Chapter 22, Problem 76AP

(a)

To determine

The total energy entering the system by heat per cycle.

Answer to Problem 76AP

The total energy entering the system by heat per cycle is $10.5nR{T}_i$.

Explanation of Solution

Apply ideal gas equation at the point A.

  $P_i{V}_i=nR{T}_i$                                                                                                                (I)

Here, $P_i$ is the initial pressure, $V_i$ is the initial volume, $n$ is the number of moles, $R$ is the universal gas constant and $T_i$ is the initial temperature.

Apply ideal gas equation at the point B.

  $3P_i{V}_i=nR{T}_B$                                                                                                            (II)

Here, $T_B$ is the temperature at $B$.

Substitute $nR{T}_i$ for $P_i{V}_i$ in equation (II) to get $T_B$.

  $\begin{array}{c}3nR{T}_i=nR{T}_B\\ T_B=3T_i\end{array}$                                                                                                          (III)

Apply ideal gas equation at the point C.

  $\begin{array}{c}\left(3P_i\right)\left(2V_i\right)=nR{T}_C\\ 6P_i{V}_i=nR{T}_C\end{array}$                                                                                                  (IV)

Here, $T_C$ is the temperature at $C$.

Substitute $nR{T}_i$ for $P_i{V}_i$ in equation (IV) to get $T_C$.

  $\begin{array}{c}6nR{T}_i=nR{T}_C\\ T_C=6T_i\end{array}$                                                                                                          (V)

Apply ideal gas equation at the point D.

  $\begin{array}{c}{P}_i\left(2V_i\right)=nR{T}_D\\ 2P_i{V}_i=nR{T}_D\end{array}$                                                                                                       (VI)

Here, $T_D$ is the temperature at $D$.

Substitute $nR{T}_i$ for $P_i{V}_i$ in equation (VI) to get $T_D$.

  $\begin{array}{c}2nR{T}_i=nR{T}_D\\ T_D=2T_i\end{array}$                                                                                                        (VII)

From figure P22.76, the curve $AB$ is isochoric process.

Write the expression for the heat change during process defined by curve $AB$.

  $Q_{AB}=n{C}_V\left(T_B-T_i\right)$                                                                                              (VIII)

Here, $Q_{AB}$ is the heat change during process defined by curve $AB$ and $C_V$ is the specific heat capacity at constant volume.

From figure P22.76, the curve $BC$ is isobaric process.

Write the expression for the heat change during process defined by curve $BC$.

  $Q_{BC}=n{C}_P\left(T_C-T_B\right)$                                                                                                (IX)

Here, $Q_{BC}$ is the heat change during process defined by curve $BC$ and $C_P$ is the specific heat capacity at constant pressure.

From figure P22.76, the curve $CD$ is isochoric process.

Write the expression for the heat change during process defined by curve $CD$.

  $Q_{CD}=n{C}_V\left(T_D-T_C\right)$                                                                                                (X)

Here, $Q_{CD}$ is the heat change during process defined by curve $CD$.

From figure P22.76, the curve $DA$ is isobaric process.

Write the expression for the heat change during process defined by curve $DA$.

  $Q_{DA}=n{C}_P\left(T_i-T_D\right)$                                                                                                (XI)

Here, $Q_{DA}$ is the heat change during process defined by curve $DA$.

In this cycle heat is entering through process $AB$ and $BC$.

Write the expression for the total heat entering the cycle.

  $Q_{\text{entering}}=Q_{AB}+Q_{BC}$                                                                                                (XII)

Here, $Q_{\text{entering }}$ is the heat entering into the gas.

This heat entering must be equal to heat absorbed from hot reservoir by the gas.

  $Q_{\text{entering}}=\left|Q_h\right|$                                                                                                         (XIII)

Here, $\left|Q_h\right|$ is the heat absorbed from hot reservoir by the gas.

Conclusion:

The specific heat capacity at constant volume of the monoatomic gas is $C_V=\frac{3R}{2}$ and specific heat capacity at constant pressure of the monoatomic gas is $C_P=\frac{5R}{2}$.

Substitute $\frac{3R}{2}$ for $C_V$ and $3T_i$ for $T_B$  in equation (VIII) to get $Q_{AB}$.

  $\begin{array}{c}{Q}_{AB}=n\frac{3R}{2T_i}\left(3T_i-T_i\right)\\ =n\frac{3R}{2T_i}\left(2T_i\right)\\ =3nR{T}_i\end{array}$

Substitute $\frac{5R}{2}$ for $C_P$ , $6T_i$ for $T_C$ and $3T_i$ for $T_B$ in equation (IX) to get $Q_{BC}$.

