Answer:
- Work done is 2.39 kJ
- heat transfer is 20.23 kJ/kg
Explanation:
Given the data in the question;
First we obtain for specific volumes and specific enthalpy from "Table Properties Refrigerant 134a;
Specific Volume v₁ = 0.02547 m³/kg
Specific enthalpy u₁ = 243.78 kJ/kg
Specific Volume V₂ = 0.02846 m³/kg
Specific enthalpy u₂ = 261.62 kJ/kg
p = 8 bar = 800 kPa
Any changes in kinetic and potential energy are negligible.
So we determine the work done by using the equation at constant pressure
]Work done W = p( v₂ - v₁ )
we substitute
W = 800 kPa( 0.02846 m³/kg - 0.02547 m³/kg )
W = 800 kPa( 0.00299 m³/kg )
W = 2.39 kJ
Therefore, Work done is 2.39 kJ
Heat transfer;
using equation at constant pressure
Heat transfer Q = W + ( u₂ - u₁ )
so we substitute
Q = 2.392 kJ + ( 261.62 kJ/kg - 243.78 kJ/kg )
Q = 2.392 kJ + 17.84 kJ/kg )
Q = 20.23 kJ/kg
Therefore, heat transfer is 20.23 kJ/kg
