Chapter 7.13, Problem 146P

a)

To determine

The rate of heat removal from the chicken.

Answer to Problem 146P

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

Explanation of Solution

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

${\dot{E}}_{in}-{\dot{E}}_{out}=\Delta {\dot{E}}_{system}$   $chicken$ (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 $\Delta {\dot{E}}_{system}$.

Write the expression to calculate the mass flow of the chicken.

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

Here, average mass of the chicken is $m_{chicken}$ and number of chickens dropped into the chiller is $n$.

Conclusion:

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

Substitute 0 for $\Delta {\dot{E}}_{system}$ in Equation (1).

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

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 $T_2$.

From Equation (II) write the expression to 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)

Here, the mass flow rate of chicken is ${\dot{m}}_{chicken}$, rate of heat removal from the chicken is ${\dot{Q}}_{chicken}$ and specific heat at constant

pressure for chicken is $c_{p,chicken}$.

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 $4.18\text{kJ}/\text{kg}\cdot °\text{C}$.

Substitute $2.2\text{kg}/\text{chicken}$ for $m_{chicken}$ and $250\text{chicken}/\text{hr}$ for $n$ in Equation (II).

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

Substitute $0.1528\text{kg}/\text{s}$ for ${\dot{m}}_{chicken}$, $3.54\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $15°\text{C}$ for $T_1$, and $3°\text{C}$ for $T_2$ in Equation (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.

Answer to Problem 146P

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

Explanation of Solution

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 $\Delta {\dot{S}}_{system}$.

Write the expression to calculate the total rate of heat gained by the water.

${\dot{Q}}_{water}={\dot{Q}}_{chicken}+{\dot{Q}}_{heatgain}$ (VI)

Here, total rate of heat gained by the water is ${\dot{Q}}_{water}$.

Write the expression to calculate the total rate of heat gained by the water ${\dot{Q}}_{water}$.

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

Here, mass flow rate of water is ${\dot{m}}_{water}$, specific heat at constant pressure for water is $c_{p,water}$ and change in temperature of water is $\Delta T_{water}$.

Conclusion:

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

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

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

Here, Mass flow rate at stage 1 and 2 are chicken and stage 3 and 4 are water , entropy at stage 1 is $s_1$, entropy at stage 2 is $s_2$, entropy at stage 3 is $s_3$, entropy at stage 4 is $s_4$, mass flow rate of chicken is ${\dot{m}}_{chicken}$, mass flow rate of water is ${\dot{m}}_{water}$, initial chiller temperature $T_3$ , final chiller temperature $T_4$.

Substitute 6.49 kW for ${\dot{Q}}_{chicken}$ and $150\text{kJ}/\text{hr}$ for ${\dot{Q}}_{heatgain}$ in Equation (VI).

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

Substitute $\text{6}\text{.532}\text{kW}$ for ${\dot{Q}}_{water}$, $4.18\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_{p,water}$ and $2°\text{C}$ for $\Delta T$ in Equation (VII).

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

Substitute $0.1528\text{kg}/\text{s}$ for ${\dot{m}}_{chicken}$, $3.54\text{kJ}/\text{kg}\cdot \text{K}$ for $c_{p,chicken}$, $3°\text{C}$ for $T_2$, $15°\text{C}$ for $T_1$.$0.781\text{kg}/\text{s}$ for ${\dot{m}}_{water}$, $4.18\text{kJ}/\text{kg}\cdot \text{K}$ for $c_{p,water}$, $0.5°\text{C}$ for $T_3$, $2.5°\text{C}$ for $T_4$, 0.0417 kW for ${\dot{Q}}_{in}$ and $25°\text{C}$ for $T_{surr}$ in Equation (VIII).

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

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

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