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Otto Cycle: net work done

  1. Jun 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that the net work done by the otto engine per cycle is
    Cv(Tc - Tb)(1 - Ta/Tb)

    PV diagram:
    [PLAIN]http://www.antonine-education.co.uk/physics_a2/options/module_7/topic_4/Otto_indic.gif [Broken]

    2. Relevant equations
    Work done by a kmole of gas expanding irreversibly and adiabatically from
    P1 V1 T1 to P2 V2 T2:
    = (P1V1 - P2V2)/(γ - 1) = R(T1 - T2)/(γ - 1)

    3. The attempt at a solution
    The only work done by the otto engine is during the adiabatic expansions/compressions. So the work done by the gas from point A to point B is
    = R(Ta - Tb)/(γ - 1)

    Likewise, the work done by the gas from point C to point D is
    = R(Tc - Td)/(γ - 1)

    So the net work done by the gas will be
    = R(Ta - Tb)/(γ - 1) + R(Tc - Td)/(γ - 1)
    = (R/(γ - 1))(Ta - Tb + Tc - Td)
    = Cv[(Tc - Tb) + Ta - Td]

    Here I get stuck. I don't see why the answer doesn't have a Td term. Have I gone wrong somewhere?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jun 18, 2011 #2

    Andrew Mason

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    Since it is adiabatic, [itex]W = -\Delta U = -nC_v\Delta T[/itex]
    [itex] = -nC_v(T_b - T_a)[/itex]

    [itex] = -nC_v(T_d - T_c)[/itex]
    Your work is correct. Can you express Td in terms of the other temperatures?

    AM
     
    Last edited: Jun 18, 2011
  4. Jun 18, 2011 #3
    I'm sorry, but I still can't see how to find Td... Your equations seem essentially the same as mine.
     
  5. Jun 19, 2011 #4

    Andrew Mason

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    Think of W as Qin-Qout or Qh-Qc. What are Qh and Qc in terms of Ta, Tb, Tc and Td?

    AM
     
  6. Jun 21, 2011 #5

    Andrew Mason

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    What you have to do here is express Td in terms of Ta, Tb and Tc

    To do that you must first determine the ratio of Ta to Tb (it depends on the ratio of V1/V2 and gamma). Then determine the ratio of Td to Tc (try to express it in a similar way). Once you do that you will be able to express Td in terms of the other temperatures.

    For this cycle, the work can be calculated easily using W = Qh-Qc. Since a-b and c-d are adiabatic, heat flows only from b-c and d-a. Since it is isochoric, [itex]\Delta Q_h = nC_v\Delta T_{b-c} \text{ and } \Delta Q_c = nC_v\Delta T_{d-a}[/itex].

    AM
     
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