Determine the density of air at the indoor conditions.
ρo=PoRTo (I)
Here, the house maintain a pressure is Po, the universal gas constant is R, and the house maintain a temperature is T0.
Determine interior volume of the house per hour
ν˙air=w×l (II)
Here, the width of the house is w and the length of the house is l.
Determine the mass flow rate of air vented out.
m˙air=ρoν˙air (III)
Determine the rate of energy loss by the ventilating fans.
Q˙loss,fan=m˙aircp(Tindoor−Toutdoor) (IV)
Here, the specific heat of air at room temperature is cP, the indoor air vented out at temperature is Tindoor, and the outdoor air temperature is Toutdoor.
Determine the amount of fuel energy loss by “vented out”.
Fuel energy loss=Q˙loss,fan×Δt/ηfurnace (V)
Here, the fan takes time is Δt and the efficiency of the furnace is ηfurnace.
Determine the amount of cost of the heat “vented out” per hour.
Money loss=(Fuel energy loss)×(Unit cost of energy) (VI)
Conclusion:
From the Table A-1, “Molar mass, gas constant, and critical-point properties” to obtain the value of gas constant of air as 0.287 kPa⋅m3/kg⋅K.
From the Table A-2a, “Ideal-gas specific heats of various common gases” to obtain the value of specific heat of air at room temperature as 1.0 kJ/kg⋅°C.
Substitute 92 kPa for Po, 0.287 kPa⋅m3/kg⋅K for R, and 22°C for To in Equation (I).
ρo=92 kPa(0.287 kPa⋅m3/kg⋅K)×(22°C)=92 kPa(0.287 kPa⋅m3/kg⋅K)×(22°C+273)=92 kPa84.665 kPa⋅m3/kg =1.0866 kg/m3
≅1.087 kg/m3
Substitute 200 m2/h for w and 2.8 m/h for l in Equation (II).
ν˙air=(200 m2/h)×(2.8 m/h)=560 m3/h
Substitute 1.087 kg/m3 for ρo and 560 m3 for ν˙air in Equation (III).
m˙air=(1.087 kg/m3)×(560 m3/h)=608.72 kg/h=608.72 kg/h×(1 kg/s3600 kg/h)=0.169089 kg/s
Substitute 0.169089 kg/s for m˙air, 1.0 kJ/kg⋅°C for cp, 22°C for Tindoor and 5°C for Toutdoor in Equation (IV).
Q˙loss,fan=(0.169089 kg/s)×(1.0 kJ/kg⋅°C)×(22°C−5°C)=(0.169089 kg/s)×(1.0 kJ/kg⋅°C)×(17°C)=2.8745 kJ/s=2.8745 kJ/s×(1 kW1 kJ/s)
=2.8745 kW
Substitute 2.8745 kW for Q˙loss,fan, 1 h for Δt, and 0.96 for ηfurnace in Equation (V).
Fuel energy loss=(2.8745 kW)×(1 h)/(0.96)=(2.8745 kW)×(1.041667 h)=2.994 kWh
Substitute 2.994 kWh for fuel energy loss, $1.20/therm for unit cost of energy in Equation (VI).
Money loss=(2.994 kWh)×($1.20/therm)=(2.994 kWh)×($1.20/therm)×(1 therm29.3 kWh)=$0.1226≅$0.123
Thus, the cost of energy “vented out” by the fans in 1 h is $0.123_.