Heat Chap12 088

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Chapter 12Radiation Heat Transfer

Review Problems

12-88 The temperature of air in a duct is measured by athermocouple. The radiation effect on the temperaturemeasurement is to be quantified, and the actual airtemperature is to be determined.

Assumptions The surfaces are opaque, diffuse, and gray.

PropertiesThe emissivity of thermocouple is given to be=0.6.AnalysisThe actual temperature of the air can bedetermined from

K1111=

+=

+=

CW/m60

])K500()K850)[(KW/m1067.5)(6.0(K850

)(

2

44428

44

h

TTTT wthththf

12-89 The temperature of hot gases in a duct is measured by a thermocouple. The actual temperature ofthe gas is to be determined, and compared with that without a radiation shield.

Assumptions The surfaces are opaque, diffuse, and gray.

PropertiesThe emissivity of the thermocouple is given to be =0.7.AnalysisAssuming the area of the shield to be very close to the sensorof the thermometer, the radiation heat transfer from the sensor isdetermined from

2

44428

21

42

41

sensorfromrad,

W/m9.257

115.0

121

7.0

1

])K380()K530)[(KW/m1067.5(

11

211

)(

=

+

=

+

=

TTQ

Then the actual temperature of the gas can be determined from a heat transfer balance to be

K532==

=

=

ff

thf

TT

TTh

qq

22

2

sensorfromconv,sensortoconv,

W/m9.257)530C(W/m120

W/m9.257)(

Without the shield the temperature of the gas would be

K549.2=

+=

+=

CW/m120

])K380()K530)[(KW/m1067.5)(7.0(K530

)(

2

44428

44

hTTTT wthththf

12-68

Air, Tf

Tw

= 500 K

ThermocoupleT

th= 850 K

= 0.6

Air, Tf

Tw

= 380 K

ThermocoupleT

th= 530 K

1

= 0.7

2

= 0.15

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12-90E A sealed electronic box is placed in a vacuum chamber. The highest temperature at which thesurrounding surfaces must be kept if this box is cooled by radiation alone is to be determined.

Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3Convection heat transfer is not considered. 4 Heat transfer from the bottom surface of the box isnegligible.

PropertiesThe emissivity of the outer surface of the box is = 0.95.

AnalysisThe total surface area is2

ft67.3)11()12/18(4 =+=sA

Then the temperature of the surrounding surfaces is determined to be

F43==

=

=

R503

])R590)[(RBtu/h.ft101714.0)(m67.3)(95.0(Btu/h)41214.3100(

)(

444282

44

surr

surr

surrssrad

T

T

TTAQ

12-91 A double-walled spherical tank is used to store iced water. The air space between the two walls is

evacuated. The rate of heat transfer to the iced water and the amount of ice that melts a 24-h period are tobe determined.

Assumptions 1 Steady operating conditions exist 2 Thesurfaces are opaque, diffuse, and gray.

PropertiesThe emissivities of both surfaces are given to be 1= 2 = 0.15.

Analysis(a) Assuming the conduction resistance s of the wallsto be negligible, the rate of heat transfer to the iced water inthe tank is determined to be

m m2A D1 12 22 01 12 69= = = ( . ) .

W107.4=

+

++=

+

=

2

444282

2

2

1

2

2

1

4142112

04.2

01.2

15.0

15.01

15.0

1

])K2730()K27320)[(KW/m1067.5)(m69.12(

11

)(

D

D

TTAQ

(b) The amount of heat transfer during a 24-hour period is

kJ9275)s360024)(kJ/s1074.0( === tQQ

The amount of ice that melts during this period then becomes

kg27.8====kJ/kg7.333

kJ9275

if

ifh

QmmhQ

12-69

D2

= 2.04 m

T2

= 20C

2= 0.15

D1

= 2.01 m

T1

= 0C

1= 0.15

Vacuum

Icedwater0 C

Tsurr

8 in

100 W = 0.95

Ts=130F

12 in

12 in

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12-92 Two concentric spheres which are maintained at uniform temperatures are separated by air at 1 atmpressure. The rate of heat transfer between the two spheres by natural convection and radiation is to bedetermined.

Assumptions1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Air is anideal gas with constant properties.

