Chapter 8, Problem 142P

a)

To determine

The rate of heat removal from the chicken.

Explanation of Solution

Given:

The average mass of the chicken $\left(m_{chicken}\right)$ is $2.2\text{kg}/\text{chicken}$.

The number of chickens dropped into the chiller $\left(n\right)$ is $250\text{chicken}/\text{hr}$.

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

The final temperature of the chicken $\left(T_2\right)$ is $3°\text{C}$.

The specific heat of chicken $\left(c_p\right)$ is $3.54\text{kJ}/\text{kg}\cdot \text{K}$.

Calculation:

Refer TableA-3, “Properties of common liquids, solids, and foods”, select the specific heat at constant pressure $\left(c_{p,water}\right)$ for water as $4.18\text{kJ}/\text{kg}\cdot °\text{C}$.

Calculate the mass flow of the chicken.

  ${\dot{m}}_{chicken}=n\times m_{chicken}$

  $\begin{array}{c}{\dot{m}}_{chicken}=\left(250\text{chicken}/\text{hr}\right)\left(2.2\text{kg}/\text{chicken}\right)\\ =550\text{kg}/\text{hr}\left(\frac{\text{1}\text{hr}}{\text{3600}\text{s}}\right)\\ =0.1528\text{kg}/\text{s}\end{array}$

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

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

Here, rate of net energy transfer in to 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 $\Delta {\dot{E}}_{system}$.

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

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

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

Here, mass flow rate is $\dot{m}$, enthalpy at initial state is $h_1$, enthalpy at final state is $h_2$, rate of heat transfer input is ${\dot{Q}}_{in}$, entry temperature is $T_1$ and exit temperature is $T_2$.

Refer Equation (II),

Calculate the rate of heat removal from the chicken.

  ${\dot{Q}}_{chicken}={\dot{m}}_{chicken}{c}_{p,chicken}\left(T_1-T_2\right)$        (IV)

  $\begin{array}{c}{\dot{Q}}_{chicken}=\left(0.1528\text{kg}/\text{s}\right)\left(3.54\text{kJ}/\text{kg}\cdot \text{K}\right)\left(15°\text{C}-3°\text{C}\right)\\ =\left(0.1528\text{kg}/\text{s}\right)\left(3.54\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\left(15+273\right)\text{K}-\left(3+273\right)\text{K}\right)\\ =6.49\text{kJ}/\text{s}\left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =6.49\text{kW}\end{array}$

Thus, the rate of heat removal from the chicken is $6.49\text{kW}$.

b)

To determine

The rate of entropy generation during the process.

Explanation of Solution

Calculate the total rate of heat gained by the water $\left({\dot{Q}}_{water}\right)$.

  $\begin{array}{c}{\dot{Q}}_{water}={\dot{Q}}_{chicken}+{\dot{Q}}_{heatgain}\\ {\dot{Q}}_{water}=6.49kW+150\text{kJ}/\text{hr}\\ =6.49kW+150\text{kJ}/\text{hr}\left(\frac{\text{1}\text{hr}}{\text{3600}\text{s}}\right)\\ =6.49\text{kW}+0.041\text{67}\text{kJ}/\text{s}\left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =6.532\text{kW}\end{array}$

Calculate mass flow rate of water $\left({\dot{m}}_{water}\right)$.

  ${\dot{Q}}_{water}={\dot{m}}_{water}\left(c_{p,water}\Delta T_{water}\right)$

  $\begin{array}{l}\text{6}\text{.532}\text{kW}={\dot{m}}_{water}\left(4.18\text{kJ}/\text{kg}\cdot °\text{C}\times 2°\text{C}\right)\\ \text{6}\text{.532}\text{kW}\left(\frac{1\text{kJ}/\text{s}}{1\text{kW}}\right)={\dot{m}}_{water}\left(4.18\text{kJ}/\text{kg}\cdot °\text{C}\times 2°\text{C}\right)\\ {\dot{m}}_{water}=0.781\text{kg}/\text{s}\end{array}$

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

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

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 ${\dot{S}}_{in}={\dot{m}}_1{s}_1+{\dot{m}}_3{s}_3+\frac{{\dot{Q}}_{in}}{T_{surr}}$, ${\dot{S}}_{out}={\dot{m}}_2{s}_2+{\dot{m}}_4{s}_4$, and $\Delta {\dot{S}}_{system}=0$ in Equation (V).

  $\begin{array}{l}{\dot{m}}_1{s}_1+{\dot{m}}_3{s}_3+\frac{{\dot{Q}}_{in}}{T_{surr}}-\left({\dot{m}}_2{s}_2+{\dot{m}}_4{s}_4\right)+{\dot{S}}_{gen}=0\\ {\dot{m}}_1{s}_1+{\dot{m}}_3{s}_3+\frac{{\dot{Q}}_{in}}{T_{surr}}-{\dot{m}}_2{s}_2-{\dot{m}}_4{s}_4+{\dot{S}}_{gen}=0\end{array}$

  $\begin{array}{c}{\dot{S}}_{gen}=-{\dot{m}}_{chicken}{s}_1-{\dot{m}}_{water}{s}_3-\frac{{\dot{Q}}_{in}}{T_{surr}}+{\dot{m}}_{chicken}{s}_2+{\dot{m}}_{water}{s}_4\\ ={\dot{m}}_{chicken}\left(s_2-s_1\right)+{\dot{m}}_{water}\left(s_4-s_3\right)-\frac{{\dot{Q}}_{in}}{T_{surr}}\\ ={\dot{m}}_{chicken}{c}_{p,chicken}\ln \left(\frac{T_2}{T_1}\right)+{\dot{m}}_{water}{c}_{p,water}\ln \left(\frac{T_4}{T_3}\right)-\frac{{\dot{Q}}_{in}}{T_{surr}}\end{array}$

  ${\dot{S}}_{gen}=\left\{\begin{array}{l}\left(0.1528\text{kg}/\text{s}\right)\left(3.54\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{3°\text{C}}{15°\text{C}}\right)\\ +\left(0.781\text{kg}/\text{s}\right)\left(4.18\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{2.5°\text{C}}{0.5°\text{C}}\right)\\ -\frac{0.0417kW}{25°\text{C}}\end{array}\right\}$

  $\begin{array}{c}{\dot{S}}_{gen}=\left\{\begin{array}{l}\left(0.1528\text{kg}/\text{s}\right)\left(3.54\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{\left(3+273\right)\text{K}}{\left(15+273\right)\text{K}}\right)\\ +\left(0.781\text{kg}/\text{s}\right)\left(4.18\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{\left(2.5+273\right)\text{K}}{\left(0.5+273\right)\text{K}}\right)\\ -\frac{0.0417kW}{\left(25+273\right)\text{K}}\end{array}\right\}\\ =0.000625\text{kJ}/\text{s}\cdot \text{K}\left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =0.000625\text{kW}/\text{K}\end{array}$

Thus, the rate of entropy generation during the process is $0.000625\text{kW}/\text{K}$.