Chapter 21, Problem 15P

An electric generating station is designed to have an electric output power of 1.40 MW using a turbine with two-thirds the efficiency of a Carnot engine. The exhaust energy is transferred by heat into a cooling tower at 110°C. (a) Find the rate at which the station exhausts energy by heat as a function of the fuel combustion temperature T_h. (b) If the firebox is modified to run hotter by using more advanced combustion technology, how does the amount of energy exhaust change? (c) Find the exhaust power for T_h = 800°C. (d) Find the value of T_h for which the exhaust power would be only half as large as in part (c). (e) Find the value of T_h for which the exhaust power would be one-fourth as large as in part (c).

To determine

The rate at which the station exhaust energy by heat as a function of the fuel combustion temperature $T_H$.

Answer to Problem 15P

The rate at which the station exhaust energy by heat as a function of the fuel combustion temperature $T_H$ is $1.40\left(\frac{0.5T_H+383}{T_H-383}\right)$.

Explanation of Solution

The rate of work output of the engine is $1.40\text{MW}$, thetemperature into the cooling tower is $110°\text{C}$.

Write the formula to calculate the carnot efficiency of the engine.

    $\begin{array}{c}\eta =\left(\frac{T_H-T_C}{T_H}\right)\\ =\left(1-\frac{T_C}{T_H}\right)\end{array}$

Here, $\eta$ is the carnot efficiency of the engine, $T_C$ is the temperature into the cooling tower and $T_H$ is the fuel combustion temperature.

The actual efficiency of the engine is equal to two-thirds of the efficiency of the carnot engine.

    ${\eta }_a=\frac{2}{3}\eta$                                                                                (I)

Here, ${\eta }_a$ is the actual efficiency of the engine.

Substitute $\left(1-\frac{T_C}{T_H}\right)$ for $\eta$ in equation (I) to find ${\eta }_a$,

    $\begin{array}{c}{\eta }_a=\frac{2}{3}\left(1-\frac{T_C}{T_H}\right)\\ =\frac{2}{3}\left(\frac{T_H-T_C}{T_H}\right)\end{array}$

Write the formula to calculate the rate of heat input to the engine.

    $\begin{array}{c}{\eta }_a=\frac{W}{Q}\\ Q=\frac{W}{{\eta }_a}\end{array}$

Here, $W$ is the rate of work output of the engine and $Q$ is the rate of heat input to the engine.

Write the formula to calculate the rate at which the station exhaust energy by heat as a function of the fuel combustion temperature $T_H$.

    $\frac{Q_e}{\Delta t}=\frac{Q-W}{\Delta t}$                                                                                       (II)

Here, $Q_e$ is the energy expelled by the engine, $\Delta t$ is the time interval and $\frac{Q_e}{\Delta t}$ is the exhaust power.

Substitute $\frac{W}{{\eta }_a}$ for $Q$ in equation (II) to find $\frac{Q_e}{\Delta t}$,

    $\begin{array}{c}\frac{Q_e}{\Delta t}=\left(\frac{W}{{\eta }_a}-W\right)/\Delta t\\ =\frac{W}{\Delta t}\left(\frac{1}{{\eta }_a}-1\right)\end{array}$                                                                          (III)

Substitute $\frac{2}{3}\left(\frac{T_H-T_C}{T_H}\right)$ for ${\eta }_a$ in equation (3) to find $\frac{Q_e}{\Delta t}$,

    $\begin{array}{c}\frac{Q_e}{\Delta t}=\frac{W}{\Delta t}\left(\frac{1}{\frac{2}{3}\left(\frac{T_H-T_C}{T_H}\right)}-1\right)\\ =\frac{W}{\Delta t}\left(\frac{3}{2}\left(\frac{T_H}{T_H-T_C}\right)-1\right)\\ =\frac{W}{\Delta t}\left(\frac{T_H+2T_C}{2\left(T_H-T_C\right)}\right)\\ =\frac{W}{\Delta t}\left(\frac{0.5T_H+T_C}{T_H-T_C}\right)\end{array}$                                                                   (IV)

Conclusion:

Substitute $1.40\text{MW}$ for $\frac{W}{\Delta t}$ and $110°\text{C}$ for $T_C$ in equation (IV) to find $\frac{Q_e}{\Delta t}$,

