are given to be k = 29 Btu/h.ft.°F and α = 0.61 ft2/h.

Analysis After 5 minutes

2ro

Plate: First the Biot number is calculated to be

01006.0)FBtu/h.ft. 29(

)ft 12/5.0)(F.Btu/h.ft 7( 2=

°°

==k

hLBi

2roThe constants λ1 and A1 corresponding to this Biot number are, from Table 4-1,

0017.1 and 0998.0 11 == Aλ

2L


2.028.29ft) 12/5.0(

min/h) min/60 /h)(5ft 61.0(2

2

2>===

Ltατ

Then the center temperature of the plate becomes

F318°=⎯→⎯==−−

⎯→⎯=−−

= −−

∞

∞0

)28.29()0998.0(01

0,0 748.0)0017.1(

7540075 22

1 TeT

eATTTT

iwall

τλθ

Cylinder:

Bi = 0 01. 0025.1 and 1412.0 1114 ==⎯⎯⎯ →⎯ − ATable λ

F257°=⎯→⎯==−−

⎯→⎯=−−

= −−

∞

∞0

)28.29()1412.0(01

0,0 559.0)0025.1(

7540075 22

1 TeT

eATTTT

icyl

τλθ

Sphere:

Bi = 0 01. 0030.1 and 1730.0 1114 ==⎯⎯⎯ →⎯ − ATable λ

F211°=⎯→⎯==−−

⎯→⎯=−−

= −−

∞

∞0

)28.29()1730.0(01

0,0 418.0)0030.1(

7540075 22

1 TeT

eATTTT

isph

τλθ

After 10 minutes

2.056.58ft) 12/5.0(

min/h) min/60 /h)(10ft 61.0(2

2

2>===

Ltατ

Plate:

F257°=⎯→⎯==−−

⎯→⎯=−−

= −−

∞

∞0

)56.58()0998.0(01

0,0 559.0)0017.1(

7540075 22

1 TeT

eATTTT

iwall

τλθ

4-94


Cylinder:

F176°=⎯→⎯==−−

⎯→⎯=−−

= −−

∞

∞0

)56.58()1412.0(01

0,0 312.0)0025.1(

7540075 22

1 TeT

eATTTT

icyl

τλθ

Sphere:

F132°=⎯→⎯==−−

⎯→⎯=−−

= −−

∞

∞0

)56.58()1730.0(01

0,0 174.0)0030.1(

7540075 22

1 TeT

eATTTT

isph

τλθ

After 30 minutes

2.068.175ft) 12/5.0(

min/h) min/60 /h)(30ft 61.0(2

2

2>===

Ltατ

Plate:

F132°=⎯→⎯==−−

⎯→⎯=−−

= −−

∞

∞0

)68.175()0998.0(01

0,0 174.0)0017.1(

7540075 22

1 TeT

eATTTT

iwall

τλθ

Cylinder:

F84.8°=⎯→⎯==−−

⎯→⎯=−−

= −−

∞

∞0

)68.175()1412.0(01

0,0 030.0)0025.1(

7540075 22

1 TeT

eATTTT

icyl

τλθ

Sphere:

F76.7°=⎯→⎯==−−

⎯→⎯=−−

= −−

∞

∞0

)68.175()1730.0(01

0,0 0052.0)0030.1(

7540075 22

1 TeT

eATTTT

isph

τλθ

The sphere has the largest surface area through which heat is transferred per unit volume, and thus the highest rate of heat transfer. Consequently, the center temperature of the sphere is always the lowest.

