Chapter 9.12, Problem 149P

(a)

To determine

The maximum temperature in the cycle.

Answer to Problem 149P

The maximum temperature in the cycle is $\underset{\_}{\text{2308}\text{K}}$.

Explanation of Solution

Draw the $P-v$ diagram of four stroke as in Figure (I).

Refer to Table A-2b, obtain the properties of air at 1000 K.

$\begin{array}{l}\text{constant}\text{pressure}\text{specific}\text{heat},c_p=1.142\text{kJ/kg}\cdot \text{K}\\ \text{constant}\text{volume}\text{specific}\text{heat},c_v=0.855\text{kJ/kg}\cdot \text{K}\\ \text{Gas}\text{constant},R=0.287\text{kJ/kg}\cdot \text{K}\\ \text{specific heat}\text{ratio,}k=1.336\end{array}$

Express the compression ratio.

$r=\frac{v_c+v_d}{v_c}$ (I)

Here, clearance volume is $v_c$ and displacement volume is $v_d$.

Express the total volume of the engine at the beginning of compression process (state 1).

$v_1=v_c+v_d$ (II)

Process 1-2: Isentropic compression

Calculate the temperature at state 2.

$\begin{array}{c}{T}_2=T_1{\left(\frac{v_1}{v_2}\right)}^{k-1}\\ =T_1{\left(r\right)}^{k-1}\end{array}$ (III)

Here, temperature at state 1 is $T_1$, volume at state 1 and 2 is $v_1\text{and}{v}_2$ respectively.

Calculate the pressure at state 2.

$\begin{array}{c}{P}_2=P_1{\left(\frac{v_1}{v_2}\right)}^k\\ =P_1{\left(r\right)}^k\end{array}$ (IV)

Here, pressure at state 1 is $P_1$.

Process 2-x and x-3: Constant-volume and constant pressure heat addition processes

Calculate the temperature at x state.

$T_x=T_2\frac{P_x}{P_2}$ (V)

Here, pressure at x state is $P_x$.

Calculate the heat addition to the process 2-x.

$q_{2-x}=c_v\left(T_x-T_2\right)$ (VI)

Here, constant volume specific heat is $c_v$.

Calculate the heat addition to the process x-3.

$\begin{array}{l}{q}_{2-x}=q_{x-3}\\ q_{2-x}=c_p\left(T_3-T_x\right)\end{array}$ (VII)

Here, constant pressure specific heat is $c_p$.

Conclusion:

Substitute 16 for r and 1.8 L for $v_d$ in Equation (I).

$\begin{array}{l}16=\frac{v_c+1.8\text{L}}{v_c}\\ 16=\frac{v_c+1.8\text{L}\times \frac{0.001{\text{m}}^{\text{3}}}{\text{L}}}{v_c}\\ v_c=0.00012{\text{m}}^{\text{3}}\end{array}$

The value of $v_c=v_2=v_x$.

Substitute $0.00012{\text{m}}^{\text{3}}$ for $v_c$ and 1.8 L for $v_d$ in Equation (II).

$\begin{array}{c}{v}_1=0.00012{\text{m}}^{\text{3}}+1.8\text{L}\\ =0.00012{\text{m}}^{\text{3}}+1.8\text{L}\times \frac{\text{0}\text{.001}{\text{m}}^{\text{3}}}{1\text{L}}\\ =0.00192{\text{m}}^{\text{3}}\\ =v_4\end{array}$

Substitute 343 K for $T_1$, 16 for r, and 1.336 for k in Equation (III).

$\begin{array}{c}{T}_2=\left(343\text{K}\right){\left(16\right)}^{1.336-1}\\ =870.7\text{K}\end{array}$

Substitute 95 kPa for $P_1$, 16 for r, and 1.336 for k in Equation (IV).

$\begin{array}{c}{P}_2=\left(95\text{kPa}\right){\left(16\right)}^{1.336}\\ =3859\text{kPa}\end{array}$

Substitute 3859 kPa for $P_2$, 7500 kPa for $P_x$, and 870.7 K for $T_2$ in Equation (V).

$\begin{array}{c}{T}_x=\left(870.7\text{K}\right)\frac{7500\text{kPa}}{3859\text{kPa}}\\ =1692\text{K}\end{array}$

Substitute $0.855\text{kJ/kg}\cdot \text{K}$ for $c_v$, 1692 K for $T_x$, and 870.7 K for $T_2$ in Equation (VI).

