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Carnot engine

  1. Apr 6, 2014 #1

    Maylis

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    Hello,

    I have the problem statement attached in the thumbnail. I am confused about this question. I looked at the solution manual, and they are saying that the pressure at states 3 and 4 are the same, as well as at states 1 and 2. I don't see how this is. Looking at the carnot cycle PV diagram, clearly all 4 states are at different pressures.

    Anyways, I did the calculation with the same pressures. Now I am working on part C, and I thought since step 1-->2 is an isotherm, I could use the equation Q=RT ln(P2/P1). However, if P2 and P1 are the same, then Q is zero. What is going on here?


    ImageUploadedByPhysics Forums1396769557.990116.jpg

    ImageUploadedByPhysics Forums1396769606.310728.jpg
     

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  3. Apr 6, 2014 #2

    Simon Bridge

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    Do we not need figure 8.2 as well?
    I can see how a carnot cycle can operate between three pressures but not just two.
    Are you sure you are reading the correct key - what you describe sounds like a simple steam engine cycle.
     
  4. Apr 6, 2014 #3

    Maylis

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    My apologies but figure 8.2 is a TS diagram of the Carnot cycle which I drew in my attempt, I am certain I am reading the correct key
     
    Last edited: Apr 6, 2014
  5. Apr 6, 2014 #4

    Maylis

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    Here is a photo of the part of the solution regarding the pressure. Table F.1 is a saturated steam table

    ImageUploadedByPhysics Forums1396810485.305649.jpg
     
  6. Apr 6, 2014 #5

    Simon Bridge

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    Oh the states include a liquid state and a bunch of saturated vapor states.
    Looks like changes of phase are happening during the cycle.
    You would be expected to account for this in your analysis.

    Of course it may just be that the pressures are the same to 3dp (i.e. the isotherms are almost flat).
    Or that the answer key is just wrong.
     
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