13–82
13–94 Prob. 13-93 is reconsidered. The effect of the percent reduction in the net rate of radiation heat transfer between
the plates on the emissivity of the radiation shields is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
13–83
13–95 A large ASTM B152 copper plate and a large ceramic plate are placed in parallel near each other. The temperature
of the ceramic plate and the net radiation heat flux between the two plates are known. The emissivity of a radiation shield
𝑄̇12, 1-shield =𝜎𝐴(𝑇14−𝑇2
4)
𝑞̇12, 1-shield =𝑄̇12, 1-shield
4)
𝜀3=2[𝜎(𝑇14−𝑇2
4)
13–85
13–97 A coaxial radiation shield is placed between two coaxial cylinders which are maintained at uniform temperatures. The
net rate of radiation heat transfer between the two cylinders is to be determined and compared with that without the shield.
Assumptions 1 Steady operating conditions exist 2 The surfaces are
opaque, diffuse, and gray. 3 Convection heat transfer is not considered.
2
33shield
22outerpipe,
m 628.0)m 1)(m 2.0(
====
LDAA
The net rate of radiation heat transfer between the two
D2 = 0.5 m
D1 = 0.1 m
13–87
3
shield12,1
Q
[W]
0.05
0.07
0.09
0.11
0.13
0.15
0.17
0.19
0.21
0.23
0.25
0.27
0.29
0.31
0.33
0.35
213.1
291.5
366.5
438.3
507
572.9
636
696.7
755
811.1
865
917
967.1
1015
1062
1107
0.05 0.1 0.15 0.2 0.25 0.3 0.35
200
300
400
500
600
700
800
900
1000
1100
3
Q12,1shield [W]
13–88
13–99 A long cylindrical fuel rod is enclosed by a concentric stainless steel tube. The surface temperature of the fuel rod
and the radiation heat transfer rate from the fuel rod to the stainless steel tube are known. (a) The emissivity of a concentric
𝑄̇12, 1-shield =𝜎(𝑇14−𝑇2
4)
1−𝜀1
𝜀3=2[𝜎𝜋𝐷3(𝑇14−𝑇2
4)
𝑄̇12 =𝜎(𝑇14−𝑇2
4)
13–90
13-100 Two very large plates are maintained at uniform temperatures. The number of thin aluminum sheets that will reduce
the net rate of radiation heat transfer between the two plates to one-fifth is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is
not considered.
2
44428
21
4
2
4
1
shield ,12
W/m134,23
1
5.0
1
5.0
1
])K 700()K 1100)[(K W/m1067.5(
1
11
)(
=
−+
−
=
−+
−
=
−
TT
Qno
The number of sheets that need to be inserted in order to
reduce the net rate of heat transfer between the two plates to
2=
−++
−+
−
=
−++
−+
−
=
−
48.1
1
1.0
1
1.0
1
1
5.0
1
5.0
1
])K 700()K 1100)[(K W/m1067.5(
) W/m(23,134
5
1
1
11
1
11
)(
shield
shield
44428
2
2,31,3
shield
21
4
2
4
1
shields,12
N
N
N
TT
Q
That is, only two sheets are more than enough to reduce heat transfer to one-fifth.
The number of sheets that need to be inserted in order to reduce the net rate of heat transfer between the two plates
to one-fifth can be determined from
1
=
−++
−+
−
=
−++
−+
−
=
−
632.0
1
1.0
1
1.0
1
1
5.0
1
5.0
1
])K 800()K 1000)[(K W/m1067.5(
) W/m(11,159
5
1
1
11
1
11
)(
shield
shield
44428
2
2,31,3
shield
21
4
2
4
1
shields,12
N
N
N
TT
Q
That is, only one sheet with a low emissivity is more than enough to reduce heat transfer to one-fifth.
T2 = 700 K
2 = 0.5
T1 = 1100 K
1 = 0.5
Radiation shields
3 = 0.1
13–92
13–102 Prob. 13-101 is reconsidered. The effects of the number of the aluminum sheets and the emissivities of the
plates on the net rate of radiation heat transfer between the two plates are to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
N=5
13–94
13–104 An engine cover is made of two parallel plates. The number of radiation shields necessary to keep the top plate
below 150°C, to prevent fire hazards in the event of oil leakage, is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is
not considered.
