Chapter 11, Problem 5SP

Suppose that an oil-fired power plant is designed to produce 125 MW (megawatts) of electrical power. The turbine operates between temperatures of 740°C and 380°C and has an efficiency that is 80% of the ideal Carnot efficiency for these temperatures.

a.    What is the Carnot efficiency for these temperatures?

b.    What is the efficiency of the actual oil-fired turbines?

c.    How many kilowatt-hours (kW·h) of electrical energy does the plant generate in 1 h? (The kilowatt-hour is an energy unit equal to 1 kW of power multiplied by 1 h.)

d.    How many kilowatt-hours of heat must be obtained from the oil in each hour?

e.    If one barrel of oil yields 1700 kW·h of heat, how much oil is used by the plant each hour?

To determine

The Carnot efficiency.

Answer to Problem 5SP

The Carnot efficiency is $35.5\%$.

Explanation of Solution

Given info:

Temperature of hot reservoir is $740°\text{C}$ and temperature of cold reservoir is $380°\text{C}$.

Write an expression to calculate the efficiency.

$\eta =1-\frac{T_C}{T_H}$

Here,

$\eta$ is the efficiency of the Carnot engine

$T_C$ is the temperature of cold reservoir

$T_H$ is the temperature of hot reservoir

Substitute $380°\text{C}$ for $T_C$ and $740°\text{C}$ for $T_H$ to find $\eta$.

$\begin{array}{c}\eta =1-\frac{\left(380+273\right)\text{K}}{\left(740+273\right)\text{K}}\\ =0.355\\ =\left(0.355\right)\left(100\%\right)\\ =35.5\%\end{array}$

Thus, the Carnot efficiency is $35.5\%$.

Conclusion:

The Carnot efficiency is $35.5\%$.

To determine

The efficiency of the actual oil-fired turbines.

Answer to Problem 5SP

The efficiency of the actual oil-fired turbines is $28.4\%$.

Explanation of Solution

Given info:

The efficiency of the oil-fired power plant is 80% of the efficiency of the Carnot efficiency.

Write an expression for efficiency of the actual oil-fired turbines.

$\eta \text{'}=80\%\eta$

Here,

$\eta \text{'}$ is the efficiency of the oil-fired power plant.

Substitute $0.355$ for $\eta$ to find $\eta \text{'}$.

$\begin{array}{c}W=\left(80\%\right)\left(0.355\right)\\ =28.4\%\end{array}$

Thus, the efficiency of the actual oil-fired turbines is $28.4\%$.

Conclusion:

The efficiency of the actual oil-fired turbines is $28.4\%$.

To determine

The electrical energy generated by the plant in one hour.

Answer to Problem 5SP

The electrical energy generated by the plant in one hour is $125000\text{kW}\cdot \text{h}$.

Explanation of Solution

Given info:

The power of the oil-fired power plant is $125\text{MW}$ and the time is $1\text{h}$.

Write an expression for electrical energy generated by the plant in one hour.

$E=Pt$

Here,

$E$ is the energy

$P$ is the energy

$t$ is the energy

Substitute $125\text{MW}$ for $P$ and $1\text{h}$ for $t$ to find $E$.

$\begin{array}{c}E=\left(125\text{MW}\right)\left(\frac{{10}^3\text{kW}}{1\text{MW}}\right)\left(1\text{h}\right)\\ =125000\text{kW}\cdot \text{h}\end{array}$

Thus, the electrical energy generated by the plant in one hour is $125000\text{kW}\cdot \text{h}$.

Conclusion:

The electrical energy generated by the plant in one hour is $125000\text{kW}\cdot \text{h}$.

To determine

The heat obtained from oil in each hour.

Answer to Problem 5SP

The heat obtained from oil in each hour is $440000\text{kW}\cdot \text{h}$.

Explanation of Solution

Write an expression for efficiency of the actual oil-fired turbines.

$E\text{'}=\frac{E}{\eta \text{'}}$

Here,

$E\text{'}$ is the heat obtained from oil in each hour.

Substitute $0.284$ for $\eta \text{'}$ and $1.25\times {10}^5\text{kW}\cdot \text{h}$ for $E$ to find $\eta \text{'}$.

$\begin{array}{c}E\text{'}=\frac{1.25\times {10}^5\text{kW}\cdot \text{h}}{0.284}\\ =440000\text{kW}\cdot \text{h}\end{array}$

Thus, the heat obtained from oil in each hour is $440000\text{kW}\cdot \text{h}$.

Conclusion:

The heat obtained from oil in each hour is $440000\text{kW}\cdot \text{h}$.

To determine

The amount of oil required to produce $1700\text{kW}\cdot \text{h}$ in each hour.

Answer to Problem 5SP

The amount of oil required to produce $1700\text{kW}\cdot \text{h}$ in each hour is $259$ barrels of oil.

Explanation of Solution

Given info:

The energy released is $1700\text{kW}\cdot \text{h}$.

Write an expression for the amount of oil required to produce $1700\text{kW}\cdot \text{h}$ in an hour.

$n=\frac{E\text{'}}{E_h}$

Here,

$E\text{'}$ is the energy released per hour

$E_h$ is the energy released from one barrel oil per hour

$n$ is the  amount of oil required to produce $1700\text{kW}\cdot \text{h}$ in an hour

Substitute $4.40\times {10}^5\text{kW}\cdot \text{h}$ for $E\text{'}$ and $1700\text{kW}\cdot \text{h}$ for $E_h$ to find $n$.

$\begin{array}{c}n=\frac{4.40\times {10}^5\text{kW}\cdot \text{h}}{1700\text{kW}\cdot \text{h/barrel}}\\ =259\text{barrels}\end{array}$

Thus, the amount of oil required to produce $1700\text{kW}\cdot \text{h}$ in each hour is $259$ barrels of oil

Conclusion:

The amount of oil required to produce $1700\text{kW}\cdot \text{h}$ in each hour is $259$ barrels of oil.