Question: An ideal heat pump is used to maintain the inside temperature of a house at \({T_{{\rm{in}}}} = 22{\rm{^\circ C}}\) when the outside temperature is \({T_{{\rm{out}}}}\). Assume that when it is operating, the heat pump does work at a rate of 1500 W. Also assume that the house loses heat via conduction through its walls and other surfaces at a rate given by \(\left( {650\;{{\rm{W}} \mathord{\left/

{\vphantom {{\rm{W}} {{\rm{^\circ C}}}}} \right.} {{\rm{^\circ C}}}}} \right)\left( {{T_{{\rm{in}}}} - {T_{{\rm{out}}}}} \right)\). (a) For what outside temperature would the heat pump have to operate all the time in order to maintain the house at an inside temperature of 22°C? (b) If the outside temperature is 8°C, what percentage of the time does the heat pump have to operate in order to maintain the house at an inside temperature of 22°C?

A heat pump is a machine that moves heat energy from a lower-temperature reservoir to a higher-temperature reservoir by doing work.

The inside temperature of the house is \({T_{{\rm{in}}}} = 22{\rm{^\circ C}}\).

The outside temperature of the house is \({T_{{\rm{out}}}}\).

The rate of work done or power is \(P = 1500\;{\rm{W}}\).

The heat rejection rate from the house is \({Q_{\rm{R}}} = \left( {650\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{\rm{^\circ C}}}}} \right.} {{\rm{^\circ C}}}}} \right)\left( {{T_{{\rm{in}}}} - {T_{{\rm{out}}}}} \right)\).

(a)

The expression for the power is

\(P = \frac{W}{{\Delta t}}\).

Here, \(W\) is the work done and \(\Delta t\) is the time.

The expression for the rate of heat input to the house is

\({Q_{\rm{S}}} = \frac{{{Q_{\rm{L}}} + W}}{{\Delta t}}\).

Here, \({Q_{\rm{L}}}\) is the heat rejected from the house.

From the law of conservation of energy, the rate of energy supplied is equal to the rate of energy lost for a given interval of time.

\(\begin{aligned}{c}{Q_{\rm{S}}} &= {Q_{\rm{R}}}\\\frac{{{Q_{\rm{L}}} + W}}{{\Delta t}} &= \left( {650\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{\rm{^\circ C}}}}} \right.} {{\rm{^\circ C}}}}} \right)\left( {{T_{{\rm{in}}}} - {T_{{\rm{out}}}}} \right)\\{Q_{\rm{L}}} + W &= \left( {\left( {650\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{\rm{^\circ C}}}}} \right.} {{\rm{^\circ C}}}}} \right)\left( {{T_{{\rm{in}}}} - {T_{{\rm{out}}}}} \right)} \right)\end{aligned}\)

The expression for the efficiency of the heat pump is

\(e = 1 - \frac{{{T_{{\rm{out}}}}}}{{{T_{{\rm{in}}}}}}\). … (i)

The expression for the efficiency of the heat pump in terms of work done and heat rejected is

\(e = \frac{W}{{{Q_{\rm{L}}} + W}}\). … (ii)

Equate equations (i) and (ii).

\(\begin{aligned}{c}\frac{W}{{{Q_{\rm{L}}} + W}} &= 1 - \frac{{{T_{{\rm{out}}}}}}{{{T_{{\rm{in}}}}}}\\{Q_{\rm{L}}} + W &= \left( {\frac{{{T_{{\rm{in}}}}}}{{{T_{{\rm{in}}}} - {T_{{\rm{out}}}}}}} \right)W\\\left( {\left( {650\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{\rm{^\circ C}}}}} \right.} {{\rm{^\circ C}}}}} \right)\left( {{T_{{\rm{in}}}} - {T_{{\rm{out}}}}} \right)} \right)\Delta t &= \left( {\frac{{{T_{{\rm{in}}}}}}{{{T_{{\rm{in}}}} - {T_{{\rm{out}}}}}}} \right)W\\\left( {\left( {650\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{\rm{^\circ C}}}}} \right.} {{\rm{^\circ C}}}}} \right)\left( {{T_{{\rm{in}}}} - {T_{{\rm{out}}}}} \right)} \right) &= \frac{W}{{\Delta t}}\left( {\frac{{{T_{{\rm{in}}}}}}{{{T_{{\rm{in}}}} - {T_{{\rm{out}}}}}}} \right)\end{aligned}\)

