Chapter 9.12, Problem 164RP

a)

To determine

The net power output of the cycle.

Answer to Problem 164RP

The net power output of the cycle is $6488\text{kW}$.

Explanation of Solution

Draw the $P-\nu$ for an ideal Otto cycle as shown in Figure (1).

Write the expression for compression ratio to calculate the clearance volume or one cylinder.

$r=\frac{{\nu }_c+{\nu }_d/4}{{\nu }_c}$ (I)

Here, clearance volume is ${\nu }_c$, displacement volume for four cylinders ${\nu }_d$.

Write the expression to calculate the volume at state 1.

${\nu }_1={\nu }_c+\frac{{\nu }_d}{4}$ (II)

Write the expression to calculate the mass of the air.

$\begin{array}{l}{P}_1{\nu }_1=mR{T}_1\\ m=\frac{P_1{\nu }_1}{R{T}_1}\end{array}$ (III)

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

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

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

$\begin{array}{l}\frac{P_2{v}_2}{T_2}=\frac{P_1{v}_1}{T_1}\\ P_2=\left(\frac{v_1}{v_2}\right)\frac{T_2}{T_1}\times P_1\end{array}$ (V)

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

Write the expression for heat addition process 2-3$\left(Q_{23,in}\right)$.

$Q_{23,in}=m{c}_v\left(T_3-T_2\right)$ (VI)

Here, temperature at state 3 is $T_3$ and specific heat at constant volume is $c_v$.

Write the temperature, pressure, and specific volume relation for the constant volume heat addition process 2-3.

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

For process 2-3, $v_2$ and $v_3$ are equal.

$P_3=\frac{T_3}{T_2}{P}_2$ (VII)

Conclusion:

From Table A-1, “Ideal-gas specific heats of various common gases”, obtain the value of gas constant $\left(R\right)$ of air as $0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$.

Substitute $11$ for $r$ and $1.8\text{L}$ for ${\nu }_d$ in Equation (I).

$\begin{array}{l}11=\frac{{\nu }_c+\frac{1.8\text{L}}{4}}{{\nu }_c}\\ 11=\frac{{\nu }_c+\frac{1.8\text{L}\left(\frac{1{\text{m}}^3}{1000\text{L}}\right)}{4}}{{\nu }_c}\\ 11=\frac{{\nu }_c+\frac{0.0018{\text{m}}^3}{4}}{{\nu }_c}\\ {\nu }_c={\nu }_2=0.000045{\text{m}}^3\end{array}$

Substitute $0.000045{\text{m}}^3$ for ${\nu }_c$ and $1.8\text{L}$ for ${\nu }_d$ in Equation (II).

$\begin{array}{c}{\nu }_1=0.000045{\text{m}}^3+\frac{1.8\text{L}}{4}\\ =0.000045{\text{m}}^3+\frac{1.8\text{L}}{4}\left(\frac{1{\text{m}}^3}{1000\text{L}}\right)\\ =0.000495{\text{m}}^3\end{array}$

Substitute $90\text{kPa}$ for $P_1$ , $0.000495{\text{m}}^3$ for ${\nu }_1$ , $0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R$ and $50°\text{C}$ for $T_1$ in Equation (III).

$\begin{array}{c}m=\frac{90\text{kPa}\times 0.000495{\text{m}}^3}{0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\times 50°\text{C}}\\ =\frac{90\text{kPa}\times 0.000495{\text{m}}^3}{0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\times \left(50+273\right)\text{K}}\\ =0.0004805\text{kg}\end{array}$

Substitute $50°\text{C}$ for, $11$ for $\left(\frac{v_1}{v_2}\right)$, and $0.35$ for $k$ in Equation (IV).

$\begin{array}{c}{T}_2=50°\text{C}{\left(11\right)}^{1.35-1}\\ =\left(50+273\right)\text{K}{\left(11\right)}^{1.35-1}\\ =747.6\text{K}\end{array}$

Substitute $11$ for $\left(\frac{v_1}{v_2}\right)$, $90\text{kPa}$ for $P_1$, $747.6\text{K}$ for $T_2$, and $50°\text{C}$ for $T_1$ to find $P_2$ in Equation (V).

$\begin{array}{c}{P}_2=\left(11\right)\left(\frac{747.6\text{K}}{50°\text{C}}\right)\left(90\text{kPa}\right)\\ =\left(11\right)\left(\frac{747.6\text{K}}{\left(50+273\right)\text{K}}\right)\left(90\text{kPa}\right)\\ =2292\text{kPa}\end{array}$

Substitute $0.5\text{kJ}$ for $Q_{23,in}$, $0.0004805\text{kg}$ for $m$ , $0.821\text{kJ}/\text{kg}\cdot \text{K}$ for $c_v$, $747.6\text{K}$ for $T_2$ in Equation(VI).

$\begin{array}{l}0.5\text{kJ}=\left(0.0004805\text{kg}\right)\left(0.821\text{kJ}/\text{kg}\cdot \text{K}\right)\left(T_3-747.6\text{K}\right)\\ T_3=2015\text{K}\end{array}$

Substitute $2015\text{K}$ for $T_3$, $747.6\text{K}$ for $T_2$, and $2292\text{kPa}$ for $P_2$ in Equation (VIII).

$\begin{array}{c}{P}_3=\frac{2015\text{K}}{747.6\text{K}}\left(2292\text{kPa}\right)\\ =6176\text{kPa}\end{array}$

b)

To determine

The net work per cycle per cylinder and the thermal efficiency of the cycle.

Answer to Problem 164RP

The net work per cycle per cylinder is $\text{0}\text{.284}\text{kJ}$.

