Thermodynamics HW Solutions 604

# Thermodynamics HW Solutions 604 - Chapter 7 External Forced...

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Chapter 7 External Forced Convection 7-70 Air is cooled by an evaporating refrigerator. The refrigeration capacity and the pressure drop across the tube bank are to be determined. Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the tubes is equal to the temperature of refrigerant. Properties The exit temperature of air, and thus the mean temperature, is not known. We evaluate the air properties at the assumed mean temperature of -5 ° C (will be checked later) and 1 atm (Table A-15): k = 0.02326 W/m-K ρ = 1.316 kg/m 3 C p =1.006 kJ/kg-K Pr = 0.7375 μ = 1.705 × 10 -5 kg/m-s Pr s = Pr @ Ts = 0.7408 Also, the density of air at the inlet temperature of 0 ° C (for use in the mass flow rate calculation at the inlet) is ρ i = 1.292 kg/m 3 . Analysis It is given that D = 0.008 m, S L = S T = 0.015 m, and V = 4 m/s. Then the maximum velocity and the Reynolds number S L S T V=4 m/s T i =0 ° C T s =-20 ° C D based on the maximum velocity become m/s 571 . 8 m/s)
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## This note was uploaded on 01/22/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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