Assign4-1 - Problem Statement: 46 Refrigerant 134a at...

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Problem Statement: 46 Refrigerant 134a at 600.000 kPa and 150.00 0.30 is cooled until its temperature is -30.00 0.10 Schematic: State 1 State 2 Assumptions/Approximations: The thermal energy stored in the cylinder itself is negligible Physical Laws: [Since KE = PE = 0] Property (x,y) = Property(lower) + (Property (higher) - Property(lower))*(Property(given values) - Property(lower))/(Property(higher) - (Property(lower)) Property (x,y) = Property(lower) + x*(Property (higher) - Property(lower)) Properties/Sources: Conversions: 1.00 kJ = 1.00 Use Table A-13, then A-11 State P (kPa) u (kJ/kg) x (Quality) 1 150.00 600.000 0.055522 357.96 2 -30.00 -30.00 84.430 84.430 0.000720 0.225800 0.018507 12.59 213.11 28.44 7.90% Given Table Calculated - See equations above o C is contained in a spring-loaded, piston-cylinder device with an initial volum o C and its volume is m 3 . Determine the heat transferred to and the work produced by the refrigera A*P ATM A*P ATM F S A*P WATER1 A*P WATER2 m PISTON *g m PISTON *g The compression or expansion process is quasi-equilibrium. E IN - E OUT = E SYSTEM [E IN - E OUT = Net energy transfer by heat, work and mass; E SYSTEM = Change in internal, kinetic, potential and other ener
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This note was uploaded on 06/20/2009 for the course THERMO n/a taught by Professor N/a during the Spring '09 term at Abraham Baldwin Agricultural College.

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Assign4-1 - Problem Statement: 46 Refrigerant 134a at...

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