Solution manual heat and mass transfer a practical approach 3rd edition cengel CH04 3

4 The heat transfer coefficient is constant and uniform over the entire surface.. 4 The heat transfer coefficient is constant and uniform over the entire surface.. 2 The heat transfer c

Review Problems

4-110 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

1.0019.0)

C W/m

43(

)m02.0)(

C W/m40(

m02.0

L

c s c

h c

s

)t s 000544 0 (

1 - 3

6 2

1 -

5015

500)

(

s000544.0m)(0.02)C.J/m10675.3(

C W/m40ρ

ρ V

/sm1017.1

C W/m

.05015

500

6.52019.0

11

2 0

m)02.0)(

15(

2 5

2 2

α

τr o

t

The difference is due to the reading error of the chart

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4-111 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 2.5 h using the analytical solution,

6°C 42°C

30 cm Kiln wall

T

T t

x

T

i s

i

α2

)s/h3600h5.2)(

/sm1023.0(2

m3.06

4-112 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

Properties The thermal properties of the soil are given

Analysis The depth at which the temperature drops

to 0°C in 75 days is determined using the analytical

T

T t

x

T

i s

i

α2

)s/h3600h/day24day75)(

/sm104.1(215

10

15

0

2 5

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

4-113 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 r o = D/2 = 1 cm The Biot numbers and corresponding constants

are first determined to be

37.47)

C W/m

76.0(

)m06.0)(

C W/m600

C W/m

76.0(

)m01.0)(

C W/m600

α

τ=

2105.0)

01.0(

)102()1249.2(exp)5514.1

(

)06.0(

)102()5380.1(exp)2726

2

7 2

1 1

2 2

e A e

A t

01.0(

s)/s)(244m

102(

2

2 7

o cyl

r

t

α

τ

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

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4-114 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 steel plate 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 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

1.0036.0C

W/m

5.60

)m0025.0)(

C W/m860(

m0025.0

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

s36min6.0m/min15

m9velocity

lengthtime

s10092.0m)C)(0.0025J/kg

434)(

kg/m(7854

C W/m

s

L c

h c

hA b

ρ

ρ V

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

C 65.5°

45)()

t T e

t T e

kg/m7854

=J/min 10048.1C)455.65)(

CJ/kg

434)(

kg/min1178(])(

=m c T t T∞

Q& & p

4-115E 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

ft3545.04

)ft1867.0(34

33

4

ft1867.0lbm/ft75

lbm14

3

3 3

3

3 3

π

ρρ

VV

VV

o

r

m m

ft)3545.0(

h)/h)(5ft105.3(

2

2 3

o r

t

ατ

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

1 1

333.0

)333.0sin(

491.0325

),()

,

(

2 2

λλλ

λθ

λ

τ λ

r r

r r e

A T

T

T t x T t

x

o

o i

sph

By trial and error, it is determined from Table 4-2 that the equation above is satisfied when Bi = 20

corresponding to λ1=2.9857 and A1 =1.9781 Then the heat transfer coefficient can be determined from

)ft3545.0(

)20)(

FBtu/h.ft

26.0(

o

o

r

kBi h k

02953.09857.2

)9857.2sin(

)9781.1(/

)/sin(

32540

325)

,

1

1 1

2 2

t r T

e r

r

r r e

A t

r

T

o

o o

o o o

λ

λ

τ λ

(c) The maximum possible heat transfer is

Btu3910

=F)40325)(

FBtu/lbm

98.0)(

lbm14()(

Q

Then the actual amount of heat transfer becomes

Btu 3240

828.0(828

.0

828.0)

9857.2(

)9857.2cos(

)9857.2()9857.2sin(

)491.0(31)cos(

)sin(

31

max

3 3

1

1 1 1 ,

max

Q Q

Q

Q

sph o

λ

λλλθ

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

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4-116 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

D = 0.2 m

Tree

Ti = 30°C Hot

gases

T∞ = 520°C

)C W/m

17.0(

)m1.0)(

C W/m65

The constants λ1andA1corresponding to this Biot

number are, from Table 4-2,

5989.1 and 3420

1.0(

s/h)6003h/s)(4m1028.1(

2

2 7

>

),(01935

.0)0332.0()

5989.1(52030

520)

,

(

)/()

,(),(

) 184 0 ( ) 3420 2 (

1 0 1

2 2

t r T e

t

r

T

r r J e A T

T

T t r T t r

o o

o i

o cyl

Therefore, the trees will ignite (Note:J0 is read from Table 4-3)

4-117 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/ρc p = 0.146×10-6 m2/s (Table A-9)

Analysis We use the transient chart in Fig 4-29 in this

case for convenience (instead of the analytic solution),

10

2

595.0)12(25

)12(31)

x

T T

T t x

T

αξ

C) W/m

607.0()

