Chapter 13, Problem 19P

To determine

Find the amount of natural gas required in ${\text{ft}}^{\text{3}}$ and $\text{lbm}$ for generating for electricity for each year.

Answer to Problem 19P

The amount of natural gas required in ${\text{ft}}^{\text{3}}$ and $\text{lbm}$ for generating for electricity for each year is calculated and tabulated in Table 1.

Explanation of Solution

Given data:

Refer to problem 13-19 accompanying table in the textbook.

The average efficiency of the power plant is 35%.

The heating value of the natural gas is $1000\frac{\text{Btu}}{{\text{ft}}^{\text{3}}}\left(22000\frac{\text{Btu}}{\text{lbm}}\right)$.

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 Btu to $\text{lbm}$,

$\begin{array}{c}1\text{Btu=}\frac{1}{22000}\text{lbm}\left[\therefore 22000\frac{\text{Btu}}{\text{lbm}}=1\right]\\ =4.5454\times {10}^{-5}\text{lbm}\end{array}$ (2)

Convert $\text{lbm}$ to ${\text{ft}}^{\text{3}}$,

$\begin{array}{l}1000\frac{\text{Btu}}{{\text{ft}}^{\text{3}}}=22000\frac{\text{Btu}}{\text{lbm}}\\ 1000\frac{1}{{\text{ft}}^{\text{3}}}=22000\frac{1}{\text{lbm}}\\ \frac{1}{{\text{ft}}^{\text{3}}}=22\frac{1}{\text{lbm}}\\ 1\text{lbm}=22{\text{ft}}^{\text{3}}\end{array}$ (3)

The value of the $1\frac{\text{Btu}}{\text{s}}$ is,

$1\frac{\text{Btu}}{\text{s}}=1.055\text{kW}$ (4)

Convert 1 hr into seconds,

$1\text{hr=3600}\text{s}$

Rearrange the equation,

$\text{1}\text{s=}\frac{1}{3600}\text{hr}$ (5)

Substitute equation (5) in equation (4),

$\begin{array}{l}1\frac{\text{Btu}}{\frac{1}{3600}\text{hr}}=1.055\text{kW}\\ \text{1}\text{Btu}=\frac{1.055}{3600}\text{kWhr}\\ \text{1}\text{Btu=2}\text{.93}\times {\text{10}}^{-4}\text{kWhr}\end{array}$

Rearrange the equation,

$\begin{array}{c}\\ \text{1}\text{kWhr}=\frac{1}{\text{2}\text{.93}\times {\text{10}}^{-4}}\text{Btu}\\ \text{=3412}\text{.96}\text{Btu}\end{array}$

Calculation:

Converting the given data from kWh to Btu:

Substitute $\text{3412}\text{.96}\text{Btu}$ for $\text{1}\text{kWhr}$ in $346.2399\times {10}^9\text{kWhr}$ for the year 1980,

$\begin{array}{c}346.2399\times {10}^9\text{kWhr}=\left(346.2399\times {10}^9\right)\left(\text{3412}\text{.96}\text{Btu}\right)\\ \text{=1}\text{.181702929}\times {10}^{15}\text{Btu}\\ =1181702.929\times {10}^9\text{Btu}\end{array}$

Substitute $\text{3412}\text{.96}\text{Btu}$ for $\text{1}\text{kWhr}$ in $372.7652\times {10}^9\text{kWhr}$ for the year 1990,

$\begin{array}{c}372.7652\times {10}^9\text{kWhr}=\left(372.7652\times {10}^9\right)\left(\text{3412}\text{.96}\text{Btu}\right)\\ \text{=1}\text{.272232717}\times {10}^{15}\text{Btu}\\ =\text{1272232}\text{.717}\times {10}^9\text{Btu}\end{array}$

Substitute $\text{3412}\text{.96}\text{Btu}$ for $\text{1}\text{kWhr}$ in $601.0382\times {10}^9\text{kWhr}$ for the year 2000,

$\begin{array}{c}601.0382\times {10}^9\text{kWhr}=\left(601.0382\times {10}^9\right)\left(\text{3412}\text{.96}\text{Btu}\right)\\ \text{=2}\text{.051319335}\times {10}^{15}\text{Btu}\\ =\text{2051319}\text{.335}\times {10}^9\text{Btu}\end{array}$

Substitute $\text{3412}\text{.96}\text{Btu}$ for $\text{1}\text{kWhr}$ in $751.8189\times {10}^9\text{kWhr}$ for the year 2005,

$\begin{array}{c}751.8189\times {10}^9\text{kWhr}=\left(751.8189\times {10}^9\text{kWhr}\right)\left(\text{3412}\text{.96}\text{Btu}\right)\\ \text{=2}\text{.565927833}\times {10}^{15}\text{Btu}\\ =\text{2565927}\text{.833}\times {10}^9\text{Btu}\end{array}$