4-95


4-117E "!PROBLEM 4-117E" "GIVEN" 2*L=1/12 "[ft]" 2*r_o_c=1/12 "[ft], c stands for cylinder" 2*r_o_s=1/12 "[ft], s stands for sphere" T_i=400 "[F]" T_infinity=75 "[F]" h=7 "[Btu/h-ft^2-F]" "time=5 [min], parameter to be varied" "PROPERTIES" k=15 "[Btu/h-ft-F]" alpha=0.333*Convert(ft^2/h, ft^2/min) "[ft^2/min]" "ANALYSIS" "For plane wall" Bi_w=(h*L)/k "From Table 4-1 corresponding to this Bi number, we read" lambda_1_w=0.1410 A_1_w=1.0033 tau_w=(alpha*time)/L^2 (T_o_w-T_infinity)/(T_i-T_infinity)=A_1_w*exp(-lambda_1_w^2*tau_w) "For long cylinder" Bi_c=(h*r_o_c)/k "From Table 4-1 corresponding to this Bi number, we read" lambda_1_c=0.1995 A_1_c=1.0050 tau_c=(alpha*time)/r_o_c^2 (T_o_c-T_infinity)/(T_i-T_infinity)=A_1_c*exp(-lambda_1_c^2*tau_c) "For sphere" Bi_s=(h*r_o_s)/k "From Table 4-1 corresponding to this Bi number, we read" lambda_1_s=0.2445 A_1_s=1.0060 tau_s=(alpha*time)/r_o_s^2 (T_o_s-T_infinity)/(T_i-T_infinity)=A_1_s*exp(-lambda_1_s^2*tau_s)

time [min] To,w [F] To,c [F] To,s [F] 5 312.3 247.9 200.7 10 247.7 166.5 123.4 15 200.7 123.4 93.6 20 166.5 100.6 82.15 25 141.6 88.57 77.75 30 123.4 82.18 76.06 35 110.3 78.8 75.41 40 100.7 77.01 75.16 45 93.67 76.07 75.06 50 88.59 75.56 75.02 55 84.89 75.3 75.01 60 82.2 75.16 75

4-96


0 10 20 30 40 50 6050

100

150

200

250

300

350

50

100

150

200

250

300

350

time [min]

T o [F

]

wall

cylinder

sphere

4-97


4-118 Internal combustion engine valves are quenched in a large oil bath. The time it takes for the valve temperature to drop to specified temperatures and the maximum heat transfer are to be determined.

Assumptions 1 The thermal properties of the valves are constant. 2 The heat transfer coefficient is constant and uniform over the entire surface. 3 Depending on the size of the oil bath, the oil bath temperature will increase during quenching. However, an average canstant temperature as specified in the problem will be used. 4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be verified).

Properties The thermal conductivity, density, and specific heat of the balls are given to be k = 48 W/m.°C, ρ = 7840 kg/m3, and Cp = 440 J/kg.°C.

Oil T∞ = 45°C

Engine valveTi = 800°C

Analysis (a) The characteristic length of the balls and the Biot number are

1.0 024.0)CW/m. 48(

)m 0018.0)(C.W/m 650(

m 0018.08

m) 008.0(8.188.1

2)4/(8.1

2

2

<=°

°==

=====

khL

Bi

DDL

LDAVL

c

sc π

π

Therefore, we can use lumped system analysis. Then the time for a final valve temperature of 400°C becomes

s 7.2=⎯→⎯=

−−

⎯→⎯=−−

=°

°===

−−

∞

∞ teeTTTtT

DCh

VChA

b

bt

i

pp

s

)ts 10468.0(

1-3

2

-1

4580045400)(

s 10468.0m) C)(0.008J/kg. 440)( kg/m1.8(7840

C).W/m 650(88.1

8ρρ

(b) The time for a final valve temperature of 200°C is

s 15.1=⎯→⎯=−−

⎯→⎯=−− −−

∞

∞ teeTTTtT bt

i

)ts 10468.0( -1

4580045200)(

(c) The time for a final valve temperature of 46°C is

s 63.3=⎯→⎯=−−

⎯→⎯=−− −−

∞

∞ teeTTTtT bt

i

)ts 10468.0( -1

458004546)(

(d) The maximum amount of heat transfer from a single valve is determined from

)(per valve = J 564,23C)45800)(CJ/kg. 440)(kg 0709.0(][

kg 0709.04

m) 10.0(m) 008.0(8.1)kg/m 7840(4

8.1 23

2

kJ 23.56=°−°=−=

=π

=π

ρ=ρ=

ifp TTmCQ

LDVm

4-98


4-119 A watermelon is placed into a lake to cool it. The heat transfer coefficient at the surface of the watermelon and the temperature of the outer surface of the watermelon are to be determined.

