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I How to derive this heat input equation?

  1. Sep 18, 2016 #1
    There's one equation that I've seen being used already, which by itself is quite simple, but I can't understand where it comes from.

    The context is as follows: suppose we have one heat engine operating between systems [itex]A[/itex] and [itex]B[/itex] whose temperatures are [itex]T_A>T_B[/itex]. Let's suppose further that the system [itex]A[/itex] has heat capacity [itex]C[/itex].

    I've seem many times people using that under these conditions the heat input is:

    [tex]Q_1 = \int_{T_A}^{T_B} C dT.[/tex]

    Now I really can't see why this is true and I'll make my point. Let's suppose that the heat input is [itex]Q_1[/itex], the heat output is [itex]Q_2[/itex] and the work done by the working substance is [itex]W[/itex]. In that case all we know is that we have

    [tex]Q_1 = W + Q_2.[/tex]

    The equation I presented would be true if two conditions were met:
    1. During one cycle, the heat rejected by system [itex]A[/itex] is entirely transfered into the working substance;
    2. During one cycle the system [itex]A[/itex] undergoes a process that takes it from temperature [itex]T_A[/itex] to temperature [itex]T_B[/itex];
    Now I don't see why this would happen. Especially condition two, it doesn't make any sense. System [itex]A[/itex] would need to cool until it got to the same temperature as system [itex]B[/itex].

    I don't see any reason why this process would occur. Furthermore, if it really occured it would be really strange, because at the end of the cycle, the two systems [itex]A[/itex] and [itex]B[/itex] would be at the same temperature!

    So it doesn't make any sense. So how this equation is true? Why the heat input can be given by it anyway? How can one actually make sense of this?
  2. jcsd
  3. Sep 20, 2016 #2
    Can you cite a specific reference where the first equation is used to solve a heat engine problem, and can you summarize the analysis?
  4. Sep 20, 2016 #3
    I don't have one specific book reference, indeed I searched for some example of this in books, but couldn't find any one. I've seem this being done in some thermodynamics lectures, in some quite simple examples.

    To give one example, summarizing the analysis:
    In that example, the analysis was to use that if the machine has efficiency [itex]\eta[/itex], then [itex]W = \eta Q[/itex]. If a Carnot engine operating between the reservoirs has an effiency [itex]\eta_C[/itex] then [itex]W = \eta Q \leq \eta_C Q[/itex], so that the maximum work would be [itex]W_{\mathrm{max}} = \eta_C Q[/itex]. In the end this heat is found through that equation. For this special case it becomes [itex]Q = c \rho V \Delta T[/itex], where [itex]c[/itex] is the specific heat of water and [itex] \rho[/itex] its density.

    Again, I can't see why [itex]Q[/itex] is given like that, with that equation and that specific temperature interval.

    This example makes clearer the kind of analysis I'm in doubt?
  5. Sep 20, 2016 #4
    I can help you with this. Let's work the problem.

    What is the final temperature of the tank? What is the change in entropy of the tank? If ##Q_A## is the heat transferred to the river, what is the change in entropy of the river (considered an infinite reservoir)?

  6. Sep 20, 2016 #5
    Thanks for the aid.

    Now, with regards to the final temperature of the tank, I believe that's my greatest doubt. I mean, I thought that all heat engines operated between heat reservoirs, that is, their temperature were fixed. Are we dropping this assumption here? If so, my next guess would be that the tank would reach thermal equilibrium with the working susbtance at some temperature [itex]T'[/itex], but I don't know how we could find this temperature here, because we don't know the initial temperature of the working substance.

    Indeed by the wording of the problem there seems to be three systems here: the river (system [itex]A[/itex]), the water tank (system [itex]B[/itex]) and the working substance. If there was no working substance, the tank would reach thermal equilibrium with the river and since the temperature of the river is fixed this would mean [itex]T' = T_A[/itex].

    The change in entropy of the tank, in this particular case is simple, because we have constant heat capacity [itex]C = \rho V c[/itex] where [itex]c[/itex] is the specific heat of water at constant volume. In that case

    [tex]\Delta S_B = \int_{T_B}^{T'} \dfrac{dQ}{T} = \int_{T_B}^{T'} C\dfrac{dT}{T} = C \ln \left(\dfrac{T'}{T_B}\right).[/tex]

    Now, if [itex]Q_A[/itex] is the heat transfered to the river, being at constant temperature we would have [itex]T_A\Delta S_A = Q_A[/itex].
  7. Sep 21, 2016 #6

    Philip Wood

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    Gold Member

    Hope it's not rude of me to chip in, but I wouldn't worry about the temperature of the working substance. In a Carnot cycle the ws almost reaches the temperature of the high temperature heat reservoir because of (the correct amount of) adiabatic compression, so along the 'top' isothermal there is negligible difference between temperature of the reservoir and that of the ws. Don't forget, this is an IDEAL engine! The same applies along the bottom isothermal, whose temperature the ws reaches by adiabatic expansion.

    If I may say so, I think your main difficulty arose because the question didn't say that during an individual cycle there isn't enough heat taken in for the top reservoir to change its temperature significantly. For each individual cycle the engine can be treated as operating between reservoirs of 'fixed' temperature. But over the course of many cycles, the temperature of the top reservoir falls, so the efficiency of the engine keeps falling. This is the argument I pursued in order to answer the question (without using the idea of entropy), but I'll now get out of the way.
  8. Sep 21, 2016 #7
    Excellent analysis!!!

    What you are missing here is that we are not going to put the working substance through just one pass of a cycle (say, a Carnot cycle), and the cyclic passes that we put the working substance through are not going to all be identical. But the change in entropy for the working substance over each of the passes is going to be zero.

    Over each individual cycle, we are only going to remove a tiny amount of heat from tank B and transfer a tiny (smaller) amount of heat to the river A. So the Carnot cycles are going to start out long and skinny on a P-V plot, with the adiabatic expansion and compression legs very long, and the isothermal expansion and compression legs very short. In the first cycle, the temperature of the tank (hot reservoir) will drop slightly, and, in each subsequent cycle it will drop a little more. As the temperature of the tank decreases from cycle-to-cycle, the adiabatic expansion and compression legs will get shorter until, in the end, the temperature of tank B approaches the temperature of the river A. At that point, we will have "run out of gas" (i.e., run out of driving force), and we will no longer be able to do any work. So, based on this, the final temperature T' that you are looking for will be ##T_A##.

    You now have enough information to calculate the maximum work.
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