1(

2 6 - 2

2

2 2

2 2

α

h

k t

Freezer

T∞ = -12°C

Watermelon

Ti = 25°C

4-118 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 4-16b we have

25.01

1

28.010075

10093

o o

hr

k Bi r

r r

.010025

10075

25.01

0.756

/s m 10

)(920kg/m/s)(3700m

1022.2(

s/min60min3

m)01.0)(

40.0(40

0

3 2

7

2 2

p p

o

c k c

k

t r

αρα

α

α

4-119 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

kg0000141

0m)0015.0(3

4)kg/m1000(34

kg0000654

0m)0025.0(3

4)kg/m1000(34

3 3

3

3 3

ρρ

ππ

ρ

ρ

f f

f

i i

i

r m

r m

00000654

=C)518(m10341.5(C) W/m400()(

m10341.52

m)(0.0015+m)0025.0[(

42

)(

4

2 5 2

2 5 2

2 2

Q

r r A

i s

f i s

&

ππ

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

min 7.7

=

=

=

=Δ

⎯→

⎯Δ

J/s0.2777

J8.127

Q

Q t t

Q

Q

&

&

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4-120E A plate, a long cylinder, and a sphere are exposed to cool air The center temperature of each

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

2r o

)FBtu/h.ft

15(

)ft12/5.0)(

F.Btu/h.ft7

min/h)min/60/h)(5ft333

98 15 ( 1387 0 ( 0

1 0

75400

2

T e

T e

A T T

T T

98 15 ( 1962 0 ( 0

1 ,

75400

2

T e

T e

A T T

T T i

98 15 ( 2405 0 ( 0

1 ,

75400

2

T e

T e

A T T

T T

12/5.0(

min/h)min/60/h)(10ft333.0(

97 31 ( 1387 0 ( 0

1 0

,

75400

2

T e

T e

A T T

T T

97 31 ( 1962 0 ( 0

1 ,

75400

2

T e

T e

A T

97 31 ( ) 2405 0 ( 0

1 0

,

75400

2

T e

T e

A T

T

T T

i

θ

After 30 minutes

2.09.95ft)

12/5.0(

min/h)min/60/h)(30ft333.0(

9 95 ( 1387 0 ( 0

1 0

,

75400

2

T e

T e

A T T

T T

9 95 ( 1962 0 ( 0

1 ,

75400

2

T e

T e

A T

9 95 ( 2405 0 ( 0

1 ,

75400

2

T e

T e

A T

T

T T

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4-121E A plate, a long cylinder, and a sphere are exposed to cool air The center temperature of each

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

2r o

Plate: First the Biot number is calculated to be

)FBtu/h.ft

29(

)ft12/5.0)(

F.Btu/h.ft7

min/h)min/60/h)(5ft61

28 29 ( 0998 0 ( 0

1 0

,

75400

2

T e

T e

A T T

T T

28 29 ( ) 1412 0 ( 0

1

0 ,

75400

2

T e

T e

A T T

T T i

28 29 ( 1730 0 ( 0

1 ,

75400

2

T e

T e

A T T

T T

12/5.0(

min/h)min/60/h)(10ft61.0(

56 58 ( 0998 0 ( 0

1 0

,

75400

2

T e

T e

A T T

T T

56 58 ( ) 1412 0 ( 0

2

T e

T e

A T

56 58 ( ) 1730 0 ( 0

1 0

,

75400

2

T e

T e

A T

T

T T

i

θ

After 30 minutes

2.068.175ft)

12/5.0(

min/h)min/60/h)(30ft61.0(

68 175 ( 0998 0 ( 0

1 0

,

75400

2

T e

T e

A T T

T T

68 175 ( ) 1412 0 ( 0

2

T e

T e

A T

68 175 ( 1730 0 ( 0

1 0

,

75400

2

T e

T e

A T

T

T T

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4-122E EES Prob 4-120E is reconsidered The center temperature of each geometry as a function of the

cooling time is to be plotted

Analysis The problem is solved using EES, and the solution is given below

"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"

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4-123 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 c p = 440 J/kg.°C

Oil

T∞ = 50°C Engine valveTi = 800°C

Analysis (a) The characteristic length of the

balls and the Biot number are

1.003.0C

W/m

48

)m0018.0)(

C W/m800

(

m0018.08

m)008.0(8.18

8.12

)4/(

8

1

2 2

L D A

Therefore, we can use lumped system analysis Then the

time for a final valve temperature of 400°C becomes

s 5.9

h c

hA

b

bt i

p p

s

)t s 1288 0 (

1 - 3

2

1 -

50800

50400)

(

s1288.0m)C)(0.008J/kg

440)(

kg/m1.8(7840

C) W/m800(88

.1

8ρ

ρ V

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

s 12.5

50800

50200)

(

(c) The time for a final valve temperature of 51°C is

s 51.4

50800

5051)

