Chapter 6, Problem 143RQ

(a)

To determine

The rate of heat removal from the eggs, in $\text{Btu/h}$.

Explanation of Solution

Given:

The Density of egg $\left(\rho \right)$ is $\text{67}{\text{.4 lbm/ft}}^{\text{3}}$.

The Specific heat of egg $\left(c_{p,egg}\right)$ is $\text{0}\text{.80 Btu/lbm·°F}$.

The average mass of egg $\left(m\right)$ is 0.14 lbm.

The initial temperature of egg $\left(T_1\right)$ is $\text{90°F}$.

The final average temperature of egg $\left(T_2\right)$ is 50°F.

The change in temperature of air $\left(T\right)$ is $34°\text{F}$.

The flow rate of egg is 3000 egg/hour.

Calculation:

Consider the system is in steady state. Hence, the inlet and exit mass flow rates are equal.

The mass flow rate of eggs are as follows.

  ${\dot{m}}_1={\dot{m}}_2={\dot{m}}_{\text{eggs}}$

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

  $\begin{array}{c}{\dot{m}}_{eggs}=\left(3000\text{egg/h}\right)\left(0.14\text{lbm/egg}\right)\\ =420\text{lbm/h}\end{array}$

Write the energy rate balance equation for one inlet and one outlet system.

  $\left[{\dot{Q}}_{\text{1}}+{\dot{W}}_{\text{1}}+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+g{z}_1\right)\right]-\left[{\dot{Q}}_{\text{2}}+{\dot{W}}_{\text{2}}+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+g{z}_2\right)\right]=\Delta {\dot{E}}_{\text{system}}$        (I)

Here, the rate of heat transfer is $\dot{Q}$, the rate of work transfer is $\dot{W}$, the enthalpy is $h$ and the velocity is $V$, the gravitational acceleration is $g$, the elevation from the datum is $z$ and the rate of change in net energy of the system is $\Delta {\dot{E}}_{\text{system}}$; the suffixes 1 and 2 indicates the inlet and outlet of the system.

Consider the system is at steady state. Hence, the rate of change in net energy of the system becomes zero.

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

Neglect the work transfer, kinetic, and potential energy changes. The heat transfer occurs from eggs. The eggs cooled by the air i.e. the heat removed from the eggs.

Consider, the eggs alone enters the chiller. Here, the mass flow rate of the eggs is ${\dot{m}}_c$.

The Equations (I) reduced as follows.

  $\begin{array}{l}\left[0+0+{\dot{m}}_{\text{eggs}}\left(h_1+0+0\right)\right]-\left[{\dot{Q}}_2+0+{\dot{m}}_{\text{eggs}}\left(h_2+0+0\right)\right]=0\\ {\dot{m}}_{\text{eggs}}{h}_1-\left({\dot{Q}}_2+{\dot{m}}_{\text{eggs}}{h}_2\right)=0\\ {\dot{m}}_{\text{eggs}}{h}_1-{\dot{Q}}_2-{\dot{m}}_{\text{eggs}}{h}_2\\ {\dot{Q}}_2={\dot{m}}_{\text{eggs}}{h}_1-{\dot{m}}_{\text{eggs}}{h}_2\end{array}$

  ${\dot{Q}}_2={\dot{m}}_{\text{eggs}}\left(h_1-h_2\right)$        (II)

Write the formula for change in enthalpy $\left(h_1-h_2\right)$.

  $h_1-h_2=c_{p,egg}\left(T_1-T_2\right)$

Here, the specific heat of egg at constant pressure is $c_{p,egg}$, the temperature is $T$.

Substitute $c_{p,egg}\left(T_1-T_2\right)$ for $\left(h_1-h_2\right)$ in Equation (II).

  $\begin{array}{c}{\dot{Q}}_2={\dot{m}}_{egg}\left[c_{p,egg}\left(T_1-T_2\right)\right]\\ ={\dot{m}}_{egg}{c}_{p,egg}\left(T_1-T_2\right)\end{array}$        (III)

Substitute $420\text{lbm/h}$ for ${\dot{m}}_{eggs}$, $0.80\text{Btu/lbm}\cdot \text{°F}$ for $c_{p,egg}$, $90°\text{F}$ for $T_1$ and $50°\text{F}$ for $T_2$ in Equation (III).

  $\begin{array}{c}{\dot{Q}}_2=\left(420\text{lbm/h}\right)\left(0.80\text{Btu/lbm}\cdot \text{°F}\right)\left(90°\text{F}-50°\text{F}\right)\\ =\left(420\text{lbm/h}\right)\left(0.80\text{Btu/lbm}\cdot \text{°F}\right)\left(40°\text{F}\right)\\ =13440\text{Btu/h}\end{array}$

Here, ${\dot{Q}}_2$ is the heat removed from the egg and equal to the heat gained by the air.

${\dot{Q}}_2={\dot{Q}}_{\text{egg}}$

Thus, the rate of heat removal from the eggs is $13440\text{Btu/h}$.

(b)

To determine

The volumetric flow rate of air, in ${\text{ft}}^3\text{/h}$, if the temperature rise of air is not to exceed $10\text{}°\text{F}$.

Explanation of Solution

The heat gained by the air is equal to the total of heat loss by the eggs

The total heat gained by the air is as follows.

  ${\dot{Q}}_{air}={\dot{Q}}_{\text{eggs}}$

Write formula the mass flow rate of air $\left({\dot{m}}_{air}\right)$.

  ${\dot{m}}_{air}=\frac{{\dot{Q}}_{air}}{c_{p,air}\Delta T}$

  $\begin{array}{c}{\dot{m}}_{air}=\frac{13440\text{Btu/h}}{\left(0.24\text{Btu/lbm}\cdot \text{°F}\right)10°\text{F}}\\ =5600\text{lbm/h}\end{array}$

Refer Table A-2E, “Ideal-gas specific heats of various common gases”.

The specific heat $\left(c_{p,air}\right)$ and gas constant $\left(R\right)$ of air is $0.24\text{Btu/lbm}\cdot \text{°F}$ and $0.3704\text{psia}\cdot {\text{ft}}^3\text{/lbm}\cdot \text{R}\left(0.06855\text{Btu/lbm}\cdot \text{R}\right)$ respectively.

The atmospheric pressure of air is $14.7\text{psia}$.

Write the formula for density of air.

  ${\rho }_{air}=\frac{P}{RT}$

  $\begin{array}{c}{\rho }_{air}=\frac{14.7\text{psia}}{\left(0.3704\text{psia}\cdot {\text{ft}}^3\text{/lbm}\cdot \text{R}\right)\left(34°\text{F}\right)}\\ =\frac{14.7\text{psia}}{\left(0.3704\text{psia}\cdot {\text{ft}}^3\text{/lbm}\cdot \text{R}\right)\left(34+460\right)\text{R}}\\ =0.08034{\text{lbm/ft}}^3\end{array}$

Write the formula for volumetric flow rate of air.

  ${\dot{V}}_{air}=\frac{{\dot{m}}_{air}}{{\rho }_{air}}$

  $\begin{array}{c}{\dot{V}}_{air}=\frac{5600\text{lbm/h}}{0.08034{\text{lbm/ft}}^3}\\ =69705.7524{\text{ft}}^3\text{/h}\\ \simeq 69700{\text{ft}}^3\text{/h}\end{array}$

Thus, the volumetric flow rate of air, in ${\text{ft}}^3\text{/h}$, if the temperature rise of air is not to exceed $10\text{}°\text{F}$ is $69700{\text{ft}}^3\text{/h}$.