Chapter 18, Problem 16P

(a)

To determine

The unknown pressure, volume and temperature.

Answer to Problem 16P

The unknown parameters have been tabulated.

State	$P\left(\text{kPa}\right)$	$V\left(\text{L}\right)$	$T\left(\text{K}\right)$
A	$1400$	$10$	$720$
B	$875$	$16$	$720$
C	$445$	$24$	$549$
D	$712$	$15$	$549$

Explanation of Solution

Consider the first adiabatic process along D through A

Write the equation for adiabatic process,

    $\begin{array}{c}P{V}^{\gamma }=\text{constant}\\ P_D{V}_D{}^{\gamma }=P_A{V}_A{}^{\gamma }\end{array}$        (I)

Here $P_A$ is the pressure at A, $P_D$ is the pressure at D, $V_A$ is the volume at A, $V_D$ is the volume at D and $\gamma$ is the specific heat ratio.

Rearrange (I) in terms of $P_D$

    $P_D=P_A{\left(\frac{V_A}{V_D}\right)}^{\gamma }$ in        (II)

Write the ideal gas equation

    $\begin{array}{l}{P}_A{V}_A=nR{T}_A\\ P_D{V}_D=nR{T}_D\end{array}$        (III)

Here $R$ is the real gas constant, $n$ is the number of moles, $T_A$ and $T_D$ is the temperature at A and D respectively.

Substitute for $P_D$ and $P_A$ from (III) in (I)

    $\begin{array}{c}\left(\frac{nR{T}_A}{V_A}\right)V_A{}^{\gamma }=\left(\frac{nR{T}_D}{V_D}\right)V_D{}^{\gamma }\\ T_D=T_A{\left(\frac{V_A}{V_D}\right)}^{\gamma -1}\end{array}$        (IV)

Consider the isothermal process along C through D.

Therefore,

    $T_D=T_C$

Here, $T_C$ is the temperature at C.

Write down the equation for isothermal process.

    $\begin{array}{c}{P}_C{V}_C=P_D{V}_D\\ P_C=P_D\left(\frac{V_D}{V_C}\right)\end{array}$        (V)

Here $P_C$ is the pressure at C, $V_C$ is the volume at C.

Substitute (II) in (V)

    $\begin{array}{c}{P}_C=\left[P_A{\left(\frac{V_A}{V_D}\right)}^{\gamma }\right]\left(\frac{V_D}{V_C}\right)\\ =\frac{P_A{V}_A^{\gamma }}{V_C{V}_D{}^{\gamma -1}}\end{array}$        (VI)

Consider the adiabatic process along B through C.

  $P_B{V}_B{}^{\gamma }=P_C{V}_C{}^{\gamma }$        (VII)

Here $P_B$ is the pressure at B, $V_B$ is the volume at B.

Substitute (VI) in (VII)

    $P_B{V}_B{}^{\gamma }=\left(\frac{P_A{V}_A^{\gamma }}{V_C{V}_D{}^{\gamma -1}}\right)V_C{}^{\gamma }$        (VIII)

Consider the isothermal process along A through B

    $\begin{array}{c}{P}_A{V}_A=P_B{V}_B\\ P_B=P_A\left(\frac{V_A}{V_B}\right)\end{array}$        (IX)

Substitute (IX) in (VIII)

    $\begin{array}{c}{P}_A\left(\frac{V_A}{V_B}\right)V_B{}^{\gamma }=\left(\frac{P_A{V}_A^{\gamma }}{V_C{V}_D{}^{\gamma -1}}\right)V_C{}^{\gamma }\\ V_B=\frac{V_A{V}_C}{V_D}\end{array}$        (X)

Conclusion:

Substitute $1400\text{kPa}$ for $P_A$, $10\text{L}$ for $V_A$, $15\text{L}$ for $V_D$ and $5/3$ for $\gamma$ in (II)

    $\begin{array}{c}{P}_D=1400\text{kPa}{\left(\frac{10\text{L}}{15\text{L}}\right)}^{\frac{5}{3}}\\ =712\text{kPa}\end{array}$

