# Thermodynamics - Carnot efficiency

1. Feb 14, 2010

### Wellesley

1. The problem statement, all variables and given/known data
On a hot day a house is kept cool by an air conditioner. The outside temperature is 32°C and the inside temperature is 21°C. Heat leaks into the house at the rate of 9000 kcal/h. If the air conditioner has the efficiency of a Carnot engine, what is the mechanical power that it requires to hold the inside temperature constant?

3. The attempt at a solution

e= 1-294/305 = 0.036066

e=W/QH

Power = 9000 kCal/hour --> 37656000 J/hour --> 10460 J/s

.036066=W/10460 ---> W (really power)=377.426 J/s

This problem seems pretty straightforward...but the back of the book has an answer of 390 W, which is ~13 more J/s than my answer, so I must have made a mistake....yet there are only two equations, and one conversion! Is this a rounding error from the book?

2. Feb 14, 2010

### kuruman

I think you're right.

3. Feb 14, 2010

### Andrew Mason

The Coefficient of Performance is defined as:

$$COP = \frac{Q_c}{W} = \frac{Q_c}{Q_h-Q_c}$$

which, for a Carnot refrigerator is:

$$COP = \frac{T_c}{T_h-T_c}$$

In this case COP = 294/11 = 26.7

Since Qc = 9000 Kcal/hr (ie. heat removed from the cold reservoir) or 37680 kJ/hr or 10470 watts, W = Qc/COP = 10470/26.7 = 392 watts

AM

4. Feb 16, 2010

### Wellesley

I looked through the thermodynamics whole chapter, and I couldn't find a single reference to this equation. Is it derived from the Carnot engine efficiency -> e=W/Q? Your solution makes sense, I'm just puzzled why the equation is not even in the book.

Thanks for responding. I'm glad I didn't submit the assignments yet...

5. Feb 16, 2010

### Andrew Mason

Performance measurement depends on the function of the device.

Efficiency (output/input or W/Qh) is used for heat engines since it produces work from heat input. For heat pumps, which deliver heat by moving it from colder to warmer reservoirs, a coefficient of performance (COP) is used of Qh/W (output/input being heat output to work input). For a refrigerator, which inputs work to remove heat from a cold reservoir, the performance is measured by Qc/W (output/input or heat removed to work input).

But you don't really need to use COP explicity. In all cases: W = Qh-Qc and for Carnot cycles, $\Delta S = Q_c/T_c - Q_h/T_h = 0$ so Qc/Qh = Tc/Th. Note W = Qh-Qc = Qc(Qh/Qc - 1) = Qc(Th/Tc-1)

In this case you are given the amount of heat that must be removed from the cold reservoir (Qc). You can just use the above line to find W from Qc.

AM