Chapter 18, Problem 8P

(a)

To determine

To show: The overall efficiency of the two engine is $e=e_1+e_2-e_1{e}_2$.

Answer to Problem 8P

The overall efficiency of the two engine is $e=e_1+e_2-e_1{e}_2$.

Explanation of Solution

Given info: The efficiency of the two engine is $e_1$ and $e_2$. The lower and higher temperature of the engine $1$ is $T_{\text{i}}$ and $T_{\text{h}}$ respectively. The lower and the higher temperature of the engine $2$ is $T_{\text{c}}$ and $T_{\text{i}}$ respectively.

Write the expression of the efficiency of the engine $1$.

$\begin{array}{c}{e}_1=\frac{W_{\text{eng}\text{1}}}{Q_{1\text{h}}}\\ W_{\text{eng}\text{1}}=e_1{Q}_{1\text{h}}\end{array}$

Here,

$W_{\text{eng}\text{1}}$ is the work done by the engine $1$.

$Q_{\text{1h}}$ is the heat supplied to the engine $1$.

Write the expression of the efficiency of the engine $2$.

$\begin{array}{c}{e}_2=\frac{W_{\text{eng}2}}{Q_{\text{2h}}}\\ W_{\text{eng}2}=e_2{Q}_{\text{2h}}\end{array}$

Here,

$W_{\text{eng}2}$ is the work done by the engine $2$.

$Q_{\text{2h}}$ is the heat supplied to the engine $2$.

The exhaust heat of engine $1$ is supplied to the engine $2$.

$\begin{array}{c}{Q}_{\text{2h}}=Q_{1\text{c}}\\ =Q_{1\text{h}}-W_{\text{eng}\text{1}}\end{array}$

Here,

$Q_{1\text{c}}$ is the exhaust heat of engine $1$.

Substitute $e_1{Q}_{1\text{h}}$ for $W_{\text{eng}\text{1}}$ in above equation.

$Q_{\text{2h}}=Q_{1\text{h}}-e_1{Q}_{1\text{h}}$

Write the expression of the efficiency of the two engine device.

$e=\frac{W_{\text{eng}\text{1}}+W_{\text{eng}\text{2}}}{Q_{\text{1h}}}$

Substitute $e_1{Q}_{1\text{h}}$ for $W_{\text{eng}\text{1}}$ and $e_2{Q}_{\text{2h}}$ for $e_2{Q}_{\text{2h}}$ in above equation.

$e=\frac{e_1{Q}_{1\text{h}}+e_2{Q}_{\text{2h}}}{Q_{\text{1h}}}$

Substitute $Q_{1\text{h}}-e_1{Q}_{1\text{h}}$ for $Q_{\text{2h}}$ in above equation.

$\begin{array}{c}e=\frac{e_1{Q}_{1\text{h}}+e_2\left(Q_{1\text{h}}-e_1{Q}_{1\text{h}}\right)}{Q_{\text{1h}}}\\ =\frac{e_1{Q}_{1\text{h}}+e_2{Q}_{1\text{h}}-e_1{e}_2{Q}_{1\text{h}}}{Q_{1\text{h}}}\\ =e_1+e_2-e_1{e}_2\end{array}$

Conclusion:

Therefore, the overall efficiency of the two engine is $e=e_1+e_2-e_1{e}_2$.

(b)

To determine

The efficiency of the combination engine.

Answer to Problem 8P

The efficiency of the combination engine is $1-\frac{T_{\text{c}}}{T_{\text{h}}}$.

Explanation of Solution

Given info: The efficiency of the two engine is $e_1$ and $e_2$. The lower and higher temperature of the engine $1$ is $T_{\text{i}}$ and $T_{\text{h}}$ respectively. The lower and the higher temperature of the engine $2$ is $T_{\text{c}}$ and $T_{\text{i}}$ respectively.

The overall efficiency of the two engine is,

$e=e_1+e_2-e_1{e}_2$ (1)

Write the expression of the Carnot efficiency of the engine $1$.

$e_1=1-\frac{T_{\text{i}}}{T_{\text{h}}}$

Write the expression of the Carnot efficiency of the engine $2$.

$e_2=1-\frac{T_{\text{c}}}{T_{\text{i}}}$

Substitute $1-\frac{T_{\text{i}}}{T_{\text{h}}}$ for $e_1$ and $1-\frac{T_{\text{c}}}{T_{\text{i}}}$ for $e_2$ in equation (1).

$\begin{array}{c}e=\left(1-\frac{T_{\text{i}}}{T_{\text{h}}}\right)+\left(1-\frac{T_{\text{c}}}{T_{\text{i}}}\right)-\left(1-\frac{T_{\text{i}}}{T_{\text{h}}}\right)\left(1-\frac{T_{\text{c}}}{T_{\text{i}}}\right)\\ =2-\frac{T_{\text{i}}}{T_{\text{h}}}-\frac{T_{\text{c}}}{T_{\text{i}}}-1+\frac{T_{\text{c}}}{T_{\text{i}}}+\frac{T_{\text{i}}}{T_{\text{h}}}-\frac{T_{\text{i}}}{T_{\text{h}}}\times \frac{T_{\text{c}}}{T_{\text{i}}}\\ =1-\frac{T_{\text{c}}}{T_{\text{h}}}\end{array}$

Conclusion:

Therefore, the efficiency of the combination engine is $1-\frac{T_{\text{c}}}{T_{\text{h}}}$.

