13-1
Solutions Manual
for
Heat and Mass Transfer: Fundamentals & Applications
Fourth Edition
Yunus A. Cengel & Afshin J. Ghajar
McGraw-Hill, 2011
Chapter 13
RADIATION HEAT TRANSFER
PROPRIETARY AND CONFIDENTIAL
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13-2
View Factors
13-1C The view factor Fi → j represents the fraction of the radiation leaving surface i that strikes surface j directly. The view
factor from a surface to itself is non-zero for concave surfaces.
13-2C The cross-string method is applicable to geometries which are very long in one direction relative to the other
directions. By attaching strings between corners the Crossed-Strings Method is expressed as
Fi → j =
∑ Crossed strings − ∑ Uncrossed strings
2 × string on surface i
N
13-3C The summation rule for an enclosure and is expressed as
∑F
i→ j
= 1 where N is the number of surfaces of the
j =1
enclosure. It states that the sum of the view factors from surface i of an enclosure to all surfaces of the enclosure, including
to itself must be equal to unity.
The superposition rule is stated as the view factor from a surface i to a surface j is equal to the sum of the view
factors from surface i to the parts of surface j, F1→( 2,3) = F1→2 + F1→3 .
13-4C The pair of view factors Fi → j and F j →i are related to each other by the reciprocity rule Ai Fij = A j F ji where Ai is
the area of the surface i and Aj is the area of the surface j. Therefore,
A1 F12 = A2 F21 ⎯
⎯→ F12 =
A2
F21
A1
13-5 An enclosure consisting of five surfaces is considered. The number of
view factors this geometry involves and the number of these view factors that
can be determined by the application of the reciprocity and summation rules
are to be determined.
Analysis A five surface enclosure (N=5) involves N = 5 = 25 view
N ( N − 1) 5(5 − 1)
factors and we need to determine
=
= 10 view factors
2
2
directly. The remaining 25-10 = 15 of the view factors can be determined by
the application of the reciprocity and summation rules.
2
1
2
2
5
4
3
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13-3
13-6 An enclosure consisting of thirteen surfaces is considered.
The number of view factors this geometry involves and the
number of these view factors that can be determined by the
application of the reciprocity and summation rules are to be
determined.
Analysis A thirteen surface enclosure (N = 13) involves
N 2 = 13 2 = 169 view factors and we need to determine
N ( N − 1) 13(13 − 1)
=
= 78 view factors directly. The
2
2
remaining 169 - 78 = 91 of the view factors can be determined
by the application of the reciprocity and summation rules.
4
2
5
3
6
1
7
13
8
12
10
9
11
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13-4
13-7 A cylindrical enclosure is considered. (a) The expression for the view factor between the base and the side surface F13
in terms of K and (b) the value of the view factor F13 for L = D are to be determined.
Assumptions 1 The surfaces are diffuse emitters and reflectors.
Analysis (a) The surfaces are designated as follows: Base surface as A1, top surface as A2, and side surface as A3
Applying the summation rule to A1, we have
(where F11 = 0 )
F11 + F12 + F13 = 1
or
F13 = 1 − F12
(1)
For coaxial parallel disks, from Table 13-1, with i = 1, j = 2,
1/ 2 ⎫
⎧
2
⎡
⎛ D2 ⎞ ⎤ ⎪ 1
1⎪
2
⎟⎟ ⎥ ⎬ = [ S − ( S 2 − 4 )1 / 2 ]
F12 = ⎨S − ⎢ S − 4 ⎜⎜
D
2⎪
⎢
⎝ 1 ⎠ ⎥⎦ ⎪ 2
⎣
⎭
⎩
S = 1+
1 + R22
R12
= 2+
1
R
2
= 2+
4
( D / L)
2
= 2 + 4K 2
(2)
(3)
where
R1 = R2 = R =
D
1
=
2L 2K
Substituting Eq. (3) into Eq. (2), we get
1
{2 + 4 K 2 − [(2 + 4 K 2 ) 2 − 4 ]1 / 2 }
2
1
= [2 + 4 K 2 − (16 K 4 + 16 K 2 )1 / 2 ]
2
1
= [2 + 4 K 2 − 4 K ( K 2 + 1)1 / 2 ]
2
= 1 + 2 K 2 − 2 K ( K 2 + 1)1 / 2
F12 =
Substituting the above expression for F12 into Eq. (1) yields the expression for F13:
F13 = 1 − [1 + 2 K 2 − 2 K ( K 2 + 1)1 / 2 ]
Hence,
F13 = 2 K ( K 2 + 1)1 / 2 − 2 K 2
(b) The value of the view factor F13 for L = D (i.e., K = 1) is
F13 = 2(1)(12 + 1)1 / 2 − 2(1) 2 = 2 2 − 2 = 0.828
Discussion If the cylinder has a length and diameter of L = 2D, then from the expression for F13 we have
F13 = 2(2)(2 2 + 1)1 / 2 − 2(2) 2 = 0.944
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13-5
13-8 The expression for the view factor F12 of two infinitely long parallel plates is to be determined using the Hottel’s
crossed-strings method.
Assumptions 1 The surfaces are diffuse emitters and reflectors.
Analysis From the Hottel’s crossed-strings method, we have
Fi → j =
Σ (Crossed strings) − Σ ( Uncrossed strings)
2 × (String on surface i )
For uncrossed strings, we have
L1 = L2 = ( w 2 + w 2 )1 / 2 = ( w 2 + w 2 )1 / 2 = 2 w
For crossed strings, we have
L3 = ( w 2 + 4w 2 )1 / 2 = 5 w
and
L4 = w
Applying the Hottel’s crossed-strings method, we get F12 as
F12 =
( L3 + L4 ) − ( L1 + L2 )
2w
( 5 w + w) − ( 2 w + 2 w)
2w
= 0.204
=
Discussion The Hottel’s crossed-string method is applicable only to surfaces that are very long, such that they can be
considered to be two-dimensional and radiation interaction through the end surfaces is negligible.
13-9 The view factors between the rectangular surfaces shown in the figure are to be determined.
Assumptions The surfaces are diffuse emitters and reflectors.
Analysis From Fig. 13-6,
W=4m
L3 1.2
⎫
=
= 0.3⎪
⎪
W
4
⎬ F31 = 0.26
L1 1.2
=
= 0.3 ⎪
⎪⎭
W
4
L2 = 1.2 m
L1 = 1.2 m
and
L3 1.2
⎫
=
= 0.3
⎪⎪
W
4
⎬ F3→(1+ 2) = 0.32
L1 + L2 2.4
=
= 0.6⎪
⎪⎭
W
4
A2
(2)
A1
(1)
L3 = 1.2 m
A3
(3)
We note that A1 = A3. Then the reciprocity and superposition rules gives
A1 F13 = A3 F31 ⎯
⎯→ F13 = F31 = 0.26
F3→(1+ 2) = F31 + F32 ⎯
⎯→
Finally,
0.32 = 0.26 + F32 ⎯
⎯→ F32 = 0.06
A2 = A3 ⎯
⎯→ F23 = F32 = 0.06
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13-6
13-10 A cylindrical enclosure is considered. The view factor from the side surface of this cylindrical enclosure to its base
surface is to be determined.
Assumptions The surfaces are diffuse emitters and reflectors.
Analysis We designate the surfaces as follows:
Base surface by (1),
(2)
top surface by (2), and
side surface by (3).
Then from Fig. 13-7
(3)
⎫
⎪
⎪
⎬ F12 = F21 = 0.05
r2
r2
=
= 0.25⎪
⎪⎭
L 4r2
L 4r1
=
=4
r1
r1
L=2D=4r
(1)
D=2r
summation rule : F11 + F12 + F13 = 1
0 + 0.05 + F13 = 1 ⎯
⎯→ F13 = 0.95
reciprocity rule : A1 F13 = A3 F31 ⎯
⎯→ F31 =
A1
πr 2
πr 2
1
F13 = 1 F13 = 1 2 F13 = (0.95) = 0.119
A3
2πr1 L
8
8πr1
Discussion This problem can be solved more accurately by using the view factor relation from Table 13-1 to be
R1 =
r1
r
= 1 = 0.25
L 4r1
R2 =
r2
r
= 2 = 0.25
L 4r2
S = 1+
F12
1 + R 22
R12
= 1+
1 + 0.25 2
0.25 2
= 18
0.5 ⎫
⎧
2
⎡
⎛ R2 ⎞ ⎤ ⎪
⎪
2
⎟ ⎥ ⎬=
= ⎨S − ⎢ S − 4⎜⎜
R1 ⎟⎠ ⎥ ⎪
⎢
⎝
⎪
⎦ ⎭
⎣
⎩
1
2
1
2
0.5 ⎫
⎧
2
⎡ 2
⎛1⎞ ⎤ ⎪
⎪
⎨18 − ⎢18 − 4⎜ ⎟ ⎥ ⎬ = 0.056
⎝ 1 ⎠ ⎦⎥ ⎪
⎪
⎣⎢
⎩
⎭
F13 = 1 − F12 = 1 − 0.056 = 0.944
reciprocity rule : A1 F13 = A3 F31 ⎯
⎯→ F31 =
A1
πr 2
πr 2
1
F13 = 1 F13 = 1 2 F13 = (0.944) = 0.118
2πr1 L
8
A3
8πr1
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13-7
13-11 A semispherical furnace is considered. The view factor from the dome of this furnace to its flat base is to be
determined.
Assumptions The surfaces are diffuse emitters and reflectors.
(2)
Analysis We number the surfaces as follows:
(1): circular base surface
(1)
(2): dome surface
Surface (1) is flat, and thus F11 = 0 .
D
Summation rule : F11 + F12 = 1 → F12 = 1
πD 2
reciprocity rule : A 1 F12 = A2 F21 ⎯
⎯→ F21 =
A1
A
1
F12 = 1 (1) = 4 2 = = 0.5
2
A2
A2
πD
2
13-12 Three infinitely long cylinders are located parallel to each other. The view
factor between the cylinder in the middle and the surroundings is to be
determined.
Assumptions The cylinder surfaces are diffuse emitters and reflectors.
(surr)
Analysis The view factor between two cylinder facing each other
is, from Prob. 13-17,
F1− 2
D
2⎛⎜ s 2 + D 2 − s ⎞⎟
⎠
⎝
=
πD
Noting that the radiation leaving cylinder 1 that does not strike the
cylinder will strike the surroundings, and this is also the case for
the other half of the cylinder, the view factor between the cylinder
in the middle and the surroundings becomes
F1− surr = 1 − 2 F1− 2
D
(2)
D
(1)
s
(2)
s
4⎛⎜ s 2 + D 2 − s ⎞⎟
⎝
⎠
= 1−
πD
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13-8
13-13 The view factors from the base of a cube to each of the other five surfaces
are to be determined.
(2)
Assumptions The surfaces are diffuse emitters and reflectors.
Analysis Noting that L1 / D = L 2 / D = 1 , from Fig. 13-6 we read
(3), (4), (5), (6)
side surfaces
F12 = 0.2
Because of symmetry, we have
F12 = F13 = F14 = F15 = F16 = 0.2
(1)
13-14 The view factor from the conical side surface to a hole located at the center of the base of a conical enclosure is to be
determined.
Assumptions The conical side surface is diffuse emitter and reflector.
Analysis We number different surfaces as
the hole located at the center of the base (1)
the base of conical enclosure
(2)
conical side surface
(3)
h
(3)
Surfaces 1 and 2 are flat, and they have no direct view of each other. Therefore,
F11 = F22 = F12 = F21 = 0
(2)
summation rule : F11 + F12 + F13 = 1 ⎯
⎯→ F13 = 1
⎯→
reciprocity rule : A 1 F13 = A3 F31 ⎯
πd
4
2
(1) =
(1)
d
πDh
2
⎯→ F31 =
F31 ⎯
2
d
2Dh
D
13-15 The four view factors associated with an enclosure formed by two very long concentric cylinders are to be determined.
Assumptions 1 The surfaces are diffuse emitters and reflectors. 2 End effects are neglected.
Analysis We number different surfaces as
(2)
the outer surface of the inner cylinder (1)
(1)
the inner surface of the outer cylinder (2)
No radiation leaving surface 1 strikes itself and thus F11 = 0
D2
D1
All radiation leaving surface 1 strikes surface 2 and thus F12 = 1
reciprocity rule : A 1 F12 = A2 F21 ⎯
⎯→ F21 =
A1
D
πD1 h
(1) = 1
F12 =
D2
A2
πD 2 h
summation rule : F21 + F22 = 1 ⎯
⎯→ F22 = 1 − F21 = 1 −
D1
D2
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13-9
13-16 The view factors between the rectangular surfaces shown in the figure are to be determined.
Assumptions The surfaces are diffuse emitters and reflectors.
Analysis We designate the different surfaces as follows:
4m
shaded part of perpendicular surface by (1),
bottom part of perpendicular surface by (3),
1m
(1)
shaded part of horizontal surface by (2), and
1m
(3)
front part of horizontal surface by (4).
(2)
1m
(4)
1m
(a) From Fig.13-6
L2 1
L2 2
⎫
⎫
= = 0.25⎪
= = 0.5 ⎪
⎪
⎪
W 4
W 4
⎬ F23 = 0.26 and
⎬ F2→(1+ 3) = 0.33
L1 1
L
1
1
= = 0.25⎪
= = 0.25 ⎪
⎪⎭
⎪⎭
W 4
W 4
superposit ion rule : F2→(1+3) = F21 + F23 ⎯
⎯→ F21 = F2→(1+3) − F23 = 0.33 − 0.26 = 0.07
reciprocity rule : A1 = A2 ⎯
⎯→ A1 F12 = A2 F21 ⎯
⎯→ F12 = F21 = 0.07
(b) From Fig.13-6,
L2 1
⎫
= = 0.25⎪
⎪
W 4
⎬ F( 4+ 2)→3 = 0.16
L1 2
= = 0 .5 ⎪
⎪⎭
W 4
and
L2 2
⎫
= = 0.5 ⎪
⎪
W 4
⎬ F( 4+ 2)→(1+3) = 0.24
L1 2
= = 0.5 ⎪
⎪⎭
W 4
superposit ion rule : F( 4+ 2)→(1+3) = F( 4 + 2)→1 + F( 4 + 2)→3 ⎯
⎯→ F( 4 + 2)→1 = 0.24 − 0.16 = 0.08
reciprocity rule : A( 4+ 2) F( 4+ 2)→1 = A1 F1→( 4+ 2)
⎯
⎯→ F1→( 4+ 2) =
A( 4+ 2)
A1
F( 4+ 2)→1 =
8
(0.08) = 0.16
4
4m
1m
(1)
superposition rule : F1→( 4+ 2) = F14 + F12
⎯
⎯→ F14 = 0.16 − 0.08 = 0.08
since F12 = 0.07 (from part a). Note that F14 in part (b) is
equivalent to F12 in part (a).
(3)
1m
(4)
1m
1m
(2)
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13-10
(c) We designate
shaded part of top surface by (1),
3m
(1)
(3)
remaining part of top surface by (3),
remaining part of bottom surface by (4), and
shaded part of bottom surface by (2).
3m
1m
1m
From Fig.13-5,
L2 3
⎫
= =1 ⎪
⎪
D 3
⎬ F( 2 + 4)→(1+3) = 0.15
L1 2
= = 0.67⎪
⎪⎭
D 3
1m
1m
and
(4)
(2)
L2 3
⎫
= =1 ⎪
⎪
D 3
⎬ F14 = 0.082
L1 1
= = 0.33⎪
⎪⎭
D 3
superposition rule : F( 2+ 4)→(1+3) = F( 2+ 4)→1 + F( 2+ 4)→3
symmetry rule : F( 2+ 4)→1 = F( 2+ 4)→3
Substituting symmetry rule gives
F( 2+ 4)→1 = F( 2 + 4)→3 =
F( 2+ 4)→(1+3)
2
=
0.15
= 0.075
2
reciprocity rule : A1 F1→( 2+ 4) = A( 2+ 4) F( 2+ 4)→1 ⎯
⎯→(2) F1→( 2+ 4) = (4)(0.075) ⎯
⎯→ F1→( 2+ 4) = 0.15
superposition rule : F1→( 2 + 4) = F12 + F14 ⎯
⎯→ 0.15 = F12 + 0.082 ⎯
⎯→ F12 = 0.15 − 0.082 = 0.068
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13-11
13-17 The view factor between the two infinitely long parallel cylinders located a distance s apart from each other is to be
determined.
Assumptions The surfaces are diffuse emitters and reflectors.
Analysis Using the crossed-strings method, the view factor between two
cylinders facing each other for s/D > 3 is determined to be
F1− 2 =
=
or
F1− 2
D
∑ Crossed strings − ∑ Uncrossed strings
D
(2)
2 × String on surface 1
2 s + D − 2s
2(πD / 2)
2
(1)
2
s
2⎛⎜ s 2 + D 2 − s ⎞⎟
⎝
⎠
=
πD
13-18 View factors from the very long grooves shown in the figure to the surroundings are to be determined.
Assumptions 1 The surfaces are diffuse emitters and reflectors. 2 End effects are neglected.
Analysis (a) We designate the circular dome surface by (1) and the imaginary flat top surface by (2). Noting that (2) is flat,
F22 = 0
D
summation rule : F21 + F22 = 1 ⎯
⎯→ F21 = 1
(2)
A2
2
D
(1) = = 0.64
F21 =
πD
π
A1
2
reciprocity rule : A 1 F12 = A2 F21 ⎯
⎯→ F12 =
(1)
(b) We designate the two identical surfaces of length b by (1) and (3), and the imaginary flat top surface by (2). Noting that
(2) is flat,
F22 = 0
a
summation rule : F21 + F22 + F23 = 1 ⎯
⎯→ F21 = F23 = 0.5 (symmetry)
(2)
summation rule : F22 + F2→(1+3) = 1 ⎯
⎯→ F2→(1+3) = 1
b
reciprocity rule : A 2 F2→(1+ 3) = A(1+3) F(1+ 3)→ 2
⎯
⎯→ F(1+ 3) → 2 = F(1+ 3)→ surr =
(3)
(1)
b
A2
a
(1) =
A(1+ 3)
2b
(c) We designate the bottom surface by (1), the side surfaces by (2) and (3), and
the imaginary top surface by (4). Surface 4 is flat and is completely surrounded
by other surfaces. Therefore, F44 = 0 and F4→(1+ 2+3) = 1 .
reciprocity rule : A 4 F4→(1+ 2 + 3) = A(1+ 2 + 3) F(1+ 2 + 3)→ 4
⎯
⎯→ F(1+ 2 + 3)→ 4 = F(1+ 2 + 3) → surr =
A4
A(1+ 2 +3)
a
(1) =
a + 2b
(4)
b
b
(2)
(3)
(1)
a
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13-12
13-19 A circular cone is positioned on a common axis with a circular disk, the values of F11 and F12 for specified L and D are
to be determined.
Assumptions 1 The surfaces are diffuse emitters and reflectors. 2 The surface A1 is treated as a single surface.
Analysis The area for A1, A2, and A3 are
πD ⎡
2
⎛D⎞ ⎤
A1 =
⎢L + ⎜ ⎟ ⎥
2 ⎢⎣
⎝ 2 ⎠ ⎥⎦
1/ 2
5 D 2π
4
=
2
and
A2 = A3 =
πD 2
4
The surfaces A1 and A3 can be treated as an enclosure, and using the summation rule yields
F11 + F13 = 1
Applying reciprocity relation between A1 and A3, we have
→
A1 F13 = A3 F31
F11 = 1 − F13 = 1 − ( A3 / A1 ) F31
Note that
→
F31 + F33 = 1
F31 = 1
(since F33 = 0 )
Hence, the view factor F11 is
F11 = 1 − ( A3 / A1 ) = 1 −
πD 2
5 D 2π
= 0.553
The radiation leaving A2 is intercepted by A1 and A3 equally,
F21 = F23
Applying reciprocity relation between A1 and A2, we have
→
A1 F12 = A2 F21
F12 = ( A2 / A1 ) F21 = ( A2 / A1 ) F23 =
F23
5
For coaxial parallel disks, the view factor F23 is evaluated from Table 13-1 as
F23
⎧
⎡
⎛D
1⎪
= ⎨S − ⎢ S 2 − 4 ⎜⎜ 3
2⎪
⎢
⎝ D2
⎣
⎩
⎞
⎟⎟
⎠
2⎤
⎥
⎥
⎦
1/ 2 ⎫
1
⎪ 1
2
1/ 2
2
1/ 2
⎬ = [ S − ( S − 4 ) ] = [6 − (6 − 4 ) ] = 0.1716
2
2
⎪
⎭
where
R1 = R2 = R =
S = 1+
1 + R22
R12
D 1
=
2L 2
= 2+
1
R2
= 2+4 = 6
Hence, the view factor F12 is
F12 =
F23
5
=
0.1716
= 0.0767
5
Discussion As long as L = D, the view factors are constant at F11 = 0.553 and F12 = 0.0767 (i.e., F11 and F12 are independent
of L and D).
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13-13
13-20 A cylindrical surface and a disk are oriented coaxially a distance L apart. The view factor F12 between them is to be
determined.
Assumptions The surfaces are diffuse emitters and reflectors.
Analysis The end surfaces A3 and A4 are treated as hypothetical
surfaces. The summation rule for surfaces facing the disk can be
expressed as
→
F23 = F21 + F24
F21 = F23 − F24
(1)
From reciprocity relation, we have
A2 F21 = A1 F12
(2)
Substituting Eq. (2) in to Eq. (1) and rearranging give
→
( A1 / A2 ) F12 = F23 − F24
F12 = ( A2 / A1 )( F23 − F24 )
(3)
The view factors F23 and F24 can be determined from the relation in Table 13-1 by treating the two surfaces as coaxial
parallel disks of identical diameters:
Fi → j = F j →i = 1 +
1 − 4R 2 + 1
2R 2
where R =
r
L
For surface 3 (L = 2D):
R=
r D/2 D/2
=
=
= 0.25
L
L
2D
and
F23 = 1 +
1 − 4 × 0.25 2 + 1
2 × 0.25 2
For surface 4 (L = 4D): R =
= 0.05573
r D/2 D/2
=
=
= 0.125
L
L
4D
and
F24 = 1 +
1 − 4 × 0.125 2 + 1
2 × 0.125 2
= 0.01515
Substituting these values of F23 and F24 into Eq. (3), and noting that L = 2D, the view factor between the cylindrical surface
and the disk is determined to be
F12 =
=
A2
πD 2 / 4
πD 2 / 4
( F23 − F24 ) =
( F23 − F24 ) =
( F23 − F24 )
πDL
πD(2 D)
A1
1
1
( F23 − F24 ) = (0.05573 − 0.01515) = 0.00507
8
8
Discussion The view factors F23 and F24 can also be determined using Fig. 13-7, but this would involve reading error.
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13-14
13-21 A cylindrical surface and a disk are oriented coaxially a distance L apart. The view factor F12 between them is to be
determined.
Assumptions The surfaces are diffuse emitters and reflectors.
Analysis The end surfaces A3 and A4 are treated as hypothetical
surfaces. The summation rule for surfaces facing the disk can be
expressed as
F23 = F21 + F24
→
F21 = F23 − F24
(1)
From reciprocity relation, we have
A2 F21 = A1 F12
(2)
Substituting Eq. (2) in to Eq. (1) and rearranging give
→
( A1 / A2 ) F12 = F23 − F24
F12 = ( A2 / A1 )( F23 − F24 )
(3)
The view factors F23 and F24 can be determined from the relation in Table 13-1 by treating the two surfaces as coaxial
parallel disks of identical diameters:
Fi → j = F j →i = 1 +
1 − 4R 2 + 1
2R 2
where R =
r
L
For surface 3 (L = D):
R=
r D/2 D/2
=
=
= 0.5
L
L
D
and
F23 = 1 +
1 − 4 × 0.5 2 + 1
2 × 0.5 2
= 0.1716
For surface 4 (L = 2D):
R=
r D/2 D/2
=
=
= 0.25
L
L
2D
and
F24 = 1 +
1 − 4 × 0.25 2 + 1
2 × 0.25 2
= 0.05573
Substituting these values of F23 and F24 into Eq. (3), and noting that L = D, the view factor between the cylindrical surface
and the disk is determined to be
F12 =
A2
1
πD 2 / 4
( F23 − F24 ) =
( F23 − F24 ) = (0.1716 − 0.05573) = 0.0290
πDL
A1
4
Discussion The view factors F23 and F24 can also be determined using Fig. 13-7, but this would involve reading error.
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13-15
Radiation Heat Transfer between Surfaces
13-22C The two methods used in radiation analysis are the matrix and network methods. In matrix method, equations 13-34
and 13-35 give N linear algebraic equations for the determination of the N unknown radiosities for an N -surface enclosure.
Once the radiosities are available, the unknown surface temperatures and heat transfer rates can be determined from these
equations respectively. This method involves the use of matrices especially when there are a large number of surfaces.
Therefore this method requires some knowledge of linear algebra.
The network method involves drawing a surface resistance associated with each surface of an enclosure and
connecting them with space resistances. Then the radiation problem is solved by treating it as an electrical network problem
where the radiation heat transfer replaces the current and the radiosity replaces the potential. The network method is not
practical for enclosures with more than three or four surfaces due to the increased complexity of the network.
13-23C Radiosity is the total radiation energy leaving a surface per unit time and per unit area. Radiosity includes the
emitted radiation energy as well as reflected energy. Radiosity and emitted energy are equal for blackbodies since a
blackbody does not reflect any radiation.
1− ε i
and it represents the resistance of a surface to the emission of
Ai ε i
radiation. It is zero for black surfaces. The space resistance is the radiation resistance between two surfaces and is expressed
1
as Rij =
Ai Fij
13-24C Radiation surface resistance is given as Ri =
13-25C The analysis of radiation exchange between black surfaces is relatively easy because of the absence of reflection.
The rate of radiation heat transfer between two surfaces in this case is expressed as Q& = A1 F12 σ (T1 4 − T2 4 ) where A1 is the
surface area, F12 is the view factor, and T1 and T2 are the temperatures of two surfaces.
13-26C Some surfaces encountered in numerous practical heat transfer applications are modeled as being adiabatic as the
back sides of these surfaces are well insulated and net heat transfer through these surfaces is zero. When the convection
effects on the front (heat transfer) side of such a surface is negligible and steady-state conditions are reached, the surface
must lose as much radiation energy as it receives. Such a surface is called reradiating surface. In radiation analysis, the
surface resistance of a reradiating surface is taken to be zero since there is no heat transfer through it.
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13-16
13-27 Two aligned parallel rectangles are apart by a distance of 2 m. The percentage of change in radiation heat transfer rate
when the rectangles are moved apart from 2 m to 8 m is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are black. 3 Convection heat transfer is not considered.
Analysis For D = 2 m, L1 = 6 m and L2 = 8 m, we have
L1 6
= =3
D 2
L2 8
= =4
D 2
and
From Fig. 13-5, we get
F12 ≈ 0.58
(for D = 2m)
For D = 8 m, L1 = 6 m and L2 = 8 m, we have
L1 6
= = 0.75
D 8
and
L2 8
= =1
D 8
From Fig. 13-5, we get
F12 ≈ 0.165
(for D = 8 m)
The rate of radiation heat transfer between the two rectangles is
Q&12 = AF12σ (T14 − T24 )
Hence, the percentage of change in radiation heat transfer rate is
% Change =
Q&12 ( D = 2 m) − Q&12 ( D = 8 m) F12 ( D = 2 m) − F12 ( D = 8 m)
=
F12 ( D = 2 m)
Q& ( D = 2 m)
12
0.58 − 0.165
=
0.58
= 0.716 (or 71.6%)
Discussion By moving the distance between the two parallel rectangles from 2 m to 8 m, there is about 72% reduction in
radiation heat transfer rate.
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13-17
13-28 A solid sphere is placed in an evacuated equilateral triangular enclosure. The view factor from the enclosure to the
sphere and the emissivity of the enclosure are 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 sphere is given to be ε1 = 0.45.
Analysis (a) We take the sphere to be surface 1 and the surrounding
enclosure to be surface 2. The view factor from surface 2 to surface
1 is determined from reciprocity relation:
A1 = πD = π (1 m) = 3.142 m
2
2
A2 = 3 L − D
2
2
T1 = 600 K
ε1 = 0.45
2m
T2 = 420 K
ε2 = ?
2
2m
2m
L
= 3 (2 m) 2 − (1 m) 2
= 5.196 m 2
2
2
1m
A1 F12 = A2 F21
(3.142)(1) = (5.196) F21
2m
F21 = 0.605
(b) The net rate of radiation heat transfer can be expressed for this two-surface enclosure to yield the emissivity of the
enclosure:
Q& =
3100 W =
(
σ T1 4 − T2 4
)
1 − ε1
1−ε2
1
+
+
A1ε 1
A1 F12
A2 ε 2
[
]
(5.67 × 10 −8 W/m 2 ⋅ K 4 ) (600 K )4 − (420 K )4
1− ε2
1 − 0.45
1
+
+
2
2
(3.142 m )(0.45) (3.142 m )(1) (5.196 m 2 )ε 2
ε 2 = 0.150
13-29 A long cylindrical rod coated with a new material is placed in an evacuated long cylindrical enclosure which is
maintained at a uniform temperature. The emissivity of the coating on the rod is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are
opaque, diffuse, and gray.