PropertiesThe emissivities of the surfaces are given to be1 = 2 = 0.5. The properties of air at 1 atm and theaverage temperature of (T1+T2)/2 = (350+275)/2 = 312.5 K= 39.5C are (Table A-15)

1-

25

K0032.0K5.312

1

7256.0Pr

/sm10697.1

CW/m.02658.0

==

==

=

k

Analysis(a) Noting thatDi = D1 andDo = D2, the characteristic length is

m0.05m)0.15m25.0(2

1)(

2

1=== ioc DDL

Then

5

225

3-12

2

321 10415.7)7256.0(

)/sm10697.1(

)m05.0)(K275350)(K003200.0)(m/s81.9(Pr

)(=

=

=

cLTTgRa

The effective thermal conductivity is

[ ] [ ]005900.0

m)25.0(m)15.0(m)25.0(m)15.0(

m05.0

)()(57/5-7/5-455/75/74

sph =+

=+

=

oioi

c

DDDD

LF

[ ] CW/m.1315)10415.7)(00590.0(7256.0861.07256.0C)W/m.02658.0(74.0

)(Pr861.0

Pr74.0

4/154/1

4/14/1

eff

=

+=

+= RaFkk sph

Then the rate of heat transfer between the spheres becomes

W23.3=

=

= K)275350(

)m05.0(

)m25.0)(m15.0()CW/m.1315.0()(eff oi

c

oi TTL

DDkQ

(b) The rate of heat transfer by radiation is determined from

W32.3=

+

=

+

=

===

2

444282

2

2

1

2

2

1

41

421

12

22211

25.0

15.0

9.0

9.01

9.0

1

])K275()K350)[(KW/m1067.5)(m0707.0(

11

)(

m0707.0)m15.0(

D

D

TTAQ

DA

12-70

D2

= 25 cm

T2 = 275 K

2= 0.5

D1 = 15 cmT

1= 350 K

1

= 0.9

Lc

=5 cm

AIR1 atm

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12-93 A solar collector is considered. The absorber plate and the glass cover are maintained at uniformtemperatures, and are separated by air. The rate of heat loss from the absorber plate by natural convectionand radiation is to be determined.

Assumptions 1 Steady operating conditions exist 2 Thesurfaces are opaque, diffuse, and gray. 3 Air is an ideal gaswith constant properties.

Properties The emissivities of surfaces are given to be 1 = 0.9for glass and 2 = 0.8 for the absorber plate. The properties ofair at 1 atm and the average temperature of (T1+T2)/2 =(80+32)/2 = 56C are (Table A-15)

1-

25

K003040.0K)27356(

11

7212.0Pr

/sm10857.1

CW/m.02779.0

=+

==

==

=

fT

k

AnalysisFor = 0 , we have horizontal rectangular enclosure.The characteristic length in this case is the distance between the

two glassesLc = L = 0.03 m Then,

4

225

3-12

2

321 10083.8)7212.0(

)/sm10857.1(

)m03.0)(K3280)(K00304.0)(m/s81.9(Pr

)(=

=

=

LTTgRa

2m5.4m)3(m)5.1( === WHAs

[ ] [ ]

747.3

118

)20cos()10083.8(

)20cos()10083.8(

)208.1sin(17081

)20cos()10083.8(

1708144.11

118

)cosRa(

cosRa

)8.1(sin17081

cosRa

1708144.11Nu

3/14

4

6.1

4

3/16.1

=

+

+=

+

+=

++

++

W750=

=

=m03.0

C)3280()m5.4)(747.3)(CW/m.02779.0(

221

L

TTkNuAQ s

Neglecting the end effects, the rate of heat transfer by radiation is determined from

W1289=

+

++=

+

=

19.0

1

8.0

1

])K27332()K27380)[(KW/m1067.5)(m5.4(

111

)( 444282

21

42

41

rad

TTAQ

s

DiscussionThe rates of heat loss by natural convection for the horizontal and vertical cases would be asfollows (Note that the Ra number remains the same):

Horizontal:

812.3118

)10083.8(

10083.8

1708144.111

18

Ra

Ra

1708144.11Nu

3/14

4

3/1

=

+

+=

+

+=

++++

W1017=

=

=m03.0

C)3280()m6)(812.3)(CW/m.02779.0(

221

L

TTkNuAQ s

Vertical:

001.2m03.0

m2)7212.0()10083.8(42.0Pr42.0

3.0012.04/14

3.0012.04/1 =

=

=

L

HRaNu

12-71

Solar

radiation

= 20Insulation

Absorber plateT

1=80C

1

= 0.8

Glass cover,T

2= 32C

2

= 0.9

1.5 m

L = 3 cm

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W534=

=

=m03.0

C)3280()m6)(001.2)(CW/m.02779.0(

221

L

TTkNuAQ s

12-72

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12-94E The circulating pump of a solar collector that consists of a horizontal tube and its glass coverfails. The equilibrium temperature of the tube is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Thetube and its cover are isothermal. 3 Air is an ideal gas. 4The surfaces are opaque, diffuse, and gray for infraredradiation. 5 The glass cover is transparent to solarradiation.

