Chapter 6, Problem 153RQ

(a)

To determine

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

Explanation of Solution

Given:

The maximum flow rate $\left({\dot{V}}_{\max }\right)$ is $13.3\text{ L/min}$.

The reduced flow rate $\left({\dot{V}}_{\text{Red}}\right)$ is $10.5\text{ L/min}$.

The temperature of the hot stream $\left(T_1\right)$ is $15°\text{C}$.

The temperature of the hot stream $\left(T_2\right)$ is $55°\text{C}$.

The temperature of the mixed stream $\left(T_3\right)$ is $42°\text{C}$.

The duration of the shower $\left(t\right)$ is 5 min/person.

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

The number of person $\left(n\right)$ is 4 persons.

Calculation:

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).

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

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

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

(b)

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.

Explanation of Solution

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 $1\text{kg/L}$.

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.

Calculate the volume flow rated saved.

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

Calculate the total volume flow rate saved by the family per year.

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

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

Calculate the total mass of water saved per year.

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

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

Calculate the energy saved due to the installation of lower shower heads.

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

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

  $\simeq 641\text{kWh/year}$

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 $641\text{kWh/year}$.