1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Thermodynamics - efficiency from pv diagram

  1. Sep 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Determine the efficiency for the cycle shown in the figure, using the definition. Assume that the gas in the cycle is ideal monatomic gas.

    Compare with the efficiency of a Carnot engine operating between the same temperature extremes.

    see attachment

    2. Relevant equations

    the cycle is composed of two isobaric and two isovolumetric paths.

    isovolumetric/constant volume:
    work, W = 0
    heat, Q = deltaU = nC_v(deltaT) where deltaU is change in internal energy, n is number of moles, C_v is constant volume (monotomic so = 3R/2 where R = 8.314), delta T is change in temp

    isobaric/constant pressure
    work, W = p(V_2 - V_1) where p is pressure in pascals, V_2 is final volume, V_1 is initial volume
    heat, Q = deltaU + W = nC_p(deltaT) where C_p is constant pressure = C_v + R

    ideal gas law: PV = nRT

    efficiency, e = W/Q_h = (Q_h - Q_c)/Q_h = 1 - (Q_c/Q_h) where W is work, Q_h is heat energy (high), and Q_c is cool heat energy (low)

    3. The attempt at a solution

    i intend to use efficiency, e = W/Q_h, since W = 0 for each of the two isovolumetric paths, the work only comes from the two isobaric paths, i calculated:

    top horizontal path to be W = 2431.8 joules, and the bottom horizontal path to be W = 1215.9 joules, so net work would be W_net = 3647.7 joules

    now in order to solve for Q_h or even Q_c for that matter I need to be able to solve for heat Q, in order for that i need to know the number of moles and the temp in kelvin.

    since i am only concerned with Q_h, do i only need to know the heat energy Q from the top horizontal (isobaric) path?

    so Q = nC_p(deltaT) and using ideal gas law and solving for T and subsituting in for T is the constant pressure heat equation i get:

    Q = nC_p(PV/nR) so the n's cancel out, and C_P = C_v + R = 3(8.314)/2 + 8.314 = 20.785, here is where i am confused. since the top horizontal path is isobaric, the volume changes thus what volume do i use for V?

    so currently my equation for Q would be Q = nC_p(PV/nR) = 20.785[((607950Pa)(V))/8.314]

    any help/clarification appreciated
     

    Attached Files:

    • img.JPG
      img.JPG
      File size:
      3.8 KB
      Views:
      546
  2. jcsd
  3. Sep 13, 2008 #2

    Ygggdrasil

    User Avatar
    Science Advisor
    2015 Award

    Remember you're looking for ΔT. You need to find the temperature at each point using V1 and V2 then take the difference in temperature.
     
  4. Sep 13, 2008 #3
    i cant see the attachment for some reason, but im pretty sure its the same problem i have already.
    *i think you calculated the work wrong - the work for the second (bottom isobaric path) should be negative, so the total work will end up being 1215 J.

    Then you need to find Qh (heat absorbed) in order to get the efficiency; e=W/Qh
     
    Last edited: Sep 13, 2008
  5. Sep 13, 2008 #4
    oh and for finding Q (for the top isobaric path) you would use the equation Q=n*Cp*deltaT
    You know it's a monatomic ideal gas, so Cp = (5/2)R
    Since you don't have n or deltaT, you use the ideal gas law: P*deltaV=n*R*deltaT
    Solve for deltaT and plug in the Q equation (the n, and R will end up canceling out)
    The rest should follow the same way... I got the right answers, but for the other Q values, I think I assumed some things and I just happened to get lucky with the other numbers - so I cant' really help you out on the rest of it =/
     
  6. Sep 13, 2008 #5
    i tried what you said and got Q = 6079.5 joules for top isobaric, and Q = -1823.85 joules for bottom isobaric

    so e = W/Q_h = 1215/(6079.5-1823.85) = 1215/4255.65 = 28.5% which was wrong, what now?
     
