Problem 13‐28 from Heat Transfer, 4th Edition by Cengel & Ghajar The Problem 13‐28 statement states, on p. 781, that ”A solid sphere of 1 m diameter … is kept in an evacuated, long, equilateral triangular enclosure whose sides are 2 m long.” That (“long”) implies “infinite,” channel‐like equilateral triangular duct (of infinite Area A2) with a ball inside (of finite area A1=πD2., thus view factors are, F12=1, F21=0, and F22=1. Furthermore, the Fig. P13‐28 implies (thus unfortunately confusing) that the triangular enclosure’s cross‐section is “touching the sphere” that would require √3 side length, not arbitrarily given 2 m long, although any length bigger than √3 is valid, but not touching with all sides. The solution (attached below) calculates the enclosure area as being equilateral tetrahedron of three sides and illustrations with the three, 2 m sides are touching 1 m diameter ball, the latter being impossible, see above, and also tetrahedron has four equilateral triangles, one base (if so considered) and three additional towards its apex, see http://en.wikipedia.org/wiki/Tetrahedron for more details. If that is the case (although conflicting with the problem statement, see above), than the touching tetrahedron’s edge length should be √6=2.45 m, thus would not fit in a tetrahedron with 2 m triangular length (and cross‐section cut would be quite different). In addition the enclosure area would be 4 times, not 3 times, the equilateral triangle area, i.e., 4
√
√3. It appears that the P13‐28 solution is wrong for either case scenario. As stated in the Problem statement, see above, the solution, using the same given procedure, should simply be: 0andε anyvalue, since the result does not depend of ε . 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.
T1 = 600 K
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:
2
2
A1 = πD = π (1 m) = 3.142 m
2
A2 = 3 L − D
2
2
L
2m
= 3 (2 m) 2 − (1 m) 2
= 5.196 m 2
2
2
ε1 = 0.45
2m
2m
T2 = 420 K
ε2 = ?
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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