Assumptions 1 The watermelon is a homogeneous spherical object. 2 Heat conduction in the watermelon is one-dimensional because of symmetry about the midpoint. 3 The thermal properties of the watermelon are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The properties of the watermelon are given to be k = 0.618 W/m.°C, α = 0.15×10-6 m2/s, ρ = 995 kg/m3 and Cp = 4.18 kJ/kg.°C.

Analysis The Fourier number is

[ ]252.0

m) 10.0(s/min 60min) 4060(4/s)m 1015.0(

2

26

2=

×+××=

α=τ

−

ort

Water melon

Ti = 35°C

Lake 15°C

which is greater than 0.2. Then the one-term solution can be written in the form

)252.0(11

0sph0,

21

21 25.0

15351520 λ−τλ−

∞

∞ ==−−

⎯→⎯=−−

=θ eAeATTTT

i

It is determined from Table 4-1 by trial and error that this equation is satisfied when Bi = 10, which corresponds to λ1 12 8363 19249= =. and A . . Then the heat transfer coefficient can be determined from

C. W/m61.8 2 °=°

==⎯→⎯=)m 10.0(

)10)(C W/m.618.0(

o

o

rkBih

khr

Bi

The temperature at the surface of the watermelon is

C 15.5 °=⎯→⎯=

−−

=λλ

=−−

=θ −τλ−

∞

∞

),(0269.01535

15),(8363.2

)rad 8363.2sin()9249.1(/

)/sin(),(),( )252.0()8363.2(

1

11

221

trTtrT

err

rreA

TTTtrT

tr

oo

oo

oo

i

ospho

4-99


4-120 Large food slabs are cooled in a refrigeration room. Center temperatures are to be determined for different foods. Assumptions 1 Heat conduction in the slabs is one-dimensional since the slab is large relative to its thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the slabs are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of foods are given to be k = 0.233 W/m.°C and α = 0.11×10-6 m2/s for margarine, k = 0.082 W/m.°C and α = 0.10×10-6 m2/s for white cake, and k = 0.106 W/m.°C and α = 0.12×10-6 m2/s for chocolate cake. Analysis (a) In the case of margarine, the Biot number is Air

T∞ = 0°C

Margarine, Ti = 30°C

365.5)CW/m. 233.0(

)m 05.0)(C.W/m 25( 2=

°°

==k

hLBi

The constants λ1 and A1 corresponding to this Biot number are, from Table 4-1, 2431.1and 3269.1 11 == Aλ

The Fourier number is 2.09504.0m) 05.0(

s/h) 6003 h/s)(6m 1011.0(2

26

2 >=××

==−

Ltατ

Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the temperature at the center of the box if the box contains margarine becomes

