Download - 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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Chapter 12Radiation Heat Transfer
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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Chapter 12Radiation Heat Transfer
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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Chapter 12Radiation Heat Transfer
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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Chapter 12Radiation Heat Transfer
W534=
=
=m03.0
C)3280()m6)(001.2)(CW/m.02779.0(
221
L
TTkNuAQ s
12-72
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Chapter 12Radiation Heat Transfer
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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Chapter 12Radiation Heat Transfer
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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Chapter 12Radiation Heat Transfer
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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Chapter 12Radiation Heat Transfer
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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Chapter 12Radiation Heat Transfer
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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Chapter 12Radiation Heat Transfer
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
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7/30/2019 Heat Chap12 088
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Chapter 12Radiation Heat Transfer
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.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3Convection heat transfer is not considered.
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
14/18
Chapter 12Radiation Heat Transfer
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.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3Convection heat transfer is not considered.
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
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7/30/2019 Heat Chap12 088
15/18
Chapter 12Radiation Heat Transfer
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
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7/30/2019 Heat Chap12 088
16/18
Chapter 12Radiation Heat Transfer
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(
atmft0.018atmm00545.0K790
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
Chapter 12Radiation Heat Transfer
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
Chapter 12Radiation Heat Transfer
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(
atmft0.018atmm00545.0K790
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