Heat Chap12-041

Heat Chap12-041 - Chapter 12 Radiation Heat Transfer 12-41...

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Chapter 12 Radiation Heat Transfer 12-41 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. 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 12-1), 75 . 0 m 0.20 m 15 . 0 = = = = L r R R i j i 7778 . 3 0.75 75 . 0 1 1 1 2 2 2 2 = + = + + = i j R R S 2864 . 0 75 . 0 75 . 0 4 7778 . 3 7778 . 3 2 1 4 2 1 2 / 1 2 2 2 / 1 2 2 12 = - - = - - = = i j ij R R S S F F (It can also be determined from Fig. 12-7). Then the initial rate of radiation heat transfer from the coal bricks to the stakes becomes W 1674 = - × = - = - ] ) K 278 ( ) K 1100 )[( K W/m 10 67 . 5 ]( 4 / m) 3 . 0 ( )[ 2864 . 0 ( ) ( 4 4 4 2 8 2 4 2 4 1 1 12 12 π σ T T A F Q 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 room can be determined from 1 23 13 12 2 1 1 1 1 - + + - = R R R E E Q b b where W / m K K W / m W / m K K W / m 4 2 2 4 2 2 E T E T b b 1 1 8 4 4 2 2 8 4 4 67 10 1100 83 015 67 10 18 273 407 = = × = = = × + = - - σ σ (5. . )( ) , . )( ) and A A 1 2 2 0 3 007069 = = = π ( . ) . m 4 m 2 2 - 2 13 1 23 13 2 - 2 12 1 12 m 82 . 19 ) 2864 . 0 1 )( m 07069 . 0 ( 1 1 m 39 . 49 ) 2864 . 0 )( m 07069 . 0 ( 1 1 = - = = = = = = F A R R F A R Substituting, W   3757 = + - = - 1 2 - 2 - 2 12 ) m 82 . 19 ( 2 1 m 39 . 49 1 W/m ) 407 015 , 83 ( Q 12-31 Steaks, T 2 = 278 K, ε 2 = 1 Coal bricks, T 1 = 1100 K, ε 1 = 1 0.20 m
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Chapter 12 Radiation Heat Transfer 12-42E A room is heated by lectric 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 2 1 23 13 12 2 1 1 1 1 R R R R E E Q b b +
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This note was uploaded on 08/25/2009 for the course AET AET432 taught by Professor Rajadas during the Spring '06 term at ASU.

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Heat Chap12-041 - Chapter 12 Radiation Heat Transfer 12-41...

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