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Heat transferred in a cyclic process

  1. Oct 23, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Assume that a gas obeys the VDW Virial expansion Pv = RT + (b-a/RT)P to first order in P and u = 3RT - aP/RT to first order in P, where v and u are molar quantities.
    In the following cycle (see attachment), the heat transferred to the gas is transferred by direct thermal contact with a reservoir at T1 (for the const pressure process at P1 and isotherm at T1) and heat removed is transferred via a reservoir in thermal contact at T2 (for the const pressure process P2 and isotherm at T2). Derive an algebraic expression for the following sum around the cycle $$\sum_i \frac{Q_i}{T_{res_i}}$$ and show that the sum is less than zero.

    2. Relevant equations
    First law in terms of change in entropy: du = TdS + Pdv

    3. The attempt at a solution
    I need to find 4 expressions for the heat in/out of the working substance. For an isobaric process (P1), ##dQ_{1,in} = \left(\frac{\partial H}{\partial T}\right)_PdT = (4R + \frac{2aP_1}{RT^2})dT,## by differentiating u and v above since H=u+Pv. Integrate this from T2 to T1 gives heat in at constant pressure P1.

    For heat in during the isotherm, use the first law so that $$Q_{2,in} = TdS = -\frac{a}{RT_1}(P_2-P_1)-RT_1\ln(P_2/P_1)$$ using ##dS =1/T_1 \left(-\frac{a}{RT_1}(P_2-P_1) - RT_1\ln(P_2/P_1)\right)##that I derived earlier.

    For heat out during the constant pressure P2, same as above in the other constant pressure process but reverse the indices and put a negative at the front (since heat in is defined +ve) => $$Q_{3,out} = -4R(T_2-T_1) + \frac{2aP_2}{R}(\frac{1}{T_2}-\frac{1}{T_1})$$

    For heat out during the isotherm T2, same as above in the other isotherm process,but reverse indices and put in a negative: ##Q_{4,out} = a/RT_2 (P_1-P_2) + RT_2\ln(P_1/P_2)##

    When I sum all these terms and then analyse each term independently, I see that every term is positive and so the sum cannot possibly be less than zero.

    Many thanks.
     

    Attached Files:

  2. jcsd
  3. Oct 23, 2013 #2

    Andrew Mason

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    This is not correct. The fundamental thermodynamic relationship is: TdS = dU + PdV or dU = TdS - PdV.

    Also, the sign of the heat out must be negative. If all heats were positive, there would be no cycle.

    AM
     
    Last edited: Oct 23, 2013
  4. Oct 23, 2013 #3

    CAF123

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    I think this makes sense; If we do work on the system by surroundings, the work done dW is -PdV. However, the only part of this equation I used in my attempt was the fact that dQ=TdS.
    Yes, I have used this in my attempt.
    Thanks for reply, can you see where I went wrong in my attempt?
     
  5. Oct 23, 2013 #4
    I solved this problem also, and got the exact same results that you did for the four heat loads. I'm very confident that they are correct. (In doing the constant temperature segments, I did it entirely by the first law, and did not invoke the entropy. I simply evaluated ΔU, and added the integral of Pdv. But, as might be expected, our results are the same.)

    When I took my sum, the terms were not all positive. Check your signs in the summations. Some of your Q's were for heat leaving, and some were for heat entering. If I just focus on the ideal gas contributions to the summation, I end up with:

    [tex]4R\left(1-\frac{T_2}{T_1}\right)+4R\left(1-\frac{T_1}{T_2}\right)[/tex]

    These two terms clearly don't have the same sign, and their sum is negative.

    Chet
     
  6. Oct 23, 2013 #5

    CAF123

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    Hi Chet,
    If we just focus on ##Q_{1,in}## (heat in during constant process P1) and ##Q_{3,out}## (heat out during constant pressure P2), then these expressions are of the same form but they have different indices for the Ti and Pi and the latter has a negative in front. So, $$Q_{1,in} = 4R(T_1-T_2) - \frac{2aP_1}{R}(\frac{1}{T_1}-\frac{1}{T_2})$$, while $$Q_{3,out} = - (4R(T_2-T_1) - \frac{2aP_2}{R}(\frac{1}{T_2}-\frac{1}{T_1}))$$ When you add these two terms, should you not get a term $$4R(T_1-T_2) - 4R(T_2-T_1)?$$ instead?
    EDIT: To be clear, I reliase why you have 1-T2/T1 and 1-T1/T2, it was just the sign in between these two terms that I have differently.
     