(c)

To determine

Whether the efficiency is improved by using two engine instead of one.

Answer to Problem 8P

The efficiency is remains same even after combining the two engine.

Explanation of Solution

Given info: The efficiency of the two engine is $e_1$ and $e_2$. The lower and higher temperature of the engine $1$ is $T_{\text{i}}$ and $T_{\text{h}}$ respectively. The lower and the higher temperature of the engine $2$ is $T_{\text{c}}$ and $T_{\text{i}}$ respectively.

The Carnot engine has the maximum efficiency and no engine can have the efficiency more than the Carnot engine’s efficiency.

If the two Carnot engine is combined together than the efficiency is remain same because the combined engine is also a Carnot engine. So there is no requirement to use two engines simultaneously instead of one engine to improve the net efficiency. The work output increases but on the other hand the heat supplied also increase.

Conclusion:

Therefore, the efficiency is remains same even after combining the two engine.

(d)

To determine

The value of the intermediate temperature $T_{\text{i}}$ due to which the both the engine produce same work when connects in series.

Answer to Problem 8P

The value of the intermediate temperature $T_{\text{i}}$ due to which the both the engine produce same work when connects in series is $\frac{1}{2}\left(T_{\text{h}}+T_{\text{c}}\right)$.

Explanation of Solution

Given info: The efficiency of the two engine is $e_1$ and $e_2$. The lower and higher temperature of the engine $1$ is $T_{\text{i}}$ and $T_{\text{h}}$ respectively. The lower and the higher temperature of the engine $2$ is $T_{\text{c}}$ and $T_{\text{i}}$ respectively.

The work done by both the engine is same.

$W_{\text{eng}\text{2}}=W_{\text{eng}\text{1}}$

Write the expression of the efficiency of the two engine device.

$e=\frac{W_{\text{eng}\text{1}}+W_{\text{eng}2}}{Q_{\text{1h}}}$

Substitute $W_{\text{eng}\text{1}}$ for $W_{\text{eng}2}$ in above equation.

$\begin{array}{c}e=\frac{W_{\text{eng}\text{1}}+W_{\text{eng}1}}{Q_{\text{1h}}}\\ =\frac{2W_{\text{eng}\text{1}}}{Q_{\text{1h}}}\end{array}$

Substitute $e_1$ for $\frac{W_{\text{eng}\text{1}}}{Q_{\text{1h}}}$ in above equation.

$e=2e_1$

Substitute $1-\frac{T_{\text{c}}}{T_{\text{h}}}$ for $e$ and $1-\frac{T_{\text{i}}}{T_{\text{h}}}$ for $e_1$ in above equation.

$\begin{array}{c}1-\frac{T_{\text{c}}}{T_{\text{h}}}=2\left(1-\frac{T_{\text{i}}}{T_{\text{h}}}\right)\\ 2T_{\text{i}}=T_{\text{h}}+T_{\text{c}}\\ T_{\text{i}}=\frac{1}{2}\left(T_{\text{h}}+T_{\text{c}}\right)\end{array}$

Conclusion:

Therefore, the value of the intermediate temperature $T_{\text{i}}$ due to which the both the engine produce same work when connects in series is $\frac{1}{2}\left(T_{\text{h}}+T_{\text{c}}\right)$.

(e)

To determine

The value of the intermediate temperature $T_{\text{i}}$ due to which the efficiency of both the engine is same when connects in series.

Answer to Problem 8P

The value of the intermediate temperature $T_{\text{i}}$ due to which the efficiency of both the engine is same when connects in series is $\sqrt{T_{\text{h}}\cdot T_{\text{c}}}$.

Explanation of Solution

Given info: The efficiency of the two engine is $e_1$ and $e_2$. The lower and higher temperature of the engine $1$ is $T_{\text{i}}$ and $T_{\text{h}}$ respectively. The lower and the higher temperature of the engine $2$ is $T_{\text{c}}$ and $T_{\text{i}}$ respectively.

The efficiency of both the engine is same.

$e_1=e_2$

Substitute $1-\frac{T_{\text{i}}}{T_{\text{h}}}$ for $e_1$ and $1-\frac{T_{\text{c}}}{T_{\text{i}}}$ for $e_2$ in above equation.

$\begin{array}{c}1-\frac{T_{\text{i}}}{T_{\text{h}}}=1-\frac{T_{\text{c}}}{T_{\text{i}}}\\ T_{\text{i}}=\sqrt{T_{\text{h}}\cdot T_{\text{c}}}\end{array}$

Conclusion:

Therefore, the value of the intermediate temperature $T_{\text{i}}$ due to which the efficiency of both the engine is same when connects in series is $\sqrt{T_{\text{h}}\cdot T_{\text{c}}}$.