A quantity of air undergoes a thermodynamic cycle consisting of three internally reversible processes in series. Assume that the air behaves as an ideal gas. This may not be a good assumption, but let's work with it here anyway.
Step 1 - 2 : Isothermal expansion at 250 K from 4.75 bar to 1.0 bar.
Step 2 - 3 : Adiabatic compression to 4.75 bar.
Step 3 - 1 : Isobaric cooling.
a.) Sketch the cycle on a PV diagram.
b.) Sketch the cycle on a TS diagram.
c.) Determine T3 in Kelvin
d.) If the cycle is a power cycle, determine its thermal efficiency. If the cycle is a refrigeration cycle, determine its COP.
6 comments:
For W23, I want to say that it is equivalent to the change in internal energy (U2-U3). Is this correct? Factored in with the rest of the W's, I obtain a COP that is ~3.96, not 3.9.
(Ofcourse I am doing this by adding up W's, not defining W=Qh-Qc)
I'm a little confused. I can get all the changes in entropy for each step, but I have no idea how to get the volume and temperature at 3. Any hint on what I'm missing?
Hi Anon,
The way I did it is by using the equation T1*P1*((1-gamma)/gamma) = T2*P2*((1-gamma)/gamma) on page 155 under "Isentropic Processes." In our case, use 2 and 3 instead of 1 and 2, then solve for T3. Gamma air = 1.4. However, the result that I got is 546K instead of 540K like in the answer. Professor, if you see this blog please verify my explanation. Also, I use the formula for a reversible refrigeration cycle to calculate my COP using T3 as my Th and T1 as Tc. Unfortunately, my COP is 1.79, which is nowhere close to the answer given. I must have done something wrong. Please help!
Anon 10:29:
This formula:
T1*P1*((1-gamma)/gamma) = T2*P2*((1-gamma)/gamma)
only applies if
1- Ideal gas
2- Isentropic
3- gamma = constant.
That is not the case in this problem. Gamma is not constant. Use Snought in the Ideal Gas Property Table in the TCD notebook appendix.
You cannot use COPR = fxn(TH, TC) because we do not know TH. It is not T3. Instead, evaluate COPR = QC/Wcycle = QC/(QH-QC). QC + T1*(S2-S1). Get QH from Q31 - W31 = deltaU13. Evaluate W31 = INT(P dV) because it is internally reversible.
Best of luck !
Hi professor,
First of all I don't know how to find S2 and S1(b/c there's not Sat Temp and Sat Pres for air, only Ideal Gas Table) to use in the equation
QC = T1*(S2-S1), instead I use the equation
delta S12 = Snought2 - Snought1 -RLn(P2/P1)
Snought 1 and 2 will disappear b/c T is constant. Once I got delta S12, I set it equal to Q12/T (T=T1=T2=350K) to solve for Q12. Is this method correct or am I doing something wrong again?
Secondly, I don't know how to solve for the equation
Q31 - W31 = deltaU13
I can find W31 by evaluating INT(PdV) with Vhat = RT/P. However, I again find U3 and U1 of AIR. I really don't know what to do. Please help me!
Anon 11:46 PM
Correct, use:
delta S12 = Snought2 - Snought1 -(R/MW)*Ln(P2/P1)
Correct again: Q12 = T1 * deltaS12
Good work.
Look up U1 and U3 of air in the ideal gas property table for air. They are fxns of T only and you know T1 (given) and T3 (from part a).
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