$\begin{array}{c}{q}_{2-x}=0.855\text{kJ/kg}\cdot \text{K}\left(1692\text{K}-870.7\text{K}\right)\\ =702.6\text{kJ/kg}\end{array}$

Substitute $1.142\text{kJ/kg}\cdot \text{K}$ for $c_p$, 1692 K for $T_x$, and 702.6 kJ/kg for $q_{2-x}$ in Equation (VII).

$\begin{array}{l}702.6\text{kJ/kg}=1.142\text{kJ/kg}\cdot \text{K}\left(T_3-1692\text{K}\right)\\ T_3=2308\text{K}\end{array}$

Thus, the maximum temperature in the cycle is $\underset{\_}{\text{2308}\text{K}}$.

(b)

To determine

The net work output.

The thermal efficiency.

Answer to Problem 149P

The net work output is $\underset{\_}{\text{835}\text{.8}\text{kJ/kg}}$.

The thermal efficiency is $\underset{\_}{59.5\%}$.

Explanation of Solution

Express the total heat addition to the process.

$q_{\text{in}}=q_{2-x}+q_{x-3}$ (VIII)

Calculate the volume at state 3.

$v_3=v_x\frac{T_3}{T_x}$ (IX)

Here, volume at state x is $v_x$, temperature at state 3 and x are $T_3\text{and}{T}_x$ respectively.

Process 3-4: Isentropic expansion

Calculate the temperature at state 4.

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

Here, volume at state 3 and 4 are $v_3$ and $v_4$ respectively.

Calculate the pressure at state 4.

$P_4=P_3{\left(\frac{v_3}{v_4}\right)}^k$ (XI)

Here, pressure at state 3 and 4 are $P_3$ and $P_4$ respectively.

Process 4-1: Constant volume heat rejection

Calculate the heat rejection.

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

Calculate the net work output.

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

Calculate the thermal efficiency.

${\eta }_{\text{th}}=\frac{w_{\text{net,}\text{out}}}{q_{\text{in}}}$ (XIV)

Conclusion:

Substitute 702.6 kJ/kg for $q_{2-x}$ and 702.6 kJ/kg for $q_{x-3}$ in Equation (VIII).

$\begin{array}{c}{q}_{\text{in}}=\text{702}\text{.6}\text{kJ/kg}+\text{702}\text{.6}\text{kJ/kg}\\ =1405\text{kJ/kg}\end{array}$

Substitute $0.00012{\text{m}}^{\text{3}}$ for $v_x$, 2308 K for $T_3$, and 1692 K for $T_x$ in Equation (IX).

$\begin{array}{c}{v}_3=\left(0.00012{\text{m}}^{\text{3}}\right)\frac{2308\text{K}}{1692\text{K}}\\ =0.0001636{\text{m}}^{\text{3}}\end{array}$

Substitute 2308 K for $T_3$, $0.0001636{\text{m}}^{\text{3}}$ for $v_3$, 1.336 for k, and $0.00192{\text{m}}^{\text{3}}$ for $v_4$ in Equation (X).

$\begin{array}{c}{T}_4=2308\text{K}{\left(\frac{0.0001636{\text{m}}^{\text{3}}}{0.00192{\text{m}}^{\text{3}}}\right)}^{1.336-1}\\ =1.009\text{K}\end{array}$

Substitute 7500 kPa for $P_3$, $0.0001636{\text{m}}^{\text{3}}$ for $v_3$, 1.336 for k, and $0.00192{\text{m}}^{\text{3}}$ for $v_4$ in Equation (XI).

$\begin{array}{c}{P}_4=7500\text{kPa}{\left(\frac{0.0001636{\text{m}}^{\text{3}}}{0.00192{\text{m}}^{\text{3}}}\right)}^{1.336}\\ =279.4\text{kPa}\end{array}$

Substitute $0.855\text{kJ/kg}\cdot \text{K}$ for $c_v$, 1009 K for $T_4$, and 343 K for $T_1$ in Equation (XII).