Properties The emissivity of the enclosure is given to be ε2 = 0.95.
D2 = 0.1 m
T2 = 200 K
ε2 = 0.95
D1 = 0.01 m
T1 = 600 K
ε1 = ?
Analysis The emissivity of the coating on the rod is determined from
A σ (T 4 − T2 4 )
Q& 12 = 1 1
1 1 − ε 2 ⎛ r1 ⎞
⎜ ⎟
+
ε1
ε 2 ⎜⎝ r2 ⎟⎠
12 W =
[π (0.01 m)(1 m)](5.67 × 10 −8 W/m 2 ⋅ K 4 )[(600 K )4 − (200 K )4 ]
1 1 − 0.95 ⎛ 1 ⎞
+
⎜ ⎟
ε1
0.95 ⎝ 10 ⎠
Vacuum
which gives
ε1 = 0.0527
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13-18
13-30E Top and side surfaces of a cubical furnace are black, and are maintained at uniform temperatures. Net radiation heat
transfer rate to the base from the top and side surfaces are 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 emissivities are given to be ε = 0.4 for the bottom surface and 1 for other surfaces.
Analysis We consider the base surface to be surface 1, the top surface to be surface 2 and the side surfaces to be surface 3.
The cubical furnace can be considered to be three-surface enclosure. The areas and blackbody emissive powers of surfaces
are
A1 = A2 = (10 ft ) 2 = 100 ft 2
E b1 = σT1 = (0.1714 × 10
4
T2 = 1600 R
−8
Btu/h.ft .R )(800 R ) = 702 Btu/h.ft
−8
Btu/h.ft 2 .R 4 )(1600 R ) 4 = 11,233 Btu/h.ft 2
E b 2 = σT2 = (0.1714 × 10
4
A3 = 4(10 ft ) 2 = 400 ft 2
2
4
4
ε2 = 1
2
E b3 = σT3 4 = (0.1714 × 10 −8 Btu/h.ft 2 .R 4 )(2400 R ) 4 = 56,866 Btu/h.ft 2
T3 = 2400 R
ε3 = 1
The view factor from the base to the top surface of the cube is F12 = 0.2 . From the
summation rule, the view factor from the base or top to the side surfaces is
F11 + F12 + F13 = 1 ⎯
⎯→ F13 = 1 − F12 = 1 − 0.2 = 0.8
since the base surface is flat and thus F11 = 0 . Then the radiation resistances become
1 − ε1
1 − 0. 4
=
= 0.015 ft -2
A1ε 1 (100 ft 2 )(0.4)
1
1
=
=
= 0.0125 ft -2
A1 F13 (100 ft 2 )(0.8)
R1 =
R13
R12 =
T1 = 800 R
ε1 = 0.4
1
1
=
= 0.0500 ft -2
A1 F12 (100 ft 2 )(0.2)
Note that the side and the top surfaces are black, and thus their radiosities are equal to their emissive powers. The radiosity of
the base surface is determined
E b1 − J 1 E b 2 − J 1 E b 3 − J 1
+
+
=0
R1
R12
R13
Substituting,
702 − J 1 11,233 − J 1 56,866 − J 1
+
+
=0⎯
⎯→ J 1 = 28,925 W/m 2
0.015
0.05
0.0125
(a) The net rate of radiation heat transfer between the base and the side surfaces is
E − J 1 (56,866 − 28,915) Btu/h.ft 2
Q& 31 = b3
=
= 2.235 × 10 6 Btu/h
R13
0.0125 ft -2
(b) The net rate of radiation heat transfer between the base and the top surfaces is
J − Eb 2 (28,925 − 11,233) Btu/h.ft 2
Q&12 = 1
=
= 3.538 × 10 5 Btu/h
R12
0.05 ft -2
The net rate of radiation heat transfer to the base surface is finally determined from
Q&1 = Q& 21 + Q& 31 = −3.538 × 10 5 + 2.235 × 10 6 = 1.882 × 10 6 Btu/h
Discussion The same result can be found form
J − Eb1 (28,925 − 702) Btu/h.ft 2
Q&1 = 1
=
= 1.882 × 10 6 Btu/h
-2
R1
0.015 ft
The result is the same as expected.
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13-19
Prob. 13-30E is reconsidered. The effect of base surface emissivity on the net rates of radiation heat transfer
13-31E
between the base and the side surfaces, between the base and top surfaces, and to the base surface is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
a=10 [ft]
epsilon_1=0.4
T_1=800 [R]
T_2=1600 [R]
T_3=2400 [R]
"ANALYSIS"
sigma=0.1714E-8 [Btu/h-ft^2-R^4] “Stefan-Boltzmann constant"
"Consider the base surface 1, the top surface 2, and the side surface 3"
E_b1=sigma*T_1^4
E_b2=sigma*T_2^4
E_b3=sigma*T_3^4
A_1=a^2
A_2=A_1
A_3=4*a^2
F_12=0.2 "view factor from the base to the top of a cube"
F_11+F_12+F_13=1 "summation rule"
F_11=0 "since the base surface is flat"
R_1=(1-epsilon_1)/(A_1*epsilon_1) "surface resistance"
R_12=1/(A_1*F_12) "space resistance"
R_13=1/(A_1*F_13) "space resistance"
(E_b1-J_1)/R_1+(E_b2-J_1)/R_12+(E_b3-J_1)/R_13=0 "J_1 : radiosity of base surface"
"(a)"
Q_dot_31=(E_b3-J_1)/R_13
"(b)"
Q_dot_12=(J_1-E_b2)/R_12
Q_dot_21=-Q_dot_12
Q_dot_1=Q_dot_21+Q_dot_31
ε1
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
Q& 31
[Btu/h]
1.106E+06
1.295E+06
1.483E+06
1.671E+06
1.859E+06
2.047E+06
2.235E+06
2.423E+06
2.612E+06
2.800E+06
2.988E+06
3.176E+06
3.364E+06
3.552E+06
3.741E+06
3.929E+06
4.117E+06
Q& 12
[Btu/h]
636061
589024
541986
494948
447911
400873
353835
306798
259760
212722
165685
118647
71610
24572
-22466
-69503
-116541
Q&1
[Btu/h]
470376
705565
940753
1.176E+06
1.411E+06
1.646E+06
1.882E+06
2.117E+06
2.352E+06
2.587E+06
2.822E+06
3.057E+06
3.293E+06
3.528E+06
3.763E+06
3.998E+06
4.233E+06
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13-20
4.5x 106
4.0x 106
Q31 [Btu/h]
3.5x 106
3.0x 106
2.5x 106
2.0x 106
1.5x 106
1.0x 106
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.6
0.7
0.8
0.9
0.7
0.8
0.9
ε1
700000
600000
500000
Q12 [Btu/h]
400000
300000
200000
100000
0
-100000
-200000
0.1
0.2
0.3
0.4
0.5
ε1
4.5x 106
4.0x 106
Q1 [Btu/h]
3.5x 106
3.0x 106
2.5x 106
2.0x 106
1.5x 106
1.0x 106
5.0x 105
0.0x 100
0.1
0.2
0.3
0.4
0.5
0.6
ε1
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13-21
13-32 Two very large parallel plates are maintained at uniform temperatures. The net rate of radiation heat transfer between
the two plates 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.
T1 = 600 K
ε1 = 0.5
Properties The emissivities ε of the plates are given to be
0.5 and 0.9.
Analysis The net rate of radiation heat transfer between the
two surfaces per unit area of the plates is determined
directly from
T2 = 400 K
ε2 = 0.9
Q& 12 σ (T1 4 − T2 4 ) (5.67 × 10 −8 W/m 2 ⋅ K 4 )[(600 K ) 4 − (400 K ) 4 ]
=
= 2793 W/m 2
=
1
1
1
1
As
+
−1
+
−1
ε1 ε 2
0. 5 0. 9
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13-22
Prob. 13-32 is reconsidered. The effects of the temperature and the emissivity of the hot plate on the net rate of
13-33
radiation heat transfer between the plates are to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
T_1=600 [K]
T_2=400 [K]
epsilon_1=0.5
epsilon_2=0.9
sigma=5.67E-8 [W/m^2-K^4] “Stefan-Boltzmann constant"
"ANALYSIS"
q_dot_12=(sigma*(T_1^4-T_2^4))/(1/epsilon_1+1/epsilon_2-1)
T1 [K]
q&12 [W/m2]
500
525
550
575
600
625
650
675
700
725
750
775
800
825
850
875
900
925
950
975
1000
991.1
1353
1770
2248
2793
3411
4107
4888
5761
6733
7810
9001
10313
11754
13332
15056
16934
18975
21188
23584
26170
ε1
q&12 [W/m2]
5000
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
583.2
870
1154
1434
1712
1987
2258
2527
2793
3056
3317
3575
3830
4082
4332
4580
4825
4500
30000
25000
15000
2
q 12 [W /m ]
20000
10000
5000
0
500
600
700
800
900
1000
T 1 [K]
4000
3500
2
q 12 [W /m ]
3000
2500
2000
1500
1000
500
0.1
0.2
0.3
0.4
0.5
ε1
0.6
0.7
0.8
0.9
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13-23
13-34 The radiation heat flux between two infinitely long parallel plates of specified surface temperatures is to be
determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are black. 3 Convection heat transfer is not considered. 4
The surface temperatures are uniform.
Analysis From the Hottel’s crossed-strings method, we have
Fi → j =
Σ (Crossed strings ) − Σ ( Uncrossed strings )
2 × (String on surface i )
For uncrossed strings, we have
L1 = L2 = ( w 2 + w 2 )1 / 2 = ( w 2 + w 2 )1 / 2 = 2 w
For crossed strings, we have
L3 = ( w 2 + 4w 2 )1 / 2 = 5 w
and
L4 = w
Applying the Hottel’s crossed-strings method, we get F12 as
F12 =
( L3 + L4 ) − ( L1 + L2 )
2w
( 5 w + w) − ( 2 w + 2 w)
2w
= 0.204
=
The radiation heat flux between the two surfaces is
q&12 = F12σ (T14 − T24 )
= (0.204)(5.67 × 10 −8 W/m 2 ⋅ K 4 )(700 4 − 300 4 ) K 4
= 2680 W/m 2
Discussion The Hottel’s crossed-string method is applicable only to surfaces that are very long, such that they can be
considered to be two-dimensional and radiation interaction through the end surfaces is negligible.
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13-24
13-35 The base and the dome of a long semi-cylindrical dryer are maintained at uniform temperatures. The drying rate per
unit length experienced by the base surface is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are black. 3 Convection heat transfer is not considered. 4
The dryer is well insulated from heat loss to the surrounding.
Properties The latent heat of vaporization for water is hfg = 2257 kJ/kg (Table A-2)
Analysis The view factor from the dome to the base is determined from
→
F11 + F12 = 1
F12 = 1
(where F11 = 0 )
Hence, from reciprocity relation, we get
F21 =
A1
DL
2
F12 =
=
A2
πDL / 2 π
Applying energy balance on the base surface, we have
Q& 21 = Q& evap = m& h fg
Hence,
m& h fg = A2 F21σ (T24 − T14 ) =
πDL
2
F21σ (T24 − T14 )
or
πD
πD 2
m&
σ (T24 − T14 )
=
F21σ (T24 − T14 ) =
2h fg π
L 2h fg
=
D
σ (T24 − T14 )
h fg
=
(1.5 m)
(2257 × 10 J/kg)
= 0.0370 kg/s ⋅ m
3
(5.67 × 10 −8 W/m 2 ⋅ K 4 )(1000 4 − 370 4 ) K 4
Discussion The view factor from the dome to the base is constant F21 = 2/π, which implies that the view factor is
independent of the dryer dimensions.
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13-25
13-36 The base, top, and side surfaces of a furnace of cylindrical shape are black, and are maintained at uniform
temperatures. The net rate of radiation heat transfer to or from the top surface is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are black.
3 Convection heat transfer is not considered.
T1 = 700 K
ε1 = 1
R=3m
Properties The emissivity of all surfaces are ε = 1 since they are black.
Analysis We consider the top surface to be surface 1, the base surface to be
surface 2 and the side surfaces to be surface 3. The cylindrical furnace can
be considered to be three-surface enclosure. We assume that steady-state
conditions exist. Since all surfaces are black, the radiosities are equal to the
emissive power of surfaces, and the net rate of radiation heat transfer from
the top surface can be determined from
Q& = A1 F12σ (T1 4 − T2 4 ) + A1 F13σ (T1 4 − T3 4 )
and
A1 = πR 2 = π (3 m) 2 = 28.27 m 2
H=3m
T3 = 500 K
ε3 = 1
T2 = 1400 K
ε2 = 1
R=3m
The view factor from the base to the top surface of the cylinder is F12 = 0.38 (From Figure 13-7). The view factor from the
base to the side surfaces is determined by applying the summation rule to be
F11 + F12 + F13 = 1 ⎯
⎯→ F13 = 1 − F12 = 1 − 0.38 = 0.62
Substituting,
Q& = A1 F12σ (T1 4 − T2 4 ) + A1 F13σ (T1 4 − T3 4 )
= (28.27 m 2 )(0.38)(5.67 × 10 -8 W/m 2 .K 4 )(700 K 4 - 500 K 4 )
+ (28.27 m 2 )(0.62)(5.67 × 10 -8 W/m 2 .K 4 )(700 K 4 - 1400 K 4 )
= −3.579 × 10 6 W = −3579 kW
Discussion The negative sign indicates that net heat transfer is to the top surface.
PROPRIETARY MATERIAL. © 2011 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.
13-26
13-37E A room is heated by electric resistance heaters placed on the ceiling which is maintained at a uniform temperature.
The rate of heat loss from the room through the floor 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. 4 There is no heat loss through the side surfaces.
Properties The emissivities are ε = 1 for the ceiling and ε = 0.8 for the floor. The emissivity of insulated (or reradiating)
surfaces is also 1.
Analysis The room can be considered to be three-surface enclosure with the ceiling
surface 1, the floor surface 2 and the side surfaces surface 3. We assume steady-state
conditions exist. Since the side surfaces are reradiating, there is no heat transfer through
them, and the entire heat lost by the ceiling must be gained by the floor. Then the rate of
heat loss from the room through its floor can be determined from
Q& 1 =
Ceiling: 12 ft × 12 ft
T1 = 90°F
ε1 = 1
E b1 − E b 2
⎛ 1
1
⎜
⎜R + R +R
13
23
⎝ 12
⎞
⎟
⎟
⎠
−1
+ R2
9 ft
Insulated side
surfacess
where
E b1 = σT1 4 = (0.1714 ×10 −8 Btu/h.ft 2 .R 4 )(90 + 460 R ) 4 = 157 Btu/h.ft 2
E b 2 = σT2 4 = (0.1714 ×10 −8 Btu/h.ft 2 .R 4 )(65 + 460 R ) 4 = 130 Btu/h.ft 2
T2 = 65°F
ε2 = 0.8
and
A1 = A2 = (12 ft ) 2 = 144 ft 2
The view factor from the floor to the ceiling of the room is F12 = 0.27 (From Figure 13-5). The view factor from the ceiling
or the floor to the side surfaces is determined by applying the summation rule to be
F11 + F12 + F13 = 1 ⎯
⎯→ F13 = 1 − F12 = 1 − 0.27 = 0.73
since the ceiling is flat and thus F11 = 0 . Then the radiation resistances which appear in the equation above become
1− ε 2
1 − 0.8
=
= 0.00174 ft -2
2
A2 ε 2 (144 ft )(0.8)
1
1
=
=
= 0.02572 ft - 2
A1 F12 (144 ft 2 )(0.27)
R2 =
R12
R13 = R 23 =
1
1
=
= 0.009513 ft - 2
2
A1 F13 (144 ft )(0.73)
Substituting,
Q& 12 =
(157 − 130) Btu/h.ft 2
⎛
⎞
1
1
⎜
⎟
+
⎜ 0.02572 ft -2 2(0.009513 ft -2 ) ⎟
⎝
⎠
= 2130 Btu/h
−1
+ 0.00174 ft -2
PROPRIETARY MATERIAL. © 2011 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.
13-27
13-38 Two very long concentric cylinders are maintained at uniform temperatures. The net rate of radiation heat transfer
between the two cylinders 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 emissivities of surfaces are given to be ε1 = 1 and ε2 = 0.55.
Analysis The net rate of radiation heat transfer between the two cylinders
per unit length of the cylinders is determined from
D2 = 0.5 m
T2 = 500 K
ε2 = 0.55
D1 = 0.35 m
T1 = 950 K
ε1 = 1
A σ (T1 4 − T2 4 )
Q& 12 = 1
1 1 − ε 2 ⎛ r1 ⎞
⎜ ⎟
+
ε1
ε 2 ⎜⎝ r2 ⎟⎠
[π (0.35 m)(1 m)](5.67 × 10 −8 W/m 2 ⋅ K 4 )[(950 K) 4 − (500 K ) 4 ]
1 1 − 0.55 ⎛ 3.5 ⎞
+
⎟
⎜
1
0.55 ⎝ 5 ⎠
= 29,810 W = 29.81 kW
=
Vacuum
PROPRIETARY MATERIAL. © 2011 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.
13-28
13-39 Radiation heat transfer occurs between a sphere and a circular disk. The view factors and the net rate of radiation heat
transfer for the existing and modified cases are 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 emissivities of sphere and disk are given to be ε1 = 0.9 and ε2 = 0.5, respectively.
Analysis (a) We take the sphere to be surface 1 and the disk to be surface 2. The view factor from surface 1 to surface 2 is
determined from
F12
⎧ ⎡
⎛r
⎪
= 0.5⎨1 − ⎢1 + ⎜⎜ 2
⎪ ⎢⎣ ⎝ h
⎩
⎞
⎟⎟
⎠
2⎤
−0.5 ⎫
⎥
⎥⎦
−0.5 ⎫
⎧ ⎡
2
⎪
⎪
⎛ 1.2 m ⎞ ⎤
⎪
⎟ ⎥
⎬ = 0.2764
⎬ = 0.5⎨1 − ⎢1 + ⎜
⎪ ⎢⎣ ⎝ 0.60 m ⎠ ⎥⎦
⎪
⎪
⎭
⎩
⎭
The view factor from surface 2 to surface 1 is determined from
reciprocity relation:
A1 = 4πr12 = 4π (0.3 m) 2
A2 = πr2 2 = π (1.2 m) 2 =
= 1.131 m
r1
2
h
4.524 m 2
r2
A1 F12 = A2 F21
(1.131)(0.2764) = (4.524) F21
F21 = 0.0691
(b) The net rate of radiation heat transfer between the surfaces can be determined from
Q& =
(
σ T1 4 − T2 4
)
1− ε1
1− ε 2
1
+
+
A1ε 1 A1 F12 A2 ε 2
=
[
]
(5.67 × 10 −8 W/m 2 ⋅ K 4 ) (873 K )4 − (473 K )4
= 8550 W
1 − 0.9
1
1 − 0.5
+
+
(1.131 m 2 )(0.9) (1.131 m 2 )(0.2764) (4.524 m 2 )(0.5)
(c) The best values are ε 1 = ε 2 = 1 and h = r1 = 0.3 m . Then the view factor becomes
F12
−0.5 ⎫
−0.5 ⎫
⎧ ⎡
⎧ ⎡
2
2
⎛ r2 ⎞ ⎤
⎪
⎪
⎛ 1.2 m ⎞ ⎤
⎪
⎪
0
.
5
1
1
=
−
+
= 0.5⎨1 − ⎢1 + ⎜⎜ ⎟⎟ ⎥
⎟ ⎥
⎨ ⎢ ⎜
⎬ = 0.3787
⎬
h
0
.
30
m
⎠ ⎥⎦
⎪ ⎢⎣ ⎝
⎪
⎪ ⎢⎣ ⎝ ⎠ ⎥⎦
⎪
⎩
⎭
⎩
⎭
The net rate of radiation heat transfer in this case is
(
)
[
]
Q& = A1 F12σ T1 4 − T2 4 = (1.131 m 2 )(0.3787)(5.67 × 10 −8 W/m 2 ⋅ K 4 ) (873 K )4 − (473 K )4 = 12,890 W
PROPRIETARY MATERIAL. © 2011 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.
13-29
13-40 Two parallel disks whose back sides are insulated are black, and are maintained at a uniform temperature. The net rate
of radiation heat transfer from the disks to the environment 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 emissivities of all surfaces are ε = 1 since they are black.
Disk 1, T1 = 450 K, ε1 = 1
Analysis Both disks possess same properties and they are black. Noting
that environment can also be considered to be blackbody, we can treat this
geometry as a three surface enclosure. We consider the two disks to be
surfaces 1 and 2 and the environment to be surface 3. Then from Figure
13-7, we read
0.40 m
F12 = F21 = 0.26
F13 = 1 − 0.26 = 0.74
D = 0.6 m
(summation rule)
Environment
T3 =300 K
ε1 = 1
Disk 2, T2 = 450 K, ε2 = 1
The net rate of radiation heat transfer from the disks into the environment
then becomes
Q& 3 = Q& 13 + Q& 23 = 2Q& 13
Q& 3 = 2 F13 A1σ (T1 4 − T3 4 )
= 2(0.74)[π (0.3 m) 2 ](5.67 × 10 −8 W/m 2 ⋅ K 4 )[(450 K )4 − (300 K )4 ]
= 781 W
13-41E Two black parallel rectangles are spaced apart by a distance of 1 ft, the temperature of the bottom rectangle is to be
determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are black. 3 Convection heat transfer is not considered.
Analysis For D = 1 ft, L1 = 3 ft and L2 = 5 ft, we have
L1 3
= =3
D 1
and
L2 5
= =5
D 1
From Fig. 13-5, we get
F12 ≈ 0.60
The rate of radiation heat transfer between the two rectangles is
Q&12 = AF12σ (T14 − T24 ) = L1 L2 F12σ (T14 − T24 )
Hence
1/ 4
⎤
⎡ Q&12
+ T24 ⎥
T1 = ⎢
⎦
⎣ L1 L2 F12σ
⎤
⎡
180000 Btu/h
=⎢
+ (60 + 460) 4 R 4 ⎥
2
4
−8
⎥⎦
⎣⎢ (3 ft )(5 ft )(0.60)(0.1714 × 10 Btu/h ⋅ ft ⋅ R )
1/ 4
T1 = 1851 R
Discussion If T1 and T2 are constant, increasing the distance between the two rectangles will decrease the view factor F12,
thereby decreasing the radiation heat transfer rate received by the top rectangle.
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13-30
13-42 Two coaxial parallel disks of equal diameter 1 m are originally placed at a distance of 1 m apart. The new distance
between the disks such that there is a 75% reduction in radiation heat transfer rate from the original distance of 1 m is to be
determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are black. 3 Convection heat transfer is not considered.
Analysis For coaxial parallel disks, from Table 13-1, with i = 1, j = 2,
1/ 2 ⎫
⎧
2
⎡
⎛ D2 ⎞ ⎤ ⎪ 1
1⎪
2
⎟⎟ ⎥ ⎬ = [ S − ( S 2 − 4 )1 / 2 ]
F12 = ⎨S − ⎢ S − 4 ⎜⎜
2⎪
D
⎢
1
⎠ ⎥⎦ ⎪ 2
⎝
⎣
⎭
⎩
S = 1+
where
1 + R22
R12
R1 = R2 = R =
= 2+
1
R2
= 2 + 4( L / D) 2
(1)
(2)
D
2L
Substituting Eq. (2) into Eq. (1), we get
1
(2 + 4( L / D) 2 − {[ 2 + 4( L / D) 2 ] 2 − 4 }1 / 2 )
2
1
= {2 + 4( L / D) 2 − [16( L / D) 4 + 16( L / D) 2 ]1 / 2 }
2
1
= {2 + 4( L / D) 2 − 4( L / D)[( L / D) 2 + 1]1 / 2 }
2
F12 =
F12 = 1 + 2( L / D) 2 − 2( L / D)[( L / D) 2 + 1]1 / 2
Hence, for D = 1 m we have
F12 = 1 + 2 L2 − 2 L[ L2 + 1]1 / 2
(3)
From Eq. (3), the view factor at the original distance L = 1 m (with D = l m) is
F12, old = 1 + 2(1) 2 − 2(1)[(1) 2 + 1]1 / 2 = 3 − 2 2 = 0.1716
The rate of radiation heat transfer between the two surfaces is
Q&12 = AF12σ (T14 − T24 )
The percentage of reduction in radiation heat transfer rate can be expressed as
% Change =
Q& 12, old − Q&12, new F12, old − F12, new
=
F12, old
Q&
12, old
For the two surfaces to experience 75% reduction in radiation heat transfer rate, the new view factor should be
F12, new = 0.25 F12, old = 0.0429
Then, substituting the value of F12 , new into Eq. (3), the new distance Lnew such that the two surfaces experience 75%
reduction in radiation heat transfer rate can be calculated as
F12, new = 1 + 2 L2new − 2 Lnew [ L2new + 1]1 / 2
0.0429 = 1 + 2 L2new − 2 Lnew [ L2new + 1]1 / 2
Hence, Lnew = 2.31 m
Discussion Increasing the distance between the disks would decrease the view factor F12, thereby reducing the radiation heat
& .
transfer rate Q
12
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preparation. If you are a student using this Manual, you are using it without permission.
13-31
13-43 A furnace shaped like a long equilateral-triangular duct is considered. The temperature of the base surface 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. 4 End effects are neglected.
Properties The emissivities of surfaces are given to be ε1 = 0.8 and ε2 = 0.4.
Analysis This geometry can be treated as a two surface enclosure since
two surfaces have identical properties. We consider base surface to be
surface 1 and other two surface to be surface 2. Then the view factor
between the two becomes F12 = 1 . The temperature of the base surface is
determined from
Q& 12 =
σ (T1 4 − T2 4 )
1 − ε1
1− ε2
1
+
+
A1ε 1
A1 F12
A2 ε 2
(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(T1 )4 − (600 K )4 ]
1 − 0.8
1
1 − 0.4
+
+
2
2
(1 m )(0.8) (1 m )(1) (2 m 2 )(0.4)
T1 = 630 K
800 W =
T2 = 600 K
ε2 = 0.4
q1 = 800 W/m2
ε1 = 0.8
b=2m
Note that A1 = 1 m 2 and A2 = 2 m 2 .
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preparation. If you are a student using this Manual, you are using it without permission.
13-32
Prob. 13-43 is reconsidered. The effects of the rate of the heat transfer at the base surface and the temperature
13-44
of the side surfaces on the temperature of the base surface are to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
a=2 [m]
epsilon_1=0.8
epsilon_2=0.4
Q_dot_12=800 [W]
T_2=600 [K]
sigma=5.67E-8 [W/m^2-K^4]
"ANALYSIS"
"Consider the base surface to be surface 1, the side surfaces to be surface 2"
Q_dot_12=(sigma*(T_1^4-T_2^4))/((1-epsilon_1)/(A_1*epsilon_1)+1/(A_1*F_12)+(1epsilon_2)/(A_2*epsilon_2))
F_12=1
A_1=1 "[m^2], since rate of heat supply is given per meter square area"
A_2=2*A_1
T1
[K]
619.4
620.4
621.3
622.2
623.1
624
624.9
625.8
626.7
627.6
628.5
629.4
630.3
631.2
632
632.9
633.8
634.6
635.5
636.4
637.2
637.5
633.5
T1 [K]
Q& 12
[W]
500
525
550
575
600
625
650
675
700
725
750
775
800
825
850
875
900
925
950
975
1000
629.5
625.5
621.5
617.5
500
600
700
800
900
1000
Q12 [W]
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13-33
T1
[K]
436.5
445.5
456
468.1
481.7
496.7
512.9
530.4
548.8
568.1
588.2
609
630.3
652.1
674.3
696.9
719.7
750
700
650
T1 [K]
T2
[K]
300
325
350
375
400
425
450
475
500
525
550
575
600
625
650
675
700
600
550
500
450
400
300
350
400
450
500
550
600
650
700
T2 [K]
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
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13-34
13-45 The floor and the ceiling of a cubical furnace are maintained at uniform temperatures. The net rate of radiation heat
transfer between the floor and the ceiling 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 emissivities of all surfaces are ε = 1 since they are black or reradiating.
Analysis We consider the ceiling to be surface 1, the floor to be surface 2 and the side surfaces to be surface 3. The furnace
can be considered to be three-surface enclosure. We assume that steady-state conditions exist. Since the side surfaces are
reradiating, there is no heat transfer through them, and the entire heat lost by the ceiling must be gained by the floor. The
view factor from the ceiling to the floor of the furnace is F12 = 0.2 . Then the rate of heat loss from the ceiling can be
determined from
Q& 1 =
E b1 − E b 2
⎛ 1
1
⎜
⎜R + R +R
13
23
⎝ 12
⎞
⎟
⎟
⎠
a=4m
−1
T1 = 1100 K
ε1 = 1
where
E b1 = σT1 4 = (5.67 × 10 −8 W/m 2 .K 4 )(1100 K ) 4 = 83,015 W/m 2
Reradiating side
surfacess
E b 2 = σT2 4 = (5.67 × 10 −8 W/m 2 .K 4 )(550 K ) 4 = 5188 W/m 2
and
T2 = 550 K
ε2 = 1
A1 = A2 = (4 m) 2 = 16 m 2
1
1
=
= 0.3125 m -2
A1 F12 (16 m 2 )(0.2)
1
1
= R 23 =
=
= 0.078125 m -2
2
A1 F13 (16 m )(0.8)
R12 =
R13
Substituting,
Q& 12 =
(83,015 − 5188) W/m 2
⎞
⎛
1
1
⎜
⎟
+
⎜ 0.3125 m -2 2(0.078125 m -2 ) ⎟
⎝
⎠
−1
= 7.47 ×10 5 W = 747 kW
PROPRIETARY MATERIAL. © 2011 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.
