Refrigerant-134a is throttled from the saturated liquid state at 700 kPa to a pressure of 157.38 kPa. Determine the temperature drop during this process and the final specific volume of the refrigerant. The inlet enthalpy of R-134a is 88.82 kJ/kg and saturation temperature is 26.69°C.

temperature drop = 42.69°C

final specific volume=0.03523m^3/kg

Explanation:

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ...)

through prior knowledge of two other properties

We will call state 1 at the entrance of the expansion valve, and 2 at the exit.

using thermodynamic tables we first find the coolant temperature in state 1 using enthalpy and pressure

T1 (h=88.82kJ/kg, p=700kPa) = 26.69°C

an expansion valve is characterized by the fact that its process is done at constant enthalpy, therefore we use this value and the pressure to find the specific volume and temperature in state 2

T2 (h=88.82kJ/kg, p=157.38 kPa) = -16°C

V2 (h=88.82kJ/kg, p=157.38 kPa) = 0.03523m^3/kg

finally we find the temperature dropby subtracting the temperatures in state 1 and 2

ΔT=T1-T2=26.69 - (-16) = 42.69°C.