How many cubic feet can 1 Peltier diode cool efficiently?

Question

I want to build a mini fridge, however, I want to build it bigger than the ones shown in the tutorials I've watched, so I assume I need more than one Peltier diode to cool it.

However, for that I need to know what the maximum amount of space 1 Peltier diode is good for.

I want to make it 6 cubic feet.

How many Peltier diodes would I need for that, and how much voltage do I need to power all of them?

EDIT: I plan on using acrylic panels as its outside, laced with Auto Car Heat Sound Deadener Insulation and styrofoam on the inside, including gaskets wherever needed to make sure no air escapes from the box.

In order to cool the heatsinks on the hot side I want to use several 12 cm computer fans.

The average temperature in the room the coolbox will be in is about 30°C.

I need the cooling inside to average at 0-3 °C at most.

My plan is to use more than one cooler, in order for easier cooling as both will work at half the power, and thus will be less noisy.

Answer 1

Peltiers are very, very inefficient, compared to a compressor at least. With that volume box, you need to play hard-ball with physics, to get any useful performance at sensible power.

1) Insulate the box properly. Use PU panels (Celotex, Kingspan, that sort of thing), the thickest you can aesthetically use (my 20L coldbox uses PU 70mm thick). Rubber foam sound insulation will not provide anything like that performance.

Once you've figured out the surface area of the box, and the thickness of PU you're using, you can calculate how many watts you'll need to keep it below ambient, and so calculate how many Peltiers you'll need.

Here's an excellent online calculator, which given the power and the delta T you want to achieve, will give you figures for the lowest power, and the smallest (unfortunately not the same) way of achieving that.

2) Reduce the hot side temperature of the Peltiers by using an evaporation cooled water circuit (look up Bong cooler) to waterblocks to reduce the delta T the Peltier has to provide. Depending on the humidity, you will be able to get many degrees below the actual 'dry bulb' air temperature, approaching the 'wet bulb' temperature.

Better still, use a compressor solution, which is head and shoulders above Peltiers for efficiency and capacity.

Answer 2

It's hard to be sure how much benefit the heat & sound deadining insulation will provide. The manufacturer gives no specs, and it isn't clear what that material is. So I'll ignore any benefit this provides.

Assuming your 6 cubic feet is a cuboid roughly 500mm * 500mm * 750mm (which seems the mostly likely kind of shape even though it isn't the most efficient, which would be to have it as a sphere of 550mm diameter), it'll have a surface area of roughly 2m^2.

Expanded polystyrene insulation has a thermal conductivity of about 0.036W/mK. With a thickness of 100mm (which is about as thick as you could realistically use here) that's 0.36W/m^2K, so 0.72W/K. With the ~30 degree temperature gradient you want, that's therefore about 21W of heat ingress you'll need to move.

TECs are more efficient when there is a lower thermal gradient, so minimizing the gradient is key. However, realistically, it will be difficult to keep the hot side of your TECs cooler than approximately 50 degrees. Also, the cold side will need to be cooler than the average temperature inside your box in order to absorb energy from their environment. Having internal fans may help, but will also put significant extra power into the internal environment. Experimentation will be required to figure out what the best approach there is. However, I would guess your cool side is going to need to be somewhere around -10C in order to have effective cooling of the box.

This means your coolers will be operating with about a 60 degree gradient. For a TEC1-12715 that means it's running close to its maximum gradient (79 degrees gradient @ 50 degrees hot side). Running it at 12V (which is the highest voltage you can safely run it at if you aren't using a current controlled source -- see on the datasheet 'performance curves at Th=50C, voltage vs DT' and look for the max current line) you'll seed approximately 20W of cooling capacity out of each TEC (find current on the same chart as above when DT=60, then use the QC vs DT chart above to find the cooling capacity).

On top of the 21W you'll need to maintain temperature, you'll also need to bring the temperature down to your target in the first place. Assuming your cooler will be used for cooling of water, and that it'll be about 25% full most of the time, you're looking at chilling about 40 litres of water by 30 degrees = about 5,000 kJ of energy you need to move. If you want this done in 12 hours, that's an additional 120W of cooling capacity you'll need. Although during the period this is happening the interior of the chiller will be warmer, so your TECs will operate with higher efficiency, probably producing an average of 30W cooling capacity each.

However, none of this analysis includes information about heat that leaks from the hot side of the TEC directly across the interface and into your container. It will be difficult to insulate directly where the TECs are installed, and therefore around each you'll see a thermal bridge. With a gradient of 60 degrees, this is likely to be a significant inefficiency, so I'd allow at least another 20W for this.

This suggests that, at least with the assumptions I'm making, you'll probably want at least 5 of the type of cooler I've linked to. Obviously there are other types, but the information I used here is all available in the datasheets, so it should be possible to see how it changes with different types.

Answer 3

Two aspects have to looked into when using a thermoelectric device to cool something in this manner.