Properties The properties of air should be evaluated at theaverage temperature. But we do not know the exittemperature of the air in the duct, and thus we cannotdetermine the bulk fluid and glass cover temperatures atthis point, and thus we cannot evaluate the averagetemperatures. Therefore, we will assume the glasstemperature to be 85F, and use properties at an anticipatedaverage temperature of (75+85)/2 =80F (Table A-15E),

s

k

/ft101.697/hft6110.0

FftBtu/h01481.0

24-2 ==

=

R540

11

7290.0Pr

ave

==

=

T

Analysis We have a horizontal cylindrical enclosure filled with air at 0.5 atm pressure. The problem

involves heat transfer from the aluminum tube to the glass cover and from the outer surface of the glasscover to the surrounding ambient air. When steady operation is reached, these two heat transfer rates mustequal the rate of heat gain. That is,

Btu/h30gainsolarambient-glassglass-tube === QQQ (per foot of tube)

The heat transfer surface area of the glass cover is

2ft309.1ft)1)(ft12/5()( ==== WDAA oglasso (per foot of tube)

To determine the Rayleigh number, we need to know the surface temperature of the glass, which is notavailable. Therefore, solution will require a trial-and-error approach. Assuming the glass covertemperature to be 85F, the Rayleigh number, the Nusselt number, the convection heat transfercoefficient, and the rate of natural convection heat transfer from the glass cover to the ambient air are

determined to be

6

224

32

2

3

10092.1)7290.0()/sft10675.1(

)ft12/5)(R7585](R)540/(1)[ft/s2.32(

Pr)(

Ra

=

=

=

ooDo

DTTg

( )[ ] ( )[ ]95.14

7290.0/559.01

)10092.1(387.06.0

Pr/559.01

Ra387.06.0Nu

2

27/816/9

6/162

27/816/9

1/6D

=

+

+=

++=

FftBtu/h5315.0)95.14(

ft12/5

FftBtu/h01481.0Nu

2

0

=

==D

kho

Btu/h96.6F)7585)(ft309.1)(FftBtu/h5315.0()(22

conv, === TTAhQ oooo

Also,

[ ]Btu/h5.30

R)535(R)545()ft309.1)(RftBtu/h101714.0)(9.0(

)(

442428

4sky

4rad,

==

=

TTAQ oooo

Then the total rate of heat loss from the glass cover becomes

Btu/h5.375.300.7rad,conv,total, =+=+= ooo QQQ

12-73

D2=5 in

Aluminum tube

D1=2.5 in,T

1

1= 0.9

Air space0.5 atm

Plastic cover,

2= 0.9, T

2

Water

T= 75F

Tsky

= 60F

30 Btu/h.ft

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which is more than 30 Btu/h. Therefore, the assumed temperature of 85F for the glass cover is high.Repeating the calculations with lower temperatures (including the evaluation of properties), the glasscover temperature corresponding to 30 Btu/h is determined to be 81.5F.

The temperature of the aluminum tube is determined in a similar manner using the naturalconvection and radiation relations for two horizontal concentric cylinders. The characteristic length inthis case is the distance between the two cylinders, which is

ft1.25/12in25.12/)5.25(2/)( ====ioc

DDL

Also,2

ft6545.0ft)1)(ft12/5.2()( ==== WDAA itubei (per foot of tube)

We start the calculations by assuming the tube temperature to be 118.5F, and thus an averagetemperature of (81.5+118.5)/2 = 100F=640 R. Using properties at 100F,

4

224

32

2

3

10334.1)726.0()/sft10809.1(

)ft12/25.1)(R5.815.118](R)640/(1)[ft/s2.32(Pr

)(Ra =

=

=

LTTg oiL

The effective thermal conductivity is

1466.0]ft)12/5(ft)12/5.2[(ft)(1.25/12

)]5.2/5[ln(

)(

)]/[ln(53/5-3/5-3

4

55/35/33

4

cyc =+

=+

= oic

io

DDL

DDF

FftBtu/h03227.0

)10334.11466.0(0.7260.861

0.726F)ftBtu/h01529.0(386.0

)Ra(Pr861.0

Pr

386.0

4/14

4/1

cyc

4/1

eff

=

+=

+= LFkk

Then the rate of heat transfer between the cylinders by convection becomes

Btu/h8.10F)5.815.118(ln(5/2.5)

