Chapter 13, Problem 17P

To determine

Find the amount of coal (in kg) required for generating for each year.

Answer to Problem 17P

The amount of coal (in kg) required for generating electricity for each year are calculated and tabulated in Table 1.

Explanation of Solution

Given data:

Refer to problem 13-17 accompanying table in the textbook, the average efficiency of the power plants is 35%.

The heating value of the coal is $7.5\frac{\text{MJ}}{\text{kg}}$.

Formula used:

Formula to calculate the power plant efficiency is,

$\text{Power}\text{plant}\text{effeicency}=\frac{\text{Energy}\text{generated}}{\text{Energy}\text{input}\text{from}\text{fuel}}$

Rearrange the equation,

$\text{Energy}\text{input}\text{from}\text{fuel}=\frac{\text{Energy}\text{generated}}{\text{power}\text{plant}\text{effeicency}}$ (1)

Convert kWh to MJ,

$1\text{kWh}=3.6\text{MJ}$ (2)

Formula to calculate the amount of coal required for generating one year is,

$\text{amount}\text{of}\text{coal}\text{required=}\frac{\text{Energy}\text{input}\text{from}\text{fuel}}{\text{Heating}\text{value}\text{of}\text{the}\text{coal}}$ (3)

Calculation:

Find the energy input from the fuel in each year:

Substitute $1161.562\times {10}^9\text{kWh}$ for energy generated and $0.35$ (35%) for power plant efficiency in equation (1) to find energy input from the fuel in 1980.

$\begin{array}{c}\text{Energy}\text{input}\text{from}\text{fuel}=\frac{1161.562\times {10}^9\text{kWh}}{0.35}\\ =3.318748571\times {10}^{12}\text{kWh}\\ \text{Energy}\text{input}\text{from}\text{fuel}\cong 3318.748571\times {10}^9\text{kWh}\end{array}$

Substitute $1594.011\times {10}^9\text{kWh}$ for energy generated and $0.35$ (35%) for power plant efficiency in equation (1) to find energy input from the fuel in 1990.

$\begin{array}{c}\text{Energy}\text{input}\text{from}\text{fuel}=\frac{1594.011\times {10}^9\text{kWh}}{0.35}\\ =4.554317143\times {10}^{12}\text{kWh}\\ \text{Energy}\text{input}\text{from}\text{fuel}\cong \text{4554}\text{.317143}\times {10}^9\text{kWh}\end{array}$

Substitute $1966.265\times {10}^9\text{kWh}$ for energy generated and $0.35$ for power plant efficiency in equation (1) to find energy input from the fuel in 2000.

$\begin{array}{c}\text{Energy}\text{input}\text{from}\text{fuel}=\frac{1966.265\times {10}^9\text{kWh}}{0.35}\\ =5.6179\times {10}^{12}\text{kWh}\\ \text{Energy}\text{input}\text{from}\text{fuel}\cong 5617.9\times {10}^9\text{kWh}\end{array}$

Substitute $2040.913\times {10}^9\text{kWh}$ for energy generated and $0.35$ for power plant efficiency in equation (1) to find energy input from the fuel in 2005.

$\begin{array}{c}\text{Energy}\text{input}\text{from}\text{fuel}=\frac{2040.913\times {10}^9\text{kWh}}{0.35}\\ =5.8318\times {10}^{12}\text{kWh}\\ \text{Energy}\text{input}\text{from}\text{fuel}\cong 5831.18\times {10}^9\text{kWh}\end{array}$

Substitute $2217.555\times {10}^9\text{kWh}$ for energy generated and $0.35$ for power plant efficiency in equation (1) to find energy input from the fuel in 2010.

$\begin{array}{c}\text{Energy}\text{input}\text{from}\text{fuel}=\frac{2217.555\times {10}^9\text{kWh}}{0.35}\\ =6.335871429\times {10}^{12}\text{kWh}\\ \text{Energy}\text{input}\text{from}\text{fuel}\cong 6335.871429\times {10}^9\text{kWh}\end{array}$

Substitute $2504.786\times {10}^9\text{kWh}$ for energy generated and $0.35$ for power plant efficiency in equation (1) to find energy input from the fuel in 2020.

