Chapter 7.13, Problem 209RP

1. (a)   Water flows through a shower head steadily at a rate of 10 L/min. An electric resistance heater placed in the water pipe heats the water from 16 to 43°C. Taking the density of water to be 1 kg/L, determine the electric power input to the heater in kW and the rate of entropy generation during this process in kW/K.

FIGURE P7–209

1. (b)   In an effort to conserve energy, it is proposed to pass the drained warm water at a temperature of 39°C through a heat exchanger to preheat the incoming cold water. If the heat exchanger has an effectiveness of 0.50 (that is, it recovers only half of the energy that can possibly be transferred from the drained water to incoming cold water), determine the electric power input required in this case and the reduction in the rate of entropy generation in the resistance heating section.

To determine

The electric power input to the heater and the rate of entropy generation during the process.

Answer to Problem 209RP

The electric power input to the heater is $18.8\text{kW}$.

The rate of entropy generation during the process is $0.0622\text{kJ}/\text{K}$.

Explanation of Solution

Write the expression for the energy balance of steady flow system.

${\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=\Delta {\dot{E}}_{\text{system}}$ (I).

Here, rate of net energy transfer in to the control volume is ${\dot{E}}_{\text{in}}$, rate of net energy transfer exit from the control volume is ${\dot{E}}_{\text{out}}$ and rate of change in internal energy of system is $\Delta {\dot{E}}_{\text{system}}$.

Write the expression to calculate the mass flow rate $\left(\dot{m}\right)$ of water.

$\dot{m}=\rho \dot{\nu }$ (II).

Here, density of water at room temperature is $\rho$ and volume flow rate of water is $\dot{\nu }$.

Write the expression for the entropy balance equation of the system for steady flow process.

${\dot{S}}_{in}-{\dot{S}}_{out}+{\dot{S}}_{gen}=\Delta {\dot{S}}_{system}$ (III).

Here, rate of net entropy in is ${\dot{S}}_{in}$, rate of net entropy out is ${\dot{S}}_{out}$, rate of entropy generation is ${\dot{S}}_{gen}$ and rate of change of entropy of the system is $\Delta {\dot{S}}_{system}$

Conclusion:

There is only one exit and one inlet, write the equation for the mass balance of steady flow system as,

$\begin{array}{c}{\dot{m}}_1={\dot{m}}_2\\ =\dot{m}\end{array}$

Here, mass flow rate of water at inlet is ${\dot{m}}_1$, mass flow rate of water at exit is ${\dot{m}}_2$ and mass flow rate of water is $\dot{m}$.

The rate of change in internal energy of system inside the system is zero at steady state,

Substitute 0 for $\Delta {\dot{E}}_{\text{system}}$ in Equation (I).

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\dot{W}}_{\text{e,in}}+\dot{m}{h}_1=\dot{m}{h}_2\\ {\dot{W}}_{\text{e,in}}=\dot{m}\left(h_2-h_1\right)\end{array}$

$=\dot{m}c\left(T_2-T_1\right)$ (IV).

Here, electric power input to the heater is ${\dot{W}}_{\text{e,in}}$, initial enthalpy of water is $h_1$ , final enthalpy of water is $h_2$, specific heat of water is $c$, final temperature is $T_2$ , mass flow rate of water is $\dot{m}$ and initial temperature is $T_1$.

From Table A-3 “Properties of common liquids, solids and foods”, the value for the density $\left(\rho \right)$ of water at room temperature is $\text{1}\text{kg}/\text{L}$ and specific heat of water $\left(c\right)$ at room temperature is $\text{4}\text{.18}\text{kJ}/\text{kg}\cdot \text{°C}$.

Substitute $\text{1}\text{kg}/\text{L}$ for $\rho$ and $\text{10}\text{L}/\text{min}$ for $\dot{\nu }$ in Equation (II).

$\begin{array}{c}\dot{m}=\left(\text{1}\text{kg}/\text{L}\right)\left(\text{10}\text{L}/\text{min}\right)\\ =10\text{kg}/\text{min}\end{array}$

Substitute $10\text{kg}/\text{min}$ for $\dot{m}$, $\text{4}\text{.18}\text{kJ}/\text{kg}\cdot \text{°C}$ for $c$, $\text{43°C}$ for $T_2$ and $\text{16°C}$ for $T_1$ in Equation (IV).

