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Peltier requirements for fastest cooling...

 Post subject: Re: Peltier requirements for fastest cooling...
PostPosted: Mon May 02, 2016 5:16 pm 
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Joined: Fri Jan 23, 2015 7:42 am
Posts: 18
Location: Sanford, FL.
The specific heat of water is 4.197 Joules per gram per degree K (or C). At 250ml, you have 250 grams of water since water is the basis of the definition. To take your water from room temperature (say 23degC) to 0degC will require the removal of 4.197 * 250 * 23 = 24,133 Joules. And that's assuming your container is perfectly insulated which is, of course, impossible.

Since a Watt is one Joule per second, and it's taking you 4 minutes (240 seconds) to move the temp to 0degC, you are cooling at a rate of 24,133 / 240 = 100.55 Watts ... again, if your insulation were perfect.

I assume you're also pumping the water to circulate it through the cold side heat exchanger. The act of pumping using a centrifugal pump will heat the water. So you're fighting that as well as heat infiltration from the surrounding environment. Of course, it looks like your pump is actually submerged in the cooled water container so in that case you can calculate the power being delivered to your pump (volts x amps = Watts) as being all added back to the water and not just that transferred by the pump impeller.

Also, getting to 0degC with regular water presents some challenges. Since you're pulling the hot side of the TEC to near 10degC, your cold side is probably well below freezing. I would bet that you're getting ice buildup inside of your cold side block and it's is restricting your flow and your efficiency.

In any case, at 11v and 8.5 amps you're driving the TECs each with 93.5 Watts. At that power level, not having the datasheets handy, I suspect you're getting 60 Watts of cooling or so from each TEC. Unless you're saying that 8.5 amps is shared between them. Then you're only driving 93.5 Watts in total, which seems low to be able to get the water to 0degC in 4 minutes. Unless, of course, it started at something lower than room temp.

I would be hesitant to drive the TEC's too much harder. The COP will go down and you'll be left with diminishing returns. In general, I wouldn't drive a TEC to more than 85% of it's rated power due to the rolloff in efficiency. But each situation varies. You could probably get them to 13V and still be drawing good heat from the water.

Given the icing in your pictures, I'm thinking the inside of your cold side block is nearly frozen closed.

Do you have the TEC's stacked together between your heat exchangers? Holy cow, that's going to be difficult to calculate. You'd be way better off running them side by side than stacked. You don't get to double the cooling power by stacking them. You get a higher deltaT but at a much lower efficiency.

Get two more blocks and split the water flow between them with each TEC doing 1/2 the water.


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 Post subject: Re: Peltier requirements for fastest cooling...
PostPosted: Tue May 03, 2016 2:35 am 
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Joined: Fri Jan 23, 2015 7:42 am
Posts: 18
Location: Sanford, FL.
Oh, I see. You have two hot side blocks with a TEC between each and the one cold block. So ignore my last paragraph above. It was hard to make out at first due to all of the ice buildup.

So yes, I suspect you're getting about 120 watts of cooling and you're losing some of it to a lack of insulation and what looks like the pump submerged in the liquid ... a small pump, maybe 5 Watts. I think my estimates hold up in that case.


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 Post subject: Re: Peltier requirements for fastest cooling...
PostPosted: Sun May 08, 2016 6:33 pm 
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Joined: Fri Jan 23, 2015 7:42 am
Posts: 18
Location: Sanford, FL.
Those are nice, just make sure you're power supply can push them up there. Seems like a fun experiment.


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