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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Textbook Question
Chapter 6.11, Problem 111P

A Carnot heat pump is to be used to heat a house and maintain it at 25°C in winter. On a day when the average outdoor temperature remains at about 2°C, the house is estimated to lose heat at a rate of 55,000 kJ/h. If the heat pump consumes 4.8 kW of power while operating, determine (a) how long the heat pump ran on that day; (b) the total heating costs, assuming an average price of $0.11/kWh for electricity; and (c) the heating cost for the same day if resistance heating is used instead of a heat pump.

FIGURE P6–111

Chapter 6.11, Problem 111P, A Carnot heat pump is to be used to heat a house and maintain it at 25C in winter. On a day when the

(a)

Expert Solution
Check Mark
To determine

The actual running time of the heat pump in a day.

Answer to Problem 111P

The actual running time of the heat pump in a day is 5.90h_.

Explanation of Solution

Determine the coefficient of performance of the Carnot heat pump depends on the temperature limits in the cycle.

COPHP=11(TL/TH) (I)

Here, the temperature inside the house is TH, and the temperature outside the house is TL.

Determine the total amount of heat lost by the house.

QH=Q˙H(perday) (II)

Here, the rate of heat gain per unit degree is Q˙H.

Determine the work input of a Carnot heat pump.

COPHP=QHWnet,inWnet,in=QHCOPHP (III)

Here the power input required by heat pump is Wnet,in.

Determine amount of time the heat pump ran.

Δt=Wnet,inW˙net,in (IV)

Here, the rate of work input of a Carnot heat pump is W˙net,in

Conclusion:

Substitute 2°C for TL and 25°C for TH in Equation (I).

COPHP=11((2°C+273)/(25°C+273))=11((275K)/(298K))=12.956K12.96K

Substitute 55000kJ/h for Q˙H and 1 day for per day in Equation (II).

QH=(55000kJ/h)×(1day)=(55000kJ/h)×(1day)×(24hr1day)=1,320,000kJ

Substitute 1,320,000kJ for QH and 12.96 for COPHP in Equation (III).

Wnet,in=(1,320,000kJ)(12.96)=101851.852kJ

Substitute 101851.852kJ for Wnet,in and 4.8kW for W˙net,in in Equation (IV).

Δt=(101851.852kJ)(4.8kW×(1kJ/s1kW))=21219.1s×(1hr3600s)=5.894hr5.90hr

Thus, the actual running time of the heat pump in a day is 5.90h_.

(b)

Expert Solution
Check Mark
To determine

The total heating cost that day.

Answer to Problem 111P

The total heating cost that day is $3.11_.

Explanation of Solution

Determine the total heating cost that day.

Cost=W×(price)=(W˙net,in×Δt)×(price) (V)

Conclusion:

Substitute 4.8kW for W˙net,in, 5.894 hr for Δt, 0.11$/kWh for price in Equation (V).

Cost=((4.8kW)×(5.894h))×(0.11$/kWh)=(28.272kWh)×(0.11$/kWh)=$3.1099$3.11

Thus, the total heating cost that day is $3.11_.

(c)

Expert Solution
Check Mark
To determine

The amount of cost if resistance heating is used instead of heat pump.

Answer to Problem 111P

The amount of cost if resistance heating is used instead of heat pump is $40.3_.

Explanation of Solution

Determine the amount of cost if resistance heating is used instead of heat pump.

NewCost=QH×(price) (VI)

Conclusion:

Substitute 1,320,000kJ for QH, and 0.11$/kWh for price in Equation (VI).

NewCost=(1,320,000kJ)×(0.11$/kWh)=((1,320,000kJ)×(1kWh3600kJ))×(0.11$/kWh)=$40.3

Thus, the amount of cost if resistance heating is used instead of heat pump is $40.3_.

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Chapter 6 Solutions

Thermodynamics: An Engineering Approach

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