Chapter 15, Problem 97P

(a)

To determine

The maximum possible efficiency of the plant.

Answer to Problem 97P

The maximum possible efficiency of the plant is $39.6\%$.

Explanation of Solution

Write an expression for the maximum possible efficiency of the plant.

$e_r=1-\frac{T_C}{T_H}$ (I)

Here, $e_r$ is the maximum possible efficiency of the plant, $T_C$ is the temperature of cold reservoir and $T_H$ is the temperature of hot reservoir.

Conclusion:

Substitute $323\text{K}$ for $T_C$ and $535\text{K}$ for $T_H$ in equation (I) to find $e_r$.

$\begin{array}{c}{e}_r=1-\frac{323\text{K}}{535\text{K}}\\ =\left(0.396\right)\left(100\%\right)\\ =39.6\%\end{array}$

Thus, the maximum possible efficiency of the plant is $39.6\%$.

(b)

To determine

The rate at which the heat removed.

Answer to Problem 97P

The rate at which the heat removed is $4.98\times {10}^8\text{W}$.

Explanation of Solution

Write an expression for the heat removed.

$\begin{array}{c}\frac{Q_C}{\Delta t}=\frac{W_{net}}{e\Delta t}-\frac{W_{net}}{\Delta t}\\ =\frac{W_{net}}{\Delta t}\left(\frac{1}{e}-1\right)\end{array}$ (II)

Here, $Q_C$ is the heat removed, $\Delta t$ is the length of interval, $W_{net}$ is the net work done and $e$ is the efficiency.

Conclusion:

Substitute $1.23\times {10}^8\text{W}$ for $W_{net}/\Delta t$, $50.0\%e_r$ for $e$ and $39.6\%$ for $e_r$ in equation (II) to find $Q_C/\Delta t$.

$\begin{array}{c}{Q}_C/\Delta t=\left(1.23\times {10}^8\text{W}\right)\left(\frac{1}{\left(50.0\%\right)\left(e_r\right)}-1\right)\\ =\left(1.23\times {10}^8\text{W}\right)\left(\frac{1}{\left(50.0\%\right)\left(39.6\%\right)}-1\right)\\ =\left(1.23\times {10}^8\text{W}\right)\left(\frac{1}{\left(50.0\%\right)\left(\frac{1}{100\%}\right)\left(39.6\%\right)\left(\frac{1}{100\%}\right)}-1\right)\\ \left(1.23\times {10}^8\text{W}\right)\left(\frac{1}{\left(0.50\right)\left(0.396\right)}-1\right)\\ =4.98\times {10}^8\text{W}\end{array}$

Thus, the rate at which the heat removed is $4.98\times {10}^8\text{W}$.

(c)

To determine

The maximum possible rate at which water can be pumped uphill.

Answer to Problem 97P

The maximum possible rate at which water can be pumped uphill is $33.0{\text{m}}^3\text{/s}$.

Explanation of Solution

Write an expression for the rate of work done.

$\begin{array}{c}\frac{W_{net}}{\Delta t}=\frac{mgh}{\Delta t}\\ =\left(m/\Delta t\right)gh\end{array}$ (III)

Here, $m$ is the mass, $g$ is the acceleration due to gravity and $h$ is the height.

Rewrite the equation (III) to find $m/\Delta t$.

$\left(m/\Delta t\right)=\frac{\left(W_{net}/\Delta t\right)}{gh}$ (IV)

Conclusion:

Substitute $1.23\times {10}^8\text{W}$ for $W_{net}/\Delta t$, $9.80{\text{m/s}}^2$ for $g$ and $380\text{m}$ for $h$ in equation (IV) to find $m/\Delta t$.

$\begin{array}{c}\left(m/\Delta t\right)=\frac{\left(1.23\times {10}^8\text{W}\right)}{\left(9.80{\text{m/s}}^2\right)\left(380\text{m}\right)}\\ =\frac{1.23\times {10}^8\text{W}}{3724{\text{m}}^2{\text{/s}}^2}\\ =3.30\times {10}^4\text{kg/s}\end{array}$

Thus, the maximum possible rate at which water can be pumped uphill is $33.0{\text{m}}^3\text{/s}$.

Thus, calculate the volume flow rate.

$\begin{array}{c}\left(V/\Delta t\right)=\left(m/\Delta t\right)\left(\frac{1{\text{m}}^3}{1000\text{kg}}\right)\\ =33.0{\text{m}}^3\text{/s}\end{array}$

Here, $\left(V/\Delta t\right)$ is the volume flow rate.

Thus, the maximum possible rate at which water can be pumped uphill is $33.0{\text{m}}^3\text{/s}$.