Chapter 9, Problem 53P

a)

To determine

The thermal efficiency of the diesel engine.

Explanation of Solution

Given:

Compression ratio $\left(r\right)$ is $20$.

Initial temperature of the cycle $\left(T_1\right)$ is $20°\text{C}$.

Initial pressure of the cycle $\left(P_1\right)$ is $\text{95}\text{kPa}$.

Maximum temperature of the cycle $\left(T_3\right)$ is $2200\text{K}$.

Calculation:

Draw the $P-v$ diagram of the cycle as in Figure (1).

Refer Table A-2, “Ideal-gas specific heats of various common gases”, obtain the gas constant of the air.

  $R=0.287\text{kJ}/\text{kg}\cdot \text{K}$

Refer table A-21, “Ideal gas properties of the air”, obtain the enthalpy and relative specific volume of air at the temperature of $293\text{K}$ .

  $\begin{array}{l}{u}_1=\text{209}\text{.06}\text{kJ}/\text{kg}\\ v_{r_1}=659.2\end{array}$

Calculate the relative specific volume at state 2$\left(v_{r_2}\right)$.

  $v_{r_2}=\frac{V_2}{V_1}{v}_{r_1}$

  $\begin{array}{c}{v}_{r_2}=\frac{1}{r}{v}_{r_1}\\ =\frac{1}{20}\left(659.2\right)\\ =32.96\end{array}$

Refer table A-21, “Ideal gas properties of the air”, obtain the temperature and enthalpy of air at the relative specific volume of $32.96$.

  $\begin{array}{l}{T}_2=\text{912}\text{.7}\text{K}\\ h_2=947.17\text{kJ}/\text{kg}\end{array}$

Calculate the cut-off ratio $\left(r_c\right)$.

  $\frac{P_3{v}_3}{T_3}=\frac{P_2{v}_2}{T_2}$

  $\begin{array}{c}{T}_3=\frac{v_3}{v_2}{T}_2\\ =\left(r_c\right)T_2\end{array}$

  $\begin{array}{c}{r}_c=\frac{T_3}{T_2}\\ =\frac{2200\text{K}}{912.7\text{K}}\\ =1.854\end{array}$

Refer table A-21, “Ideal gas properties of the air”, obtain the internal energy and enthalpy of air at the temperature of $2200\text{K}$.

  $\begin{array}{l}{u}_3=1872.4\text{kJ}/\text{kg}\\ h_3=2503.2\text{kJ}/\text{kg}\end{array}$

Calculate the heat addition in the constant pressure heat addition process 2-3$\left(q_{\text{in}}\right)$.

  $\begin{array}{c}{q}_{\text{in}}=h_3-h_2\\ =2503.2\text{kJ}/\text{kg}-941.17\text{kJ}/\text{kg}\\ =1556\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Calculate the temperature at state 4$\left(T_4\right)$.

  $\begin{array}{c}{T}_4=T_3{\left(\frac{V_3}{V_4}\right)}^{n-1}\\ =T_3{\left(\frac{2.410V_2}{V_4}\right)}^{n-1}\\ =T_3{\left(\frac{2.410}{r}\right)}^{n-1}\end{array}$

  $\begin{array}{c}=\left(2200\text{K}\right){\left(\frac{2.410}{20}\right)}^{1.35-1}\\ =1049\text{K}\end{array}$

Calculate the work output for the polytrophic process 3-4$\left(w_{34,\text{out}}\right)$.

  $\begin{array}{c}{w}_{34,\text{out}}=\frac{R\left(T_4-T_3\right)}{1-n}\\ =\frac{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(1049\text{K}-2200\text{K}\right)}{1-1.35}\\ =943.8\text{kJ}/\text{kg}\end{array}$

Calculate the heat rejected for the polytrophic process 3-4$\left(q_{34,\text{out}}\right)$.

  $\begin{array}{c}{q}_{34,\text{out}}=-w_{34,\text{out}}-\left(u_4-u_3\right)\\ =-\left(943.8\text{kJ}/\text{kg}\right)-\left(801.13\text{kJ}/\text{kg}-1872.4\text{kJ}/\text{kg}\right)\\ =127.5\text{kJ}/\text{kg}\end{array}$

Calculate the heat rejected by the constant volume heat rejection process 4-1$\left(q_{41,\text{out}}\right)$.

  $\begin{array}{c}{q}_{41,\text{out}}=u_4-u_1\\ =801.13\text{kJ}/\text{kg}-209.06\text{kJ}/\text{kg}\\ =592.1\text{kJ}/\text{kg}\end{array}$

Calculate the total heat rejected by the cycle $\left(q_{\text{out}}\right)$.

  $\begin{array}{c}{q}_{\text{out}}=q_{34,\text{out}}+q_{41,\text{out}}\\ =127.5\text{kJ}/\text{kg}+592.1\text{kJ}/\text{kg}\\ =719.6\text{kJ}/\text{kg}\end{array}$

Calculate the thermal efficiency of the diesel engine $\left({\eta }_{\text{th}}\right)$.

  $\begin{array}{c}{\eta }_{\text{th}}=1-\frac{q_{\text{out}}}{q_{\text{in}}}\\ =1-\frac{719.6\text{kJ}/\text{kg}}{1556\text{kJ}/\text{kg}\cdot \text{K}}\\ =0.5375\\ =53.75\%\end{array}$

Thus, the thermal efficiency of the diesel engine is $63.5\%$.

b)

To determine

The mean effective pressure.

Explanation of Solution

Calculation:

Calculate the specific volume at state 1$\left(v_1\right)$.

  $\begin{array}{c}{v}_1=\frac{R{T}_1}{P_1}\\ =\frac{\left(0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\left(293\text{K}\right)}{95\text{kPa}}\\ =0.885{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Calculate the mean effective pressure $\left(\text{MEP}\right)$.

  $\begin{array}{c}\text{MEP}=\frac{w_{\text{net,out}}}{v_1-v_2}\\ =\frac{q_{\text{in}}-q_{\text{out}}}{v_1\left(1-1/r\right)}\\ =\frac{1556\text{kJ}/\text{kg}\cdot \text{K}-719.6\text{kJ}/\text{kg}\cdot \text{K}}{\left(0.885{\text{m}}^{\text{3}}/\text{kg}\right)\left(1-1/20\right)}\end{array}$

  $\begin{array}{c}=\frac{836.4\text{kJ}/\text{kg}\cdot \text{K}}{\left(0.885{\text{m}}^{\text{3}}/\text{kg}\right)\left(1-1/20\right)}\left(\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kJ}}\right)\\ =995\text{kPa}\end{array}$

Thus, the mean effective pressure is $\text{933}\text{kPa}$.