Chapter 7, Problem 127RQ

(a)

To determine

The rate of heat removal from the breads.

Explanation of Solution

Given:

The number of loaves per hours is $1200\text{breads}/\text{hr}$.

The average mass per bread is $350\text{g}/\text{bread}$.

The mass flow rate of breads loves $\left(\dot{m}\right)$ is $420\text{kg}/\text{h}$.

The specific heat of bread loves $\left(c_p\right)$ is $2.93\text{kJ}/\text{kg}\cdot °\text{C}$.

The latent heat of bread loves $\left(c_p\right)$ is $109.3\text{kJ}/\text{kg}$.

The initial temperature of the bread $\left(T_1\right)$ is $30\text{°C}$.

The final temperature of the bread $\left(T_2\right)$ is $-10°\text{C}$.

The COP of the refrigerator $\left(\text{COP}\right)$ is 1.2.

Calculation:

Calculate the mass flow rate of the bread.

  ${\dot{m}}_{bread}=\left(\text{no}\text{.}\text{of}\text{loaves}\text{per}\text{hours}\right)\times \left(\text{average}\text{mass}\text{per bread}\right)$

  $\begin{array}{c}{\dot{m}}_{\text{bread}}=\left(1200\text{breads}/\text{hr}\right)\times \left(350\text{g}/\text{bread}\right)\\ =\left(1200\text{breads}/\text{hr}\right)\times \left(350\text{g}/\text{bread}\right)\times \left(\frac{1\text{kg}}{1000\text{g}}\right)\\ =420\text{kg}/\text{h}\times \left(\frac{1\text{kg}/\text{s}}{3600}\right)\\ =0.1167\text{kg}/\text{s}\end{array}$

Calculate the average temperature of bread.

  $T_{\text{avg}}=\frac{\left(T_1+T_2\right)}{2}$

  $\begin{array}{c}{T}_{\text{avg}}=\frac{\left(-30+(-22)\right)°\text{C}}{2}\\ =-26°\text{C}\\ =-26°\text{C+273}\\ =\text{247}\text{K}\end{array}$

Refer the Table A-1, “Molar mass, gas constant, and critical-point properties” to obtain value of gas constant of air as $0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$.

Calculate the rate of removal of heat from the breads.

  $\begin{array}{c}{\dot{Q}}_{\text{bread}}=\dot{m}{c}_p\Delta T\\ =\dot{m}{c}_p\left(T_1-T_2\right)\end{array}$

  $\begin{array}{c}{\dot{Q}}_{\text{bread}}=\left(420\text{kg}/\text{h}\right)\left(2.93\text{kJ}/\text{kg}\cdot °\text{C}\right)\left(30-\left(-10\right)\right)°\text{C}\\ =\left(1230.6\text{kJ}/\text{h}\cdot \text{°C}\right)\left(40°\text{C}\right)\\ =49224\text{kJ}/\text{h}\end{array}$

Calculate the rate of cooling of bread.

  ${\dot{Q}}_{\text{freezing}}={\left(\dot{m}{h}_{\text{latent}}\right)}_{\text{bread}}$

  $\begin{array}{c}{\dot{Q}}_{\text{freezing}}=\left(420\text{kg}/\text{h}\right)\times \left(109.3\text{kJ}/\text{kg}\right)\\ =45,906\text{kJ}/\text{h}\end{array}$

Calculate the total rate of heat removal from bread.

  ${\dot{Q}}_{\text{total}}={\dot{Q}}_{\text{bread}}+{\dot{Q}}_{\text{freezing}}$

  $\begin{array}{c}{\dot{Q}}_{\text{total}}=\left(49224\text{kJ}/\text{h}\right)+\left(45,906\text{kJ}/\text{h}\right)\\ =95,310\text{kJ}/\text{h}\end{array}$

Thus, the rate of heat removal from the breads is $\underset{\_}{95,130\text{kJ}/\text{min}}$.

(b)

To determine

The required volume flow rate of air.

Explanation of Solution

Refer the Table A-2, “Ideal-gas specific heats of various common gases” to obtain value of specific heat of pressure of air at approx. $250\text{K}$ temperature as $1.003\text{kJ}/\text{kg}\cdot \text{K}$.

Calculate the mass flow rate of air.

  ${\dot{m}}_{\text{air}}=\frac{{\dot{Q}}_{\text{air}}}{{\left(c_p\Delta T\right)}_{\text{air}}}$

  $\begin{array}{c}{\dot{m}}_{\text{air}}=\frac{\left(95130\text{kJ}/\text{h}\right)}{\left(1.0\text{kJ}/\text{kg}\cdot \text{K}\right)\left(8°\text{C}\right)}\\ =\frac{\left(95130\text{kJ}/\text{h}\right)}{\left(1.0\text{kJ}/\text{kg}\cdot °\text{C}\right)\left(8°\text{C}\right)}\\ =\frac{\left(95130\text{kJ}/\text{h}\right)}{\left(8\text{kJ}/\text{kg}\right)}\\ =11,891\text{kg}/\text{h}\end{array}$

Calculate the density of the air.

  $\rho =\frac{P}{RT}$

  $\begin{array}{c}\rho =\frac{\left(101.3\text{kPa}\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\left(-30°\text{C}+273\right)}\\ =\frac{\left(101.3\text{kPa}\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\left(243\text{K}\right)}\\ =\frac{\left(101.3\text{kPa}\right)}{\left(69.741\text{kPa}\cdot {\text{m}}^3/\text{kg}\right)}\\ =1.4525\text{kg}/{\text{m}}^{\text{3}}\end{array}$

     $\cong 1.4525\text{kg}/{\text{m}}^{\text{3}}$

Calculate the volume flow rate of air.

  ${\dot{\nu }}_{\text{air}}=\frac{{\dot{m}}_{\text{air}}}{{\rho }_{\text{air}}}$

  $\begin{array}{c}{\dot{\nu }}_{\text{air}}=\frac{\left(11,891\text{kg}/\text{h}\right)}{\left(1.4525\text{kg}/{\text{m}}^{\text{3}}\right)}\\ =8186.5{\text{m}}^{\text{3}}/\text{h}\\ \cong 8187{\text{m}}^{\text{3}}/\text{h}\end{array}$

Thus, the required volume flow rate of air is $\underset{\_}{8187{\text{m}}^3/\text{h}}$.

(c)

To determine

The size of the compressor of the refrigeration system.

Explanation of Solution

Calculate the size of the compressor of the refrigeration system.

  ${\dot{W}}_{\text{R}}=\frac{{\dot{Q}}_{\text{R}}}{\text{COP}}$

  $\begin{array}{c}{\dot{W}}_{\text{R}}=\frac{\left(95130\text{kJ}/\text{h}\right)}{\left(1.2\right)}\\ =79275\text{kJ}/\text{h}\times \left(\frac{0.00027778\text{kW}}{1\text{kJ}/\text{h}}\right)\\ =22.02\text{kW}\end{array}$

Thus, the size of the compressor of the refrigeration system is $\underset{\_}{22.02\text{kW}}$.