F)ftBtu/h03227.0(2)(

)/ln(

2 effconv, =

==

oi

io

i TTDD

kQ

Also,

[ ]Btu/h0.25

in5

in5.2

9.0

9.01

9.0

1

R)5.541(R)5.578()ft6545.0)(RftBtu/h101714.0(

11

)( 442428

i

4o

4

rad, =

+

=

+

=

o

i

o

o

iii

D

D

TTAQ

Then the total rate of heat loss from the glass cover becomes

Btu/h8.350.258.10rad,conv,total, =+=+= iii QQQ

which is more than 30 Btu/h. Therefore, the assumed temperature of 118.5F for the tube is high. Bytrying other values, the tube temperature corresponding to 30 Btu/h is determined to be 113.2F.Therefore, the tube will reach an equilibrium temperature of 113.2F when the pump fails.

12-74

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12-95 A double-pane window consists of two sheets of glass separated by an air space. The rates of heattransfer through the window by natural convection and radiation are to be determined.

Assumptions1 Steady operating conditions exist 2 The surfaces areopaque, diffuse, and gray. 3 Air is an ideal gas with constant specificheats. 4 Heat transfer through the window is one-dimensional and theedge effects are negligible.

Properties The emissivities of glass surfaces are given to be 1 = 2 =0.9. The properties of air at 0.3 atm and the average temperature of(T1+T2)/2 = (15+5)/2 = 10C are (Table A-15)

1-

2551

K003534.0K)27310(

1

7336.0Pr

/sm10753.4/0.310426.13.0/

CW/m.02439.0

=+

=

====

=

atm

k

AnalysisThe characteristic length in this case is the distance between the glasses, m05.0== LLc

4

225

3-12

2

321 10918.1)7336.0(

)/sm10753.4(

)m05.0(K)515)(K003534.0)(m/s81.9(Pr

)(=

=

=

LTTgRa

539.105.0

2)10918.1(197.0197.0

9/14/14

9/14/1 =

=

=

L

HRaNu

2m6)m3)(m2( ==sA

Then the rate of heat transfer by natural convection becomes

W45.0=

=

=m05.0

C)515()m6)(539.1)(CW/m.02439.0(

221

L

TTkNuAQ sconv

The rate of heat transfer by radiation is determined from

( ) ( )

W252=

+

++=

+

=

19.0

1

9.0

1

]K2735K27315)[KW/m1067.5)(m6(

111

)(

444282

21

4

2

4

1rad

TTAQ s

Then the rate of total heat transfer becomes

W297=+=+= 25245radconvtotal QQQ

DiscussionNote that heat transfer through the window is mostly by radiation.

12-75

5C15CL = 5 cm

H= 2 m

QAir

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12-96 A simple solar collector is built by placing a clear plastic tube around a garden hose. The rate ofheat loss from the water in the hose by natural convection and radiation is to be determined.

Assumptions1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Air is anideal gas with constant specific heats.

Properties The emissivities of surfaces are given to be 1 = 2 = 0.9. The properties of air are at 1 atmand the film temperature of (Ts+T)/2 = (40+25)/2 = 32.5C are (Table A-15)

1-

25

K003273.0K)2735.32(

1

7275.0Pr

/sm10632.1

CW/m.02607.0

=+

=

==

=

k

Analysis Under steady conditions, the heat transfer ratefrom the water in the hose equals to the rate of heat lossfrom the clear plastic tube to the surroundings by naturalconvection and radiation. The characteristic length in thiscase is the diameter of the plastic tube,

m06.02 === DDL plasticc .

5

225

3-12

2

32 10842.2)7275.0(

)/sm10632.1(

)m06.0(K)2540)(K003273.0)(m/s81.9(Pr

)(=

=

=

DTTgRa s

( )[ ] ( )[ ]30.10

7241.0/559.01

)10842.2(387.06.0

Pr/559.01

Ra387.06.0Nu

2

27/816/9

6/152

27/816/9

1/6D =

+

+=

++=

2

22

2

2

m1885.0m)m)(106.0(

C.W/m475.4)30.10(m06.0

CW/m.02607.0

====

=

==

LDAA

NuD

kh

plastic

Then the rate of heat transfer from the outer surface by natural convection becomes

W12.7=== C)2540)(m1885.0)(C.W/m475.4()( 222 TThAQ sconvThe rate of heat transfer by radiation from the outer surface is determined from

W26.2=++=

= ]K)27315()K27340)[(KW/m1067.5)(m1885.0)(90.0(

)(

444282

442rad skys TTAQ

Finally,

W9.382.267.12, =+=losstotalQ

DiscussionNote that heat transfer is mostly by radiation.

