# A glass bottle washing facility uses a well-agitated hot-water bath at 50 C that is placed on the...

## Question:

A glass bottle washing facility uses a well-agitated hot-water bath at 50{eq}^{\circ} {/eq}C that is placed on the ground. The bottles enter at a rate of 450 per minute at an ambient temperature of 20{eq}^{\circ} {/eq}C and leave at the water temperature. Each bottle has a mass of 150 g and removes 0.2 g of water as it leaves the bath wet. Make-up water is supplied at 15{eq}^{\circ} {/eq}C. Disregarding any heat losses from the outer surfaces of the bath, determine the rate at which

(a) water and

(b) heat must be supplied to maintain steady operation.

## Total heat transfer:

Total heat transfer is the summation of heat transfer from various processes either the heat is adding in a process or getting rejected from the process. It is also named as the overall rate of heat transfer.

## Answer and Explanation:

**Given data**

- Bottle leaving temperature {eq}{T_1} = 50\;{\rm{^\circ C}} {/eq}

- Ambient temperature {eq}{T_2} = 20\;{\rm{^\circ C}}. {/eq}

- Water bath temperature {eq}{T_w} = 50\;{\rm{^\circ C}} {/eq}

- Rate of water entering {eq}N = {{450} {\left/ {\vphantom {{450} {{\rm{min}}}}} \right. } {{\rm{min}}}}. {/eq}

- Mass of bottle {eq}m = 150\;g {/eq}

- Mass carried by one bottle {eq}{m_{\left( {carried} \right)}} = 0.2\;g. {/eq}

- Make up temperature {eq}{T_{\left( {make - up} \right)}} = 15\;{\rm{^\circ C}}. {/eq}

Expression for mass flow rate is,

{eq}\dot m = mN {/eq}

Substituting the values,

{eq}\begin{align*} \dot m &= 0.15\;{\rm{kg}} \times \dfrac{{{\rm{450}}}}{{60}}\\ \dot m &= 1.125\;{{{\rm{kg}}} {\left/ {\vphantom {{{\rm{kg}}} {\rm{s}}}} \right. } {\rm{s}}} \end{align*} {/eq}

The amount of water getting out from the bottle,

{eq}{\dot m_{\left( {out} \right)}} = N \times {m_{\left( {carried} \right)}} {/eq}

Substituting the values,

{eq}\begin{align*} {{\dot m}_{\left( {out} \right)}} &= \dfrac{{450}}{{60}} \times 0.0002\\ {{\dot m}_{\left( {out} \right)}} &= 0.0015\;{{{\rm{kg}}} {\left/ {\vphantom {{{\rm{kg}}} {\rm{s}}}} \right. } {\rm{s}}} \end{align*} {/eq}

The expression for heat added to the bottle is,

{eq}{Q_{\left( {bottle} \right)}} = {\dot m_{\left( {bottle} \right)}} \times {c_g} \times \left( {{T_2} - {T_1}} \right) {/eq}

Here, {eq}{c_g} = 0.8\;{{{\rm{kJ}}} {\left/ {\vphantom {{{\rm{kJ}}} {{\rm{kg}} \cdot {\rm{K}}}}} \right. } {{\rm{kg}} \cdot {\rm{K}}}} {/eq}

Substituting the values,

{eq}\begin{align*} {{\dot Q}_{\left( {bottle} \right)}} &= 1.125 \times 0.8 \times \left( {50 - 20} \right)\\ {{\dot Q}_{\left( {bottle} \right)}} &= 27\;{\rm{kW}} \end{align*} {/eq}

The expression for heat rejected by makeup water is,

{eq}{\dot Q_{\left( {out} \right)}} = {\dot m_{\left( {out} \right)}} \times {c_w} \times \left( {{T_2} - {T_{(makeup}}} \right) {/eq}

Substituting the values,

{eq}\begin{align*} {{\dot Q}_{\left( {out} \right)}} &= 0.0015 \times 4.18 \times \left( {50 - 15} \right)\\ {{\dot Q}_{\left( {out} \right)}} &= 0.219\;{\rm{kW}} \end{align*} {/eq}

Total heat rate is,

{eq}{\dot Q_{\left( {total} \right)}} = {\dot Q_{\left( {out} \right)}} + {\dot Q_{\left( {bottle} \right)}} {/eq}

Substituting the values,

{eq}\begin{align*} {{\dot Q}_{\left( {total} \right)}} &= 27\;{\rm{kW}} + 0.219\;{\rm{kW}}\\ {{\dot Q}_{\left( {total} \right)}} &= 27.219\;{\rm{kW}} \end{align*} {/eq}

Thus, the rate of water leaving is {eq}0.0015\;{{{\rm{kg}}} {\left/ {\vphantom {{{\rm{kg}}} {\rm{s}}}} \right. } {\rm{s}}} {/eq} and the total heat supplied is {eq}27.219\;{\rm{kW}}. {/eq}