Chapter 9, Problem 49P

To determine

Calculate the quantity of available water to unit $\text{-}$4.

Answer to Problem 49P

The quantity of available water to unit 4 in $\frac{\text{gallons}}{\min }$, $\frac{{\text{ft}}^3}{\text{s}}$, $\frac{{\text{m}}^3}{\text{s}}$, $\frac{\text{lbm}}{\text{s}}$, and $\frac{\text{kg}}{\text{s}}$ are $54.7\underset{\_}{\frac{\text{gallons}}{\min }}$, $0.122\frac{{\text{ft}}^3}{\text{s}},0.00345\frac{{\text{m}}^3}{\text{s}},7.6\frac{\text{lbm}}{\text{s}},\text{and}3.45\frac{\text{kg}}{\text{s}}$ respectively.

Explanation of Solution

Calculation:

Consider that the density of the water is generally known as $62.4\frac{\text{lbm}}{{\text{ft}}^{\text{3}}}$.

$\text{density}=62.4\frac{\text{lbm}}{{\text{ft}}^{\text{3}}}$

Convert the unit from $\frac{\text{lbm}}{{\text{ft}}^{\text{3}}}$ to $\frac{\text{lbm}}{\text{gallon}}$.

$\begin{array}{c}\text{density}=62.4\frac{\text{lbm}}{{\text{ft}}^{\text{3}}}\left(\frac{1{\text{ft}}^3}{\text{7}\text{.48052}\text{gallons}}\right)\left[\therefore 1{\text{ft}}^3=\text{7}\text{.48052}\text{gallons}\right]\\ =8.34\frac{\text{lbm}}{\text{gallon}}\end{array}$

Consider that the unit-1 water consumption rate.

$\text{Unit-1 consumption rate}=18\frac{\text{lbm}}{\text{s}}$

Convert the unit from $\frac{\text{lbm}}{\text{s}}$ to $\frac{\text{gallons}}{\text{min}}$.

$\begin{array}{c}\text{Unit-1 consumption rate}=18\frac{\text{lbm}}{\text{s}}\left(\frac{60\text{s}}{1\min }\right)\left(\frac{1\text{gallons}}{8.34\text{lbm}}\right)\left[\begin{array}{l}\therefore 1\text{min}=60\text{s},\\ \text{and}1\text{gallons}=8.34\text{lbm}\end{array}\right]\\ =129.5\frac{\text{gallons}}{\min }\end{array}$

Consider that the unit-2 water consumption rate.

$\text{Unit-2 consumption rate}=18\frac{\text{lbm}}{\text{s}}$

Convert the unit from $\frac{\text{lbm}}{\text{s}}$ to $\frac{\text{gallons}}{\text{min}}$.

$\begin{array}{c}\text{Unit-2 consumption rate}=14\frac{\text{lbm}}{\text{s}}\left(\frac{60\text{s}}{1\min }\right)\left(\frac{1\text{gallons}}{8.34\text{lbm}}\right)\left[\begin{array}{l}\therefore 1\text{min}=60\text{s},\\ \text{and}1\text{gallons}=8.34\text{lbm}\end{array}\right]\\ =100.7\frac{\text{gallons}}{\min }\end{array}$

Consider that the unit-3 water consumption rate.

$\text{Unit-3 consumption rate}=18\frac{\text{lbm}}{\text{s}}$

Convert the unit from $\frac{\text{lbm}}{\text{s}}$ to $\frac{\text{gallons}}{\text{min}}$.