13-35
13-46 Two concentric spheres are maintained at uniform temperatures. The net rate of radiation heat transfer between the
two spheres and the convection heat transfer coefficient at the outer surface are to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are
opaque, diffuse, and gray.
Properties The emissivities of surfaces are given to be ε1 = 0.5
and ε2 = 0.7.
D2 = 0.6 m
T2 = 500 K
ε2 = 0.7
Analysis The net rate of radiation heat transfer between the two
spheres is
Q& 12 =
(
A1σ T1 4 − T2 4
1
ε1
=
+
)
D1 = 0.3 m
T1 = 800 K
ε1 = 0.5
ε = 0.35
⎞
⎟
⎟
⎠
1 − ε 2 ⎛⎜ r1 2
ε 2 ⎜⎝ r2 2
Tsurr =30°C
T∞ = 30°C
[π (0.3 m) ](5.67 × 10
2
−8
)[
W/m 2 ⋅ K 4 (800 K )4 − (500 K )4
1 1 − 0.7 ⎛ 0.15 m ⎞
+
⎜
⎟
0.5
0.7 ⎝ 0.3 m ⎠
]
2
= 2641 W
Radiation heat transfer rate from the outer sphere to the surrounding surfaces are
Q& rad = εFA2σ (T2 4 − Tsurr 4 )
= (0.35)(1)[π (0.6 m) 2 ](5.67 × 10 −8 W/m 2 ⋅ K 4 )[(500 K ) 4 − (30 + 273 K ) 4 ]
= 1214 W
The convection heat transfer rate at the outer surface of the cylinder is determined from requirement that heat transferred
from the inner sphere to the outer sphere must be equal to the heat transfer from the outer surface of the outer sphere to the
environment by convection and radiation. That is,
Q& conv = Q&12 − Q& rad = 2641 − 1214 = 1427 W
Then the convection heat transfer coefficient becomes
Q& conv. = hA2 (T2 − T∞ )
[
]
1427 W = h π (0.6 m) 2 (500 K - 303 K)
h = 6.40 W/m ⋅ °C
2
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preparation. If you are a student using this Manual, you are using it without permission.
13-36
13-47 A spherical tank filled with liquid nitrogen is kept in an evacuated cubic enclosure. The net rate of radiation heat
transfer to the liquid nitrogen 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. 4 The thermal resistance of the tank is negligible.
Properties The emissivities of surfaces are given to be ε1 = 0.1 and ε2 = 0.8.
Analysis We take the sphere to be surface 1 and the surrounding cubic
enclosure to be surface 2. Noting that F12 = 1 , for this two-surface enclosure,
the net rate of radiation heat transfer to liquid nitrogen can be determined from
(
)
A σ T 4 − T2 4
Q& 21 = −Q& 12 = − 1 1
1 1 − ε 2 ⎛ A1
⎜
+
ε1
ε 2 ⎜⎝ A2
[π (2 m) ](5.67 ×10
=−
2
−8
⎞
⎟⎟
⎠
)[
W/m 2 ⋅ K 4 (100 K )4 − (240 K )4
1 1 − 0.8 ⎡ π (2 m) 2 ⎤
+
⎢
⎥
0.1
0.8 ⎢⎣ 6(3 m) 2 ⎥⎦
Cube, a =3 m
T2 = 240 K
ε2 = 0.8
D1 = 2 m
T1 = 100 K
ε1 = 0.1
Liquid
N2
]
Vacuum
= 228 W
13-48 A spherical tank filled with liquid nitrogen is kept in an evacuated spherical enclosure. The net rate of radiation heat
transfer to the liquid nitrogen 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. 4 The thermal resistance of the tank is negligible.
Properties The emissivities of surfaces are given to be ε1 = 0.1 and
ε2 = 0.8.
Analysis The net rate of radiation heat transfer to liquid nitrogen can be
determined from
Q& 12 =
(
A1σ T1 4 − T2 4
1
ε1
=
)
1 − ε 2 ⎛⎜ r1
ε 2 ⎜⎝ r2 2
2
+
⎞
⎟
⎟
⎠
[π (2 m) ](5.67 ×10
2
D1 = 2 m
T1 = 100 K
ε1 = 0.1
D2 = 3 m
T2 = 240 K
ε2 = 0.8
−8
)[
W/m 2 ⋅ K 4 (240 K )4 − (100 K )4
1 1 − 0.8 ⎡ (1 m) ⎤
+
⎢
⎥
0.1
0.8 ⎣⎢ (1.5 m) 2 ⎦⎥
]
Liquid
N2
2
Vacuum
= 227 W
PROPRIETARY MATERIAL. © 2011 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.
13-37
Prob. 13-47 is reconsidered. The effects of the side length and the emissivity of the cubic enclosure, and the
13-49
emissivity of the spherical tank on the net rate of radiation heat transfer are to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
D=2 [m]
a=3 [m]
T_1=100 [K]
T_2=240 [K]
epsilon_1=0.1
epsilon_2=0.8
sigma=5.67E-8 [W/m^2-K^4] “Stefan-Boltzmann constant"
"ANALYSIS"
"Consider the sphere to be surface 1, the surrounding cubic enclosure to be surface 2"
Q_dot_12=(A_1*sigma*(T_1^4-T_2^4))/(1/epsilon_1+(1-epsilon_2)/epsilon_2*(A_1/A_2))
Q_dot_21=-Q_dot_12
A_1=pi*D^2
A_2=6*a^2
2
2.5
3
3.5
4
4.5
5
5.5
6
6.5
7
7.5
8
8.5
9
9.5
10
10.5
11
11.5
12
Q& 21
[W]
226.3
227.4
227.9
228.3
228.5
228.7
228.8
228.9
228.9
229
229
229.1
229.1
229.1
229.1
229.1
229.1
229.2
229.2
229.2
229.2
229.5
229
228.5
Q21 [W]
a
[m]
228
227.5
227
226.5
226
2
4
6
8
10
12
a [m]
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13-38
ε2
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
Q& 21
[W]
189.6
202.6
209.7
214.3
217.5
219.8
221.5
222.9
224.1
225
225.8
226.4
227
227.5
227.9
228.3
228.7
2000
1800
1600
1400
Q21 [W]
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
Q& 21
[W]
227.9
340.9
453.3
565
676
786.4
896.2
1005
1114
1222
1329
1436
1542
1648
1753
1857
1961
1200
1000
800
600
400
200
0.1
0.2
0.3
0.4
0.5
ε1
0.6
0.7
0.8
0.9
230
225
220
215
Q21 [W]
ε1
210
205
200
195
190
185
0.1
0.2
0.3
0.4
0.5
ε2
0.6
0.7
0.8
0.9
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13-39
13-50 A circular grill is considered. The bottom of the grill is covered with hot coal bricks, while the wire mesh on top of the
grill is covered with steaks. The initial rate of radiation heat transfer from coal bricks to the steaks is to be determined for
two cases.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque,
diffuse, and gray. 3 Convection heat transfer is not considered.
Steaks, T2 = 278 K, ε2 = 1
Properties The emissivities are ε = 1 for all surfaces since they are black or
reradiating.
Analysis We consider the coal bricks to be surface 1, the steaks to be surface
2 and the side surfaces to be surface 3. First we determine the view factor
between the bricks and the steaks (Table 13-1),
Ri = R j =
S = 1+
F12
ri 0.15 m
=
= 0.75
L 0.20 m
1+ R j 2
Ri 2
=
1 + 0.75 2
0.75 2
0.20 m
Coal bricks, T1 = 950 K, ε1 = 1
= 3.7778
1/ 2 ⎫
1/ 2 ⎫
⎧
2
⎧
2⎤
⎡
⎡
⎛ Rj ⎞ ⎤ ⎪ 1 ⎪
1⎪
0
.
75
⎛
⎞
⎪
2
2
⎟ ⎥ ⎬ = ⎨3.7778 − ⎢3.7778 − 4⎜
= Fij = ⎨S − ⎢ S − 4⎜⎜
⎟ ⎥ ⎬ = 0.2864
⎟ ⎥
⎢
2⎪
2
0
.
75
R
⎝
⎠
⎢
⎥
⎪
⎝ i ⎠ ⎦ ⎪
⎣
⎦ ⎪⎭
⎣
⎩
⎭
⎩
(It can also be determined from Fig. 13-7).
Then the initial rate of radiation heat transfer from the coal bricks to the stakes becomes
Q& 12 = F12 A1σ (T1 4 − T2 4 )
= (0.2864)[π (0.3 m) 2 / 4](5.67 × 10 −8 W/m 2 ⋅ K 4 )[(950 K ) 4 − (278 K ) 4 ]
= 928 W
When the side opening is closed with aluminum foil, the entire heat lost by the coal bricks must be gained by the stakes since
there will be no heat transfer through a reradiating surface. The grill can be considered to be three-surface enclosure. Then
the rate of heat loss from the coal bricks can be determined from
Q& 1 =
⎛ 1
1
⎜
⎜R + R +R
13
23
⎝ 12
⎞
⎟
⎟
⎠
−1
E b1 = σT1 4 = (5.67 × 10 −8 W/m 2 .K 4 )(950 K ) 4 = 46,183 W/m 2
where
and
E b1 − E b 2
E b 2 = σT2 4 = (5.67 × 10 −8 W/m 2 .K 4 )(5 + 273 K ) 4 = 339 W/m 2
A1 = A2 =
π (0.3 m) 2
4
= 0.07069 m 2
1
1
=
= 49.39 m -2
A1 F12 (0.07069 m 2 )(0.2864)
1
1
= R 23 =
=
= 19.82 m - 2
A1 F13 (0.07069 m 2 )(1 − 0.2864)
R12 =
R13
Substituting,
Q& 12 =
(46,183 − 339) W/m 2
⎞
⎛
1
1
⎜
⎟
+
⎜ 49.39 m -2 2(19.82 m -2 ) ⎟
⎝
⎠
−1
= 2085 W
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13-40
13-51 The base and the dome of a hemispherical furnace are maintained at uniform temperatures. The net rate of radiation
heat transfer from the dome to the base surface 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.
T2 = 1000 K
ε2 = 1
T1 = 400 K
ε1 = 0.7
Analysis The view factor is first determined from
F11 = 0 (flat surface)
F11 + F12 = 1 → F12 = 1 (summation rule)
Noting that the dome is black, net rate of radiation heat transfer
from dome to the base surface can be determined from
D=5m
Q& 21 = −Q& 12 = −εA1 F12σ (T1 4 − T2 4 )
= −(0.7)[π (5 m) 2 /4 ](1)(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(400 K ) 4 − (1000 K ) 4 ]
= 7.594 × 10 5 W
= 759 kW
The positive sign indicates that the net heat transfer is from the dome to the base surface, as expected.
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13-41
13-52 Two perpendicular rectangular surfaces with a common edge are maintained at specified temperatures. The net rate of
radiation heat transfers between the two surfaces and between the horizontal surface and the surroundings are 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 emissivities of the horizontal rectangle and the surroundings are ε = 0.75 and ε = 0.85, respectively.
Analysis We consider the horizontal rectangle to be surface 1, the vertical rectangle to be surface 2 and the surroundings to
be surface 3. This system can be considered to be a three-surface enclosure. The view factor from surface 1 to surface 2 is
determined from
L1 0.8
⎫
=
= 0 .5 ⎪
⎪
W 1.6
⎬ F12 = 0.27 (Fig. 13-6)
L 2 1 .2
=
= 0.75⎪
⎪⎭
W 1 .6
T2 = 700 K
ε2 = 1
W = 1.6 m
L2 = 1.2 m
A2
The surface areas are
L1 = 0.8 m
A1 = (0.8 m)(1.6 m) = 1.28 m 2
A2 = (1.2 m)(1.6 m) = 1.92 m 2
A1
(2)
(3)
T3 = 290 K
ε3 = 0.85
(1)
T1 =450 K
ε1 =0.75
1.2 × 0.8
+ 0.8 2 + 1.2 2 × 1.6 = 3.268 m 2
2
Note that the surface area of the surroundings is determined assuming that surroundings forms flat surfaces at all openings to
form an enclosure. Then other view factors are determined to be
A3 = 2 ×
A1F12 = A2 F21 ⎯
⎯→(1.28)(0.27) = (1.92) F21 ⎯
⎯→ F21 = 0.18
(reciprocity rule)
F11 + F12 + F13 = 1 ⎯
⎯→ 0 + 0.27 + F13 = 1 ⎯
⎯→ F13 = 0.73
(summation rule)
F21 + F22 + F23 = 1 ⎯
⎯→ 0.18 + 0 + F23 = 1 ⎯
⎯→ F23 = 0.82
(summation rule)
A1 F13 = A3 F31 ⎯
⎯→(1.28)(0.73) = (3.268) F31 ⎯
⎯→ F31 = 0.29
(reciprocity rule)
A2 F23 = A3 F32 ⎯
⎯→(1.92)(0.82) = (3.268) F32 ⎯
⎯→ F32 = 0.48
(reciprocity rule)
We now apply Eq. 13-35b to each surface to determine the radiosities.
σT1 4 = J 1 +
Surface 1:
(5.67 × 10 −8 W/m 2 .K 4 )(450 K ) 4 = J 1 +
1 − ε1
ε1
[F12 ( J 1 − J 2 ) + F13 ( J 1 − J 3 )]
1 − 0.75
[0.27( J 1 − J 2 ) + 0.73( J 1 − J 3 )]
0.75
σT2 4 = J 2 ⎯
⎯→(5.67 × 10 −8 W/m 2 .K 4 )(700 K ) 4 = J 2
Surface 2:
σT3 4 = J 3 +
Surface 3:
(5.67 × 10 −8 W/m 2 .K 4 )(290 K ) 4 = J 3 +
1− ε3
ε3
[F31 ( J 3 − J 1 ) + F32 ( J 3 − J 2 )]
1 − 0.85
[0.29( J 3 − J 1 ) + 0.48( J 3 − J 2 )]
0.85
Solving the above equations, we find
J 1 = 2937 W/m 2 , J 2 = 13,614 W/m 2 , J 3 = 1501 W/m 2
Then the net rate of radiation heat transfers between the two surfaces and between the horizontal surface and the
surroundings are determined to be
Q& = −Q& = − A F ( J − J ) = −(1.28 m 2 )(0.27)(2937 − 13,614) W/m 2 = 3690 W
21
12
1 12
1
2
Q&13 = A1 F13 ( J 1 − J 3 ) = (1.28 m 2 )(0.73)(2937 − 1501) W/m 2 = 1342 W
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13-42
13-53 Two long parallel cylinders are maintained at specified temperatures. The rates of radiation heat transfer between the
cylinders and between the hot cylinder and the surroundings are to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are black. 3 Convection heat transfer is not considered.
Analysis We consider the hot cylinder to be surface 1, cold cylinder to be surface 2, and the surroundings to be surface 3.
Using the crossed-strings method, the view factor between two cylinders facing each other is determined to be
F1− 2 =
=
∑ Crossed strings − ∑ Uncrossed strings
2 × String on surface 1
2 s + D − 2s
2(πD / 2)
2
2
T2 = 275 K
ε2 = 1
D
(3)
T3 = 300 K
ε3 = 1
or
F1− 2
2⎛⎜ s 2 + D 2 − s ⎞⎟
⎝
⎠
=
πD
2⎛⎜ 0.3 2 + 0.20 2 − 0.5 ⎞⎟
⎠
⎝
=
π (0.20)
= 0.444
D
(2)
s
(1)
T1 = 425 K
ε1 = 1
The view factor between the hot cylinder and the surroundings is
F13 = 1 − F12 = 1 − 0.444 = 0.556 (summation rule)
The rate of radiation heat transfer between the cylinders per meter length is
A = πDL / 2 = π (0.20 m)(1 m) / 2 = 0.3142 m 2
Q& 12 = AF12σ (T1 4 − T2 4 )
= (0.3142 m 2 )(0.444)(5.67 × 10 −8 W/m 2 .°C)(425 4 − 275 4 )K 4
= 212.8 W
Note that half of the surface area of the cylinder is used, which is the only area that faces the other cylinder. The rate of
radiation heat transfer between the hot cylinder and the surroundings per meter length of the cylinder is
A1 = πDL = π (0.20 m)(1 m) = 0.6283 m 2
Q& 13 = A1 F13σ (T1 4 − T3 4 )
= (0.6283 m 2 )(0.556)(5.67 × 10 −8 W/m 2 .°C)(425 4 − 300 4 )K 4
= 485.8 W
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13-43
13-54 A long semi-cylindrical duct with specified temperature on the side surface is considered. The temperature of the base
surface for a specified heat transfer rate 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 the side surface is ε = 0.4.
Analysis We consider the base surface to be surface 1, the side surface to be surface 2. This system is a two-surface
enclosure, and we consider a unit length of the duct. The surface areas and the view factor are determined as
A1 = (1.0 m)(1.0 m) = 1.0 m 2
A2 = πDL / 2 = π (1.0 m)(1 m) / 2 = 1.571 m 2
T2 = 650 K
ε2 = 0.4
F11 + F12 = 1 ⎯
⎯→ 0 + F12 = 1 ⎯
⎯→ F12 = 1 (summation rule)
T1 = ?
ε1 = 1
The temperature of the base surface is determined from
σ (T1 4 − T2 4 )
Q& 12 =
1− ε 2
1
+
A1 F12 A2 ε 2
D=1m
(5.67 × 10 −8 W/m 2 ⋅ K 4 )[T1 4 − (650 K) 4 ]
1
1 − 0.4
+
(1.0 m 2 )(1) (1.571 m 2 )(0.4)
T1 = 684.8 K
1200 W =
13-55 A hemisphere with specified base and dome temperatures and heat transfer rate is considered. The emissivity of the
dome 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 the base surface is ε = 0.55.
Analysis We consider the base surface to be surface 1, the dome surface to be surface 2. This system is a two-surface
enclosure. The surface areas and the view factor are determined as
A1 = πD 2 / 4 = π (0.3 m) 2 / 4 = 0.07069 m 2
A2 = πD 2 / 2 = π (0.3 m) 2 / 2 = 0.1414 m 2
F11 + F12 = 1 ⎯
⎯→ 0 + F12 = 1 ⎯
⎯→ F12 = 1 (summation rule)
The emissivity of the dome is determined from
Q& 21 = −Q& 12 = −
65 W =
σ (T1 4 − T2 4 )
T2 = 600 K
ε2 = ?
T1 = 400 K
ε1 = 0.55
D = 0.3 m
1 − ε1
1− ε2
1
+
+
A1ε 1
A1 F12
A2 ε 2
(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(400 K )4 − (600 K )4 ]
⎯
⎯→ ε 2 = 0.0981
1− ε2
1 − 0.55
1
+
+
(0.07069 m 2 )(0.55) (0.07069 m 2 )(1) (0.1414 m 2 )ε 2
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13-44
13-56 A hot cylindrical surface is placed coaxially with a disk at a distance L apart. The radiation heat transfer rate from the
cylindrical surface to the disk is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces
are black. 3 Convection heat transfer is not considered. 4 Outer
surface of the cylinder is well insulated.
Analysis The end surfaces A3 and A4 are treated as hypothetical
surfaces. Applying the summation rule, we have
→
F23 = F21 + F24
F21 = F23 − F24
(1)
From reciprocity relation, we have
A2 F21 = A1 F12
(2)
Substituting Eq. (2) in to Eq. (1), we get
→
( A1 / A2 ) F12 = F23 − F24
F12 = ( A2 / A1 )( F23 − F24 )
(3)
The view factors F23 and F24 can be determined by treating them as view factors for coaxial parallel disks using Table 13-1:
With
R2 =
r2 D / 2
=
= 0 .5
L
L
R3 = R2 =
and
r3 D / 2
=
= 0.5
L
L
We get
S = 1+
F23
With
1 + R32
R22
= 1+
1 + (0.5) 2
(0.5) 2
=6
1/ 2 ⎫
⎧
2
⎡
⎛ D3 ⎞ ⎤ ⎪ 1
1⎪
2
⎟⎟ ⎥ ⎬ = [6 − (6 2 − 4 )1 / 2 ] = 0.1716
= ⎨S − ⎢ S − 4 ⎜⎜
D
2⎪
⎢
⎝ 2 ⎠ ⎥⎦ ⎪ 2
⎣
⎩
⎭
R2 =
D/2
= 0.25
2L
and
R2 = R4 = 0.25
We get
S = 1+
F24
1 + R42
R22
= 1+
1 + (0.25) 2
(0.25) 2
= 18
1/ 2 ⎫
⎧
2
⎡
⎛ D4 ⎞ ⎤ ⎪ 1
1⎪
2
⎟⎟ ⎥ ⎬ = [18 − (18 2 − 4 )1 / 2 ] = 0.05573
= ⎨S − ⎢ S − 4 ⎜⎜
D
2⎪
⎢
⎝ 2 ⎠ ⎥⎦ ⎪ 2
⎣
⎭
⎩
Substituting the values of F23 and F24 into Eq. (3), we have
F12 = ( A2 / A1 )( F23 − F24 ) = ( A2 / A1 )(0.1716 − 0.05573) = 0.1159( A2 / A1 )
The rate of heat transfer by radiation is then
Q&12 = A1 F12σ (T14 − T24 )
= 0.1159 A1 ( A2 / A1 )σ (T14 − T24 )
= 0.1159 A2σ (T14 − T24 )
= 0.1159(πD 2 / 4)σ (T14 − T24 )
= 0.1159
π (0.2 m) 2
4
(5.67 × 10 −8 W/m 2 ⋅ K 4 )(1000 4 − 300 4 ) K 4
= 205 W
Discussion The view factors F23 and F24 can also be determined using Fig. 13-7.
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preparation. If you are a student using this Manual, you are using it without permission.
13-45
13-57 Two parallel black disks are positioned coaxially, where the lower disk is heated electrically. The temperature of the
upper disk is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are black. 3 Convection heat transfer is not considered.
Analysis For coaxial parallel disks, from Table 13-1, with i = 1, j = 2,
R1 =
D1 / 2 0.2 / 2
=
= 0.4
L
0.25
S = 1+
1 + R22
R12
1 + (0.8) 2
= 1+
(0.4) 2
⎧
⎡
⎛D
1⎪
F12 = ⎨S − ⎢ S 2 − 4 ⎜⎜ 2
2⎪
⎢
⎝ D1
⎣
⎩
R2 =
and
⎞
⎟⎟
⎠
2⎤
⎥
⎥
⎦
D2 / 2 0.4 / 2
=
= 0.8
L
0.25
= 11.25
1/ 2 ⎫
1/ 2 ⎫
⎧
2
⎡
⎪ 1⎪
⎛ 0.4 ⎞ ⎤ ⎪
2
⎟ ⎥ ⎬ = 0.3676
⎬ = ⎨11.25 − ⎢(11.25) − 4 ⎜
⎝ 0.2 ⎠ ⎥⎦ ⎪
⎢⎣
⎪ 2⎪
⎭
⎩
⎭
Then, using the summation rule,
F12 + F1surr = 1
→
F1surr = 1 − F12 = 0.6324
The net radiation heat transfer rate leaving the lower surface can be expressed as
4
Q& elec = Q&12 + Q&1surr = A1 F12σ (T14 − T24 ) + A1 F1surrσ (T14 − Tsurr
)
4
Q& elec = A1σ [ F12 (T14 − T24 ) + F1surr (T14 − Tsurr
)]
Hence
⎡
Q& elec
F
4 ⎤
)⎥
T2 = ⎢T14 −
+ 1surr (T14 − Tsurr
A1 F12σ
F12
⎦⎥
⎣⎢
1/ 4
1/ 4
⎡
⎤
4Q& elec
F1surr 4
4
= ⎢T14 −
+
−
(
)
T
T
1
surr ⎥
F12
πD12 F12σ
⎣⎢
⎦⎥
⎡
4(100 W )
T2 = ⎢(500 K ) 4 −
2
π (0.2 m) (0.3676)(5.67 × 10 −8 W/m 2 ⋅ K 4 )
⎣⎢
+
0.6324
⎤
(500 4 − 300 4 ) K 4 ⎥
0.3676
⎦
1/ 4
T2 = 241 K
Discussion The view factor F12 can also be determined using Fig. 13-7 to be
F12 ≈ 0.36
with
L / r1 = 2.5
and
r2 / L = 0.8
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preparation. If you are a student using this Manual, you are using it without permission.
13-46
13-58 Two phase gas-liquid oxygen is stored in a spherical tank this is enclosed by a concentric spherical surface at 273 K.
The heat transfer rate at the spherical tank surface is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are opaque, diffuse, and gray. 3 Heat transfer by radiation
only.
Properties The emissivity of the two surfaces is given as ε1 = ε2 = 0.01.
The normal boiling point of oxygen is −183°C (Table A-2).
Analysis For concentric spheres, the rate of radiation heat transfer at
the inner surface is (from Table 13-3),
Q&12 =
A1σ (T14 − T24 )
1 − ε 2 ⎛ D1
⎜
+
ε1
ε 2 ⎜⎝ D2
1
⎞
⎟⎟
⎠
2
Note that the spherical tank surface has the same temperature as the
oxygen at normal boiling point, T1 = −183°C = 90 K. Hence,
π (1 m) 2 (5.67 × 10 −8 W/m 2 ⋅ K 4 )(90 4 − 273 4 ) K 4
Q&12 =
= −7.05 W
2
1
1 − 0.01 ⎛ 1 ⎞
+
⎜
⎟
0.01
0.01 ⎝ 1.6 ⎠
Discussion The negative value of Q& 12 indicates that heat is being added to the oxygen. As long as the oxygen is maintained
in the two-phase gas-liquid state, its temperature will remain constant at the normal boiling point.
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13-47
Radiation Shields and the Radiation Effect
13-59C A person who feels fine in a room at a specified temperature may feel chilly in another room at the same temperature
as a result of radiation effect if the walls of second room are at a considerably lower temperature. For example most people
feel comfortable in a room at 22°C if the walls of the room are also roughly at that temperature. When the wall temperature
drops to 5°C for some reason, the interior temperature of the room must be raised to at least 27°C to maintain the same level
of comfort. Also, people sitting near the windows of a room in winter will feel colder because of the radiation exchange
between the person and the cold windows.
13-60C The influence of radiation on heat transfer or temperature of a surface is called the radiation effect. The radiation
exchange between the sensor and the surroundings may cause the thermometer to indicate a different reading for the medium
temperature. To minimize the radiation effect, the sensor should be coated with a material of high reflectivity (low
emissivity).
13-61C Radiation heat transfer between two surfaces can be reduced greatly by inserting a thin, high reflectivity(low
emissivity) sheet of material between the two surfaces. Such highly reflective thin plates or shells are known as radiation
shields. Multilayer radiation shields constructed of about 20 shields per cm. thickness separated by evacuated space are
commonly used in cryogenic and space applications to minimize heat transfer. Radiation shields are also used in temperature
measurements of fluids to reduce the error caused by the radiation effect.
PROPRIETARY MATERIAL. © 2011 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.
13-48
13-62 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.
Properties The emissivities of surfaces are given to be ε1 = 0.5, ε2 = 0.5, and ε3 = 0.1.
Analysis The net rate of radiation heat transfer between the plates in the case of no shield is
σ (T1 4 − T2 4 )
Q& 12,no shield =
⎞
⎛ 1
1
⎜⎜ +
− 1⎟⎟
ε
ε
2
⎠
⎝ 1
=
T1 = 1100 K
ε1 = 0.5
(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(1100 K ) 4 − (700 K ) 4 ]
1
⎛ 1
⎞
+
− 1⎟
⎜
0
.
5
0
.
5
⎝
⎠
= 23,134 W/m 2
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
Q& 12,shields =
σ (T1 4 − T2 4 )
⎛ 1
⎞
⎛ 1
⎞
1
1
⎜⎜ +
− 1⎟⎟ + N shield ⎜
+
− 1⎟
⎜ε
⎟
⎝ ε1 ε 2
⎠
⎝ 3,1 ε 3, 2
⎠
T2 = 700 K
ε2 = 0.5
Radiation shields
ε3 = 0.1
(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(1100 K ) 4 − (700 K ) 4 ]
1
(23,134 W/m 2 ) =
1
1
5
⎞
⎛ 1
⎞
⎛ 1
+
− 1⎟ + N shield ⎜
+
− 1⎟
⎜
0
.
1
0
.
1
0
.
5
0
.
5
⎠
⎝
⎠
⎝
N shield = 1.48 ≅ 2
That is, only two sheets are more than enough to reduce heat transfer to one-fifth.
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preparation. If you are a student using this Manual, you are using it without permission.
13-49
13-63 Air is flowing between two infinitely large parallel plates. The convection heat transfer coefficient associated with the
air is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are opaque, diffuse, and gray. 3 The surface temperatures
are uniform.