    $\begin{array}{c}\frac{Q_e}{\Delta t}=1.40\text{ MW}\left(\frac{0.5T_H+\left(110°\text{C}+273\right)\text{K}}{T_H-\left(110°\text{C}+273\right)\text{K}}\right)\\ =1.40\left(\frac{0.5T_H+383}{T_H-383}\right)\end{array}$

Thus, the rate at which the station exhaust energy by heat as a function of the fuel combustion temperature $T_H$ is $1.40\left(\frac{0.5T_H+383}{T_H-383}\right)$. The energy exhaust rate is in megawatts and the fuel temperature is in Kelvin.

To determine

The effect on the amount of the energy if the firebox is modified to run hotter by using more advanced combustion technology.

Answer to Problem 15P

The  amount of the energy exhaust decreases as the fire box temperature increases.

Explanation of Solution

If the firebox is modified to run hotter by using more advanced combustion technology, the exhaust power increases by a factor of $0.5T_H+383$ and decreases by a factor of $T_H-383$. Thus, there is an overall decrease in power exhaust. Since power exhaust decreases the corresponding energy exhaust also should be decreasing.

Conclusion:

The  amount of the energy exhaust decreases as the fire box temperature increases.

To determine

The exhaust power for $T_H=800°\text{C}$.

Answer to Problem 15P

The exhaust power for $T_H=800°\text{C}$ is $1.87\text{MW}$.

Explanation of Solution

 The rate of work output of the engine is $1.40\text{MW}$, fuel combustion temperature is $800°\text{C}$, the temperature into the cooling tower is $110°\text{C}$.

From equation (IV), write the formula to calculate the exhaust power for $T_H=800°\text{C}$.

    $\frac{Q_e}{\Delta t}=W\left(\frac{0.5T_H+T_C}{T_H-T_C}\right)$

Conclusion:

Substitute $1.40\text{MW}$ for $W$, $800°\text{C}$ for $T_H$, $110°\text{C}$ for $T_C$ in equation (4) to find $\frac{Q_e}{\Delta t}$,

    $\begin{array}{c}\frac{Q_e}{\Delta t}=1.40\text{MW}\left(\frac{0.5\left(800°\text{C}+273\right)\text{K}+\left(110°\text{C}+273\right)\text{K}}{\left(800°\text{C}+273\right)\text{K}-\left(110°\text{C}+273\right)\text{K}}\right)\\ =1.40\text{MW}\left(\frac{0.5\left(1073\right)\text{K}+\left(383\right)\text{K}}{\left(1073\right)\text{K}-\left(383\right)\text{K}}\right)\\ =1.8656\text{MW}\\ \approx 1.87\text{MW}\end{array}$

Thus, the exhaust power for $T_H=800°\text{C}$ is $1.87\text{MW}$.

To determine

The value of $T_H$ for which the exhaust power would be only half as large as in part (c).

Answer to Problem 15P

The value of $T_H$ for which the exhaust power would be only half as large as in part (c) is $3.84\times {10}^3\text{K}$.

Explanation of Solution

 The rate of work output of the engine is $1.40\text{MW}$, the temperature into the cooling tower is $110°\text{C}$.

Write the expression for the exhaust power whuch would be only half as large as in part (c).

    ${\left(\frac{Q_e}{\Delta t}\right)}^{\prime }=\frac{1}{2}\left(\frac{Q_e}{\Delta t}\right)$                                                                             (V)

Here, ${\left(\frac{Q_e}{\Delta t}\right)}^{\prime }$ is the exhaust power whuch would be only half as large as in part (c).

Substitute $1.86\text{MW}$ for $\left(\frac{Q_e}{\Delta t}\right)$ in equation (5) to find ${\left(\frac{Q_e}{\Delta t}\right)}^{\prime }$,

    $\begin{array}{c}{\left(\frac{Q_e}{\Delta t}\right)}^{\prime }=\frac{1}{2}\left(1.86\text{MW}\right)\\ =0.933\text{MW}\end{array}$

Thus, the exhaust power whuch would be only half as large as in part (c) is $0.933\text{MW}$.

From equation (IV), Write the formula to calculate the value of $T_H$ for which the exhaust power would be only half as large as in part (c).

