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I need to know some details about Peltier coolers before moving into making an air-conditioner.

I have watched several videos on YouTube that demonstrate making air-conditioners with Peltier thermo-electric coolers. They show that there is a decrease in temperature, on the cool side, which is about -10 °C to -20 °C.

We also know that Peltier coolers have an avg delta Tmax of ~65 °C.

But I am a bit confused about this Tmax. I think it is generally about 30-40 °C because in practice absolute conditions can't be maintained.

So, does the hot side temperature depend on the surrounding temperature?How much temperature decrease, would I get if the surrounding temperature is 40 °C with 12 Volts?

I also want to know if it is efficient for cooling a 10 ft x 10 ft room. I think using 5/6 Peltier cooler can handle that, but is it efficient? Or should I use a compressor?

And should I use only a heatsink with a fan, or a heatsink + fan + watercooler, for cooling the hot side?

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    \$\begingroup\$ Not suitable for this \$\endgroup\$ – Sunnyskyguy EE75 Jul 21 '18 at 2:31
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    \$\begingroup\$ Efficiency is terrible (2-5%). You'd need about 100 of those drawing perhaps 5kW to equal a dirt-cheap 6000 BTU window air conditioner. \$\endgroup\$ – Spehro Pefhany Jul 21 '18 at 3:12
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We'll start with the easy one first: Do not try to cool a room with TECs. It simply will not work. Forget it. Don't even think about it. Cooling loads for room air conditioning will run in the hundreds of watts.

OK, with that out of the way. Yes, hot side temperature will depend on ambient. Assuming you're using a heat sink of some sort, the best hot side temperature you can get will be ambient. And since, in order to shed heat, the heat sink has to get above ambient, that's a lower limit. Typically, a TEC heat sink will be cooled via forced air, which reduces the heat sink temperature.

Second, the temperature differential will be closely linked to the amount of heat you're trying to remove. The 65 C number you've run across is for essentially no heat flow. Any increase in heat removed will reduce the temperature difference.

I suggest you play around with a TEC calculator such as this one. Consider the following starting point.

1) Ambient temperature is 40 C 2) Desired room temperature is 20 C 3) the heat sink is maintained at 60 C (Note: this is roughly the "Ouch!" point where you can't keep your finger on it.) 4) The heat load is 500 watts. For instance, that's the heat put out by 5 resting adults. Unless the room is very well insulated, you might get that amount of power just from the external temperature coming through the walls and air leakage.

Run the calculator. See Qmax? A 12706 has a Qmax of about 60 watts, so you'd need about 32 modules. Each one would be run at about 3 amps. From the 12706 data sheet you'd want to drive your modules at about 9 volts. Total dissipation would be about 860 watts. And before you think that's not unreasonable, you'll need to price out a heat sink which will mount 32 modules and provide a thermal impedance of .023 deg C/watt, which is the temperature difference between hot side and ambient (60 - 40) divided by the dissipation (860 watts) at 860 watts total. Trust me, this is a monster.

Get an air conditioner.

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  • \$\begingroup\$ SO is it okay for personal small cooler? \$\endgroup\$ – Tick Twitch Jul 22 '18 at 0:25
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    \$\begingroup\$ @TickTwitch - I have no idea. I don't know how big or how well-insulated. I don't know what temperature you want to cool to. I don't know anything about your heat sink or its ventilation. In other words, all of the issues I brought up about cooling a room apply, and you haven't addressed any of them. In principle, the answer is yes, but only because you can buy TEC portable coolers. \$\endgroup\$ – WhatRoughBeast Jul 22 '18 at 3:13
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does the hot side temperature depend on the surrounding temperature?

Yes. The hot side is being heated with the heat transferred from the cold side, plus the I x V that is driving the Peltier.

That heat needs to be transferred away from the hot side by some mechanism: conduction, convection, or radiation. The rate of all of those mechanisms will depend on the temperature of the material around the hot side.

How much temperature decrease, would I get if the surrounding temperature is 40 °C with 12 Volts?

This isn't answerable, because the surrounding temperature isn't the only factor that determines the temperature.

You also need to know what kind of heat sink is attached to the hot side, and what's its effective thermal resistance. This will itself depend on whether you're providing forced air over the heat sink or just relying on natural convection.

Of course you also need to know the heat being transferred from the cold side, which depends on the temperature of the cold side.

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