Properties The emissivity of the upper plate is given as ε1 = 0.7. The lower plate surface is black, ε2 = 1.
Analysis For infinitely large parallel plates, the rate of radiation heat transfer is (from Table 13-3),
Aσ (T14 − T24 )
Q&12 =
1
1
+
−1
ε1
ε2
Applying energy balance on the lower plate, we have
Q& 12 = Q& conv
hA(T2 − T∞ ) =
Aσ (T14 − T24 )
1
1
+
−1
ε1
h=
σ
1
ε1
=
+
1
ε2
−1
ε2
(T14 − T24 )
(T2 − T∞ )
(5.67 × 10 −8 W/m 2 ⋅ K 4 ) (500 4 − 330 4 ) K 4
1
(330 − 290) K
+1−1
0.7
h = 50.3 W/m 2 ⋅ K
Discussion The calculated value of the convection heat transfer coefficient (h = 50.3 W/m2·K) is typical for forced
convection of gases (see Table 1-5)
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13-50
13-64 A thin aluminum sheet is placed between two very large parallel plates that are maintained at uniform temperatures.
The net rate of radiation heat transfer between the two plates is to be determined for the cases of with and without the shield.
Assumptions 1 Steady operating conditions exist 2 The surfaces are
opaque, diffuse, and gray. 3 Convection heat transfer is not
considered.
Properties The emissivities of surfaces are given to be ε1 = 0.5,
ε2 = 0.8, and ε3 = 0.15.
Analysis The net rate of radiation heat transfer with a thin aluminum
shield per unit area of the plates is
Q& 12,one shield =
=
σ (T1 4 − T2 4 )
⎞
⎞ ⎛ 1
⎛ 1
1
1
⎜⎜ +
+
− 1⎟
− 1⎟⎟ + ⎜
⎟
⎜
⎠ ⎝ ε 3,1 ε 3, 2
⎝ ε1 ε 2
⎠
T1 = 900 K
ε1 = 0.5
T2 = 650 K
ε2 = 0.8
Radiation
shield
0 15
(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(900 K ) 4 − (650 K ) 4 ]
1
1
⎞
⎞ ⎛ 1
⎛ 1
+
− 1⎟
+
− 1⎟ + ⎜
⎜
⎝ 0.5 0.8 ⎠ ⎝ 0.15 0.15 ⎠
= 1857 W/m 2
The net rate of radiation heat transfer between the plates in the case of no shield is
σ (T14 − T2 4 ) (5.67 × 10 −8 W/m 2 ⋅ K 4 )[(900 K ) 4 − (650 K ) 4 ]
= 12,035 W/m 2
Q&12, no shield =
=
1
⎞
⎛1
⎞
⎛ 1
1
+
− 1⎟
⎜
⎜⎜ +
− 1⎟⎟
⎝ 0.5 0.8 ⎠
⎝ ε1 ε 2
⎠
Then the ratio of radiation heat transfer for the two cases becomes
Q& 12,one shield
1
1857 W
=
≅
&
Q12,no shield 12,035 W 6
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preparation. If you are a student using this Manual, you are using it without permission.
13-51
Prob. 13-64 is reconsidered. The net rate of radiation heat transfer between the two plates as a function of the
13-65
emissivity of the aluminum sheet is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
epsilon_3=0.15
T_1=900 [K]
T_2=650 [K]
epsilon_1=0.5
epsilon_2=0.8
"ANALYSIS"
sigma=5.67E-8 [W/m^2-K^4]
Q_dot_12_1shield=(sigma*(T_1^4-T_2^4))/((1/epsilon_1+1/epsilon_2-1)+(1/epsilon_3+1/epsilon_3-1))
0.05
0.06
0.07
0.08
0.09
0.1
0.11
0.12
0.13
0.14
0.15
0.16
0.17
0.18
0.19
0.2
0.21
0.22
0.23
0.24
0.25
[W/m2]
656.5
783
908.1
1032
1154
1274
1394
1511
1628
1743
1857
1969
2081
2191
2299
2407
2513
2619
2723
2826
2928
3000
2500
2
Q& 12,1 shield
Q12;1shield [W/m ]
ε3
2000
1500
1000
500
0,05
0,1
0,15
0,2
0,25
ε3
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
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13-52
13-66 The rate of heat loss from a person by radiation in a large room whose walls are maintained at a uniform temperature
is to be determined for two cases.
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 the person is given to be ε1 = 0.85.
Analysis (a) Noting that the view factor from the person to the walls
F12 = 1 , the rate of heat loss from that person to the walls at a large
room which are at a temperature of 295 K is
ROOM
T2
T1 = 30°C
ε1 = 0.85
A = 1.9 m2
Qrad
Q& 12 = ε 1 F12 A1σ (T1 4 − T2 4 )
= (0.85)(1)(1.9 m 2 )(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(303 K ) 4 − (295 K ) 4 ]
= 78.3 W
(b) When the walls are at a temperature of 260 K,
Q& 12 = ε 1 F12 A1σ (T1 4 − T2 4 )
= (0.85)(1)(1.9 m 2 )(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(303 K ) 4 − (260 K ) 4 ]
= 353 W
13-67 Five identical thin aluminum sheets are placed between two very large parallel plates which are maintained at uniform
temperatures. The net rate of radiation heat transfer between the two plates 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.
T1 = 800 K
ε1 = 0.1
Properties The emissivities of surfaces are given to be ε1 = ε2 = 0.1
and ε3 = 0.1.
Analysis Since the plates and the sheets have the same emissivity
value, the net rate of radiation heat transfer with 5 thin aluminum
shield can be determined from
Q& 12,5 shield =
=
4
4
1 &
1 σ (T1 − T2 )
Q12, no shield =
N +1
N +1 ⎛ 1
⎞
1
⎜⎜ +
− 1⎟⎟
⎠
⎝ ε1 ε 2
1 (5.67 × 10
5 +1
−8
W/m ⋅ K )[(800 K ) − (450 K ) ]
= 183 W/m 2
1
1
⎛
⎞
+
− 1⎟
⎜
⎝ 0.1 0.1 ⎠
2
4
4
4
T2 = 450 K
ε2 = 0.1
Radiation shields
ε3 = 0.1
The net rate of radiation heat transfer without the shield is
Q& 12,5 shield =
1 &
Q12, no shield ⎯
⎯→ Q& 12, no shield = ( N + 1)Q& 12,5 shield = 6 × 183 W = 1098 W
N +1
PROPRIETARY MATERIAL. © 2011 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.
13-53
Prob. 13-67 is reconsidered. The effects of the number of the aluminum sheets and the emissivities of the
13-68
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
epsilon_3=0.1
epsilon_1=0.1
epsilon_2=epsilon_1
T_1=800 [K]
T_2=450 [K]
"ANALYSIS"
sigma=5.67E-8 [W/m^2-K^4] “Stefan-Boltzmann constant"
Q_dot_12_shields=1/(N+1)*Q_dot_12_NoShield
Q_dot_12_NoShield=(sigma*(T_1^4-T_2^4))/(1/epsilon_1+1/epsilon_2-1)
600
Q&12,shields
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
[W/m2]
183.3
282.4
387
497.6
614.7
738.9
870.8
1011
1161
1321
1493
1677
1876
2090
2322
2575
2850
400
300
200
100
0
1
2
3
4
5
6
7
8
0,5
0,6
0,7
9
10
N
3000
2500
Q12;shields [W/m ]
ε1
500
2
[W/m2]
550
366.7
275
220
183.3
157.1
137.5
122.2
110
100
Q12;shields [W/m ]
1
2
3
4
5
6
7
8
9
10
Q&12,shields
2
N
2000
1500
1000
500
0
0,1
0,2
0,3
0,4
ε1
0,8
0,9
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13-54
13-69 Liquid nitrogen is stored in a spherical tank this is enclosed by a concentric spherical surface at 273 K. The rate of
vaporization for the liquid nitrogen is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are opaque, diffuse, and gray. 3 Heat transfer by radiation
only.
Properties The emissivity of the two surfaces is given
as ε1 = ε2 = 0.01. The latent heat of vaporization for
nitrogen is hfg = 198.6 kJ/kg (Table A-2).
Analysis For concentric spheres, the rate of radiation heat
transfer at the inner surface is (from Table 13-3),
Q&12 =
A1σ (T14 − T24 )
1 − ε 2 ⎛ D1
⎜
+
ε1
ε 2 ⎜⎝ D2
1
⎞
⎟⎟
⎠
2
Hence,
π (1 m) 2 (5.67 × 10 −8 W/m 2 ⋅ K 4 )(80 4 − 273 4 ) K 4
Q&12 =
= −7.082 W
2
1
1 − 0.01 ⎛ 1 ⎞
+
⎜
⎟
0.01
0.01 ⎝ 1.6 ⎠
The rate of vaporization can be determined using
− Q& 12 = m& h fg
m& = −
Q&12
− 7.082 W
=−
h fg
198.6 × 10 3 J/kg
m& = 3.57 × 10 −5 kg/s
Discussion The rate of vaporization can be reduced by placing a radiation shield midway between the inner and outer
spherical surfaces.
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preparation. If you are a student using this Manual, you are using it without permission.
13-55
13-70E A radiation shield is placed between two parallel disks which are maintained at uniform temperatures. The net rate of
radiation heat transfer through the shields is to be determined.
T1 = 1350 R, ε1 = 1
Assumptions 1 Steady operating conditions exist 2 The
surfaces are black. 3 Convection heat transfer is not
considered.
Properties The emissivities of surfaces are given to be
ε1 = ε2 = 1 and ε3 = 0.15.
Analysis From Fig. 13-7 we have F32 = F13 = 0.52 . Then
F34 = 1 − 0.52 = 0.48 . The disk in the middle is surrounded by
black surfaces on both sides. Therefore, heat transfer between
the top surface of the middle disk and its black surroundings
can expressed as
T∞ = 540 R
ε3 = 1
1 ft
ε3 = 0.15
1 ft
T2 = 650 R, ε2 = 1
Q& 3 = εA3σ [ F31 (T34 − T14 )] + εA3σ [ F32 (T34 − T24 )]
= 0.15(7.069 ft 2 )(0.1714 × 10 −8 Btu/h.ft 2 ⋅ R 4 ){0.52[(T34 − (1350 R ) 4 ] + 0.48[T34 − (540 R ) 4 ]}
where A3 = π (3 ft ) 2 / 4 = 7.069 ft 2 . Similarly, for the bottom surface of the middle disk, we have
− Q& 3 = εA3σ [ F32 (T24 − T44 )] + εA3σ [ F34 (T34 − T44 )]
= 0.15(7.069 ft 2 )(0.1714 × 10 −8 Btu/h.ft 2 ⋅ R 4 ){0.48[(T34 − (650 R ) 4 ] + 0.52[T34 − (540 R ) 4 ]}
Combining the equations above, the rate of heat transfer between the disks through the radiation shield (the middle disk) is
determined to be
Q& = 1490 Btu/h
and
T3 = 987 R
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13-56
13-71 A radiation shield is placed between two large parallel plates which are maintained at uniform temperatures. The
emissivity of the radiation shield is to be determined if the radiation heat transfer between the plates is reduced to 15% of that
without the radiation shield.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is
not considered.
Properties The emissivities of surfaces are given to be ε1 = 0.6 and ε2 = 0.9.
Analysis First, the net rate of radiation heat transfer between the two large parallel plates per unit area without a shield is
σ (T14 − T24 ) (5.67 × 10−8 W/m2 ⋅ K 4 )[(650 K ) 4 − (400 K )4 ]
Q&12, no shield =
=
= 4877 W/m2
1
1
1 1
+
−1
+
−1
ε1 ε 2
0.6 0.9
The radiation heat transfer in the case of one shield is
Q&12, one shield = 0.15 × Q&12, no shield
= 0.15 × 4877 W/m 2 = 731.6 W/m 2
T1 = 650 K
ε1 = 0.6
Then the emissivity of the radiation shield becomes
Q&12,one shield =
731.6 W/m 2 =
σ (T14 − T2 4 )
T2 = 400 K
ε2 = 0.9
Radiation
shield
⎞
⎞ ⎛ 1
⎛1 1
1
⎟
⎜ + − 1⎟ + ⎜
⎟ ⎜ ε + ε − 1⎟
⎜ε ε
2
3, 2
⎠ ⎝ 3,1
⎝ 1
⎠
(5.67 × 10−8 W/m 2 ⋅ K 4 )[(650 K ) 4 − (400 K ) 4 ]
⎞
1
⎛ 1
⎞ ⎛ 2
+
− 1⎟ + ⎜⎜ − 1⎟⎟
⎜
⎝ 0.6 0.9 ⎠ ⎝ ε 3 ⎠
which gives ε 3 = 0.18
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13-57
Prob. 13-71 is reconsidered. The effect of the percent reduction in the net rate of radiation heat transfer
13-72
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"
T_1=650 [K]
T_2=400 [K]
epsilon_1=0.6
epsilon_2=0.9
PercentReduction=85 “[%]”
"ANALYSIS"
sigma=5.67E-8 [W/m^2-K^4] “Stefan-Boltzmann constant"
Q_dot_12_NoShield=(sigma*(T_1^4-T_2^4))/(1/epsilon_1+1/epsilon_2-1)
Q_dot_12_1shield=(sigma*(T_1^4-T_2^4))/((1/epsilon_1+1/epsilon_2-1)+(1/epsilon_3+1/epsilon_3-1))
Q_dot_12_1shield=(1-PercentReduction/100)*Q_dot_12_NoShield
ε3
0.9153
0.8148
0.72
0.6304
0.5455
0.4649
0.3885
0.3158
0.2466
0.1806
0.1176
0.05751
1
0.8
0.6
ε3
Percent Reduction
[%]
40
45
50
55
60
65
70
75
80
85
90
95
0.4
0.2
0
40
50
60
70
80
90
100
PercentReduction [%]
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13-58
13-73 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.
Properties The emissivities of surfaces are given to be
ε1 = 0.7, ε2 = 0.4. and ε3 = 0.2.
D2 = 0.5 m
T2 = 500 K
ε2 = 0.4
D1 = 0.1 m
T1 = 750 K
ε1 = 0.7
Analysis The surface areas of the cylinders and the
shield per unit length are
Apipe,inner = A1 = πD1 L = π (0.1 m)(1 m) = 0.314 m 2
Apipe,outer = A2 = πD 2 L = π (0.5 m)(1 m) = 1.571 m 2
Ashield = A3 = πD3 L = π (0.2 m)(1 m) = 0.628 m 2
The net rate of radiation heat transfer between the two
cylinders with a shield per unit length is
Q& 12,one shield =
σ (T1 4 − T2 4 )
1 − ε 3,1 1 − ε 3,2
1 − ε1
1−ε2
1
1
+
+
+
+
+
A1ε 1
A1 F13
A3ε 3,1
A3ε 3, 2
A3 F3, 2
A2 ε 2
Radiation shield
D3 = 0.2 m
ε3 = 0.2
(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(750 K ) 4 − (500 K ) 4 ]
1 − 0.7
1
1 − 0.2
1
1 − 0.4
+
+2
+
+
(0.314)(0.7) (0.314)(1)
(0.628)(0.2) (0.628)(1) (1.571)(0.4)
= 726 W
=
If there was no shield,
Q&12,no shield =
σ (T1 4 − T2 4 )
1 1 − ε 2 ⎛ D1
⎜
+
ε1
ε 2 ⎜⎝ D2
⎞
⎟⎟
⎠
=
(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(750 K ) 4 − (500 K ) 4 ]
= 8329 W
1 1 − 0.4 ⎛ 0.1 ⎞
+
⎟
⎜
0.7
0.4 ⎝ 0.5 ⎠
Then their ratio becomes
Q& 12,one shield
726 W
=
= 0.0872
&
8329
W
Q12,no shield
PROPRIETARY MATERIAL. © 2011 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.
13-59
Prob. 13-73 is reconsidered. The effects of the diameter of the outer cylinder and the emissivity of the
13-74
radiation shield on the net rate of radiation heat transfer between the two cylinders are to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
D_1=0.10 [m]
D_2=0.50 [m]
D_3=0.20 [m]
epsilon_1=0.7
epsilon_2=0.4
epsilon_3=0.2
T_1=750 [K]
T_2=500 [K]
"ANALYSIS"
sigma=5.67E-8 [W/m^2-K^4] “Stefan-Boltzmann constant"
L=1 [m] “a unit length of the cylinders is considered"
A_1=pi*D_1*L
A_2=pi*D_2*L
A_3=pi*D_3*L
F_13=1
F_32=1
Q_dot_12_1shield=(sigma*(T_1^4-T_2^4))/((1-epsilon_1)/(A_1*epsilon_1)+1/(A_1*F_13)+(1epsilon_3)/(A_3*epsilon_3)+(1-epsilon_3)/(A_3*epsilon_3)+1/(A_3*F_32)+(1-epsilon_2)/(A_2*epsilon_2))
0.25
0.275
0.3
0.325
0.35
0.375
0.4
0.425
0.45
0.475
0.5
Q& 12,1 shield
730
[W]
692.8
698.6
703.5
707.8
711.4
714.7
717.5
720
722.3
724.3
726.1
725
720
Q12,1shield [W]
D2
[m]
715
710
705
700
695
690
0.25
0.3
0.35
0.4
0.45
0.5
D2 [m]
PROPRIETARY MATERIAL. © 2011 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.
13-60
ε3
Q& 12,1 shield
1100
[W]
213.1
291.5
366.5
438.3
507
572.9
636
696.7
755
811.1
865
917
967.1
1015
1062
1107
1000
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
Q12,1shield [W]
900
800
700
600
500
400
300
200
0.05
0.1
0.15
0.2
ε3
0.25
0.3
0.35
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13-61
13-75 A long cylindrical black surface fuel rod is shielded by a concentric surface that has a uniform temperature. The
surface temperature of the fuel rod is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The fuel rod surface is black. 3 The shield is opaque, diffuse, and gray. 4
The fuel rod and shield form an infinitely long concentric cylinder.
Properties The emissivity of the shield is given as ε2 = 0.05. The fuel rod surface is black, ε1 = 1.
Analysis For infinitely long concentric cylinder, the rate of radiation heat transfer at the fuel rod surface is (from Table 13-3),
Q&12 =
A1σ (T14 − T24 )
1
ε1
+
1 − ε 2 ⎛ D1
⎜
ε 2 ⎜⎝ D2
⎞
⎟⎟
⎠
=
A1σ (T14 − T24 )
= 0.09524 A1σ (T14 − T24 )
1 1 − 0.05 ⎛ 25 ⎞
+
⎜ ⎟
1
0.05 ⎝ 50 ⎠
Applying energy balance on the shield, we have the following expression:
4
Q&12 = Q& conv + Q& rad = hA2 (T2 − T∞ ) + ε 2 A2σ (T24 − Tsurr
)
Hence
T14 =
4
h(T2 − T∞ ) + ε 2σ (T24 − Tsurr
) ⎛ A2
⎜⎜
0.09524σ
⎝ A1
⎞
⎟⎟ + T24
⎠
=
4
h(T2 − T∞ ) + ε 2σ (T24 − Tsurr
) ⎛ D2
⎜⎜
0.09524σ
⎝ D1
⎞
⎟⎟ + T24
⎠
or
4
⎡ h(T − T∞ ) + ε 2σ (T24 − Tsurr
) ⎛ D2
⎜⎜
T1 = ⎢ 2
0.09524σ
⎝ D1
⎣⎢
⎤
⎞
⎟⎟ + T24 ⎥
⎠
⎦⎥
1/ 4
⎡ (15)(320 − 300) + (0.05)(5.67 × 10 −8 )(320 4 − 300 4 ) ⎛ 50 ⎞ 4
⎤
4
T1 = ⎢
⎜ ⎟ K + (320 K ) ⎥
−8
0.09524(5.67 × 10 )
⎝ 25 ⎠
⎢⎣
⎥⎦
1/ 4
T1 = 594 K
Discussion The use of absolute temperatures is necessary for calculations involving radiation heat transfer.
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13-62
Radiation Exchange with Absorbing and Emitting Gases
13-76C 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.
13-77C Gases emit and absorb radiation at a number of narrow wavelength bands. The emissivity-wavelength charts of gases
typically involve various peaks and dips together with discontinuities, and show clearly the band nature of absorption and the
strong nongray characteristics. This is in contrast to solids, which emit and absorb radiation over the entire spectrum.
13-78C Using Kirchhoff’s law, the spectral emissivity of a medium of thickness L in terms of the spectral absorption
coefficient is expressed as ε λ = α λ = 1 − e −κ λ L .
13-79C Spectral transmissivity of a medium of thickness L is the ratio of the intensity of radiation leaving the medium to that
I λ ,L
= e −κ λ L and τλ =1 - αλ .
entering the medium, and is expressed as τ λ =
I λ ,0
13-80 An equimolar mixture of CO2 and O2 gases at 800 K and a total pressure of 0.5 atm is considered. The emissivity of
the gas is to be determined.
Assumptions All the gases in the mixture are ideal gases.
Analysis Volumetric fractions are equal to pressure fractions. Therefore, the partial pressure of CO2 is
Pc = y CO 2 P = 0.5(0.5 atm) = 0.25 atm
Then,
Pc L = (0.25 atm)(1.2 m) = 0.30 m ⋅ atm = 0.98 ft ⋅ atm
The emissivity of CO2 corresponding to this value at the gas temperature of Tg = 800 K and 1 atm is, from Fig. 13-36,
ε c , 1 atm = 0.15
This is the base emissivity value at 1 atm, and it needs to be corrected for the 0.5 atm total pressure. The pressure correction
factor is, from Fig. 13-37,
Cc = 0.90
Then the effective emissivity of the gas becomes
ε g = C c ε c, 1 atm = 0.90 × 0.15 = 0.135
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13-63
13-81 The temperature, pressure, and composition of combustion gases flowing inside long tubes are given. The rate of heat
transfer from combustion gases to tube wall is to be determined.
Assumptions All the gases in the mixture are ideal gases.
Analysis The mean beam length for an infinite cicrcular cylinder is, from
Ts = 500 K
Table 13-4,
L = 0.95(0.10 m) = 0.095 m
D = 10 cm
Then,
Pc L = (0.12 atm)(0.095 m) = 0.0114 m ⋅ atm = 0.037 ft ⋅ atm
Pw L = (0.18 atm)(0.095 m) = 0.0171 m ⋅ atm = 0.056 ft ⋅ atm
Combustion
gases, 1 atm
Tg = 1000 K
The emissivities of CO2 and H2O corresponding to these values at the gas
temperature of Tg = 1000 K and 1atm are, from Fig. 13-36,
ε c, 1 atm = 0.055 and ε w, 1 atm = 0.039
Both CO2 and H2O are present in the same mixture, and we need to correct for the overlap of emission bands. The emissivity
correction factor at T = Tg = 1000 K is, from Fig. 13-38,
Pc L + Pw L = 0.037 + 0.056 = 0.093
Pw
0.18
=
= 0. 6
Pw + Pc 0.18 + 0.12
⎫
⎪
⎬ ∆ε = 0.0
⎪⎭
Then the effective emissivity of the combustion gases becomes
ε g = C c ε c , 1 atm + C w ε w, 1 atm − ∆ε = 1 × 0.055 + 1 × 0.039 − 0.0 = 0.094
Note that the pressure correction factor is 1 for both gases since the total pressure is 1 atm. For a source temperature of Ts =
500 K, the absorptivity of the gas is again determined using the emissivity charts as follows:
Pc L
Ts
500 K
= (0.12 atm)(0.095 m)
= 0.007125 m ⋅ atm = 0.023 ft ⋅ atm
800 K
Tg
Pw L
Ts
500 K
= (0.18 atm)(0.095 m)
= 0.01069 m ⋅ atm = 0.035 ft ⋅ atm
800 K
Tg
The emissivities of CO2 and H2O corresponding to these values at a temperature of Ts = 500 K and 1atm are, from Fig. 1336,
ε c, 1 atm = 0.042 and ε w, 1 atm = 0.050
Then the absorptivities of CO2 and H2O become
⎛ Tg
α c = C c ⎜⎜
⎝ Ts
αw
⎛ Tg
= C w ⎜⎜
⎝ Ts
⎞
⎟
⎟
⎠
0.65
⎞
⎟
⎟
⎠
⎛ 800 K ⎞
⎟
⎝ 500 K ⎠
0.65
ε c , 1 atm = (1)⎜
0.45
⎛ 800 K ⎞
⎟
⎝ 500 K ⎠
ε w, 1 atm = (1)⎜
(0.042) = 0.057
0.45
(0.050) = 0.062
Also ∆α = ∆ε, but the emissivity correction factor is to be evaluated from Fig. 13-38 at T = Ts = 500 K instead of Tg = 1000
K. There is no chart for 500 K in the figure, but we can read ∆ε values at 400 K and 800 K, and interpolate. At Pw/(Pw+ Pc) =
0.6 and PcL +PwL = 0.093 we read ∆ε = 0.0. Then the absorptivity of the combustion gases becomes
α g = α c + α w − ∆α = 0.057 + 0.062 − 0.0 = 0.119
The surface area of the pipe is
As = πDL = π (0.10 m)(6 m) = 1.885 m 2
Then the net rate of radiation heat transfer from the combustion gases to the walls of the tube becomes
Q& = A σ (ε T 4 − α T 4 )
net
s
g g
g s
= (1.885 m )(5.67 × 10 −8 W/m 2 ⋅ K 4 )[0.094(1000 K ) 4 − 0.119(500 K ) 4 ]
= 9252 W
2
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13-64
13-82 A mixture of CO2 and N2 gases at 600 K and a total pressure of 1 atm are contained in a cylindrical container. The rate
of radiation heat transfer between the gas and the container walls is to be determined.
Assumptions All the gases in the mixture are ideal gases.
Analysis The mean beam length is, from Table 13-4
8m
L = 0.60D = 0.60(8 m) = 4.8 m
Then,
Pc L = (0.15 atm)(4.8 m) = 0.72 m ⋅ atm = 2.36 ft ⋅ atm
8m
Tg = 600 K
Ts = 450 K
The emissivity of CO2 corresponding to this value at the gas
temperature of Tg = 600 K and 1 atm is, from Fig. 13-36,
ε c , 1 atm = 0.16
For a source temperature of Ts = 450 K, the absorptivity of the gas is again determined using the emissivity charts as follows:
Pc L
Ts
450 K
= (0.15 atm)(4.8 m)
= 0.54 m ⋅ atm = 1.77 ft ⋅ atm
Tg
600 K
The emissivity of CO2 corresponding to this value at a temperature of Ts = 450 K and 1atm are, from Fig. 13-36,
ε c , 1 atm = 0.14
The absorptivity of CO2 is determined from
⎛ Tg ⎞
⎟
⎟
⎝ Ts ⎠
0.65
α c = C c ⎜⎜
⎛ 600 K ⎞
⎟
⎝ 450 K ⎠
ε c, 1 atm = (1)⎜
0.65
(0.14) = 0.17
The surface area of the cylindrical surface is
As = πDH + 2
πD 2
4
= π (8 m)(8 m) + 2
π (8 m) 2
4
= 301.6 m 2
Then the net rate of radiation heat transfer from the gas mixture to the walls of the furnace becomes
Q& net = As σ (ε g T g4 − α g Ts4 )
= (301.6 m 2 )(5.67 × 10 −8 W/m 2 ⋅ K 4 )[0.16(600 K ) 4 − 0.17(450 K ) 4 ]
= 2.35 × 10 5 W
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13-65
13-83 A mixture of H2O and N2 gases at 600 K and a total pressure of 1 atm are contained in a cylindrical container. The rate
of radiation heat transfer between the gas and the container walls is to be determined.
Assumptions All the gases in the mixture are ideal gases.
8m
Analysis The mean beam length is, from Table 13-4
L = 0.60D = 0.60(8 m) = 4.8 m
Then,
8m
Pw L = (0.15 atm)(4.8 m) = 0.72 m ⋅ atm = 2.36 ft ⋅ atm
Tg = 600 K
Ts = 450 K
The emissivity of H2O corresponding to this value at the gas temperature of Tg = 600 K
and 1 atm is, from Fig. 13-36,
ε w 1 atm = 0.36
For a source temperature of Ts = 450 K, the absorptivity of the gas is again determined using the emissivity charts as follows:
Pw L
Ts
450 K
= (0.15 atm)(4.8 m)
= 0.54 m ⋅ atm = 1.77 ft ⋅ atm
Tg
600 K
The emissivity of H2O corresponding to this value at a temperature of Ts = 450 K and 1atm are, from Fig. 13-36,
ε w, 1 atm = 0.34
The absorptivity of H2O is determined from
⎛ Tg ⎞
⎟
⎟
⎝ Ts ⎠
0.65
α w = C w ⎜⎜
⎛ 600 K ⎞
⎟
⎝ 450 K ⎠
0.45
ε w, 1 atm = (1)⎜
(0.34) = 0.39
The surface area of the cylindrical surface is
As = πDH + 2
πD 2
4
= π (8 m)(8 m) + 2
π (8 m) 2
4
= 301.6 m 2
Then the net rate of radiation heat transfer from the gas mixture to the walls of the furnace becomes
Q& net = As σ (ε g T g4 − α g Ts4 )
= (301.6 m 2 )(5.67 × 10 −8 W/m 2 ⋅ K 4 )[0.36(600 K ) 4 − 0.39(450 K ) 4 ]
= 5.244 × 10 5 W
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13-66
13-84 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
rate of radiation heat transfer between the gas mixture and furnace walls is to be determined.