    Last edited: Oct 23, 2013
  7. Oct 23, 2013 #6
    This is exactly what I thought you did. Q3out=-Q3in. All the Q's in the summation are supposed to be heat added to the system from the surroundings , i.e., Qin's. Wherever you had a Qout in your summation, you need to change the sign. You also know this because, for an ideal gas, the Q's for the two constant pressure segments have to add up to zero. This is because they are determined by ΔH, which, in turn, is a function only of the temperature differences.

    Chet
     
  8. Oct 23, 2013 #7

    CAF123

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    I am not sure I follow - is this not what I did? My summation looks like : $$\sum_i \frac{Q_i}{T_{res_i}} = \frac{Q_{1,in}}{T_1} +\frac{Q_{2,in}}{T_2} - \frac{Q_{3,out}}{T_3} - \frac{Q_{4,out}}{T_4}$$where I put a negative in front of the terms where heat is leaving the working substance.
     
  9. Oct 23, 2013 #8
    Look at you own equation for Q3out. If you substitute that exact relationship into this expression, what do you get?

    Chet
     
  10. Oct 24, 2013 #9

    CAF123

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    So the expression becomes $$4R\left(1-\frac{T_2}{T_1}\right) + 4R\left(1-\frac{T_1}{T_2}\right) - \frac{2aP_1}{RT_1}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) - \frac{2aP_2}{T_2R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) - \frac{a}{RT_1^2}\left(P_2-P_1\right) - \frac{a}{RT_2^2}\left(P_1-P_2\right)$$

    I thought initially that the negative's I had in front of the terms ##Q_{3,out}## and ##Q_{4,out}## were precisely those negatives that appeared in Q1/T1 + Q2/T2 - Q3/T3 - Q4/T4, however, this does not seem to be the case. So in my expression for Q3, I already put a negative in front of the expression since it is a Q_out and then when I substitute it into the Clausius' inequality, I take it's negative again? This does not seem right.

    I have proved that the sum of the first two terms and the last two terms are less than zero, but I am a little unsure of the middle two terms - I can't factor it out.
     
    Last edited: Oct 24, 2013
  11. Oct 24, 2013 #10
    Your difficulties began in post #5, when you wrote, "For heat out during the constant pressure P2, same as above in the other constant pressure process but reverse the indices and put a negative at the front." You would have been better off not putting the negative sign out in front, and just leaving it as Q3,in. Then you would at least have been OK with all positives in the summation for dQ/T. But you could still have recovered from this by doing the algebra correctly using the version with the minus signs for Q3,out and Q4,out.

    To convince yourself that what we are saying is correct, just calculate Q1,in+Q3,in=Q1,in-Q3,out for the ideal gas case. This sum should be zero in the case of an ideal gas since the change in enthalpy around the cycle is zero.
    I want to think about this part a little more.

    Chet
     
  12. Oct 24, 2013 #11

    CAF123

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    I see. ##Q_{3,in} = 4R(T_1-T_2) - \frac{2aP_2}{R} (\frac{1}{T_1} - \frac{1}{T_2})## so ##Q_{3,out} = -Q_{3,in} = 4R(T_2-T_1) - \frac{2aP_2}{R}(\frac{1}{T_2}-\frac{1}{T_1})##. My expression before should have been ##\sum_i \frac{Q_i}{T_{res,i}} = \frac{Q_{1,in}}{T_1} + \frac{Q_{2,in}}{T_2} + \frac{Q_{3,out}}{T_3} + \frac{Q_{4,out}}{T_4}##, I think.

    Also, just to check the terms ##Q_{3,in}## are imaginary here, because the heat is removed from the gas in this part of the cycle.

    The next part of the question asks to compute the increase in entropy of the universe (system + surr). The entropy of the system around the closed cycle is zero. The entropy of the surr decreases when heat is removed in process 1 and 2 and increases in processes 3 and 4. I have expressions for the changes in entropy of the gas in each of these processes, so can I say ##\Delta S_{gas} = -\Delta S_{surr}## and just add up my expressions?
     
    Last edited: Oct 24, 2013
  13. Oct 24, 2013 #12
    I determined the sum of all the four terms with the minus signs, and got:

    [tex]-\frac{a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)^2(P_1+P_2)[/tex]

    This is, of course, negative. I think I did the algebra right, but I'd feel better if you checked this result.

    Chet
     
  14. Oct 24, 2013 #13
    This should read:
    ##Q_{3,out} = -Q_{3,in} = -4R(T_2-T_1) +\frac{2aP_2}{R}(\frac{1}{T_2}-\frac{1}{T_1})##

    It isn't imaginary. It is negative. As you said, negative Q represents heat flow from the system to the surroundings.

