Chapter 5.5, Problem 156RP

(a)

To determine

The rate of water must supplied to maintain steady operation.

Answer to Problem 156RP

The rate of water must supplied to maintain steady operation is $0.0015\text{kg/s}$.

Explanation of Solution

The rate of water must be supplied to maintain the steady state operation is equal to the rate of water removed by the bottles.

The rater water removed by the bottles is expressed as follows.

${\dot{m}}_{\text{water,out}}=\left(\text{Flow rate of bottle}\right)\left(\text{water removed per bottle}\right)$ (I)

Conclusion:

Substitute $450\text{bottle/min}$ for Flow rate of bottle and $0.2\text{g/bottle}$ for water removed per bottle in Equation (I).

$\begin{array}{c}{\dot{m}}_{\text{water,out}}=\left(450\text{bottle/min}\right)\left(0.2\text{g/bottle}\right)\\ =\left(450\frac{\text{bottle}}{\text{min}}\times \frac{1\text{min}}{60\text{s}}\right)\left(0.2\frac{\text{g}}{\text{bottle}}\times \frac{1\text{kg}}{1000\text{g}}\right)\\ =0.0015\text{kg/s}\end{array}$

Thus, The rate of water must supplied to maintain steady operation is $0.0015\text{kg/s}$.

(b)

To determine

The rate of heat must supplied to maintain steady operation.

Answer to Problem 156RP

The rate of heat must supplied to maintain steady operation is $27.22\text{}\text{kW}$.

Explanation of Solution

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

The mass flow rate of bottles are as follows.

${\dot{m}}_1={\dot{m}}_2={\dot{m}}_b$

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

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

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 water bath to bottles. The bottles heated by the hot water bath i.e. the heat gained by the bottles.

The Equations (II) reduced as follows for bottles.

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

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

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

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

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

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

Here, ${\dot{Q}}_2$ is the heat removed from the hot water bath by the glass bottles.

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

Write the formula for heat removed by the water that is carried by the bottle.

${\dot{Q}}_{\text{water, out}}={\dot{m}}_{\text{water,out}}{c}_{p,w}\left(T_{2,w}-T_{1,w}\right)$ (V)

Here, the specific heat of water is $c_{p,w}$ and the initial and final temperatures of water is $T_{1,w}\text{and }{T}_{2,w}$.

The rate of heat must supplied to maintain steady operation is equal to the total heat removed by the glass bottles and water carried by the bottles.

The total heat removed from the hot water bath is expressed as follows.

${\dot{Q}}_{\text{total, removed}}={\dot{Q}}_{\text{bottle}}+{\dot{Q}}_{\text{water, out}}$ (VI)

Refer Table A-3(a), “Properties of common liquids, solids, and foods”.

The specific heat of water $\left(c_{p,w}\right)$ is $4.18\text{kJ/kg}\cdot \text{°C}$.

Refer Table A-3(b), “Properties of common liquids, solids, and foods”.

The specific heat corresponding to glass, window (glass bottles) $\left(c_{p,b}\right)$ is $0.8\text{kJ/kg}\cdot \text{°C}$.

Conclusion:

Here, the bottles enters the hot water bath at the rate of 450 bottle per minute and each bottle weighs $150\text{}\text{g}$.

Thus, the mass flow rate of bottles $\left({\dot{m}}_b\right)$ is expressed as follows.

$\begin{array}{c}{\dot{m}}_b=\left(450\text{bottle/min}\right)\left(150\text{g/bottle}\right)\\ =\left(450\frac{\text{bottle}}{\text{min}}\times \frac{1\text{min}}{60\text{s}}\right)\left(150\frac{\text{g}}{\text{bottle}}\times \frac{1\text{}\text{kg}}{1000\text{g}}\right)\\ =1.125\text{kg/s}\end{array}$

Substitute $1.125\text{kg/s}$ for ${\dot{m}}_b$, $0.8\text{kJ/kg}\cdot \text{°C}$ for $c_{p,b}$, $50°\text{C}$ for $T_2$ and $20°\text{C}$ for $T_1$ in Equation (IV).

$\begin{array}{c}{\dot{Q}}_2=\left(1.125\text{kg/s}\right)\left(0.8\text{kJ/kg}\cdot \text{°C}\right)\left(50°\text{C}-20°\text{C}\right)\\ =\left(1.125\text{kg/s}\right)\left(0.8\text{kJ/kg}\cdot \text{°C}\right)\left(30°\text{C}\right)\\ =27\text{kJ/s}\times \frac{1\text{kW}}{1\text{kJ/s}}\\ =27\text{kW}\end{array}$

The heat removed by the bottles is,

${\dot{Q}}_2={\dot{Q}}_{\text{bottle}}=27\text{kW}$

Substitute $0.0015\text{kg/s}$ for ${\dot{m}}_{\text{water,out}}$, $4.18\text{kJ/kg}\cdot \text{°C}$ for $c_{p,w}$, $50°\text{C}$ for $T_{2,w}$ and $15°\text{C}$ for $T_{1,w}$ in Equation (V).

$\begin{array}{c}{\dot{Q}}_{\text{water, out}}=\left(0.0015\text{kg/s}\right)\left(4.18\text{kJ/kg}\cdot \text{°C}\right)\left(50°\text{C}-15°\text{C}\right)\\ =\left(0.0015\text{kg/s}\right)\left(4.18\text{kJ/kg}\cdot \text{°C}\right)\left(35°\text{C}\right)\\ =0.2195\text{kJ/s}\times \frac{1\text{kW}}{1\text{kJ/s}}\\ =0.2195\text{kW}\end{array}$

Substitute $27\text{kW}$ for ${\dot{Q}}_{\text{bottle}}$ and $0.2195\text{kW}$ for ${\dot{Q}}_{\text{water, out}}$ in Equation (VI).

$\begin{array}{c}{\dot{Q}}_{\text{total, removed}}=27\text{kW}+0.2195\text{kW}\\ =27.2195\text{kW}\\ \simeq \text{27}\text{.22}\text{kW}\end{array}$

Thus, the rate of heat must supplied to maintain steady operation is $27.22\text{}\text{kW}$.