(

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

)(per valve

=J400,23C)50800)(

CJ/kg

440)(

kg0709.0(][

kg0709.04

m)10.0(m)008.0(8.1)kg/m7840(4

8

kJ 23.4

p T T mc

Q

L D

4-124 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 c p = 4.18 kJ/kg.°C

Analysis The Fourier number is

m)10.0(

s/min60min) 4060(4/s)m1015

Ti = 35 °C

Lake

15 °C

which is greater than 0.2 Then the one-term solution can be

written in the form

1535

15

τ λ

It is determined from Table 4-2 by trial and error that this equation is satisfied when Bi = 10, which

corresponds to λ1=2.8363 and A1 =1.9249 Then the heat transfer coefficient can be determined from

C W/m 61.8 2 °

)10)(

C W/m

618.0(

o

o

r

kBi h k

hr

Bi

The temperature at the surface of the watermelon is

C 15.5°

.015

35

15)

,

(

8363.2

)rad8363.2sin(

)9249.1(/

)/sin(

),()

,

1

1 1

2 2

t r T t

r

T

e r

r

r r e

A T

T

T t r T t

r

o o

o o

o o i

o sph

λ

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4-125 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)C W/m

233.0(

)m05.0)(

C W/m25

The constants λ1andA1corresponding to this Biot

number are, from Table 4-2,

2431.1 and 3269

s/h)6003h/s)(6m1011.0(

2

2 6

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°

.00

30

0)

,

0

(

)2431.1()

,0()

t T t

T

e e

A T T

T t T t

082.0(

)m05.0)(

C W/m25

05.0(

s/h)6003h/s)(6m1010.0(

2

2 6

.00

30

0)

,

0

(

)2661.1()

,0()

,

0

t T t

T

e e

A T T

T t T t

106.0(

)m05.0)(

C W/m25

05.0(

s/h)6003h/s)(6m1012.0(

2

2 6

.00

30

0)

,

0

(

)2634.1()

,0()

,

0

t T t

T

e e

A T T

T t T t

i

θ

4-126 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/m3and cp = 0.84 kJ/kg.°C

30 cm

Column

16 °C

Air 28°C

Analysis (a) The Biot number is

658.2)C W/m

79.0(

)m15.0)(

C W/m14

The constants λ1andA1corresponding to this

Biot number are, from Table 4-2,

3915.1 and 7240

λ

Once the constant =0.3841 is determined from Table 4-3

corresponding to the constant

2827)

/()

,

1 0

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 time it will take for the column surface temperature to rise to 27°C becomes

hours 7.1

m)15.0)(

6771.0(

2 7

2 2

α

τr o

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 5320

kg4.452)]

m4(m)15.0()[

kg/m1600(

max

2 3

2

i p

o

T T mc Q

L r

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

determine

1860.0)

3915.1()

T

i

τ λ

Once the constant J1 = 0.5787 is determined from Table 4-3 corresponding to the constant λ1, the amount

of heat transfer becomes

kJ 4660

875.07240.1

5787.01860.021)(2

1

max

1

1 1 0

cyl max

Q

Q Q

J T T

T T Q

Q

λ

PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and

educators for course preparation If you are a student using this Manual, you are using it without permission.

4-127 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

350°C

10 m/min

Air 30°C

Aluminum wire

Analysis (a) The characteristic length of

the wire and the Biot number are

1.000011.0C

W/m

236

)m00075.0)(

C W/m

35

(

m00075.02

m0015.022

2 2

h c

hA b

t bt

i

c p p

s

) s 0193 0 (

1 - 3

2

1 -

30350

3050)

(

s0193.0m)C)(0.00075J/kg

896)(

kg/m(2702

C W/m35ρ

ρ V

(b) The wire travels a distance of

m 24

=

=

→

= length (10/60m/s)(144s)time

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/min191.0m/min)10(m)0015.0()kg/m2702()

=kJ/min51.3

=C)50350)(

CkJ/kg

896.0)(

kg/min191.0(])(

=m c T t T∞

Q& & p

4-128 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

.0000068.0C

W/m

386

)m00075.0)(

C W/m35

(

m00075.02

m0015.022

2 2

L r

A

L

c

o o

h c

hA

b

t bt

i

c p p

s

) s 0136 0 (

1 - 3

2

1 -

30350

3050)

(

s0136.0m)C)(0.00075J/kg

383)(

kg/m(8950

C W/m35ρ

ρ V

(b) The wire travels a distance of

m 34

m/min10lengthtime

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/min633.0m/min)10(m)0015.0()kg/m8950()

=kJ/min72.7

=C)50350)(

CkJ/kg

383.0)(

kg/min633.0(])(

=m c T t T∞

Q& & p

PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and

educators for course preparation If you are a student using this Manual, you are using it without permission.