Assumptions All the gases in the mixture are ideal gases.
Analysis The mean beam length is, from Table 13-4
L = 0.65D = 0.65(5 m) =3.25 m
The mole fraction is equal to pressure fraction. Then,
Pc L = (0.15 atm)(3.25 m) = 0.4875 m ⋅ atm = 1.60 ft ⋅ atm
The emissivity of CO2 corresponding to this value at the gas temperature of Tg = 1200 K
and 1 atm is, from Fig. 13-36,
5m
Tg = 1200 K
Ts = 600 K
ε c , 1 atm = 0.17
For a source temperature of Ts = 600 K, the absorptivity of the gas is again determined using the emissivity charts as follows:
Pc L
Ts
600 K
= (0.15 atm)(3.25 m)
= 0.244 m ⋅ atm = 0.800 ft ⋅ atm
Tg
1200 K
The emissivity of CO2 corresponding to this value at a temperature of Ts = 600 K and 1atm are, from Fig. 13-36,
ε c , 1 atm = 0.12
The absorptivity of CO2 is determined from
⎛ Tg ⎞
⎟
⎟
⎝ Ts ⎠
α c = C c ⎜⎜
0.65
⎛ 1200 K ⎞
⎟
⎝ 600 K ⎠
ε c , 1 atm = (1)⎜
0.65
(0.12) = 0.1883
The surface area of the sphere is
As = πD 2 = π (5 m) 2 = 78.54 m 2
Then the net rate of radiation heat transfer from the gas mixture to the walls of the furnace becomes
Q& net = As σ (ε g T g4 − α g Ts4 )
= (78.54 m 2 )(5.67 × 10 −8 W/m 2 ⋅ K 4 )[0.17(1200 K ) 4 − 0.1883(600 K ) 4 ]
= 1.46 × 10 6 W
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13-67
13-85 The temperature, pressure, and composition of combustion gases flowing inside long tubes are given. The rate of heat
transfer from combustion gases to tube wall is to be determined.
Assumptions All the gases in the mixture are ideal gases.
Analysis The volumetric analysis of a gas mixture gives the mole fractions yi of the components, which are equivalent to
pressure fractions for an ideal gas mixture. Therefore, the partial pressures of CO2 and H2O are
Pc = y CO 2 P = 0.06(1 atm) = 0.06 atm
Ts = 600 K
P =y
P = 0.09(1 atm) = 0.09 atm
w
H 2O
The mean beam length for an infinite cicrcular cylinder is, from
Table 13-4,
L = 0.95(0.15 m) = 0.1425 m
Then,
Pc L = (0.06 atm)(0.1425 m) = 0.00855 m ⋅ atm = 0.028 ft ⋅ atm
D = 15 cm
Combustion
gases, 1 atm
Tg = 1500 K
Pw L = (0.09 atm)(0.1425 m) = 0.0128 m ⋅ atm = 0.042 ft ⋅ atm
The emissivities of CO2 and H2O corresponding to these values at the gas temperature of Tg = 1500 K and 1atm are, from
Fig. 13-36,
ε c, 1 atm = 0.034 and ε w, 1 atm = 0.016
Both CO2 and H2O are present in the same mixture, and we need to correct for the overlap of emission bands. The emissivity
correction factor at T = Tg = 1500 K is, from Fig. 13-38,
Pc L + Pw L = 0.028 + 0.042 = 0.07 ⎫
⎪
Pw
0.09
⎬ ∆ε = 0.0
=
= 0.6
⎪⎭
Pw + Pc 0.09 + 0.06
Then the effective emissivity of the combustion gases becomes
ε g = C c ε c, 1 atm + C w ε w, 1 atm − ∆ε = 1× 0.034 + 1× 0.016 − 0.0 = 0.05
Note that the pressure correction factor is 1 for both gases since the total pressure is 1 atm. For a source temperature of Ts =
600 K, the absorptivity of the gas is again determined using the emissivity charts as follows:
T
600 K
= 0.00342 m ⋅ atm = 0.011 ft ⋅ atm
Pc L s = (0.06 atm)(0.1425 m)
1500 K
Tg
Pw L
Ts
600 K
= (0.09 atm)(0.1425 m)
= 0.00513 m ⋅ atm = 0.017 ft ⋅ atm
1500 K
Tg
The emissivities of CO2 and H2O corresponding to these values at a temperature of Ts = 600 K and 1atm are, from Fig. 1336,
ε c , 1 atm = 0.031 and ε w, 1 atm = 0.027
Then the absorptivities of CO2 and H2O become
⎛ Tg ⎞
⎟
⎟
⎝ Ts ⎠
0.65
α c = C c ⎜⎜
⎛ 1500 K ⎞
⎟
⎝ 600 K ⎠
0.65
ε c, 1 atm = (1)⎜
(0.031) = 0.056
0.45
0.45
⎛ Tg ⎞
⎛ 1500 K ⎞
⎟
⎜
α w = C w ⎜ ⎟ ε w, 1 atm = (1)⎜
⎟ (0.027) = 0.041
⎝ 600 K ⎠
⎝ Ts ⎠
Also ∆α = ∆ε, but the emissivity correction factor is to be evaluated from Fig. 13-38 at T = Ts = 600 K instead of Tg = 1500
K. There is no chart for 600 K in the figure, but we can read ∆ε values at 400 K and 800 K, and take their average. At
Pw/(Pw+ Pc) = 0.6 and PcL +PwL = 0.07 we read ∆ε = 0.0. Then the absorptivity of the combustion gases becomes
α g = α c + α w − ∆α = 0.056 + 0.041 − 0.0 = 0.097
The surface area of the pipe per m length of tube is
As = πDL = π (0.15 m)(1 m) = 0.4712 m 2
Then the net rate of radiation heat transfer from the combustion gases to the walls of the furnace becomes
Q& = A σ (ε T 4 − α T 4 )
net
s
g g
g s
= (0.4712 m )(5.67 ×10 −8 W/m 2 ⋅ K 4 )[0.05(1500 K ) 4 − 0.097(600 K ) 4 ] = 6427 W
2
PROPRIETARY MATERIAL. © 2011 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.
13-68
13-86 The temperature, pressure, and composition of combustion gases flowing inside long tubes are given. The rate of heat
transfer from combustion gases to tube wall is to be determined.
Assumptions All the gases in the mixture are ideal gases.
Analysis The volumetric analysis of a gas mixture gives the mole fractions yi of the components, which are equivalent to
pressure fractions for an ideal gas mixture. Therefore, the partial pressures of CO2 and H2O are
Pc = y CO 2 P = 0.06(1 atm) = 0.06 atm
Ts = 600 K
Pw = y H O P = 0.09(1 atm) = 0.09 atm
2
The mean beam length for an infinite cicrcular cylinder is, from
Table 13-4,
L = 0.95(0.15 m) = 0.1425 m
Then,
Pc L = (0.06 atm)(0.1425 m) = 0.00855 m ⋅ atm = 0.028 ft ⋅ atm
D = 15 cm
Combustion
gases, 3 atm
Tg = 1500 K
Pw L = (0.09 atm)(0.1425 m) = 0.0128 m ⋅ atm = 0.042 ft ⋅ atm
The emissivities of CO2 and H2O corresponding to these values at the gas temperature of Tg = 1500 K and 1atm are, from
Fig. 13-36,
ε c, 1 atm = 0.034 and ε w, 1 atm = 0.016
These are base emissivity values at 1 atm, and they need to be corrected for the 3 atm total pressure. Noting that (Pw+P)/2 =
(0.09+3)/2 = 1.545 atm, the pressure correction factors are, from Fig. 13-37,
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 emission bands. The emissivity
correction factor at T = Tg = 1500 K is, from Fig. 13-38,
Pc L + Pw L = 0.028 + 0.042 = 0.07 ⎫
⎪
Pw
0.09
⎬ ∆ε = 0.0
=
= 0.6
⎪⎭
Pw + Pc 0.09 + 0.06
Then the effective emissivity of the combustion gases becomes
ε g = C c ε c, 1 atm + C w ε w, 1 atm − ∆ε = 1.5 × 0.034 + 1.8 × 0.016 − 0.0 = 0.080
For a source temperature of Ts = 600 K, the absorptivity of the gas is again determined using the emissivity charts as follows:
T
600 K
= 0.00342 m ⋅ atm = 0.011 ft ⋅ atm
Pc L s = (0.06 atm)(0.1425 m)
1500 K
Tg
Pw L
Ts
600 K
= (0.09 atm)(0.1425 m)
= 0.00513 m ⋅ atm = 0.017 ft ⋅ atm
1500 K
Tg
The emissivities of CO2 and H2O corresponding to these values at a temperature of Ts = 600 K and 1atm are, from Fig. 1336,
ε c , 1 atm = 0.031 and ε w, 1 atm = 0.027
Then the absorptivities of CO2 and H2O become
⎛ Tg ⎞
⎟
⎟
⎝ Ts ⎠
α c = C c ⎜⎜
0.65
⎛ 1500 K ⎞
⎟
⎝ 600 K ⎠
ε c , 1 atm = (1.5)⎜
0.65
(0.031) = 0.084
0.45
0.45
⎛ Tg ⎞
⎛ 1500 K ⎞
⎜
⎟
α w = C w ⎜ ⎟ ε w, 1 atm = (1.8)⎜
⎟ (0.027) = 0.073
⎝ 600 K ⎠
⎝ Ts ⎠
Also ∆α = ∆ε, but the emissivity correction factor is to be evaluated from Fig. 13-38 at T = Ts = 600 K instead of Tg = 1500
K. There is no chart for 600 K in the figure, but we can read ∆ε values at 400 K and 800 K, and take their average. At
Pw/(Pw+ Pc) = 0.6 and PcL +PwL = 0.07 we read ∆ε = 0.0. Then the absorptivity of the combustion gases becomes
α g = α c + α w − ∆α = 0.084 + 0.073 − 0.0 = 0.157
The surface area of the pipe per m length of tube is
As = πDL = π (0.15 m)(1 m) = 0.4712 m 2
Then the net rate of radiation heat transfer from the combustion gases to the walls of the furnace becomes
Q& net = As σ (ε g T g4 − α g Ts4 )
= (0.4712 m 2 )(5.67 ×10 −8 W/m 2 ⋅ K 4 )[0.08(1500 K ) 4 − 0.157(600 K ) 4 ] = 10,280 W
PROPRIETARY MATERIAL. © 2011 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.
13-69
13-87 The temperature, pressure, and composition of a gas mixture is given. The emissivity of the mixture is to be
determined.
Assumptions 1 All the gases in the mixture are ideal gases. 2 The emissivity determined is the mean emissivity for radiation
emitted to all surfaces of the cubical enclosure.
Analysis The volumetric analysis of a gas mixture gives the mole fractions yi of the components, which are equivalent to
pressure fractions for an ideal gas mixture. Therefore, the partial pressures of CO2 and H2O are
Pc = y CO 2 P = 0.10(1 atm) = 0.10 atm
Pw = y H 2O P = 0.09(1 atm) = 0.09 atm
6m
The mean beam length for a cube of side length 6 m for radiation
emitted to all surfaces is, from Table 13-4,
Combustion
gases
1200 K
L = 0.66(6 m) = 3.96 m
Then,
Pc L = (0.10 atm)(3.96 m) = 0.396 m ⋅ atm = 1.30 ft ⋅ atm
Pw L = (0.09 atm)(3.96 m) = 0.36 m ⋅ atm = 1.18 ft ⋅ atm
The emissivities of CO2 and H2O corresponding to these values at the gas temperature of Tg = 1000 K and 1atm are, from
Fig. 13-36,
ε c , 1 atm = 0.17 and
ε w, 1 atm = 0.16
Both CO2 and H2O are present in the same mixture, and we need to correct for the overlap of emission bands. The emissivity
correction factor at T = Tg = 1200 K is, from Fig. 13-38,
Pc L + Pw L = 1.30 + 1.18 = 2.48
Pw
0.09
=
= 0.474
Pw + Pc 0.09 + 0.10
⎫
⎪
⎬ ∆ε = 0.049
⎪⎭
Then the effective emissivity of the combustion gases becomes
ε g = C c ε c , 1 atm + C w ε w, 1 atm − ∆ε = 1 × 0.17 + 1 × 0.16 − 0.049 = 0.281
Note that the pressure correction factor is 1 for both gases since the total pressure is 1 atm.
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13-70
13-88 The temperature, pressure, and composition of combustion gases flowing inside long tubes are given. The rate of heat
transfer from combustion gases to tube wall is to be determined.
Assumptions All the gases in the mixture are ideal gases.
Analysis The volumetric analysis of a gas mixture gives the mole fractions yi of the components, which are equivalent to
pressure fractions for an ideal gas mixture. Therefore, the partial pressures of CO2 and H2O are
Pc = y CO 2 P = 0.10(1 atm) = 0.10 atm
Ts = 600 K
Pw = y H 2O P = 0.10(1 atm) = 0.10 atm
The mean beam length for this geometry is, from Table 13-4,
20 cm
L = 3.6V/As = 1.8D = 1.8(0.20 m) = 0.36 m
where D is the distance between the plates. Then,
Combustion
gases, 1 atm
Tg = 1200 K
Pc L = Pw L = (0.10 atm)(0.36 m) = 0.036 m ⋅ atm = 0.118 ft ⋅ atm
The emissivities of CO2 and H2O corresponding to these values at the gas temperature of Tg = 1200 K and 1atm are, from
Fig. 13-36,
ε c, 1 atm = 0.080 and
ε w, 1 atm = 0.055
Both CO2 and H2O are present in the same mixture, and we need to correct for the overlap of emission bands. The emissivity
correction factor at T = Tg = 1200 K is, from Fig. 13-38,
Pc L + Pw L = 0.118 + 0.118 = 0.236
Pw
0.10
=
= 0. 5
Pw + Pc 0.10 + 0.10
⎫
⎪
⎬ ∆ε = 0.0025
⎪⎭
Then the effective emissivity of the combustion gases becomes
ε g = C c ε c, 1 atm + C w ε w, 1 atm − ∆ε = 1× 0.080 + 1× 0.055 − 0.0025 = 0.1325
Note that the pressure correction factor is 1 for both gases since the total pressure is 1 atm. For a source temperature of Ts =
600 K, the absorptivity of the gas is again determined using the emissivity charts as follows:
Pc L
Ts
T
600 K
= Pw L s = (0.10 atm)(0.36 m)
= 0.018 m ⋅ atm = 0.059 ft ⋅ atm
Tg
Tg
1200 K
The emissivities of CO2 and H2O corresponding to these values at a temperature of Ts = 600 K and 1atm are, from Fig. 1336,
ε c, 1 atm = 0.060 and
ε w, 1 atm = 0.067
Then the absorptivities of CO2 and H2O become
⎛ Tg ⎞
⎟
⎟
⎝ Ts ⎠
0.65
α c = C c ⎜⎜
αw
⎛ Tg
= C w ⎜⎜
⎝ Ts
⎞
⎟
⎟
⎠
⎛ 1200 K ⎞
⎟
⎝ 600 K ⎠
0.65
ε c, 1 atm = (1)⎜
0.45
⎛ 1200 K ⎞
⎟
⎝ 600 K ⎠
ε w, 1 atm = (1)⎜
(0.060) = 0.090
0.45
(0.067) = 0.092
Also ∆α = ∆ε, but the emissivity correction factor is to be evaluated from Fig. 13-38 at T = Ts = 600 K instead of Tg = 1200
K. There is no chart for 600 K in the figure, but we can read ∆ε values at 400 K and 800 K, and take their average. At
Pw/(Pw+ Pc) = 0.5 and PcL +PwL = 0.236 we read ∆ε = 0.00125. Then the absorptivity of the combustion gases becomes
α g = α c + α w − ∆α = 0.090 + 0.092 − 0.00125 = 0.1808
Then the net rate of radiation heat transfer from the gas to each plate per unit surface area becomes
Q& net = As σ (ε g T g4 − α g Ts4 )
= (1 m 2 )(5.67 × 10 −8 W/m 2 ⋅ K 4 )[0.1325(1200 K ) 4 − 0.1808(600 K ) 4 ]
= 1.42 × 10 4 W
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13-71
Special Topic: Heat Transfer from the Human Body
13-89C (a) Heat is lost through the skin by convection, radiation, and evaporation. (b) The body loses both sensible heat by
convection and latent heat by evaporation from the lungs, but there is no heat transfer in the lungs by radiation.
13-90C Sensible heat is the energy associated with a temperature change. The sensible heat loss from a human body
increases as (a) the skin temperature increases, (b) the environment temperature decreases, and (c) the air motion (and thus
the convection heat transfer coefficient) increases.
13-91C Latent heat is the energy released as water vapor condenses on cold surfaces, or the energy absorbed from a warm
surface as liquid water evaporates. The latent heat loss from a human body increases as (a) the skin wettedness increases and
(b) the relative humidity of the environment decreases. The rate of evaporation from the body is related to the rate of latent
heat loss by Q& latent = m& vapor h fg where hfg is the latent heat of vaporization of water at the skin temperature.
13-92C The insulating effect of clothing is expressed in the unit clo with 1 clo = 0.155 m2.°C/W = 0.880 ft2.°F.h/Btu.
Clothing serves as insulation, and thus reduces heat loss from the body by convection, radiation, and evaporation by serving
as a resistance against heat flow and vapor flow. Clothing decreases heat gain from the sun by serving as a radiation shield.
13-93C Yes, roughly one-third of the metabolic heat generated by a person who is resting or doing light work is dissipated to
the environment by convection, one-third by evaporation, and the remaining one-third by radiation.
13-94C The operative temperature Toperative is the average of the mean radiant and ambient temperatures weighed by their
respective convection and radiation heat transfer coefficients, and is expressed as
Toperative =
hconv Tambient + hrad Tsurr Tambient + Tsurr
≅
hconv + hrad
2
When the convection and radiation heat transfer coefficients are equal to each other, the operative temperature becomes the
arithmetic average of the ambient and surrounding surface temperatures. Another environmental index used in thermal
comfort analysis is the effective temperature, which combines the effects of temperature and humidity.
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13-72
13-95 The convection heat transfer coefficient for a clothed person while walking in still air at a velocity of 0.5 to 2 m/s is
given by h = 8.6V 0.53 where V is in m/s and h is in W/m2.°C. The convection coefficients in that range vary from 5.96
W/m2.°C at 0.5 m/s to 12.42 W/m2.°C at 2 m/s. Therefore, at low velocities, the radiation and convection heat transfer
coefficients are comparable in magnitude. But at high velocities, the convection coefficient is much larger than the radiation
heat transfer coefficient.
h = 8.6V0.53
W/m2.°C
5.96
7.38
8.60
9.68
10.66
11.57
12.42
1 3 .0
1 2 .0
1 1 .0
1 0 .0
9 .0
h
Velocity,
m/s
0.50
0.75
1.00
1.25
1.50
1.75
2.00
8 .0
7 .0
6 .0
5 .0
0 .4
0 .6
0 .8
1 .0
1 .2
V
1 .4
1 .6
1 .8
2 .0
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13-73
13-96 A man wearing summer clothes feels comfortable in a room at 20°C. The room temperature at which this man would
feel thermally comfortable when unclothed is to be determined.
Assumptions 1 Steady conditions exist. 2 The latent heat loss from the person remains the same. 3 The heat transfer
coefficients remain the same. 4 The air in the room is still (there are no winds or running fans). 5 The surface areas of the
clothed and unclothed person are the same.
Analysis At low air velocities, the convection heat transfer coefficient for a standing
man is given in Table 13-5 to be 4.0 W/m2.°C. The radiation heat transfer coefficient at
typical indoor conditions is 4.7 W/m2.°C. Therefore, the heat transfer coefficient for a
standing person for combined convection and radiation is
Troom= 20°C
Tskin= 33°C
hcombined = hconv + hrad = 4.0 + 4.7 = 8.7 W/m 2 .°C
The thermal resistance of the clothing is given to be
Rcloth = 1.1 clo = 1.1× 0.155 m 2 .°C/W = 0.171 m 2 .°C/W
Clothed
person
Noting that the surface area of an average man is 1.8 m2, the sensible heat loss from this
person when clothed is determined to be
− Tambient )
A (T
(1.8 m 2 )(33 − 20)°C
=
= 82 W
Q& sensible,clothed = s skin
1
1
2
0.171 m .°C/W +
Rcloth +
hcombined
8.7 W/m 2 .°C
From heat transfer point of view, taking the clothes off is equivalent to removing the clothing insulation or setting Rcloth = 0.
The heat transfer in this case can be expressed as
A (T
− Tambient ) (1.8 m 2 )(33 − Tambient )°C
Q& sensible,unclothed = s skin
=
1
1
hcombined
8.7 W/m 2 .°C
To maintain thermal comfort after taking the clothes off, the skin temperature of the person and the rate of heat transfer from
him must remain the same. Then setting the equation above equal to 82 W gives
Tambient = 27.8°C
Therefore, the air temperature needs to be raised from 22 to 27.8°C to ensure that the person will feel comfortable in the
room after he takes his clothes off. Note that the effect of clothing on latent heat is assumed to be negligible in the solution
above. We also assumed the surface area of the clothed and unclothed person to be the same for simplicity, and these two
effects should counteract each other.
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13-74
13-97E An average person produces 0.50 lbm of moisture while taking a shower.
The contribution of showers of a family of four to the latent heat load of the airconditioner per day is to be determined.
Moisture
0.5 lbm
Assumptions All the water vapor from the shower is condensed by the airconditioning system.
Properties The latent heat of vaporization of water is given to be 1050 Btu/lbm.
Analysis The amount of moisture produced per day is
mvapor = (Moisture produced per person)(No. of persons)
= (0.5 lbm/person)(4 persons/day) = 2 lbm/day
Then the latent heat load due to showers becomes
Qlatent = mvaporhfg = (2 lbm/day)(1050 Btu/lbm) = 2100 Btu/day
13-98 A car mechanic is working in a shop heated by radiant heaters in winter. The lowest ambient temperature the worker
can work in comfortably is to be determined.
Assumptions 1 The air motion in the room is negligible, and the mechanic is standing. 2 The average clothing and exposed
skin temperature of the mechanic is 33°C.
Properties The emissivity and absorptivity of the person is given to be 0.95. The convection heat transfer coefficient from a
standing body in still air or air moving with a velocity under 0.2 m/s is hconv = 4.0 W/m2⋅°C (Table 13-5).
Analysis The equivalent thermal resistance of clothing is
Rcloth = 0.7 clo = 0.7 × 0.155 m 2 .°C/W = 0.1085 m 2 .°C/W
Radiation from the heaters incident on the person and the rate of
sensible heat generation by the person are
Radiant
heater
Q& rad, incident = 0.05 × Q& rad, total = 0.05(4 kW) = 0.2 kW = 200 W
Q& gen, sensible = 0.5 × Q& gen, total = 0.5(350 W) = 175 W
Under steady conditions, and energy balance on the body can be
expressed as
E& in − E& out + E& gen = 0
Q& rad from heater − Q& conv + rad from body + Q& gen, sensible = 0
or
4
αQ& rad, incident − hconv As (Ts − Tsurr ) − εAs σ (Ts4 − Tsurr
) + Q& gen, sensible = 0
0.95(200 W) − (4.0 W/m 2 ⋅ K)(1.8 m 2 )(306 − Tsurr )
4
) + 175 W = 0
− 0.95(1.8 m 2 )(5.67 × 10 -8 W/m 2 ⋅ K 4 )[(306 K) 4 − Tsurr
Solving the equation above gives
T surr = 284.8 K = 11.8°C
Therefore, the mechanic can work comfortably at temperatures as low as 12°C.
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13-75
13-99 Chilled air is to cool a room by removing the heat generated in a large insulated classroom by lights and students. The
required flow rate of air that needs to be supplied to the room is to be determined.
Assumptions 1 The moisture produced by the bodies leave the room as vapor without any condensing, and thus the
classroom has no latent heat load. 2 Heat gain through the walls and the roof is negligible.
Properties The specific heat of air at room temperature is 1.00 kJ/kg⋅°C (Table A-15). The average rate of metabolic heat
generation by a person sitting or doing light work is 115 W (70 W sensible, and 45 W latent).
Analysis The rate of sensible heat generation by the people in the
room and the total rate of sensible internal heat generation are
Return
air
Q& gen, sensible = q& gen, sensible (No. of people)
= (70 W/person)(75 persons) = 5250 W
&
Qtotal, sensible = Q& gen, sensible + Q& lighting
= 5250 + 2000 = 7250 W
Then the required mass flow rate of chilled air becomes
m& air =
=
Chilled
air
15°C
Lights
2 kW
25°C
75 Students
Q& total, sensible
c p ∆T
7.25 kJ/s
= 0.725 kg/s
(1.0 kJ/kg ⋅ °C)(25 − 15)°C
Discussion The latent heat will be removed by the air-conditioning system as the moisture condenses outside the cooling
coils.
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13-100 The average mean radiation temperature during a cold day drops to 18°C. The required rise in the indoor air
temperature to maintain the same level of comfort in the same clothing is to be determined.
Assumptions 1 Air motion in the room is negligible. 2 The average clothing and exposed skin temperature remains the same.
3 The latent heat loss from the body remains constant. 4 Heat transfer through the lungs remain constant.
Properties The emissivity of the person is 0.95 (from Appendix tables). The convection heat transfer coefficient from the
body in still air or air moving with a velocity under 0.2 m/s is hconv = 3.1 W/m2⋅°C (Table 13-5).
Analysis The total rate of heat transfer from the body is the sum of the rates of heat loss by convection, radiation, and
evaporation,
Q& body, total = Q& sensible + Q& latent + Q& lungs = (Q& conv + Qrad ) + Q& latent + Q& lungs
Noting that heat transfer from the skin by evaporation and from the lungs
remains constant, the sum of the convection and radiation heat transfer from
the person must remain constant.
22°C
22°C
4
Q& sensible,old = hAs (Ts − Tair, old ) + εAs σ (Ts4 − Tsurr,
old )
= hAs (Ts − 22) + 0.95 As σ [(Ts + 273) 4 − (22 + 273) 4 ]
4
Q& sensible,new = hAs (Ts − Tair, new ) + εAs σ (Ts4 − Tsurr,
new )
= hAs (Ts − Tair, new ) + 0.95 As σ [(Ts + 273) 4 − (18 + 273) 4 ]
Setting the two relations above equal to each other, canceling the surface
area As, and simplifying gives
− 22h − 0.95σ (22 + 273)4 = − hTair, new − 0.95σ (18 + 273) 4
3.1(Tair, new − 22) + 0.95 × 5.67 × 10−8 (2914 − 2954 ) = 0
Solving for the new air temperature gives
Tair, new = 29.0°C
Therefore, the air temperature must be raised to 29°C to counteract the increase in heat transfer by radiation.
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13-101 The average mean radiation temperature during a cold day drops to 10°C. The required rise in the indoor air
temperature to maintain the same level of comfort in the same clothing is to be determined.
Assumptions 1 Air motion in the room is negligible. 2 The average clothing and exposed skin temperature remains the same.
3 The latent heat loss from the body remains constant. 4 Heat transfer through the lungs remain constant.
Properties The emissivity of the person is 0.95 (from Appendix tables). The convection heat transfer coefficient from the
body in still air or air moving with a velocity under 0.2 m/s is hconv = 3.1 W/m2⋅°C (Table 13-5).
Analysis The total rate of heat transfer from the body is the sum of the rates of heat loss by convection, radiation, and
evaporation,
Q& body, total = Q& sensible + Q& latent + Q& lungs = (Q& conv + Q rad ) + Q& latent + Q& lungs
Noting that heat transfer from the skin by evaporation and from the lungs
remains constant, the sum of the convection and radiation heat transfer from the
person must remain constant.
22°C
22°C
4
Q& sensible,old = hAs (Ts − Tair, old ) + εAs σ (Ts4 − Tsurr,
old )
= hAs (Ts − 22) + 0.95 As σ [(Ts + 273) 4 − (22 + 273) 4 ]
4
Q& sensible,new = hAs (Ts − Tair, new ) + εAs σ (Ts4 − Tsurr,
new )
= hAs (Ts − Tair, new ) + 0.95 As σ [(Ts + 273) 4 − (10 + 273) 4 ]
Setting the two relations above equal to each other, canceling the surface
area As, and simplifying gives
− 22h − 0.95σ (22 + 273) 4 = − hTair, new − 0.95σ (10 + 273) 4
3.1(Tair, new − 22) + 0.95 × 5.67 × 10 −8 (283 4 − 295 4 ) = 0
Solving for the new air temperature gives
Tair, new = 42.1°C
Therefore, the air temperature must be raised to 42.11°C to counteract the increase in heat transfer by radiation.
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13-102 There are 100 chickens in a breeding room. The rate of total heat generation and the rate of moisture production in
the room are to be determined.
Assumptions All the moisture from the chickens is condensed by the air-conditioning system.
Properties The latent heat of vaporization of water is given to be 2430 kJ/kg. The average metabolic rate of chicken during
normal activity is 10.2 W (3.78 W sensible and 6.42 W latent).
