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Heat engines: how can we yield work?

  1. May 25, 2012 #1
    So, I'm studying for the MCAT. There..I've said it. :) But, it's not a homework question: I just don't understand the concept.

    My study book's thermodynamics section includes a very rough sketch of a heat engine to describe the 2nd Law of Thermodynamics. A side note in the text says that if this section is confusing, just ignore the heat engine and simply know the 2nd Law of Thermodynamics (...that heat cannot be turned into work 100% in a cyclical process); however, I am interested in the heat engine (mainly because I don't understand how it works, haha).

    My lapse in understanding: how can it take less work to compress a piston to its original state than how much work the piston did? It takes work to compress the piston, but how can that be any less than what work was done by the system through expansion?

    I get the first "part" of the heat engine. We have a piston and it's in an isothermal environment. We remove some force holding the piston and thus the volume expands and pressure decreases. Since it is isothermal, however, heat transfers energy into the system to keep it at the same temperature (ΔE = q + w where ΔE = 0) via a hot reservoir. The heat we put into the system = the work we get out. Yes, this makes sense.

    But, in all the websites and textbooks I've looked at, then somehow we can compress the piston back to its original state by doing work to the system--but we use less work!

    Here's what I thought should be happening: So, let's say the gas expands and does 10J of work in the first "part" of the engine. To keep it isothermal, a hot reservoir transfers 10J of energy into the system. So, now we have an expanded cylinder and we've let some gas expand isothermally by adding some heat (producing a 100% conversion to work). I would think, though, that to "reset" the piston back to its normal state, we're going to do work to the system (10J of work) and a cold reservoir will heat up by 10J (to keep it isothermal).

    No net work. We have to use just as much work to "reset" the piston. I don't see how we can reset the piston by somehow using less work (and thus have positive net work done by the system).


    I realize this is pretty elementary. If you have any links I can peruse, let me know. I've looked at a fair few, but maybe if I look at them and tell you precisely where I don't understand what happened, we can sort through this conundrum. :(
  2. jcsd
  3. May 26, 2012 #2

    Andrew Mason

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    The work done by the gas between two states is ∫PdV. If you understand that, you can understand how a heat engine produces net work over a cycle.

    The area under a path on a PV diagram is equal to ∫PdV. That is just math.

    The simple answer is that the area under the forward part of the cycle (positive work done by the gas) is greater than the area under the return part of the cycle (negative work done by the gas).

    This occurs because the expansion occurs at high temperatures and compression occurs at lower temperatures and, therefore, at lower pressures. A simple cycle would be 1. constant volume (V1) heat flow into gas from P1 to P2 2. continued heat flow into gas and constant pressure expansion at P2 to V2. 3. constant volume (V2) heat flow out of gas from P2 to P1. 4. continued heat flow out and constant pressure compression at P1 to V1. Repeat.

    The expansion occurs at high pressure P2. The compression occurs at low pressure P1. Net work is P2(V2-V1) - P1(V2-V1) > 0

  4. May 26, 2012 #3


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    By adding heat after compression and removing heat after expansion.
    The heat isn't added to keep it isothermal, it is just added in an isothermal way. The heat is added to increase the pressure and enable the creation of mechanical work.
  5. May 26, 2012 #4


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    Try starting with the idealized Carnot cycle. There are four steps in the cycle, not two. Two of the steps are isothermal and two are adiabatic.


    In "real" heat engines the separation of the four steps is not so clear cut, but the same general princple applies.
  6. May 26, 2012 #5
    Thank you all for your replies: I'm getting closer in understanding!

    Right! :D The area under the PV curve is equal to the work done: I do get that. What I didn't get was why the second curve (for the compression) is a different curve than the original one.

    Thank you, Andrew Mason: I now understand the different pressures and work is definitely P*ΔV. And the heat loss in step three is LESS than the heat in step one because I'm at a larger volume in step three, right? And that difference there is my work, right? So, I put 50J of heat into the system and I get 50J of work out (heating and expansion steps). But, then to reset, I put in a cold reservoir that removes 10J of heat (reducing pressure) and it will take then also 10J of work to compress, right? So, I have a net of 40J work done by the system?

    However, in your example the expansion occurs at a constant pressure. But, in my study book and in the linked Carnot cycle, pressure actually decreases through expansion. I guess this is a complication on top of the simple cycle, right?



    OK, but why when I add heat adiabatically do I have a greater pressure increase than a volume decrease? Why couldn't I have had a greater volume increase than pressure decrease? Why is there a preference?

    I understand why heat added in a isothermal way causes a volume increase greater than a pressure decrease.

    ΔE = q + w = 0
    ΔE = 10 + x = 0
    x = -10J

    And work done by the system can only occur with a volume increase.


    Thanks! I hadn't found Wikipedia's Carnot cycle (just their heat engine and 2nd law page). I'll look at this real hard now...
  7. May 27, 2012 #6

    Andrew Mason

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    If the pressure decreased at the same rate as volume increased, then PV = constant. But PV = nRT so that would mean that T would be constant so there is no change in U. If no heat flow occurs (Q=0), this means that no work can be done (Q = ΔU + W). Since work is being done, T must decrease and this means that PV must decrease. Since V is increasing, P must decrease at a greater rate than V increases.

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