12-76

D2=6 cm

Garden hose

D1=2 cm,T

1

1

= 0.9

Air space

Plastic cover,

2= 0.9, T

2=40C

Water

T= 25C

Tsky

= 15C

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12-98 A cylindrical furnace with specified top and bottom surface temperatures and specified heat transferrate at the bottom surface is considered. The temperature of the side surface and the net rates of heattransfer between the top and the bottom surfaces, and between the bottom and the side surfaces are to bedetermined.Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3Convection heat transfer is not considered.PropertiesThe emissivities of the top, bottom, and side surfaces are 0.70, 0.50, and 0.40, respectively.Analysis We consider the top surface to be surface 1, the bottom

surface to be surface 2, and the side surface to be surface 3. Thissystem is a three-surface enclosure. The view factor from surface 1 tosurface 2 is determined from

17.0

5.02.1

6.0

26.0

2.1

12 =

==

==F

L

r

r

L

(Fig. 12-7)

The surface areas are

23

22221

m524.4)m2.1)(m2.1(

m131.14/)m2.1(4/

===

====

DLA

DAA

Then other view factors are determined to be

17.02112 ==FF83.0117.001 1313131211 ==++=++ FFFFF (summation rule), 83.01323 ==FF

21.0)524.4()83.0)(131.1( 3131313131 === FFFAFA (reciprocity rule), 21.03132 ==FF

We now apply Eq. 12-35 to each surface

Surface 1:

[ ]

[ ])(83.0)(17.070.0

70.01)K500)(K.W/m1067.5(

)()(1

312114428

311321121

11

41

JJJJJ

JJFJJFJT

+

+=

+

+=

Surface 2:

[ ]

[ ])(83.0)(17.050.050.01

)K500)(K.W/m1067.5(

)()(1

32122

4428

322312212

22

42

JJJJJ

JJFJJFJT

+

+=

+

+=

Surface 3:

[ ]

[ ])(21.0)(21.040.0

40.01)K.W/m1067.5(

)()(1

312134

3428

233213313

33

43

JJJJJT

JJFJJFJT

+

+=

+

+=

We now apply Eq. 12-34 to surface 2

[ ] [ ])(83.0)(17.0)m131.1()()( 32122

3223122122 JJJJJJFJJFAQ +=+=

Solving the above four equations, we find2

32

22

13 W/m8193,W/m8883,W/m4974, ==== JJJT K631The rate of heat transfer between the bottom and the top surface is

W751.6=== 221221221 W/m)49748883)(17.0)(m131.1()( JJFAQ

The rate of heat transfer between the bottom and the side surface is

W644.0=== 223223223 W/m)81978883)(83.0)(m131.1()( JJFAQ

Discussion The sum of these two heat transfer rates are 751.6 + 644 = 1395.6 W, which is practicallyequal to 1400 W heat supply rate from surface 2. This must be satisfied to maintain the surfaces at thespecified temperatures under steady operation. Note that the difference is due to round-off error.

12-78

T1 = 500 K

1

= 0.70

r1

= 0.6 m

T3

= ?

3

= 0.40

h = 1.2 m

T2

= 650 K

2

= 0.50

r2

= 0.6 m

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12-99 A cylindrical furnace with specified top and bottom surface temperatures and specified heat transferrate at the bottom surface is considered. The emissivity of the top surface and the net rates of heat transfer

between the top and the bottom surfaces, and between the bottom and the side surfaces are to bedetermined.

Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3Convection heat transfer is not considered.

PropertiesThe emissivity of the bottom surface is 0.90.

AnalysisWe consider the top surface to be surface 1, the base surface tobe surface 2, and the side surface to be surface 3. This system is a three-surface enclosure. The view factor from the base to the top surface of the

cube is F12 0 2= . . The view factor from the base or the top to the sidesurfaces is determined by applying the summation rule to be

F F F F F11 12 13 13 121 1 1 0 2 08+ + = = = =. .

since the base surface is flat and thus F11 0= . Other view factors are

20.0,80.0,20.0 323113231221 ====== FFFFFF

We now apply Eq. 9-35 to each surface

Surface 1:

[ ]

[ ])(80.0)(20.01

)K700)(K.W/m1067.5(

)()(1

31211

11

4428

311321121

1

1

4

1

JJJJJ

JJFJJFJT

+

+=

+

+=

Surface 2:

[ ]

