Thermodynamics Question 4 Follow Link Below for question ple

Thermodynamics

Question 4: Follow Link Below for question please.

http://imgur.com/pZ67f58

Solution

a)

This is a constant volume process because the container is rigid.

Specific volume is

V₁=V₂=V/m =0.012/10 =0.0012 m³/kg

From Table A-12 of R134a ,we obtain T₁ at 120 Kpa

T₁=-22.36⁰C

V_f=0.0007323 m³/kg

V_g=0.1614

and

X₁=(V₁-V_f)/(V_g-V_f) = (0.0012-0.0007323)/(0.1614-0.0007323) =0.002911

h₁=h_f+x₁hf_g=21.32+0.002911*212.54=21.94 KJ/Kg

The Total enthalpy is

H₁=mh₁=10*21.94 =219.4 KJ

b)

The Final state is also saturated mixture ,at 650Kpa

T₂=21.58^oC

V_f=0.0008196 m³/kg

V_g=0.0341

X₂=(0.0012-0.0008196)/(0.0341-0.0008196) =0.01143

h₂=h_f+x₂h_fg =79.48+0.01143*179.71 =81.53 KJ/kg

H₂=mh₂=10*81.53 =815.3 KJ