Peltier --> project design

Hi to all.

I'm planning to buy one 100W peltier module that should be used as heater / cooler at the same time. Both sides of the peltier module will have aluminum plates of around 5mm, and each plate will be in separate housing. The space that I want to cool is 40 x 60 x 25 cm and heat part consist of small box like container with dimension of 15 x 15 x 15 cm .

The cold plate will have additional fan and the warm plate will be used as water container.

Question:

- can i cool down this "cooler" plate to some 2 degrees Celsius ( cooler box should have constant 3-4 degrees Celsius )

GM

Reply to
gm
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Depends on the heat input to the cold side, the heat sink, the fan, and the ambient temperature. If the heat input is small and the ambient is room temperature, you can probably figure out some heat sinking arrangement that will work.

You'll probably be disappointed at how big a heat sink and how much air flow you need.

Cheers

Phil Hobbs

Reply to
Phil Hobbs

Not at the same time. Similar to the old riddle about the blonde with the Thermos, except you're on the right track. :^)

Even with very good insulation around the cold box (so that the heat flow is minimal), the cold side will never be below so-and-so degrees less than the hot side.

Suppose the TEC is a thermal current source. Then, it's a Norton source, with its thermal resistance in parallel. After all, it's made of solid materials, and doesn't magically stop being a conductor just because you dump a stream of amps through it!

You're better off using a stupidly large heatsink (so that the hot side can be as near to RT as possible), and still maybe needing two stages (which means way more than 100W, more like 600W!) to get the temperature low enough. (But hey, you'll have more than enough heat for that little hot compartment...)

Needless to say, you're /way/ better off using a compressor type system! If nothing else, get a regular cube fridge and an automotive inverter to run it?

Tim

--
Seven Transistor Labs, LLC 
Electrical Engineering Consultation and Contract Design 
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Reply to
Tim Williams

I'm confused, what is on the hot plate side? It's sitting in a water bath? How does the heat leave the water bath? How hot is the water? (Making tea?)

The problem with a peltier is the element is a good heat conductor. so if the hot plate is too hot, then heat leaks back into the cold side at a rate that is faster than the peltier-action can push it out. The peltier element also makes it's own heat... result.. thermal run away.

George H.

Reply to
George Herold

Relatively common ones will just about support 70C temperature difference between their hot and cold sides at their limit.

Taking ballpark guesstimates and ambient of 27C (aka 300K) a 32W device will pump 20W and so run flat out for deltaT=70C would just about support cold side 3C and hot side 70C but it would be struggling.

eg

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You will be operating it close to the 75C graphs on the last page.

My experience with them has generally been a bit disappointing. The cold side needs to be very well insulated and the hot side on a big heatsink to get anything like the datasheet advertised performance.

I don't think it need be that extreme if the cold side only has to be around freezing point. They do need stacking for cooled CCDs @ -40C.

However, it isn't going to be much good at cooling warm things down - only at just about maintaining an already cold volume of space.

Silent with no moving parts is an attraction of Peltier devices.

He would be well advised to make a 1/3 power scale model of what he intends with a cheap 30W unit. I suspect it would be just about OK for keeping already chilled things cold but pretty well useless as a fridge.

The hot side will get plenty hot though and might well need more cooling! (otherwise thermal conduction of heat back to the cold side will be faster than the poor device can pump it away again)

--
Regards, 
Martin Brown
Reply to
Martin Brown

See

Sloman A.W., Buggs P., Molloy J., and Stewart D. ?A microcontroller

-based driver to stabilise the temperature of an optical stage to 1mK in th e range 4C to 38C, using a Peltier heat pump and a thermistor sensor? ? Measurement Science and Technology, 7 1653-64 (1996)

Appendix A to that paper says

"The relationship between the current driven through a Peltier junction and the heat transferred is approximately parabolic. The manufacturers characterize their functions in terms of three parameters: (i) Imax , the maximum current that can usefully be driven through the junction; (ii) Tmax , the temperature difference across the junction at this current with no thermal load; and (iii) Qmax , the heat transferred by this current with no temperature difference across the junction. An expression for temperature difference as a function of current and heat load in terms of these parameters can be derived:

delta Tp = Tmax.(2I/Imax ? I^2/Imax^2? Qp/Qmax) in degre es K

where "delta Tp" is the temperature difference between the input side of the Peltier junction and the exhaust side, which is positive when the input is cooler; I is the current flowing through the Peltier junction in amperes; and Qp is the heat flowing into the input side of the junction, in watts. The equation can be re-written to give Qp as a function of

1Tp for either face of the junction (with appropriate sign changes).