[ ])(80.0)(20.090.0

90.01)K950)(K.W/m1067.5(

)()(1

321224428

322312212

22

42

JJJJJ

JJFJJFJT

+

+=

+

+=

Surface 3:3

4428

34

3

)K450)(K.W/m1067.5(

J

JT

=

=

We now apply Eq. 9-34 to surface 2

[ ] [ ])(80.0)(20.0)m9()()( 32122

3223122122 JJJJJJFJJFAQ +=+=

Solving the above four equations, we find

23

22

211 W/m2325,W/m985,41,W/m736,11, ==== JJJ0.44

The rate of heat transfer between the bottom and the top surface is

2221 m9)m3( ===AA

kW54.4=== 221221221 W/m)736,11985,41)(20.0)(m9()( JJFAQ

The rate of heat transfer between the bottom and the side surface is

2213 m36)m9(44 === AA

kW285.6=== 223223223 W/m)2325985,41)(8.0)(m9()( JJFAQ

Discussion The sum of these two heat transfer rates are 54.4 + 285.6 = 340 kW, which is equal to 340 kWheat supply rate from surface 2.

12-79

T1

= 700 K

1

= ?

T2

= 950 K

2

= 0.90

T3

= 450 K

3

= 1

3 m

7/30/2019 Heat Chap12 088

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12-100 A thin aluminum sheet is placed between two very large parallel plates that are maintained atuniform temperatures. The net rate of radiation heat transfer between the plates and the temperature of theradiation shield are to be determined.


PropertiesThe emissivities of surfaces are given to be 1 = 0.8, 2 = 0.9, and 3 = 0.12.

AnalysisThe net rate of radiation heat transfer with athin aluminum shield per unit area of the plates is

2W/m748.9=

++

+

=

+

+

+

=

112.0

1

12.0

11

9.0

1

8.0

1

])K550()K750)[(KW/m1067.5(

111

111

)(

44428

2,31,321

42

41

shieldone,12

TTQ

The equilibrium temperature of the radiation shield is determined from

K671.3=

+

=

+

=

3

43

44282

31

43

41

13

112.0

1

8.0

1

])K750)[(KW/m1067.5(W/m9.748

111

)(

TT

TTQ

12-80

T2

= 550 K

2

= 0.9

T1

= 750 K

1

= 0.8

Radiation shield

3= 0.12

7/30/2019 Heat Chap12 088

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12-101 Two thin radiation shields are placed between two large parallel plates that are maintained atuniform temperatures. The net rate of radiation heat transfer between the plates with and without theshields, and the temperatures of radiation shields are to be determined.


PropertiesThe emissivities of surfaces are given to be 1 = 0.6, 2 = 0.7, 3 = 0.10, and 4 = 0.15.

AnalysisThe net rate of radiation heat transfer withoutthe shields per unit area of the plates is

2W/m3288=

+

=

+

=

17.0

1

6.0

1

])K300()K600)[(KW/m1067.5(

111

)(

44428

21

42

41

shieldno12,

TTQ

The net rate of radiation heat transfer with two thinradiation shields per unit area of the plates is

2W/m206=

++

++

+

=

+

+

+

+

+

=

115.0

1

15.0

11

10.0

1

10.0

11

7.0

1

6.0

1

])K300()K600)[(KW/m1067.5(

111

111

111

)(

44428

443321

42

41

shieldstwo12,

TTQ

The equilibrium temperatures of the radiation shields are determined from

K549=

+

=

+

=

3

43

44282

31

43

41

13

110.0

1

6.0

1

])K600)[(KW/m1067.5(W/m206

111

)(

TT

TT

Q

K429=

+

=

+

=

4

444

4282

24

42

44

42

17.0

1

15.0

1

])K300()[KW/m1067.5(

W/m206

111

)(

T

T

TTQ

12-81

T2

= 300 K

2

= 0.7

T1

= 600 K

1

= 0.6

3

= 0.10

4 = 0.15

7/30/2019 Heat Chap12 088

15/18


12-102 Combustion gases flow inside a tube in a boiler. The rates of heat transfer by convection andradiation and the rate of evaporation of water are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The inner surfaces of the duct are smooth. 3Combustion gases are assumed to have the properties of air, which is an ideal gas with constant

properties.