C 7.0 °=⎯→⎯=

−−

==−−

=θ −τλ−

∞

∞

),0(233.0030

0),0(

)2431.1(),0(

),0( )9504.0()3269.1(1

221

tTtT

eeATT

TtTt

iwall

(b) Repeating the calculations for white cake,

⎯→⎯=°

°== 24.15

)CW/m. 082.0()m 05.0)(C.W/m 25( 2

khLBi 2661.1and 4641.1 11 == Aλ

2.0864.0m) 05.0(

s/h) 6003 h/s)(6m 1010.0(2

26

2 >=××

==−

Ltατ

C 6.0 °=⎯→⎯=

−−

==−−

=θ −τλ−

∞

∞

),0(199.0030

0),0(

)2661.1(),0(

),0( )864.0()4641.1(1

221

tTtT

eeATT

TtTt

iwall

(c) Repeating the calculations for chocolate cake,

⎯→⎯=°

°== 79.11

)CW/m. 106.0()m 05.0)(C.W/m 25( 2

khLBi 2634.1and 4356.1 11 == Aλ

2.00368.1m) 05.0(

s/h) 6003 h/s)(6m 1012.0(2

26

2 >=××

==−

Ltατ

C 4.5 °=⎯→⎯=

−−

==−−

=θ −τλ−

∞

∞

),0(149.0030

0),0(

)2634.1(),0(

),0( )0368.1()4356.1(1

221

tTtT

eeATT

TtTt

iwall

4-100


4-121 A cold cylindrical concrete column is exposed to warm ambient air during the day. The time it will take for the surface temperature to rise to a specified value, the amounts of heat transfer for specified values of center and surface temperatures are to be determined. Assumptions 1 Heat conduction in the column is one-dimensional since it is long and it has thermal symmetry about the center line. 2 The thermal properties of the column are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of concrete are given to be k = 0.79 W/m.°C, α = 5.94×10-7 m2/s, ρ = 1600 kg/m3 and Cp = 0.84 kJ/kg.°C Analysis (a) The Biot number is

658.2)C W/m.79.0(

)m 15.0)(C. W/m14( 2=

°°

==k

hrBi o

The constants λ1 and A1 corresponding to this Biot number are, from Table 4-1, 3915.1 and 7240.1 11 ==λ A

Once the constant =0.3841 is determined from Table 4-2 corresponding to the constant

J0

λ1 , the Fourier number is determined to be

6253.0)3841.0()3915.1(28162827)/(

),( 221 )7240.1(

101 =τ⎯→⎯=−−

⎯→⎯λ=−− τ−τλ−

∞

∞ errJeATT

TtrToo

i

o

30 cm

Column 16°C

Air 28°C

which is above the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can be used. Then the time it will take for the column surface temperature to rise to 27°C becomes

hours 6.6==×

=ατ

=−

s 685,23/s)m 1094.5(

m) 15.0)(6253.0(27

22or

t

(b) The heat transfer to the column will stop when the center temperature of column reaches to the ambient temperature, which is 28°C. That is, we are asked to determine the maximum heat transfer between the ambient air and the column.

kJ 3990=°−°=−=

=π=ρπ=ρ=

∞ C)1628)(CkJ/kg. 84.0)(kg 8.395(][kg 8.395)]m 5.3(m) 15.0()[kg/m 1600(

max

232

ip

o

TTmCQLrVm

(c) To determine the amount of heat transfer until the surface temperature reaches to 27°C, we first determine

2169.0)3915.1(),0( )6253.0()7240.1(

122

1 ===−− −τλ−

∞

∞ eeATT

TtT

i

Once the constant J1 = 0.5787 is determined from Table 4-2 corresponding to the constant λ1 , the amount of heat transfer becomes

kJ 3409==

=

=××−=λλ

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=⎟⎟⎠

⎞⎜⎜⎝

⎛

∞

∞

)kJ 3990(854.00854

854.07240.15787.02169.021

)(21

max

1

110

cylmax

QQQ

JTTTT

QQ

i

4-101


4-122 Long aluminum wires are extruded and exposed to atmospheric air. The time it will take for the wire to cool, the distance the wire travels, and the rate of heat transfer from the wire are to be determined.

Assumptions 1 Heat conduction in the wires is one-dimensional in the radial direction. 2 The thermal properties of the aluminum are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be verified).

Properties The properties of aluminum are given to be k = 236 W/m.°C, ρ = 2702 kg/m3, Cp = 0.896 kJ/kg.°C, and α = 9.75×10-5 m2/s.

