Refrigerant 134a is throttled from the standard liquid state

Refrigerant- 134a is throttled from the standard liquid state at 800 kPa to a pressure of 200 kPa. Determine the temp drop during this process and the final specific volume of the refrigerant

Solution

>> Using Property Table for refrigerant R134-a,

>> At P1 = 800 kPa , Saturated Liquid State,

From Table,

h1 = Enthalpy = hf = 95.48 KJ/Kg

T1 = Temperature = 31.31 C

>> Now, As Throttling Process Occurs upto P2 = 200 kPa

As, Under Throttling, Enthalpy remains constant

=> h2 = h1 = 95.48 KJ/Kg

>> Now, Using Table or P2 = 200 kPa and h2 = 95.48 KJ/Kg

T2 = - 10.09 C

and, as hf = 38.41 KJ/Kg

and, hg = 244.50 KJ/Kg

As, h2 = hf + x*(hg - hf)

=> 95.48 = 38.41 + x*(244.50 - 38.41)

=> x = 0.2769

>> Now, Vg = 0.09995 m³/Kg

and, Vf = 0.753*10^-3 m³/Kg

>> So, V2 = Vf + x*(Vg - Vf)

=> V2 = 0.753*10^-3 + 0.2769*(0.9995 - 0.753*10^-3 )

=> V2 = Final Specific Volume of the Refrigerant = 0.2773 m³/Kg ........ANSWER....

and, temperature Drop during this process = T1 - T2 = 31.31 - (- 10.09) = 41.40 C ......ANSWER....