Properties The properties of air at 1200 K = 927C and 1 atm are (Table A-15)

/sm10586.1

CW/m.07574.0

kg/m2944.0

25-

3

=

==

k 7221.0Pr

CJ/kg.1173

==pC

Analysis(a) The Reynolds number is

373,28/sm10586.1

m)m/s)(0.15(3Re

25=

=

=

DmV

which is greater than 10,000. Therefore, the flow isturbulent and the entry lengths in this case are roughly

m5.1m)15.0(1010 == DLL th

which is much shorter than the total length of the duct. Therefore, we can assume fully developedturbulent flow in the entire duct, and determine the Nusselt number from

14.76)7221.0()373,28(023.0PrRe023.03.08.03.08.0 ====

k

hDNu h

Heat transfer coefficient is

C.W/m45.38)14.76(m15.0

CW/m.07574.0 2 =

== NuD

kh

Next we determine the exit temperature of air

kg/s0.01561=)m67m/s)(0.017)(3kg/m2944.0(

m0.01767=/4m)15.0(4/

m2.827=m)m)(615.0(

23

222

2

==

==

==

c

c

VAm

DA

DLA

C107.2)927105(105)( )1173)(01561.0()827.2)(45.38(

)/( ===

eeTTTT p

CmhAisse

Then the rate of heat transfer by convection becomes

W15,010=== C)2.107927)(CJ/kg.3kg/s)(11701561.0()(conv eip TTCmQ

Next, we determine the emissivity of combustion gases. First, the mean beam length for an infinitecircular cylinder is, from Table 12-4,

L = 0.95(0.15 m) = 0.1425 m

Then,

atmft0.075atmm0228.0m)5atm)(0.14216.0(

atmft.0370atmm0114.0m)5atm)(0.14208.0(

======

LP

LP

w

c

The emissivities of CO2 and H2O corresponding to these values at the average gas temperature ofTg=(Tg+Tg)/2 = (927+107.2)/2 = 517.1C = 790 K and 1atm are, from Fig. 12-36,

055.0atm1, =c and 062.0atm1, = w

Both CO2 and H2O are present in the same mixture, and we need to correct for the overlap of emissionbands. The emissivity correction factor at T = Tg = 800 K is, from Fig. 12-38,

12-82

Ts

= 105C

D = 15 cm

Combustiongases, 1 atmT

i= 1200 K

3 m/s

7/30/2019 Heat Chap12 088

16/18


0.067.0

08.016.0

16.0

112.0075.0037.0

=

=+

=+

=+=+

cw

w

wc

PP

P

LPLP

Then the effective emissivity of the combustion gases becomes

0.1170.0062.01055.01atm1,atm1, =+=+= wwccg CC

Note that the pressure correction factor is 1 for both gases since the total pressure is 1 atm. For a sourcetemperature ofTs = 105C = 378 K, the absorptivity of the gas is again determined using the emissivitycharts as follows:

atmft0.036atmm0109.0K790

K378m)5atm)(0.14216.0(


K378m)5atm)(0.14208.0(

===

===

g

sw

g

sc

T

TLP

T

TLP

The emissivities of CO2 and H2O corresponding to these values at a temperature ofTs = 378 K and 1atmare, from Fig. 12-36,

037.0atm1, =c and 062.0atm1, = w

Then the absorptivities of CO2 and H2O become

0864.0)062.0(K378

K790)1(

0597.0)037.0(K378

K790)1(

45.0

atm1,

45.0

65.0

atm1,

65.0

=

=

=

=

=

=

ws

g

ww

cs

g

cc

T

TC

T

TC

Also = , but the emissivity correction factor is to be evaluated from Fig. 12-38 at T = Ts = 378 Kinstead ofTg = 790 K. We use the chart for 400 K. At Pw/(Pw+ Pc) = 0.67 and PcL +PwL = 0.112 we read = 0.0. Then the absorptivity of the combustion gases becomes

0.1460.00864.00597.0 =+=+= wcg

The emissivity of the inner surface s of the tubes is 0.9. Then the net rate of radiation heat transfer fromthe combustion gases to the walls of the tube becomes

W6486=

+

=

+

=

])K378(146.0)K790(117.0)[KW/m1067.5)(m827.2(2

19.0

)(2

1

444282

44rad sgggs

sTTAQ

(b) The heat of vaporization of water at 1 atm is 2257 kJ/kg (Table A-9). Then rate of evaporation ofwater becomes

kg/s0.0644=

+

=

+

==+ J/kg107.333

W)6486010,15(3

radconv

evapevapradconvfg

fg h

QQ

mhmQQ

12-83

7/30/2019 Heat Chap12 088

17/18


12-103 Combustion gases flow inside a tube in a boiler. The rates of heat transfer by convection andradiation and the rate of evaporation of water are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The inner surfaces of the duct are smooth. 3Combustion gases are assumed to have the properties of air, which is an ideal gas with constant

properties.

