Chapter 8, Problem 138P

(a)

To determine

The rate of heat transfer in the heat exchanger.

Explanation of Solution

Given:

The mass flow rate of ethylene glycol $\left(\dot{m}\right)$ is $2\text{ kg/s}$.

The specific heat of ethylene glycol at constant pressure $\left(c_p\right)$ is $2.56\text{kJ}/\text{kg}\cdot \text{K}$.

The entry temperature of ethylene glycol $\left(T_1\right)$ is 80 C.

The exit temperature of ethylene glycol $\left(T_2\right)$ is 40 C.

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

The entry temperature of water $\left(T_3\right)$ is 20 C.

The exit temperature of water $\left(T_4\right)$ is 55 C.

Calculation:

For the steady flow system, rate of change in internal energy of the system is zero.

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

Write the equation for the energy balance equation for closed system.

  $\begin{array}{l}{\dot{E}}_{in}-{\dot{E}}_{out}=\Delta {\dot{E}}_{system}\\ {\dot{E}}_{in}-{\dot{E}}_{out}=0\\ \dot{m}{h}_1+{\dot{Q}}_{in}=\dot{m}{h}_2\\ {\dot{Q}}_{in}=\dot{m}\left(h_2-h_1\right)\\ =\dot{m}{c}_p\left(T_2-T_1\right)\end{array}$        (I)

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

Refer Table A-3, “properties of common liquids table”, select the specific heat at constant pressure $\left(c_p\right)$ for ethylene glycol substance as $2.56\text{kJ}/\text{kg}\cdot \text{K}$.

The rate of heat transfer from the water must be equal to the rate of heat transfer to the ethylene glycol.

  $\begin{array}{l}{\dot{Q}}_{in}={\dot{Q}}_{out}\\ {\dot{Q}}_{out}=\dot{m}{c}_p\left(T_1-T_2\right)\end{array}$

  $\begin{array}{c}{\dot{Q}}_{out}=\left(2\text{kg}/\text{s}\right)\left(2.56\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\left(80+273\right)\text{K}-\left(40+273\right)\text{K}\right)\\ =204.8\text{kJ}/\text{s}\left(\frac{1\text{kW}}{\text{1}\text{kJ}/\text{s}}\right)\\ =204.8\text{kW}\end{array}$

Thus, the rate of heat transfer in the heat exchanger is $\underset{\_}{204.8\text{ kW}}$.

(b)

To determine

The rate of entropy generation in the heat exchanger.

Explanation of Solution

Refer Table A-3, “Properties of common liquids table”, note down the specific heat at constant pressure $\left(c_p\right)$ for water substance as $4.18\text{kJ}/\text{kg}\cdot \text{K}$.

Calculate mass flow rate of water using the rate of heat from the geothermal water.

  $\begin{array}{l}{\dot{Q}}_{in}={\dot{m}}_{water}{c}_p\left(T_4-T_3\right)\\ 204.8\text{kJ}/\text{s}={\dot{m}}_{water}\left(4.18\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\left(55+273\right)\text{K}-\left(20+273\right)\text{K}\right)\\ {\dot{m}}_{water}=\frac{204.8}{146.3}\\ =1.4\text{kg}/\text{s}\end{array}$

Write the expression for the entropy balance in the heat exchanger.

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

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

Substitute ${\dot{S}}_{in}={\dot{m}}_1{s}_1+{\dot{m}}_3{s}_3$, ${\dot{S}}_{out}={\dot{m}}_2{s}_2+{\dot{m}}_4{s}_4$ and $\Delta {\dot{S}}_{system}=0$ in Equation (II).

  $\begin{array}{l}{\dot{m}}_1{s}_1+{\dot{m}}_3{s}_3-\left({\dot{m}}_2{s}_2+{\dot{m}}_4{s}_4\right)+{\dot{S}}_{gen}=0\\ {\dot{m}}_1{s}_1+{\dot{m}}_3{s}_3-{\dot{m}}_2{s}_2-{\dot{m}}_4{s}_4+{\dot{S}}_{gen}=0\\ {\dot{S}}_{gen}=-{\dot{m}}_{glycol}{s}_1-{\dot{m}}_{water}{s}_3+{\dot{m}}_{glycol}{s}_2+{\dot{m}}_{water}{s}_4\\ {\dot{S}}_{gen}={\dot{m}}_{glycol}\left(s_2-s_1\right)+{\dot{m}}_{water}\left(s_4-s_3\right)\end{array}$

  ${\dot{S}}_{gen}={\dot{m}}_{glycol}{c}_p\ln \left(\frac{T_2}{T_1}\right)+{\dot{m}}_{water}{c}_p\ln \left(\frac{T_4}{T_3}\right)$

  $\begin{array}{c}{\dot{S}}_{gen}=\left\{\begin{array}{l}\left(2\text{kg}/\text{s}\right)\left(2.56\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{\left(40°\text{C}\right)}{\left(80°\text{C}\right)}\right)\\ +\left(1.4\text{kg}/\text{s}\right)\left(4.18\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{\left(55°\text{C}\right)}{\left(20°\text{C}\right)}\right)\end{array}\right\}\\ {\dot{S}}_{gen}=\left\{\begin{array}{l}\left(2\text{kg}/\text{s}\right)\left(2.56\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{\left(40+273\right)\text{K}}{\left(80+273\right)\text{K}}\right)\\ +\left(1.4\text{kg}/\text{s}\right)\left(4.18\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{\left(55+273\right)\text{K}}{\left(20+273\right)\text{K}}\right)\end{array}\right\}\\ =0.0446\text{kJ}/\text{s}\cdot \text{K}\left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =0.0446\text{kW}/\text{K}\end{array}$

Thus, the rate of entropy generation in the heat exchanger is $\underset{\_}{0.0446\text{kW}/\text{K}}$.