Analysis The total rate of heat generation of the chickens in the breeding room is
Q& gen, total = q& gen, total (No. of chickens)
= (10.2 W/chicken)(100 chickens) = 1020 W
The latent heat generated by the chicken and the rate of moisture production
are
100
Chickens
10.2 W
Q& gen, latent = q& gen, latent (No. of chickens)
= (6.42 W/chicken)(100 chickens) = 642 W
= 0.642 kW
m& moisture =
Q& gen, latent
hfg
=
0.642 kJ/s
= 0.000264 kg/s = 0.264 g/s
2430 kJ/kg
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13-79
Review Problems
13-103 A cylindrical enclosure is considered. (a) The expression for the view from the side surface to itself F33 in terms of K
and (b) the value of the view factor F33 for L = D are to be determined.
Assumptions 1 The surfaces are diffuse emitters and reflectors.
Analysis (a) The surfaces are designated as follows: Base surface as A1, top surface as A2, and side surface as A3
Applying the summation rule to A1, we have
F11 + F12 + F13 = 1
(where F11 = 0 )
or
F13 = 1 − F12
(1)
For coaxial parallel disks, from Table 13-1, with i = 1, j = 2,
1/ 2 ⎫
⎧
2⎤
⎡
⎛
⎞
D
1⎪
⎪ 1
F12 = ⎨S − ⎢ S 2 − 4 ⎜⎜ 2 ⎟⎟ ⎥ ⎬ = [ S − ( S 2 − 4 )1 / 2 ]
(2)
2⎪
⎢
⎝ D1 ⎠ ⎥⎦ ⎪ 2
⎣
⎩
⎭
S = 1+
1 + R22
R12
= 2+
1
R
2
= 2+
4
( D / L) 2
= 2 + 4K 2
(3)
1
D
=
2L 2K
Substituting Eq. (3) into Eq. (2), we get
1
F12 = {2 + 4 K 2 − [(2 + 4 K 2 ) 2 − 4 ]1 / 2 }
2
1
= [2 + 4 K 2 − (16 K 4 + 16 K 2 )1 / 2 ]
2
1
= [2 + 4 K 2 − 4 K ( K 2 + 1)1 / 2 ]
2
= 1 + 2 K 2 − 2 K ( K 2 + 1)1 / 2
Substituting the above expression for F12 into Eq. (1) yields the expression for F13:
F13 = 1 − [1 + 2 K 2 − 2 K ( K 2 + 1)1 / 2 ] = 2 K ( K 2 + 1)1 / 2 − 2 K 2
(4)
Then, applying the summation rule to A3, we have
F31 + F32 + F33 = 1
(where F31 = F32 )
or
F33 = 1 − 2 F31
Applying reciprocity relation, we have
F33 = 1 − 2 F13 ( A1 / A3 )
where F31 = F13 ( A1 / A3 )
Also, we know that
A1
D
A1 = πD 2 / 4
→
A3 = πDL
=
and
A3 4 L
Hence, the expression for F33 becomes
1
D
F13 = 1 −
F33 = 1 − 2
F13
(5)
2K
4L
Finally, substituting Eq. (4) into Eq. (5) yields the expression for F33:
1
F33 = 1 −
[2 K ( K 2 + 1)1 / 2 − 2 K 2 ]
2K
Hence, F33 = 1 + K − ( K 2 + 1)1 / 2
(b) The value of the view factor F33 for L = D (i.e., K = 1) is
where
R1 = R2 = R =
F33 = 1 + K − ( K 2 + 1)1 / 2 = 1 + 1 − (12 + 1)1 / 2 = 2 − 2 = 0.589
Discussion If the cylinder has a length and diameter of L = 2D, then from the expression for F33 we have
F33 = 1 + 2 − (2 2 + 1)1 / 2 = 0.764
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13-80
13-104 A long cylindrical black surface fuel rod is shielded by a concentric surface that has a uniform temperature. The fuel
rod generates 0.5 MW/m3 of heat per unit length. The surface temperature of the fuel rod is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The fuel rod surface is black. 3 The shield is opaque, diffuse, and gray. 4
The fuel rod and shield formed an infinitely long concentric cylinder.
Properties The emissivity of the shield is given as ε2 = 0.05. The fuel rod surface is black, ε1 = 1.
Analysis The heat generation of the fuel rod is
e&gen =
E& gen
V
=
4 E& gen
πD12 L
πD12 L
E& gen = e&genV = e&gen
4
→
Hence, the total heat generation rate per unit length (L = 1 m)
by the fuel rod is
πD12 L
π (0.025 m) 2 (1 m)
E& gen = e&gen
= (0.5 × 10 6 W/m 3 )
= 245.4 W
4
4
For infinitely long concentric cylinder, the rate of radiation
heat transfer at the fuel rod surface is (from Table 13-3),
E& gen = Q& 12 =
A1σ (T14 − T24 )
1
ε1
+
⎧⎪⎡ 1 1 − ε 2
T1 = ⎨⎢ +
ε2
⎪⎩⎣⎢ ε 1
1 − ε 2 ⎛ D1
⎜
ε 2 ⎜⎝ D2
⎛ D1
⎜⎜
⎝ D2
⎞
⎟⎟
⎠
1/ 4
⎫⎪
⎞⎤ E& gen
⎟⎟⎥
+ T24 ⎬
⎪⎭
⎠⎦⎥ A1σ
⎧⎪⎡ 1 1 − ε 2
= ⎨⎢ +
ε2
⎪⎩⎣⎢ ε 1
⎛ D1
⎜⎜
⎝ D2
1/ 4
⎫⎪
⎞⎤ E& gen
⎟⎟⎥
+ T24 ⎬
⎪⎭
⎠⎦⎥ πD1 Lσ
1/ 4
⎧⎪⎡1 1 − 0.05 ⎛ 25 ⎞⎤
⎫⎪
245.4 W
= ⎨⎢ +
+ (320 K ) 4 ⎬
⎜ ⎟⎥
8
2
4
−
0.05 ⎝ 50 ⎠⎦ π (0.025 m)(1 m)(5.67 × 10 W/m ⋅ K )
⎪⎩⎣1
⎪⎭
T1 = 876 K
Discussion The use of absolute temperatures is necessary for calculations involving radiation heat transfer.
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13-81
13-105 The base and the dome of a long semi-cylindrical dryer are maintained at uniform temperatures. The drying rate per
unit length experienced by the base surface 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. 4 The dryer is
well insulated from heat loss to the surrounding.
Properties The latent heat of vaporization for water is
hfg = 2257 kJ/kg (Table A-2)
Analysis The view factor from the dome to the base is determined from
F11 + F12 = 1
→
F12 = 1
(where F11 = 0 )
Hence, from reciprocity relation, we get
F21 =
A1
DL
2
F12 =
=
A2
πDL / 2 π
Applying energy balance on the base surface, we have
Q& 21 = Q& evap = m& h fg =
σ (T24 − T14 )
1 − ε1
1− ε2
1
+
+
A1ε 1
A2 F21 A2 ε 2
→
m& =
(σ / h fg )(T24 − T14 )
1 − ε1
2(1 − ε 2 )
2
+
+
DLε 1 πDLF21
πDLε 2
Hence
D(σ / h fg )(T24 − T14 )
m&
=
2(1 − ε 2 )
2
L 1 − ε1
+
+
ε1
πF21
πε 2
D(σ / h fg )(T24 − T14 )
=
1 − ε1
2(1 − ε 2 )
+1+
ε1
πε 2
(1.5 m)(1000 − 370 4 ) K 4
2(1 − 0.8)
1 − 0.5
+1+
0.8π
0.5
= 0.0171 kg/s ⋅ m
=
4
⎛ 5.67 × 10 −8 W/m 2 ⋅ K 4
⎜
⎜
2257 × 10 3 J/kg
⎝
⎞
⎟
⎟
⎠
Discussion The view factor from the dome to the base is constant F21 = 2/π, which implies that the view factor is
independent of the dryer dimensions.
PROPRIETARY MATERIAL. © 2011 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.
13-82
13-106 Two concentric spheres which are maintained at uniform temperatures are separated by air at 1 atm pressure. The rate
of heat transfer between the two spheres by natural convection and radiation is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Air is an ideal gas with
constant properties.
Properties The emissivities of the surfaces are given to be ε1 = ε2 = 0.75.
The properties of air at 1 atm and the average temperature of (T1+T2)/2 =
(350+275)/2 = 312.5 K = 39.5°C are (Table A-15)
D2 = 25 cm
T2 = 275 K
ε2 = 0.75
D1 = 15 cm
T1 = 350 K
ε1 = 0.75
k = 0.02658 W/m.°C
ν = 1.697 × 10 −5 m 2 /s
Pr = 0.7256
1
β=
= 0.0032 K -1
312.5 K
Lc =5 cm
Analysis (a) Noting that Di = D1 and Do = D2 , the characteristic length is
Lc =
AIR
1 atm
1
1
( Do − Di ) = (0.25 m − 0.15 m) = 0.05 m
2
2
Then
Ra =
gβ (T1 − T2 ) L3c
ν2
Pr =
(9.81 m/s 2 )(0.003200 K -1 )(350 − 275 K )(0.05 m) 3
(1.697 × 10 −5 m 2 /s) 2
(0.7256) = 7.415 × 10 5 The effective
thermal conductivity is
Fsph =
4
( Di D o ) ( D i
Lc
−7 / 5
+ Do
Pr
⎛
⎞
k eff = 0.74k ⎜
⎟
⎝ 0.861 + Pr ⎠
−7 / 5 5
=
)
0.05 m
[(0.15 m)(0.25 m)]4 [(0.15 m) -7/5 + (0.25 m) -7/5 ]
5
= 0.005900
1/ 4
( Fsph Ra )1 / 4
0.7256
⎞
⎛
= 0.74(0.02658 W/m.°C)⎜
⎟
⎝ 0.861 + 0.7256 ⎠
= 0.1315 W/m.°C
1/ 4
[(0.00590)(7.415 ×10 )]
5
1/ 4
Then the rate of heat transfer between the spheres becomes
⎛D D
Q& = k eff π ⎜⎜ i o
⎝ Lc
⎞
⎡ (0.15 m)(0.25 m) ⎤
⎟(Ti − To ) = (0.1315 W/m.°C)π ⎢
⎥ (350 − 275)K = 23.2 W
⎟
(0.05 m)
⎦
⎣
⎠
(b) The rate of heat transfer by radiation is determined from
A1 = πD1 2 = π (0.15 m) 2 = 0.0707 m 2
Q& 12 =
A1σ (T2 4 − T1 4 )
1 − ε 2 ⎛ D1
⎜
+
ε1
ε 2 ⎜⎝ D 2
1
⎞
⎟⎟
⎠
2
=
(0.0707 m 2 )(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(350 K ) 4 − (275 K ) 4 ]
1
1 − 0.75 ⎛ 0.15 ⎞
+
⎜
⎟
0.75
0.75 ⎝ 0.25 ⎠
2
= 25.6 W
PROPRIETARY MATERIAL. © 2011 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.
13-83
13-107 Radiation heat transfer occurs between two parallel coaxial disks. The view factors and the rate of radiation heat
transfer for the existing and modified cases are 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 emissivities of disk a and b are given to be εa = 0.60 and εb = 0.8, respectively.
Analysis (a) The view factor from surface a to surface b is determined as follows
a
0.20
a
=
=1
2 L 2(0.10)
0.40
b
=
=2
B=
2 L 2(0.10)
A=
C = 1+
Fab
1+ A2
B2
= 1+
L
b
1 + 12
22
= 1.5
0.5 ⎫
0.5 ⎫
2⎧
2
2⎧
2
⎡ 2
⎡ 2
⎛B⎞ ⎪
⎛2⎞ ⎪
⎛1⎞ ⎤ ⎪
⎛ A⎞ ⎤ ⎪
= 0.5⎜ ⎟ ⎨C − ⎢C − 4⎜ ⎟ ⎥ ⎬ = 0.5⎜ ⎟ ⎨1.5 − ⎢1.5 − 4⎜ ⎟ ⎥ ⎬ = 0.764
⎝ A⎠ ⎪
⎝1⎠ ⎪
⎝ 2 ⎠ ⎥⎦ ⎪
⎝ B ⎠ ⎥⎦ ⎪
⎢⎣
⎢⎣
⎭
⎩
⎭
⎩
The view factor from surface b to surface a is determined from reciprocity relation:
Aa =
πa 2
π (0.2 m) 2
= 0.0314 m 2
4
πb 2 π (0.4 m) 2
=
= 0.1257 m 2
Ab =
4
4
Aa Fab = Ab Fba
4
=
(0.0314)(0.764) = (0.1257) Fba ⎯
⎯→ Fba = 0.191
(b) The net rate of radiation heat transfer between the surfaces can be determined from
Q& ab =
(
σ Ta 4 − Tb 4
)
1− ε a
1− ε b
1
+
+
Aa ε a Aa Fab Ab ε b
=
[
]
(5.67 × 10 −8 W/m 2 ⋅ K 4 ) (873 K )4 − (473 K )4
= 464 W
1 − 0 .6
1
1 − 0.8
+
+
(0.0314 m 2 )(0.6) (0.0314 m 2 )(0.764) (0.1257 m 2 )(0.8)
(c) In this case we have ε c = 0.7, Ac → ∞, Fac = Fbc = 1 and Q& ac = Q& cb = Q& bc . An energy balance gives
(
σ T a 4 − Tc 4
)
1− ε a
1− ε c
1
+
+
Aa ε a Aa Fac Ac ε c
=
(
σ Tc 4 − Tb 4
)
1− ε c
1− ε b
1
+
+
Ac ε c Ac Fcb Ab ε b
T a 4 − Tc 4
Tc 4 − Tb 4
=
1− ε a
1− ε b
1
1
+
+
+0 0+
Aa ε a Aa Fac
Ab Fbc Ab ε b
(873) 4 − Tc 4
Tc 4 − 473 4
=
1 − 0.6
1
1
1 − 0.8
+
+
2
2
2
(0.0314 m )(0.6) (0.0314 m )(1) (0.1257 m )(1) (0.1257 m 2 )(0.8)
⎯
⎯→ Tc = 605 K
Then
Q& bc = Q& ac =
(
σ T a 4 − Tc 4
)
1− ε a
1− ε c
1
+
+
Aa ε a Aa Fac Ac ε c
=
[
]
(5.67 × 10 −8 W/m 2 ⋅ K 4 ) (873 K )4 − (605 K )4
= 477 W
1 − 0.6
1
+
+0
(0.0314 m 2 )(0.6) (0.0314 m 2 )(1)
Discussion The rate of heat transfer is higher in part (c) because the large disk c is able to collect all radiation emitted by
disk a. It is not acting as a shield.
PROPRIETARY MATERIAL. © 2011 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.
13-84
13-108 Radiation heat transfer occurs between a tube-bank and a wall. The view factors, the net rate of radiation heat
transfer, and the temperature of tube surface are to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 The tube wall thickness and
convection from the outer surface are negligible.
Properties The emissivities of the wall and tube bank are given to be εi = 0.8 and εj = 0.9, respectively.
Analysis (a) We take the wall to be surface i and the tube bank to be
surface j. The view factor from surface i to surface j is determined from
⎡ ⎛ D ⎞2 ⎤
Fij = 1 − ⎢1 − ⎜ ⎟ ⎥
⎢⎣ ⎝ s ⎠ ⎥⎦
0.5
⎡ ⎛ 1.5 ⎞ 2 ⎤
= 1 − ⎢1 − ⎜
⎟ ⎥
⎢⎣ ⎝ 3 ⎠ ⎥⎦
0.5
⎧
2
⎤ ⎫⎪
⎛ D ⎞⎪ −1 ⎡⎛ s ⎞
+ ⎜ ⎟⎨tan ⎢⎜ ⎟ − 1⎥ ⎬
⎝ s ⎠⎪
⎢⎣⎝ D ⎠
⎥⎦ ⎪
⎩
⎭
0.5
0.5
⎧
2
⎤ ⎫⎪
⎛ 1.5 ⎞⎪ −1 ⎡⎛ 3 ⎞
+⎜
⎟ − 1⎥ ⎬ = 0.658
⎟⎨tan ⎢⎜
⎝ 3 ⎠⎪
⎥⎦ ⎪
⎢⎣⎝ 1.5 ⎠
⎩
⎭
The view factor from surface j to surface i is determined from reciprocity relation. Taking s to be the width of the wall
Ai Fij = A j F ji ⎯
⎯→ F ji =
Ai
sL
s
3
Fij =
Fij =
Fij =
(0.658) = 0.419
Aj
πDL
πD
π (1.5)
(b) The net rate of radiation heat transfer between the surfaces can be determined from
q& =
=
(
σ Ti 4 − T j 4
⎛ 1− ε i
⎜
⎜ ε
⎝ i
)
⎛ 1− ε j
⎞ 1
1
⎜
⎟
⎟ A + A F +⎜ ε
i ij
⎠ i
⎝ j
[
⎞ 1
⎟
⎟ Aj
⎠
=
(
σ Ti 4 − T j 4
1− ε i
εi
+
)
1 ⎛⎜ 1 − ε j
+
Fij ⎜⎝ ε j
]
⎞ Ai
⎟
⎟ Aj
⎠
(5.67 × 10 −8 W/m 2 ⋅ K 4 ) (1173 K )4 − (333 K )4
= 57,900 W/m 2
1 − 0.8
1
⎛ 1 − 0.9 ⎞ (0.03 m)
+
+⎜
⎟
0.8
0.658 ⎝ 0.9 ⎠ π (0.015 m)
(c) Under steady conditions, the rate of radiation heat transfer from the wall to the tube surface is equal to the rate of
convection heat transfer from the tube wall to the fluid. Denoting Tw to be the wall temperature,
(
σ Ti − Tw
1− ε i
εi
4
4
)
1 ⎛⎜ 1 − ε j
+
+
Fij ⎜⎝ ε j
[
q& rad = q& conv
⎞ Ai
⎟
⎟ Aj
⎠
= hA j (Tw − T j )
]
(5.67 × 10 −8 W/m 2 ⋅ K 4 ) (1173 K )4 − Tw4
⎡ π (0.015 m) ⎤
[Tw − (40 + 273 K )]
= (2000 W/m 2 ⋅ K ) ⎢
(
0
.
03
m
)
0.03 m ⎥⎦
−
1 − 0.8
1
1
0
.
9
⎛
⎞
⎣
+
+⎜
⎟
0.8
0.658 ⎝ 0.9 ⎠ π (0.015 m)
Solving this equation by an equation solver such as EES, we obtain
Tw = 331.4 K = 58.4°C
PROPRIETARY MATERIAL. © 2011 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.
13-85
13-109 The temperature of hot gases in a duct is measured by a thermocouple. The actual temperature of the gas is to be
determined, and compared with that without a radiation shield.
Assumptions The surfaces are opaque, diffuse, and gray.
Properties The emissivity of the thermocouple is given to be ε =0.7.
Analysis Assuming the area of the shield to be very close to the sensor of the thermometer, the radiation heat transfer from
the sensor is determined from
Q& rad, from sensor =
σ (T1 4 − T2 4 )
⎞
⎞ ⎛ 1
⎛ 1
⎜⎜ − 1⎟⎟ + ⎜⎜ 2
− 1⎟⎟
ε
ε
⎠
⎠ ⎝ 2
⎝ 1
=
(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(490 K ) 4 − (320 K ) 4 ]
= 209.5 W/m 2
1
1
⎛
⎞ ⎛
⎞
− 1⎟ + ⎜ 2
− 1⎟
⎜
⎝ 0.7
⎠ ⎝ 0.15 ⎠
Then the actual temperature of the gas can be determined from a
heat transfer balance to be
q& conv,to sensor = q& conv,from sensor
h(T f − Tth ) = 257.9 W/m 2
120 W/m 2 ⋅ °C(T f − 490) = 209.5 W/m 2
⎯
⎯→ T f = 492 K
Without the shield the temperature of the gas would be
Tw = 320 K
Thermocouple
Tth = 490 K
ε1 = 0.7
ε2 = 0.15
ε thσ (Tth − Tw )
4
T f = Tth +
Air, Tf
= 490 K +
4
h
(0.7)(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(490 K ) 4 − (320 K ) 4 ]
120 W/m 2 ⋅ °C
= 506 K
PROPRIETARY MATERIAL. © 2011 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.
13-86
A simple solar collector is built by placing a clear plastic tube around a garden hose. The rate of heat loss from
13-110
the water in the hose by natural convection and radiation is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Air is an ideal gas with
constant specific heats.
Properties The emissivities of surfaces are given to be ε1 = ε2 = 0.9. The properties of air are at 1 atm and the film
temperature of (Ts+T∞)/2 = (40+25)/2 = 32.5°C are (Table A-15)
k = 0.02607 W/m.°C
ν = 1.632 × 10
−5
T∞ = 25°C
Tsky = 15°C
2
m /s
Pr = 0.7275
β=
1
= 0.003273 K -1
(32.5 + 273) K
Water
D2 =6 cm
Analysis Under steady conditions, the heat transfer rate from the
water in the hose equals to the rate of heat loss from the clear plastic
tube to the surroundings by natural convection and radiation. The
characteristic length in this case is the diameter of the plastic tube,
Lc = D plastic = D2 = 0.06 m .
Ra =
gβ (Ts − T∞ ) D 23
ν
2
Pr =
[
h=
Air space
Garden hose
D1 =2 cm, T1
ε1 = 0.9
(9.81 m/s 2 )(0.003273 K -1 )(40 − 25)K (0.06 m) 3
⎧
0.387 Ra 1/6
⎪
D
Nu = ⎨0.6 +
9 / 16
⎪⎩
1 + (0.559 / Pr )
Plastic cover,
ε2 = 0.9, T2 =40°C
(1.632 × 10
−5
2
m /s)
2
(0.7275) = 2.842 × 10 5
2
2
⎧
⎫
0.387(2.842 × 10 5 )1 / 6 ⎫⎪
⎪
⎪
=
+
0
.
6
= 10.30
⎨
8 / 27 ⎬
8 / 27 ⎬
⎪⎭
⎪⎩
⎪⎭
1 + (0.559 / 0.7241)9 / 16
]
[
]
k
0.02607 W/m.°C
Nu =
(10.30) = 4.475 W/m 2 .°C
D2
0.06 m
A plastic = A2 = πD 2 L = π (0.06 m)(1 m) = 0.1885 m 2
Then the rate of heat transfer from the outer surface by natural convection becomes
Q& conv = hA2 (Ts − T∞ ) = (4.475 W/m 2 .°C)(0.1885 m 2 )(40 − 25)°C = 12.7 W
The rate of heat transfer by radiation from the outer surface is determined from
Q& rad = εA2σ (Ts 4 − Tsky 4 )
= (0.90)(0.1885 m 2 )(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(40 + 273 K ) 4 − (15 + 273 K) 4 ]
= 26.1 W
Finally,
Q& total ,loss = 12.7 + 26.1 = 38.8 W
Discussion Note that heat transfer is mostly by radiation.
PROPRIETARY MATERIAL. © 2011 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.
13-87
13-111 The temperature of air in a duct is measured by a thermocouple. The radiation effect on the temperature measurement
is to be quantified, 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 ε=0.6.
Analysis The actual temperature of the air can be determined from
T f = Tth +
ε thσ (Tth 4 − Tw 4 )
= 850 K +
h
(0.6)(5.67 × 10
−8
W/m ⋅ K )[(850 K ) − (500 K ) ]
2
4
75 W/m 2 ⋅ °C
4
4
Air, Tf
Tw = 500 K
Thermocouple
Tth = 850 K
ε = 0.6
= 1058 K
PROPRIETARY MATERIAL. © 2011 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.
13-88
13-112 A solar collector is considered. The absorber plate and the glass cover are maintained at uniform temperatures, and
are separated by air. The rate of heat loss from the absorber plate by natural convection and radiation is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Air is an ideal gas with
constant properties.
Properties The emissivities of surfaces are given to be ε1 = 0.9 for glass
and ε2 = 0.8 for the absorber plate. The properties of air at 1 atm and the
average temperature of (T1+T2)/2 = (80+32)/2 = 56°C are (Table A-15)
Absorber plate
T1 = 80°C
ε1 = 0.8
Solar
radiation
k = 0.02779 W/m.°C
ν = 1.857 × 10 −5 m 2 /s
Pr = 0.7212
1.5 m
1
1
=
= 0.003040 K -1
β=
Tf
(56 + 273)K
Analysis For θ = 0° , we have horizontal rectangular
enclosure. The characteristic length in this case is the distance
between the two glasses Lc = L = 0.03 m Then,
Ra =
gβ (T1 − T2 ) L3
ν2
Pr =
L = 3 cm
Glass cover,
T2 = 32°C
ε2 = 0.9
(9.81 m/s 2 )(0.00304 K -1 )(80 − 32 K )(0.03 m) 3
(1.857 × 10 −5 m 2 /s) 2
θ = 20°
Insulation
(0.7212) = 8.083 × 10 4
As = H × W = (1.5 m)(3 m) = 4.5 m 2
+
1708 ⎤ ⎡ 1708(sin 1.8θ )1.6 ⎤ ⎡ (Ra cos θ )1 / 3 ⎤
⎡
Nu = 1 + 1.44⎢1 −
− 1⎥
⎥+⎢
⎥ ⎢1 −
Ra cos θ
18
⎣ Ra cos θ ⎦ ⎣⎢
⎥⎦
⎦⎥ ⎣⎢
+
[
]
+
1/ 3
⎤
⎡
⎤ ⎡ 1708[sin(1.8 × 20)]1.6 ⎤ ⎡ (8.083 × 10 4 ) cos(20)
1708
⎢
1
+
− 1⎥
−
= 1 + 1.44⎢1 −
⎥
⎢
⎥
4
4
18
⎥
⎣⎢ (8.083 × 10 ) cos(20) ⎦⎥ ⎣⎢ (8.083 × 10 ) cos(20) ⎦⎥ ⎢⎣
⎦
= 3.747
T − T2
(80 − 32)°C
Q& = kNuAs 1
= (0.02779 W/m.°C)(3.747)(4.5 m 2 )
= 750 W
L
0.03 m
+
Neglecting the end effects, the rate of heat transfer by radiation is determined from
A σ (T1 4 − T2 4 ) (4.5 m 2 )(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(80 + 273 K ) 4 − (32 + 273 K ) 4 ]
Q& rad = s
= 1289 W
=
1
1
1
1
+
−1
+
−1
0.8 0.9
ε1 ε 2
Discussion The rates of heat loss by natural convection for the horizontal and vertical cases would be as follows (Note that
the Ra number remains the same):
Horizontal:
+
+
+
+
⎡ (8.083 × 10 4 )1 / 3 ⎤
⎡ Ra 1 / 3 ⎤
1708 ⎤
⎡
⎡ 1708 ⎤
=
+
−
+
−
1
1
1
.
44
1
+⎢
− 1⎥ = 3.812
Nu = 1 + 1.44 ⎢1 −
⎥
⎢
⎢
⎥
⎥
4
18
Ra ⎦
⎣
⎣ 8.083 × 10 ⎦
⎥⎦
⎢⎣
⎣⎢ 18
⎦⎥
T − T2
(80 − 32)°C
= (0.02779 W/m.°C)(3.812)(6 m 2 )
= 1017 W
Q& = kNuAs 1
L
0.03 m
Vertical:
⎛H⎞
Nu = 0.42 Ra 1 / 4 Pr 0.012 ⎜ ⎟
⎝L⎠
−0.3
⎛ 2m ⎞
= 0.42(8.083 × 10 4 )1 / 4 (0.7212) 0.012 ⎜
⎟
⎝ 0.03 m ⎠
−0.3
= 2.001
T − T2
(80 − 32)°C
Q& = kNuAs 1
= (0.02779 W/m.°C)(2.001)(6 m 2 )
= 534 W
L
0.03 m
PROPRIETARY MATERIAL. © 2011 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.
13-89
The circulating pump of a solar collector that consists of a horizontal tube and its glass cover fails. The
13-113E
equilibrium temperature of the tube is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The tube and its cover are isothermal. 3 Air is an ideal gas. 4 The
surfaces are opaque, diffuse, and gray for infrared radiation. 5 The glass cover is transparent to solar radiation.
Properties The properties of air should be evaluated at the average
temperature. But we do not know the exit temperature of the air in the duct,
and thus we cannot determine the bulk fluid and glass cover temperatures at
this point, and thus we cannot evaluate the average temperatures. Therefore,
we will assume the glass temperature to be 85°F, and use properties at an
anticipated average temperature of (75+85)/2 =80°F (Table A-15E),
40 Btu/h.ft
T∞ = 75°F
Tsky = 60°F
k = 0.01481 Btu/h ⋅ ft ⋅ °F
ν = 1.697 × 10 -4 ft 2 / s
Plastic cover,
ε2 = 0.9, T2
Water
D2 =5 in
Pr = 0.7290
1
1
=
β=
Tave 540 R
Air space
0.5 atm
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 glass cover to the surrounding
ambient air. When steady operation is reached, these two heat transfer rates
must equal the rate of heat gain. That is,
Q& tube-glass = Q& glass-ambient = Q& solar gain = 40 Btu/h
Aluminum tube
D1 =2.5 in, T1
ε1 = 0.9
(per foot of tube)
The heat transfer surface area of the glass cover is
Ao = Aglass = (πD oW ) = π (5 / 12 ft )(1 ft) = 1.309 ft 2 (per foot of tube)
To determine the Rayleigh number, we need to know the surface temperature of the glass, which is not available. Therefore,
solution will require a trial-and-error approach. Assuming the glass cover temperature to be 85°F, the Rayleigh number, the
Nusselt number, the convection heat transfer coefficient, and the rate of natural convection heat transfer from the glass cover
to the ambient air are determined to be
Ra Do =
=
gβ (To − T∞ ) Do3
ν2
Pr
(32.2 ft/s 2 )[1 /(540 R)](85 − 75 R )(5 / 12 ft ) 3
(1.697 × 10 − 4 ft 2 /s) 2
⎧
0.387 Ra 1/6
⎪
D
Nu = ⎨0.6 +
9 / 16
⎪⎩
1 + (0.559 / Pr )
= 14.95
[
ho =
(0.7290) = 1.092 × 10 6
2
⎫
⎧
⎫
0.387(1.092 × 10 6 )1 / 6
⎪
⎪
⎪
=
+
0
.
