Carnot engine ( Thermodynamics )

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SUMMARY

The discussion centers on calculating the mechanical power required for an air conditioner operating as a Carnot engine, given outside and inside temperatures of 32°C and 21°C, respectively, and a heat leakage of 9000 kcal/hr. The calculated efficiency of the Carnot engine is 0.03607, leading to a work requirement of 290 kW, which contradicts the expected answer of 36 kW. Participants clarify that the efficiency should be interpreted as the Coefficient of Performance (COP) for a Carnot refrigerator, prompting a reevaluation of the calculations using the correct formulas.

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Zamze
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Homework Statement


8. A house is cooled by an air conditioner. The outside temperature is 32°C and the inside temperature is 21°C. The heat leakage into the house is 9000 kcal/hr. If the air conditioner has the efficiency of a Carnot engine, what is the mechanical power required to keep the inside temperature constant? Answer given to this problem is [36kW]


Homework Equations



efficiency= 1 - Temperature cold/ Temperature hot for carnot engine in Kelvins.
Also efficiency=Temperature cold/ work done.
I believe this should be enough.

The Attempt at a Solution


Using the first given equation.
Efficiency=1-(21+273)/(32+273)=0.03607
Using the second equation W=9000kcal/hr/0.03607
Converting properly 9000kcal/hr=9000*4.184*60^2 gives us the work
Thus W=9000*4.184/(0.03607*60^2)=290KW
Can anyone explain to me why I am not getting 36KW?

Thank you!
 
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I'd have used K=\frac{|Q_C|}{W}=\frac{Ht}{Pt}=\frac{H}{P}

Target variable is P

K=K_{\text{carnot}}=\frac{T_C}{T_H-T_C}

P=\frac{H}{K}

This doesn't seem to give the answer you're after though! Hopefully that has helped a little bit.
 
Zamze said:

Homework Statement


8. A house is cooled by an air conditioner. The outside temperature is 32°C and the inside temperature is 21°C. The heat leakage into the house is 9000 kcal/hr. If the air conditioner has the efficiency of a Carnot engine, what is the mechanical power required to keep the inside temperature constant? Answer given to this problem is [36kW]
The heat removed is related to the work done by the Coefficient of Performance (COP):

COP = output/input = Q_c/W = Q_c/(Q_h-Q_c)

For a Carnot refrigerator:

Q_c/(Q_h-Q_c) = T_c/(T_h-T_c)

Determine the amount of heat removed in one unit of time (Qc) and plug that and the value for COP into the first equation above to determine the rate of work (hint: use one second as the unit of time - be careful about converting calories to watts).

Unless I am misunderstanding what the question is asking (I am interpreting the "efficiency" to mean the COP and the reference to "Carnot engine" as "Carnot refrigerator") the given answer is wrong.

AM
 
Last edited:

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