Chapter 9.12, Problem 161RP

a)

To determine

The thermal efficiency of an ideal diesel cycle using constant specific heats.

Answer to Problem 161RP

The thermal efficiency of ideal diesel cycle is $67.9\%$.

Explanation of Solution

Draw $P-v$ diagram for an Ideal diesel cycle as shown in Figure (1).

Assuming constant specific heats

Write the temperature and specific volume relation for the isentropic compression process 1-2.

$T_2=T_1{\left(\frac{v_1}{v_2}\right)}^{k-1}$ (I)

Here, the specific heat ratio is $k$, specific volume at state 1 is $v_1$ , specific volume at state 2 is $v_2$, temperature at state 1 is $T_1$ and temperature at state 2 is $T_2$.

Write the ideal gas relation for the constant pressure heat addition process 2-3.

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

For the process 2-3, $P_2=P_3$.

$\begin{array}{c}\frac{v_3}{T_3}=\frac{v_2}{T_2}\\ \frac{v_3}{v_2}=\frac{T_3}{T_2}\end{array}$ (II)

Here, temperature at state 3 is $T_3$, pressure at state 3 is $P_3$ , pressure at state 2 is $P_2$ and specific volume at state 3 is $v_3$.

Write the expression of heat input to the cycle $\left(q_{in}\right)$.

$q_{in}=h_3-h_2$

$q_{in}=c_p\left(T_3-T_2\right)$ (III)

Here, the specific heat at constant pressure is $c_p$, enthalpy at state 3 is $h_3$ and enthalpy at state 2 is $h_2$.

Write the temperature and specific volume relation for isentropic expansion process 3-4.

$T_4=T_3{\left(\frac{v_3}{v_4}\right)}^{k-1}$ (IV)

Write the expression of heat rejected for constant volume heat rejection process 4-1 $\left(q_{out}\right)$.

$q_{out}=u_4-u_1$

$q_{out}=c_v\left(T_4-T_1\right)$ (V)

Here, specific heat at constant volume is $c_v$, internal energy at state 4 is $u_4$ , internal energy at state 1 is $u_1$, temperature at state 4 is $T_4$ and temperature at state 4 is $T_1$.

Write the expression to calculate the net work output of the engine $\left(w_{\text{net,out}}\right)$.

$w_{\text{net,out}}=q_{in}-q_{out}$ . (VI)

Write the expression of thermal efficiency of the ideal  diesel cycle $\left({\eta }_{th}\right)$.

${\eta }_{th}=1-\frac{q_{out}}{q_{in}}$ . (VII)

Conclusion:

From Table A-2a, “Ideal-gas specific heats of various common gases”, obtain the following properties of air at room temperature.

$\begin{array}{l}{c}_p=1.005\text{kJ}/\text{kg}\cdot \text{K}\\ c_v=0.718\text{kJ}/\text{kg}\cdot \text{K}\\ R=0.287\text{kJ}/\text{kg}\cdot \text{K}\\ k=1.4\end{array}$

Substitute $15°\text{C}$ for $T_1$, $22$ for $\left(\frac{v_1}{v_2}\right)$, and 1.4 for $k$ in Equation (I).

$\begin{array}{c}{T}_2=\left(15°\text{C}\right){\left(22\right)}^{1.4-1}\\ =\left(15+273\right)\text{K}{\left(22\right)}^{1.4-1}\\ =991.7\text{K}\end{array}$

Substitute $1200°\text{C}$ for $T_3$, and $991.7\text{K}$ for $T_2$ in Equation (II).

$\begin{array}{c}\frac{v_3}{v_2}=\frac{1200°\text{C}}{991.7\text{K}}\\ =\frac{\left(2000+273\right)\text{K}}{991.7\text{K}}\\ =1.485\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, and $2000°\text{C}$ for $T_3$, and $991.7\text{K}$ for $T_2$ in Equation (III).

$\begin{array}{c}{q}_{in}=\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(2000°\text{C}-991.7\text{K}\right)\\ =483.7\text{kJ}/\text{kg}\end{array}$

Substitute $1200°\text{C}$ for $T_3$, and 1.4 for $k$, $1.485v_2$ for $v_3$, $22$ for $r$, and $v_1$ for $v_4$ in Equation (IV).

$\begin{array}{c}{T}_4=\left(1200°\text{C}\right){\left(\frac{1.485v_2}{v_4}\right)}^{1.4-1}\\ =\left(1200+273\right)\text{K}{\left(\frac{1.485v_2}{v_4}\right)}^{1.4-1}\\ =\left(1473\text{K}\right){\left(\frac{1.485}{r}\right)}^{0.4}\\ =501.1\text{K}\end{array}$

Substitute $0.718\text{kJ}/\text{kg}\cdot \text{K}$ for $c_v$, $501.1\text{K}$ for $T_4$, and $15°\text{C}$ for $T_1$ in Equation (V).

$\begin{array}{c}{q}_{out}=\left(0.718\text{kJ}/\text{kg}\cdot \text{K}\right)\left(501.1\text{K}-15°\text{C}\right)\\ =\left(0.718\text{kJ}/\text{kg}\cdot \text{K}\right)\left(501.1\text{K}-\left(15+273\right)\text{K}\right)\\ =\text{153}\text{kJ}/\text{kg}\end{array}$

Substitute $\text{153}\text{kJ}/\text{kg}$ for $q_{out}$ and $483.7\text{kJ}/\text{kg}$ for $q_{in}$ in Equation (VI).

