Chapter 21, Problem 47CP

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 blanks of the table to follow the states of the gas.

Answer to Problem 47CP

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

Given info: 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 $1.4$.

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.

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.

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

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.

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

The compression ratio is,

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

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:

$V_{\text{C}}=V_{\text{B}}$

Write the expression to calculate the pressure at point C.

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

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 $V_{\text{C}}=V_{\text{B}}$

Therefore, the compression ratio is,

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

Write the expression to calculate the pressure at point D.

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

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 blanks of the table to follow the processes.

Answer to Problem 47CP

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

Given info: 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 $1.4$.

Write the expression to calculate the 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.

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

The net work done is,

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

The net heat energy is,

$Q_{\text{net}}=-W_{\text{eng}}$

Substitute $84.3\text{J}$ for $W_{\text{eng}}$ in above equation.

$Q_{\text{net}}=-84.3\text{J}$

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 is,

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 47CP

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

Explanation of Solution

Given info: 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 $1.4$.

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 47CP

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

Explanation of Solution

Given info: 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 $1.4$.

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 47CP

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

Explanation of Solution

Given info: 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 $1.4$.

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 47CP

The thermal efficiency is $0.565$.

Explanation of Solution

Given info: 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 $1.4$.

Write the expression to calculate the thermal efficiency.

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

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

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

Thus, the thermal efficiency is $0.565$.

Conclusion:

Therefore, the thermal efficiency is $0.565$.

To determine

The number of crankshaft revolution per minute.

Answer to Problem 47CP

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

Explanation of Solution

Given info: 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 $1.4$.

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

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

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