Thermodynamics Question 4 Follow Link Below for question ple

Thermodynamics

Question 4: Follow Link Below for question please.

http://imgur.com/jycAnV4

Round all answers to two decimal places. A 17 L rigid container contains 10-kg of R134a at 240 kPa. Determine the temperature and total enthalpy in the container. The container is now heated until the pressure is 750 kPa. Determine the temperature and total enthalpy when the heating is completed.

Solution

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

The specific volume is = V/m

                                         = 0.017m³/10 Kg

                                       = 0.0017 m³/ Kg

Use R-134a properties table (It is available in the heat transfer book)

Saturate temperature at 240 Kpa= -5.37 deg C

   X₁=V₁-V_f/V_fg

  

X₁ =(0.0017-0.0007618)/(0.0834-0.0007618)=0.0113

h₁=h_f+X₁h_fg

h₁=42.95+0.0113x201.14=45.22

Total enthalpy H=mh₁ = 10x45.22=452.2 KJ

Final state is also saturated mixture    

Saturate temperature at 750 Kpa= 28.71 deg C

   X₂=V₂-V_f/V_fg

X₂ =(0.0017-0.0008358)/(0.0292-0.0008358)=0.0304

h₂=h_f+X₂h_fg

h₂=86.78+0.0304x175.07=92.10

Total enthalpy H=mh₂ =92.10x10=921KJ