$\begin{array}{c}{q}_{\text{out}}=0.855\text{kJ/kg}\cdot \text{K}\left(1009\text{K}-343\text{K}\right)\\ =569.3\text{kJ/kg}\end{array}$

Substitute 569.3 kJ/kg for $q_{\text{out}}$ and 1405 kJ/kg for $q_{\text{in}}$ in Equation (XIII).

$\begin{array}{c}{w}_{\text{net,out}}=1405\text{kJ/kg}-569.3\text{kJ/kg}\\ =835.8\text{kJ/kg}\end{array}$

Thus, the net work output is $\underset{\_}{\text{835}\text{.8}\text{kJ/kg}}$.

Substitute 835.8 kJ/kg for $w_{\text{net,}\text{out}}$ and 1405 kJ/kg for $q_{\text{in}}$ in Equation (XIV).

$\begin{array}{c}{\eta }_{\text{th}}=\frac{835.8\text{kJ/kg}}{1405\text{kJ/kg}}\\ =0.5948\\ =59.5\%\end{array}$

Thus, the thermal efficiency is $\underset{\_}{59.5\%}$.

(c)

To determine

The mean effective pressure.

Answer to Problem 149P

The mean effective pressure is $\underset{\_}{\text{860}\text{.4}\text{kPa}}$.

Explanation of Solution

Calculate the mass.

$m=\frac{P_1{v}_1}{R{T}_1}$ (XV)

Calculate the mean effective pressure.

$\text{MEP}=\frac{m{w}_{\text{net,}\text{out}}}{v_1-v_2}$ (XVI)

Conclusion:

Substitute 95 kPa for $P_1$, $0.00192{\text{m}}^{\text{3}}$ for $v_1$, $0.287\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$ for R, and 343 K for $T_1$ in Equation (XV).

$\begin{array}{c}m=\frac{95\text{kPa}\left(0.00192{\text{m}}^{\text{3}}\right)}{\left(0.287\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)343\text{K}}\\ =0.001853\text{kg}\end{array}$

Substitute 0.001853 kg for m, $835.8\text{kJ/kg}$ for $w_{\text{net,}\text{out}}$, $0.00192{\text{m}}^{\text{3}}$ for $v_1$, and $0.00012{\text{m}}^{\text{3}}$ for $v_2$ in Equation (XVI).

$\begin{array}{c}\text{MEP}=\frac{0.001853\text{kg}\left(835.8\text{kJ/kg}\right)}{0.00192{\text{m}}^{\text{3}}-0.00012{\text{m}}^{\text{3}}}\\ =\frac{0.001853\text{kg}\left(835.8\text{kJ/kg}\right)}{0.00192{\text{m}}^{\text{3}}-0.00012{\text{m}}^{\text{3}}}\left(\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kJ}}\right)\\ =860.4\text{kPa}\end{array}$

Thus, the mean effective pressure is $\underset{\_}{\text{860}\text{.4}\text{kPa}}$.

(d)

To determine

The power for engine speed of 3500 rpm.

Answer to Problem 149P

The power for engine speed of 3500 rpm is $\underset{\_}{\text{28}\text{.39}\text{kW}}$

Explanation of Solution

Calculate the power for engine speed of 3500 rpm.

${\dot{W}}_{\text{net}}=m{w}_{\text{net}}\frac{\dot{n}}{2}$ (XVII)

Here, engine speed is $\dot{n}$.

Conclusion:

Substitute 2200 rev/min for $\dot{n}$, 0.001853 kg for m, and $835.8\text{kJ/kg}$ for $w_{\text{net,}\text{out}}$ in Equation (XVII).

$\begin{array}{c}{\dot{W}}_{\text{net}}=0.001853\text{kg}\left(835.8\text{kJ/kg}\right)\frac{2200\text{rev/min}}{2}\\ =0.001853\text{kg}\left(835.8\text{kJ/kg}\right)\frac{2200\text{rev/min}}{2}\left(\frac{1\min }{60\text{s}}\right)\\ =28.39\text{kW}\end{array}$

Thus, the power for engine speed of 3500 rpm is $\underset{\_}{\text{28}\text{.39}\text{kW}}$.

(e)

To determine

The second law efficiency of the cycle.

The rate of exergy output with the exhaust gases.

Answer to Problem 149P

The second law efficiency of the cycle is $\underset{\_}{68.3\%}$.

