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Homework Help: Efficiency of an Engine

  1. May 13, 2007 #1
    1. The problem statement, all variables and given/known data

    Suppose 1.0 mol of a monatomic ideal gas initially at 10L and 302K is heated at constant volume to 604K, allowed to expand isothermally to its initial pressure, and finally compressed at constant pressure to its original volume, pressure, and temperature.

    (c)What is the net energy entering the system (the gas) as heat during the cycle?
    (d)What is the net work done by the gas during the cycle?
    (e)What is the efficiency of the cycle?

    2. Relevant equations

    [tex]\varepsilon = \frac{W}{Q_H}[/tex]

    3. The attempt at a solution

    This is a previous test that I took. I got the correct answers for parts c and d (both are 970 J), but I'm stumped for part e; I did not get the correct answser at the time, and my work makes no sense to me.
    Looking at it afresh, I don't understand why the answer isn't 1: after all, doesn't W=Q_H? But I have down that the correct answer should be 0.1338 :confused:
  2. jcsd
  3. May 14, 2007 #2

    Andrew Mason

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    No. W = Qh - Qc. The work done is always less than the heat flow into the system. The efficiency is simply the answer to part d) divided by the answer to part c). [Note: The answers to c) and d) cannot be the same]

  4. May 14, 2007 #3
    It's a cycle, though. So [tex]\Delta E = Q - W = 0 \Longrightarrow Q=W[/tex].

    I checked with my friend (who got full credit for this question), but I really don't understand what she did.
    What I remember is [tex]\varepsilon = \frac{W}{Q_H} = \frac{W}{Q_{?}+Q_{?*}}[/tex]
    where ? and ?* are two of the three stages of the cycle. That formula turned into something terribly confusing with two terms, one with a ln and one with a volume difference.

    Maybe I'll ask my friend for her test and post her work up here; this is from the first test of the semester and none of us understand why her formula works. Doubtful if we understood it at the time, either... Sigh.
  5. May 14, 2007 #4

    Andrew Mason

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    Since it is a thermodynamic cycle, some of the heat flowing from the hot reservoir energy cannot be used to do work. So there is a loss of energy each cycle.

    The system returns to its original state so there is no change in internal energy over the cycle. In the forward part of the cycle (heat absorbed and work done):

    [tex]Q_h = \Delta U_{fwd} + W_{fwd}[/tex]

    In the return part of the cycle (heat expelled, work consumed):

    [tex]Q_c = \Delta U_{ret} + W_{ret}[/tex]

    So over the whole cycle:

    [tex]Q_h + Q_c = \Delta U_{fwd} + \Delta U_{ret} + W_{fwd} + W_{ret}[/tex]

    Since [itex]\Delta U_{fwd} + \Delta U_{ret} = 0[/itex], (no change in internal energy),

    [itex]Q_h + Q_c = Qh -|Qc| = W_{fwd} + W_{ret} = W_{net}[/itex]

    Last edited: May 14, 2007
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