Carnot engine ( Thermodynamics )

In summary: PERE's LAWIn summary, the conversation discusses using the equations for efficiency and work done to calculate the mechanical power required to keep the inside temperature of a house constant while being cooled by an air conditioner. The outside and inside temperatures and heat leakage into the house are given, and the efficiency of a Carnot engine is used in the calculations. However, the given answer of 36kW is incorrect, with the actual answer being 290KW.
  • #1
Zamze
11
0

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!
 
Physics news on Phys.org
  • #2
I'd have used [tex] K=\frac{|Q_C|}{W}=\frac{Ht}{Pt}=\frac{H}{P} [/tex]

Target variable is P

[tex] K=K_{\text{carnot}}=\frac{T_C}{T_H-T_C} [/tex]

[tex] P=\frac{H}{K} [/tex]

This doesn't seem to give the answer you're after though! Hopefully that has helped a little bit.
 
  • #3
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):

[tex]COP = output/input = Q_c/W = Q_c/(Q_h-Q_c)[/tex]

For a Carnot refrigerator:

[tex]Q_c/(Q_h-Q_c) = T_c/(T_h-T_c)[/tex]

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:

What is a Carnot engine?

A Carnot engine is a theoretical engine that operates on the principles of thermodynamics. It consists of a heat source, a working substance, and a heat sink, and it works by converting heat energy into mechanical work.

How does a Carnot engine work?

A Carnot engine works by taking in heat from a hot reservoir, converting some of it into work, and releasing the remaining heat into a cold reservoir. This process is known as the Carnot cycle and it follows the laws of thermodynamics.

What is the efficiency of a Carnot engine?

The efficiency of a Carnot engine is given by the equation: Efficiency = (Th - Tc) / Th, where Th is the temperature of the hot reservoir and Tc is the temperature of the cold reservoir. This means that the efficiency of a Carnot engine is dependent on the temperature difference between the two reservoirs.

What are the limitations of a Carnot engine?

A Carnot engine is a theoretical construct and thus it is not possible to build a perfect Carnot engine in reality. In addition, the efficiency of a Carnot engine is limited by the temperature difference between the hot and cold reservoirs, and it cannot operate if the temperature difference is zero.

What are the real-world applications of the Carnot engine?

Although a Carnot engine is not practical to build, the theoretical principles behind it are used in many real-world applications such as power plants, refrigerators, and heat pumps. It serves as a benchmark for the maximum possible efficiency of these systems.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
727
  • Introductory Physics Homework Help
Replies
3
Views
874
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
8K
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
5K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
20
Views
2K
Back
Top