Chapter 22, Problem 22.82CP

The compression ratio of an Otto cycle as shown in Figure 21.12 is V_A/V_B = 8.00. At the beginning A of the compression process, 500 cm³ of gas is at 100 kPa and 20.0°C. At the beginning of the adiabatic expansion, the temperature is T_C = 750°C. Model the working fluid as an ideal gas with γ = 1.40. (a) Fill in this table to follow the states of the gas:

(b) Fill in this table to follow the processes:

(c) Identify the energy input |Q_h|, (d) the energy exhaust |Q_c|, and (e) the net output work W_eng. (f) Calculate the efficiency. (g) Find the number of crankshaft revolutions per minute required for a one-cylinder engine to have an output power of 1.00 kW = 1.34 hp. Note: The thermodynamic cycle involves four piston strokes.

To determine

The states of the gas during the Otto cycle.

Answer to Problem 22.82CP

The complete table is shown below.

State	$T\left(\text{K}\right)$	$P\left(\text{kPa}\right)$	$V\left({\text{cm}}^{\text{3}}\right)$
A	293	100	500
B	673	$1.84\times {10}^3$	62.5
C	1023	$2.79\times {10}^3$	62.5
D	445	152	500

Explanation of Solution

The compression ratio of an Otto cycle is $\frac{V_{\text{A}}}{V_{\text{B}}}=8.00$, at the beginning of $A$, the initial volume of the gas is $500{\text{cm}}^{\text{3}}$ and the initial temperature and pressure of gas is $20.0\text{°C}$ and $100\text{kPa}$. At the beginning of the adiabatic expansion, the temperature is $T_{\text{C}}=750\text{°C}$. The adiabatic index is $1.4$.

In Otto cycle, the process $A\to B$ is adiabatic compression, $B\to C$ is isovolumetric heat addition, $C\to D$ is adiabatic expansion and $D\to A$ is isovolumetric heat removal.

Write the expression to calculate the quantity of the gas.

    $n=\frac{P_{\text{A}}{V}_{\text{A}}}{R{T}_{\text{A}}}$

Here, $P_{\text{A}}$ is the pressure of the gas at point A, $V_{\text{A}}$ is the volume of the gas at point A, $T_{\text{A}}$ is the temperature at point A, $R$ is the universal gas constant and $n$ is the number of moles.

Substitute $100\text{kPa}$ for $P_{\text{A}}$, $500{\text{cm}}^{\text{3}}$ for $V_{\text{A}}$, $20.0\text{°C}$ for $T_{\text{A}}$ and $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$ in above equation to find $n$.

  $\begin{array}{c}n=\frac{\left(100\text{kPa}\times \frac{1000\text{Pa}}{1\text{kPa}}\right)\left(500{\text{cm}}^{\text{3}}\times \frac{{10}^{-6}{\text{m}}^{\text{3}}}{1{\text{cm}}^{\text{3}}}\right)}{\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(20.0\text{°C}+273\right)\text{K}}\\ =0.0205\text{mol}\end{array}$

In process $A\to B$,

Write the expression to calculate the pressure at point B.

    $P_{\text{B}}=P_{\text{A}}{\left(\frac{V_{\text{A}}}{V_{\text{B}}}\right)}^{\gamma }$

Here, $P_{\text{B}}$ is the pressure of the gas at point B, $V_{\text{B}}$ is the volume of the gas at point B and $\gamma$ is the specific heat ratio.

Substitute $8$  for $\frac{V_{\text{A}}}{V_{\text{B}}}$, $1.40$ for $\gamma$ and $100\text{kPa}$ for $P_{\text{A}}$ in above equation.

    $\begin{array}{c}{P}_{\text{B}}=\left(100\text{kPa}\times \frac{1000\text{Pa}}{1\text{kPa}}\right){\left(8\right)}^{1.40}\\ =1.84\times {10}^6\text{Pa}\\ \text{=}1.84\times {10}^6\text{Pa}\times \frac{{10}^{-3}\text{kPa}}{1\text{Pa}}\\ =1.84\times {10}^3\text{kPa}\end{array}$

Write the expression for the compression ratio

    $\frac{V_{\text{A}}}{V_{\text{B}}}=8.00$

Substitute $500{\text{cm}}^{\text{3}}$ for $V_{\text{A}}$ in above equation.

    $\begin{array}{l}\frac{V_{\text{A}}}{V_{\text{B}}}=8.00\\ V_{\text{B}}=62.5{\text{cm}}^{\text{3}}\end{array}$

Write the expression to calculate the temperature at point B.

