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Fundamentals of Heat and Mass Transfer [Frank P.Incropera - David P.DeWitt] Solution Manual - CH12

Fundamentals of Heat and Mass Transfer [Frank P.Incropera - David P.DeWitt] Solution Manual - CH12

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PROBLEM 12.1 KNOWN: Rate at which radiation is intercepted by each of three surfaces (see (Example 12.1). FIND: Irradiation, G[W/m 2 ], at each of the three surfaces. SCHEMATIC: ANALYSIS: The irradiation at a surface is the rate at which radiation is incident on a surface per unit area of the surface. The irradiation at surface j due to emission from surface 1 is 1 j j j q G . A - = With A 1 = A 2 = A 3 = A 4 = 10 -3 m 2 and the incident radiation rates q 1-j from the results of Example 12.1, find 3 2 2 3 2 12.1 10 W G 12.1W/m 10 m - - × = = < 3 2 3 3 2 28.0 10 W G 28.0 W / m 10 m - - × = = < 3 2 4 3 2 19.8 10 W G 19.8 W / m . 10 m - - × = = < COMMENTS: The irradiation could also be computed from Eq. 12.15, which, for the present situation, takes the form j 1 j 1 j G I cos q w - = where I 1 = I = 7000 W/m 2 sr and ϖ 1-j is the solid angle subtended by surface 1 with respect to j. For example, 2 1 2 1 2 G I cos q w - = 2 2 G 7000 W / m sr = × ( 29 3 2 2 10 m cos60 cos 30 0.5m - × ° ° 2 2 G 12.1W/m . = Note that, since A 1 is a diffuse radiator, the intensity I is independent of direction.
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PROBLEM 12.2 KNOWN: A diffuse surface of area A 1 = 10 -4 m 2 emits diffusely with total emissive power E = 5 × 10 4 W / m 2 . FIND: (a) Rate this emission is intercepted by small surface of area A 2 = 5 × 10 -4 m 2 at a prescribed location and orientation, (b) Irradiation G 2 on A 2 , and (c) Compute and plot G 2 as a function of the separation distance r 2 for the range 0.25 r 2 1.0 m for zenith angles θ 2 = 0, 30 and 60 ° . SCHEMATIC: ASSUMPTIONS: (1) Surface A 1 emits diffusely, (2) A 1 may be approximated as a differential surface area and that 2 2 2 A r << 1. ANALYSIS: (a) The rate at which emission from A 1 is intercepted by A 2 follows from Eq. 12.5 written on a total rather than spectral basis. ( ) 1 2 e,1 1 1 2 1 q I , A cos d θ φ θ ω = . (1) Since the surface A 1 is diffuse, it follows from Eq. 12.13 that ( ) e,1 e,1 1 I , I E θ φ π = = . (2) The solid angle subtended by A 2 with respect to A 1 is 2 2 1 2 2 2 d A cos r ω θ . (3) Substituting Eqs. (2) and (3) into Eq. (1) with numerical values gives 1 2 2 1 2 1 1 2 2 E A cos q A cos r θ θ π = ( ) ( ) 4 2 4 2 4 2 2 5 10 W m 5 10 m cos30 10 m cos60 sr sr 0.5m π × × × = × × × (4) ( ) 2 5 2 3 3 1 2 q 15,915W m sr 5 10 m 1.732 10 sr 1.378 10 W = × × × × = × . < (b) From section 12, 2.3, the irradiation is the rate at which radiation is incident upon the surface per unit surface area, 3 2 1 2 2 4 2 2 q 1.378 10 W G 2.76 W m A 5 10 m × = = = × (5) < (c) Using the IHT workspace with the foregoing equations, the G 2 was computed as a function of the separation distance for selected zenith angles. The results are plotted below. Continued...
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PROBLEM 12.2 (Cont.) 0.2 0.4 0.6 0.8 1 Separation distance, r2 (m) 0 5 10 Irradiation, G2 (W/m^2) theta2 = 0 deg theta2 = 30 deg theta2 = 60 deg For all zenith angles, G 2 decreases with increasing separation distance r 2 . From Eq. (3), note that d ω 2-1 and, hence G 2 , vary inversely as the square of the separation distance. For any fixed separation distance, G 2 is a maximum when θ 2 = 0 ° and decreases with increasing θ 2 , proportional to cos θ 2 .
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