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Carnot cycle

  1. Nov 26, 2014 #1
    1. The problem statement, all variables and given/known data

    A heat engine containing an ideal gas has a reversible cycle which consists of 2 constant
    volume segments with V = 1ℓ and V = 3ℓ and two constant pressure segments with P = 1 atm and
    P = 2 atm (see figure below). The temperature at point “c” is Tc = 273 K.

    A) Is the path of the cycle clockwise or counter-clockwise? Explain.
    B) How much work is done by the engine in one cycle?
    C) What is the efficiency of the engine? Notice that this is a reversible engine, so recall
    the efficiency of a Carnot engine and use the ideal gas law.

    2. Relevant equations
    W = PdV
    W = VdP

    3. The attempt at a solution
    A) since at point c the engine is at the low temp Tc then since it is a heat engine the cycle must be going counter clockwise to go from high to low temp (please check my understanding on this)
    B) this is where I am stuck
    W = Wac + Wcd + Wdb + Wba
    W = VdeltaP(ac) + PdeltaV(cd) + VdeltaP(db) + PdeltaV(ba)
    W = (1L)(1atm - 2atm) + (1atm)(3L-1L) + (3L)(2atm-1atm) + (2atm)(1L-3L)
    W = -1atmL + 2atmL + 3atmL - 4atmL
    W = 0
    this leads me to believe i did something wrong. please help :(

    Attached Files:

  2. jcsd
  3. Nov 26, 2014 #2

    Simon Bridge

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    A. an "engine" usually has to do work on it's surroundings - which leg has to do positive work?
    B. the work done in a process is the area under the P-V diagram.
    ... so at constant volume, the work is zero.

    Note: this is not a Carnot cycle.
  4. Nov 26, 2014 #3
    im not entirely sure what you mean in A or what you mean by which leg.
    but for part B) the cycle has a height of 1 atm and a width of 2L so it has an area of 2atmL which means its does 2 atmL of work?
  5. Nov 26, 2014 #4

    Simon Bridge

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    A. One leg of the cycle would be going from a to b, or from b to a ... it's an analogy from the terms used for a foot race. Perhaps you'd prefer "stage" or maybe "process" since each leg has only one process in it?

    B. well done - though the person marking may prefer you put work in Joules.
  6. Nov 26, 2014 #5
    so it would be 203 joules.
    for part A) the expansions would do the positive work?
    for part c) for a reversible engine
    e = 1-Tc/Th
    whe have Tc we need Th
    using the ideal gas law we find number of moles by using the known temp at point c
    n = PV/RT = (1atm)(1L)/(.0821 atmL/molK)(273K) = .0446 mol
    now we can find the temps at the other points
    point a) T = PV/nR = (2 atm)(1L)/(.00366atmL/K) = 546 K
    point b) T = PV/nR = (2atm)(3L)/(.00366 atmL/K) = 1639 K
    point d) T = PV/nR = (1atm)(3L)/(.00366atmL/K) = 820 K
    Th = 1639 K
    plugging in
    e = 1 - Tc/Th = 1 - 273/1639 = .833
    e = 83.3%
  7. Nov 26, 2014 #6

    Simon Bridge

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    Well done: - $$\int_{V_1}^{V_2}P(v)\;\text{d}v > 0 : V_2>V_1$$ ... we have a definition for work where work done by the engine is positive.
    (Some people rework the equations the other way - but the equation is the same with a different sign.)

    C) the carnot effciency is the maximum efficiency.
    You can tell which temperatures are lowest and highest by sketching isotherms on the graph - which would have saves you about 2/3 of that work ;)
  8. Nov 26, 2014 #7
    thanks for your help.
    so it goes clockwise because that way V2>V1 and the work is positive?
    for C) how would I sketch isotherms on a PV graph?
  9. Nov 27, 2014 #8

    Simon Bridge

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    Look at the PV diagrams for isothermal processes.
  10. Nov 27, 2014 #9

    Andrew Mason

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    A heat engine performs net positive mechanical work. Work is done in only two sections a-b and d-c. Net work done BY the system can only be positive if Wa-b (ie. the area under a-b) is positive. This means the system is expanding from a-b. So it is being compressed from d-c.

    I am not sure where you get W = VdP. This is a reversible cycle so [itex]W = \int PdV[/itex] where W is the work done BY the system, P is the pressure of the system and V is its volume. There is ZERO work done from c-a and from b-d.

    To determine the net work done per cycle, take the area under a-b (work done by the system in expansion) and subtract the area under d-c (work done on the gas in compression).

  11. Nov 27, 2014 #10
    yeah i knew the W = VdP didnt seem right. So it would still be W = PdV but since dV = 0 then W = 0.
    W = Wac + Wcd + Wdb + Wba
    W = PdeltaV(ab) + PdeltaV(bd) + PdeltaV(dc) + PdeltaV(ca)
    W = PdeltaV(ab) + 0 + PdeltaV(dc) + 0
    W = (2atm)(3L-1L) + (1atm)(1L-3L)
    W = 4atmL + (-2atmL)
    W = 2 atmL
  12. Nov 28, 2014 #11

    Andrew Mason

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    Good. Now express that in Joules.

  13. Nov 29, 2014 #12
    oh right. thats about 203 J
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