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Development of Clausius Inequality

  1. Sep 2, 2015 #1
    I am facing some doubts trying to understand the illustration my textbook has adopted for the development of the Clausius inequality for thermodynamic cycles.I have attached an image of the content from my textbook.

    As one could see the author has assumed a closed system connected to a thermal energy reservoir at a constant thermodynamic temperature of TR through a reversible cyclic device.The cyclic device receives heat δQR from the reservoir and supplies heat δQ to the system whose temperature at that part of the boundary is T (a variable) while producing work δWrev.The system produces work δWsys as a result of this heat transfer.

    In the course of analysis he again assumes, "We now let the system undergo a cycle while the cyclic device undergoes an integral number of cycles."This implies that he is assuming the closed system is undergoing a cycle receiving heat δQ and producing a net amount of work δWsys without rejecting any heat.This is an impractical assumption itself.From the arrangement as shown the system can only produce work in a process absorbing heat δQ and producing a net work of δWsys and not produce work in a continuous cycle.To undergo a cycle it must reject some heat which by assumption is only possible for the given arrangement if the direction of δQ is reversed.Again doing so would require the direction of operation of the reversible cyclic device to be reversed and thus for a cycle of the system to be accomplished, the reversible cyclic device must be reversed.

    Thus the statement "the system undergoes a cycle while the cyclic device undergoes an integral number of cycles" seems very confusing.

    Can somebody come up with an explanation behind it?

    Attached Files:

  2. jcsd
  3. Sep 3, 2015 #2


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    No it does not imply that. In one cycle, ##\delta Q## goes in (as you deduce correctly) and, on the same footing, ##\delta W_{\rm rev} ## and ##\delta W_{\rm sys}## go out of the system. The system is NOT a closed system !

    [edit] I see what you mean. In my fourth edition of Cengel it's figure 6-1. Let me check what is stated.

    The claim is this can't work and W has to be 0 at best, so you got it all right. I think the didactics of this section are weak (at least in 4th ed. Perhaps YC improved on it, but your quandary demonstrates that it's still open for further improvement Close inspection of your foggy picture on my clean screen shows it is exactly the same text. Oh well, it teaches us to read on before jumping up and crying out) .
    Last edited: Sep 3, 2015
  4. Sep 7, 2015 #3
    Yes your statement is correct if δQ is the differential heat transfer from the reservoir to the "combined system".In my text it is "δQR" at TR.
    The combined system could be either an open or a closed system.Why cannot it be a closed system?

    The Kelvin-Planck statement only enforces the restriction that for the combined system to operate in a cycle as assumed (δQR as the net heat transfer over a cycle and δWC=δWrev+δWsys as the net work transfer over a cycle) it can be δWC=δWrev+δWsys=0 at best consequently δQR=0 for δWC=0(for a reversible cycle undergone).Thus the net work and heat transfers for the combined cycle could be zero at best or could be negative (net work input δWC<0 and consequently δQR<0 as δWC=δQR for a cycle).The irreversible case implies there would be a net work input required in order to achieve the cycle and an equal amount of heat would be rejected to the reservoir.This is expected as some additional work is required to overcome irreversibilities in the cycle which is dissipated into a less useful form of energy(thermal energy).This case also implies complete conversion of work into heat which is perfectly possible.

    This should be true regardless of the combined system being taken as a closed or an open system.

    Please correct me if I am wrong.

    My confusion is somewhat different and let me explain: As you could check in the picture it has been shown a case of reversible heat transfer through a reversible cyclic device to another "system" which has been been shown to be a closed system (a piston cylinder device).For integral number of cycles of the reversible cyclic device it is assumed that this closed system undergoes a cycle.The closed system is also assumed to develop "net" work δWsys over a cycle exchanging "net" heat δQ in the cycle which is impossible again by Kelvin-Planck statement as is true and similar for the combined system considered.Thus if the "closed system" is a reversible one it could be δWsys=0=δQ at best which means the system consumes exactly the same work as it develops in the cycle and it rejects exactly the same amount of heat as it absorbs over the cycle.The directions of δQ and δWsys must be opposite to that shown in the figure in case of an irreversible operation of the closed system to undergo a cycle.This means that for the combined cycle to occur if the closed system is an irreversible one both δQ and δWsys and necessarily δQR and δWrev should be in reverse directions as assumed in the figure so that "net" work and heat transfers for the combined cycle and the cycle undergone by the closed system are negative.

    This is my depiction of the situation that I wish to verify with others.

    Attached Files:

    Last edited: Sep 7, 2015
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