$\begin{array}{c}\text{Energy}\text{input}\text{from}\text{fuel}=\frac{2504.786\times {10}^9\text{kWh}}{0.35}\\ =7.156531429\times {10}^{12}\text{kWh}\\ \text{Energy}\text{input}\text{from}\text{fuel}\cong 7156.531429\times {10}^9\text{kWh}\end{array}$

Substitute $3380.674\times {10}^9\text{kWh}$ for energy generated and $0.35$ for power plant efficiency in equation (1) to find energy input from the fuel in 2030.

$\begin{array}{c}\text{Energy}\text{input}\text{from}\text{fuel}=\frac{3380.674\times {10}^9\text{kWh}}{0.35}\\ =9.65906857\times {10}^{12}\text{kWh}\\ \text{Energy}\text{input}\text{from}\text{fuel}\cong 9659.06857\times {10}^9\text{kWh}\end{array}$

Now convert all the values from kWh to MJ:

Substitute $3318.74857\times {10}^9\text{kWh}$ for $1\text{kWh}$ in equation (2) for the year 1980,

$\begin{array}{c}3318.748571\times {10}^9\text{kWh}=\left(3.6\text{MJ}\right)\times \left(3318.748571\times {10}^9\right)\left[\therefore \text{1}\text{kWh=3}\text{.6}\text{MJ}\right]\\ \text{=1}\text{.194749486}\times {10}^{13}\text{MJ}\\ =11947.49486\times {10}^9\text{MJ}\end{array}$

Substitute $4554.317143\times {10}^9\text{kWh}$ for $1\text{kWh}$ in equation (2) for the year 1990,

$\begin{array}{c}4554.317143\times {10}^9\text{kWh}=3.6\text{MJ}\times 4554.317143\times {10}^9\left[\therefore \text{1}\text{kWh=3}\text{.6}\text{MJ}\right]\\ \text{=1}\text{.639554171}\times {10}^{13}\text{MJ}\\ =16395.54171\times {10}^9\text{MJ}\end{array}$

Substitute $5617.9\times {10}^9\text{kWh}$ for $1\text{kWh}$ in equation (2) for the year 2000,

$\begin{array}{c}5617.9\times {10}^9\text{kWh}=\left(3.6\text{MJ}\right)\left(5617.9\times {10}^9\right)\left[\therefore \text{1}\text{kWh=3}\text{.6}\text{MJ}\right]\\ \text{=2}\text{.022444}\times {10}^{13}\text{MJ}\\ =20224.44\times {10}^9\text{MJ}\end{array}$

Substitute $5831.18\times {10}^9\text{kWh}$ for $1\text{kWh}$ in equation (2) for the year 2005,

$\begin{array}{c}5831.18\times {10}^9\text{kWh}=\left(3.6\text{MJ}\right)\left(5831.18\times {10}^9\right)\left[\therefore \text{1}\text{kWh=3}\text{.6}\text{MJ}\right]\\ \text{=2}\text{.0992248}\times {10}^{13}\text{MJ}\\ =20992.248\times {10}^9\text{MJ}\end{array}$

Substitute $6335.871429\times {10}^9\text{kWh}$ for $1\text{kWh}$ in equation (2) for the year 2010,

$\begin{array}{c}6335.871429\times {10}^9\text{kWh}=\left(3.6\text{MJ}\right)\left(6335.871429\times {10}^9\right)\left[\therefore \text{1}\text{kWh=3}\text{.6}\text{MJ}\right]\\ =2.280913714\times {10}^{13}\text{MJ}\\ =22809.13714\times {10}^9\text{MJ}\end{array}$

Substitute $7156.531429\times {10}^9\text{kWh}$ for $1\text{kWh}$ in equation (2) for the year 2020,

$\begin{array}{c}7156.531429\times {10}^9\text{kWh}=\left(3.6\text{MJ}\right)\left(7156.531429\times {10}^9\right)\left[\therefore \text{1}\text{kWh=3}\text{.6}\text{MJ}\right]\\ =2.576351314\times {10}^{13}\text{MJ}\\ =25763.51314\times {10}^9\text{MJ}\end{array}$

Substitute $9659.068571\times {10}^9\text{kWh}$ for $1\text{kWh}$ in equation (2) for the year 2030,

Now find the amount of coal required in every year:

Substitute $11947.49486\times {10}^9\text{MJ}$ for energy input from fuel and $7.5\frac{\text{MJ}}{\text{kg}}$ for heating value of the coal in equation (3) to find amount of coal is required in 1980,

