Chapter 6.11, Problem 141RP

A refrigeration system is to cool bread loaves with an average mass of 350 g from 30 to −10°C at a rate of 1200 loaves per hour with refrigerated air at −30°C. Taking the average specific and latent heats of bread to be 2.93 kJ/kg·°C and 109.3 kJ/kg, respectively, determine (a) the rate of heat removal from the breads, in kJ/h; (b) the required volume flow rate of air, in m³/h, if the temperature rise of air is not to exceed 8°C; and (c) the size of the compressor of the refrigeration system, in kW, for a COP of 1.2 for the refrigeration system.

To determine

The rate of heat removal from the breads.

Answer to Problem 141RP

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

Explanation of Solution

Determine the mass flow rate of the bread.

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

Determine the average temperature of bread.

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

Here, the initial temperature of the bread is $T_1$ and the final temperature of the bread is $T_2$.

Determine the rate of removal of heat from the breads.

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

Here, the mass flow rate of breads is $\dot{m}$ and the specific heat of bread is $c_p$.

Determine the rate of cooling of bread.

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

Here, the latent enthalpy of the bread is $h_{\text{latent}}$.

Determine the total rate of heat removal from bread.

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

Conclusion:

Substitute $1200\text{breads}/\text{hr}$ for no. of loaves per hours and $350\text{g}/\text{bread}$ for average mass per bread in Equation (I).

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

Substitute $-30°\text{C}$ for $T_1$ and $-22°\text{C}$ for $T_2$ in Equation (II).

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

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

Substitute $420\text{kg}/\text{h}$ for $\dot{m}$, $2.93\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_p$, $30\text{°C}$ for $T_2$, and $-10°\text{C}$ for $T_1$ in Equation (III)

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

Substitute $109.3\text{kJ}/\text{kg}$ for $h_{\text{latent,bread}}$ and $420\text{kg}/\text{h}$ for $\dot{m}$ in Equation (IV).

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

Substitute $49224\text{kJ}/\text{h}$ for ${\dot{Q}}_{\text{bread}}$ and $45,906\text{kJ}/\text{h}$ for ${\dot{Q}}_{\text{freezing}}$ in Equation (V).

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

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

To determine

The required volume flow rate of air.

Answer to Problem 141RP

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

Explanation of Solution

Determine the mass flow rate of air.

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

Here, the rate of heat removal from air is ${\dot{Q}}_{\text{air}}$, the specific heat of air is $c_p$, and the temperature of air is $\Delta T_{\text{air}}$.

Determine the density of the air.

$\rho =\frac{P}{RT}$ (VII)

Here, the pressure in the air is $P$, the universal gas constant is $R$, and the temperature of the air is $T$.

Determine the volume flow rate of air.

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

Conclusion:

From 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 $1.003\text{kJ}/\text{kg}\cdot \text{K}$.

Substitute $95130\text{kJ}/\text{h}$ for ${\dot{Q}}_{\text{air}}$, $1.003\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $8°\text{C}$ for $\Delta T_{\text{air}}$ in Equation (VI).

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

Substitute 101.3 kPa for $P$, $0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R$, $-30°\text{C}$ for $T$ in Equation (VII).

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

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

Substitute $11,891\text{kg}/\text{h}$ for ${\dot{m}}_{\text{air}}$ and $1.4525\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ in Equation (VIII).

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

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

To determine

The size of the compressor of the refrigeration system.

Answer to Problem 141RP

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

Explanation of Solution

Determine the size of the compressor of the refrigeration system.

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

Conclusion:

Substitute $95130\text{kJ}/\text{h}$ for ${\dot{Q}}_{\text{R}}$ and 1.2 for COP in Equation (IX).

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

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