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Efficiency of an infinite series of Carnot cycles

  1. May 19, 2005 #1
    Dear people of Physics Forums, I would like to have your opinion about the following problem: Suppose we are interested to evaluate the average efficiency ( e ) of an infinite series of carnot cycles. The temperature of the hot reservoir of each carnot Cycle of the series, Ta, is at 990K, but the cold reservoir of each cycle is at different temperatures ( Tb ) ranging from 295 K to 752K , in such a way that, the cold reservoir of the first carnot cycle is at 295 K and the cold reservoir of the last is at 752 K. ¿ What is the average efficiency of these series of Carnot cycles?. Now, we know that the efficiency of each Carnot cycle can be determined with the equation e= 1-Tb/Ta. Since we know the temperatures we can calculate the efficiency of each Carnot cycle of the series ranging from 295 to 752 K(e1, e2, e3, e4, e5, e6,e7............. ). ?How can we get the average efficiency from these set of individual values?. I notice that if I plot the efficiency of each Carnot Cycle versus Tb ( the cold resrvoir temperature of each cycle) I get an straight line. Then I applied the mean value theorem to get the average efficiency. I found that the average efficiency is 0.47. ? Is this procedure correct?.
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  3. May 20, 2005 #2

    Andrew Mason

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    The average efficiency would be the total work (output) divided by the total heat input:

    Qh is the same for each cycle. Qc is proportional to Tc (Qc/Tc = Qh/Th). Since, W = Qh - Qc,

    [tex]\eta = \frac{W}{Q_h} = \frac{\sum_{i=1}^n Q_h - Q_c(i)}{nQ_h}[/tex]

    [tex]\eta = \frac{\sum_{i=1}^n 1 - Q_c(i)/Q_h}{n}[/tex]

    [tex]\eta = \frac{\sum_{i=1}^n 1 - T_c(i)/T_h}{n}[/tex]

    [tex]\eta = \frac{1}{n}\sum_{i=1}^n 1 - T_c(i)/T_h[/tex]

    [tex]\eta = \frac{1}{n}\sum_{i=1}^n \eta(i)[/tex]

    So the total efficiency over n cycles is 1/n times the sum of the efficiencies of each cycle.

  4. May 20, 2005 #3
    Thanks Andrew for your help, but I still have a question because I have to sum infinite cycles and divide by n. ¿How can I get a numerical answer?. Because of this I use the mean value theorem of calculus. ¿Is that correct?
  5. May 20, 2005 #4
    Again, Andrew, ¿Why do you assume that each cycle receive the same amount of heat Qa? ¿ Will the average efficiency change if each cycle receive a different amount of heat, Qa1, Qa2, Qa3,......?. Thanks
  6. May 20, 2005 #5

    Andrew Mason

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    You would have to work out the series and determine the limit as [itex]n \rightarrow \infty[/itex]

    [tex]\eta = \frac{1}{n}\sum_{i=1}^n 1 - T_c(i)/T_h = \frac{1}{n}(n - \frac{1}{T_h}\sum_{i=1}^n T_c(i))[/tex]

    Now if Tc is linear and the increments of Tc are 1 degree at a time then n = [itex]T_{cend} - T_{cbeg} + 1[/itex] so

    [tex]\sum_{i=1}^n T_c(i) = \frac{(T_{cend} + T_{cbeg})(T_{cend} - T_{cbeg} + 1)}{2}[/tex]


    [tex]\eta = \frac{1}{n}(n - \frac{1}{T_h}\frac{(T_{cend} + T_{cbeg})(T_{cend} - T_{cbeg} + 1)}{2}) = (1 - \frac{(T_{cend} + T_{cbeg})(T_{cend} - T_{cbeg} + 1)}{2nT_h})[/tex]

    Substitute for [itex]n = T_{cend} - T_{cbeg} + 1[/itex]:

    [tex]\eta = (1 - \frac{(T_{cend} + T_{cbeg})(T_{cend} - T_{cbeg} + 1)}{2T_h(T_{cend} - T_{cbeg} + 1)}) = (1 - \frac{(T_{cend} + T_{cbeg})}{2T_h})[/tex]

    As you increase n and decrease the increments, the series should converge to something very very close to this (if it is linear).

