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Efficieny of a diesel engine

  1. Jan 25, 2010 #1
    1. The problem statement, all variables and given/known data
    The diagram below shows the thermodynamic cycle of a diesel engine. The compression ratio is the ratio of maximum to minimum volume; [itex]r=\frac{V_{1}}{V_{2}}[/itex]. In addition, the so-called cutoff ratio is defined by [itex]r_{c} = \frac{V_{3}}{V_{2}}[/itex]. Find an expression for the engine's efficiency, in terms of [itex]r[/itex], [itex]r_{c}[/itex] and the specific heat ratio [itex]\gamma[/itex].
    2. Relevant equations
    [tex]PV=nRT[/tex]
    PV^gamma = constant
    [tex]W_{adiabatic} = \frac{P_{1}V_{1} - P_{2}V_{2}}{\gamma-1}[/tex]
    [tex]\epsilon = \frac{W_{net}}{Q_h}[/tex]
    [tex]P_{2} = P_{3}[/tex]
    3. The attempt at a solution
    [tex]W_{net} = W_{12} + W_{23} + W_{34} + W_{41}[/tex]
    And I think I must only use negative values for Q_h because the efficiency is the ratio of work done to heat absorbed. That is a little bit against what I would think because it would make more sense to me if Q_h was just the sum of Q_12, Q_23, Q_34, Q_41.
    First of all
    [tex] W_{net} = \frac{P_{1}V_{1}-P_{1}r^\gamma V_{2}}{\gamma-1}+P_{1}r^\gamma(V_{3}-V_{2}) + \frac{P_{1}r^\gamma V_{3} - P_{1}r_{c}^\gamma V_{1}}{\gamma-1}[/tex]
    The only negative Q_h I think is Q_41 which is
    [tex]Q_{h} = Q_{41} = n C_{v} (T_{1}-T_{4}) = \frac{P_{1}V_{1}C_{v}}{R}(1-r_{c}^\gamma)[/tex]
    And when I calculate [itex]\epsilon[/itex], low and behold out pops the wrong answer (and a huge mess of V1's V2's and so on) which I cant seem to get rid of. So it seems I have plucked out the wrong values to calculate Q_h or W_net but I'm not sure which combination to pull out.

    The book manages to squeeze out
    [tex]\epsilon = 1-r^{1-\gamma}(\frac{r_{c}^\gamma-1}{\gamma(r_{c}-1)})[/tex]
     

    Attached Files:

  2. jcsd
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