Refrigerant 134a is compressed with no heat transfer in a piston–cylinder assembly from 30 lbf/in.2 , 20F to 160 lbf/in.2 The mass of refrigerant is 0.04 lb. For the refrigerant as the system, W 0.56 Btu. Kinetic and potential energy effects are negligible. Determine the final temperature, in F

The final temperature is 26.96 F

Explanation:

Initial pressure = 30 lbf/in^2 = 30×1.01325/14.696 = 2.1 bar

Initial temperature = 20 F = (20 - 32)/1.8 + 273 = 266.33 K

From steam table, entropy at 2.1 bar = 7.111 kJ/kg.K

Initial enthalpy = 7.111×266.33 = 1893.9 kJ/kg

Final pressure = 60 lbf/in^2 = 60×1.01325/14.696 = 4.1 bar

From steam table, entropy at 4.1 bar = 6.889 kJ/kg.K

Work = 0.5 Btu = 0.5×1.05506 = 0.5908 kJ

Change in enthalpy (∆H) = work/mass

mass of refrigerant = 0.04 lb = 0.04×0.45359 = 0.0181 kg

∆H = 0.5908/0.0181 = 32.64 kJ/kg

H2 = H1 - ∆H = 1893.9 - 32.64 = 1861.26 kJ/kg

Final temperature = final enthalpy ÷ final entropy = 1861.26 kJ/kg ÷ 6.889 kJ/kg.K = 270.2 K = 1.8(270.2 - 273) + 32 = -5.04 + 32 = 26.96 F

answer:

50 vo; ts

explanation:

answer:

yes that is right have a good day