Chapter 11, Problem 41P

To determine

Calculate the amount of natural gas required to heat the water in a heater.

Answer to Problem 41P

The amount of natural gas required to heat the water in a heater is $72\frac{{\text{ft}}^{\text{3}}}{\text{day}}$.

Explanation of Solution

Given data:

Volume of water used per day is $80\text{gallons}$,

Initial temperature of water is, $T_{\text{initial}}=55\text{°F}$,

Final temperature of water is, $T_{\text{final}}=140\text{°F}$,

Efficiency of heater is $80\%$,

From Table 11.13 in the textbook, the specific heat of water is, $c=4180\frac{\text{J}}{\text{kg}\cdot \text{K}}$,

Standard density of water is $1000\frac{\text{kg}}{{\text{m}}^{\text{3}}}$.

Formula used:

The relationship between degree Fahrenheit and degree Celsius is,

$T\left(\text{°C}\right)=\frac{5}{9}\left[T\left(°F\right)-32\right]$ (1)

Here,

$T\left(\text{°C}\right)$ is the temperature in degree Celsius,

$T\left(°F\right)$ is the temperature in degree Fahrenheit.

The formula to find thermal energy required to heat the water is,

$E_{\text{thermal}}=mc\left(T_{\text{final}}-T_{\text{initial}}\right)$ (2)

Here,

$m$ is the mass of the water,

$c$ is the specific heat of water,

$T_{\text{initial}}$ is the initial temperature of water,

$T_{\text{final}}$ is the final temperature of water.

Calculation:

Convert the unit of volume of the water,

$\begin{array}{c}\text{Volume}=80\frac{\text{gallons}}{\text{day}}\left[\therefore 1\text{gallon}=3.7854\times {10}^{-3}{\text{m}}^{\text{3}}\right]\\ =\left(80\frac{\text{gallons}}{\text{day}}\right)\left(\frac{3.7854\times {10}^{-3}{\text{m}}^{\text{3}}}{1\text{gallon}}\right)\\ =\left(80\times 3.7854\times {10}^{-3}\right)\frac{{\text{m}}^{\text{3}}}{\text{day}}\\ \text{Volume}=0.3028\frac{{\text{m}}^{\text{3}}}{\text{day}}\end{array}$

Substitute $55\text{°F}$ for $T\left(°F\right)$ in equation (1) to get the initial temperature in degree Celsius,

$\begin{array}{c}{T}_{\text{initial}}\left(\text{°C}\right)=\frac{5}{9}\left[55-32\right]\\ =12.77\text{°C}\end{array}$

Substitute $140\text{°F}$ for $T\left(°F\right)$ in equation (1) to get the final temperature in degree Celsius,

$\begin{array}{c}{T}_{\text{final}}\left(\text{°C}\right)=\frac{5}{9}\left[140-32\right]\\ =60\text{°C}\end{array}$

The expression to find the mass of the water is,

$\begin{array}{c}\text{Mass}=\text{Volume}\times \text{density}\\ \text{=}\left(0.3028\frac{{\text{m}}^{\text{3}}}{\text{day}}\right)\left(1000\frac{\text{kg}}{{\text{m}}^{\text{3}}}\right)\\ \text{Mass}=302.8\frac{\text{kg}}{\text{day}}\end{array}$

Substitute $302.8\frac{\text{kg}}{\text{day}}$ for $m$, $4180\frac{\text{J}}{\text{kg}\cdot \text{K}}$ for $c$, $12.77\text{°C}$ for $T_{\text{initial}}$, and $60\text{°C}$ for $T_{\text{final}}$ in equation (2) to find $E_{\text{thermal}}$,

$\begin{array}{c}{E}_{\text{thermal}}=\left(302.8\frac{\text{kg}}{\text{day}}\right)\left(4180\frac{\text{J}}{\text{kg}\cdot \text{K}}\right)\left(60\text{°C}-12.77\text{°C}\right)\left[\therefore \text{ °C}=\text{K}\right]\\ =\left(302.8\frac{\text{kg}}{\text{day}}\right)\left(4180\frac{\text{J}}{\text{kg}\cdot \text{K}}\right)\left(60-12.77\right)\text{K}\end{array}$

$E_{\text{thermal}}\text{=59}\text{.7792}\times {\text{10}}^6\frac{\text{J}}{\text{day}}$

The unit conversion on above result is,

$\begin{array}{c}{E}_{\text{thermal}}\text{=59}\text{.7792}\times {\text{10}}^6\frac{\text{J}}{\text{day}}\left[\therefore 1\text{day}=24\text{h}\right]\\ =\text{59}\text{.7792}\times {\text{10}}^6\left(\frac{\text{J}}{\text{day}}\times \frac{1\text{day}}{24\text{h}}\right)\left[\therefore 1\text{h}=3600\text{s}\right]\\ =\left(2.491\times {\text{10}}^6\frac{\text{J}}{\text{h}}\right)\left(\frac{1\text{h}}{3600\text{s}}\right)\left[\therefore 1\frac{\text{J}}{\text{s}}=1\text{W}\right]\\ E_{\text{thermal}}=693\text{W}\end{array}$

Do the unit conversion on above result,

$\begin{array}{c}{E}_{\text{thermal}}=693\text{W}\left[\therefore 1\text{W=3}\text{.412}\frac{\text{Btu}}{\text{h}}\right]\\ =\left(693\text{W}\right)\left(\frac{\text{3}\text{.412}\frac{\text{Btu}}{\text{h}}}{1\text{W}}\right)\\ E_{\text{thermal}}=2364.6\frac{\text{Btu}}{\text{h}}\end{array}$

At standard $77\text{°F}$ temperature, the heating value of methane is $978\frac{\text{Btu}}{{\text{ft}}^{\text{3}}}$. Since the major composition of natural gas is methane, consider the heating value of natural gas is $978\frac{\text{Btu}}{{\text{ft}}^{\text{3}}}$, as the range of temperature given as $55\text{°F}$ to $140\text{°F}$ considered for $77\text{°F}$.

Then the amount of natural gas required to heat the water in heater can be determined from the below expression:

$\begin{array}{c}\text{Volume}\text{of}\text{gas}=\frac{E_{\text{thermal}}}{\text{Efficiency×Heating}\text{value}\text{of}\text{gas}}\\ =\frac{2364.6\frac{\text{Btu}}{\text{h}}}{0.8\left(978\frac{\text{Btu}}{{\text{ft}}^{\text{3}}}\right)}\left[\therefore 1\text{day}=24\text{h}\right]\\ =\left(3\frac{{\text{ft}}^{\text{3}}}{\text{h}}\right)\left(\frac{24\text{h}}{1\text{day}}\right)\\ \text{Volume}\text{of}\text{gas}=72\frac{{\text{ft}}^{\text{3}}}{\text{day}}\end{array}$

Therefore, the amount of natural gas required to heat the water in a heater is $72\frac{{\text{ft}}^{\text{3}}}{\text{day}}$.

Conclusion:

Hence, the amount of natural gas required to heat the water in a heater is $72\frac{{\text{ft}}^{\text{3}}}{\text{day}}$.