Question

Question: An ideal air conditioner keeps the temperature inside a room at 21°C when the outside temperature is 32°C. If 4.8 kW of power enters a room through the windows the in form of direct radiation from the Sun, how much electrical power would be saved if the windows were shaded so only 500 W enters?

Short answer

The electric power that would be saved is \(161\;{\rm{W}}\).

Step-by-step solution

Determination of heat input to the heat engine

The heat input to the heat engine is the sum of the work performed by the engine and the heat rejected from the engine.

Given information

The inside temperature of the room is\({T_{\rm{L}}} = 21{\rm{^\circ C}}\).

The outside temperature of the room is \({T_{\rm{H}}} = 32{\rm{^\circ C}}\).

Evaluation of the amount of electric power that would save

The expression for the heat input to the system is

\({Q_{\rm{H}}} = W + {Q_{\rm{L}}}\).

Here, \(W\) is the work done and \({Q_{\rm{L}}}\) is heat output from the system.

The expression for the efficiency of the heat engine is

\(\begin{aligned}{c}e &= \frac{W}{{{Q_{\rm{H}}}}}\\e &= \frac{W}{{W + {Q_{\rm{L}}}}}.\end{aligned}\) … (i)

The expression for the efficiency of the heat engine in terms of temperature is

\(e = 1 - \frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}}}}\). … (ii)

Equate equations (i) and (ii).

\(\begin{aligned}{c}\frac{W}{{W + {Q_{\rm{L}}}}} &= 1 - \frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}}}}\\W &= {Q_{\rm{L}}}\left( {\frac{{{T_{\rm{H}}}}}{{{T_{\rm{L}}}}} - 1} \right)\\\frac{W}{t} &= \frac{{{Q_{\rm{L}}}}}{t}\left( {\frac{{{T_{\rm{H}}}}}{{{T_{\rm{L}}}}} - 1} \right)\end{aligned}\)

The electric power required for \(4.8\;{\rm{kW}}\) radiation is calculated below:

\(\begin{aligned}{l}{\left( {\frac{W}{t}} \right)_{4.8\;{\rm{kW}}}} &= \left( {4.8\;{\rm{kW}}} \right)\left( {\frac{{{{10}^3}\;{\rm{W}}}}{{1\;{\rm{kW}}}}} \right)\left( {\left( {\frac{{\left\{ {\left( {32{\rm{^\circ C}} + 273} \right)} \right\}\;{\rm{K}}}}{{\left\{ {\left( {{\rm{21^\circ C}} + 273} \right)} \right\}\;{\rm{K}}}}} \right) - 1} \right)\\{\left( {\frac{W}{t}} \right)_{4.8\;{\rm{kW}}}} &= 179.59\;{\rm{W}}\end{aligned}\)

The electric power required for \(500\;{\rm{W}}\) radiation is calculated below:

\(\begin{aligned}{l}{\left( {\frac{W}{t}} \right)_{500\;{\rm{W}}}} &= \left( {500\;{\rm{W}}} \right)\left( {\left( {\frac{{\left\{ {\left( {32{\rm{^\circ C}} + 273} \right)} \right\}\;{\rm{K}}}}{{\left\{ {\left( {{\rm{21^\circ C}} + 273} \right)} \right\}\;{\rm{K}}}}} \right) - 1} \right)\\{\left( {\frac{W}{t}} \right)_{500\;{\rm{W}}}} &= 18.7\;{\rm{W}}\end{aligned}\)

The amount of electric power that would be saved is calculated below:

\(\begin{aligned}{l}W &= {\left( {\frac{W}{t}} \right)_{4.8\;{\rm{kW}}}} - {\left( {\frac{W}{t}} \right)_{500\;{\rm{W}}}}\\W &= \left( {179.59\;{\rm{W}}} \right) - \left( {18.7\;{\rm{W}}} \right)\\W &= 161\;{\rm{W}}\end{aligned}\)

Thus, the amount of electric power that would be saved is \(161\;{\rm{W}}\).