Firstly, the energy load been presented. Using some of the parameters mentioned, like 6 cubic feet (170 litre) volume with 20 l of water, ignoring thermal ingress from the outside. External ambient temperature is 30°C. Internal is a fridge environment

That amount of air weighs about 191 grams which needs 192.6 joules of energy per degree of temperature change; the 20 litre of water weighs about 20 kg which needs 83740 joules of energy per degree of temperature change. The combined energy load per degree is about 83,930 joules per degree.

Using a 12 V power supply,the 12706 device at 35°C (hot side) draws 5 amps (12 volts by 5 amps = 60 watts) (heat flux); this device can transfer about 20 joules per second (going by the data sheets and the cold side would be at -10°C). At that rate, it'll take about 4200 seconds to cool that volume by 1°C. The total hot side heat load will be 21 + 60 = 81 watts. Also, remember no allowance has been made for thermal ingress. ("We're gonna need a bigger boat." :) )

Next aspect is, the hot side temperature and the heat sink dissipation. As the hotter this side is, the less heat that will be moved from the cool side. If the hot side is at 35°C (for the above conditions) a large heat-sink and fan assembly in the 0.06 degrees per watt range would be a good choice (for just one device). This would keep the device about 5°C above the ambient of 30°C.

A more efficient configuration running 2 (or more) of those devices on a 3 amp supply would each transfer about 13 watts (joules per second). So for 26 watts of cooling only 45 watts is used instead of 60 watts with the single device. (This is going into the COP of the device) From the Data sheet, at 3 A, a device temperature differential of 40°C occurs. Fridge internal temperature is about 3 degrees normally. A heat sink for each device is (3 - -5) = 8°C per 13 watts or 0.61 degrees per watt (should be readily available). Also, good volume fans as well as would be required. Alternatively, using a larger condenser plate may achieve the same effect.

Using multiple devices can lower the overall ohmic waste heat load and achieve higher heat transfer. If the devices is supplied with 2 amps, the ohmic heat is 10 watts with 6 watts transferred. If four devices were used at this current, then there is 40 watts of waste heat used to pump out 24 watts of heat. External heat sink rated at 0.315 °C/W, internal heat sink at 1.33 °C/W. Probably need four more to compensate for external heat ingress and fan heat.

For this device, a current of 1 amp will not produce enough device temperature differential to meet the initial requirements.

A good thermostat controlled temperature controller would be a good thing too, if it is a power controller then it may apply a wider range of power the devices to more closely control the internal temperature. But, when all is said and done, for a lot of running time, the cooler only needs to compensate to heat ingress, for the size unit hopefully less than 20 Watts. The time to cool an item is dependent on how many more cooling devices are added and that determines how much power is drawn in that cooling period. Those extra devices won't be running once the desired temperature is reached.

To handle the thermal ingress issue, probably a test and measure method would be the best, but if you assume 20 watts like a previous author then that is simply two more of these cooling assemblies. Although, a 120 cm fan may introduce a 2.5 watt heat load and may be considered for increased air circulation but would decrease the available heat pump capacity. Or you may want the devices on all the time but running at a reduced power to compensate for heat ingress.

But, from an insulation viewpoint and heat ingress, if the 100mm thick polystyrene sheet foam insulation is used then for the 170 liter internal volume, the expected energy required to offset the leakage is 16 Watts. The surface area of a cuboid to contain that volume is 1.84 square meters. You just have to specify the thermal conductance of your insulating material in Watts per square meter per degree and its thickness. (Can you get that information from the AliExpress people on that product?)

To compare with a 170 liter fridge using a compressor/refrigerant, that consumes 150 watts (from what I could find). That could mean 15 of the 2 amp units, producing 90 watts of cooling and bring the initial cooling time figure from 66.61 minutes to nearly 16 minutes. But physically mounting that many devices maybe problematic.

Some musings.
But, if you wanted just one device, use something like a 12740, using 30 amps (ohmic waste heat of 360 watts) generating a cooling effect of 150 watts needing 510 watts dissipated by the external heat sink whilst maintaining the hot side at 50°C. Two of these could make a nice little room heater.

Did you notice that I was referring to current? Well, the data sheets give the performance graphs for a constant current. Those graphs highlight that when the temperature differential is lower that more cooling occurs; meaning that the cooling performance can be much greater when the fridge is "warmer". So, initially, a 4 device configuration of 12706 will have at 2 amp will be about 105 Watts and a waste heat of 36 Watts. Only as the temperature differential increases does the cooling effect lessen. So, the better performing the heat sink is the quicker the cooling; water cooled external heat sink/radiator? Just a thought

Also, the air will cool much faster than the 20 l mass for a number of reasons. The air will be mobile and hence coming in contact with the cold surfaces and will take about 3 minutes to cool down from 30 °C. However, 20 l of water will take more than a day to chill. A 750 ml bottle of wine chills in about an hour

I hope that this helps.

Answer 4

As a first approximation, the answer to

How many peltiers would I need for that? And how much voltage do I need to power all of them?

is

1) A whole metric buttload.

