Thermodynamics HW Solutions 625

Thermodynamics HW Solutions 625 - 10 702 1 m(0.22 m/s 60...

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Chapter 7 External Forced Convection 7-95 The heat generated by four transistors mounted on a thin vertical plate is dissipated by air blown over the plate on both surfaces. The temperature of the aluminum plate is to be determined. Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Re cr = 5 × 10 5 . 3 Radiation effects are negligible. 4 The entire plate is nearly isothermal. 5 The exposed surface area of the transistor is taken to be equal to its base area. 6 Air is an ideal gas with constant properties. 7 The pressure of air is 1 atm. Properties Assuming a film temperature of 40 ° C, the properties of air are evaluated to be (Table A-15) V = 250 m/min T = 20 ° C 7255 . 0 Pr /s m 10 702 . 1 C W/m. 02662 . 0 2 5 - = × = υ ° = k 12 W T s Analysis The Reynolds number in this case is L = 22 cm [] 4 2 5 10 386 . 5 /s m
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Unformatted text preview: 10 702 . 1 m) (0.22 m/s ) 60 / 250 ( Re × = × = υ = − ∞ L V L which is smaller than the critical Reynolds number. Thus we have laminar flow. Using the proper relation for Nusselt number, heat transfer coefficient is determined to be C . W/m 75 . 16 ) 5 . 138 ( m 22 . C W/m. 02662 . 5 . 138 ) 7255 . ( ) 10 386 . 5 ( 664 . Pr Re 664 . 2 3 / 1 5 . 4 3 / 1 5 . ° = ° = = = × = = = Nu L k h k hL Nu L The temperature of aluminum plate then becomes C 50.0 ° = ° × + ° = + = ⎯→ ⎯ − = ∞ ∞ ] ) m 22 . ( 2 )[ C . W/m 75 . 16 ( W ) 12 4 ( C 20 ) ( 2 2 s s s s hA Q T T T T hA Q & & Discussion In reality, the heat transfer coefficient will be higher since the transistors will cause turbulence in the air. 7-80...
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