Wednesday, May 10, 2006
HW #8, P#1 - Shaft Work for a Compressor vs. a Pump - 6 pts
Saturated refrigerant-134a vapor at 15 psia is compressed reversibly in an adiabatic compressor to 80 psia. Determine the work input to the compressor. What would your answer be if the refrigerant were first condensed at constant pressure before it was compressed?
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9 comments:
so if its reversible that means deltaS=0 right? is that how i find H2? sorry im kinda confused between several different properties right now...
Yayiye 9:48 PM
Reversible means Sgen = 0
Adiabatic means Q = 0
deltaS = INT{dQ/T} + Sgen
So if a process is both adiabatic and reversible:
deltaS = 0 and the process is isentropic.
In this case, S2 = S1. Then, knowing both S2 and P2 you can interpolate or use the NIST Webbook to determine H2.
I hope this clears up some of the confusion. Best of luck.
I got the first part, but when it is condensed at constant pressure, does this just mean that there is no H in the problem? So no state change, so it's just some kind of boundary work???
Tyler 11:26
In this problem, we are comparing the shaft work required to raise the pressure from 15 to 80 psia on a sat vapor (it becomes a superheated vapor) to the shaft work required to raise the pressure from 15 to 80 psia on a sat LIQUID (it becomes a subcooled liquid).
Use Wsh = -V^ (Pout - Pin) for the incompressible liquid, but DO NOT do this for the vapor. It is not incompressible.
Graham 4:40 PM
Yes, that is a typo. I fixed it on the HW assignment page. Thank you for posting this so everyone sees it.
I am still not getting part B correct. I gather that we are to use the V given in the tables for Sat Liq at 15psia and then times by 80-15, but that doesn't work. I don't think I understand where this work is comming from.
from Lan
test
From Lan
katie, did you check the the unit conversion of 1btu=5.4 psia.ft3
Katie 7:03 PM
You did this correctly, but I suspect the units are not correct. You cannot multiply ft^3/lbm by lbf/ft^2 and then add them to Btu/lbm.
This comes from the fact that Wsh = INT{Vhat dP}. For an incompressible liquid, Vhat ~ constant, so it pops out of the integral. INT{dP} = deltaP. Therefore, Wsh = Vhat * deltaP.
I hope this clears things up and helps you reach the right answer.
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