    $\begin{array}{c}{\left(\frac{Q_e}{\Delta t}\right)}^{\prime }=W\left(\frac{0.5T_H+T_C}{T_H-T_C}\right)\\ \left(T_H-T_C\right){\left(\frac{Q_e}{\Delta t}\right)}^{\prime }\times \frac{1}{W}=0.5T_H+T_C\\ T_H\left(\left({\left(\frac{Q_e}{\Delta t}\right)}^{\prime }\times \frac{1}{W}\right)-0.5\right)=T_C\left(1+{\left(\frac{Q_e}{\Delta t}\right)}^{\prime }\times \frac{1}{W}\right)\\ T_H=\frac{T_C\left(1+{\left(\frac{Q_e}{\Delta t}\right)}^{\prime }\times \frac{1}{W}\right)}{\left(\left({\left(\frac{Q_e}{\Delta t}\right)}^{\prime }\times \frac{1}{W}\right)-0.5\right)}\end{array}$                                  (VI)

Conclusion:

Substitute $1.40\text{MW}$ for $W$, $0.933\text{MW}$ for ${\left(\frac{Q_e}{\Delta t}\right)}^{\prime }$, $110°\text{C}$ for $T_C$ in equation (6) to find $T_H$,

    $\begin{array}{c}{T}_H=\frac{\left(110°\text{C}+273\right)\text{K}\times \left(1+\left(0.933\text{MW}\times \frac{1}{1.40\text{MW}}\right)\right)}{\left(\left(0.933\text{MW}\times \frac{1}{1.40\text{MW}}\right)-0.5\right)}\\ =\frac{\left(383\right)\text{K}\times \left(1+\left(0.666\text{MW}\right)\right)}{\left(\left(0.666\text{MW}\right)-0.5\right)}\\ =3844.832\text{K}\\ \approx 3.84\times {10}^3\text{K}\end{array}$

Thus, the value of $T_H$ for which the exhaust power would be only half as large as in part (c) is $3.84\times {10}^3\text{K}$.

To determine

The value of $T_H$ for which the exhaust power would be one-fourth as large as in part (c).

Answer to Problem 15P

No temperature value will provide an exhaust power of one-fourth of the value in part (c).

Explanation of Solution

The rate of work output of the engine is $1.40\text{MW}$, the temperature into the cooling tower is $110°\text{C}$.

Write the expression for the exhaust power whuch would be one-fourth as large as in part (c).

    ${\left(\frac{Q_e}{\Delta t}\right)}^{\prime \text{}\prime }=\frac{1}{4}\left(\frac{Q_e}{\Delta t}\right)$                                                                  (VII)

Here, ${\left(\frac{Q_e}{\Delta t}\right)}^{\prime \text{}\prime }$ is the exhaust power whuch would be one-fourth as large as in part (c).

Substitute $1.86\text{MW}$ for $\left(\frac{Q_e}{\Delta t}\right)$ in equation (7) to find ${\left(\frac{Q_e}{\Delta t}\right)}^{\prime \text{}\prime }$,

    $\begin{array}{c}{\left(\frac{Q_e}{\Delta t}\right)}^{\prime \text{}\prime }=\frac{1}{4}\left(1.86\text{MW}\right)\\ =0.466\text{MW}\end{array}$

Thus, the exhaust power whuch would be one-fourth as large as in part (c) is $0.466\text{MW}$ which is too small.

Conclusion:

From equation (IV), Write the formula to calculate the value of $T_H$ for which the exhaust power would be one-fourth as large as in part (c).

    $\begin{array}{c}\underset{T_H\to \infty }{\lim }\left(\frac{Q_e}{\Delta t}\right)=\underset{T_H\to \infty }{\lim }1.40\text{MW}\left(\frac{0.5+\frac{383}{T_H}}{\frac{T_H}{T_H}-\frac{383}{T_H}}\right)\\ =1.40\text{MW}\left(\frac{0.5}{1}\right)\\ =0.700\text{MW}\end{array}$          (VIII)

Thus, the given exhaust power is lesser than the minimum possible exhaust power $0.700\text{MW}$. Since it is impossible to obtain a lesser exhaust power, no anwer exists.