Analysis (a) The characteristic length of the wire and the Biot number are

350°C 10 m/min

Air 30°C

Aluminum wire

1.0 00011.0CW/m. 236

)m 00075.0)(C.W/m 35(

m 00075.02

m 0015.0222

2

<=°

°==

=====

khL

Bi

rLrLr

AVL

c

o

o

o

sc π

π

Since the lumped system analysis is applicable. Then, Bi < 0.1,

s 144=⎯→⎯=

−−

⎯→⎯=−−

=°°

===

−−

∞

∞ teeTTTtT

LCh

VChA

b

tbt

i

cpp

s

)s 0193.0(

1-3

2

-1

303503050)(

s 0193.0m) C)(0.00075J/kg. 896)( kg/m(2702

C.W/m 35ρρ

(b) The wire travels a distance of

m 24==→= s) m/s)(144 60/10(lengthtime

lengthvelocity

This distance can be reduced by cooling the wire in a water or oil bath.

(c) The mass flow rate of the extruded wire through the air is

kg/min 191.0m/min) 10(m) 0015.0()kg/m 2702()4/( 2320 ==== ππρρ VrVm &&

Then the rate of heat transfer from the wire to the air becomes

W856 =kJ/min 51.3=C)50350)(CkJ/kg. 896.0)(kg/min 191.0(])([ °−°=−= ∞TtTCmQ p&&

4-102


4-123 Long copper wires are extruded and exposed to atmospheric air. The time it will take for the wire to cool, the distance the wire travels, and the rate of heat transfer from the wire are to be determined.

Assumptions 1 Heat conduction in the wires is one-dimensional in the radial direction. 2 The thermal properties of the copper are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be verified).

Properties The properties of copper are given to be k = 386 W/m.°C, ρ = 8950 kg/m3, Cp = 0.383 kJ/kg.°C, and α = 1.13×10-4 m2/s.

Analysis (a) The characteristic length of the wire and the Biot number are

350°C 10 m/min

Air 30°C

Copper wire

1.0 000068.0CW/m. 386

)m 00075.0)(C.W/m 35(

m 00075.02

m 0015.0222

2

<=°

°==

=====

khL

Bi

rLrLr

AVL

c

o

o

o

sc π

π

Since the lumped system analysis is applicable. Then, Bi < 0.1

s 204=⎯→⎯=

−−

⎯→⎯=−−

=°°

===

−−

∞

∞ teeTTTtT

LCh

VChA

b

tbt

i

cpp

s

)s 0136.0(

1-3

2

-1

303503050)(

s 0136.0m) C)(0.00075J/kg. 383)( kg/m(8950

C.W/m 35ρρ

(b) The wire travels a distance of

m 34=⎟⎠⎞

⎜⎝⎛=⎯→⎯= s) (204

s/min 60m/min 10length

timelength

velocity

This distance can be reduced by cooling the wire in a water or oil bath.

(c) The mass flow rate of the extruded wire through the air is

kg/min 633.0m/min) 10(m) 0015.0()kg/m 8950()4/( 2320 ==== ππρρ VrVm &&

Then the rate of heat transfer from the wire to the air becomes

W1212 =kJ/min 72.7=C)50350)(CkJ/kg. 383.0)(kg/min 633.0(])([ °−°=−= ∞TtTCmQ p&&

4-103


4-124 A brick house made of brick that was initially cold is exposed to warm atmospheric air at the outer surfaces. The time it will take for the temperature of the inner surfaces of the house to start changing is to be determined.

Assumptions 1 The temperature in the wall is affected by the thermal conditions at outer surfaces only, and thus the wall can be considered to be a semi-infinite medium with a specified outer surface temperature of 18°C. 2 The thermal properties of the brick wall are constant.

Properties The thermal properties of the brick are given to be k = 0.72 W/m.°C and α = 0.45×10-4 m2/s.

Analysis The exact analytical solution to this problem is

Ti = 5°C 15°C

30 cm

Wall

x0

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−−

txerfc

TTTtxT

is

i

α

),(

Substituting,

)/sm 1045.0(2

m 3.001.051551.5

26 ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

×==

−−

− terfc

Noting from Table 4-3 that 0.01 = erfc(1.8215), the time is determined to be

min 251==⎯→⎯=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

× −s 070,15 8215.1

)/sm 1045.0(2

m 3.026

tt

4-104


4-125 A thick wall is exposed to cold outside air. The wall temperatures at distances 15, 30, and 40 cm from the outer surface at the end of 2-hour cooling period are to be determined.