$\begin{array}{c}{\dot{W}}_{\text{e,in}}=10\text{kg}/\text{min}\left(\text{4}\text{.18}\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(\text{43°C}-16\text{°C}\right)\\ =10\text{kg}/\text{min}\left(\text{4}\text{.18}\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(\text{27°C}\right)\\ =18.8\text{kW}\end{array}$

Thus, the electric power input to the heater is $18.8\text{kW}$.

Substitute $\dot{m}{s}_1$ for $S_{in}$, $\dot{m}{s}_2$ for $S_{out}$ and $0$ for $\Delta S_{system}$ in Equation (III)

$\begin{array}{l}\dot{m}{s}_1-\dot{m}{s}_2+{\dot{S}}_{gen}=0\\ {\dot{S}}_{gen}=\dot{m}\left(s_2-s_1\right)\end{array}$

Since, water is incompressible substance,

${\dot{S}}_{gen,1}=\dot{m}c\ln \frac{T_2}{T_1}$ (V).

Here, rate of entropy generation at stage 1 is ${\dot{S}}_{gen,1}$, initial entropy is $s_1$ and final entropy is $s_2$.

Substitute $10\text{kg}/\text{min}$ for $\dot{m}$, $\text{4}\text{.18}\text{kJ}/\text{kg}\cdot \text{°C}$ for $c$, $\text{43°C}$ for $T_2$ and $\text{16°C}$ for $T_1$ in Equation (V).

$\begin{array}{c}{\dot{S}}_{gen,1}=\left(10\text{kg}/\text{min}\right)\left(\text{4}\text{.18}\text{kJ}/\text{kg}\cdot \text{°C}\right)\ln \frac{\text{43°C}}{\text{16°C}}\\ =\left(10\frac{\text{kg}}{\text{min}}\right)\left(\frac{1\min }{60\text{s}}\right)\left(\text{4}\text{.18}\text{kJ}/\text{kg}\cdot \text{°C}\right)\ln \frac{\left(43+273\right)\text{K}}{\left(16+273\right)\text{K}}\\ =0.0622\text{kJ}/\text{K}\end{array}$

Thus, the rate of entropy generation during the process is $0.0622\text{kJ}/\text{K}$.

To determine

The electric power input required and the reduction in the rate of entropy generation in the resistance heating section.

Answer to Problem 209RP

The electric power input required is $8\text{kW}$.

The reduction in the rate of entropy generation in the resistance heating section is $0.0272\text{kJ}/\text{K}$.

Explanation of Solution

Write the expression to calculate the energy saved $\left({\dot{Q}}_{\text{saved}}\right)$ by the heat exchanger.

${\dot{Q}}_{\text{saved}}=\epsilon \dot{m}c\left(T_{\max }-T_{\min }\right)$ (VI).

Here, effectiveness of heat exchanger is $\epsilon$, temperature of warm water is $T_{\max }$ and minimum temperature is $T_{\min }$.

Write the expression to calculate the required electric power $\left({\dot{W}}_{\text{in,new}}\right)$.

${\dot{W}}_{\text{in,new}}={\dot{W}}_{\text{e,in}}-{\dot{Q}}_{\text{saved}}$ (VII).

Here, electric power input to the heater is ${\dot{W}}_{\text{e,in}}$.

Write the expression to calculate the temperature at which the cold water leaves heat exchanger.

$\dot{Q}=\dot{m}c\left(T_{c,out}-T_{c,in}\right)$ (VIII).

Here, the energy saved is $\dot{Q}$ , the mass flow rate is $\dot{m}$, the cold water temperature at inlet is $T_{c,in}$ and the cold water temperature at outlet is $T_{c,out}$.

Write the expression to calulate the entropy generation at stage 2.

${\dot{S}}_{gen,2}=\dot{m}c\ln \frac{T_2}{T_1}$ (IX).

Here, rate of entropy generation at stage 2 is ${\dot{S}}_{gen,2}$.

Write the expression to calculate the reduction in the rate of entropy generation within the heating section

${\dot{S}}_{reduction}={\dot{S}}_{gen,1}-{\dot{S}}_{gen,2}$ (X).

