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Homework Help: Otto cycle temperatures

  1. Sep 13, 2008 #1
    1. The problem statement, all variables and given/known data

    Gasoline engines operate approximately on the Otto cycle, consisting of two adiabatic and two constant-volume segments. The Otto cycle for a particular engine is shown in figure below. see atchmt

    1. Find the maximum temperature in terms of the minimum temperature T_min. (gamma should be part of answer, where gamma = Cp/Cv -->Cp = 5R/2,Cv = 3R/2 --> R = 8.314

    2. How does the efficiency compare with that of a Carnot engine operating between the same two temperature extremes? Note: Figure neglects the intake of fuel-air and the exhaust of combustion products, which together involve essentially no net work.

    e_otto < e_carnot or e_otto = e_carnot or e_otto > e_carnot?

    2. Relevant equations

    ideal gas law, PV = nRT

    otto cycle efficiency, e = 1 - (Cp/Cv)^-1 where Cp = 5R/2, Cv = 3R/2 where R = 8.314 constant

    for otto cycle TV^(gamma -1) = constant where gamma = Cp/Cv

    3. The attempt at a solution

    for the problem about the temperature i used the TV^(gamma - 1) = constant equation and used T_min for T and thus used the figure to determine the corresponding volume

    i used a rough calculation with PV = nRT to get a general idea of temp magnitude
    pt3: P = 3P_2, V = V_1/5
    pt4: P = 2P_2, V = V_1
    pt1: P = P_2/2, V = V_1
    pt2: P = P_2, V = V_1/5

    and using T = PV/nR and subbing in each pressure and volume, and holding P_2, V_1, n and R = 1, i found pt4 = T = 2, pt3 = T = 3/5, pt2 = T = 1/5, pt1 = T = 1/2, so i assumed T_min is at pt2, and that T_max is at pt4.

    how do i use the TV equation using T_min and gamma to describe T_max?
     

    Attached Files:

  2. jcsd
  3. Sep 14, 2008 #2
    no ideas?

    i had another look and Tmax is at 3, and Tmin is at 1

    using PV= nRT:
    Tmax = PV/nR = [3P_2(V_1/5)]/nR assume n, R, P_2 and V_1 = 1
    Tmin = PV/nR = [P_2/4(V_1)]/nR

    so assuming n, R, P_2, V_1 = 1, then Tmax = 3/5 and Tmin = 1/4

    so Tmax in terms of Tmin ---> (3/5)x = 1/4 --> x = 5/12 so Tmax = (5/12)Tmin

    i'm not sure if i am correct because i need gamma in the answer, gamma = Cp/Cv
     
    Last edited: Sep 14, 2008
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