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Average thermodynamic efficiency

  1. Jun 10, 2006 #1
    Appreciated members of this forum:
    Suppose we have an infinite set of n carnot cycles and we know the efficiency of each cycle of the set: E1,E2,E3,E4,E5,,,,,En ,
    where En is the efficiency of the nth cycle.

    E1= 1-Ta/Tb1 , E2= 1-Ta/Tb2, E3= 1-Ta/Tb ..... En=1-Ta/ Tbn

    Ta is the temperature of the hot reservoir which is constant and equal to 800K
    Tbn is the temperature of the nth cold reservoir that varies from 100 to 500K.



    ¿ is the average efficiency equal to ( E1+E2+E3+E4+....+ En) / n ?


    If we plot the values of each efficiency versus the values of each Tb we get a straight line.
     
  2. jcsd
  3. Jun 10, 2006 #2
    Please, excuse me, but I made a mistake writing the equations of the efficiency. The correct form is the following :

    E1= 1-Tb1/Ta E2= 1-Tb2/Ta E3=1-Tb3/Ta En=1-Tbn/Ta
     
  4. Jun 11, 2006 #3
    Iraides,

    Your question is not very clear for me, could you try to reformulate it?
    If it is an exercise, it would be better to re-type it exactly. Then we can discuss on equal foot.
    I think it is not useful to consider an infinite number (n) of cycles. Simply deal with two cycles.
    Generalisation to 'n' cycles and then to infinite is simple math mechanics, not so interresting.
    Also, it is not clear for me what you mean by "average". What do you call average in this n-machines machine?

    I am also surprised that all your machines apparently are in contact with the same hot source (Ta).
    This is of course possible. Why not? The 'n' machines can be arranged and used freely for an experiment.
    But I don't see what could be learned by such a calculation.
    And what should be the assumption regarding the heat used by each such machine? All the same?
    An alternative would be that the work produced by each machine is the same. But then less efficient machines would need more heat.

    I wait for your clarification.

    And just for the fun, I consider a standard calculation that can done with Carnot machines working in series in between a cold and a hot source.

    To be clear, assume 3 temperatures T1(cold), T2(intermediate) and T3(hot).
    Assume a machine M32 takes heat Q3 from T3, produces work W32 and releases heat Q2 at temperature 2.
    Assume now machine M21 takes that heat Q2 from T2, produces work W21 and releases Q1 at temperature 1.

    The thermodynamic relations then are:
    W32 = (1-T2/T3) Q3
    Q2 = T2/T3 Q3
    W21 = (1-T1/T2) Q2
    Q1 = T1/T2 Q2​
    If you calculate now W31 = W32+W21 , you should not be surprised by the result.
    You could even re-word it as a nice tiny theorem !

    Michel
     
    Last edited: Jun 12, 2006
  5. Jun 12, 2006 #4
    lalbatros,
    let me rewrite the problem: we have a set of n paralell carnot cycles: the first carnot cycle takes an amount of heat Qa1 from a hot reservoir at Ta and expells an amount of heat Qb1 to a cold reservoir at Tb1 doing a work equal to W1.
    The second carnot cycle takes an amount of heat Qa2 from the hot reservoir at Ta and expells an amount of heat Qb2 to a cold reservoir at T2 doing a work W2.
    The nth carnot cycle takes an amount of heat Qan from the hot reservoir at Ta and expells an amount of heat Qbn to a cold reservoir at Tbn doing a work equal to Wn.
    Then the average efficiency, Eav, of the whole set will be

    Eav= (W1+W2+W3+.....+ Wn)/( Qa1+Qa2+Qa3+....+Qan) or

    Eav= (W1+W2+W3+.....+ Wn)/ Qa

    where Qa= (Qa1+Qa2+Qa3+....+Qan)

    but W1= Qa1( 1- Tb1/Ta) W2= Qa2(1-Tb2/Ta) Wn= Qan(1-Tbn/Ta)

    Eav=Qa1/Qa( 1- Tb1/Ta) +Qa2/Qa( 1- Tb2/Ta)+...+Qan/Qa( 1- Tbn/Ta)

    To get Eav we have to find the limit of this series for a huge number of carnot cycles, but we need to know the weights Qa1/Qa , Qa2/ Qa, Qan/ Qa

    At this point we may ask what weigths to use to estimate Eav, and I think, that to avoid any preferences, we must use equal weights for the estimation or

    Qa1/Qa = Qa2/ Qa= Qan/ Qa = 1/n where n is the number of

    carnot cycles.


    With this assumption we get my original question

    Eav= (E1+E2+E3+ En) / n

    ¿ Is above assumption reasonable from statiscal and physical arguments?
     
  6. Jun 12, 2006 #5

    Andrew Mason

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    This appears to be the same question that you posed in May of last year: See here

    Is there a particular reason you want us to go through this again?

    The average efficiency is the arithmetic mean of the efficiencies of each cycle only if the same amount of heat is transferred in each cycle.

