Question

Problem 3: Refrigerant-134a is throttled from 10 bar and 36 o C to 240 kPa. Heat is lost from the refrigerant in the amount of 1 kJ/kg to the surroundings. Determine the exit temperature of the refrigerant. Assume steady state and ignore kinetic and potential energy effects.

Answer 1

`T_(out) = -5.38\,^(\textdegree)C`

Step-by-step explanation:

By applying the First Law of Thermodynamics, the model for the throttle valve is:

`-q_(out) + h_(in) - h_(out) = 0`

The specific enthalpy at outlet is:

`h_(out) = -q_(out)+h_(in)`

The state of the refrigerant 134a at inlet is a subcooled liquid. Then, specific enthalpy is:

`h_(in) \approx 137.34\,(kJ)/(kg)`

The specific enthalpy at outlet is:

`h_(out) = -1\,(kJ)/(kg)+137.34\,(kJ)/(kg)`

`h_(out) = 136.34\,(kJ)/(kg)`

The substance at outlet is a liquid-vapor mixture. Then, exit temperature is:

`T_(out) = -5.38\,^(\textdegree)C`