Properties The emissivity of all the surfaces is
1
11
12 −+
The temperature of the top plate of the engine
cover is
1
11
4/1
12
4
12
−+−=
q
TT
13–95
13-105 The temperature of air in a duct is measured. Accounting for the radiation effect, and the actual air temperature is to
be determined.
Assumptions The surfaces are opaque, diffuse, and gray.
Properties The emissivity of thermocouple is given to be = 0.6
Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill
Education.
13–106C A nonparticipating medium is completely transparent to thermal radiation, and thus it does not emit, absorb, or
scatter radiation. A participating medium, on the other hand, emits and absorbs radiation throughout its entire volume.
13–107C Spectral transmissivity of a medium of thickness L is the ratio of the intensity of radiation leaving the medium to
L
Le
I
I
−
0,
,
13–108C Gases emit and absorb radiation at a number of narrow wavelength bands. The emissivity-wavelength charts of
13–109C Using Kirchhoff’s law, the spectral emissivity of a medium of thickness L in terms of the spectral absorption
L
−
13–110 An equimolar mixture of CO2 and O2 gases at 800 K and a total pressure of 0.5 atm is considered. The emissivity of
atm 25.0atm) 5.0(5.0
2
CO === PyPc
15.0
atm 1 , =
c
Cc = 0.90
13–97
13–111 A mixture of CO2 and N2 gases at 600 K and a total pressure of 1 atm are contained in a cylindrical container. The
L = 0.60D = 0.60(8 m) = 4.8 m
16.0
atm 1 , =
c
14.0
atm 1 , =
c
17.0)14.0(
K 450
K 600
)1(
65.0
atm 1 ,
65.0
=
=
=c
s
g
cc T
T
C
W102.35 5
=
8 m
13–98
13–112 A mixture of H2O and N2 gases at 600 K and a total pressure of 1 atm are contained in a cylindrical container. The
L = 0.60D = 0.60(8 m) = 4.8 m
Then,
atmft .362atmm 72.0m) atm)(4.8 15.0( ===LPw
The emissivity of H2O corresponding to this value at the gas temperature
of Tg = 600 K and 1 atm is, from Fig. 13–36,
36.0
atm 1 =
w
For a source temperature of Ts = 450 K, the absorptivity of the gas is again
determined using the emissivity charts as follows:
atmft 1.77atmm 54.0
K 600
K 450
m) atm)(4.8 15.0( ===
g
s
wT
T
LP
The emissivity of H2O corresponding to this value at a temperature of Ts = 450 K and 1atm are, from Fig. 13-36,
34.0
atm 1 , =
w
39.0)34.0(
K 450
K 600
)1(
45.0
atm 1 ,
65.0
=
=
=w
s
g
ww T
T
C
8 m
8 m
Tg = 600 K
Ts = 450 K
13–99
13-113 A mixture of CO2 and N2 gases at 1200 K and a total pressure of 1 atm are contained in a spherical furnace. The net
L = 0.65D = 0.65(5 m) =3.25 m
The mole fraction is equal to pressure fraction. Then,
atmft .601atmm 4875.0m) atm)(3.25 15.0( ===LP
c
12.0
atm 1 , =
c
1883.0)12.0(
K 600
K 1200
)1(
65.0
atm 1 ,
65.0
=
=
=c
s
g
cc T
T
C
5 m
13-100
13-114 The temperature, pressure, and composition of a gas mixture is given. The emissivity of the mixture is to be
determined.
atm 09.0atm) 1(09.0
atm 10.0atm) 1(10.0
2
2
H
CO
===
===
PyP
PyP
Ow
c
17.0
atm 1 , =
c
and
16.0
atm 1 , =
w
049.0
474.0
10.009.0
09.0
48.218.130.1
=
=
+
=
+
=+=+
cw
w
wc
PP
P
LPLP
Then the effective emissivity of the combustion gases becomes
6 m