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Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 20oC and exits at 3.0 bar. The refrigerant undergoes a throttling process. Determine the temperature, in oC, and the quality of the refrigerant at the exit of the expansion valve. Step 1 Determine the temperature of the refrigerant at the exit, in oC.

Respuesta :

Answer:

[tex]T_{2}[/tex] = -9.24 °C

x = 0.1057

Explanation:

The tables used in this answer and explanation come from Fundamentals of Engineering Thermodynamics 9th Edition.

Using Table A-14: Properties of Saturated Ammonia (Liquid-Vapor): Pressure Table and the given [tex]P_{2}[/tex], [tex]T_{2}[/tex] can be determined by finding the temperature that corresponds with [tex]P_{2}[/tex] on the table. In this case, [tex]T_{2}[/tex] = -9.24 °C.

The quality of the refrigerant can be determined by using data from the same table and [tex]h_{2} =274.26[/tex] kJ/kg.

Necessary data (P=3bar):

[tex]h_{f}=137.42[/tex] kJ/kg

[tex]h_{g}=1431.47[/tex] kJ/kg

The formula to calculate quality is [tex]h_{2} =h_{f}+x(h_{g}-h_{f})[/tex].

Rearranging for x:

[tex]x=\frac{h_{2}-h_{f} }{h_{g}-h_{f} }= \frac{274.26-137.42}{1431.47-137.42}=0.1057[/tex]