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Friday, April 14, 2006
HW #5 - P #5 - Temperature Drop in an Adiabatic Throttling Valve - 6 pts
Refrigerant-134a is throttled from the saturated liquid state at 700 kPa to a pressure of 160 kPa. Determine the temperature drop during this process and the final specific volume of the refrigerant.
7 comments:
Anonymous
said...
The pressure drops, therefore it should be a mixture, and by the saturation tables, I'm also getting -42C. But then if it's a mixture and all you have is pressure, how can you find the spec.vol? Something is wrong here...
Graham 3:56 PM Oh, no. You are not missing anything. My fault. I made a typo in the "Answers" on the website. DeltaT=-42 degC is correct. I apologize and I will fix it now.
Steve, because this is a throttle, the specific enthalphy doesn't change. Since it's initially a saturated liquid, you can simply look up the specific enthalpy. Then, that enthalpy will be the same after the pressure drop. So, specific enthalpy = x * specific enthalpy of the saturated vapor = (1 - x) * specific enthalpy of saturated liquid (use NIST to avoid a lot of interpolations). Then, just use that x to find the specific volume.
you can ask to determine the change of entropy. because in adiabatic process the enthalpy remains constant, from the equation that Anon wrote you can find the x. After that you can find the change in entropy by Ssatmix = x Ssatvap + (1-x) Ssatliq and then Satmix-S(at 0.8 MPa for saturated liquid)....and you have the change.
7 comments:
The pressure drops, therefore it should be a mixture, and by the saturation tables, I'm also getting -42C. But then if it's a mixture and all you have is pressure, how can you find the spec.vol? Something is wrong here...
Graham 3:56 PM
Oh, no. You are not missing anything. My fault. I made a typo in the "Answers" on the website. DeltaT=-42 degC is correct.
I apologize and I will fix it now.
Steve 8:00 PM
You are correct. See my recent reply to Graham.
Steve, because this is a throttle, the specific enthalphy doesn't change. Since it's initially a saturated liquid, you can simply look up the specific enthalpy. Then, that enthalpy will be the same after the pressure drop. So, specific enthalpy = x * specific enthalpy of the saturated vapor = (1 - x) * specific enthalpy of saturated liquid (use NIST to avoid a lot of interpolations). Then, just use that x to find the specific volume.
Anon 2:17 PM
Thank you for the explanation. Just to clarify....
Hsatmix = x Hsatvap + (1-x) Hsatliq
becomes...
x = {Hsatmix-Hsatliq} / {Hsatvap-Hsatliq}
you can ask to determine the change of entropy. because in adiabatic process the enthalpy remains constant, from the equation that Anon wrote you can find the x. After that you can find the change in entropy by Ssatmix = x Ssatvap + (1-x) Ssatliq and then
Satmix-S(at 0.8 MPa for saturated liquid)....and you have the change.
Anon 4:35 PM,
Yes, precisely. Good work.
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