Answer:
[tex]T_{out} = -5.38\,^{\textdegree}C[/tex]
Explanation:
By applying the First Law of Thermodynamics, the model for the throttle valve is:
[tex]-q_{out} + h_{in} - h_{out} = 0[/tex]
The specific enthalpy at outlet is:
[tex]h_{out} = -q_{out}+h_{in}[/tex]
The state of the refrigerant 134a at inlet is a subcooled liquid. Then, specific enthalpy is:
[tex]h_{in} \approx 137.34\,\frac{kJ}{kg}[/tex]
The specific enthalpy at outlet is:
[tex]h_{out} = -1\,\frac{kJ}{kg}+137.34\,\frac{kJ}{kg}[/tex]
[tex]h_{out} = 136.34\,\frac{kJ}{kg}[/tex]
The substance at outlet is a liquid-vapor mixture. Then, exit temperature is:
[tex]T_{out} = -5.38\,^{\textdegree}C[/tex]
