Refrigerant 134a vapor in a piston-cylinder assembly undergoes a process at constant pressure from an initial state at 8 bar and 50°C to a final state at which the refrigerant is saturated vapor. For the refrigerant, determine the work and heat transfer, per unit mass, each in kJ/kg. Any changes in kinetic and potential energy are negligible.

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nuhulawal20 nuhulawal20 · Answer 1 · 2021-04-25T12:55:01+02:00

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