$\begin{array}{c}{w}_{\text{net,out}}=483.7\text{kJ}/\text{kg}-\text{153}\text{kJ}/\text{kg}\\ =330.7\text{kJ}/\text{kg}\end{array}$

Substitute $\text{153}\text{kJ}/\text{kg}$ for $q_{out}$ and $330.7\text{kJ}/\text{kg}$ for $w_{\text{net,out}}$ in Equation (VII).

$\begin{array}{c}{\eta }_{th}=\frac{330.7\text{kJ}/\text{kg}}{483.7\text{kJ}/\text{kg}}\\ =0.698\times 100\%\\ =69.8\%\end{array}$

Thus, the thermal efficiency of an ideal diesel cycle is $69.8\%$.

b)

To determine

The thermal efficiency of ideal diesel cycle using variable specific heats.

Answer to Problem 161RP

The thermal efficiency of an ideal diesel cycle is $63.9\%$.

Explanation of Solution

Assuming variable specific heats

Write the specific volume and relative specific volume relation for the isentropic compression process 1-2.

$v_{r_2}=\frac{v_2}{v_1}{v}_{r_1}$

$v_{r_2}=\frac{1}{r}{v}_{r_1}$ (VIII)

Here, the compression ratio is r, relative specific volume at state 1 is $v_{r_1}$, relative specific volume at state 2 is $v_{r_2}$.

Write the pressure, temperature, and specific volume relation for isentropic compression process 2-3.

$\begin{array}{l}\frac{P_3{v}_3}{T_3}=\frac{P_2{v}_2}{T_2}\\ \end{array}$

For process 2-3, $P_2=P_3$.

$\frac{v_3}{v_2}=\frac{T_3}{T_2}$ (IX)

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

$q_{\text{in}}=h_3-h_2$ (X)

Write the specific volume and relative specific volume relation for the isentropic expansion process 3-4.

$v_{r_4}=\frac{v_4}{v_3}{v}_{r_3}$ (XI)

Here, relative specific volume at state 4 is $v_{r_4}$ and relative specific volume at state 3 is $v_{r_3}$.

Write the expression of heat rejected for constant volume heat rejection process 4-1 $\left(q_{\text{out}}\right)$.

$q_{\text{out}}=u_4-u_1$ (XII)

Write the expression of thermal efficiency of an deal diesel cycle $\left({\eta }_{\text{th}}\right)$.

${\eta }_{\text{th}}=1-\frac{q_{\text{out}}}{q_{\text{in}}}$ (XIII)

Conclusion:

Refer Table A-17, “Ideal gas properties of air”, obtain the following properties of air at temperature $\left(T_1\right)$ of 505 R.

$\begin{array}{l}{u}_1=205.48\text{kJ}/\text{kg}\\ v_{r_1}=688.1\end{array}$

Substitute $688.1$ for $v_{r_1}$, and $22$  for r in Equation (VIII).

$\begin{array}{c}{v}_{r_2}=\frac{1}{22}\left(688.1\right)\\ =31.28\end{array}$

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at $31.28$ relative specific volume on state 3 $\left(v_{r_2}\right)$, which gives $929.2\text{K}$ for $T_2$, and $965.73\text{kJ}/\text{kg}$ for $h_2$.

Substitute $1200°\text{C}$ for $T_3$ and $929.2\text{K}$ for $T_2$ in Equation (IX).

$\begin{array}{c}\frac{v_3}{v_2}=\frac{1200°\text{C}}{929.2\text{K}}\\ =\frac{\left(1200+273\right)\text{K}}{929.2\text{K}}\\ =1.585\end{array}$

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at $1473\text{K}$ $\left(T_3\right)$, which gives $1603.33\text{kJ}/\text{kg}$ for $h_3$, and $7.585$ for $v_{r_3}$.

Substitute $1603.33\text{kJ}/\text{kg}$ for $h_3$ and $965.73\text{kJ}/\text{kg}$ for $h_2$ in Equation (X).

$\begin{array}{c}{q}_{\text{in}}=1603.33\text{kJ}/\text{kg}-965.73\text{kJ}/\text{kg}\\ =637.6\text{kJ}/\text{kg}\end{array}$

Substitute $v_1$ for $v_4$, $1.585v_2$ for $v_3$, $22$ for r, and $7.585$ for $v_{r_3}$ in Equation (XI).

$\begin{array}{c}{v}_{r_4}=\frac{v_1}{1.585v_2}{v}_{r_3}\\ =\frac{r}{1.585}{v}_{r_3}\\ =\left(\frac{22}{1.585}\right)\left(7.585\right)\\ =105.3\end{array}$

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at $105.3$ $\left(v_{r_4}\right)$, which gives $435.61\text{kJ}/\text{kg}$ for $u_4$.

Substitute $435.61\text{kJ}/\text{kg}$ for $u_4$, and $205.48\text{kJ}/\text{kg}$ for $u_1$ in Equation (XII).

$\begin{array}{c}{q}_{\text{out}}=435.61\text{kJ}/\text{kg}-205.48\text{kJ}/\text{kg}\\ =230.13\text{kJ}/\text{kg}\end{array}$

Substitute $230.13\text{kJ}/\text{kg}$ for $q_{\text{out}}$, and $637.6\text{kJ}/\text{kg}$ for $q_{\text{in}}$ in Equation (XIII).

$\begin{array}{c}{\eta }_{\text{th}}=1-\frac{230.13\text{kJ}/\text{kg}}{637.6\text{kJ}/\text{kg}}\\ =0.639\times 100\%\\ =63.9\%\end{array}$

Thus, the thermal efficiency of an ideal diesel cycle is $63.9\%$.