Assumptions 1 The temperature in the wall is affected by the thermal conditions at outer surfaces only. Therefore, the wall can be considered to be a semi-infinite medium 2 The thermal properties of the wall are constant.

Properties The thermal properties of the brick are given to be k = 0.72 W/m.°C and α = 1.6×10-7 m2/s.

Analysis For a 15 cm distance from the outer surface, from Fig. 4-23 we have

25.0170.0

)s 36002)(s/m 10(1.62

m 15.02

98.2C W/m.0.72

)s 36002)(s/m 10(1.6C). W/m20(

26-

26-2

=−−

−

⎪⎪

⎭

⎪⎪

⎬

⎫

=××α

=ξ

=°

××°=

α

∞

∞

TTTT

tx

kth

i

Air 2°C

L =40 cm

Wall 18°C

C14.0°=⎯→⎯=−−

− TT 25.021821

For a 30 cm distance from the outer surface, from Fig. 4-23 we have

038.0140.1

)s 36002)(s/m 10(1.62

m 3.02

98.2C W/m.0.72

)s 36002)(s/m 10(1.6C). W/m20(

26-

26-2

=−−

−

⎪⎪

⎭

⎪⎪

⎬

⎫

=××α

=ξ

=°

××°=

α

∞

∞

TTTT

tx

kth

i

C17.4°=⎯→⎯=−−

− TT 038.021821

For a 40 cm distance from the outer surface, that is for the inner surface, from Fig. 4-23 we have

0187.1

)s 36002)(s/m 10(1.62

m 4.02

98.2C W/m.0.72

)s 36002)(s/m 10(1.6C). W/m20(

26-

26-2

=−−

−

⎪⎪

⎭

⎪⎪

⎬

⎫

=××α

=ξ

=°

××°=

α

∞

∞

TTTT

tx

kth

i

C18.0°=⎯→⎯=−−

− TT 021821

Discussion This last result shows that the semi-infinite medium assumption is a valid one.

4-105


4-126 The engine block of a car is allowed to cool in atmospheric air. The temperatures at the center of the top surface and at the corner after a specified period of cooling are to be determined.

Assumptions 1 Heat conduction in the block is three-dimensional, and thus the temperature varies in all three directions. 2 The thermal properties of the block are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The thermal properties of cast iron are given to be k = 52 W/m.°C and α = 1.7×10-5 m2/s.

Analysis This rectangular block can physically be formed by by the intersection of two infinite plane walls of thickness 2L = 40 cm (call planes A and B) and an infinite plane wall of thickness 2L = 80 cm (call plane C). We measure x from the center of the block.

(a) The Biot number is calculated for each of the plane wall to be

0231.0)C W/m.52(

)m 2.0)(C. W/m6( 2

BA =°

°===

khLBiBi Air

17°C Engine block

150°C 0462.0

)C W/m.52()m 4.0)(C. W/m6( 2

C =°

°==

khLBi

The constants λ1 and A1 corresponding to these Biot numbers are, from Table 4-1,

0038.1 and 150.0 B)A,(1B)A,(1 ==λ A

0076.1 and 212.0 (C)1(C)1 ==λ A

The Fourier numbers are

2.01475.1m) 2.0(

s/min) 60min /s)(45m 1070.1(2

25

2BA, >=××

==−

Ltατ

2.02869.0m) 4.0(

s/min) 60min /s)(45m 1070.1(2

25

2C >=××

==−

Ltατ

The center of the top surface of the block (whose sides are 80 cm and 40 cm) is at the center of the plane wall with 2L = 80 cm, at the center of the plane wall with 2L = 40 cm, and at the surface of the plane wall with 2L = 40 cm. The dimensionless temperatures are

