## Essential University Physics: Volume 1 (3rd Edition)

a) In problem number 56, we found: We know that the work is equal to the sum of all of the works done over each of the processes. Thus, we find: 1. The first step is adiabatic, so we find: $P_1V_1^{\gamma}=P_2V_2^{\gamma}$ Using substitution for $\frac{V_1}{V_2}$, it follows: $P_2=\frac{P_1}{r}$ This means: $W=\frac{\frac{P_1}{r}V_2-P_1V_1}{\gamma-1}$ 2. Pressure is constant, so this is isobaric. Thus, we find: $W = -\frac{P_1}{r}(V_3-V_2)$ 3. This step is adiabatic, so we find: $P_1V_1^{\gamma}=P_2V_2^{\gamma}$ $P_f=\frac{P_1V_3}{V_1r}$ Thus, it follows: $W=\frac{\frac{P_1V_3}{V_1r}V_1-V_3\frac{P_1}{r}}{\gamma-1}$ 4. This step is constant volume, so no work is done. Thus, adding all of the individual works together and plugging in $\alpha=\frac{V_3}{V_2}$, we obtain: $W = \frac{P_1V_1(r^{1-\gamma}(\alpha-1)\gamma-\alpha^{\gamma}+1)}{(\gamma-1)}$ Using this, we find that the net heat flow in is: $Q_{in} = \frac{P_1V_1r^{1-\gamma}(\alpha-1)\gamma}{(\gamma-1)}$ b) An engine's efficiency is given by: $e=\frac{W}{Q_h}$ Thus, using substitution, it follows: $e_{diesel}=1-\frac{r^{1-\gamma}(\alpha^{\gamma}-1)}{(\alpha-1)\gamma}$