The thermal efficiency of the cycle is $56.8\%$.

Explanation of Solution

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

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

Here, temperature at state 4 is $T_4$, and specific volume at state 4 is $v_4$.

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

$Q_{41,}{}_{out}=m{c}_v\left(T_4-T_1\right)$ (IX)

Write the expression to calculate the net power output $W_{net}$.

$W_{net}=Q_{23,in}-Q_{41,}{}_{out}$ (X)

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

${\eta }_{th}=\frac{W_{\text{net}}}{Q_{\text{in}}}$ (XI)

Conclusion:

Substitute $1734\text{K}$ for $T_3$ , $\frac{1}{11}$ for $\left(\frac{v_3}{v_4}\right)$, and $1.35$ for $k$ in Equation (VIII).

$\begin{array}{c}{T}_4=\left(1734\text{K}\right){\left(\frac{1}{11}\right)}^{1.35-1}\\ =1734\text{K}\times {\left(\frac{1}{11}\right)}^{0.35}\\ =870.4\text{K}\end{array}$

Substitute $0.0004805\text{kg}$ for $m$ , $0.821\text{kJ}/\text{kg}\cdot \text{K}$ for $c_v$, $870.5\text{K}$ for $T_4$, $50°\text{C}$ for $T_1$ in Equation(IX).

$\begin{array}{c}{Q}_{41,}{}_{out}=0.0004805\text{kg}\times 0.821\text{kJ}/\text{kg}\cdot \text{K}\left(870.5\text{K}-50°\text{C}\right)\\ =0.0004805\text{kg}\times 0.821\text{kJ}/\text{kg}\cdot \text{K}\left(870.5\text{K}-\left(50+273\right)\text{K}\right)\\ =0.216\text{kJ}\end{array}$

Substitute $0.5\text{kJ}$ for $Q_{23,in}$ and $0.216\text{kJ}$ for $Q_{41,}{}_{out}$ in Equation (X).

$\begin{array}{c}{W}_{net}=0.5\text{kJ}-0.216\text{kJ}\\ \text{=0}\text{.284}\text{kJ}\end{array}$

Thus, the net work per cycle per cylinder is $\text{0}\text{.284}\text{kJ}$.

Substitute $\text{0}\text{.284}\text{kJ}$ for $W_{\text{net}}$, and $0.5\text{kJ}$ for $Q_{\text{in}}$ in Equation(XI).

$\begin{array}{c}{\eta }_{th}=\frac{\text{0}\text{.284}\text{kJ}}{0.5\text{kJ}}\\ =0.568\times 100\%\\ =56.8\%\end{array}$

Thus, the thermal efficiency of the cycle is $56.8\%$.

c)

To determine

The mean effective pressure of the cycle.

Answer to Problem 164RP

The mean effective pressure of the cycle is $631.1\text{kPa}$.

Explanation of Solution

Write the expression to calculate the mean effective pressure for an ideal otto cycle $\left(\text{MEP}\right)$.

$\text{MEP}=\frac{W_{\text{net}}}{v_1-v_2}$ (XII)

Here, the compression ratio is $r$.

Conclusion:

Substitute, $\text{0}\text{.284}\text{kJ}$ for $W_{\text{net}}$, $0.000045{\text{m}}^3$ for ${\nu }_2$, and $0.000495{\text{m}}^3$ for $v_1$ in Equation (XII).

$\begin{array}{c}\text{MEP}=\frac{\text{0}\text{.284}\text{kJ}}{0.000495{\text{m}}^3-0.000045{\text{m}}^3}\\ =\frac{\text{0}\text{.284}\text{kJ}\left(\frac{1\text{kPa}\cdot {\text{m}}^3}{\text{kJ}}\right)}{0.000495{\text{m}}^3-0.000045{\text{m}}^3}\\ =631.1\text{kPa}\end{array}$

Thus, the mean effective pressure of the cycle is $631.1\text{kPa}$.

d)

To determine

The power output for an engine speed of 3000 rpm.

Answer to Problem 164RP

The power output for an engine speed of 3000 rpm is $28.4\text{W}$.

Explanation of Solution

Write the expression to calculate the power produced by the engine $\left({\dot{W}}_{\text{net}}\right)$.

${\dot{W}}_{\text{net}}=\frac{\left(n_{cyl}\times W_{\text{net}}\right)\dot{n}}{n_{\text{rev}}}$ (XIII)

Here, speed of the engine is $\dot{n}$, and number of revolutions per cycle in a four-stroke engine is $n_{\text{rev}}$.

Here, the compression ratio is $r$.

Conclusion:

In one cycle there are two revolutions in four stroke engines.

Substitute $\text{4}\text{cylinder}$ for $n_{cyl}$ , $0.284\text{kJ/cylinder-cycle}$ for $W_{\text{net}}$ , $\text{3000}\text{rev/min}$ for $\dot{n}$ and $\text{2}\text{rev/cycle}$ in Equation (XIII).

$\begin{array}{c}{\dot{W}}_{\text{net}}=\frac{\left(\text{4}\text{cylinder}\times 0.284\text{kJ/cylinder-cycle}\right)\text{3000}\text{rev/min}}{\text{2}\text{rev/cycle}}\\ =\frac{\left(\text{4}\text{cylinder}\times 0.284\text{kJ/cylinder-cycle}\right)\text{3000}\text{rev/min}\left(\frac{1\text{min}}{60\text{s}}\right)}{\text{2}\text{rev/cycle}}\\ =28.4\text{W}\end{array}$

Thus, the power output for an engine speed of 3000 rpm is $28.4\text{W}$.