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}


Learn more about this topic:

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Thermal Expansion & Heat Transfer

from High School Physics: Help and Review

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