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Mastering Physics: Heat Engine Problem

  1. Jun 3, 2013 #1
    The figure shows the cycle for a heat engine that uses a gas having gamma =1.25. The initial temperature is T1=300K, and this engine operates at 20 cycles per second.

    a.) What is the power output of the engine?

    b.) What is the engine's thermal efficiency?

    The link to this image is here:
    http://session.masteringphysics.com/problemAsset/1074111/4/knight_Figure_19_54.jpg

    This is how I attempted to solve this problem:

    Given: T1 = T2 = 300K
    P2 = P3
    P1 = 1 atm = 1.013 * 10^(5) Pa
    gamma = 1.25
    V1 = V3 = 600 cm^(3)
    V2 = 200 cm^(3)
    R = 8.31 J/mol*K or 0.08206 atm *L/mol*K

    Relevant equations:

    PV = nRT

    η = Wout/QH

    Power = Wnet/time

    Q_12 = W_12

    Q_23 = nc_pΔT

    Q_31 = nc_vΔT


    Finding number of moles

    n = PV/RT

    n = [1.013 * 10^(5) Pa(0.0006 m^(3))]/[(8.31J/(mol*K))(300K)]= 0.024 mol

    Finding P2 and P3

    P1V1 = P2V2

    P2 = [P1V1]/V2 = [1atm(0.6L)]/0.2L

    P2 = 3atm

    P2 = P3

    Finding T3

    P3V3 = nRT3

    T3 = P3V3/nR = 900K

    Finding W_12

    W_12 = nRT(ln(vf/vi)) = 0.024mol(8.31J/mol*K) (300K)(ln(0.0006m^(3)/0.0002m^(3))) = 65.73 J

    Finding W_23

    W_23 = P(vf-vi) = 3atm[(0.0006 - 0.0002)m^(3)] * 101325J = 121.59 J

    W_31 = 0 because it is an isochoric process

    Wnet = 187.32J

    Power

    P = Wnet/time = 187.32J(20(s^(-1))) = 3746.4 W = 3.746 kW

    I got it wrong. Why was gamma given? I thought it was only used for adiabatic processes. I also don't know which c_p and c_v to use since the type of gas is not given.
     
    Last edited by a moderator: Jun 3, 2013
  2. jcsd
  3. Jun 3, 2013 #2

    TSny

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    Hello ricebowl07 and welcome to PF!

    Did you consider the sign of the work for the isothermal part? (In other words, what is Vi and what is Vf for that part?)

    Also, you might want to keep one more significant figure in your calculation of the number of moles.

    [You should be able to find Cv amd Cp from gamma.]
     
    Last edited: Jun 3, 2013
  4. Jun 3, 2013 #3
    Thank you! I finally got the right answer.
     
  5. Dec 9, 2013 #4
    Hello, I have worked my way through this problem, except with a different value (16 cycles per second vs 20) and I am having trouble finding my thermal efficiency. You said find if from gamma=1.25, and I know that gamma=C_p/C_v, but I'm not sure where to go from there.
     
  6. Dec 9, 2013 #5

    TSny

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    Hello, cluke95. Welcome to PF!

    There is a simple relation between CP and CV for any ideal gas. It has the form: CP = CV + _ ? _ .

    You can combine this equation with the equation γ = CP/CV to determine the values of CP and CV. Then you can calculate the heat QH to determine the efficiency.
     
  7. Dec 9, 2013 #6

    rude man

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    My take:
    You don't need either C_p or C_V.
    (a): what is net work on a p-V diagram?
    (b): use the 1st law and the ideal gas law.
    Hint: what is ΔU for a complete cycle?
     
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