Chapter 6.11, Problem 139RP

To determine

The amount of oil a family of four will save per year by replacing the standard shower heads by the low-flow ones.

The amount of money a family of four will save per year by replacing the standard shower heads by the low-flow ones.

Answer to Problem 139RP

The amount of oil a family of four will save per year by replacing the standard shower heads by the low-flow ones is $\underset{\_}{29.1\text{gal}/\text{year}}$.

The amount of money a family of four will save per year by replacing the standard shower heads by the low-flow ones is $\underset{\_}{\text{\$81}\text{.5}/\text{year}}$.

Explanation of Solution

Determine the rate of water saved volume of the low-flow shower head a family of four will save per year.

${\dot{\nu }}_{\text{saved}}=\left[\begin{array}{l}\left({\nu }_1-{\nu }_2\right)\times \left(t/\left(\text{person}\cdot \text{day}\right)\right)\\ \times \left(n\right)\times \left(\text{number of days per year}\right)\end{array}\right]$ (I)

Here, the initial volume of the low-flow shower head is ${\nu }_1$, the final volume of the low-flow shower head is ${\nu }_2$, the time is $t$, and the number person in family is $n$.

Determine the mass flow rate of water of the low-flow shower head.

${\dot{m}}_{\text{saved}}={\rho }_{\text{water}}\times {\dot{\nu }}_{\text{saved}}$ (II)

Here, the density of water is ${\rho }_{\text{water}}$.

Determine the amount of energy saved in the low-flow shower head.

$\begin{array}{c}{E}_{\text{saved}}=\left({\dot{m}}_{\text{water}}{c}_p\Delta T\right)\\ ={\dot{m}}_{\text{water}}{c}_p\left(T_1-T_2\right)\end{array}$ (III)

Here, the specific heat of water is $c_p$, the initial temperature of the low-flow shower head is $T_1$, and the final temperature of the low-flow shower head is $T_2$.

Determine the amount fuel saved in the low-flow shower head.

$\text{Fuel}\text{saved}=\frac{E_{\text{saved}}}{\left({\eta }_{\text{fuel}}\right)\left(\text{heating}\text{value}\text{of fuel}\right)}$ (IV)

Here, the efficiency of the fuel in the low-flow shower head is ${\eta }_{\text{fuel}}$.

Determine the amount of money saved in the low-flow shower head.

$\text{Money}\text{saved}=\left(\text{fuel saved}\right)\times \left(\text{unit cost of fuel}\right)$ (V)

Conclusion:

Substitute $13.3\text{L}/\text{min}$ for ${\nu }_1$, $10.5\text{L}/\text{min}$ for ${\nu }_2$, 6 min for $t$, 4 for $n$, 356 for number of days per year in Equation (I).

$\begin{array}{c}{\dot{\nu }}_{\text{saved}}=\left[\begin{array}{l}\left(13.3-10.5\right)\text{L}/\text{min}\times \left(6\min /\left(\text{person}\cdot \text{day}\right)\right)\\ \times \left(4\text{persons}\right)\times \left(\text{365}\text{days}/\text{year}\right)\end{array}\right]\\ =\left[\begin{array}{l}2.8\text{L}/\text{min}\times \left(6\min /\left(\text{person}\cdot \text{day}\right)\right)\\ \times \left(4\text{persons}\right)\times \left(\text{365}\text{days}/\text{year}\right)\end{array}\right]\\ =24,528\text{L}/\text{year}\end{array}$

Substitute $1\text{kg}/\text{L}$ for $\rho$ and $24,528\text{L}/\text{year}$ for ${\dot{\nu }}_{\text{saved}}$ in Equation (II).

$\begin{array}{c}{\dot{m}}_{saved}=\left(1\text{kg}/\text{L}\right)\times \left(24,528\text{L}/\text{year}\right)\\ =24,528\text{kg}/\text{year}\end{array}$

Substitute $24,528\text{kg}/\text{year}$ for ${\dot{m}}_{saved}$, $4.18\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_p$, $42°\text{C}$ for $T_1$, and $15°\text{C}$ for $T_2$ in the Equation (III).

$\begin{array}{c}{E}_{\text{saved}}=\left(24,528\text{kg}/\text{year}\right)\times \left(4.18\text{kJ}/\text{kg}\cdot °\text{C}\right)\times \left(42°\text{C}-15°\text{C}\right)\\ =\left(24,528\text{kg}/\text{year}\right)\times \left(4.18\text{kJ}/\text{kg}\cdot °\text{C}\right)\times \left(27°\text{C}\right)\\ =2768230\text{kJ}/\text{year}\end{array}$

Substitute $2768230\text{kJ}/\text{year}$ for $E_{\text{saved}}$, 0.65 for ${\eta }_{\text{fuel}}$, $146,300\text{kJ}/\text{gal}$ for heating value of fuel in Equation (IV).

$\begin{array}{c}\text{Fuel}\text{saved}=\frac{\left(2768230\text{kJ}/\text{year}\right)}{\left(0.65\right)\times \left(146,300\text{kJ}/\text{gal}\right)}\\ =\frac{\left(2768230\text{kJ}/\text{year}\right)}{\left(0.65\right)\times \left(146,300\text{kJ}/\text{gal}\right)}\\ =\frac{\left(2768230\text{kJ}/\text{year}\right)}{\left(95095\text{kJ}/\text{gal}\right)}\\ =29.1\text{gal}/\text{year}\end{array}$

Thus, the amount of oil a family of four will save per year by replacing the standard shower heads by the low-flow ones is $\underset{\_}{29.1\text{gal}/\text{year}}$.

Substitute $29.1\text{gal}/\text{year}$ for fuel saved and $\text{\$2}\text{.80}/\text{gal}$ for unit cost of fuel in Equation (V).

$\begin{array}{c}\text{Money}\text{saved}=\left(29.1\text{gal}/\text{year}\right)\times \left(\text{\$2}\text{.80}/\text{gal}\right)\\ =\text{\$81}\text{.5}/\text{year}\end{array}$

Thus, the amount of money a family of four will save per year by replacing the standard shower heads by the low-flow ones is $\underset{\_}{\text{\$81}\text{.5}/\text{year}}$.