How Is Diesel Engine Efficiency Calculated Using Compression and Cutoff Ratios?

[vladimir69]

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
$$\epsilon = 1-r^{1-\gamma}(\frac{r_{c}^\gamma-1}{\gamma(r_{c}-1)})$$

 

[Nate810]

where did this come from? I'm thinking I have the wrong values for Q_h and W_net but I'm not sure which ones to use. Any insight would be appreciated. Thank you.

 

[kerimek]

which seems to be a function of r,r_c, and \gamma which is what we are looking for.

I would approach this problem by first understanding the definitions and concepts involved. The compression ratio and cutoff ratio are important factors in determining the efficiency of a diesel engine. The compression ratio is the ratio of the maximum to minimum volume in the engine, while the cutoff ratio is the ratio of the volume at the end of the combustion process to the minimum volume.

To find the efficiency of the engine, we need to calculate the net work done by the engine and the heat absorbed. The net work is given by the sum of the work done in each stage of the thermodynamic cycle, which can be calculated using the adiabatic work equation. The heat absorbed can be calculated using the specific heat capacity and the temperature difference between the beginning and end of the process.

After setting up the equations and substituting the given values, it is important to carefully consider the signs of the values. In this case, negative values for heat absorbed should be used since we are looking at the efficiency of the engine, which is the ratio of work done to heat absorbed. The book's solution also follows this approach and uses the correct signs for the heat absorbed.

I would also suggest checking the calculations and double-checking the values used to ensure accuracy. It is possible that a small error in calculation or using the wrong values could result in a different answer.

In conclusion, the efficiency of a diesel engine can be calculated using the compression ratio, cutoff ratio, and specific heat ratio. Careful attention to the definitions and signs of values is important in obtaining the correct answer.