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Efficiency of an idealgas cycle. 
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#1
Oct1010, 11:22 PM

P: 6

1. The problem statement, all variables and given/known data
A possible idealgas cycle operates as follows: (i) from an initial state (p1,V1) the gas is cooled at constant pressure to (p1,V2); (ii) the gas is heated at constant volume to (p2,V2); (iii) the gas expands adiabatically back to (p1,V1). Assuming constant heat capacities, show that the thermal eficiency is: 1ɣ[((V1/V2)  1) / ((p2/p1) 1)] 2. Relevant equations I used first law of thermal dynamics, n = W/Qh, differential forms of Cv and Cp. change in energy for the complete cycle = 0, therefor W = Qh + Ql. 3. The attempt at a solution Using efficiency = W / heat absorbed. I attempted to find the W and Q of the 3 processes separately. I was using for Isobaric: dW = Pdv but im pretty sure this is incorrect because no where does it state it is a reversible process. But the book for the class only shows reversible processes for examples and does not actually show a isobaric or isochoric process in any example. I tried using the differential form of Cv and Cp and rearranging them to find dQ, but failed. I have used 4 or 5 sheets of paper, trying every way i could think of to work this problem out. I know im missing something and if someone could just point me in the right direction I would gladly appreciate it and work out the problem on my own. Thanks 


#2
Oct1010, 11:35 PM

Mentor
P: 12,069

Welcome to Physics Forums.
You'll also need Q for one leg of the cycle (do you know which leg that is?). 


#3
Oct1010, 11:56 PM

P: 6

Using Qh for heat absorbed and Ql for heat lost I was using the equation as:
n = w/Qh = (Qh +Ql)/Qh = 1 + Ql/Qh. Ql being from (i) and Qh being from (ii). Q = 0 from (iii). Qh = ∫Cv dT Ql = ∫Cp dt + ∫pdV For adiabatic ΔQ = 0 so ΔU = W. Since, you are ending where you started ΔU for the complete cycle should = 0 correct? so ΔU for adiabatic should equal ΔU for isochoric + isobaric. Isobaric: ΔU = Ql + pdV Isochoric: ΔU = Qh ΔU for adiabatic = Ql + Qh  ∫pdV. ΔU = W so W = Ql + Qh  ∫pdV and now i lost myself. Maybe i just need some sleep. Thanks for trying. 


#4
Oct1110, 09:24 AM

P: 6

Efficiency of an idealgas cycle.
I stated that I wasnt sure how to relate the change in T to p and V, but I forgot about the ideal gas equation. So I just solved for Q's like before and put in pV/R in place of T. Doing that I ended up with n = Qin  Qout / Qin dQin = Cv/R(Vdp + pdV) and pdV = 0 dQout = Cp/R(Vdp + pdV) and Vdp = 0 n = Cv(V2(p2p1))  Cp(p1(V2V1)) / Cv(V2(p2p1)) because R's cancel out. n = 1 + (Cp/Cv) *((V2V1)/V2) * (p1(p2p1)) sub in ɣ for Cp/Cv and clean up the rest and that shows that n = 1+ɣ[((V1/V2)  1) / ((p2/p1) 1)] It is very close, only missing a  sign. I am not sure If i did a calculation wrong or that it is purely coincidental that I got this close. Now that part that scares me and why I didnt do this in the beginning is that I complete ignore the adiabatic leg, and all work done. I just simple used heat. It is close to the right answer but I am not convinced I did it correctly. Thanks for your help so far. 


#5
Oct1110, 11:12 AM

Sci Advisor
HW Helper
P: 6,671

If you are using 1  Qout/Qin for efficiency you have to take the absolute value of Qout: [tex]Q_{out} = C_pP_1(V_1  V_2)[/tex] That is where you got the signs mixed up. AM 


#6
Oct1110, 11:31 AM

P: 6

thanks, that makes me feel better.
can you explain to me exactly why we take the absolute value? Just so i can understand it and apply it to future problems. thanks again. I think I understand. Work is actually equal to the sum of the heat. By me stating Qin  Qout I am taking the negative out of Qout already. To do it the way I did I should have used n = (Qin + Qout)/ Qin 


#7
Oct1110, 03:22 PM

Sci Advisor
HW Helper
P: 6,671

AM 


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