For the following question I was able to find the heat transfer and the entropy but I am having trouble finding the work done. When i solved delta E = Qnet - Wnet I got Win = Wout but Im not sure that is correct. I think that i need to solve for the final temp and pressure but again I am stuck there. P1V1/T=P2V2/T2 wont work and since it is air and not an ideal gas PV=RT won't work either. Does anone have an idea?
3.1 One kilogram of air in a piston-cylinder device undegoes a thermodynamic cycle composed of the following reversible processes: 1�¨2 air at 2.4 bar and 250 K is expanded
isothermally to 1.2 bar; 2�¨3 air is then compressed adiabatically back to its initial volume; 3�¨1 air is finally cooled at constant volume back to its initial state. Account for variable
specific heats.
c) Calculated the work, heat transfer, and entropy change for process 2�¨3 (57.3, 0, 0)
I don't understand the meaning of "2 air, "1 air, "3 air etc
You should be able to treat air as a perfect gas, even if the specific heat is not constant.
The work done will be the sum of the integral of P dV for steps 1 and 2. Use the gas law to make P a function of V.
For the cycle, Qin - Qout = Wout
those should be arrows. I think when i used the cut and paste function it messed up!
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