thermodynamics - Heat rejected by a refrigerator?

What is the amount of heat rejected by the condenser of a vapour compression refrigeration system, typically those found in households? Something in the nature of a 25W compressor, for example

Answer

If you have a 25 W compressor, then the net heat flux into the environment will be 25 W. This is made up of the heat extracted from the inside of the refrigerator, plus the heat of losses in the compressor, minus the heat that travels from the environment back to the inside from the fridge (which is why the compressor needs to keep running).

If we concentrate just on the compressor as a heat pump, we can make some simple assumptions about the COP (coefficient of performance) of the compressor to come up with an estimate.

Assuming the inside of the fridge is at 5°C and the environment at 22°C, the compressor is pumping against a 17°C gradient. A perfect Carnot engine operating between 278 K and 295 K would have an efficiency of

$$\eta = 1 - \frac{T_l}{T_h} = 5.7\%$$

Conversely, a perfect heat pump can pump more heat than the energy you put into it (that's why they are sometimes used for heating homes, when a suitable source of nearby "heat" is available, e.g. groundwater). The ratio of heat moved vs work done is $\frac{1}{\eta} = 17.4$.

That number is the ratio between the heat rejected and the power consumed - so for an "ideal" refrigerator compressor operating between the limits given, the heat rejected would be about 430 W. Now a refrigerator pump is rarely as efficient as you would like - certainly it doesn't get close to the efficiency of the Carnot cycle.

There's a good paper on the measurement of refrigerator efficiency that probably has more information than you ever wanted to know about the subject. It includes measurements of the thermal insulation of a typical refrigerator cabinet (1.21 W/K for the freezer, 0.88 W/K for the refrigerated compartment). From Table 3 of that publication, I find a COP of about 1.7 - suggesting that for our example 1.7*25 W = 42 W of heat is extracted, and 42W+25W =

You can compare these numbers to the ones from the article, and find they are comparable - although the refrigerator in their example had a given energy consumption of 28 kWh/month which suggests about 39 W average power used. Still, that's the same ball park as 25 W.

Afterthought - reading my solution again, I would not normally make an estimate like this and quote a number that seems to be precise to two figures. It would be more reasonable to say "roughly three times the power use is rejected". 1+1.7=2.7 which is approximately 3. Probably a better approach if you are estimating and making tons of assumptions.