Chapter 7, Problem 125RQ

To determine

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

Explanation of Solution

Given:

The initial volume of the low-flow shower head $\left({\nu }_1\right)$ is $13.3\text{L}/\text{min}$.

The final volume of the low-flow shower head $\left({\nu }_2\right)$ is $10.5\text{L}/\text{min}$.

The time $\left(t\right)$ is 6 min.

The number person in family $\left(n\right)$ is 4.

The density of water $\left({\rho }_{\text{water}}\right)$ is $1\text{kg}/\text{L}$

The specific heat of water $\left(c_p\right)$ is $4.18\text{kJ}/\text{kg}\cdot °\text{C}$.

The initial temperature of the low-flow shower head $\left(T_1\right)$ is $15°\text{C}$.

The final temperature of the low-flow shower head $\left(T_2\right)$ is $42°\text{C}$.

The efficiency of the fuel in the low-flow shower head $\left({\eta }_{\text{fuel}}\right)$ is 0.65.

The heating value of fuel $\left(HV\right)$ is $146,300\text{kJ}/\text{gal}$.

The unit cost of fuel is $\text{\$2}\text{.80}/\text{gal}$

Calculation:

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

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

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

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

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

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

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

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

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

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

  $\text{Fuel}\text{saved}=\frac{E_{\text{saved}}}{\left({\eta }_{\text{fuel}}\right)\left(HV\right)}$

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

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 $\underset{\_}{29.1\text{gal}/\text{year}}$.

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

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

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

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 $\underset{\_}{\text{\$81}\text{.5}/\text{year}}$.