Respuesta :
Answer:
The the quality of the refrigerant at the exit of the expansion valve is 0.179.
Explanation:
Given that,
Initial pressure = 10 bar
Temperature = 22°C
Final pressure = 2.0 bar
We using the value of h
[tex]h = 293.4\ kJ/kg[/tex]
The refrigerant during expansion undergoes a throttling process
Therefore, [tex]h_{1}=h_{2}[/tex]
We need to calculate the quality of the refrigerant at the exit of the expansion valve
At 2.0 bar,
The property of ammonia
[tex]h_{f}=47.8 kJ/kg[/tex]
[tex]h_{g}=1417.7 kJ/kg[/tex]
Using formula
[tex]h_{2}=h_{f}+x(h_{g}-h_{f})[/tex]
Put the value into the formula
[tex]293.4 =47.8+x(1417.7-47.8)[/tex]
[tex]x=\dfrac{293.4-47.8}{1417.7-47.8}[/tex]
[tex]x=0.179[/tex]
Hence, The the quality of the refrigerant at the exit of the expansion valve is 0.179.
The value of quality of the refrigerant at the exit of the expansion valve is 0.179.
What is a throttling process?
Throttling process is the process in which the high pressure fliud is transformed into the low pressure fluid using the throttling valve.
Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 22oC and exits at 2.0 bar.
The refrigerant undergoes a throttling process.
For the ammonia, the value of enthalpy of saturated liquid is,
[tex]h_f=47.8\rm kJ/kg[/tex]
The value of enthalpy of saturated vapor is,
[tex]h_g=1417.7\rm kJ/kg[/tex]
For the enthalpy in two phase regions,
[tex]h=u+pv\\h=(1-x)h_f+xh_g\\h=h_f+x(h_g-h_f)[/tex]
Put the values given in the problem as,
[tex]293.4=47.8+x(1417.7-47.8)\\x=0.179[/tex]
Thus, the quality of the refrigerant at the exit of the expansion valve is 0.179.
Learn more about the throttling process here;
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