Here, reduction in the rate of entropy generation is ${\dot{S}}_{reduction}$.

Conclusion:

Substitute 0.5 for $\epsilon$, $10\text{kg}/\text{min}$ for $\dot{m}$, $\text{4}\text{.18}\text{kJ}/\text{kg}\cdot \text{°C}$ for $c$, $\text{39°C}$ for $T_{\max }$ and $\text{16°C}$ for $T_{\min }$ in Equation (VI).

$\begin{array}{c}{\dot{Q}}_{\text{saved}}=\left(0.5\right)\left(10\text{kg}/\text{min}\right)\left(\text{4}\text{.18}\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(\text{39°C}-16\text{°C}\right)\\ =\left(0.5\right)\left[\left(10\text{kg}/\text{min}\right)\left(\frac{1\min }{60\text{s}}\right)\right]\left(\text{4}\text{.18}\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(\text{23°C}\right)\\ =\left(0.5\right)\left(0.1666\text{kg}/\text{s}\right)\left(\text{4}\text{.18}\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(\text{23°C}\right)\end{array}$

$\begin{array}{l}=8\text{kJ}/\text{s}\left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =8\text{kW}\end{array}$

Substitute $18.8\text{kW}$ for ${\dot{W}}_{\text{e,in}}$ and $\text{8}\text{kW}$ for ${\dot{Q}}_{\text{saved}}$ in Equation (VII).

$\begin{array}{c}{\dot{W}}_{\text{in,new}}=18.8\text{kW}-8\text{kW}\\ =\text{10}\text{.8}\text{kW}\end{array}$

Substitute $8\text{kJ}/\text{s}$ for $\dot{Q}$ , $10\text{kg}/\text{min}$ for $\dot{m}$, $\text{4}\text{.18}\text{kJ}/\text{kg}\cdot \text{°C}$ for $c$, $\text{39°C}$ for $T_{\max }$ and $\text{16°C}$ for $T_{\min }$ in Equation (VIII).

$\begin{array}{c}8\text{kJ}/\text{s}=\left(10\text{kg}/\text{min}\right)\left(\text{4}\text{.18}\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(T_{c,out}-16\text{°C}\right)\\ =\left[\left(10\text{kg}/\text{min}\right)\left(\frac{1\min }{60\text{s}}\right)\right]\left(\text{4}\text{.18}\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(T_{c,out}-16\text{°C}\right)\\ =27.5°\text{C}\\ \text{=}\left(\text{27}\text{.5+273}\right)\text{K}\\ \text{=300}\text{.5}\text{K}\end{array}$

Substitute $10\text{kg}/\text{min}$ for $\dot{m}$, $\text{4}\text{.18}\text{kJ}/\text{kg}\cdot \text{°C}$ for $c$, $\text{43°C}$ for $T_2$ and $\text{16°C}$ for $T_1$ in Equation (IX).

$\begin{array}{c}{\dot{S}}_{gen,2}=\left(10\text{kg}/\text{min}\right)\left(\text{4}\text{.18}\text{kJ}/\text{kg}\cdot \text{°C}\right)\ln \frac{\text{43°C}}{\text{300}\text{.5}\text{K}}\\ =\left(10\frac{\text{kg}}{\text{min}}\right)\left(\frac{1\min }{60\text{s}}\right)\left(\text{4}\text{.18}\text{kJ}/\text{kg}\cdot \text{°C}\right)\ln \frac{\left(43+273\right)\text{K}}{\text{300}\text{.5}\text{K}}\\ =0.0350\text{kJ}/\text{K}\end{array}$

Substitute $0.0622\text{kJ}/\text{K}$ for ${\dot{S}}_{gen,1}$ and $0.0350\text{kJ}/\text{K}$ for ${\dot{S}}_{gen,2}$ in Equation (X).

$\begin{array}{c}{\dot{S}}_{reduction}=0.0622\text{kJ}/\text{K}-0.0350\text{kJ}/\text{K}\\ =0.0272\text{kJ}/\text{K}\end{array}$

Thus, the reduction in the rate of entropy generation in the resistance heating section is $0.0272\text{kJ}/\text{K}$.