Properties The properties of air at 1200 K = 927C and 3 atm are (Table A-15)

/sm105287.0

/s)/3m10586.1(

CW/m.07574.0

kg/m2944.0

25-

25-

3

=

==

=

k 7221.0Pr

CJ/kg.1173

=

=pC

Analysis(a) The Reynolds number is

114,85/sm105287.0

m)m/s)(0.15(3Re

25=

=

=

DmV

which is greater than 10,000. Therefore, the flow isturbulent and the entry lengths in this case are roughly

m5.1m)15.0(1010 == DLL th which is much shorter than the total length of the duct. Therefore, we can assume fully developed

turbulent flow in the entire duct, and determine the Nusselt number from

4.183)7221.0()114,85(023.0PrRe023.0 3.08.03.08.0 ====k

hDNu h

Heat transfer coefficient is

C.W/m59.92)4.183(m15.0

CW/m.07574.0 2 =

== NuD

kh

Next we determine the exit temperature of air

kg/s0.01561=)m67m/s)(0.017)(3kg/m2944.0(

m0.01767=/4m)15.0(4/

m2.827=m)m)(615.0(

23

222

2

==

==

==

c

c

VAm

DA

DLA

C105.0)927105(105)( )1173)(01561.0()827.2)(59.92(

)/( === eeTTTT pCmhAisse

Then the rate of heat transfer by convection becomes

W15,050=== C)0.105927)(CJ/kg.kg/s)(117301561.0()(conv eip TTCmQ

Next, we determine the emissivity of combustion gases. First, the mean beam length for an infinitecircular cylinder is, from Table 12-4,

L = 0.95(0.15 m) = 0.1425 m

Then,atmft0.075atmm0228.0m)5atm)(0.14216.0(

atmft.0370atmm0114.0m)5atm)(0.14208.0(

======

LP

LP

w

c

The emissivities of CO2 and H2O corresponding to these values at the average gas temperature ofTg=(Tg+Tg)/2 = (927+105)/2 = 516C = 790 K and 1atm are, from Fig. 12-36,

055.0atm1, =c and 062.0atm1, = wThese are the base emissivity values at 1 atm, and they need to be corrected for the 3 atm total pressure.

Noting that (Pw+P)/2 = (0.16+3)/2 = 1.58 atm, the pressure correction factors are, from Fig. 12-37,

Cc = 1.5 and Cw = 1.8

Both CO2 and H2O are present in the same mixture, and we need to correct for the overlap of emissionbands. The emissivity correction factor at T = Tg = 800 K is, from Fig. 12-38,

0.067.0

08.016.0

16.0

112.0075.0037.0

=

=+

=+

=+=+

cw

w

wc

PP

P

LPLP

12-84

Ts

= 105C

D = 15 cm

Combustiongases, 3 atmT

i= 1200 K

3 m/s

7/30/2019 Heat Chap12 088

18/18


Then the effective emissivity of the combustion gases becomes

0.1940.0062.08.1055.05.1atm1,atm1, =+=+= wwccg CC

For a source temperature ofTs = 105C = 378 K, the absorptivity of the gas is again determined using theemissivity charts as follows:

atmft0.036atmm0109.0K790K378m)5atm)(0.14216.0(


K378m)5atm)(0.14208.0(

===

===

g

sw

g

sc

TTLP

T

TLP

The emissivities of CO2 and H2O corresponding to these values at a temperature ofTs = 378 K and 1atmare, from Fig. 12-36,

037.0atm1, =c and 062.0atm1, = wThen the absorptivities of CO2 and H2O become

1555.0)062.0(K378

K790)8.1(

0896.0)037.0(K378

K790)5.1(

45.0

atm1,

45.0

65.0

atm1,

65.0

=

=

=

=

=

=

ws

g

ww

cs

g

cc

T

TC

T

TC

Also = , but the emissivity correction factor is to be evaluated from Fig. 12-38 at T = Ts = 378 Kinstead ofTg = 790 K. We use the chart for 400 K. At Pw/(Pw+ Pc) = 0.67 and PcL +PwL = 0.112 we read = 0.0. Then the absorptivity of the combustion gases becomes

0.2450.01555.00896.0 =+=+= wcg

The emissivity of the inner surfaces of the tubes is 0.9. Then the net rate of radiation heat transfer fromthe combustion gases to the walls of the tube becomes

W10,745=

+

=

+

=

])K378(245.0)K790(194.0)[KW/m1067.5)(m827.2(2

19.0

)(2

1

444282

44rad sgggs

sTTAQ

(b) The heat of vaporization of water at 1 atm is 2257 kJ/kg (Table A-9). Then rate of evaporation ofwater becomes

kg/s0.0773=

+=

+==+

J/kg107.333

W)745,10050,15(3

radconvevapevapradconv

fg

fgh

QQmhmQQ

12-104 .. 12-106 Design and Essay Problems

12-85

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Heat Chap12 088

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