    $T_{\text{B}}=\frac{P_{\text{B}}{V}_{\text{B}}}{nR}$

Substitute $1.84\times {10}^6\text{Pa}$ for $P_{\text{B}}$, $62.5{\text{cm}}^{\text{3}}$ for $V_{\text{B}}$, $0.0205\text{mol}$ for $n$ and $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$ in above equation.

    $\begin{array}{c}{T}_{\text{B}}=\frac{\left(1.84\times {10}^6\text{Pa}\right)\left(62.5{\text{cm}}^{\text{3}}\times \frac{{10}^{-6}{\text{m}}^{\text{3}}}{1{\text{cm}}^{\text{3}}}\right)}{0.0205\text{mol}\times \left(8.314\text{J}/\text{mol}\cdot \text{K}\right)}\\ =673\text{K}\end{array}$

At state C:

    $V_{\text{C}}=V_{\text{B}}$

Here, $V_{\text{C}}$ is the volume of the gas at point C.

Write the expression to calculate the pressure at point C.

    $P_{\text{C}}=\frac{nR{T}_{\text{C}}}{V_{\text{C}}}$

Here, $P_{\text{C}}$ is the pressure of the gas at point C and $T_{\text{C}}$ is the temperature at point C.

Substitute $62.5{\text{cm}}^{\text{3}}$ for $V_{\text{C}}$, $0.0205\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$ and $750\text{°C}$ for $T_{\text{C}}$ in above equation.

    $\begin{array}{c}{P}_{\text{C}}=\frac{\left(0.0205\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\times \left(750\text{°C+273}\right)\text{K}}{\left(62.5{\text{cm}}^{\text{3}}\times \frac{{10}^{-6}{\text{m}}^{\text{3}}}{1{\text{cm}}^{\text{3}}}\right)}\\ =2.79\times {10}^6\text{Pa}\\ =2.79\times {10}^6\text{Pa}\times \frac{{10}^{-3}\text{kPa}}{1\text{Pa}}\\ =2.79\times {10}^3\text{kPa}\end{array}$

State D:

    $V_{\text{D}}=V_{\text{A}}$ and $V_{\text{C}}=V_{\text{B}}$

Here, $V_{\text{D}}$ is the volume of the gas at point D.

Therefore, the compression ratio

    $\begin{array}{c}\left(\frac{V_{\text{C}}}{V_{\text{D}}}\right)=\frac{V_{\text{B}}}{V_{\text{A}}}\\ =\frac{1}{8}\end{array}$

Write the expression to calculate the pressure at point D.

    $P_{\text{D}}=P_{\text{C}}{\left(\frac{V_{\text{C}}}{V_{\text{D}}}\right)}^{\gamma }$

Here, $P_{\text{D}}$ is the pressure of the gas at point D and $T_{\text{D}}$ is the temperature at point D.

Substitute $2.79\times {10}^6\text{Pa}$ for $P_{\text{C}}$ and $\frac{1}{8}$ for $\left(\frac{V_{\text{C}}}{V_{\text{D}}}\right)$ in above equation.

    $\begin{array}{c}{P}_{\text{D}}=\left(2.79\times {10}^6\text{Pa}\right){\left(\frac{1}{8}\right)}^{1.4}\\ =1.52\times {10}^5\text{Pa}\\ \text{=}1.52\times {10}^5\text{Pa}\times \frac{{10}^{-3}\text{kPa}}{1\text{Pa}}\\ =152\text{kPa}\end{array}$

Write the expression to calculate the temperature at point D.

    $T_{\text{D}}=\frac{P_{\text{D}}{V}_{\text{D}}}{nR}$

Substitute $1.52\times {10}^5\text{Pa}$ for $P_{\text{D}}$, $500{\text{cm}}^{\text{3}}$ for $V_{\text{D}}$, $0.0205\text{mol}$ for $n$ and $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$ in above equation.

    $\begin{array}{c}{T}_{\text{D}}=\frac{\left(1.52\times {10}^5\text{Pa}\right)\left(500{\text{cm}}^{\text{3}}\times \frac{{10}^{-6}{\text{m}}^{\text{3}}}{1{\text{cm}}^{\text{3}}}\right)}{\left(0.0205\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)}\\ =445\text{K}\end{array}$

From the above explanation, the complete table is given below.

State	$T\left(\text{K}\right)$	$P\left(\text{kPa}\right)$	$V\left({\text{cm}}^{\text{3}}\right)$
A	293	100	500
B	673	$1.84\times {10}^3$	62.5
C	1023	$2.79\times {10}^3$	62.5
D	445	152	500

Conclusion:

Therefore, the complete table is given below.