6
⎨
⎬
⎬
8
/
27
8 / 27
⎪⎩
⎪⎭
⎪⎭
1 + (0.559 / 0.7290 )9 / 16
]
[
2
]
k
0.01481 Btu/h ⋅ ft ⋅ °F
(14.95) = 0.5315 Btu/h ⋅ ft 2 ⋅ °F
Nu =
D0
5 / 12 ft
Q& o,conv = ho Ao (To − T∞ ) = (0.5315 Btu/h ⋅ ft 2 ⋅ °F)(1.309 ft 2 )(85 − 75)°F = 6.96 Btu/h
Also,
4
)
Q& o,rad = ε o σAo (To4 − Tsky
[
= (0.9)(0.1714 × 10 −8 Btu/h ⋅ ft 2 ⋅ R 4 )(1.309 ft 2 ) (545 R) 4 − (520 R) 4
]
= 30.5 Btu/h
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13-90
Then the total rate of heat loss from the glass cover becomes
Q& o, total = Q& o,conv + Q& o, rad = 7.0 + 30.5 = 37.5 Btu/h
which is less than 40 Btu/h. Therefore, the assumed temperature of 85°F for the glass cover is low. Repeating the
calculations with higher temperatures (including the evaluation of properties), the glass cover temperature corresponding to
40 Btu/h is determined to be 86.2°F.
The temperature of the aluminum tube is determined in a similar manner using the natural convection and radiation
relations for two horizontal concentric cylinders. The characteristic length in this case is the distance between the two
cylinders, which is
Lc = ( Do − Di ) / 2 = (5 − 2.5) / 2 = 1.25 in = 1.25/12 ft
Also,
Ai = Atube = (πDiW ) = π (2.5 / 12 ft )(1 ft) = 0.6545 ft 2 (per foot of tube)
We start the calculations by assuming the tube temperature to be 113.8°F, and thus an average temperature of (86.2+113.8)/2
= 100°F=560 R. Using properties at 100°F,
gβ (Ti − To ) L3
Ra L =
ν
2
Pr =
(32.2 ft/s 2 )[1 /(560 R)](118.5 − 81.5 R )(1.25 / 12 ft ) 3
[(1.809 × 10
−4
2
ft /s) / 0.5]
2
(0.726) = 1.281 × 10 4
The effective thermal conductivity is
Fcyc =
[ln( Do / Di )] 4
L3c ( Di−3 / 5 + Do−3 / 5 ) 5
Pr
⎛
⎞
k eff = 0.386k ⎜
⎟
⎝ 0.861 + Pr ⎠
=
[ln(5 / 2.5)] 4
(1.25/12 ft) 3 [(2.5 / 12 ft) -3/5 + (5 / 12 ft) -3/5 ] 5
= 0.1466
1/ 4
( Fcyc Ra L )1 / 4
0.726
⎛
⎞
= 0.386(0.01529 Btu/h ⋅ ft ⋅ °F)⎜
⎟
⎝ 0.861 + 0.726 ⎠
= 0.03195 Btu/h ⋅ ft ⋅ °F
1/ 4
(0.1466 × 1.281 × 10 4 )1 / 4
Then the rate of heat transfer between the cylinders by convection becomes
Q& i , conv =
2πk eff
2π (0.03195 Btu/h ⋅ ft ⋅ °F)
(113.8 − 86.2)°F = 7.994 Btu/h
(Ti − To ) =
ln(5/2.5)
ln( Do / Di )
Also,
Q& i ,rad =
=
σAi (Ti 4 − To4 )
1 1 − ε o ⎛ Di
⎜
+
εi
ε o ⎜⎝ Do
⎞
⎟
⎟
⎠
[
]
(0.1714 × 10 −8 Btu/h ⋅ ft 2 ⋅ R 4 )(0.6545 ft 2 ) (573.8 R) 4 − (546.2 R) 4
= 18.7 Btu/h
1 1 − 0.9 ⎛ 2.5 in ⎞
+
⎟
⎜
0.9
0.9 ⎝ 5 in ⎠
Then the total rate of heat loss from the glass cover becomes
Q& i ,total = Q& i ,conv + Q& i ,rad = 7.99 + 18.7 = 26.7 Btu/h
which is less than 40 Btu/h. Therefore, the assumed temperature of 113.8°F for the tube is low. By trying other values, the
tube temperature corresponding to 40 Btu/h is determined to be 126°F. Therefore, the tube will reach an equilibrium
temperature of 126°F when the pump fails.
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13-91
13-114 A double-pane window consists of two sheets of glass separated by an air space. The rates of heat transfer through
the window by natural convection and radiation are to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque,
diffuse, and gray. 3 Air is an ideal gas with constant specific heats. 4 Heat
transfer through the window is one-dimensional and the edge effects are
negligible.
Properties The emissivities of glass surfaces are given to be ε1 = ε2 = 0.9.
The properties of air at 0.3 atm and the average temperature of (T1+T2)/2 =
(15+5)/2 = 10°C are (Table A-15)
Air
15°C
5°C
L = 3 cm
k = 0.02439 W/m.°C
ν = ν 1atm / 0.3 = 1.426 × 10 −5 /0.3 = 4.753 × 10 −5 m 2 /s
H=2m
Pr = 0.7336
β=
Q&
1
= 0.003534 K -1
(10 + 273) K
Analysis The characteristic length in this case is the distance between the glasses, Lc = L = 0.03 m
Ra =
gβ (T1 − T2 ) L3
ν2
Pr =
⎛H⎞
Nu = 0.197 Ra 1 / 4 ⎜ ⎟
⎝L⎠
(9.81 m/s 2 )(0.003534 K -1 )(15 − 5)K (0.03 m) 3
−1 / 9
(4.753 × 10 −5 m 2 /s) 2
⎛ 2 ⎞
= 0.197(3040)1 / 4 ⎜
⎟
⎝ 0.05 ⎠
(0.7336) = 3040
−1 / 9
= 0.971
Note that heat transfer through the air space is less than that by pure conduction as a result of partial evacuation of the space.
Then the rate of heat transfer through the air space becomes
As = (2 m)(5 m) = 10 m 2
T − T2
(15 − 5)°C
= (0.02439 W/m.°C)(0.971)(10 m 2 )
= 78.9 W
Q& conv = kNuAs 1
L
0.03 m
The rate of heat transfer by radiation is determined from
A σ (T1 4 − T2 4 ) (10 m 2 )(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(15 + 273 K )4 − (5 + 273 K )4 ]
Q& rad = s
= 421 W
=
1
1
1
1
+
−1
+
−1
0. 9 0 . 9
ε1 ε 2
Then the rate of total heat transfer becomes
Q& total = Q& conv + Q& rad = 79 + 421 = 500 W
Discussion Note that heat transfer through the window is mostly by radiation.
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13-92
13-115 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 to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are
opaque, diffuse, and gray.
Properties The emissivities of both surfaces are given to be
ε1 = ε2 = 0.15.
Analysis (a) Assuming the conduction resistance s of the walls to be
negligible, the rate of heat transfer to the iced water in the tank is
determined to be
A1 = πD1 2 = π (2.01 m) 2 = 12.69 m 2
A1σ (T2 4 − T1 4 )
Q& 12 =
1
ε1
=
+
1 − ε 2 ⎛ D1
⎜
ε 2 ⎜⎝ D 2
⎞
⎟⎟
⎠
2
(12.69 m 2 )(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(20 + 273 K ) 4 − (0 + 273 K ) 4 ]
1
1 − 0.15 ⎛ 2.01 ⎞
+
⎟
⎜
0.15
0.15 ⎝ 2.04 ⎠
2
= 107.4 W
(b) The amount of heat transfer during a 24-hour period is
Q = Q& ∆t = (0.1074 kJ/s)(24 × 3600 s) = 9279 kJ
The amount of ice that melts during this period then becomes
Q = mhif ⎯
⎯→ m =
Q
9279 kJ
=
= 27.8 kg
hif
333.7 kJ/kg
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13-93
13-116 A solar collector consists of a horizontal copper tube enclosed in a concentric thin glass tube. The annular space
between the copper and the glass tubes is filled with air at 1 atm. The rate of heat loss from the collector by natural
convection and radiation is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Air is an ideal gas with
constant specific heats.
Properties The emissivities of surfaces are given to be ε1 = 0.85 for the tube
surface and ε2 = 0.9 for glass cover. The properties of air at 1 atm and the
average temperature of (T1+T2)/2 = (60+40)/2 = 50°C are (Table A-15)
Plastic cover,
T2 = 40°C
ε2 = 0.9
k = 0.02735 W/m.°C
ν = 1.798 × 10 −5 m 2 /s
Water
Pr = 0.7228
D2=12 cm
1
β=
= 0.003096 K -1
(50 + 273) K
Air space
Analysis The characteristic length in this case is
Lc =
Ra =
Copper tube
D1 =5 cm, T1 = 60°C
ε1 = 0.85
1
1
( D 2 − D1 ) = (0.12 m - 0.05 m) = 0.07 m
2
2
gβ (T1 − T2 ) L3c
ν2
Pr =
(9.81 m/s 2 )(0.003096 K -1 )(60 − 40)K (0.035 m) 3
(1.798 × 10 −5 m 2 /s) 2
(0.7228) = 5.823 × 10 4
The effective thermal conductivity is
Fcyl =
[ln( Do / Di )]4
L3c ( Di −3 / 5 + Do −3 / 5 ) 5
Pr
⎞
⎛
k eff = 0.386k ⎜
⎟
⎝ 0.861 + Pr ⎠
=
[ln(0.12 / 0.05)]4
[
(0.035 m) 3 (0.05 m) -3/5 + (0.12 m) -3/5
]
5
= 0.1678
1/ 4
( Fcyl Ra )1 / 4
0.7228
⎞
⎛
= 0.386(0.02735 W/m.°C)⎜
⎟
⎝ 0.861 + 0.7228 ⎠
1/ 4
[(0.1678)(5.823 ×10 )]
4
1/ 4
= 0.08626 W/m.°C
Then the rate of heat transfer between the cylinders becomes
Q& conv =
2πk eff
2π (0.08626 W/m.°C)
(Ti − To ) =
(60 − 40)°C = 12.4 W
ln( Do / Di )
ln(0.12 / 0.05)
The rate of heat transfer by radiation is determined from
A σ (T1 4 − T2 4 )
[π (0.05 m)(1 m)](5.67 ×10 −8 W/m 2 ⋅ K 4 )[(60 + 273 K ) 4 − (40 + 273 K) 4 ]
=
Q& rad = 1
1
1 − 0.9 ⎛ 5 ⎞
1 1 − ε 2 ⎛ D1 ⎞
+
⎜ ⎟
⎜⎜
⎟⎟
+
0
.
85
0.9 ⎝ 12 ⎠
ε1
ε 2 ⎝ D2 ⎠
= 19.7 W
Finally,
Q& total ,loss = 12.4 + 19.7 = 32.1 W
(per m length)
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preparation. If you are a student using this Manual, you are using it without permission.
13-94
13-117 Radiation heat transfer occurs between two concentric disks. The view factors and the net rate of radiation heat
transfer for two cases are 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 emissivities of disk 1 and 2 are given to be εa = 0.6 and εb = 0.8, respectively.
Analysis (a) The view factor from surface 1 to surface 2 is determined using Fig. 13-7 as
L 0.10
=
= 0.5,
r1 0.20
r2 0.10
=
=1⎯
⎯→ F12 = 0.19
L 0.10
2
Using reciprocity rule,
A1 = π (0.2 m) 2 = 0.1257 m 2
A2 = π (0.1 m) = 0.0314 m
2
F21 =
L
1
2
A1
0.1257 m 2
(0.19) = 0.76
F12 =
A2
0.0314 m 2
(b) The net rate of radiation heat transfer between the surfaces can be determined from
Q& =
(
σ T1 4 − T2 4
)
1− ε1
1− ε 2
1
+
+
A1ε 1 A1 F12 A2 ε 2
=
[
]
(5.67 × 10 −8 W/m 2 ⋅ K 4 ) (1073 K )4 − (573 K )4
= 1250 W
1 − 0 .6
1
1 − 0.8
+
+
(0.1257 m 2 )(0.6) (0.1257 m 2 )(0.19) (0.0314 m 2 )(0.8)
(c) When the space between the disks is completely surrounded by a refractory surface, the net rate of radiation heat transfer
can be determined from
Q& =
(
A1σ T1 4 − T2 4
A1 + A2 − 2 A1 F12
A2 − A1 F122
=
)
⎛ 1
⎞ A ⎛ 1
⎞
+ ⎜⎜ − 1⎟⎟ + 1 ⎜⎜
− 1⎟⎟
⎝ ε 1 ⎠ A2 ⎝ ε 2
⎠
[
]
(0.1257 m 2 )(5.67 × 10 −8 W/m 2 ⋅ K 4 ) (1073 K )4 − (573 K )4
= 1510 W
0.1257 + 0.0314 − 2(0.1257)(0.19) ⎛ 1
⎞ 0.1257 ⎛ 1
⎞
+
−
+
−
1
1
⎟
⎜
⎜
⎟
0.0314 − (0.1257)(0.19) 2
⎝ 0.6 ⎠ 0.0314 ⎝ 0.8 ⎠
Discussion The rate of heat transfer in part (c) is 21 percent higher than that in part (b).
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13-95
13-118 A cylindrical furnace with specified top and bottom surface temperatures and specified heat transfer rate 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 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 the bottom surface is 0.90.
3m
Analysis We consider the top surface to be surface 1, the base surface to be
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 from
Fig. 13-5 F12 = 0.2 . The view factor from the base or the top to the side surfaces
is determined by applying the summation rule to be
F11 + F12 + F13 = 1 ⎯
⎯→ F13 = 1 − F12 = 1 − 0.2 = 0.8
since the base surface is flat and thus F11 = 0 . Other view factors are
F21 = F12 = 0.20,
F23 = F13 = 0.80,
F31 = F32 = 0.20
T1 = 700 K
ε1 = ?
T3 = 450 K
ε3 = 1
T2 = 950 K
ε2 = 0.90
We now apply Eq. 9-35 to each surface
σT1 4 = J 1 +
Surface 1:
(5.67 × 10 −8
[F12 ( J 1 − J 2 ) + F13 ( J 1 − J 3 )]
ε1
1− ε1
W/m 2 .K 4 )(700 K ) 4 = J 1 +
[0.20( J 1 − J 2 ) + 0.80( J 1 − J 3 )]
ε1
σT2 4 = J 2 +
Surface 2:
(5.67 × 10 −8 W/m 2 .K 4 )(950 K ) 4 = J 2 +
Surface 3:
1− ε1
1− ε 2
ε2
[F21 ( J 2 − J 1 ) + F23 ( J 2 − J 3 )]
1 − 0.90
[0.20( J 2 − J 1 ) + 0.80( J 2 − J 3 )]
0.90
σT3 4 = J 3
(5.67 × 10 −8 W/m 2 .K 4 )(450 K ) 4 = J 3
We now apply Eq. 9-34 to surface 2
Q& 2 = A2 [F21 ( J 2 − J 1 ) + F23 ( J 2 − J 3 )] = (9 m 2 )[0.20( J 2 − J 1 ) + 0.80( J 2 − J 3 )]
Solving the above four equations, we find
ε 1 = 0.44, J 1 = 11,736 W/m 2 , J 2 = 41,985 W/m 2 , J 3 = 2325 W/m 2
The rate of heat transfer between the bottom and the top surface is
A1 = A2 = (3 m) 2 = 9 m 2
Q& 21 = A2 F21 ( J 2 − J 1 ) = (9 m 2 )(0.20)(41,985 − 11,736) W/m 2 = 54.4 kW
The rate of heat transfer between the bottom and the side surface is
A3 = 4 A1 = 4(9 m 2 ) = 36 m 2
Q& 23 = A2 F23 ( J 2 − J 3 ) = (9 m 2 )(0.8)(41,985 − 2325) W/m 2 = 285.6 kW
Discussion The sum of these two heat transfer rates are 54.4 + 285.6 = 340 kW, which is equal to 340 kW heat supply rate
from surface 2.
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13-96
13-119 A thin aluminum sheet is placed between two very large parallel plates that are maintained at uniform temperatures.
The net rate of radiation heat transfer between the plates and the temperature of the radiation shield are 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 emissivities of surfaces are given to be ε1 = 0.8, ε2 = 0.7, and ε3 = 0.12.
Analysis The net rate of radiation heat transfer with a thin aluminum shield per unit area of the plates is
Q& 12,one shield =
=
σ (T1 4 − T2 4 )
⎞
⎛ 1
⎞ ⎛ 1
1
1
⎜⎜ +
− 1⎟⎟ + ⎜
+
− 1⎟
⎟
⎜
⎝ ε1 ε 2
⎠ ⎝ ε 3,1 ε 3, 2
⎠
T1 = 750 K
ε1 = 0.8
(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(750 K ) 4 − (400 K ) 4 ]
1
1
⎞ ⎛ 1
⎛ 1
⎞
+
− 1⎟ + ⎜
+
− 1⎟
⎜
⎝ 0.8 0.7 ⎠ ⎝ 0.12 0.12 ⎠
T2 = 400 K
ε2 = 0.7
Radiation shield
ε3 = 0.12
= 951 W/m 2
The equilibrium temperature of the radiation shield is determined from
(5.67 × 10 −8
σ (T1 4 − T3 4 )
Q& 13 =
⎯
⎯→ 951 W/m 2 =
⎛ 1
⎞
1
⎜ +
− 1⎟⎟
⎜ε
⎝ 1 ε3
⎠
W/m 2 ⋅ K 4 )[(750 K ) 4 − T3 4 ]
1
⎛ 1
⎞
+
− 1⎟
⎜
⎝ 0.8 0.12 ⎠
⎯
⎯→ T3 = 644 K
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
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13-97
13-120 Two thin radiation shields are placed between two large parallel plates that are maintained at uniform temperatures.
The net rate of radiation heat transfer between the plates with and without the shields, and the temperatures of radiation
shields are 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 emissivities of surfaces are given to be ε1 = 0.6, ε2 = 0.7, ε3 = 0.10, and ε4 = 0.15.
Analysis The net rate of radiation heat transfer without the shields per unit
area of the plates is
σ (T1 4 − T2 4 )
Q& 12,no shield =
1
1
+
−1
ε1
T1 = 700 K
ε1 = 0.6
ε2
(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(700 K ) 4 − (350 K ) 4 ]
1
1
+
−1
0.6 0.7
= 6091 W/m 2
=
The net rate of radiation heat transfer with two thin radiation shields
per unit area of the plates is
Q& 12, two −shields =
=
ε3 = 0.10
ε4 = 0.15
T2 = 350 K
ε2 = 0.7
σ (T1 4 − T2 4 )
⎞ ⎛ 1
⎞
⎛1
⎞ ⎛ 1
1
1
1
⎜⎜ +
− 1⎟⎟ + ⎜⎜
+
− 1⎟⎟
+
− 1⎟⎟ + ⎜⎜
⎝ ε1 ε 2
⎠ ⎝ ε3 ε3
⎠
⎠ ⎝ε4 ε4
(5.67 × 10 −8 W/m 2 ⋅ K 4 )[(700 K ) 4 − (350 K ) 4 ]
1
1
1
⎛ 1
⎞ ⎛ 1
⎞ ⎛ 1
⎞
+
− 1⎟ + ⎜
+
− 1⎟ + ⎜
+
− 1⎟
⎜
⎝ 0.6 0.7 ⎠ ⎝ 0.10 0.10 ⎠ ⎝ 0.15 0.15 ⎠
= 381.8 W/m 2
The equilibrium temperatures of the radiation shields are determined from
(5.67 × 10 −8
σ (T1 4 − T3 4 )
⎯
⎯→ 381.8 W/m 2 =
Q& 13 =
⎞
⎛1
1
⎜ +
− 1⎟⎟
⎜ε
⎝ 1 ε3
⎠
W/m 2 ⋅ K 4 )[(700 K ) 4 − T3 4 ]
⎯
⎯→ T3 = 688 K
1
⎛ 1
⎞
+
− 1⎟
⎜
⎝ 0.6 0.10 ⎠
(5.67 × 10 −8
σ (T4 4 − T2 4 )
⎯
⎯→ 381.8 W/m 2 =
Q& 42 =
⎞
⎛ 1
1
⎜⎜
+
− 1⎟⎟
⎠
⎝ε4 ε2
W/m 2 ⋅ K 4 )[T4 4 − (350 K ) 4 ]
⎯
⎯→ T4 = 501 K
1
⎛ 1
⎞
+
− 1⎟
⎜
⎝ 0.15 0.7 ⎠
PROPRIETARY MATERIAL. © 2011 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.
13-98
13-121 Radiation heat transfer occurs between two square parallel plates. The view factors, the rate of radiation heat transfer
and the temperature of a third plate to be inserted are 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 emissivities of plate a, b, and c are given to be εa = 0.8, εb = 0.4, and εc = 0.1, respectively.
Analysis (a) The view factor from surface a to surface b is determined as follows
b 60
a 20
=
= 0.5, B = =
= 1.5
L 40
L 40
0.5
0.5 ⎫
0.5
0.5 ⎫
1 ⎧
1 ⎧
2
2
2
2
Fab =
⎨ (1.5 + 0.5) + 4 − (1.5 − 0.5) + 4 ⎬ = 0.592
⎨ ( B + A) + 4 − ( B − A) + 4 ⎬ =
⎭
⎭ 2(0.5) ⎩
2A ⎩
The view factor from surface b to surface a is determined from
a
reciprocity relation:
a
A=
[
] [
]
[
] [
]
Aa = (0.2 m)(0.2 m) = 0.04 m 2
b
Ab = (0.6 m)(0.6 m) = 0.36 m 2
L
b
Aa Fab = Ab Fba
⎯→ Fba = 0.0658
(0.04)(0.592) = (0.36) Fba ⎯
(b) The net rate of radiation heat transfer between the surfaces can be determined from
Q& ab =
(
σ Ta 4 − Tb 4
)
=
1− ε a
1− ε b
1
+
+
Aa ε a Aa Fab Ab ε b
[
]
(5.67 × 10 −8 W/m 2 ⋅ K 4 ) (1073 K )4 − (473 K )4
= 1374 W
1 − 0.8
1
1 − 0.4
+
+
(0.04 m 2 )(0.8) (0.04 m 2 )(0.592) (0.36 m 2 )(0.4)
(c) In this case we have
c 2.0 m
a 0.2 m
=
= 1, C = =
= 10
L 0 .2 m
L 0.2 m
0.5
0.5 ⎫
0.5
0.5 ⎫
1 ⎧
1 ⎧
2
2
2
2
=
⎨ (10 + 0.5) + 4 − (10 − 0.5) + 4 ⎬ = 0.981
⎨ (C + A) + 4 − (C − A) + 4 ⎬ =
⎭
⎭ 2(0.5) ⎩
2A ⎩
A=
Fac
[
] [
]
[
] [
]
b 0.6 m
c 2.0 m
=
= 3, C = =
= 10
L 0.2 m
L 0.2 m
0.5
0.5 ⎫
1 ⎧
2
2
=
⎨ (C + B ) + 4 − (C − B ) + 4 ⎬
⎭
2A ⎩
0
.
5
0
.
5
1 ⎧
2
2
⎫
=
⎨ (10 + 3) + 4 − (10 − 3) + 4 ⎬ = 0.979
⎭
2(3) ⎩
B=
Fbc
[
[
] [
] [
]
]
Ab Fbc = Ac Fcb
(0.36)(0.979) = (4.0) Fcb ⎯
⎯→ Fba = 0.0881
An energy balance gives
(
σ T a 4 − Tc 4
)
Q& ac = Q& cb
1− ε a
1− ε c
1
+
+
Aa ε a Aa Fac Ac ε c
=
(
σ Tc 4 − Tb 4
)
1− ε c
1− ε b
1
+
+
Ac ε c Ac Fcb Ab ε b
Tc 4 − (473 K ) 4
(1073 K ) 4 − Tc 4
=
1 − 0.8
1
1 − 0.1
1 − 0.1
1
1 − 0.4
+
+
+
+
2
2
2
2
2
(0.04 m )(0.8) (0.04 m )(0.981) (4 m )(0.1) (4 m )(0.1) (4 m )(0.0881) (0.36 m 2 )(0.4)
Solving the equation with an equation solver such as EES, we obtain Tc = 754 K = 481°C
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13-99
13-122 A cylindrical furnace with specified top and bottom surface temperatures and specified heat transfer rate at the
bottom surface is considered. The temperature of the side 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 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 emissivities of the top, bottom, and side surfaces are 0.60, 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. This system is a three-surface
T1 = 450 K
enclosure. The view factor from surface 1 to surface 2 is determined from
ε1 = 0.60
L 1.2
⎫
r1 = 0.6 m
=
=2 ⎪
⎪
r 0.6
⎬ F12 = 0.17 (Fig. 13-7)
r 0.6
h = 1.2 m
=
= 0.5⎪
⎪
L 1.2
⎭
T3 = ?
The surface areas are
ε3 = 0.40
A1 = A2 = πD 2 / 4 = π (1.2 m) 2 / 4 = 1.131 m 2
T2 = 800 K
ε2 = 0.50
r2 = 0.6 m
A3 = πDL = π (1.2 m)(1.2 m) = 4.524 m 2
Then other view factors are determined to be
F12 = F21 = 0.17
F11 + F12 + F13 = 1 ⎯
⎯→ 0 + 0.17 + F13 = 1 ⎯
⎯→ F13 = 0.83 (summation rule), F23 = F13 = 0.83
A1F13 = A3 F31 ⎯
⎯→(1.131)(0.83) = (4.524) F31 ⎯
⎯→ F31 = 0.21 (reciprocity rule), F32 = F31 = 0.21
We now apply Eq. 13-35 to each surface
σT1 4 = J 1 +
Surface 1:
(5.67 × 10 −8 W/m 2 .K 4 )(450 K ) 4 = J 1 +
1 − ε1
ε1
1 − 0.60
[0.17( J 1 − J 2 ) + 0.83( J 1 − J 3 )]
0.60
σT2 4 = J 2 +
Surface 2:
(5.67 × 10 −8 W/m 2 .K 4 )(800 K ) 4 = J 2 +
σT3 4 = J 3 +
Surface 3:
(5.67 × 10 −8 W/m 2 .K 4 )T3 4 = J 3 +
1− ε3
ε3
[F12 ( J 1 − J 2 ) + F13 ( J 1 − J 3 )]
1−ε2
ε2
[F21 ( J 2 − J 1 ) + F23 ( J 2 − J 3 )]
1 − 0.50
[0.17( J 2 − J 1 ) + 0.83( J 2 − J 3 )]
0.50
[F31 ( J 3 − J 1 ) + F32 ( J 3 − J 2 )]
1 − 0.40
[0.21( J 3 − J 1 ) + 0.21( J 3 − J 2 )]
0.40
We now apply Eq. 13-34 to surface 2
Q& 2 = A2 [F21 ( J 2 − J 1 ) + F23 ( J 2 − J 3 )] = (1.131 m 2 )[0.17( J 2 − J 1 ) + 0.83( J 2 − J 3 )]
Solving the above four equations, we find
T3 = 831 K , J 1 = 10,477 W/m 2 , J 2 = 21,986 W/m 2 , J 3 = 22,852 W/m 2
The rate of heat transfer between the bottom and the top surface is
Q& 21 = A2 F21 ( J 2 − J 1 ) = (1.131 m 2 )(0.17)(21,986 − 10,477) W/m 2 = 2213 W
The rate of heat transfer between the bottom and the side surface is
Q& 23 = A2 F23 ( J 2 − J 3 ) = (1.131 m 2 )(0.83)(21,986 − 22,852) W/m 2 = −813 W
Discussion The sum of these two heat transfer rates are 2213 +(-813) = 1400 W, which is the heat supply rate from surface
2. This must be satisfied to maintain the surfaces at the specified temperatures under steady operation. Note that the
difference is due to round-off error.
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13-100
13-123 Combustion gases flow inside a tube in a boiler. The rates of heat transfer by convection and radiation 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. 3 Combustion gases are
assumed to have the properties of air, which is an ideal gas with constant properties.
