Thermodynamics HW Solutions 369

Thermodynamics HW Solutions 369 - a rate of 2.27 kg/s, and...

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Chapter 4 Transient Heat Conduction 4-100 The chilling room of a meat plant with a capacity of 350 beef carcasses is considered. The cooling load, the air flow rate, and the heat transfer area of the evaporator are to be determined. Assumptions 1 Steady operating conditions exist. 2 Specific heats of beef carcass and air are constant. Properties The density and specific heat of air at 0 ° C are given to be 1.28 kg/m 3 and 1.0 kJ/kg ⋅° C. The specific heat of beef carcass is given to be 3.14 kJ/kg ⋅° C. Analysis ( a ) The amount of beef mass that needs to be cooled per unit time is kg/s 2.27 s) 3600 s)/(12h kg/carcas 280 (350 time) ooling cooled)/(c mass beef (Total = × × = = beef m The product refrigeration load can be viewed as the energy that needs to be removed from the beef carcass as it is cooled from 35 to 16ºC at
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Unformatted text preview: a rate of 2.27 kg/s, and is determined to be Beef 35 C 280 kg Lights, 2 kW Fans, 22 kW 11kW kW 135 C 16) C)(35 kJ/kg. kg/s)(3.14 (2.27 ) ( = = = beef p beef T C m Q & & Then the total refrigeration load of the chilling room becomes kW 170 = + + + = + + + = 11 2 22 135 room chilling total, gain heat lights fan beef Q Q Q Q Q & & & & & ( b ) Heat is transferred to air at the rate determined above, and the temperature of air rises from -2.2C to 0.5C as a result. Therefore, the mass flow rate of air is kg/s 63.0 C] 2.2) ( C)[0.5 kJ/kg. (1.0 kW 170 ) ( = = = air p air air T C Q m & & Then the volume flow rate of air becomes m/s 49.2 = = = kg/m 1.28 kg/s 63.0 air air air m V & & 4-80...
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This note was uploaded on 01/19/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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