Substitute $\text{3412}\text{.96}\text{Btu}$ for $\text{1}\text{kWhr}$ in $773.8234\times {10}^9\text{kWhr}$ for the year 2010,

$\begin{array}{c}773.8234\times {10}^9\text{kWhr}=\left(773.8234\times {10}^9\right)\left(\text{3412}\text{.96}\text{Btu}\right)\\ \text{=2}\text{.641028311}\times {10}^{15}\text{Btu}\\ =\text{2641028}\text{.311}\times {10}^9\text{Btu}\end{array}$

Substitute $\text{3412}\text{.96}\text{Btu}$ for $\text{1}\text{kWhr}$ in $1102.762\times {10}^9\text{kWhr}$ for the year 2020,

$\begin{array}{c}1102.762\times {10}^9\text{kWhr}=\left(1102.762\times {10}^9\right)\left(\text{3412}\text{.96}\text{Btu}\right)\\ \text{=3}\text{.763682596}\times {10}^{15}\text{Btu}\\ =\text{3763682}\text{.596}\times {10}^9\text{Btu}\end{array}$

Substitute $\text{3412}\text{.96}\text{Btu}$ for $\text{1}\text{kWhr}$ in $992.7706\times {10}^9\text{kWhr}$ for the year 2030,

$\begin{array}{c}992.7706\times {10}^9\text{kWhr}=\left(992.7706\times {10}^9\right)\left(\text{3412}\text{.96}\text{Btu}\right)\\ \text{=}3.388286347\times {10}^{15}\text{Btu}\\ \text{=3388286}\text{.347}\times {10}^9\text{Btu}\end{array}$

Now find the energy input fuel for each year in Btu for natural gas:

Substitute $1181702.929\times {10}^9\text{Btu}$ for energy generated and 0.35 for power plant efficiency in equation (1) to find energy input from fuel in 1980,

$\begin{array}{c}\text{Energy}\text{input}\text{from}\text{fuel}=\frac{1181702.929\times {10}^9\text{Btu}}{\text{0}\text{.35}}\\ =3.376294083\times {10}^{15}\text{Btu}\\ \text{Energy}\text{input}\text{from}\text{fuel=3376294}\text{.083}\times {10}^9\text{Btu}\end{array}$

Substitute $\text{1272232}\text{.717}\times {10}^9\text{Btu}$ for energy generated and 0.35 for power plant efficiency in equation (1) to find energy input from fuel 1990,

$\begin{array}{c}\text{Energy}\text{input}\text{from}\text{fuel}=\frac{1272232.717\times {10}^9\text{Btu}}{\text{0}\text{.35}}\\ =3.63495062\times {10}^{15}\text{Btu}\\ \text{Energy}\text{input}\text{from}\text{fuel=}3634950.62\times {10}^9\text{Btu}\end{array}$

Substitute $\text{2051319}\text{.335}\times {10}^9\text{Btu}$ for energy generated and 0.35 for power plant efficiency in equation (1) to find energy input from fuel 2000,

$\begin{array}{c}\text{Energy}\text{input}\text{from}\text{fuel}=\frac{\text{2051319}\text{.335}\times {10}^9\text{Btu}}{\text{0}\text{.35}}\\ =5.860912386\times {10}^{15}\text{Btu}\\ \text{Energy}\text{input}\text{from}\text{fuel=}5860912.386\times {10}^9\text{Btu}\end{array}$

Substitute $\text{2565927}\text{.833}\times {10}^9\text{Btu}$ for energy generated and 0.35 for power plant efficiency in equation (1) to find energy input from fuel 2005,

$\begin{array}{c}\text{Energy}\text{input}\text{from}\text{fuel}=\frac{\text{2565927}\text{.833}\times {10}^9\text{Btu}}{\text{0}\text{.35}}\\ =7.33122238\times {10}^{15}\text{Btu}\\ \text{Energy}\text{input}\text{from}\text{fuel=}7331222.38\times {10}^9\text{Btu}\end{array}$

Substitute $\text{2641028}\text{.311}\times {10}^9\text{Btu}$ for energy generated and 0.35 for power plant efficiency in equation (1) to find energy input from fuel 2010,

$\begin{array}{c}\text{Energy}\text{input}\text{from}\text{fuel}=\frac{\text{2641028}\text{.311}\times {10}^9\text{Btu}}{\text{0}\text{.35}}\\ =7.545795174\times {10}^{15}\text{Btu}\\ \text{Energy}\text{input}\text{from}\text{fuel=}7545795.174\times {10}^9\text{Btu}\end{array}$

