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Unformatted text preview: 3-37 Equations
Set 6: 3-37 A piston-cylinder assembly contains 0.04 lbm of R134a. The refrigerant is compressed from an initial state where p1 =10 psia and T1 =20 F to a final state where P2 =160 psia. During the processs, the pressure the pressure and specific volume are related by p*V=constant. Determine the work in Btu. R$ = `R134a' m = 0.04 [lbm ] P1 = 10 [psia] T1 = 20 [F] mass of R134a initial pressure initial temperature from Table A-12E (1) (2) (3) (4) (5) (6) (7) (8) (9) v1 = 4.9297 ft3 /lbm P1 v1 = C this equation determines the constant, C, in psia-ft3 /lbm P2 = 160 [psia] P2 v2 = C knowing P2 and C, v2 is now known table A-12E - note how the temperature increases during the compression process T2 = T (R$, P = P2 , v = v2 ) see the property plots W = m C ln v2 v1 0.185049705 Btu psia ft3 integrate P dV to obtain this expression (10) $Tabstops 0.2 3.0 in Solution
C = 49.3 psia-ft3 /lbm m = 0.04 [lbm ] R$ = `R134a' W = -1.012 [Btu] Arrays
Row 1 2 Pi [psia] 10 160 Ti [F] 20 122 vi ft3 /lbm 4.93 0.3081 P-v: R134a T-v: R134a ...
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- Spring '07
- Thermodynamics, psia, 0.185049705 Btu, psia-ft3 /lbm, R134a initial pressure