The rate of exergy output with the exhaust gases is $\underset{\_}{\text{10}\text{.30}\text{kW}}$.

Explanation of Solution

Express the maximum thermal efficiency of the cycle.

${\eta }_{\max }=1-\frac{T_0}{T_3}$ (XVIII)

Here, dead state temperature is $T_0$.

Express the second law efficiency of the cycle.

${\eta }_{\text{II}}=\frac{{\eta }_{\text{th}}}{{\eta }_{\max }}$ (XIX)

Calculate the rate of exergy of the exhaust gases.

$\begin{array}{c}{\dot{X}}_4=m{x}_4\frac{\dot{n}}{2}\\ =m\left[\left(u_4-u_0+P_0\left(v_4-v_0\right)-T_0\left(s_4-s_0\right)\right)\right]\frac{\dot{n}}{2}\\ =m\left[c_v\left(T_4-T_0\right)+P_0\left(v_4-v_0\right)-T_0\left[c_p\ln \frac{T_4}{T_0}-R\ln \frac{P_4}{P_0}\right]\right]\frac{\dot{n}}{2}\\ =m\left[c_v\left(T_4-T_0\right)+P_0\left(\frac{V_4}{m}-\frac{R{T}_0}{P_0}\right)-T_0\left[c_p\ln \frac{T_4}{T_0}-R\ln \frac{P_4}{P_0}\right]\right]\frac{\dot{n}}{2}\end{array}$ (XX)

Here, specific internal energy at state 4, dead state is $u_4,u_0$, specific entropy at state 4 and dead state is $s_4,s_0$, specific volume at dead state is $v_0$, and pressure at dead state is $P_0$.

Conclusion:

Substitute $25°\text{C}$ for $T_0$ and 2308 K for $T_3$ in Equation (XVIII).

$\begin{array}{c}{\eta }_{\max }=1-\frac{25°\text{C}}{2308\text{K}}\\ =1-\frac{\left(25+273\right)\text{K}}{2308\text{K}}\\ =0.8709\end{array}$

Substitute 0.8709 for ${\eta }_{\max }$ and 0.5948 for ${\eta }_{\text{th}}$ in Equation (XIX).

$\begin{array}{c}{\eta }_{\text{II}}=\frac{0.5948}{0.8709}\\ =0.683\\ =68.3\%\end{array}$

Thus, the second law efficiency of the cycle is $\underset{\_}{68.3\%}$.

Substitute $0.855\text{kJ/kg}\cdot \text{K}$ for $c_v$, 2200 rev/min for $\dot{n}$, 1009 K for $T_4$, 298 K for $T_0$, 100 kPa for $P_0$, $0.00192{\text{m}}^{\text{3}}$ for $V_4$, 0.001853 kg for m, $0.287\text{kJ/kg}\cdot \text{K}$ for R, $279.4\text{kPa}$ for $P_4$, and $1.142\text{kJ/kg}\cdot \text{K}$ for $c_p$ in Equation (XX).

$\begin{array}{c}{\dot{X}}_4=0.001853\text{kg}\left\{\begin{array}{l}0.855\text{kJ/kg}\cdot \text{K}\left(\begin{array}{l}1009\text{K}-\\ 298\text{K}\end{array}\right)+100\text{kPa}\\ \left(\begin{array}{l}\frac{0.00192{\text{m}}^{\text{3}}}{0.001853\text{kg}}-\\ \frac{\left(0.287\text{kJ/kg}\cdot \text{K}\right)298\text{K}}{100\text{kPa}}\end{array}\right)-\\ 298\text{K}\left[\begin{array}{l}1.142\text{kJ/kg}\cdot \text{K}\ln \frac{1009\text{K}}{298\text{K}}-\\ 0.287\text{kJ/kg}\cdot \text{K}\ln \frac{279.4\text{kPa}}{100\text{kPa}}\end{array}\right]\end{array}\right\}\frac{2200\text{rev/min}}{2}\\ =0.001853\text{kg}\left(303.3\text{kJ/kg}\right)\frac{2200\text{rev/min}}{2}\left(\frac{1\min }{60\text{s}}\right)\\ =10.30\text{kW}\end{array}$

Thus, the rate of exergy output with the exhaust gases is $\underset{\_}{\text{10}\text{.30}\text{kW}}$.