HW2 Solutions

HW2 Solutions - ME 364 HW 2 Solutions Problem 1: The...

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ME 364 HW 2 Solutions Problem 1: The convection heat transfer coefficient for heat transfer from an electrically heated wire to air is to be determined by measuring temperatures when steady operating conditions are reached and the electric power consumed. Assumptions 1 Steady operating conditions exist since the temperature readings do not change with time. 2 Radiation heat transfer is negligible. Analysis In steady operation, the rate of heat loss from the wire equals the rate of heat generation in the wire as a result of resistance heating. That is, 240 ° C W 330 = A) V)(3 110 ( generated = = = I E Q V & & The surface area of the wire is D =0.2 2 m 0.00880 = m) m)(1.4 002 . 0 ( ππ = = DL A s L = 1.4 Q Air, 20 ° C The Newton's law of cooling for convection heat transfer is expressed as ) ( = T T hA Q s s & Disregarding any heat transfer by radiation, the convection heat transfer coefficient is determined to be C W/m 170.5 2 ° = ° = = C ) 20 240 )( m (0.00880 W 330 ) ( 2 1 T T A Q h s & Discussion If the temperature of the surrounding surfaces is equal to the air temperature in the room, the value obtained above actually represents the combined convection and radiation heat transfer coefficient. Problem 2: An electric heater placed in a room consumes 500 W power when its surfaces are at 120 ° C. The surface temperature when the heater consumes 700 W is to be determined without and with the consideration of radiation. Assumptions 1 Steady operating conditions exist. 2 The temperature is uniform over the surface. A , ε T s q conv q rad T , h T w e W & Analysis ( a ) Neglecting radiation, the convection heat transfer coefficient is determined from () C W/m 0 2 C 20 120 ) m 25 . 0 ( W 500 ) ( 2 2 ° = ° = = T T A Q h s & The surface temperature when the heater consumes 700 W is C 160 ° = ° + ° = + = ) m 25 . 0 ( C) W/m 0 2 ( W 700 C 20 2 2 hA Q T T s & ( b ) Considering radiation, the convection heat transfer coefficient is determined from [] C W/m 58 . 12 C 20 120 ) m 25 . 0 ( K) 283 ( K) 393 ( ) K W/m 10 67 . 5 )( m 5 (0.75)(0.2 - W 500 ) ( ) ( 2 2 4 4 4 2 8 2 4 surr 4 ° = ° × = = T T A T T A Q h s s σε & Then the surface temperature becomes
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() [] C 152.9 ° = = × + = + = K 9 . 425 K) 283 ( ) 10 67 . 5 )( 5 (0.75)(0.2 ) 293 )( 25 . 0 )( 58 . 12 ( 700 ) ( 4 4 8 4 surr 4 s s s s s T T T T T A T T hA Q σε & Discussion Neglecting radiation changed T s by more than 7 ° C, so assumption is not correct in this case. Problem 3: We consider a thin spherical shell element of thickness Δ r in a sphere (see Fig. 2-17 in the text). . The density of the sphere is ρ , the specific heat is c, and the length is L. The area of the sphere normal to the direction of heat transfer at any location is where r is the value of the radius at that location .
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This note was uploaded on 03/27/2008 for the course ME 364 taught by Professor Rothamer during the Spring '08 term at Wisconsin.

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HW2 Solutions - ME 364 HW 2 Solutions Problem 1: The...

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