Substitute $\text{3763682}\text{.596}\times {10}^9\text{Btu}$ for energy generated and 0.35 for power plant efficiency in equation (1) to find energy input from fuel 2020,

$\begin{array}{c}\text{Energy}\text{input}\text{from}\text{fuel}=\frac{\text{3763682}\text{.596}\times {10}^9\text{Btu}}{\text{0}\text{.35}}\\ =1.075337885\times {10}^{16}\text{Btu}\\ \text{Energy}\text{input}\text{from}\text{fuel=}10753378.85\times {10}^9\text{Btu}\end{array}$

Substitute $3388326.347\times {10}^9\text{Btu}$ for energy generated and 0.35 for power plant efficiency in equation (1) to find energy input from fuel 2030,

$\begin{array}{c}\text{Energy}\text{input}\text{from}\text{fuel}=\frac{3388286.347\times {10}^9\text{Btu}}{\text{0}\text{.35}}\\ =9680818.134\times {10}^9\text{Btu}\end{array}$

Now convert Btu to lbm as follows by using the relation $1\text{Btu=4}\text{.5454}\times {\text{10}}^{-5}\text{lbm}$,

Substitute $\text{4}\text{.5454}\times {\text{10}}^{-5}\text{lbm}$ for 1 Btu in $\text{3376294}\text{.083}\times {10}^9\text{Btu}$ for the year 1980,

$\begin{array}{c}\text{3376294}\text{.083}\times {10}^9\text{Btu}=\left(\text{3376294}\text{.083}\times {10}^9\right)\left(\text{4}\text{.5454}\times {\text{10}}^{-5}\text{lbm}\right)\\ =1.5346607\times {10}^{11}\text{lbm}\\ \cong 1.53\times {10}^{11}\text{lbm}\end{array}$

Substitute $\text{4}\text{.5454}\times {\text{10}}^{-5}\text{lbm}$ for 1 Btu in $3634950.62\times {10}^9\text{Btu}$ for the year 1990,

$\begin{array}{c}3634950.62\times {10}^9\text{Btu}=\left(3634950.62\times {10}^9\right)\left(\text{4}\text{.5454}\times {\text{10}}^{-5}\text{lbm}\right)\\ =1.652230\times {10}^{11}\text{lbm}\\ \cong 1.65\times {10}^{11}\text{lbm}\end{array}$

Substitute $\text{4}\text{.5454}\times {\text{10}}^{-5}\text{lbm}$ for 1 Btu in $5860912.386\times {10}^9\text{Btu}$ for the year 2000,

$\begin{array}{c}5860912.386\times {10}^9\text{Btu}=\left(5860912.386\times {10}^9\right)\left(\text{4}\text{.5454}\times {\text{10}}^{-5}\text{lbm}\right)\\ =2.664019\times {10}^{11}\text{lbm}\\ \cong 2.66\times {10}^{11}\text{lbm}\end{array}$

Substitute $\text{4}\text{.5454}\times {\text{10}}^{-5}\text{lbm}$ for 1 Btu in $7331222.38\times {10}^9\text{Btu}$ for the year 2005,

$\begin{array}{c}7331222.38\times {10}^9\text{Btu}=\left(7331222.38\times {10}^9\right)\left(\text{4}\text{.5454}\times {\text{10}}^{-5}\text{lbm}\right)\\ =3.3323\times {\text{10}}^{11}\text{lbm}\\ \cong 3.33\times {\text{10}}^{11}\text{lbm}\end{array}$

Substitute $\text{4}\text{.5454}\times {\text{10}}^{-5}\text{lbm}$ for 1 Btu in $7545795.174\times {10}^9\text{Btu}$ for the year 2010,

$\begin{array}{c}7545795.174\times {10}^9\text{Btu}=\left(7545795.174\times {10}^9\right)\left(\text{4}\text{.5454}\times {\text{10}}^{-5}\text{lbm}\right)\\ =3.429865\times {10}^{11}\text{lbm}\\ \cong 3.43\times {10}^{11}\text{lbm}\end{array}$

Substitute $\text{4}\text{.5454}\times {\text{10}}^{-5}\text{lbm}$ for 1 Btu in $10753378.85\times {10}^9\text{Btu}$ for the year 2020,

$\begin{array}{c}10753378.85\times {10}^9\text{Btu}=\left(10753378.85\times {10}^9\right)\left(\text{4}\text{.5454}\times {\text{10}}^{-5}\text{lbm}\right)\\ =4.88784\times {10}^{11}\text{lbm}\\ \cong 4.89\times {10}^{11}\text{lbm}\end{array}$

Substitute $\text{4}\text{.5454}\times {\text{10}}^{-5}\text{lbm}$ for 1 Btu in $9680818.134\times {10}^9\text{Btu}$ for the year 2030,

