Question

An air-conditioning system operating on the reversed Carnot cycle is required to transfer heat from a house at a rate of \(750 \mathrm{kJ} / \mathrm{min}\) to maintain its temperature at \(24^{\circ} \mathrm{C} .\) If the outdoor air temperature is \(35^{\circ} \mathrm{C},\) determine the power required to operate this air-conditioning system.

Short answer

Answer: The power required to operate the air-conditioning system is approximately \(\frac{12.5 \times 11}{297} kW\).

Step-by-step solution

Calculate the Coefficient of Performance

For a reversed Carnot cycle, the Coefficient of Performance (COP) is given by the formula: COP_{R} = \frac{T_{L}}{T_{H} - T_{L}} Where, \(T_{L}\) is the temperature of the house which is \(297K\) (converting from \(24^{\circ}C\) to Kelvin by adding \(273\)) \(T_{H}\) is the temperature of the outdoor air which is \(308K\) (converting from \(35^{\circ}C\) to Kelvin by adding \(273\)) Plugging in the given temperatures, we get: COP_{R} = \frac{297}{308 - 297} = \frac{297}{11}

Calculate the work input required

Now, we can use COP_{R} to find the work input for the air-conditioning system. The formula is: W_{in} = \frac{Q_{L}}{COP_{R}} Where, \(Q_{L}\) is the heat removed from the house which is \(750 kJ/min\). We can convert this to seconds by dividing by \(60\): \(Q_{L} = \frac{750}{60} kJ/s = 12.5 kJ/s\) Now, plugging in the values for \(Q_{L}\) and COP_{R}, we get: W_{in} = \frac{12.5}{\frac{297}{11}} = \frac{12.5 \times 11}{297}

Calculate the power required

We can now use the work input required to find the power needed. Since work input is already in \(kJ/s\), we don't have to do any conversion and can directly express the power required in \(kW\). P = W_{in} = \frac{12.5 \times 11}{297} kW So the power required to operate the air-conditioning system is \(\frac{12.5 \times 11}{297} kW\).