$\begin{array}{c}\text{Unit-3 consumption rate}=16\frac{\text{lbm}}{\text{s}}\left(\frac{60\text{s}}{1\min }\right)\left(\frac{1\text{gallons}}{8.34\text{lbm}}\right)\left[\begin{array}{l}\therefore 1\text{min}=60\text{s},\\ \text{and}1\text{gallons}=8.34\text{lbm}\end{array}\right]\\ =115.1\frac{\text{gallons}}{\min }\end{array}$

The amount of water available to unit-4 is,

$\text{The amount of water available to unit-}4=\left\{\begin{array}{l}\text{water}\text{available}\text{rate}-\\ \text{unit-1 consumption rate}-\\ \text{unit-2 consumption rate}-\\ \text{unit-3 consumption rate}\end{array}\right\}$

Substitute $400\frac{\text{gallons}}{\text{min}}$ for water available rate, $129.5\frac{\text{gallons}}{\text{min}}$ for unit-1 consumption rate, $100.7\frac{\text{gallons}}{\text{min}}$ for unit-2 consumption rate, and $115.1\frac{\text{gallons}}{\text{min}}$ for unit-3 consumption rate.

$\text{The amount of water available to unit-}4=\left\{\left[\begin{array}{l}400\frac{\text{gallons}}{\text{min}}-129.5\frac{\text{gallons}}{\text{min}}-\\ 100.7\frac{\text{gallons}}{\text{min}}-115.1\frac{\text{gallons}}{\text{min}}\end{array}\right]\right\}$

$\text{The amount of water available to unit-}4=54.7\frac{\text{gallons}}{\min }$ (1)

Convert the unit of equation (1) from $\frac{\text{gallons}}{\min }$ to $\frac{{\text{ft}}^3}{\text{s}}$.

$\begin{array}{c}\text{density}=54.7\frac{\text{gallons}}{\min }\left(\frac{1{\text{ft}}^3}{\text{7}\text{.48052}\text{gallons}}\right)\left(\frac{1\min }{60\text{s}}\right)\left[\begin{array}{l}\therefore 1{\text{ft}}^3=\text{7}\text{.48052}\text{gallons,}\text{and}\\ 1\min =60\text{s}\end{array}\right]\\ =0.122\frac{{\text{ft}}^3}{\text{s}}\end{array}$

Convert the unit from $\frac{{\text{ft}}^3}{\text{s}}$ to $\frac{{\text{m}}^3}{\text{s}}$.

$\begin{array}{c}\text{density}=0.122\frac{{\text{ft}}^3}{\text{s}}{\left(\frac{\text{0}\text{.3048}\text{m}}{1\text{ft}}\right)}^3\left[\therefore 1\text{ft}=\text{0}\text{.3048}\text{m}\right]\\ =0.00345\frac{{\text{m}}^3}{\text{s}}\end{array}$

Convert the unit of equation (1) from $\frac{\text{gallons}}{\min }$ to $\frac{\text{lbm}}{\text{s}}$.

$\begin{array}{c}\text{density}=54.7\frac{\text{gallons}}{\min }\left(\frac{1\frac{\text{lbm}}{\text{s}}}{7.19\frac{\text{gallons}}{\min }}\right)\left[\therefore 1\frac{\text{lbm}}{\text{s}}=7.19\frac{\text{gallons}}{\min }\right]\\ =7.6\frac{\text{lbm}}{\text{s}}\end{array}$

Convert the unit from $\frac{\text{lbm}}{\text{s}}$ to $\frac{\text{kg}}{\text{s}}$.

$\begin{array}{c}\text{density}=7.6\times \text{0}\text{.453592}\frac{\text{kg}}{\text{s}}\left[\therefore 1\text{lbm}=\text{0}\text{.453592}\text{kg}\right]\\ =3.45\frac{\text{kg}}{\text{s}}\end{array}$

Conclusion:

Thus, the quantity of available water to unit 4 in $\frac{\text{gallons}}{\min }$, $\frac{{\text{ft}}^3}{\text{s}}$, $\frac{{\text{m}}^3}{\text{s}}$, $\frac{\text{lbm}}{\text{s}}$, and $\frac{\text{kg}}{\text{s}}$ are $54.7\underset{\_}{\frac{\text{gallons}}{\min }}$, $0.122\frac{{\text{ft}}^3}{\text{s}},0.00345\frac{{\text{m}}^3}{\text{s}},7.6\frac{\text{lbm}}{\text{s}},\text{and}3.45\frac{\text{kg}}{\text{s}}$ respectively.