efficiency of an infinite series of Carnot cyclesby Iraides Belandria Tags: carnot, cycles, efficiency, infinite, series 

#1
May1905, 03:44 PM

P: 55

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= 1Tb/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?.




#2
May2005, 09:00 AM

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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. AM 



#3
May2005, 10:33 AM

P: 55

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?




#4
May2005, 10:55 AM

P: 55

efficiency of an infinite series of Carnot cycles
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




#5
May2005, 04:09 PM

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[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] So: [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). AM 



#6
May2005, 04:11 PM

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AM 



#7
May2005, 10:34 PM

P: 55





#8
May2105, 10:01 AM

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AM 



#9
May2105, 11:56 AM

P: 55

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




#10
May2305, 11:04 AM

P: 55

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?.




#11
May2305, 11:17 AM

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AM 



#12
May2505, 10:27 PM

P: 55

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 ? 



#13
Jun205, 07:55 PM

P: 55

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 and summatory Tc(i)= [(Tcend + Tc beg) ( Tcend  Tc beg + 1)] / 2 Could you explain these equations with more details? Thanks 



#14
Jun205, 11:36 PM

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[tex]S = (n / 2) * [2i + (n1)][/tex] where n = Tcend  Tcbeg + 1 and i = Tcbeg eg: [tex]5 + 6 + 7 + 8 + 9 + 10[/tex] n = 6 (105+1); i = 5; S = (6/2) * (10 + 5) = 3 * 15 = 45 AM 



#15
Jun305, 12:13 AM

P: 55

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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