State	$T\left(K\right)$	$P\left(\text{kPa}\right)$	$V\left({\text{cm}}^{\text{3}}\right)$
A	293	100	500
B	673	$1.84\times {10}^3$	62.5
C	1023	$2.79\times {10}^3$	62.5
D	445	152	500

To determine

The heat transferred, work done and the change in internal energy during the different process in the Otto cycle.

Answer to Problem 22.82CP

The complete table is shown below.

Process	Q	$W_{\text{eng}}$	$\Delta E_{\mathrm{int}}$
$A\to B$	0	-162	162
$B\to C$	149	0	149
$C\to D$	0	246	-246
$D\to A$	-65	0	-65
$ABCD$	84.3	84.3	0

Explanation of Solution

The process $A\to B$ is adiabatic and hence there is no transfer of heat during this process. Thus, $Q=0$ in this process.

Let $Q$ be the heat transferred, $\Delta E$ is the change in internal energy and $W_{\text{out}}$ is the work done.

Write the expression for change in internal energy in A to B process.

    $\Delta E_{\mathrm{int}\text{A}\to \text{B}}=\frac{5}{2}nR\left(T_{\text{B}}-T_{\text{A}}\right)$

Substitute $0.0205\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $673\text{K}$ for $T_{\text{B}}$ and $293\text{K}$ for $T_{\text{A}}$ in above equation.

    $\begin{array}{c}\Delta E_{\mathrm{int}\text{A}\to \text{B}}=\frac{5}{2}\left(0.0205\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(673\text{K}-293\text{K}\right)\\ =162\text{J}\end{array}$

Write the expression of first law of thermodynamics.

    $\Delta E_{\mathrm{int}\text{A}\to \text{B}}=Q-W_{\text{out}}$

Substitute $162\text{J}$ for $\Delta E_{\mathrm{int}\text{A}\to \text{B}}$ and 0 for $Q$ in above equation.

    $\begin{array}{c}162\text{J}=0-W_{\text{out}}\\ W_{\text{out}}=-162\text{J}\end{array}$

Write the expression to calculate the energy in B to C process.

    $\Delta E_{\mathrm{int}\text{B}\to \text{C}}=\frac{5}{2}nR\left(T_{\text{C}}-T_{\text{B}}\right)$

Substitute $0.0205\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $673\text{K}$ for $T_{\text{B}}$ and $1023\text{K}$ for $T_{\text{C}}$ in above equation,

    $\begin{array}{c}\Delta E_{\mathrm{int}\text{B}\to \text{C}}=\frac{5}{2}\left(0.0205\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(1023\text{K}-673\text{K}\right)\\ =149\text{J}\end{array}$

Write the expression of first law of thermodynamics.

    $\Delta E_{\mathrm{int}\text{B}\to \text{C}}=Q-W_{\text{out}}$

Substitute $149\text{J}$ for $\Delta E_{\mathrm{int}\text{B}\to \text{C}}$ and $0$ for $W_{\text{out}}$ in above equation.

    $\begin{array}{c}149\text{J}=Q-0\\ Q=149\text{J}\end{array}$

Write the expression to calculate the energy in C to D process.

    $\Delta E_{\mathrm{int}\text{C}\to \text{D}}=\frac{5}{2}nR\left(T_{\text{D}}-T_{\text{C}}\right)$

Substitute $0.0205\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $445\text{K}$ for $T_{\text{D}}$ and $1023\text{K}$ for $T_{\text{C}}$ in above equation,

    $\begin{array}{c}\Delta E_{\mathrm{int}\text{C}\to \text{D}}=\frac{5}{2}\left(0.0205\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(445\text{K}-1023\text{K}\right)\\ =-246\text{J}\end{array}$

Write the expression of first law of thermodynamics.

    $\Delta E_{\mathrm{int}\text{C}\to \text{D}}=Q-W_{\text{out}}$

Substitute $-246\text{J}$ for $\Delta E_{\mathrm{int}\text{C}\to \text{D}}$ and 0 for $Q$ in above equation.

    $\begin{array}{c}-246\text{J}=0-W_{\text{out}}\\ W_{\text{out}}=246\text{J}\end{array}$

Write the expression to calculate the energy in D to A process.

    $\Delta E_{\mathrm{int}\text{D}\to \text{A}}=\frac{5}{2}nR\left(T_{\text{A}}-T_{\text{D}}\right)$

Substitute $0.0205\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $445\text{K}$ for $T_{\text{D}}$ and $293\text{K}$ for $T_{\text{A}}$ in above equation,

    $\begin{array}{c}\Delta E_{\mathrm{int}\text{D}\to \text{A}}=\frac{5}{2}\left(0.0205\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(293\text{K}-445\text{K}\right)\\ =-65\text{J}\end{array}$

Write the expression of first law of thermodynamics.