    No. You can't do it that way. You need to identify a reversible process that takes the surroundings from its initial state to its final state. Call the working gas System #1 and the surroundings, System #2. We already know that the change in entropy for System #1 is zero (cycle), and we also know that the Clausius Inequality is satisfied over the cycle by the system (the change in entropy is greater than the summation we calculated, which is negative). But we don't know much about the final state of the surroundings (System #2), so we can't (in my judgement) calculate its change in entropy. The best we can do is to establish a lower bound to the change in entropy for the surroundings. This lower bound can be calculated for System #2 by again using the Clausius Inequality. The heat flows to the surroundings in the process are minus those for the system. So the change in entropy for the surroundings is greater or equal to minus the summation we calculated in part 1. The combination of the system and the surroundings constitutes and isolated system, so, for our irreversible process, the change in entropy for the combination must be greater than zero. In fact, the change in entropy for the combination is equal to the change in entropy for the surroundings, which has as a lower bound the value we calculated.

    Chet
     
  15. Oct 24, 2013 #14

    CAF123

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    It makes sense up to this point. But when we sub ##Q_{3,out}## into the Clausius' inequality, we have to take the -ve of ##Q_{3,out}## which I don't quite get. As far as I understand, the negative in Clausius' inequality would mean that the heat is leaving the engine. But we already included this negative in line **. I am not saying you're wrong of course, I just want to understand.


    Is there any particular reason why? Does ##\Delta S_{system} = \Delta S_{surr}## hold only for a reversible process?

    Why reversible?

    ##\Delta S_{surr} \geq -##summation in #9.

    We know it is irreversible since the Clausius inequality for the system was not equal to zero, right?

    The change in entropy of the gas was zero, so the increase in entropy of the universe comes from the increase in entropy of the surroundings.

    Thanks.
     
  16. Oct 24, 2013 #15
    All I was doing here was point out an algebraic error in a line of your previous post. Nothing more fundamental than this.

    ##\Delta S_{system} = -\Delta S_{surr}## only for a reversible process.

    Because the only way to calculate the change in entropy between two equilibrium states of a system is to evaluate dQ/T for a reversible path between the two states. In such a process, the heat flows from the surroundings are equal and oppositein sign to those of the system, and they take place at the same temperatures.
    Yes, for this specific process.
    We know it is irreversible because, during the constant pressure steps, the gas starts out at one temperature and changes to another temperature, while the heat transferred to the reservoir (surroundings) takes place at only one of the temperatures. If these steps had been done reversibly, the surroundings temperature would have had to be changed gradually so that it was only incrementally different from gas temperature over each of these steps. What we are saying here is that, during the constant pressure steps, the gas and the reservoir were not close to thermal equilibrium during the heat transfer. All this causes the Clausius inequality to not equal zero, both for the system and for the surroundings.
    Yes.
     
  17. Oct 24, 2013 #16

    CAF123

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    What algebra error did I make?
     
  18. Oct 24, 2013 #17
    Oops. My mistake. I misread the first line of your response in #11. Sorry.

    Chet
     
  19. Oct 25, 2013 #18

    CAF123

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    Ok,but just to check, what I said in #11 is correct, right? In particular, Q_{3,in} is the direction anticlockwise around that constant P process, since Q_{3,out} is the direction clockwise. Q_{3,out} = -Q_{3,in} and this is where I took into account the minus sign. So the Clausius inequality reads $$\sum_i \frac{Q_i}{T_{res, i}}= \frac{Q_{1,in}}{T_1} + \frac{Q_{2,in}}{T_1} + \frac{Q_{3,out}}{T_2} + \frac{Q_{4,out}}{T_2}$$ I hope this is right, since it makes sense to me.
    Thanks again for your help.
     
    Last edited: Oct 25, 2013
  20. Oct 25, 2013 #19
    Uh Oh. I looked it over again, and I was right in the first place about your algebra error in the first line of #11. The the very last two terms on the right have the wrong signs.

    In terms of clockwise and anti-clockwise, the cycle that you are following is in the clockwise direction. In step 3, the temperature of the gas cools from T1 to T2. So the heat flow from the surroundings to the system is negative in magnitude; this simply means that heat flows from the system to the surrounding in this step. This is analogous to a mechanics problem where Vx is the velocity in the + x-direction, but, if Vx is negative, the particle is traveling in the -x direction. Vx still represents the velocity in the + x direction, but it is just negative. In the first law, Q is always regarded as the heat flow from the surroundings to the system, but its sign can be negative if the flow is actually in the other direction.
     
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