Exercicio resolvido de transferencia de calor
2
PROBLEM 12.8 KNOWN: Radiation from a diffuse radiant source A 1 with intensity I 1 = 1.2 × 10 5 W/m 2 ⋅sr is incident on a mirror A m , which reflects radiation onto the radiation detector A 2 . FIND: (a) Radiant power incident on A m due to emission from the source, A 1 , q 1→m (mW), (b) Intensity of radiant power leaving the perfectly reflecting, diffuse mirror A m , I m (W/m 2 ⋅sr), and (c) Radiant power incident on the detector A 2 due to the reflected radiation leaving A m , q m→2 (μW), (d) Plot the radiant power q m→2 as a function of the lateral separation distance y o for the range 0 ≤ y o ≤ 0.2 m; explain features of the resulting curve. SCHEMATIC: ASSUMPTIONS: (1) Surface A 1 emits diffusely, (2) Surface A m does not emit, but reflects perfectly and diffusely, and (3) Surface areas are much smaller than the square of their separation distances. ANALYSIS: (a) The radiant power leaving A 1 that is incident on A m is q I A cos 1 m 1 1 m-1 → = ⋅ ⋅ ⋅ 1 θ ω Δ where ω m-1 is the solid angle A m subtends with respect to A 1 , Eq. 12.7, Δω θ m-1 n 2 m m o 2 o 2 2 2 2 dA r A cos x y m cos 45 0.1 m sr ≡ = + = × ⋅ ° + = × − − 2 10 01 7 07 10 4 2 3 . . with θ θ θ m 1 and = °− = ° 90 45 1 , q W/m sr 1 10 m cos 45 7.07 10 sr 60 mW 1 m 2 -4 2 -3 → = × ⋅ × × × °× × = 12 10 5 . < (b) The intensity of radiation leaving A m , after perfect and diffuse reflection, is I q A W 2 10 m W/m sr m 1 m m -4 2 2 = = × × × = ⋅ → − / / . b g π π 60 10 955 3 (c) The radiant power leaving A m due to reflected radiation leaving A m is q q I A cos m 2 2 m m m 2 m → − = = ⋅ ⋅ ⋅ θ ω Δ where Δω 2-m is the solid angle that A 2 subtends with respect to A m , Eq. 12.7, Continued …
description
PROBLEM 12.8 do livro de transferência de Calor e Massa.Incropera, Frank P.
Transcript of Exercicio resolvido de transferencia de calor
-
PROBLEM 12.8 KNOWN: Radiation from a diffuse radiant source A1 with intensity I1 = 1.2 105 W/m2sr is incident on a mirror Am, which reflects radiation onto the radiation detector A2. FIND: (a) Radiant power incident on Am due to emission from the source, A1, q1m (mW), (b) Intensity of radiant power leaving the perfectly reflecting, diffuse mirror Am, Im (W/m
2sr), and (c) Radiant power incident on the detector A2 due to the reflected radiation leaving Am, qm2 (W), (d) Plot the radiant power qm2 as a function of the lateral separation distance yo for the range 0 yo 0.2 m; explain features of the resulting curve. SCHEMATIC:
ASSUMPTIONS: (1) Surface A1 emits diffusely, (2) Surface Am does not emit, but reflects perfectly and diffusely, and (3) Surface areas are much smaller than the square of their separation distances. ANALYSIS: (a) The radiant power leaving A1 that is incident on Am is q I A cos1 m 1 1 m-1 = 1 where m-1 is the solid angle Am subtends with respect to A1, Eq. 12.7, m-1 n2 m m
o2
o2
2
2 2dA
r
A cos
x y
m cos 45
0.1 m sr = + =
+
= 2 10
017 07 10
4
23
..
with m 1 and = = 90 451 , q W / m sr 1 10 m cos 45 7.07 10 sr 60 mW1 m
2 -4 2 -3 = =12 105. < (b) The intensity of radiation leaving Am, after perfect and diffuse reflection, is
I q A W
2 10 m W / m srm 1 m m -4 2
2= = =
/ / .b g 60 10 955
3
(c) The radiant power leaving Am due to reflected radiation leaving Am is q q I A cos m 2 2 m m m 2 m = = where 2-m is the solid angle that A2 subtends with respect to Am, Eq. 12.7,
Continued
-
PROBLEM 12.8 (Cont.)
2 24
231 10
01354 10
= +
= +
= m n2 2 2o o o
2
2
2 2dA
r
A cos
L x y
m cos 45
0.1 m sr
b g ..
with 2 = 90 - m 2 -4 2 -3m 2 2q q 95.5 W/m sr 2 10 m cos 45 3.54 10 sr 47.8 W = = = < (d) Using the foregoing equations in the IHT workspace, q2 is calculated and plotted as a function of yo for the range 0 yo 0.2 m. From the relations, note that q2 is dependent upon the geometric arrangement of the surfaces in the following manner. For small values of yo, that is, when 1 0, the cos 1 term is at a maximum, near unity. But, the solid angles m-1 and 2-m are very small. As yo increases, the cos 1 term doesnt diminish as much as the solid angles increase, causing q2 to increase. A maximum in the power is reached as the cos 1 term decreases and the solid angles increase. The maximum radiant power occurs when yo = 0.058 m which corresponds to 1 = 30.
Emitted power from A1 reflected from Am onto A2
0 0.05 0.1 0.15 0.2
yo (m)
0
20
40
60
80
100
q2 (u
W)