Solve further as shown below:

\(\begin{aligned}{c}{\left( {{T_{{\rm{in}}}} - {T_{{\rm{out}}}}} \right)^2} &= \frac{{W{T_{{\rm{in}}}}}}{{\left( {650\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{\rm{^\circ C}}}}} \right.} {{\rm{^\circ C}}}}} \right)\Delta t}}\\\left( {{T_{{\rm{in}}}} - {T_{{\rm{out}}}}} \right) &= \sqrt {\frac{{W{T_{{\rm{in}}}}}}{{\left( {650\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{\rm{^\circ C}}}}} \right.} {{\rm{^\circ C}}}}} \right)\Delta t}}} \\{T_{{\rm{out}}}} &= {T_{{\rm{in}}}} - \sqrt {\frac{{W{T_{{\rm{in}}}}}}{{\left( {650\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{\rm{^\circ C}}}}} \right.} {{\rm{^\circ C}}}}} \right)\Delta t}}} \end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}{T_{{\rm{out}}}} &= \left( {\left( {22{\rm{^\circ C}} + 273} \right)\;{\rm{K}}} \right) - \sqrt {\frac{{\left( {1500\;{\rm{W}}} \right)\left( {\left( {22{\rm{^\circ C}} + 273} \right)\;{\rm{K}}} \right)}}{{\left( {650\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{\rm{^\circ C}}}}} \right.} {{\rm{^\circ C}}}}} \right)}}} \\{T_{{\rm{out}}}} &= \left( {269\;{\rm{K - 273}}} \right){\rm{^\circ C}}\\{T_{{\rm{out}}}} &= - 4.0{\rm{^\circ C}}\end{aligned}\)

Thus, the required outside temperature is \( - 4.0{\rm{^\circ C}}\).

(b)

In this part, it is given that the outside temperature is \({T_{{\rm{out}}}} = 8{\rm{^\circ C}}\).

The rate of heat rejection is calculated below:

\(\begin{aligned}{c}{Q_{\rm{R}}} &= \left( {650\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{\rm{^\circ C}}}}} \right.} {{\rm{^\circ C}}}}} \right)\left( {{T_{{\rm{in}}}} - {T_{{\rm{out}}}}} \right)\\{Q_{\rm{R}}} &= \left( {650\;{{\rm{W}} \mathord{\left/{\vphantom {{\rm{W}} {{\rm{^\circ C}}}}} \right.} {{\rm{^\circ C}}}}} \right)\left( {22{\rm{^\circ C}} - 8{\rm{^\circ C}}} \right)\\{Q_{\rm{R}}} &= 9100\;{\rm{W}}\end{aligned}\)

The rate of work done by the pump is calculated below:

\(\begin{aligned}{c}{\left( {\frac{W}{{\Delta t}}} \right)_{\rm{P}}} &= {Q_R}\left( {\frac{{{T_{{\rm{in}}}} - {T_{{\rm{out}}}}}}{{{T_{{\rm{in}}}}}}} \right)\\{\left( {\frac{W}{{\Delta t}}} \right)_{\rm{P}}} &= \left( {9100\;{\rm{W}}} \right)\left( {\frac{{\left( {\left( {22{\rm{^\circ C}} + 273} \right)\;{\rm{K}}} \right) - \left( {\left( {{\rm{8^\circ C}} + 273} \right)\;{\rm{K}}} \right)}}{{\left( {\left( {22{\rm{^\circ C}} + 273} \right)\;{\rm{K}}} \right)}}} \right)\\{\left( {\frac{W}{{\Delta t}}} \right)_{\rm{P}}} &= 432\;{\rm{W}}\end{aligned}\)

The percentage of time that the pump has to work is calculated below:

\(\begin{aligned}{c}t\left( \% \right) &= \frac{{{{\left( {\frac{W}{{\Delta t}}} \right)}_{\rm{P}}}}}{P}\\t\left( \% \right) &= \left( {\frac{{432\;{\rm{W}}}}{{1500\;{\rm{W}}}}} \right) \times 100\% \\t\left( \% \right) &= 29\% \end{aligned}\)

Thus, the pump has to work for \(29\% \) of the total time.