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Carnot Engine cycle on PV Diagram

  1. Mar 26, 2017 #1
    I posted this in the engineer / comp science thread, but I've had no one reply or help. I really could use some guidance and I don't know where else to post.

    1. The problem statement, all variables and given/known data
    I am tasked to create a PV Diagram of a Carnot Engine Cycle. I must find pressure, volume, Q, W, ΔU, and ΔS on all four points. This is what has been given to me by my teacher:

    https://lh6.googleusercontent.com/LupjMh6SJxv2M4loxPxaSIMEaU6wtMeMfz0_jjY6dgyyEewHwdgzplahrV0Xidfkjpn4dfu3h0wcjoc=w1920-h1021
    TCold = 300 K
    THot = 1700 K
    pc = 1.01*105 Pa
    vc = 0.01 m3
    Qa to b = 300 J
    γ (gamma) = 1.40

    2. Relevant equations
    (1) p1v1=p2v2
    (2) W=nRT ln(V2/V1)
    (3) p1v1γ=p2v2γ
    (4) W= p1v1-p2v2 / (γ-1)
    (5) T1V1(γ-1) = T2V2(γ-1)
    (6) W = nRT ln(V2/V1)
    (7) pv = nRT

    3. The attempt at a solution
    Using the above equations I managed to get point b's pressure and volume. What I got for point b:
    vb = 0.028 m3
    pb = 2.389*104 Pa
    First, I used the gas law equation (7) to get moles. This came out to n = 0.405 moles. Then, I used equation (5) to get volume, solving for Vb(y-1). I then used equation (3) to get pressure, solving for pb. I then got W = -852.7 J for the path b to c using equation (4). This seems kind of odd to me and I'm not sure if its correct because it's doing more work than the amount of heat it is providing. I assume Qa to b = is QH ? It seems so low though.

    I'm trying to figure out how am I going to get points d and a without knowing pressure or volume on those points. A class mate had suggested I used equation (5) for a to b and c to d, but those paths are isothermal. Isn't equation (5) adiabatic only? I don't know if I can even use the gas law because I'd need pressure or volume and I don't have that.

    There is also the idea that I have to use equation (6) and solve for volume that way. The problem is I don't know where to start. I'm a little rusty on my calculus (it's been about 4 years). I tried to break it down to W = nRT(ln (V2) - ln (V1))= nRT( 1/V2 - 1/V1), but this doesn't seem to help and I may have done it wrong.

    I'd be grateful for any kind of help.
     
    Last edited: Mar 26, 2017
  2. jcsd
  3. Mar 26, 2017 #2

    vela

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    Did you mean ##Q_{A \to B} = 300\text{ J}##? 300 K doesn't make sense.

    Try figure out what you easily can from the given information. For example, what's the efficiency of the engine? If you know that, you can figure out ##Q_{C to D}## and W and so on.
     
  4. Mar 26, 2017 #3
    Yeah, sorry I had the wrong unit there (fixed it).

    I could try using e = Th - Tc / TH to get the efficiency and then use e = W / QH in some way to get volume or pressure perhaps.

    I wonder if perhaps I have the temperatures mixed up. and Tc should be c → d ?
     
  5. Mar 26, 2017 #4

    vela

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    You can figure out QC from e and QH.

    On each leg, either Qi to f=0 (adiabatic) or ΔUi to f=0 (isothermal). Take C to D. It's an isothermal process, so you know the internal energy doesn't change. The first law then tells you that W(C to D) = ±QC (I'm not sure what sign convention your class follows). Well you found QC earlier, so you know W(C to D), so you can figure out VB using (6). Then use the ideal gas law to find pB. Just keep going like that around the cycle.
     
  6. Mar 26, 2017 #5
    Thank you you've been very helpful.

    I have just 1 last question though. How can I get VB from ln (V2/V1) ? would eW/nRT / V2 = V1 work?
     
  7. Mar 26, 2017 #6

    vela

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    Close. The units don't work out there. You have ##\rm{m^{-3}}## on one side and ##\rm{m^3}## on the other. ##V_1## should be some multiple of ##V_2##. Other than the algebra mistake, though, you have the right idea.
     
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