SHORT ANSWER: No GOAL write-up needed. But clearly show SKETCH, PROPERTY TABLE,

ID: 1861059 • Letter: S

Question

SHORT ANSWER: No GOAL write-up needed.

But clearly show SKETCH, PROPERTY TABLE, MASS & ENERGY BALANCE EQUATIONS FOR EACH.

a) A rigid tank containing 20 kg of Refrigerant 134a as saturated vapor at 1.0 bars and -26.43 oC is

heated until the temperature reaches 20 oC. The final volume of the R-134a is __________ m3

b) A rigid tank with a volume of 5.0 m3 contains Refrigerant 134a at -16 oC and mixture quality of 25

%. The total mass of the R-134a is ______________ kg.

c) A rigid tank containing 20 kg of Refrigerant R-134a at 6.0 bars and 140 oC is cooled causing the

pressure to fall to 4.0 bars. The final temperature is _____ oC.

d) Twenty kilograms (20 kg) of Refrigerant R-134a contained in a piston-cylinder device initially at

6.0 bars and 140 oC is cooled at constant pressure to the saturated liquid state. The specific

internal energy of the saturated liquid is ____________ kJ/kg.

a) At 1 bar for sat. vapor (quality = 1) we get v = 0.193 m^3/kg

Volume V = mv = 20*0.193 = 3.86 m^3

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b) At -16 deg C and quality = 0.25 we get v = 0.0319 m^3/kg

Mass m = V/v = 5 / 0.0319 = 156.74 kg

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c) At 6 bars, 140 deg , we get v = 0.054 m^3/kg

Density = 1/0.054 = 18.519 kg/m^3

At 4 bars and density = 18.519 kg/m^3, we get T = 19.2 deg C

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d) At 6 bars and quality = 0 (sat.liquid) we get u = 81000 J/kg

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