    $\Delta E_{\mathrm{int}\text{D}\to \text{A}}=Q-W_{\text{out}}$

Substitute $-65\text{J}$ for $\Delta E_{\mathrm{int}\text{D}\to \text{A}}$ and $0$ for $W_{\text{out}}$ in above equation.

    $\begin{array}{c}-65\text{J}=Q-0\\ Q=-65\text{J}\end{array}$

Add all the work done found above to find the net work done

    $\begin{array}{c}{W}_{\text{eng}}=-162\text{J+0+246}\text{.3}\text{J+0}\\ =84.3\text{J}\end{array}$

Add the heat energy transferred in the four process given above to find the net heat energy

    $\begin{array}{c}{Q}_{\text{net}}=\text{0+149}\text{J+0}-65.0\text{J}\\ =84.3\text{J}\end{array}$

The change in internal energy during a cyclic process is zero.

Thus, $\Delta E_{\mathrm{int}}=0$ for the Otto cycle.

From the above explanation, the complete table is given below.

Process	Q	$W_{\text{eng}}$	$\Delta E_{\mathrm{int}}$
$A\to B$	0	-162	162
$B\to C$	149	0	149
$C\to D$	0	246	-246
$D\to A$	-65	0	-65
$ABCD$	84.3	84.3	0

Conclusion:

Therefore, the complete table

Process	Q	$W_{\text{eng}}$	$\Delta E_{\mathrm{int}}$
$A\to B$	0	-162	162
$B\to C$	149	0	149
$C\to D$	0	246	-246
$D\to A$	-65	0	-65
$ABCD$	84.3	84.3	0

To determine

The heat input during $B\to C$.

Answer to Problem 22.82CP

The heat input during $B\to C$ is $149\text{J}$.

Explanation of Solution

From part (b), the heat input during $B\to C$, $Q_h$ is $149\text{J}$.

Thus, the heat input during $B\to C$ is $149\text{J}$.

Conclusion:

Therefore, the heat input during $B\to C$ is $149\text{J}$.

To determine

The heat exhaust during $D\to A$.

Answer to Problem 22.82CP

The heat exhaust during $D\to A$, $Q_{\text{C}}$ is $65.0\text{J}$.

Explanation of Solution

From part (b)

The heat exhaust during $D\to A$, $Q_{\text{C}}$ is $65.0\text{J}$.

Thus, the heat exhaust during $D\to A$, $Q_{\text{C}}$ is $65.0\text{J}$.

Conclusion:

Therefore, the heat exhaust during $D\to A$, $Q_{\text{C}}$ is $65.0\text{J}$.

To determine

The net work output.

Answer to Problem 22.82CP

The net work output is $W_{\text{eng}}=84.3\text{J}$.

Explanation of Solution

From part (b)

The net work output is $W_{\text{eng}}=84.3\text{J}$.

Thus, the net work output is $W_{\text{eng}}=84.3\text{J}$.

Conclusion:

Therefore, the net work output is $W_{\text{eng}}=84.3\text{J}$.

To determine

The thermal efficiency.

Answer to Problem 22.82CP

The thermal efficiency is $0.565$.

Explanation of Solution

Write the expression to calculate the thermal efficiency.

    $e=\frac{W_{\text{eng}}}{Q_{\text{h}}}$

Conclusion:

Substitute $84.3\text{J}$ for $W_E$ and $149\text{J}$ for $Q_{\text{h}}$ in above equation.

    $\begin{array}{c}e=\frac{84.3\text{J}}{149\text{J}}\\ =0.565\end{array}$

Therefore, the thermal efficiency is $0.565$.

To determine

The number of crankshaft revolution per minute.

Answer to Problem 22.82CP

The number of crankshaft revolution per minute is $1.42\times {10}^3\text{rev}/\text{min}$.

Explanation of Solution

Write the expression to calculate the output power.

    $P=\frac{f}{2}{W}_{\text{eng}}$

Here, $f$ is the angular speed of the crankshaft.

Substitute $84.3\text{J/cycle}$ for $W_{\text{eng}}$ and $1\text{kW}$ for $P$ in above equation.

    $\begin{array}{c}1\text{kW}\times \frac{1000\text{J}/\text{s}}{1\text{kW}}=\frac{f}{120}\left(84.3\text{J/cycle}\right)\\ f=1.42\times {10}^3\text{rev}/\text{min}\end{array}$

Thus, the number of crankshaft revolution per minute is $1.42\times {10}^3\text{rev}/\text{min}$.