  7. Sep 13, 2008 #6
    remember that Qh is heat absorbed. So if you got a negative number for Q of the bottom isobaric path, that would be Qc not Qh. I can't really compare my numbers with yours because I kept the units in atm. and liters. But solve Q for the isovolumetric paths (one will be negative and the other will be positive - use the positive number for Qh)
     
  8. Sep 13, 2008 #7
    i still can't seem to get it, how do i find the q for the isvolumetric paths when Q = 3/2(P*deltaV) where pressure changes over each pf the two isovolumetric paths?

    help please
     
  9. Sep 14, 2008 #8

    Ygggdrasil

    User Avatar
    Science Advisor
    2015 Award

    For an ideal gas ΔU = nCvΔT

    For a monatomic ideal gas Cv = (3/2)R

    Therefore, ΔU = (3/2)nRΔT

    Because of the ideal gas law, nRΔT = Δ(PV) if n is fixed.

    Therefore, ΔU = (3/2)Δ(PV)
     
  10. Sep 14, 2008 #9
    let 6atm, 2L = 1, 6atm,6L = 2, 3atm, 6L = 3, 3atm, 2L = 4:

    1-->2 isobaric expansion
    Q = nC_pT = n(5R/2)(PV/nR) = 5/2(PV) = 5/2(607950)(0.006-0.002) = 6079.5

    2-->3 isovolumetric compression
    Q = n(3R/2)(PV/nR) = 3/2(PV) = 3/2(303975-607950)(0.006) = -2735.8

    3-->4 isbaric compression
    Q = n(5R/2)(PV/nR) = 5/2(PV) = 5/2(303975)(0.002-0.006) = -3039.75

    4-->1 isovolumetric expansion
    Q = 3/2(PV) = 3/2(6-7950 - 303975)(0.002) = 911.93

    work, W = (607950*0.004) - (303975*0.004) = 1215.9

    e = W/Q_h = 1215.9/6079.5 = 20% which is incorrect

    now what?


    and assuming carnot efficiency e_c = 1 - T_c/T_h

    at 1, T = PV/nR = 607950(0.002) = 1215.9 holding nR constant
    at 2, T = 607950(0.006) = 3647.7 ------> T_h
    at 3, T = 303975(0.006) = 1823.85
    at 4, T = 303975(0.002) = 607.95 ----> T_c

    so carnot efficiency = 1 - (T_c/T_h) = 1 - (607.95/3647.7) = 0.83 = 83.33%

    correct??
     
    Last edited: Sep 14, 2008
  11. Sep 14, 2008 #10
    Your efficiency for the first part should come out to be 17%.
    But now I am stuck at "Compare with the efficiency of a Carnot engine operating between the same temperature extremes". How do I do that?
     
  12. Sep 14, 2008 #11
    i think it is a ratio of e_carnot/e where e = 17%

    ^^^^that ratio is correct, e_carnot/e = 83.33/17 = 4.90
     
    Last edited: Sep 14, 2008
  13. Sep 14, 2008 #12

    Ygggdrasil

    User Avatar
    Science Advisor
    2015 Award

    For qh, you didn't take into account the heat absorbed during the 4-->1 isometric heating.

    The Carnot efficiency that you calculated is correct.
     
    Last edited: Sep 14, 2008
  14. Sep 14, 2008 #13

    Ygggdrasil

    User Avatar
    Science Advisor
    2015 Award

    You can't calculate the real temperatures since you don't know n, but you can find the temperature at each point in terms of n. Since both T_h and T_c will contain the term n, it will cancel out and you can still find the Carnot efficiency.
     
  15. Sep 14, 2008 #14
    isometric?
     
  16. Sep 14, 2008 #15

    Ygggdrasil

    User Avatar
    Science Advisor
    2015 Award

    Constant volume heating from step 4->1. I mistakenly wrote expansion instead of heating.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Thermodynamics - efficiency from pv diagram
Loading...