2) You need as much voltage as you do for one. You just need more current.

OK, so first you need to understand that your "insulation" is probably crap. The fact that it is sold as a sound deadener suggests that it won't do well for thermal insulation. Plus, it's really thin, and this is Not A Good Thing. Other answers have gone into the math needed to calculate the power required, but they have typically used "reasonable" insulation values. You cannot assume that, so calculation is pretty much useless here.

Build yourself a smaller version first. Let's assume you go for a 1 foot cube. Put a single TEC on one side and drive it at full voltage (probably 12 volts). Drill a small hole in one face so you can put a thermometer inside and read it from the outside. Now let the unit run overnight. The next morning, read the internal temperature and subtract from the current ambient temperature. Make sure you do this before the temperature starts climbing, or you'll overestimate the difference. I'll pretty much guarantee that it won't be 30 C. So add another TEC and repeat. Do this until you do get a 30 C difference. You'll probably find it extremely useful to add more layers of insulation. Like, a lot of layers. The stuff you've specified is only 7 mm thick, for pity's sake. I'd try at least 10 layers. Of course, this is undoubtedly much more than you're willing to pay, but you can find out for yourself if one layer works. Trust me, it won't.

When you add TECs, wire them in parallel. The voltage you need will remain constant, probably 12 volts. Of course, the current will add up. If your TEC draws 6 amps at 12 volts, two will need 12 amps at 12 volts, and 3 will take 18 amps. Since you will need a lot of coolers, you'll need a very large power supply.

Also, when you go to multiple coolers, you must manage your waste heat exhaust properly. You cannot simply put them side by side without considering where their air is coming from and where it's going. Most particularly, you cannot have the hot exhaust of one cooler blowing into the "cool" input of the next unit in line, and the total exhaust cannot feed back into the overall input - no matter how much of a pain this is to ensure. And once you get to the final unit, you may well find that routing the exhaust air to avoid this will be a real chore. Just a warning.

Once you get a working box going, you can scale it up pretty simply. Measure your desired final unit and find the area of the interior volume. Let's say you want 2 ft x 2 ft x 1.5 ft (6 cubic feet total). The area will be 4 (top) + 4 (bottom) + 4 x 3 (sides), or 20 square feet. The openings for the coolers will reduce this a bit, but the need for overlap at the joins will increase it. Let's go with 20 for this example. Then, if you needed 4 TECs for your 1 foot unit, you'll need 20/6 times 4, or about 13 or 14 TECs. And frankly, with the insulation you've specified I'd be astounded if you get away with 4 coolers for your test unit.

As I've mentioned, as you use more units in parallel you'll need more current. If this becomes too much of a problem (and it probably will) you can connect your Peltier elements (but not the fans, probably) in series. So, for instance, you might connect 12 elements as 3 parallel strings of 4 elements in series. Using 12 volt, 6 amp TECs would then require 48 volts (4 x 12) at 18 amps (3 x 6).

As you may (or may not) have noticed, this will take a lot of power. For 12 units, each unit will need 72 watts (12 x 6), and 12 of them will need 864 watts. This is actually more than a new-model full-size refrigerator will use running continuously, although it does have the advantage of not having a turn-on surge which all compressors exhibit. You can draw your own conclusions about using Peltiers for large-scale cooling.

Build a little one first. You will not be happy with the results, but at least you won't waste as much money as you will trying for a big one first. You can reuse the coolers if you decide to go with a big unit.

Answer 5

This is a conceptually-challenged question. Peltier elements are "heat pumps", they create temperature difference between the sides of array Peltier elements for a given heat flux. The temperature difference is determined by balance of heat fluxes on both sides, including Joule dissipation from applied current flow and parasitic thermal conductivity of elements. Due to this parasitic backflow, most arrays can produce no more than 60-65 C difference for a zero heat flux.

So the low-side temperature inside your cooler will be determined by how well the cooled volume is insulated from ambient, including radiation flux, or how much heat will be entering your inner volume via insulation. If your volume is isolated using double Dewar vacuum walls (speaking hypothetically), and all enclosed into low-density Styrofoam, any small element should be able to cool the inside air down to -30C if you manage to maintain the hot side at +30C (which is impossible with 30 C ambient). To get the internal temperature at 0 C you will need to maintain the hot side at ~60 C, which is doable with massive heat sinks and strong fans.

But again, the amount of Peltier arrays will be determined by how good your insulation is. It is a challenging problem to determine the insulation quality and corresponding heat flux that will be needed to remove, and only then the number of elements can be determined.

Answer 6

I built a 1.2 cubic ft fridge using a 30 W, 40 mm module. It dropped the temp down to 41 from a 95°F ambient. The heat exchanger would freeze up on occasion so I had to make a few design changes. Heat sink design, insulation factor, surface area, and type of alloy are extremely important. Know your thermal conductivity and oxidation info! Chest type is the only way to go. Otherwise just too much loss every time you open it.