Substitute $720\text{K}$ for $T_A$, $10\text{L}$ for $V_A$, $15\text{L}$ for $V_D$ and $5/3$ for $\gamma$ in (IV)

    $\begin{array}{c}{T}_D=720\text{K}{\left(\frac{10\text{L}}{15\text{L}}\right)}^{\frac{5}{3}-1}\\ =549\text{K}\end{array}$

Therefore,

    $T_C=549\text{K}$

Substitute $1400\text{kPa}$ for $P_A$, $10\text{L}$ for $V_A$, $15\text{L}$ for $V_D$, $24\text{L}$ for $V_C$ and $5/3$ for $\gamma$ in (VI)

    $\begin{array}{c}{P}_C=\frac{\left(1400\text{kPa}\right){\left(10\text{L}\right)}^{\frac{5}{3}}}{\left(24\text{L}\right){\left(15\text{L}\right)}^{\frac{5}{3}-1}}\\ =445\text{kPa}\end{array}$

Substitute $10\text{L}$ for $V_A$, $15\text{L}$ for $V_D$, $24\text{L}$ for $V_C$ in (X)

    $\begin{array}{c}{V}_B=\frac{\left(10\text{L}\right)\left(24\text{L}\right)}{\left(15\text{L}\right)}\\ =16\text{L}\end{array}$

Substitute $1400\text{kPa}$ for $P_A$, $10\text{L}$ for $V_A$ and $16\text{L}$ for $V_B$ in (IX)

    $\begin{array}{c}{P}_B=1400\text{kPa}\left(\frac{10\text{L}}{16\text{L}}\right)\\ =875\text{kPa}\end{array}$

Therefore,

State	$P\left(\text{kPa}\right)$	$V\left(\text{L}\right)$	$T\left(\text{K}\right)$
A	$1400$	$10$	$720$
B	$875$	$16$	$720$
C	$445$	$24$	$549$
D	$712$	$15$	$549$

(b)

To determine

Energy, work done and change in internal energy involved in each steps.

Answer to Problem 16P

Energy, work done and change in internal energy has been tabulated.

Process	$Q\left(\text{kJ}\right)$	$W\left(\text{kJ}\right)$	$\Delta E\left(\text{kJ}\right)$
A to B	$6.58$	$-6.58$	$0$
B to C	$0$	$-4.98$	$-4.98$
C to D	$-5.02$	$5.02$	$0$
D to E	$0$	$4.98$	$4.98$

Explanation of Solution

Consider the isothermal process along A through B.

Write the equation for energy transfer

    $\Delta E=n{C}_V\Delta T$        (XI)

Here $\Delta E$ is the internal energy, $n$ is the number of moles, $C_V$ is the specific heat at constant volume and $\Delta T$ is the temperature.

$\Delta T=0$ for isothermal process,

Then by first law of thermodynamics

    $\begin{array}{c}Q=-W\\ =nR{T}_A\ln \left(\frac{V_B}{V_A}\right)\end{array}$        (XII)

Here $Q$ is the heat energy and $W$ is the work done

Consider the adiabatic process along B through C where $Q=0$

    $\begin{array}{c}W=\Delta E\\ =n{C}_V\left(T_C-T_B\right)\end{array}$        (XIII)

Here $C_V$ is the specific heat capacity at constant volume.

Substitute $C_V=\frac{5}{3}R$ in         (XIII)

    $\begin{array}{c}W=\Delta E\\ =n\left(\frac{5}{3}R\right)\left(T_C-T_B\right)\end{array}$        (XIV)

Consider the isothermal process along C through D

    $\begin{array}{c}Q=-W\\ =nR{T}_C\ln \left(\frac{V_D}{V_C}\right)\end{array}$        (XV)

Consider the adiabatic process along D through A.