$\begin{array}{c}\text{amount}\text{of}\text{coal}\text{required=}\frac{11947.49486\times {10}^9\text{MJ}}{7.5\frac{\text{MJ}}{\text{kg}}}\\ =1.593\times {10}^{12}\text{kg}\\ \cong 1.59\times {10}^{12}\text{kg}\end{array}$

Substitute $16395.54171\times {10}^9\text{MJ}$ for energy input from fuel and $7.5\frac{\text{MJ}}{\text{kg}}$ for heating value of the coal in equation (3) to find amount of coal is required in 1990,

$\begin{array}{c}\text{amount}\text{of}\text{coal}\text{required=}\frac{16395.54171\times {10}^9\text{MJ}}{7.5\frac{\text{MJ}}{\text{kg}}}\\ =2.18607\times {10}^{12}\text{kg}\\ \cong 2.19\times {10}^{12}\text{kg}\end{array}$

Substitute $20224.44\times {10}^9\text{MJ}$ for energy input from fuel and $7.5\frac{\text{MJ}}{\text{kg}}$ for heating value of the coal in equation (3) to find amount of coal is required in 2000,

$\begin{array}{c}\text{amount}\text{of}\text{coal}\text{required=}\frac{20224.44\times {10}^9\text{MJ}}{7.5\frac{\text{MJ}}{\text{kg}}}\\ =2.696592\times {10}^{12}\text{kg}\\ \cong 2.7\times {10}^{12}\text{kg}\end{array}$

Substitute $20992.248\times {10}^9\text{MJ}$ for energy input from fuel and $7.5\frac{\text{MJ}}{\text{kg}}$ for heating value of the coal in equation (3) to find amount of coal is required in 2005,

$\begin{array}{c}\text{amount}\text{of}\text{coal}\text{required=}\frac{20992.248\times {10}^9\text{MJ}}{7.5\frac{\text{MJ}}{\text{kg}}}\\ =2.79896\times {10}^{12}\text{kg}\\ \cong 2.8\times {10}^{12}\text{kg}\end{array}$

Substitute $22809.13714\times {10}^9\text{MJ}$ for energy input from fuel and $7.5\frac{\text{MJ}}{\text{kg}}$ for heating value of the coal in equation (3) to find amount of coal is required in 2010,

$\begin{array}{c}\text{amount}\text{of}\text{coal}\text{required=}\frac{22809.13714\times {10}^9\text{MJ}}{7.5\frac{\text{MJ}}{\text{kg}}}\\ =3.04122\times {10}^{12}\text{kg}\\ \cong 3.04\times {10}^{12}\text{kg}\end{array}$

Substitute $25763.51314\times {10}^9\text{MJ}$ for energy input from fuel and $7.5\frac{\text{MJ}}{\text{kg}}$ for heating value of the coal in equation (3) to find amount of coal is required in 2020,

$\begin{array}{c}\text{amount}\text{of}\text{coal}\text{required=}\frac{25763.51314\times {10}^9\text{MJ}}{7.5\frac{\text{MJ}}{\text{kg}}}\\ =3.43514\times {10}^{12}\text{kg}\\ \cong 3.44\times {10}^{12}\text{kg}\end{array}$

Substitute $34772.64686\times {10}^9\text{MJ}$ for energy input from fuel and $7.5\frac{\text{MJ}}{\text{kg}}$ for heating value of the coal in equation (3) to find amount of coal is required in 2030,

$\begin{array}{c}\text{amount}\text{of}\text{coal}\text{required=}\frac{34772.64686\times {10}^9\text{MJ}}{7.5\frac{\text{MJ}}{\text{kg}}}\\ =4.63635\times {10}^{12}\text{kg}\\ \cong 4.64\times {10}^{12}\text{kg}\end{array}$

Therefore, the energy input for all years for Coal, in kWh and MJ along with coal required to produce 35% of efficiency is shown in Table 1 (with approximately rounded values).

Table 1

Energy Produced (10^9 kWh)	Energy(considering efficiency) 109 kWh	Energy (considering efficiency) 109 MJ	Amount of coal needed in kg
1161.562	3318.748571	11947.49486	1.59E+12
1594.011	4554.317143	16395.54171	2.19E+12
1966.265	5617.9	20224.44	2.70E+12
2040.913	5831.18	20992.248	2.80E+12
2217.555	6335.871429	22809.13714	3.04E+12
2504.786	7156.531429	25763.51314	3.44E+12
3380.674	9569.068571	34772.64686	4.64E+12

Conclusion:

Hence, the amount of coal (in kg) required for generating electricity for each year has been calculated.