Thermodynamics Question 5 Follow Link Below for question ple

Thermodynamics

Question 5: Follow Link Below for question please.

http://imgur.com/EoizJtN

Solution

initial pressure is 60 kPa

initial volume = 33.608/100=0.33608 kg/m^3

initial temperature = -36.95 C

initial internal energy= [(209.13-3.79)/(0.311-0.710)]x[0.710-0.33608] + 3.79=196.22 kJ/kg

pressure can not remain at 60 kPa as at that pressure this becomes supersaturated refrigerant.

only at 140 kpa and 20C temp, the refrigerant will have half of its volume.

final temp = 20 C

internal energy= 248.2 kj/kg

change in energy = (248.2 - 196.22 ) x 100= 5200 KJ