Answer:
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.
