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"Lost work" and irreversibility

  1. Aug 3, 2014 #1
    I'd like some help to understand how to calculate "lost work" for irreversible heat transfer across a finite temperature difference.
    I'd also be grateful for any links to the derivation of a clear, general expression for lost work (in terms of entropy).
  2. jcsd
  3. Aug 3, 2014 #2
    That's easy. Carnot cycle is your best friend when finding how much work you can squeeze out of a given temperature difference.

    [itex]\eta[/itex] = efficiency (fraction)
    [itex]T_{L}[/itex] = low temperature sink
    [itex]T_{H}[/itex] = high temperature source
    [itex]\eta = 1 - \frac{T_{L}}{T_{H}}[/itex]

    So, eta is your fraction of useful work, so 1-eta is your fraction of wasted work. Bear in mind, this is for a Carnot engine, an impossible engine to build in any case. For any real engine, you would need to know the efficiency and the input to find the wasted work (or heat, depending on how you look at it).
    Last edited: Aug 3, 2014
  4. Aug 3, 2014 #3
    Unfortunately, that is not the definition of 'lost work', even though one cannot convert all of the heat energy into useful work and you do lose all that heat to the environment.

    Lost work is the difference in work output of engine acting reversibly, where the change in entropy S=0, to that of an real engine acting irreversibly, where the change in S>0.

    Here is a derivation

    And a simpler way of explaining it,
  5. Aug 4, 2014 #4
    In that case, the only way you could find it is if they give you the second law efficiency, defined as


    In that case, the so called "lost work" (which is not a word I've ever heard in all my years of school dealing with thermodynamics and heat transfer, but I'll have to take your word for it) would then be able to be found by


    In terms of entropy, that's going to be a little trickier. I can tell you that


    I'm not sure how you would quantify the irreversibilities of the non-Carnot cycle without having either the second law of efficiency or some terms related to entropy generated. In a Carnot cycle, the entropy is conserved (ie no entropy is generated). When you add irreversibilities, you add generated entropy. I'm pretty sure you can use an entropy balance to find how the entropy generated term comes into play and use that to derive what you need.

    Okay, I obsessed over it but I think I found the answer. Assuming two isothermal reservoirs, I believe the work lost to be this:


    I can take you through the derivation if you need, but I'm going to bed for the night so it will have to wait until tomorrow.
  6. Aug 4, 2014 #5
    Thanks very much for the rapid responses. I really appreciate this support.
    My confusion has been partly dealt with, in that I can now calculate the energy which becomes unavailable as work in an irreversible process (as a function of entropy increase). A further issue continues to nag at me:

    Does it make sense to insert a Carnot engine between two systems at different temperatures (as opposed to between two reservoirs)? ie using irreversible heat transfers to and from such an engine will see it grind to a halt, surely, after extracting some amount of work...but I'm not clear about how to apply the above general expression to calculate lost work in this non standard situation.
  7. Aug 4, 2014 #6
    Hmmm, I'm not sure what the distinction is you're trying to make between systems and reservoirs in this case. For instance, a reservoir, as I define it, is an infinite source of energy at a given temperature. How is a "system" defined in this case?

    Either way, all heat transfer processes are irreversible. Heat only flows in one direction, from high temperatures to low temperatures. You can never get energy from the low temperature reservoir going to a high temperature reservoir. Remember, the entropy of an isothermal process is:

    [itex]S_{isothermal} = \frac{Q_{transferred}}{T_{constant}}[/itex]

    I think you were getting at the fact that the entropy processes of a heat engine operating between two reservoirs are:

    [itex]S_{high temp} = \frac{Q_{in}}{T_{high}}[/itex]


    [itex]S_{low temp} = \frac{Q_{out}}{T_{low}}[/itex]

    I'm not exactly sure what you mean by "system," once again, but I think you mean something like finite sources at an initial given temperature? In this case, it's dynamic, and the lost work derived above for such a case is invalid. Basically, what would happen is this: the engine would run as long as [itex]T_{source} > T_{sink}[/itex], at which point it would stop producing work. The time it takes for that to happen depends on the heat storage capacity of both systems (cp and the mass of each system, ultimately the energy stored per unit temperature).

    Update: one more thing, the change in entropy of a system with finite heat storage is:

    [itex]S_{2} - S_{1} = C LN(\frac{T_{2}}{T_{1}})[/itex]

    where C is the heat storage in unit energy per unit temperature (cp times mass). This expression only holds for solids and liquids. Using a gas reservoir would make it far more complicated, so I'm gonna have to pass on deriving that unless your question specifically asks for that. Honestly, I'm not sure there's a straight forward way to derive that.
    Last edited: Aug 4, 2014
  8. Aug 4, 2014 #7
    "system" My understanding is that a reservoir maintains constant temperature when transferring heat, but a system of finte capacity does not.
    "all heat transfer processes are irreversible" This has me a bit confused. Surely the (idealised), quasistatic process by which eg a reservoir at T+dt transfers heat to a system at T is considered reversible? ie dS=dQ(rev)/T ? All real Qs are irrev, perhaps I misunderstood?

    The equation you mention S2-S1=C Ln(T1/T2)? I'd be grateful to know where I can find that derived please.
  9. Aug 4, 2014 #8
    You know, I'm not sure what the idealized case is where the temperature difference is a differential quantity. I guess, since dT->0, then Q->0 as well, so I'm not sure why the idealized case would be worth investigating, but some homework problems are like that, aren't they? I'm not sure what to tell you there, other than try to use the definition of entropy or a differential control volume at the boundary to try to determine what the number should be. That one is a little bit trickier.

    Here's a link to the derivation of a solid or liquid of constant heat capacitance:
  10. Aug 5, 2014 #9
    Many thanks
  11. Aug 7, 2014 #10
    To get the work lost from irreversibilties, you need to calculate the Δenergy in the isentropic scenario and the Δenergy for the real process in question. The difference between these values will be the work lost.
  12. Aug 7, 2014 #11
    That is correct. However, if you take a Carnot cycle, replace the isentropic processes with processes that increase the entropy, and calculate the loss in energy due to the entropy increase, you should end up with S_gen*T_low, where T_low is also referred to as the dead state temperature (I think). I will try to post some of the derivations if I get a chance.
  13. Aug 7, 2014 #12
    Here's the derivation, I tried some of the other ones and they were too hard for me to be confident that I did them right (like the finite source and sink).

    Anyway, here's the derivation and hope it helps a bit.

    Attached Files:

  14. Aug 10, 2014 #13
    Thank you. This is very helpful.
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