    Last edited: May 20, 2005
  7. May 20, 2005 #6

    Andrew Mason

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    Why would Qh depend upon Qc?

  8. May 20, 2005 #7
    Excuse me, I don understand your answer to the second question.¿Would the average efficiency depends upon the heat received by each carnot cycle of the series or it is independen of it?
  9. May 21, 2005 #8

    Andrew Mason

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    The average efficiency would depend on the total heat received, which is the sum of the heats received in each carnot cycle. I am assuming they are the same but they may not be. I was asking you why they would depend on Qc.

  10. May 21, 2005 #9
    I appreciate very much your interest in trying to solve my doubts about this problem. Indeed I am not sure if Qh depend upon Qc. My principal doubt at this moment is if the average efficiency of the series is independent of the amount of heat received by each Carnot Cycle. In your solution you assume that each cycle receives the same amount of heat. But what happen, if the case is more general, and we assume different values for each carnot Cycle?. Would we obtain the same value?. Thanks
  11. May 23, 2005 #10
    Other mathematical question is related to find an average for a set of infinite values of efficiencies. As I have stated at the beginning of the thread, we can estimate the numerical eficciency of each Carnot cycle of the series without problem, because I know the temperatures of each Cycle. ?How can we evaluate the average of these values of efficiencies of the infinite carnot cycles?.We also can see that these set of values of efficiency varies linearly with Tc. ¿If this is the case, would be correct to say that the average coincides with the value giving by the limit of above series?.
  12. May 23, 2005 #11

    Andrew Mason

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    As I said before, the average efficiency is the total Work done divided by the total heat input. You can't assume that the average efficiency is the average of all the individual efficiencies just as you cannot determine average speed by averaging speeds (ie. average speed it is total distance divided by total time not an average of speeds over the journey).

  13. May 25, 2005 #12
    Dear Mason, If we assume that each carnot cycle receive differents amounts of heat from the hot reservoir, Qh1, Qh2, Qh3,....Qhi, we get the following expression for the average efficiency (Using above procedure)

    n = (Qh1/Qh) n1+ (Qh2/Qh) n2 + (Qh3/Qh) n3 + .....(Qhi/Qh) ni

    Where Qh is the total heat received fom the hot reservoir equal to Qh=Qh1+Qh2+ Qh3+ ......+Qhi+.... , and ni are efficiencies of each cycle

    This equation shows that the average eficciency, n , depends on the amount of heat received by the cycles. In a certain way each Qhi/Qh is a weigth to average the efficiency of the series of cycles. In your deduction you use a uniform weigth to get the average efficiency , Qhi /Qh = N , Where N is the number of carnot cycles. ¿Is this the weigth that we must use in order to compare the performance of any cycle against a series of carnot cycles?. ¿Statiscally , what will be the best weigth to compare an specific Cycle? ? What weight is fair, just ?
  14. Jun 2, 2005 #13
    Dear Andrew Mason, I have tried to understand how you obtain the following expresions in the limit of the serie proposed by you

    n= Tcend -Tcbeg + 1

    summatory Tc(i)= [(Tcend + Tc beg) ( Tcend - Tc beg + 1)] / 2

    Could you explain these equations with more details?

  15. Jun 2, 2005 #14

    Andrew Mason

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    It is just the sum of an arithmetic series with increment of 1.

    [tex]S = (n / 2) * [2i + (n-1)][/tex] where n = Tcend - Tcbeg + 1 and i = Tcbeg


    [tex]5 + 6 + 7 + 8 + 9 + 10[/tex]

    n = 6 (10-5+1); i = 5; S = (6/2) * (10 + 5) = 3 * 15 = 45

  16. Jun 3, 2005 #15
    Andrew, Thank you, very much. I appreciate your cooperation, and I would like to say that my empirical results coincide with your proposition, as you may verify in the begining of this thread. In any case, the most important consequence of this discussion is that I have found a thermodynamic cycle, operating between the temperatures levels proposed in this post, more efficient than an infinite series of carnot cycles. Because of this reason, I have insisted so much in the solution of this problem, from differents points of view. All of the alternatives, I have explored coincide with your results. thanks to god. Now, I think I am confident on my results. I am going to try to publish an article related to this cycle, and I am going to use this thread as a bibliographycal reference. ¿ Can I do this? . In the future, I will inform you about it.
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