    $\begin{array}{c}W=\Delta E\\ =n{C}_V\left(T_A-T_D\right)\\ =n\left(\frac{5}{3}R\right)\left(T_A-T_D\right)\end{array}$        (XVI)

Conclusion:

Substitute $2.34\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $720\text{K}$ for $T_A$, $10\text{L}$ for $V_A$ and $16\text{L}$ for $V_B$ in (XII)

    $\begin{array}{c}W=-Q\\ =\left(2.34\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(720\text{K}\right)\ln \left(\frac{16\text{L}}{10\text{L}}\right)\\ =6.58\text{kJ}\end{array}$

Substitute $2.34\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $720\text{K}$ for $T_B$, $549\text{K}$ for $T_C$ in (XIV))

    $\begin{array}{c}W=\Delta E\\ =\left(2.34\text{mol}\right)\left(\frac{5}{3}\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\right)\left(549\text{K}-720\text{K}\right)\\ =-4.98\text{kJ}\end{array}$

Substitute $2.34\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $549\text{K}$ for $T_C$, $15\text{L}$ for $V_D$ and $24\text{L}$ for $V_C$ in (XV)

    $\begin{array}{c}Q=-W\\ =\left(2.34\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(549\text{K}\right)\ln \left(\frac{15\text{L}}{24\text{L}}\right)\\ =-5.02\text{kJ}\end{array}$

Substitute $2.34\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $720\text{K}$ for $T_A$, $549\text{K}$ for $T_D$ in (XVI))

    $\begin{array}{c}W=\Delta E\\ =\left(2.34\text{mol}\right)\left(\frac{5}{3}\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\right)\left(720\text{K-}549\text{K}\right)\\ =4.98\text{kJ}\end{array}$

Therefore

Process	$Q\left(\text{kJ}\right)$	$W\left(\text{kJ}\right)$	$\Delta E\left(\text{kJ}\right)$
A to B	$6.58$	$-6.58$	$0$
B to C	$0$	$-4.98$	$-4.98$
C to D	$-5.02$	$5.02$	$0$
D to E	$0$	$4.98$	$4.98$

(c)

To determine

Efficiency of the engine.

Answer to Problem 16P

Efficiency is $\underset{\_}{23.7\%}$.

Explanation of Solution

Write the equation for efficiency

    $\begin{array}{c}e=\frac{W_{\text{eng}}}{Q_h}\\ =-\frac{W_{ABCD}}{Q_A{}_{\to B}}\\ =\frac{-\left(W_A+W_B+W_C+W_D\right)}{Q_A+Q_B+Q_C+Q_D}\end{array}$        (XVII)

Here $e$ is the efficiency, $W_A,W_B,W_C\&W_D$ are work done at A, B, C and D respectively, $Q_A,Q_B,Q_C\&Q_D$ are energy by heat transfer at A, B, C and D respectively.

Conclusion:

Refer sub part b and substitute $-1.56\text{kJ}$ for $W_{ABCD}$ and $6.58\text{kJ}$ for $Q_A{}_{\to B}$ in (XVII)

    $\begin{array}{c}e=\frac{1.56\text{kJ}}{6.58\text{kJ}}\\ =0.237\\ =23.7\%\end{array}$

Efficiency is $\underset{\_}{23.7\%}$.

(d)

To determine

To prove that the efficiency is equal to Carnot efficiency.

Answer to Problem 16P

The efficiency is equal to Carnot efficiency.

Explanation of Solution

Write the equation for Carnot efficiency

    $e_C=1-\frac{T_c}{T_h}$        (XVIII)

Here $e_C$ is the Carnot efficiency, $T_c$ is the temperature of the cold reservoir and $T_h$ is the temperature of the hot reservoir.

Conclusion:

Substitute $549\text{K}$ for $T_c$ and $720\text{K}$ for $T_h$ in (XVIII)

    $\begin{array}{c}{e}_C=1-\frac{549\text{K}}{720\text{K}}\\ =0.237\\ =23.7\%\end{array}$

Thus, the efficiency is equal to Carnot efficiency