A 2.4-m-high 200-m2 house is maintained at 22°C by an air-conditioning system whose COP is 3.2. It is estimated that the kitchen, bath, and other ventilating fans of the house discharge a houseful of conditioned air once every hour. If the average outdoor temperature is 32°C, the density of air is 1.20 kg/m3, and the unit cost of electricity is $0.1/kWh, the amount of money "vented out" by the fans in 10 hours is Consider the specific heat to be 1.005 kJ/kg·°C.

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usamakld99 usamakld99 · Answer 1 · 2019-11-14T20:28:16+01:00

Answer:

Cost= 0.5 $

Explanation:

Coefficient of performance of the air conditioning system is given as

[tex]COP=\frac{Q}{W}=W=\frac{Q}{COP}[/tex]

Energy = energy discharged or vented every hour= m c ΔT

Q = heat energy (Joules, J)

m = mass of a substance (kg)

c = specific heat (units J/kg∙K)

∆ is a symbol meaning "the change in"

∆T = change in temperature (Kelvins, K)

therefore

where m = ρV = density of air x volume = 1.20 x 200 x 2.4 = 576Kg

Energy = energy discharged or vented every hour= ρVc ΔT

Q= 576 x 1.005 x 10 = 5760 KJ/hour

therefore=ore 1kw = 3600KJ/hour

Q = 1.6 Kwh

[tex]W=\frac{Q}{COP}[/tex]

Q = [tex]\frac{1.6}{3.2} = 0.5kwh[/tex]

Cost = per unit cost x energy vent x no of hours

Cost = 0.1 x 0.5 x 10

Cost= 0.5 $