Chapter 21, Problem 15P

(a)

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

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

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.

$T_H$ is the fuel combustion temperature.

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

${\eta }_a=\frac{2}{3}\eta$ (1)

Here,

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

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

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

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.

$Q$ is the rate of heat input to the engine.

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

$\frac{Q_e}{\Delta t}=Q-W$ (2)

Here,

$\frac{Q_e}{\Delta t}$ is the rate at which the station exhaust energy by heat as a function of the fuel combustion temperature $T_H$.

Substitute $\frac{W}{{\eta }_a}$ for $Q$ in equation (2) to find $\frac{Q_e}{\Delta t}$,

$\begin{array}{c}\frac{Q_e}{\Delta t}=\frac{W}{{\eta }_a}-W\\ =W\left(\frac{1}{{\eta }_a}-1\right)\end{array}$ (3)

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

$\begin{array}{c}\frac{Q_e}{\Delta t}=W\left(\frac{1}{\frac{2}{3}\left(\frac{T_H-T_C}{T_H}\right)}-1\right)\\ =W\left(\frac{3}{2}\left(\frac{T_H}{T_H-T_C}\right)-1\right)\\ =W\left(\frac{T_H+2T_C}{2\left(T_H-T_C\right)}\right)\\ =W\left(\frac{0.5T_H+T_C}{T_H-T_C}\right)\end{array}$ (4)

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

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

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

Conclusion:

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

(b)

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 change if the firebox is modified to run hotter by using more advanced combustion technology because the exhaust power 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 amount of the energy exhaust change because the exhaust power is inversely proportional to the fire box temperature. So, the exhaust power decreases as the fire box temperature increases.

Conclusion:

The amount of the energy exhaust change if the firebox is modified to run hotter by using more advanced combustion technology because the exhaust power decreases as the fire box temperature increases.

(c)

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

Given information: 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 (4), the formula to calculate the exhaust power for $T_H=800°\text{C}$.

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

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

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

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

Conclusion:

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

(d)

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

Given information: 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).

${\left(\frac{Q_e}{\Delta t}\right)}^{\prime }=\frac{1}{2}\left(\frac{Q_e}{\Delta t}\right)$ (5)

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

$\begin{array}{c}{\left(\frac{Q_e}{\Delta t}\right)}^{\prime }=W\left(\frac{0.5T_H+T_C}{T_H-T_C}\right)\\ \left(T_H-T_C\right){\left(\frac{Q_e}{\Delta t}\right)}^{\prime }\times \frac{1}{W}=0.5T_H+T_C\\ T_H\left(\left({\left(\frac{Q_e}{\Delta t}\right)}^{\prime }\times \frac{1}{W}\right)-0.5\right)=T_C\left(1+{\left(\frac{Q_e}{\Delta t}\right)}^{\prime }\times \frac{1}{W}\right)\\ T_H=\frac{T_C\left(1+{\left(\frac{Q_e}{\Delta t}\right)}^{\prime }\times \frac{1}{W}\right)}{\left(\left({\left(\frac{Q_e}{\Delta t}\right)}^{\prime }\times \frac{1}{W}\right)-0.5\right)}\end{array}$ (6)

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 $3.84\times {10}^3\text{K}$.

Conclusion:

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

(e)

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

The value of $T_H$ not exists because the exhaust energy can not be that small.

Explanation of Solution

Given information: 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).

${\left(\frac{Q_e}{\Delta t}\right)}^{\prime \text{}\prime }=\frac{1}{4}\left(\frac{Q_e}{\Delta t}\right)$ (7)

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.

From equation (4), the formula to calculate the value of $T_H$ for which the exhaust power would be one-fourth as large as in part (c).

$\begin{array}{c}{\left(\frac{Q_e}{\Delta t}\right)}^{\prime \text{}\prime }=W\left(\frac{0.5T_H+T_C}{T_H-T_C}\right)\\ \left(T_H-T_C\right){\left(\frac{Q_e}{\Delta t}\right)}^{\prime \text{}\prime }\times \frac{1}{W}=0.5T_H+T_C\\ T_H\left(\left({\left(\frac{Q_e}{\Delta t}\right)}^{\prime \text{}\prime }\times \frac{1}{W}\right)-0.5\right)=T_C\left(1+{\left(\frac{Q_e}{\Delta t}\right)}^{\prime \text{}\prime }\times \frac{1}{W}\right)\\ T_H=\frac{T_C\left(1+{\left(\frac{Q_e}{\Delta t}\right)}^{\prime \text{}\prime }\times \frac{1}{W}\right)}{\left(\left({\left(\frac{Q_e}{\Delta t}\right)}^{\prime \text{}\prime }\times \frac{1}{W}\right)-0.5\right)}\end{array}$ (8)

Substitute $1.40\text{MW}$ for $W$, $0.466\text{MW}$ for ${\left(\frac{Q_e}{\Delta t}\right)}^{\prime \text{}\prime }$, $110°\text{C}$ for $T_C$ in equation (8) to find $T_H$,

$\begin{array}{c}{T}_H=\frac{\left(110°\text{C}+273\right)\text{K}\times \left(1+\left(0.466\text{MW}\times \frac{1}{1.40\text{MW}}\right)\right)}{\left(\left(0.466\text{MW}\times \frac{1}{1.40\text{MW}}\right)-0.5\right)}\\ =\frac{\left(383\right)\text{K}\times \left(1+\left(0.333\text{MW}\right)\right)}{\left(\left(0.333\text{MW}\right)-0.5\right)}\\ =-3057.119\text{K}\\ \approx -3.05\times {10}^3\text{K}\end{array}$

Thus, the value of $T_H$ for which the exhaust power would be one-fourth as large as in part (c) is $-3.05\times {10}^3\text{K}$. In this the value of $T_H$ not exists because the exhaust energy can not be that small.

Conclusion:

Therefore, the value of $T_H$ not exists because the exhaust energy can not be that small.