0
$\begingroup$

thermodynamic cycle (adiabatic (a–b), isobaric (b–c), isochoric (c–a)

I am trying to prove that the efficiency of the above cycle is equal to:

$e = 1 - \gamma \frac{V_2 - V_1}{P_1 - P_2}$ for 1 mole of an ideal gas.

I assume I should use the equation $e \equiv \frac{\Delta W_{cycle}}{Q_H} = 1 - \frac{T_C}{T_H}$.

For the adiabatic process, the work is: $\frac{PV^\gamma (V_2^{1-\gamma} - V_1^{1-\gamma})}{(1-\gamma)}=C_V(T_2-T_1)$ and for the isobaric process, the work is $P_2(V_2 - V_1)$. The isochoric has no work done because there is no change in volume.

I guess that I would find the change in work of the cycle as the work done from the isobaric process minus the work done from the adiabatic process.

I am a little unsure of how to proceed. I don't know where I can get the $P_1$ and $P_2$ values from. What should I be looking for and how should I determine $Q_H$?

Any help would be greatly appreciated, Thanks!

Improve this question
$\endgroup$
5
  • $\begingroup$ What temperatures would you use for $T_H$ and $T_C$? $\endgroup$
    – Bob D
    Oct 17, 2019 at 7:51
  • $\begingroup$ I don't even really know how I would begin from that. I was expecting the change in work to be a better starting point but I realize that's probably not the best way. $\endgroup$
    – Put1demerde
    Oct 19, 2019 at 4:15
  • $\begingroup$ Before I can respond to this comment and the other under my post I need to know which version of the first law you are using. Are you going $\Delta U=Q-W$ or are you using $\Delta U=Q+W$? $\endgroup$
    – Bob D
    Oct 19, 2019 at 6:49
  • $\begingroup$ @Bob D, I guess $\Delta U = Q – W$. $\endgroup$
    – Put1demerde
    Oct 19, 2019 at 19:21
  • $\begingroup$ At least that's what we've been using in class... $\endgroup$
    – Put1demerde
    Oct 21, 2019 at 0:17

2 Answers 2

Reset to default
1
$\begingroup$

I will only comment on your assumption of the efficiency of the cycle.

The efficiency equation you are using assumes a reversible cycle where all the heat added $Q_H$ occurs at a single temperature $T_H$ and all the heat rejected $Q_C$ occurs at a single temperature $T_C$. In your cycle the heat added during the isochoric process and rejected during the isobaric process do not occur at single temperatures but over a range of temperatures. If you use the formula you will need to use the mean temperatures in the two processes, not the maximum and minimum.

Hope this helps.

Improve this answer
$\endgroup$
3
  • $\begingroup$ I think I might understand what you've indicated here and I can sort of make sense of the equation (although I really don't know why it is the way it is), so I'll try to explain. Sorry if I butcher the entirety of thermo in a few sentences. Since the heat added and lost in the two processes (isochoric and isobaric) don't occur at particular max and min temp values we want to look, instead, for the range in which the lower temperature is to be found (in the isobaric process) which is the heat lost at constant pressure (times the change in volume) and vice versa for the heat gained? $\endgroup$
    – Put1demerde
    Oct 19, 2019 at 4:37
  • $\begingroup$ Therefore, the heat lost in the isobaric process (lowest temperature range) is $C_{p}(V_{2}–V_{1})$ and the heat gained in the isochoric process (highest temperature range) is $+C_{v}(P_{1}–P_{2})$. Again, sorry if this is just rubbish, I am clueless when it comes to thermo. $\endgroup$
    – Put1demerde
    Oct 19, 2019 at 4:44
  • $\begingroup$ @G.P. I have to be honest with you. Based on your comments and your responses to my questions, it appears you do not have enough of a grasp of the fundamentals in thermodynamics to tackle the task you are trying to do. It is commendable that you want to try, but you need to learn more about the fundamentals first. Though tempted to walk you through this (my instinct as a teacher) but as I have often been reminded, that is not the basic purpose of this site. So I am sorry I can't go further, but I encourage you, if you haven't already, to take a basic course in thermodynamics. Good luck. $\endgroup$
    – Bob D
    Oct 20, 2019 at 21:40
0
$\begingroup$

The QH you are asking for is the net heat given to the system which is nC v( Tf - Ti ) here as heat in adiabatic process is 0 and in the isobaric process, heat is taken out from the system here ( since Tf < Ti for the isobaric process) Therefore, now you have net heat and net work ( both work positive and negative work would be considered ) I hope you can arrive at the ans now

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thank you for the hint. I am mostly unsure of how to get the P_1 and P_2 values from algebraic manipulation. $\endgroup$
    – Put1demerde
    Oct 17, 2019 at 15:35
  • $\begingroup$ @G.P. What "values" of P1 and P2 are you looking for? In what terms do you want them? $\endgroup$
    – Bob D
    Oct 17, 2019 at 17:02
  • $\begingroup$ @Bob D The efficiency of this cycle has the term $(P_{1} - P_{2})$ in the denominator. I'm confused where we are able to get the terms $P_{1}$ and $P_{2}$ from. Thermodynamic relations maybe? I guess from the adiabatic process we know $P_{1}V_{1} = P_{2}V_{2}$. But with this I get the term $\frac{\gamma P_{2}V_{1} - \gamma P_{2}V_{2} - P_{2}V_{1} - P_{1}V_{2}}{(1-\gamma)(nC_{v}(T_{f}–T_{i}))}$, this doesn't look like it will simplify to the proposed solution... $\endgroup$
    – Put1demerde
    Oct 19, 2019 at 3:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.