9782.0)0038.1( )1475.1()150.0(1

0(A) wallo,

221 ===

−−

=θ −τλ−

∞

∞ eeATTTT

i

9672.0)150.0cos()0038.1()/cos(),(

),( )1475.1()150.0(11(B) wall

221 ==λ=

−−

=θ −τλ−

∞

∞ eLLeATT

TtxTtL

i

9947.0)0076.1( )2869.0()212.0(1

0(C) wallo,

221 ===

−−

=θ −τλ−

∞

∞ eeATTTT

i

Then the center temperature of the top surface of the cylinder becomes

C142.2°=⎯→⎯=−

−

=××=θ×θ×θ=⎥⎦

⎤⎢⎣

⎡−

−

∞

∞

),0,0,(9411.017150

17),0,0,(

9411.09947.09782.09672.0),(),0,0,(

(C) wallo,(A) wallo,(B) wall

tLTtLT

tLTT

TtLT

cylindershorti

(b) The corner of the block is at the surface of each plane wall. The dimensionless temperature for the surface of the plane walls with 2L = 40 cm is determined in part (a). The dimensionless temperature for the surface of the plane wall with 2L = 80 cm is determined from

4-106


9724.0)212.0cos()0076.1()/cos(),(

),( )2869.0()212.0(11(C) wall

221 ==λ=

−−

=θ −τλ−

∞

∞ eLLeATT

TtxTtL

i

Then the corner temperature of the block becomes

C138.0°=⎯→⎯=−

−

=××=θ×θ×θ=⎥⎦

⎤⎢⎣

⎡−

−

∞

∞

),,,(9097.017150

17),,,(

9097.09672.09672.09724.0),(),(),(),,,(

Awall,Bwall,Cwall,

tLLLTtLLLT

tLtLtLTT

TtLLLT

cylindershorti

4-107


4-127 A man is found dead in a room. The time passed since his death is to be estimated.

Assumptions 1 Heat conduction in the body is two-dimensional, and thus the temperature varies in both radial r- and x- directions. 2 The thermal properties of the body are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The human body is modeled as a cylinder. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

Properties The thermal properties of body are given to be k = 0.62 W/m.°C and α = 0.15×10-6 m2/s.

Analysis A short cylinder can be formed by the intersection of a long cylinder of radius D/2 = 14 cm and a plane wall of thickness 2L = 180 cm. We measure x from the midplane. The temperature of the body is specified at a point that is at the center of the plane wall but at the surface of the cylinder. The Biot numbers and the corresponding constants are first determined to be

06.13)C W/m.62.0(

)m 90.0)(C. W/m9( 2

wall =°

°==

khLBi

⎯ →⎯ = =λ1 114495 12644. . and A

03.2)C W/m.62.0(

)m 14.0)(C. W/m9( 20

cyl =°

°==

khr

Bi

3408.1and 6052.1 11 ==⎯→⎯ Aλ

Noting that for the plane wall and for cylinder and J

τ α= t L/ 2 τ α= t r/ 02

0(1.6052)=0.4524 from Table 4-2, and assuming that τ > 0.2 in all dimensions so that the one-term approximate solution for transient heat conduction is applicable, the product solution method can be written for this problem as

D0 = 28 cm

z

Human body Ti = 36°C

r

Air T∞ = 16°C

2L=180 cm

hours 9.0==⎯→⎯

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡ ×−×

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡ ×−=

⎥⎦⎤

⎢⎣⎡ λ=

−−

θθ=θ

−−

τλ−τλ−

s 404,32

)4524.0()14.0(

)1015.0()6052.1(exp)3408.1()90.0(

)1015.0()4495.1(exp)2644.1(40.0

)/()(16361623

),(),0(),,0(

2

62

2

62

01011

cyl0wall0

21

21

t

tt

rrJeAeA

trttr block

4-128 ··· 4-131 Design and Essay Problems

4-108

Ch.4

Documents

Transcript of Ch.4