  $\begin{array}{c}{Q}_{BC}=n\frac{5R}{2}\left(6T_i-3T_i\right)\\ =n\frac{5R}{2}\left(3T_i\right)\\ =7.5nR{T}_i\end{array}$

Substitute $3nR{T}_i$ for $Q_{AB}$ and $7.5nR{T}_i$ for $Q_{BC}$ in equation (XII) to get $Q_{\text{entering}}$.

  $\begin{array}{c}{Q}_{\text{entering}}=3nR{T}_i+7.5nR{T}_i\\ =10.5nR{T}_i\end{array}$

Therefore, the total energy entering the system by heat per cycle is $10.5nR{T}_i$.

(b)

To determine

The total energy leaving the system by heat per cycle.

Answer to Problem 76AP

The total energy leaving the system by heat per cycle is $8.50nR{T}_i$.

Explanation of Solution

In this cycle heat is leaving through processes $CD$ and $DA$.

Write the expression for the heat leaving the system.

  $Q_{\text{leaving}}=\left|Q_{CD}+Q_{DA}\right|$                                                                                             (XIV)

Here, $Q_{\text{leaving}}$ is the heat leaving from the gas.

This heat leaving must be equal to heat expelled to cold reservoir by the gas.

  $Q_{\text{leaving}}=\left|Q_c\right|$                                                                                                          (XV)

Here, $\left|Q_c\right|$ is the heat expelled to cold reservoir by the gas.

Conclusion:

Substitute $\frac{3R}{2}$ for $C_V$ , $2T_i$ for $T_D$ and $6T_i$ for $T_C$ in equation (X) to get $Q_{CD}$.

  $\begin{array}{c}{Q}_{CD}=n\frac{3R}{2T_i}\left(2T_i-6T_i\right)\\ =n\frac{3R}{2T_i}\left(-4T_i\right)\\ =-6nR{T}_i\end{array}$

Substitute $\frac{5R}{2}$ for $C_P$ , $2T_i$ for $T_D$ in equation (XI) to get $Q_{DA}$.

  $\begin{array}{c}{Q}_{DA}=n\frac{5R}{2}\left(T_i-2T_i\right)\\ =n\frac{5R}{2}\left(-T_i\right)\\ =-2.50nR{T}_i\end{array}$

Substitute $-6nR{T}_i$ for $Q_{CD}$ and $-2.50nR{T}_i$ for $Q_{DA}$ in equation (XV) to get $Q_{\text{leaving}}$.

  $\begin{array}{c}{Q}_{\text{leaving}}=\left|-6nR{T}_i+-2.50nR{T}_i\right|\\ =8.50nR{T}_i\end{array}$

Therefore, the total energy leaving of an engine operating in this cycle is $8.50nR{T}_i$.

(c)

To determine

The efficiency of the engine operating in this cycle.

Answer to Problem 76AP

The efficiency of the engine operating in this cycle is $0.190$.

Explanation of Solution

Write the expression for the efficiency of the engine.

  $e=\frac{\left|Q_h\right|-\left|Q_c\right|}{\left|Q_h\right|}$                                                                                                       (XVI)

Here, $e$ is the actual efficiency of the engine.

Substitute $10.5nR{T}_i$ for $Q_{\text{entering}}$ in equation (XIII) to find $\left|Q_h\right|$ .

  $\left|Q_h\right|=10.5nR{T}_i$

Substitute $8.50nR{T}_i$ for $Q_{\text{leaving}}$ in equation (XV) to find $\left|Q_c\right|$ .

  $\left|Q_c\right|=8.50nR{T}_i$

Conclusion:

Put the above two equations in equation (XVI) to find $e$ .

  $\begin{array}{c}e=\frac{10.5nR{T}_i-8.50nR{T}_i}{10.5nR{T}_i}\\ =\frac{10.5-8.50}{10.5}\\ =0.190\end{array}$

Therefore, the efficiency of the engine operating in this cycle is $0.190$.

(d)

To determine

The comparison between actual efficiency of the engine and Carnot efficiency.

Answer to Problem 76AP

The Carnot efficiency of the engine is $0.833$ and it is much higher than actual efficiency.

Explanation of Solution

Write the expression for the Carnot efficiency.

  $e_C=1-\frac{T_c}{T_h}$

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

Conclusion:

Substitute $T_i$ for $T_c$ and $6T_i$ for $T_h$ in above equation to get $e_C$.

  $\begin{array}{c}{e}_C=1-\frac{T_i}{6T_i}\\ =0.833\end{array}$

Therefore, Carnot efficiency is $0.833$ and it is much higher than actual efficiency.