Chapter 22, Problem 24PQ

(a)

To determine

The efficiency of a diesel engine $e_{\text{diesel}}=1-\frac{1}{\gamma }\left(\frac{r^{-\gamma }-R^{-\gamma }}{r^{-1}-R^{-1}}\right)$

Answer to Problem 24PQ

The efficiency of a diesel engine is given by $e_{\text{diesel}}=1-\frac{1}{\gamma }\left(\frac{r^{-\gamma }-R^{-\gamma }}{r^{-1}-R^{-1}}\right)$ is shown.

Explanation of Solution

Write the expression to calculate efficiency of the diesel engine.

    $e_{\text{diesel}}=1-\frac{\left|Q_c\right|}{Q_h}$                                                                                              (I)

Here, $e_{\text{diesel}}$ is the efficiency of the diesel engine, $Q_c$ is the heat energy corresponding to cold reservoir, and $Q_h$ is the heat energy corresponding to hot reservoir.

Write the expression for the energy when heat enters the engine from the hot reservoir through isobaric process $BC$ .

    $Q_h=n{C}_P\left(T_C-T_B\right)$                                                                                      (II)

Here, $n$ is the number of moles, $C_p$ is the specific heat isobaric process, $T_C$ is the temperature at $C$, and $T_B$ is the temperature at $B$.

Write the expression for the energy when heat leaves the engine to the cold reservoir through isochoric process $DA$.

    $\left|Q_c\right|=n{C}_v\left(T_D-T_A\right)$                                                                                     (III)

Here, $C_v$ is the specific heat at isochoric process, $T_D$ is the temperature at $D$, and $T_A$ is the temperature at $A$.

For an adiabatic process at $AB$ ,

    $\begin{array}{c}\frac{T_A}{T_B}={\left(\frac{V_{\max }}{V_{\min }}\right)}^{\left(1-\gamma \right)}\\ =R^{1-\gamma }\\ T_A=T_B{R}^{1-\gamma }\end{array}$                                                                                    (IV)

Here, $V_{\max }$ is the maximum volume, $V_{\min }$ is the minimum volume, and $\gamma$ is the specific heat ratio.

For an adiabatic process at $CD$ ,

    $\begin{array}{c}\frac{T_D}{T_C}={\left(\frac{V_{\max }}{V_{\mathrm{int}}}\right)}^{\left(1-\gamma \right)}\\ =r^{1-\gamma }\\ T_D=T_C{r}^{1-\gamma }\end{array}$                                                                                    (V)

Here, $V_{\mathrm{int}}$ is the intermediate volume.

Write the expression for specific heat capacity.

    $\gamma =\frac{C_p}{C_v}$                                                                                              (VI)

Substitute (II), (III) in (I).

    $e_{\text{diesel}}=1-\frac{n{C}_v\left(T_D-T_A\right)}{n{C}_P\left(T_C-T_B\right)}$                                                                           (VII)

Substitute (IV), (V), (VI) in (VII).

    $e_{\text{diesel}}=1-\frac{1\left(T_C{r}^{1-\gamma }-T_B{R}^{1-\gamma }\right)}{\gamma \left(T_C-T_B\right)}$                                                                   (VIII)

At $B$ and $C$ pressure is same $P_C=P_B$, Hence

    $\begin{array}{c}\frac{T_C}{V_{\mathrm{int}}}=\frac{T_B}{V_{\min }}\\ T_C=\frac{V_{\mathrm{int}}}{V_{\min }}{T}_B\end{array}$                                                                                       (IX)

Conclusion:

Substitute (IX) in (VIII).

