Chapter 5.5, Problem 168RP

The maximum flow rate of standard shower heads is about 3.5 gpm (13.3 L/min) and can be reduced to 2.75 gpm (10.5 L/min) by switching to low-flow shower heads that are equipped with flow controllers. Consider a family of four, with each person taking a 5-min shower every morning. City water at 15°C is heated to 55°C in an electric water heater and tempered to 42°C by cold water at the T-elbow of the shower before being routed to the shower heads. Assuming a constant specific heat of 4.18 kJ/kg·°C for water, determine (a) the ratio of the flow rates of the hot and cold water as they enter the T-elbow and (b) the amount of electricity that will be saved per year, in kWh, by replacing the standard shower heads with the low-flow ones.

To determine

The ratio of flow rates of hot and cold water as they enter the T-elbow.

Answer to Problem 168RP

The ratio of flow rates of hot and cold water as they enter the T-elbow is $2.08$.

Explanation of Solution

Here, the two streams (comparatively hot and cold) of fluids are mixed in a rigid mixing chamber and operates at steady state. Hence, the inlet and exit mass flow rates are equal.

${\dot{m}}_1+{\dot{m}}_2={\dot{m}}_3$ (I)

Write the energy rate balance equation for two inlet and one outlet system.

$\left\{\begin{array}{l}\left[{\dot{Q}}_{\text{1}}+{\dot{W}}_{\text{1}}+{\dot{m}}_1\left(h_1+\frac{V_1{}^2}{2}+g{z}_1\right)\right]\\ +\left[{\dot{Q}}_{\text{2}}+{\dot{W}}_{\text{2}}+{\dot{m}}_2\left(h_2+\frac{V_2{}^2}{2}+g{z}_2\right)\right]\\ -\left[{\dot{Q}}_{\text{3}}+{\dot{W}}_{\text{3}}+{\dot{m}}_3\left(h_3+\frac{V_3{}^2}{2}+g{z}_3\right)\right]\end{array}\right\}=\Delta {\dot{E}}_{\text{system}}$ (II)

Here, the rate of heat transfer is $\dot{Q}$, the rate of work transfer is $\dot{W}$, the enthalpy is $h$ and the velocity is $V$, the gravitational acceleration is $g$, the elevation from the datum is $z$ and the rate of change in net energy of the system is $\Delta {\dot{E}}_{\text{system}}$; the suffixes 1 indicates the hot stream inlet, 2 indicates the cold stream inlet and 3 indicates the mixed water stream outlet.

The system is at steady state. Hence, the rate of change in net energy of the system becomes zero.

$\Delta {\dot{E}}_{\text{system}}=0$

Neglect the heat transfer, work transfer, kinetic and potential energies.

The Equation (II) reduced as follows.

$\begin{array}{l}{\dot{m}}_1{h}_1+{\dot{m}}_2{h}_2-{\dot{m}}_3{h}_3=0\\ {\dot{m}}_1{h}_1+{\dot{m}}_2{h}_2={\dot{m}}_3{h}_3\end{array}$ (III)

Substitute ${\dot{m}}_1+{\dot{m}}_2$ for ${\dot{m}}_3$.

$\begin{array}{l}{\dot{m}}_1{h}_1+{\dot{m}}_2{h}_2=\left({\dot{m}}_1+{\dot{m}}_2\right)h_3\\ {\dot{m}}_1{h}_1+{\dot{m}}_2{h}_2={\dot{m}}_1{h}_3+{\dot{m}}_2{h}_3\\ {\dot{m}}_2{h}_2-{\dot{m}}_2{h}_3={\dot{m}}_1{h}_3-{\dot{m}}_1{h}_1\\ {\dot{m}}_2\left(h_2-h_3\right)={\dot{m}}_1\left(h_3-h_1\right)\end{array}$

$\frac{{\dot{m}}_2}{{\dot{m}}_1}=\frac{h_3-h_1}{h_2-h_3}$ (IV)

The change in enthalpy is expressed as follow.

$\begin{array}{l}{h}_3-h_1=c_p\left(T_3-T_1\right)\\ h_2-h_3=c_p\left(T_2-T_3\right)\end{array}$

Here, the specific heat is $c_p$, the temperature is $T$ and the enthalpy is $h$.

Substitute $c_p\left(T_3-T_1\right)$ for $h_3-h_1$ and $c_p\left(T_2-T_3\right)$ for $h_2-h_3$ in Equation (IV).

