{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–4. Sign up to view the full content.

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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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...
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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}