Learning undergraduate engineering thermodynamics might be less painful with a blog. I hope that students, faculty and interested observers will share their thoughts on the laws of thermodynamics, phase and chemical equilibrium and many related topics.
Friday, April 14, 2006
HW #7 - P #10 - Compression of an Ideal Gas in an Open and a Closed System - 6 pts
Helium gas is compressed from 90 kPa and 30°C to 450 kPa in a reversible, adiabatic process. Determine the final temperature and the work done, assuming the process takes place ... a.) in a piston and cylinder device b.)in a steady-flow compressor
6 comments:
Anonymous
said...
First, shouldn't the work be negative because were doing work on the system to compress the helium? And second, how is part b different. Both parts say to use the 1st law for closed systems, which with assumptions reduces to w=-deltaU. I used the integral of Cv over the temperature range for part a, but whats different about part b that we can't use that again?
I'm not getting anywhere with this problem. I used PV=RT to determine T2 and then integrated Cv over that range, but that is not giving me the right W value. I am doing a similar thing for b, but finding delH by integrating Cp and my answers are not even close. Help
Katie 3:33 PM I am not sure how you used the IG EOS to determine T2, but I think that is the problem. Since you only know P2, you cannot use the IG EOS to determine T2.
There are several ways to determine T2, but the easiest is to use the 2nd to last eqn on page 185 of the Thermo-CD notebook. This eqn applies because the compression is isentropic in both parts of the problem. Gamma = Cp/Cv.
After you have the correct value of T2, your procedure will give the correct answers.
6 comments:
First, shouldn't the work be negative because were doing work on the system to compress the helium? And second, how is part b different. Both parts say to use the 1st law for closed systems, which with assumptions reduces to w=-deltaU. I used the integral of Cv over the temperature range for part a, but whats different about part b that we can't use that again?
Alias 7:20 PM
Yes, the answera are approximately (a) -850 kJ/kg and (b) -1400 kJ/kg.
Part (a) is a closed system with boundary and part (b) is an open system with shaft.
The hint on part (b) has a typo. It is an open system. Sorry about that :)
In part (b) you end up using integral{Cp dT} to get deltaH for the 1st Law for open systems.
I'm not getting anywhere with this problem. I used PV=RT to determine T2 and then integrated Cv over that range, but that is not giving me the right W value. I am doing a similar thing for b, but finding delH by integrating Cp and my answers are not even close. Help
Katie 3:33 PM
I am not sure how you used the IG EOS to determine T2, but I think that is the problem. Since you only know P2, you cannot use the IG EOS to determine T2.
There are several ways to determine T2, but the easiest is to use the 2nd to last eqn on page 185 of the Thermo-CD notebook. This eqn applies because the compression is isentropic in both parts of the problem. Gamma = Cp/Cv.
After you have the correct value of T2, your procedure will give the correct answers.
katie - you can use the ideal gas isentropic equation to get T2 = T1 * (P2/P1)^(g/g-1) where g i s gamma.
Once you have T as Dr. B said you are right about integraiting Cp and Cv.
Tyler 6:45 PM
Correct.
Post a Comment