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

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    $\begingroup$ What do you mean by "rejected"? $\endgroup$ – CuriousOne Jan 19 '16 at 11:32
  • $\begingroup$ @CuriousOne - "rejected" is a common phrase for the heat that is released from the high temperature end of a heat pump to the environment $\endgroup$ – Floris Jan 19 '16 at 16:33
  • $\begingroup$ @Floris: Didn't know that... some of the uses of the 1+ million words of the English language are still not in my repertoire, I guess (I would estimate the missing chunk around 95%). Thanks! $\endgroup$ – CuriousOne Jan 19 '16 at 18:30
  • $\begingroup$ @CuriousOne - well, that's one down, 949,999 to go. You're welcome. $\endgroup$ – Floris Jan 19 '16 at 18:31
  • $\begingroup$ See for example en.wikipedia.org/wiki/… $\endgroup$ – Floris Jan 19 '16 at 18:41

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 =

67 W is rejected

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.


This is an example worked out for Total Heat of Rejection, note the example is using different variables. This may be helpful to work your problem, fit in your variables.

Compressor Performance

110°F condensing temperature

10°F evaporating temperature

75°F incoming water temperature

Refrigerant R–22

Evaporating Watts = 6500

Evaporating Load: 40,200 Btu

watts x 3.4 = Heat of Compression

Heat of Compression + Evaporating Load = Total Heat of Rejection

6500 watts x 3.4 = 22,100 Btu

Heat of Compression = 22,100 Btu

Evaporating Load = 40,200Btu Total Heat of Rejection = 62,300 Btu

Total Heat of Rejection (THR) is the heat absorbed at the evaporator plus the heat picked up in the suction line plus the heat added to the refrigerant in the compressor.

  • $\begingroup$ This answer doesn't make sense. Is it BTU/hr? If so, a household refrigerator doesn't reject anywhere close to that amount of heat to the indoor environment. $\endgroup$ – David White Jan 19 '16 at 12:00
  • $\begingroup$ yes, that should be by hour. I believe the example I am using the watts are much higher then a fridge. 6500 vs 25, its a working example. $\endgroup$ – Ed Yablecki Jan 19 '16 at 12:28
  • $\begingroup$ Also, it looks like my example is water cooled, where most refridgerators are air cooled. Tho, this should not affect answer, because you are still figuring energy in equals energy out. $\endgroup$ – Ed Yablecki Jan 19 '16 at 12:34

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