Question

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.

Answer 1

`T_(2)` = -9.24 °C

x = 0.1057

Step-by-step 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
`P_(2)`,
`T_(2)` can be determined by finding the temperature that corresponds with
`P_(2)` on the table. In this case,
`T_(2)` = -9.24 °C.

The quality of the refrigerant can be determined by using data from the same table and
`h_(2) =274.26` kJ/kg.

Necessary data (P=3bar):

`h_(f)=137.42` kJ/kg

`h_(g)=1431.47` kJ/kg

The formula to calculate quality is
`h_(2) =h_(f)+x(h_(g)-h_(f))`.

Rearranging for x:

`x=(h_(2)-h_(f) )/(h_(g)-h_(f) )= (274.26-137.42)/(1431.47-137.42)=0.1057`