$\begin{array}{l}\frac{{\dot{m}}_2}{{\dot{m}}_1}=\frac{c_p\left(T_3-T_1\right)}{c_p\left(T_2-T_3\right)}\\ \frac{{\dot{m}}_2}{{\dot{m}}_1}=\frac{T_3-T_1}{T_2-T_3}\end{array}$ (V)

Conclusion:

Substitute $42°\text{C}$ for $T_3$, $15°\text{C}$ for $T_1$, and $55°\text{C}$ for $T_2$ in Equation (V).

$\begin{array}{c}\frac{{\dot{m}}_2}{{\dot{m}}_1}=\frac{42°\text{C}-15°\text{C}}{55°\text{C}-42°\text{C}}\\ =\frac{27°\text{C}}{13°\text{C}}\\ =2.0769\\ \simeq 2.08\end{array}$

Thus, the ratio of flow rates of hot and cold water as they enter the T-elbow is $2.08$.

To determine

The amount of electricity that will be saved per year, in $\text{kWh}$, by replacing the standard shower heads with the low-flow ones.

Answer to Problem 168RP

The amount of electricity that will be saved per year, in $\text{kWh}$, by replacing the standard shower heads with the low-flow ones is $641\text{kWh/year}$.

Explanation of Solution

Here, the volume flow rate in the shower heads is lowered by installing low-flow shower heads equipped with flow controllers. This reduces the volume flow rate of water that is consumed per person.

The volume flow rated saved is expressed as follows,

$\begin{array}{c}{\dot{V}}_{\text{saved}}=13.3\text{L/min}-10.5\text{L/min}\\ =2.8\text{L/min}\end{array}$

The total volume flow rate saved by the family per year is expressed as follows.

${\dot{V}}_{\text{f,saved}}={\dot{V}}_{\text{saved}}\times \left(\begin{array}{l}\text{no}\text{.of persons}\\ \text{in family}\end{array}\right)\times \left(\begin{array}{l}\text{shower time used}\\ \text{per person per day }\end{array}\right)$ (VI)

Write the formula for total mass of water saved per year.

${\dot{m}}_{\text{saved}}=\rho {\dot{V}}_{\text{f,saved}}$ (VII)

Write the formula for energy saved due to the installation of lower shower heads.

$\text{Energy saved}={\dot{m}}_{\text{saved}}{c}_p\left(T_3-T_1\right)$ (VIII)

Refer Table A-3 (a), “Properties of common liquids, solids, and foods”.

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

The density of the water $\left(\rho \right)$ is taken as $1\text{kg/L}$.

Conclusion:

Substitute $2.8\text{L/min}$ for ${\dot{V}}_{\text{saved}}$, $4\text{person}$ for no. of persons in family and $5\text{min/person}\cdot \text{day}$ for shower time used per person per day in Equation (VI).

$\begin{array}{c}{\dot{V}}_{\text{f,saved}}=2.8\text{L/min}\times \left(4\text{person}\right)\times \left(5\text{min/person}\cdot \text{day}\right)\\ =2.8\text{L/min}\times \left(4\text{person}\right)\times \left(5\frac{\text{min}}{\text{person}\cdot \text{day}}\times \frac{365\text{day}}{1\text{year}}\right)\\ =20440\text{L/year}\end{array}$

Substitute $1\text{kg/L}$ for $\rho$ and $20440\text{L/year}$ for ${\dot{V}}_{\text{f,saved}}$ in Equation (VII).

$\begin{array}{c}{\dot{m}}_{\text{saved}}=\left(1\text{kg/L}\right)\left(20440\text{L/year}\right)\\ =20440\text{kg/year}\end{array}$

Substitute $20440\text{kg/year}$ for ${\dot{m}}_{\text{saved}}$, $4.18\text{kJ/kg}\cdot \text{°C}$ for $c_p$,$42°\text{C}$ for $T_3$, and $15°\text{C}$ for $T_1$, in Equation (VIII).

$\begin{array}{c}\text{Energy saved}=\left(20440\text{kg/year}\right)\left(4.18\text{kJ/kg}\cdot \text{°C}\right)\left(42°\text{C}-15°\text{C}\right)\\ =\left(20440\text{kg/year}\right)\left(4.18\text{kJ/kg}\cdot \text{°C}\right)\left(27°\text{C}\right)\\ =2306858.4\text{kJ/year}\\ \simeq \text{2307000}\text{kJ/year}\times \frac{1\text{}\text{kWh}}{3600\text{kJ}}\end{array}$

$\simeq 641\text{kWh/year}$

Thus, the amount of electricity that will be saved per year, in $\text{kWh}$, by replacing the standard shower heads with the low-flow ones is $641\text{kWh/year}$.