Fridge thermal testingPosted: May 22, 2011
Finally, I got around to writing code for the fridge. WordPress doesn’t let me upload zip files, so let me know if anyone’s interested in the code.
I did some thermal testing of the fridge test its insulation (and also do some calculations on thermal loss). I filled the fridge with about 1.6l of water, set its output to maximum, and logged the temperature as it heated up using my tempearature logger code (the data is coming off the USB cable using USB HID).
Plugged the USB cable in. Surprising effect from the USB LED shining through the thin plastic.
Filled about 1.6l from the tap
And heated up. Bubbly!
The water from the tap was some 10 °C below ambient. Also It would seem that opening the lid to take that last photo has quite a noticeable effect on the temperatures.
Heating is almost painfully slow, taking some three hours to reach its peak. There’s a noticeable dip at the top there where I opened the lid to take the photo. This dip is very noticeable in later graphs.
The power flow diagram for the system looks something like this:
The peltier acts as a solid-state heat pump, putting electrical energy in results in a transfer of thermal energy from one side to the other. As a result, one side gets cold, while the other gets hot. However, the resulting balance is not symmetrical, since the input electrical power ALSO results in a heating effect.
The net result is that the peltier creates much more heating than it does cooling. This is good news for us because I WANT to use that extra heat, it means I get more heat than it has expended energy for (i.e. maybe 60W of heat from only 30W of electrical energy, a coefficient of performance of 200%, compared to 100% for regular resistive heaters). This is however bad news for fridge mode, and also people trying to use peltiers to cool their CPUs, as they have to dissipate much more heat than the would otherwise.
Because I know the volume of water, and its temperature rise over time, I can easily calculate the thermal energy entering the water over time, i.e. I can calculate the power gained by the water:
E = m c ?T
P = E / t
Plotting the power gained by the water over time reveals that this value starts at about 65W, and drops over time (as the water temperature increases) to about 20W.
I can later graph this power gain against temperature, which is a more useful graph. However, I first turn off the heater and let the water cool by itself, and use the same equation to work out the power lost by the water through the insulation as a function of temperature gradient (the difference in temperature between the inside of the fridge, and the outside). This is the red arrow in the power flow diagram above.
The data is a little patchy here, and incidentally it also exhibits a little jump where I opened the lid a second time to check the calibration of our PT100 probe with a meat thermometer. I must remember to stop screwing with the results like this…
I assume that the power lost is linear with temperature gradient (a valid assumption based on the physics of thermal energy flow), and so I find the power loss to be approximately 0.44W/°C.
I can add this value onto our results for Power Gained, to find the total power entering the fridge from the peltier as a function of the temperature gradient:
The results are very interesting, as it shows that the thermal power produced by the peltier at zero temperature gradient (inside of fridge at same temperature as outside of fridge) is around 60W, which is quite a bit higher than the maximum electrical power consumption of 40W (according to the sticker on the back of the fridge). This is good: it is heating the water with more energy than the amount of electrical energy put in.
I highly suspect that the fridge is using a 12706-type 60W peltier element, since such elements are extremely common and cheap (you can get them at about £3 a piece).
Digging up the datasheet for the 60W 12706 peltier elements, I can see that the specs match those of the fridge. The fridge claims a maximum electrical power consumption of 40W, which is consistent with running the peltier at 12V (it is only 60W when running at around 15V).
I can also make use of the graph of thermal power transferred against temperature difference:
I can take the 3.0A line (for this is about where we are operating), this is the green arrow on our schematic, and subtract it from the total power (the pink arrow on our schematic, and the pink data points on our graph), to find the approximate electrical energy consumed. (There is a minor flaw in this: by taking the straight 3.0A line, I’ve assumed that the electrical energy consumed is constant, which is not the case, but it is a reasonbly close estimate).
The green line is this data (above), and the purple points are the calculated electrical power consumed. Showing an actual electrical power consumption of about 30W, consistent with what I expected. (Although I would normally expect the electrical power consumed to drop as temperature gradient increases owing to the thermoelectric effect).
So I’ve proved that I’m getting somewhere between 120% and 200% of the electrical power going in for heating. The mini-fridge temperature controlled system is therefore extremely energy efficient, more-so than regular resistive heaters used in other slow cookers and sous-vide cookers. The bonus is (in the summer anyway) that my room also gets colder as the heat is pumped into the water (although probably not a noticeable amount).