- #1
vladimir69
- 124
- 0
Homework Statement
The diagram below shows the thermodynamic cycle of a diesel engine. The compression ratio is the ratio of maximum to minimum volume; r=\frac{V_{1}}{V_{2}}. In addition, the so-called cutoff ratio is defined by r_{c} = \frac{V_{3}}{V_{2}}. Find an expression for the engine's efficiency, in terms of r, r_{c} and the specific heat ratio \gamma.
Homework Equations
PV=nRT
PV^gamma = constant
W_{adiabatic} = \frac{P_{1}V_{1} - P_{2}V_{2}}{\gamma-1}
\epsilon = \frac{W_{net}}{Q_h}
P_{2} = P_{3}
The Attempt at a Solution
W_{net} = W_{12} + W_{23} + W_{34} + W_{41}
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
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}
The only negative Q_h I think is Q_41 which is
Q_{h} = Q_{41} = n C_{v} (T_{1}-T_{4}) = \frac{P_{1}V_{1}C_{v}}{R}(1-r_{c}^\gamma)
And when I calculate \epsilon, low and behold out pops the wrong answer (and a huge mess of V1's V2's and so on) which I can't 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
\epsilon = 1-r^{1-\gamma}(\frac{r_{c}^\gamma-1}{\gamma(r_{c}-1)})