$\begin{array}{c}9680818.134\times {10}^9\text{Btu}=\left(9680818.134\times {10}^9\right)\left(\text{4}\text{.5454}\times {\text{10}}^{-5}\text{lbm}\right)\\ =4.4003190\times {\text{10}}^{11}\text{lbm}\\ \cong 4.40\times {\text{10}}^{11}\text{lbm}\end{array}$

Now convert Form lbm to ${\text{ft}}^{\text{3}}$ by using the relation $\text{1lbm}=22{\text{ft}}^{\text{3}}$ as follows,

Substitute $22{\text{ft}}^{\text{3}}$ for 1 lbm in $1.5346607\times {10}^{11}\text{lbm}$ for the year 1980,

$\begin{array}{c}1.5346607\times {10}^{11}\text{lbm}=\left(1.5346607\times {10}^{11}\right)\left(22{\text{ft}}^{\text{3}}\right)\\ =3.37625\times {10}^{12}{\text{ft}}^{\text{3}}\\ \cong 3.38\times {10}^{12}{\text{ft}}^{\text{3}}\end{array}$

Substitute $22{\text{ft}}^{\text{3}}$ for 1 lbm in $1.652230\times {10}^{12}\text{lbm}$ for the year 1990,

$\begin{array}{c}1.652230\times {10}^{11}\text{lbm}=\left(1.652230\times {10}^{11}\right)\left(22{\text{ft}}^{\text{3}}\right)\\ =3.6354906\times {10}^{12}{\text{ft}}^{\text{3}}\\ \cong 3.63\times {10}^{12}{\text{ft}}^{\text{3}}\end{array}$

Substitute $22{\text{ft}}^{\text{3}}$ for 1 lbm in $2.664019\times {10}^{12}{\text{lb}}_{\text{m}}$ for the year 2000,

$\begin{array}{c}2.664019\times {10}^{11}\text{lbm}=\left(2.664019\times {10}^{11}\right)\left(22{\text{ft}}^{\text{3}}\right)\\ =5.8608418\times {10}^{12}{\text{ft}}^{\text{3}}\\ \cong 5.86\times {10}^{12}{\text{ft}}^{\text{3}}\end{array}$

Substitute $22{\text{ft}}^{\text{3}}$ for 1 lbm in $3.33233\times {10}^{10}\text{lbm}$ for the year 2005,

$\begin{array}{c}3.33233\times {10}^{11}{\text{lb}}_{\text{m}}=\left(3.33233\times {10}^{11}\right)\left(22{\text{ft}}^{\text{3}}\right)\\ =7.331126\times {10}^{12}{\text{ft}}^{\text{3}}\\ \cong 7.33\times {10}^{12}{\text{ft}}^{\text{3}}\end{array}$

Substitute $22{\text{ft}}^{\text{3}}$ for 1 lbm in $3.429865\times {10}^{11}\text{lbm}$ for the year 2010,

$\begin{array}{c}3.429865\times {10}^{11}\text{lbm}=\left(3.429865\times {10}^{11}\right)\left(22{\text{ft}}^{\text{3}}\right)\\ =7.545703\times {10}^{12}{\text{ft}}^{\text{3}}\\ \cong 7.55\times {10}^{12}{\text{ft}}^{\text{3}}\end{array}$

Substitute $22{\text{ft}}^{\text{3}}$ for 1 lbm in $4.88784\times {10}^{11}\text{lbm}$ for the year 2020,

$\begin{array}{c}4.88784\times {10}^{11}\text{lbm}=\left(4.88784\times {10}^{11}\right)\left(22{\text{ft}}^{\text{3}}\right)\\ =1.0753358\times {10}^{13}{\text{ft}}^{\text{3}}\\ \cong 1.08\times {10}^{13}{\text{ft}}^{\text{3}}\end{array}$

Substitute $22{\text{ft}}^{\text{3}}$ for 1 lbm in $4.400319\times {10}^{11}\text{lbm}$ for the year 2030,

$\begin{array}{c}4.400319\times {10}^{11}\text{lbm}=\left(4.400319\times {10}^{11}\right)\left(22{\text{ft}}^{\text{3}}\right)\\ =9.6807018\times {10}^{12}{\text{ft}}^{\text{3}}\\ \cong 9.68\times {10}^{12}{\text{ft}}^{\text{3}}\end{array}$

Therefore, the amount of natural gas in ${\text{ft}}^{\text{3}}$ and $\text{lbm}$ for generating electricity for each year required is tabulated as in Table 1 with approximately rounded values for amount of natural gas needed in lbm and amount of natural gas needed in ${\text{ft}}^3$.

Conclusion:

Hence, the amount of natural gas in ${\text{ft}}^{\text{3}}$ and $\text{lbm}$ for generating electricity for each year has been calculated.