Question 7.48: When a man returns to his well-sealed house on a summer day,......

An air conditioner with a known COP cools a house to desired temperature in 15 \mathrm{~min}. The power consumption of the air conditioner is to be determined.

Assumptions 1 The air conditioner operates steadily. 2 The house is well-sealed so that no air leaks in or out during cooling. \mathrm{3} Air is an ideal gas with constant specific heats at room temperature.

Properties The constant volume specific heat of air is given to be c_{v}=0.72 \mathrm{~kJ} / \mathrm{kg} .{ }^{\circ} \mathrm{C}.

Analysis Since the house is well-sealed (constant volume), the total amount of heat that needs to be removed from the house is

Q_{L}=\left(m c_{v} \Delta T\right)_{\text {House }}=(800 \mathrm{~kg})\left(0.72 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)(32-20)^{\circ} \mathrm{C}=6912 \mathrm{~kJ}

This heat is removed in 15 minutes. Thus the average rate of heat removal from the house is

\dot{Q}_{L}=\frac{Q_{L}}{\Delta t}=\frac{6912 \mathrm{~kJ}}{15 \times 60 \mathrm{~s}}=7.68 \mathrm{~kW}

Using the definition of the coefficient of performance, the power input to the air-conditioner is determined to be

\dot{W}_{\text {net,in }}=\frac{\dot{Q}_{L}}{\mathrm{COP}_{\mathrm{R}}}=\frac{7.68 \mathrm{~kW}}{2.5}=3.07 \mathrm{k W}