Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
9th Edition
ISBN: 9781259822674
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Question
Chapter 6.11, Problem 151RP
To determine

The cost of energy “vented out” by the fans in 1 h.

Expert Solution & Answer
Check Mark

Answer to Problem 151RP

The cost of energy “vented out” by the fans in 1 h is $0.106_.

Explanation of Solution

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(TindoorToutdoor) (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 electric energy loss by “vented out”.

Electric energy loss=Q˙loss,fan×Δt/COP (V)

Here, the fan takes time is Δt and the coefficient of performance of fan is COP.

Determine the amount of cost of the heat “vented out” per hour.

Moneyloss=(electricenergyloss)×(Unitcostofenergy) (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 0.287kPam3/kgK.

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 1.0kJ/kg°C.

Substitute 92 kPa for Po, 0.287kPam3/kgK for R, and 22°C for To in Equation (I).

ρo=92kPa(0.287kPam3/kgK)×(22°C)=92kPa(0.287kPam3/kgK)×(22°C+273)=92kPa84.665kPam3/kg=1.0866kg/m3

     1.087kg/m3

Substitute 200m2/h for w and 2.8m/h for l in Equation (II).

ν˙air=(200m2/h)×(2.8m/h)=560m3/h

Substitute 1.087kg/m3 for ρo and 560m3 for ν˙air in Equation (III).

m˙air=(1.087kg/m3)×(560m3/h)=608.72kg/h=608.72kg/h×(1kg/s3600kg/h)=0.169kg/s

Substitute 0.169kg/s for m˙air, 1.0kJ/kg°C for cp, 33°C for Tindoor and 22°C for Toutdoor in Equation (IV).

Q˙loss,fan=(0.169kg/s)×(1.0kJ/kg°C)×(33°C22°C)=(0.169kg/s)×(1.0kJ/kg°C)×(11°C)=1.859kJ/s=1.859kJ/s×(1kW1kJ/s)

          =1.859kW

Substitute 1.859kW for Q˙loss,fan, 1 h for Δt, and 2.1 for COP in Equation (V).

Electric energy loss=(1.859kW)×(1h)/(2.1)=(1.859kW)×(0.47619h)=0.885238kWh

Substitute 0.885238kWh for electric energy loss, $0.12/kWh for unit cost of energy in Equation (VI).

Moneyloss=(0.885238kWh)×($0.12/kWh)=(0.885238kWh)×($0.12/kWh)=$0.106

Thus, the cost of energy “vented out” by the fans in 1 h is $0.106_.

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Chapter 6 Solutions

Thermodynamics: An Engineering Approach

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