    $\begin{array}{c}{e}_{\text{diesel}}=1-\frac{1}{\gamma }\left[\frac{\left(\frac{V_{\mathrm{int}}}{V_{\min }}{T}_B{r}^{1-\gamma }-T_B{R}^{1-\gamma }\right)}{\left(\frac{V_{\mathrm{int}}}{V_{\min }}{T}_B-T_B\right)}\right]\\ =1-\frac{1}{\gamma }\left[\frac{\left(\frac{V_{\mathrm{int}}}{V_{\min }}r\cdot r^{-\gamma }-R\cdot R^{-\gamma }\right)}{\left(\frac{V_{\mathrm{int}}}{V_{\min }}-1\right)}\right]\\ =1-\frac{1}{\gamma }\left[\frac{\left(V_{\mathrm{int}}\frac{V_{\max }}{V_{\mathrm{int}}}\cdot r^{-\gamma }-V_{\min }\frac{V_{\max }}{V_{\min }}\cdot R^{-\gamma }\right)}{\left(V_{\mathrm{int}}-V_{\min }\right)}\right]\\ =1-\frac{1}{\gamma }\left(\frac{r^{-\gamma }-R^{-\gamma }}{\frac{V_{\mathrm{int}}}{V_{\max }}-\frac{V_{\min }}{V_{\max }}}\right)\\ =1-\frac{1}{\gamma }\left(\frac{r^{-\gamma }-R^{-\gamma }}{r^{-1}-R^{-1}}\right)\end{array}$                                         (X)

Hence, the efficiency of a diesel engine is given by $e_{\text{diesel}}=1-\frac{1}{\gamma }\left(\frac{r^{-\gamma }-R^{-\gamma }}{r^{-1}-R^{-1}}\right)$ is shown.

(b)

To determine

The efficiency of the diesel engine represented by Figure P22.24.

Answer to Problem 24PQ

The efficiency of the diesel engine is $56\%$

Explanation of Solution

Rewrite (X).

    $e_{\text{diesel}}=1-\frac{1}{\gamma }\left(\frac{r^{-\gamma }-R^{-\gamma }}{r^{-1}-R^{-1}}\right)$                                                                     (X)

Use $\frac{V_{\max }}{V_{\mathrm{int}}}$ for $r$ and $\frac{V_{\max }}{V_{\text{min}}}$ for $R$ in (X).

    $e_{\text{diesel}}=1-\frac{1}{\gamma }\left(\frac{{\left(\frac{V_{\max }}{V_{\mathrm{int}}}\right)}^{-\gamma }-{\left(\frac{V_{\max }}{V_{\text{min}}}\right)}^{-\gamma }}{{\left(\frac{V_{\max }}{V_{\mathrm{int}}}\right)}^{-1}-{\left(\frac{V_{\max }}{V_{\text{min}}}\right)}^{-1}}\right)$                                                    (XI)

Conclusion:

Substitute $1.4$ for $\gamma$, $1050{\text{cm}}^{\text{3}}$ for $V_{\max }$ , $50{\text{cm}}^{\text{3}}$ for $V_{\min }$ , and $230{\text{cm}}^{\text{3}}$ for $V_{\mathrm{int}}$ in (XI) to calculate $e_{\text{diesel}}$.

    $\begin{array}{c}{e}_{\text{diesel}}=1-\frac{1}{1.4}\left(\frac{{\left(\frac{1050{\text{cm}}^{\text{3}}}{230{\text{cm}}^{\text{3}}}\right)}^{-1.4}-{\left(\frac{1050{\text{cm}}^{\text{3}}}{50{\text{cm}}^{\text{3}}}\right)}^{-1.4}}{{\left(\frac{1050{\text{cm}}^{\text{3}}}{230{\text{cm}}^{\text{3}}}\right)}^{-1}-{\left(\frac{1050{\text{cm}}^{\text{3}}}{50{\text{cm}}^{\text{3}}}\right)}^{-1}}\right)\\ =1-\frac{1}{1.4}\left(\frac{{\left(4.57\right)}^{-1.4}-{\left(21.0\right)}^{-1.4}}{{\left(4.57\right)}^{-1}-{\left(21.0\right)}^{-1}}\right)\\ =0.56\\ =56\%\end{array}$

Therefore, the efficiency of the diesel engine is $56\%$