Kitchen fridge thermal power and COP

Question

In general, a kitchen refrigerator is essentially a heat pump. The vapor compression cycle creates a temperature difference, and the insulated compartment stays cold. But how can I determine the heat output of such a heat pump or its COP?

The user manuals only give the average energy consumption per day, but because of the insulation the compressor doesn't have to run all the time, it just turns on when the temperature inside exceeds some threshold, and stops when it reaches another threshold.

As a side question, is a kitchen refrigerator a viable option for low energy cooling? For example, if I have a device with a TDP of less than 300 watts and it needs sub-ambient temperatures, can I just put it in the refrigerator?

Answer 1

I like the answer by @Transistor better but if you don't have the option of turning off the refrigerator, you could get an idea of the COP by measuring the power consumption of the device under somewhat controlled conditions and measuring again after putting in a known thermal load.

Since COP is Q_cooling/P_in

Say, you ran the device for 24 hours and used 1500 watt-hours. If you put in a 10 watt light bulb and run it for another 24 hours and now used 1590 watt-hours, your COP would be:

$$ COP = \frac{Q_c}{P_{in}} = \frac{Q_c}{1500 Wh} = \frac{(Q_c + 24h*10W)}{1590 Wh} $$

Assuming my algebra and arithmetic are correct, $$ Q_c=4000 Wh $$ so

$$ COP=2.66 $$

As for whether it is economical to run device that puts out 300 watts in your refrigerator, you can work that out the same way.

$$ COP=2.66 =\frac{300W*24h}{x Wh} $$ solving for X, you will use about an extra 2700 Watt-Hours each 24 hour period.

At $0.10 per kWh, that's

$$ cost= (0.300 kW)(24 hours)($0.10/kWh) = $0.72 $$ to keep the device running and another: $$ cost= (2.7 kWh)($0.10/kWh) = $0.27 $$ to keep it cool.

That's assuming your refrigerator can keep up. If you exceed the cooling capacity of your device, it will run continuously, keeping it as cool as it can until it dies.

Answer 2

I think you're going to have to start by measuring the heat loss / gain / quality of insulation of the fridge. The easiest way I can think of is to put a heater (incandescent lamp) - maybe 200 W - in the unpowered fridge, measure the power input and the temperature at which the air inside the fridge stabilises.

• Let's say it stabilises at 35°C and the room is at 20°C. We know that ΔT = 35 - 20 = 15°C.
• We know power in = 200 W.
• Heat loss = 200 / 15 = 13.3 W/°C.

Next work out the heat gain in operation:

• Fridge operation point = 4°C.
• Room = 20°C.
• ΔT = 16°C.
• Heat from room into fridge = 13.3 W/°C × 16°C = 213 W.

Power up the fridge. When the fridge temperature has stabilised motor the period and duty cycle of the compressor. Let's say it runs for 3 minutes out of 5 so duty cycle = 0.6. Let's say the compressor draws 120 W.

• P = 120 × 0.6 = 72 W_avg.
• CoP = 213 W / 72 W = 3.

This should give you a feel for the efficiency of the compressor with little trouble and expense.

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... if I have a device with a TDP of less than 300 watts and it needs sub-ambient temperatures, can I just put it in the refrigerator?

I have no idea what a TDP is. "Total Dissipated Power"?

After the measurements and calculations above you'll know the input power and the CoP so you can work out the maximum thermal load it can handle in addition to the heat gain through its insulation.