    AM
     
    Last edited: Jun 12, 2006
  7. Jun 12, 2006 #6
    Iraides,

    You have calculated correctly the problem that you defined.
    But I don't see any physical consequence to be discussed.

    Assume you investigate all gas turbines in the world operating at a burner temperature of -say- 900°C.
    You can put all the results in a database and make averages of any kind (similar heat consumption, similar power, ...).
    Could you learn some new physics by making the average?
    For sure there is no physics to learn, only statistics on the turbines performances.
    These turbines do not interact in any way, they are independent. How could some physics be learned?

    I don't know what your motivation is.
    But, from my own experience, I can only suggest you to consider practical thermodynamic applications. The foundations of thermodynamics are very abstract, and -I believe- more difficult to learn than anything else in physics (without considering the mathematical aspect).
    The application of the principles to concrete systems (gas turbines, steam turbines, heat exchangers, boilers, ...) is the only way to avoid deception in a first step.

    Michel
     
    Last edited: Jun 12, 2006
  8. Jun 12, 2006 #7

    Yes, Andrew Mason, what happen if the amount of heat transferred to each cycle is different?.
    From above equations It appears that the average efficiency depends on the way we transfer heat. Above equation sugests that if we transfer a great amount of heat to the lowest cold reservoirs we obtain a higher efficiency than the value we obtain when we transfer a high amount of heat to the high temperature of the cold reservoirs. In any case, under these situaciones we have preferenced either the cold or hot reservoirs to get an average efficiency, but I am interested to know if we are able to get a value for the average efficiency independent of preferences. ¿ Stadistically and physically, is this value the one obtained when we use the same amount of heat in each cycle?. Also, if the Carnot cycle efficiency depends only on the temperatures of the heat reservoirs why the average efficiency of a set of Carnot cycles seem to depend on the heat distribution ?.
    At the end of our discussion in may, I asked you some of these questions , I think, but I did not get any reply. At that time your suggestions were very helpfull to me because they coincide with my ideas, but I still have about doubts.
     
  9. Jun 12, 2006 #8

    Andrew Mason

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    Take the simplest case. One Carnot engine operating between Ta and Tb and transferring Q from the hot reservoir. Another Carnot engine operating between Ta and Tc (Tc<Tb) but transferring kQ from the hot reservoir.

    Since [itex]\eta_1 = 1 - T_a/T_b = W_1/Q[/itex] we have:

    [tex]W_1 = Q(1 - T_a/T_b)[/tex]

    Similarly:

    [tex]W_2 = kQ(1 - T_a/T_c)[/tex]

    So the average efficiency is:

    [tex]\eta_{avg} = (W_1 + W_2)/(Q + kQ) = (Q(1 - T_a/T_b) + kQ(1 - T_a/T_c))/(1+k)Q = ((1 - T_a/T_b) + k(1 - T_a/T_c))/(1+k) \ne \bar\eta[/tex]

    So unless the heat flow is the same for each energy, you can't simply average the efficiencies of each engine to find the average efficiency.

    AM
     
  10. Jun 12, 2006 #9
    Andrew Mason,
    I understand your argument, but do you believe that the infinite set of Carnot cycles have diferent efficiencies according to the heat distribution, or , at the end, when you average all of the possible combinations, you will get only one value, equal to Eav= 1- ( Tb1+Tbn)/2Ta. This average corresponds to equal heat distribution.'?????.
     
  11. Jun 13, 2006 #10

    Andrew Mason

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    Well it is not exactly an argument. The average efficiency is the total work / total heat flow. You seem to want to define it differently.

    But the answer to your question is NO, as I have explained: it depends on the heat flow in each cycle.

    AM
     
  12. Jun 14, 2006 #11
    Iraides,

    Anyhow you will define the efficiency of a set of machines, the average will be somewhere in between the highest efficiency and the smallest efficiency. The exact value will depend on the distribution of the total heat used among the different machines (sources).

    There is little to be learned from making this average. You could play to find an equivalent temperature that matches this average, and it will be in between the extreme temperatures of your set.

    Maybe you could have some fun trying to find some statistical data on the power stations around the world. You could do that by types of stations. You could analyse the average efficiency, the typical operating temperatures, the average heat lost, ... and then try to figure out where/how investments could be the most efficient to save our planet from greenhouse gases (or to spare resources for the generations to come). But clearly, by doing the average you will not discover new physics.

    Regards,

    Michel
     
  13. Jun 17, 2006 #12
    Mason, suppose you design a new cycle that operates between the same heat reservoirs specified in the begining of this thread. ¿ Which of the infinite possible combinations of the set of Carnot Cycles should we use for comparison?
     
  14. Jun 18, 2006 #13

    Andrew Mason

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    Is this a new cycle that is not a Carnot cycle?

    AM
     
  15. Jun 19, 2006 #14
    Yes, this new cycle is not a Carnot cycle, and it works between the same heat reservoirs defined in this post : it takes heat from a hot reservoir at Ta and rejects heat to an huge number of cold reservoirs ranging from Tb1 to Tbn.
     
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