Properties The properties of air at 1000 K = 727°C and 1 atm are (Table A-15)
ρ = 0.3535 kg/m 3
k = 0.06704 W/m.°C
Ts = 105°C
ν = 1.185 × 10 m /s
-4
2
D = 15 cm
c p = 1140 J/kg.°C
Combustion
gases, 1 atm
Ti = 1000 K
3 m/s
Pr = 0.7107
Analysis (a) The Reynolds number is
Re =
Vavg D
ν
=
(3 m/s)(0.15 m)
1.185 × 10 − 4 m 2 /s
= 3797
which is higher than 2300, but much less than 10,000. We assume laminar flow. The Nusselt number in this case is
Nu =
hDh
= 3.66
k
Heat transfer coefficient is
h=
k
0.06704 W/m.°C
(3.66) = 1.636 W/m 2 .°C
Nu =
D
0.15 m
Next we determine the exit temperature of air
A = πDL = π (0.15 m)(6 m) = 2.827 m 2
Ac = πD 2 / 4 = π (0.15 m) 2 /4 = 0.01767 m 2
m& = ρVAc = (0.3535 kg/m 3 )(3 m/s)(0.01767 m 2 ) = 0.01874 kg/s
Te = Ts − (Ts − Ti )e
− hA /( m& c p )
= 105 − (105 − 727)e
−
(1.636)( 2.827 )
( 0.01874)(1140)
= 605.9°C
Then the rate of heat transfer by convection becomes
Q& conv = m& c p (Ti − Te ) = (0.01874 kg/s)(1140 J/kg.°C)(727 − 605.9)°C = 2587 W
Next, we determine the emissivity of combustion gases. First, the mean beam length for an infinite circular cylinder is, from
Table 13-4,
L = 0.95(0.15 m) = 0.1425 m
Then,
Pc L = (0.08 atm)(0.1425 m) = 0.0114 m ⋅ atm = 0.037 ft ⋅ atm
Pw L = (0.16 atm)(0.1425 m) = 0.0228 m ⋅ atm = 0.075 ft ⋅ atm
The emissivities of CO2 and H2O corresponding to these values at the average gas temperature of Tg=(Tg+Tg)/2 =
(727+606)/2 = 667°C = 940 K and 1atm are, from Fig. 13-36,
ε c, 1 atm = 0.055 and
ε w, 1 atm = 0.050
Both CO2 and H2O are present in the same mixture, and we need to correct for the overlap of emission bands. The emissivity
correction factor at T = Tg = 940 K is, from Fig. 13-38,
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13-101
Pc L + Pw L = 0.037 + 0.075 = 0.112
Pw
0.16
=
= 0.67
Pw + Pc 0.16 + 0.08
⎫
⎪
⎬ ∆ε = 0.0
⎪⎭
Then the effective emissivity of the combustion gases becomes
ε g = C c ε c , 1 atm + C w ε w, 1 atm − ∆ε = 1 × 0.055 + 1 × 0.050 − 0.0 = 0.105
Note that the pressure correction factor is 1 for both gases since the total pressure is 1 atm. For a source temperature of Ts =
105°C = 378 K, the absorptivity of the gas is again determined using the emissivity charts as follows:
Pc L
Ts
378 K
= (0.08 atm)(0.1425 m)
= 0.00458 m ⋅ atm = 0.015 ft ⋅ atm
940 K
Tg
Pw L
Ts
378 K
= (0.16 atm)(0.1425 m)
= 0.00916 m ⋅ atm = 0.030 ft ⋅ atm
940 K
Tg
The emissivities of CO2 and H2O corresponding to these values at a temperature of Ts = 378 K and 1atm are, from Fig. 1336,
ε c , 1 atm = 0.034 and
ε w, 1 atm = 0.055
Then the absorptivities of CO2 and H2O become
⎛ Tg ⎞
⎟
⎟
⎝ Ts ⎠
0.65
α c = C c ⎜⎜
⎛ Tg ⎞
⎟
⎟
⎝ Ts ⎠
α w = C w ⎜⎜
⎛ 940 K ⎞
⎟
⎝ 378 K ⎠
0.65
ε c , 1 atm = (1)⎜
0.45
⎛ 940 K ⎞
⎟
⎝ 378 K ⎠
ε w, 1 atm = (1)⎜
(0.034) = 0.0615
0.45
(0.055) = 0.0829
Also ∆α = ∆ε, but the emissivity correction factor is to be evaluated from Fig. 13-38 at T = Ts = 378 K instead of Tg = 940
K. We use the chart for 400 K. At Pw/(Pw+ Pc) = 0.67 and PcL +PwL = 0.112 we read ∆ε = 0.0. Then the absorptivity of the
combustion gases becomes
α g = α c + α w − ∆α = 0.0615 + 0.0829 − 0.0 = 0.144
The emissivity of the inner surfaces of the tubes is 0.9. Then the net rate of radiation heat transfer from the combustion gases
to the walls of the tube becomes
ε +1
Q& rad = s
As σ (ε g T g4 − α g Ts4 )
2
0.9 + 1
=
(2.827 m 2 )(5.67 × 10 −8 W/m 2 ⋅ K 4 )[0.105(940 K ) 4 − 0.144(378 K ) 4 ]
2
= 12,040 W
(b) The heat of vaporization of water at 1 atm is 2257 kJ/kg (Table A-9). Then rate of evaporation of water becomes
+ Q& rad (2587 + 12,040) W
Q&
⎯→ m& evap = conv
=
= 0.00648 kg/s
Q& conv + Q& rad = m& evap h fg ⎯
h fg
2257 × 10 3 J/kg
PROPRIETARY MATERIAL. © 2011 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.
13-102
13-124 Combustion gases flow inside a tube in a boiler. The rates of heat transfer by convection and radiation 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. 3 Combustion gases are
assumed to have the properties of air, which is an ideal gas with constant properties.
Properties The properties of air at 1000 K = 727°C and 3 atm are (Table A-15)
ρ = 0.3535 kg/m 3
k = 0.06704 W/m.°C
Ts = 105°C
ν = ν = (1.185 × 10 m /s)/3 = 0.395 × 10 m /s
-4
2
-4
2
D = 15 cm
c p = 1140 J/kg.°C
Combustion
gases, 3 atm
Ti = 1000 K
3 m/s
Pr = 0.7107
Analysis (a) The Reynolds number is
Re =
Vavg D
ν
=
(3 m/s)(0.15 m)
0.395 × 10 − 4 m 2 /s
= 11,392
which is greater than 10,000. The flow is turbulent. The entry lengths in this case are roughly
Lh ≈ Lt ≈ 10 D = 10(0.15 m) = 1.5 m
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
Nu =
hDh
= 0.023 Re 0.8 Pr 0.3 = 0.023(11,392) 0.8 (0.7107) 0.3 = 36.52
k
Heat transfer coefficient is
h=
k
0.06704 W/m.°C
(36.52) = 16.32 W/m 2 .°C
Nu =
D
0.15 m
Next we determine the exit temperature of air
A = πDL = π (0.15 m)(6 m) = 2.827 m 2
Ac = πD 2 / 4 = π (0.15 m) 2 /4 = 0.01767 m 2
m& = ρVAc = (0.3535 kg/m 3 )(3 m/s)(0.01767 m 2 ) = 0.01874 kg/s
Te = Ts − (Ts − Ti )e
− hA /( m& c p )
= 105 − (105 − 727)e
−
(16.32)( 2.827 )
( 0.01874)(1140 )
= 176.8°C
Then the rate of heat transfer by convection becomes
Q& conv = m& c p (Ti − Te ) = (0.01874 kg/s)(1140 J/kg.°C)(727 − 176.8)°C = 11,760 W
Next, we determine the emissivity of combustion gases. First, the mean beam length for an infinite circular cylinder is, from
Table 13-4,
L = 0.95(0.15 m) = 0.1425 m
Then,
Pc L = (0.08 atm)(0.1425 m) = 0.0114 m ⋅ atm = 0.037 ft ⋅ atm
Pw L = (0.16 atm)(0.1425 m) = 0.0228 m ⋅ atm = 0.075 ft ⋅ atm
The emissivities of CO2 and H2O corresponding to these values at the average gas temperature of Tg=(Tg+Tg)/2 =
(727+177)/2 = 452°C = 725 K and 1atm are, from Fig. 13-36,
ε c , 1 atm = 0.053 and
ε w, 1 atm = 0.068
These 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. 13-37,
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13-103
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 emission bands. The emissivity
correction factor at T = Tg = 725 K is, from Fig. 13-38,
Pc L + Pw L = 0.037 + 0.075 = 0.112
Pw
0.16
=
= 0.67
Pw + Pc 0.16 + 0.08
⎫
⎪
⎬ ∆ε = 0.0
⎪⎭
Then the effective emissivity of the combustion gases becomes
ε g = C c ε c , 1 atm + C w ε w, 1 atm − ∆ε = 1.5 × 0.053 + 1.8 × 0.068 − 0.0 = 0.202
For a source temperature of Ts = 105°C = 378 K, the absorptivity of the gas is again determined using the emissivity charts as
follows:
Pc L
Ts
378 K
= (0.08 atm)(0.1425 m)
= 0.00594 m ⋅ atm = 0.020 ft ⋅ atm
725 K
Tg
Pw L
Ts
378 K
= (0.16 atm)(0.1425 m)
= 0.0119 m ⋅ atm = 0.039 ft ⋅ atm
725 K
Tg
The emissivities of CO2 and H2O corresponding to these values at a temperature of Ts = 378 K and 1atm are, from Fig. 1336,
ε c , 1 atm = 0.040 and
ε w, 1 atm = 0.065
Then the absorptivities of CO2 and H2O become
⎛ Tg ⎞
⎟
⎟
⎝ Ts ⎠
0.65
α c = C c ⎜⎜
⎛ Tg ⎞
⎟
⎟
⎝ Ts ⎠
α w = C w ⎜⎜
⎛ 725 K ⎞
⎟
⎝ 378 K ⎠
0.65
ε c , 1 atm = (1.5)⎜
0.45
⎛ 725 K ⎞
⎟
⎝ 378 K ⎠
ε w, 1 atm = (1.8)⎜
(0.040) = 0.0916
0.45
(0.065) = 0.1568
Also ∆α = ∆ε, but the emissivity correction factor is to be evaluated from Fig. 13-38 at T = Ts = 378 K instead of Tg = 725
K. We use the chart for 400 K. At Pw/(Pw+ Pc) = 0.67 and PcL +PwL = 0.112 we read ∆ε = 0.0. Then the absorptivity of the
combustion gases becomes
α g = α c + α w − ∆α = 0.0916 + 0.1568 − 0.0 = 0.248
The emissivity of the inner surfaces of the tubes is 0.9. Then the net rate of radiation heat transfer from the combustion gases
to the walls of the tube becomes
ε +1
Q& rad = s
As σ (ε g T g4 − α g Ts4 )
2
0.9 + 1
=
(2.827 m 2 )(5.67 × 10 −8 W/m 2 ⋅ K 4 )[0.202(725 K ) 4 − 0.248(378 K ) 4 ]
2
= 7727 W
(b) The heat of vaporization of water at 1 atm is 2257 kJ/kg (Table A-9). Then rate of evaporation of water becomes
+ Q& rad (11,760 + 7727) W
Q&
⎯→ m& evap = conv
=
= 0.00863 kg/s
Q& conv + Q& rad = m& evap h fg ⎯
h fg
2257 × 10 3 J/kg
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preparation. If you are a student using this Manual, you are using it without permission.
13-104
13-125 Two diffuse surfaces A1 and A2 placed at an specified orientation, (a) the expression for the view factor F12 in terms
of A2 and L, and (b) the value of the view factor F12 when A2 = 0.02 m2 and L = 1 m are to be determined.
Assumptions 1 The surfaces A1 and A2 are diffuse. 2 Both A1 and A2 can be approximated as differential surfaces since both
are very small compared to the square of the distance between them.
Analysis (a) The view factor for surfaces A1 and A2 can be determined using the integral
F12 =
1
A1
∫ ∫
=
1
A1
∫
cos θ 1 cos θ 2
π r2
cos θ1 cos θ 2
A2 A1
dA1 dA2
A1 dA2
π r2
1 cos θ1 cos θ 2
A1 A2
=
A1
π r2
cos θ1 cos θ 2
=
A2
A2
π r2
From orientation of the two surfaces, we have
θ1 = θ 2
cos θ1 = cos θ 2
→
(1)
and
r = 2 L cos θ1
(2)
Substituting Eqs. (1) and (2) into the expression for F12, we get
F12 =
(cos θ1 ) 2
π (2 L cos θ1 )
2
A2 =
A2
4 L2π
(b) The value of the view factor F12 when A2 = 0.02 m2 and L = 1 m is
F12 =
A2
4L2π
=
0.02 m 2
4(1 m) 2 π
= 1.59 × 10 − 3
Discussion The view factor F21 can simply be determined with the reciprocity relation as
F21 =
A1
A
F12 = 21
A2
4L π
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13-105
13-126 Two parallel black disks are positioned coaxially, where the lower disk is heated electrically. The temperature of the
upper disk is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are black. 3 Convection heat transfer is not considered. 4
Radiation heat transfer only between the two disks. 5 No heat loss to the surrounding.
Analysis For coaxial parallel disks, from Table 13-1, with i = 1, j = 2,
R1 =
D1 / 2 0.2 / 2
=
= 0.4
L
0.25
S = 1+
1 + R22
R12
= 1+
1 + (0.8) 2
(0.4) 2
⎧
⎡
⎛D
1⎪
F12 = ⎨S − ⎢ S 2 − 4 ⎜⎜ 2
2⎪
⎢
⎝ D1
⎣
⎩
R2 =
and
⎞
⎟⎟
⎠
2⎤
D2 / 2 0.4 / 2
=
= 0.8
L
0.25
= 11.25
1/ 2 ⎫
⎥
⎥
⎦
1/ 2 ⎫
⎧
2
⎡
⎪ 1⎪
⎛ 0.4 ⎞ ⎤ ⎪
2
⎟ ⎥ ⎬ = 0.3676
⎬ = ⎨11.25 − ⎢(11.25) − 4 ⎜
⎝ 0.2 ⎠ ⎥⎦ ⎪
⎢⎣
⎪ 2⎪
⎩
⎭
⎭
The net radiation heat transfer rate leaving the lower surface can be expressed as
Q& elec = Q&12 = A1 F12σ (T14 − T24 )
→
⎛
4Q& elec
T2 = ⎜ T14 −
⎜
πD12 F12σ
⎝
⎡
⎤
4(20 W )
T2 = ⎢(500 K ) 4 −
⎥
π (0.2 m) 2 (0.3676)(5.67 × 10 −8 W/m 2 ⋅ K 4 ) ⎦⎥
⎣⎢
1/ 4
⎞
⎟
⎟
⎠
1/ 4
= 423 K
Discussion The view factor F12 can also be determined using Fig. 13-7 to be
F12 ≈ 0.36
with
L / r1 = 2.5
and
r2 / L = 0.8
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13-106
Fundamentals of Engineering (FE) Exam Problems
13-127 Consider a surface at 0ºC that may be assumed to be a blackbody in an environment at 25ºC. If 300 W/m2 radiation is
incident on the surface, the radiosity of this black surface is
(a) 0 W/m2
(b) 15 W/m2
(c) 132 W/m2
(d) 300 W/m2
(e) 315 W/m2
Answer (e) 315 W/m2
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
T=0 [C]
T_infinity=25 [C]
G=300 [W/m^2]
sigma=5.67E-8 [W/m^2-K^4]
J=sigma*(T+273)^4 "J=E_b for a blackbody"
"Some Wrong Solutions with Common Mistakes"
W1_J=sigma*T^4 "Using C unit for temperature"
W2_J=sigma*((T_infinity+273)^4-(T+273)^4) "Finding radiation exchange between the surface and the
environment"
W3_J=G "Using the incident radiation as the answer"
W4_J=J-G "Finding the difference between the emissive power and incident radiation"
13-128 Consider a 15-cm-diameter sphere placed within a cubical enclosure with a side length of 15 cm. The view factor
from any of the square cube surface to the sphere is
(a) 0.09
(b) 0.26
(c) 0.52
(d) 0.78
(e) 1
Answer (c) 0.52
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
D=0.15 [m]
s=0.15 [m]
A1=pi*D^2
A2=6*s^2
F_12=1
A1*F_12=A2*F_21 "Reciprocity relation"
"Some Wrong Solutions with Common Mistakes"
W_F_21=F_21/6 "Dividing the result by 6"
PROPRIETARY MATERIAL. © 2011 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.
13-107
13-129 A 90-cm-diameter flat black disk is placed in the center of the top surface of a 1 m × 1 m × 1 m black box. The view
factor from the entire interior surface of the box to the interior surface of the disk is
(a) 0.07
(b) 0.13
(c) 0.26
(d) 0.32
(e) 0.50
Answer (b) 0.13
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
d=0.9 [m]
A1=pi*d^2/4 [m^2]
A2=5*1*1 [m^2]
F12=1
F21=A1*F12/A2
13-130 Consider two concentric spheres forming an enclosure with diameters 12 and 18 cm and surface temperatures 300
and 500 K, respectively. Assuming that the surfaces are black, the net radiation exchange between the two spheres is
(a) 21 W
(b) 140 W
(c) 160 W
(d) 1275 W
(e) 3084 W
Answer (b) 140 W
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
D1=0.12 [m]
D2=0.18 [m]
T1=300 [K]
T2=500 [K]
sigma=5.67E-8 [W/m^2-K^4]
A1=pi*D1^2
F_12=1
Q_dot=A1*F_12*sigma*(T2^4-T1^4)
"Some Wrong Solutions with Common Mistakes"
W1_Q_dot=F_12*sigma*(T2^4-T1^4) "Ignoring surface area"
W2_Q_dot=A1*F_12*sigma*T1^4 "Emissive power of inner surface"
W3_Q_dot=A1*F_12*sigma*T2^4 "Emissive power of outer surface"
PROPRIETARY MATERIAL. © 2011 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.
13-108
13-131 Consider a vertical 2-m-diameter cylindrical furnace whose surfaces closely approximate black surfaces. The base,
top, and side surfaces of the furnace are maintained at 400 K, 600 K, and 1200 K, respectively. If the view factor from the
base surface to the top surface is 0.2, the net radiation heat transfer between the base and the side surfaces is
(a) 73 kW
(b) 126 kW
(c) 215 kW
(d) 292 kW
(e) 344 kW
Answer (d) 292 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
D=2 [m]
T1=400 [K]
T2=600 [K]
T3=1200 [K]
F_12=0.2
A1=pi*D^2/4
A2=A1
F_13=1-F_12
sigma=5.67E-8 [W/m^2-K^4]
Q_dot_13=A1*F_13*sigma*(T1^4-T3^4)
"Some Wrong Solutions with Common Mistakes"
W_Q_dot_13=A1*F_12*sigma*(T1^4-T3^4) "Using the view factor between the base and top surfaces"
13-132 Consider a vertical 2-m-diameter cylindrical furnace whose surfaces closely approximate black surfaces. The base,
top, and side surfaces of the furnace are maintained at 400 K, 600 K, and 900 K, respectively. If the view factor from the
base surface to the top surface is 0.2, the net radiation heat transfer from the bottom surface is
(a) -93.6 kW
(b) -86.1 kW
(c) 0 kW
(d) 86.1 kW
(e) 93.6 kW
Answer (a) -93.6 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
D=2 [m]
T1=400 [K]
T2=600 [K]
T3=900 [K]
A1=pi*D^2/4
A2=A1
F_12=0.2
F_13=1-F_12
sigma=5.67E-8 [W/m^2-K^4]
Q_dot_12=A1*F_13*sigma*(T1^4-T2^4)
Q_dot_13=A1*F_13*sigma*(T1^4-T3^4)
Q_dot_1=Q_dot_12+Q_dot_13
"Some Wrong Solutions with Common Mistakes"
W1_Q_dot_1=-Q_dot_1 "Using wrong sign"
W2_Q_dot_1=Q_dot_12-Q_dot_13 "Substracting heat transfer terms"
W3_Q_dot_1=Q_dot_13-Q_dot_12 "Substracting heat transfer terms"
PROPRIETARY MATERIAL. © 2011 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.
13-109
13-133 Consider an infinitely long three-sided triangular enclosure with side lengths 5 cm, 3, cm, and 4 cm. The view factor
from the 5 cm side to the 4 cm side is
(a) 0.3
(b) 0.4
(c) 0.5
(d) 0.6
(e) 0.7
Answer (d) 0.6
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
w1=5 [cm]
w2=3 [cm]
w3=4 [cm]
F_13=(w1+w3-w2)/(2*w1) "from Table 12-2"
"Some Wrong Solutions with Common Mistakes"
W_F_13=(w1+w2-w3)/(2*w1) "Using incorrect form of the equation"
13-134 The number of view factors that need to be evaluated directly for a 10-surface enclosure is
(a) 1
(b) 10
(c) 22
(d) 34
(e) 45
Answer (e) 45
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
N=10
n_viewfactors=1/2*N*(N-1)
PROPRIETARY MATERIAL. © 2011 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.
13-110
13-135 Consider a gray and opaque surface at 0ºC in an environment at 25ºC. The surface has an emissivity of 0.8. If the
radiation incident on the surface is 240 W/m2, the radiosity of the surface is
(a) 38 W/m2
(b) 132 W/m2
(c) 240 W/m2
(d) 300 W/m2
(e) 315 W/m2
Answer (d) 300 W/m2
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
T=0 [C]
T_infinity=25 [C]
epsilon=0.80
G=240 [W/m^2]
sigma=5.67E-8 [W/m^2-K^4]
J=epsilon*sigma*(T+273)^4+(1-epsilon)*G
"Some Wrong Solutions with Common Mistakes"
W1_J=sigma*(T+273)^4 "Radiosity for a black surface"
W2_J=epsilon*sigma*T^4+(1-epsilon)*G "Using C unit for temperature"
W3_J=sigma*((T_infinity+273)^4-(T+273)^4) "Finding radiation exchange between the surface and the
environment"
W4_J=G "Using the incident radiation as the answer"
13-136 Two concentric spheres are maintained at uniform temperatures T1 = 45ºC and T2 = 280ºC and have emissivities ε1 =
0.25 and ε2 = 0.7, respectively. If the ratio of the diameters is D1/D2 = 0.30, the net rate of radiation heat transfer between the
two spheres per unit surface area of the inner sphere is
(a) 86 W/m2
(b) 1169 W/m2
(c) 1181 W/m2
(d) 2510 W/m2
(e) 3306 W/m2
Answer (b) 1169 W/m2
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
T1=45 [C]
T2=280 [C]
epsilon_1=0.25
epsilon_2=0.70
D1\D2=0.30
sigma=5.67E-8 [W/m^2-K^4]
Q_dot=((sigma*((T2+273)^4-(T1+273)^4)))/((1/epsilon_1)+(1-epsilon_2)/epsilon_2*D1\D2^2)
"Some Wrong Solutions with Common Mistakes"
W1_Q_dot=((sigma*(T2^4-T1^4)))/((1/epsilon_1)+(1-epsilon_2)/epsilon_2*D1\D2^2) "Using C unit for
temperature"
W2_Q_dot=epsilon_1*sigma*((T2+273)^4-(T1+273)^4) "The equation when the outer sphere is black"
W3_Q_dot=epsilon_2*sigma*((T2+273)^4-(T1+273)^4) "The equation when the inner sphere is black"
PROPRIETARY MATERIAL. © 2011 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.
13-111
2
13-137 A solar flux of 1400 W/m directly strikes a space vehicle surface which has a solar absortivity of 0.4 and thermal
emissivity of 0.6. The equilibrium temperature of this surface in space at 0 K is
(a) 300 K
(b) 360 K
(c) 410 K
(d) 467 K
(e) 510 K
Answer (b) 360 K
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
a=0.4
e=0.6
Q=1400 [W/m^2]
a*Q=e*sigma#*T^4
13-138 Consider a 3-m × 3-m × 3-m cubical furnace. The base surface is black and has a temperature of 400 K. The
radiosities for the top and side surfaces are calculated to be 7500 W/m2 and 3200 W/m2, respectively. If the temperature of
the side surfaces is 485 K, the emissivity of the side surfaces is
(a) 0.37
(b) 0.55
(c) 0.63
(d) 0.80
(e) 0.89
Answer (e) 0.89
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
s=3 [m]
T1=400 [K]
epsilon_1=1
J2=7500 [W/m^2]
J3=3200 [W/m^2]
T3=485 [K]
sigma=5.67E-8 [W/m^2-K^4]
F_31=0.2
F_32=F_31
J1=sigma*T1^4
sigma*T3^4=J3+(1-epsilon_3)/epsilon_3*(F_31*(J3-J1)+F_32*(J3-J2))
PROPRIETARY MATERIAL. © 2011 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.
13-112
13-139 Two very large parallel plates are maintained at uniform temperatures T1 = 750 K and T2 = 500 K and have
emissivities ε1 = 0.85 and ε2 = 0.7, respectively. If a thin aluminum sheet with the same emissivity on both sides is to be
placed between the plates in order to reduce the net rate of radiation heat transfer between the plates by 90 percent, the
emissivity of the aluminum sheet must be
(a) 0.07
(b) 0.10
(c) 0.13
(d) 0.16
(e) 0.19
Answer (c) 0.13
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
T1=750 [K]
T2=500 [K]
epsilon_1=0.85
epsilon_2=0.70
f=0.9
sigma=5.67E-8 [W/m^2-K^4]
Q_dot_noshield=(sigma*(T1^4-T2^4))/((1/epsilon_1)+(1/epsilon_2)-1)
Q_dot_1shield=(1-f)*Q_dot_noshield
Q_dot_1shield=(sigma*(T1^4-T2^4))/((1/epsilon_1)+(1/epsilon_2)-1+(1/epsilon_3)+(1/epsilon_3)-1)
13-140 A 70-cm-diameter flat black disk is placed at the center of the ceiling a 1 m × 1 m × 1 m black box. If the
temperature of the box is 620oC and the temperature of the disk is 27oC, the rate of heat transfer by radiation between the
interior of the box and the disk is
(a) 2 kW
(b) 5 kW
(c) 8 kW
(d) 11 kW
(e) 14 kW
Answer (e) 14 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
d=0.7 [m]
A1=pi*d^2/4 [m^2]
A2=5*1*1 [m^2]
F12=1
T2=893 [K]
T1=300 [K]
F21=A1*F12/A2
Q=A2*F21*sigma#*(T2^4-T1^4)
PROPRIETARY MATERIAL. © 2011 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.
13-113
13-141 Two grey surfaces that form an enclosure exchange heat with one another by thermal radiation. Surface 1 has a
temperature of 400 K, an area of 0.2 m2, and a total emissivity of 0.4. Surface 2 has a temperature of 600 K, an area of 0.3
m2, and a total emissivity of 0.6. If the view factor F12 is 0.3, the rate of radiation heat transfer between the two surfaces is
(a) 135 W
(b) 223 W
(c) 296 W
(d) 342 W
(e) 422 W
Answer (b) 223 W
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
A1=0.2 [m^2]
T1=400 [K]
e1=0.4
A2=0.3 [m^2]
T2=600 [K]
e2=0.6
F12=0.3
R1=(1-e1)/(A1*e1)
R2=1/(A1*F12)
R3=(1-e2)/(A2*e2)
Q=sigma#*(T2^4-T1^4)/(R1+R2+R3)
13-142 The surfaces of a two-surface enclosure exchange heat with one another by thermal radiation. Surface 1 has a
temperature of 400 K, an area of 0.2 m2, and a total emissivity of 0.4. Surface 2 is black, has a temperature of 800 K, and an
area of 0.3 m2. If the view factor F12 is 0.3, the rate of radiation heat transfer between the two surfaces is
(a) 340 W
(b) 560 W
(c) 780 W
(d) 900 W
(e) 1160 W
Answer (d) 900 W
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
A1=0.2 [m^2]
T1=400 [K]
e1=0.4
A2=0.3 [m^2]
T2=800 [K]
F12=0.3
R1=(1-e1)/(A1*e1)
R2=1/(A1*F12)
Q=sigma#*(T2^4-T1^4)/(R1+R2)
PROPRIETARY MATERIAL. © 2011 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.
13-114
13-143 Consider a 3-m × 3-m × 3-m cubical furnace. The base surface of the furnace is black and has a temperature of 550
K. The radiosities for the top and side surfaces are calculated to be 7500 W/m2 and 3200 W/m2, respectively. The net rate of
radiation heat transfer to the bottom surface is
(a) 10 kW
(b) 54 kW
(c) 61 kW
(d) 113 kW
(e) 248 kW
Answer (a) 10 kW
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES
screen.
s=3 [m]
T1=550 [K]
epsilon_1=1
J2=7500 [W/m^2]
J3=3200 [W/m^2]
sigma=5.67E-8 [W/m^2-K^4]
A1=s^2
F_12=0.2
F_13=0.8
J1=sigma*T1^4
Q_dot_1=A1*(F_12*(J1-J2)+F_13*(J1-J3))
"Some Wrong Solutions with Common Mistakes"
W1_Q_dot_1=(F_12*(J1-J2)+F_13*(J1-J3)) "Not multiplying with area"
W2_A1=6*s^2 "Using total area"
W2_Q_dot_1=W2_A1*(F_12*(J1-J2)+F_13*(J1-J3))
13-144 ….. 13-146 Design and Essay Problems
KJ
PROPRIETARY MATERIAL. © 2011 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.