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Efficiency of heat engines

  1. Apr 13, 2016 #1
    1. The problem statement, all variables and given/known data

    Two identical tanks of water are at absolute temperatures ##T_A## and ##T_B## respectively, where ##T_{A} > T_{B}##. The tanks each have a heat capacity ##C##, and they are thermally isolated from their environment. Suppose that a heat engine is installed in contact with the two tanks, in order to extract useful work from their temperature difference.

    (a) Suppose that the engine is completely inefficient and generates no work. What will be the final temperature ##T_{f}## of the tanks at equilibrium?

    (b) Suppose instead that the engine is the most efficient engine possible. What will be the final temperature ##T_{f}## of the tanks at equilibrium?

    (c) How much work will have been generated by the efficient engine of part (b)?

    2. Relevant equations

    3. The attempt at a solution

    (a) If the engine is completely inefficient and generates no work, then heat energy is being transferred from the hot tank with temperature ##T_A## to the cold tank with temperature ##T_B##. The tanks are identical, so each of them have the same mass ##m##. Therefore,


    ##\implies T_{f}=\frac{T_{A}+T_{B}}{2}##

    Am I correct so far?
  2. jcsd
  3. Apr 13, 2016 #2
    Yes, you are correct for the part (a)
  4. Apr 13, 2016 #3
    For (b), do I have to assume that the tanks are heat baths, so the temperatures of the tanks remain unchanged?
    Last edited: Apr 13, 2016
  5. Apr 13, 2016 #4
    No that happens when they have infinite heat capacity.
  6. Apr 13, 2016 #5
    Well, if part of the heat energy (being transferred from the hot tank to the cold tank) is sucked out by the heat engine and converted to work, then only the wasted heat energy is absorbed by the cold tank. Will the transfer of the wasted amount of heat energy from the hot tank to the cold tank determine the final temperature of the two tanks?

    Now, doesn't efficiency = ##1 - \frac{T_{B}}{T_{A}}## mean that the wasted heat energy = ##\frac{T_{B}}{T_{A}}(C)(T_{f}-T_{B})##?

    Where do I go from here?

    P.S. : In my answer to part (a), the ##m##'s will be missing because ##C = mc##.
  7. Apr 13, 2016 #6
    Assume that at the end of the cycle both become Tf', which will be more than (TA + TB)/2 and the total energy available is C(TA - TB), which was wasted in part (a) of the problem with no work done. So now do you understand what is the correct expression of wasted energy for part (b)? You are using the Carnot formula for a non-Carnot engine. You need to just use the everyday idea of what is efficiency!
    Last edited: Apr 13, 2016
  8. Apr 13, 2016 #7
    Do the temperatures of the tanks necessarily have to become the unique final equilibrium temperature ##T'_{f}## at the end of the cycle, and not at some other point of the cycle?
  9. Apr 13, 2016 #8
    This will not be a reversible process so there is cycle but not reversible, in the sense that the engine might come back to its original state but te reservoirs have not come back to original temperatures at least. Just think in terms of the end results.
  10. Apr 13, 2016 #9
    I think you misunderstood my question. I am not asking about the reversibility of the heat engine cycle.

    I am asking how you knew that it's only at the end of the cycle, and not at some other point in the cycle, that the tanks reach the equilibrium temperature. Is this because, once the tanks reach equilibrium temperature, then no heat exchange between them is possible, so that the heat engine cannot extract heat energy to do useful work?
  11. Apr 13, 2016 #10
    Exactly there comes the idea of reversibility, in a reversible process system and reservoirs are all the time in equilibrium but in all other processes we consider them in equilibrium at certain points after which or before which there is no change in the macroscopic variables in this case temperature. The problem is also asking about the final temperature of the reservoirs after you have got the necessary work from the system/systems. As far as end states are concerned the thermodynamic laws hold for them.
    Last edited: Apr 13, 2016
  12. Apr 13, 2016 #11
    Why is the final equilibrium temperature of the tanks greater than ##\frac{T_{A}+T_{B}}{2}##?
  13. Apr 13, 2016 #12
    because if it is more more heat will be wasted, which the system does not have.
  14. Apr 13, 2016 #13
    I am having difficulty understanding your explanation. Would you please elaborate?
  15. Apr 18, 2016 #14
    I, as I think you do, believe the final temperature for Part (b) is less than (TA+TB)/2.

    Have you solved Parts (b) and (c)?
  16. Apr 18, 2016 #15
    I haven't.

    Can you please provide the first two to three lines of the solutions so I know exactly how to proceed?

    I am completely lost.
  17. Apr 19, 2016 #16
    First, clearly define variables:

    TA, TB, Tf, C=mc as above.
    Ta = variable temperature of tank A.
    Tb = variable temperature of tank B.
    QA = variable heat leaving tank A and entering engine.
    QB = variable heat leaving engine and entering tank B.
    W = variable work produced by engine.

    We know the engine will have Carnot efficiency as you've stated in #5, but it's a variable = (1-Tb/Ta).

    We know the efficiency is also W/QA.

    We know the efficiency changes during the course of the process, so an integration is needed.

    Do a differential energy balance on the engine.
    Combine with differential temperature changes in the tanks, e.g., dQA = -CdTa.
    Eliminate W and Q's.
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