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Carnot Engines

  1. Jun 10, 2007 #1
    Hi all,

    Sat this problem in an exam recently and it's got me wondering. First off, you have a reservoir at a lower temperature T1. You also have a mass of water at a higher temperature T2 (in this case it was 10C and 100C, I can't remember the mass but let's say it was 1, and Cv 4.2).

    First we were asked how much work can be extracted taking the water to the revervoirs temperature, which can be solved by setting up the appropriate integral

    Cv * int (1 - T1 / T2) dT2

    The next part of the question was more tricky. Instead of a revervoir at the lower temperature, we had another body of water, which also changes temperature. Now what is the maximum work we can extract?

    I have in the past solved a similar question by looking at the entropy of each body of water and considering energetics, but can anyone think of a cunning integral that could be used to solve this problem?
     
  2. jcsd
  3. Jun 15, 2007 #2

    mgb_phys

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    The efficency of a carnot cycle only depends on the temperature difference between the reservoirs.
    I imagine the second part is asking what is the end point, which is of course when the two reservoirs have the same temperature and when will you get there which depends on the heat capacity (mass and type of liquid) of the reservoirs.
     
  4. Jun 16, 2007 #3
    I should restate the queston. You have two bodies, of equal heat capacity, with temperatures T1 and T2. You run a reversable heat engine between them. How much work do you extract and what is the final temperature?

    You can calculate it by saying entropy is converved, so C ln T1 + C ln T2 = 2C ln Tf and hence Tf^2 = T1 * T2. Then the work extracted must simply be C(T1 + T2 - 2Tf)

    However, is there an integral way of doing this similar to the case with the reservoir? At each temperature of the two bodies, the efficiency is different so the elementary work depends on both T1 and T2.
     
  5. Jun 17, 2007 #4
    The first law (energy conservation) gives you [itex]V_{hot} C_V (-\Delta T_{hot}) = \Delta W + V_{cold} C_V \Delta T_{cold}[/itex] (1), or (in sum total) that [itex]W=C_V \{ V_h T_h + V_c T_c - (V_h + V_c) T_{final} \} [/itex] (2).

    The second law (Carnot efficiency) gives you [itex]\Delta W = (V_h C_V (-\Delta T_h)) (1-\frac{T_c}{T_h})[/itex], or (by combining with 1) that [itex]V_c \frac{\Delta T_c}{T_c} = V_h \frac{- \Delta T_h}{T_h}[/itex] (3).

    Having separated variables, each side of 3 can be integrated (from initial to final temperature) to show that:
    [tex]T_f = (T_c^{V_c} T_h^{V_h})^{\frac 1 {V_c + V_h}}[/tex]
    which can now be written into 2. Is that the "integral" way of finding the total work that can be extracted?
     
    Last edited: Jun 17, 2007
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