Compared with the manufacturer?s performance graphs, this expression somewhat underestimates the heat transferred at low temperature differences, probably because it does not include allowance for the change in the Peltier coefficient of the junction with absolute temperature. It was sufficiently accurate for our purposes.

The temperature difference "delta Tp" is across the Peltier junction itself. We are interested in the temperature difference " delta Tb" between the points whose temperatures we measure, the temperature-controlled block and the exhaust labyrinth/neat-sink.

Since we can calculate the net heat flow through both interfaces from the expression given above, we can estimate "delta Tb" if the combined thermal resistance R across both interfaces to the Peltier junction is known (or can be estimated)."

It goes on to give the more complicated expression. E-mail me if you want a copy.

A couple of degrees of temperature difference isn't hard to get, but - as P hil Hobbs says - it's going to take quite a bit of current, and a big heat sink to keep the thermal resistances adequately small.

--
Bill Sloman, Sydney
Reply to
bill.sloman

That's Tmax - around 60K for a single stage junction.

Peltier junctions have no moving parts, and don't vibrate - through the fan blowing the cooling air across the heat sink does.

For small and precisely controlled temperature differences Peltier junctions are hard to beat.

--
Bill Sloman, Sydney
Reply to
bill.sloman

Thnx for reply.

The main idea is to save the space and for the sake of simplicity i would go with peltier design ( and it's cheaper then conventional systems ). On the other hand it would be far simpler to implement...

Insulation will be really strong so dont worry about that :.-) Cold box will be from aluminum and it will be isolated with around 2 cm of insulation material + 2 cm of wood panel.

I found one interesting article on instructubles regarding the cooling temperature. Please take a look:

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GM

Reply to
gm

*** This water will be used for dish cleaning purposes. So i was thinking, why not to use this automatically as cooling part. I get warm water + cooling.

To solve the heat leaking problem i can do the following:

- increase the size of water container

- pump cold water in the container every half hour or so.

This water will be washed away anyway, so....

Reply to
gm

Oh, okay, so it's an RV project or something like that.

As long as you're standing over it so that the hot can't get too hot, it'll probably work fine. Do put in a thermal cutout though!

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
Principal Consultant 
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Reply to
Phil Hobbs

I think you're trying to do too many things with your Peltier device.

It's already been said, but I'll say it again: a Peltier device is only good for a certain temperature differential. And that temperature differential isn't much compared with refrigerators that have compressors. AND that temperature differential depends on how much power is being generated on the cold side.

So if you want side A to be cold, then you want to keep side B as close to ambient as you can, all the time.

(For that matter, you don't want to do what you're proposing with a conventional refrigerator -- pumping heat from a cold space to a warm space is "going uphill" thermodynamically, and takes more energy no matter what magic device is in the middle. I'm pretty sure it comes from the 2nd law of thermodynamics, but it's inescapable. You're probably no worse off, as far as power consumption is concerned, to cool your cold space with your refrigeration device, then warm your warm space with a heater.)

--
www.wescottdesign.com
Reply to
Tim Wescott

you can get more out than you put in, i.e. on the hot side you get the heat from driving "the pump" plus the heat you pumped from the cold side

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-Lasse

Reply to
Lasse Langwadt Christensen

Actually a heat pump can easily have 4 times the heat output compared to the energy input to drive the pump. That is why heat pumps are an alternative to plain heating when trying to conserve energy.

Of course those figures are for compression/evaporation pumps like found in a common refrigerator, not for a Peltier device. But even in a Peltier device, all electrical energy you put into it is appearing as heat on the warm side, and the pumped heat from the cold side is appearing additional to that. So it is always going to do better than plain heating.

Reply to
Rob

I have seen videos of Peltier units freezing water. As others have said, it is all a matter of preventing heat from getting to the cold side including through the Peltier unit itself which you can't prevent.

How will you get fresh water in? I think it will need changing more than twice per hour. Do some simple calcs of the heat the Peltier unit will produce on the hot side (heat from cold unit plus heat from Peltier unit which will be a lot more) and the heat capacity of water. That will tell you how much the temp will rise in a given time interval.

--

Rick C
Reply to
rickman

Dunno. It would be helpful if you would disclose the maker and model number of your prospective Peltier module. The problem is that the data sheet specifies the maximum temperature differential (typically about 70K) while assuming no thermal load on either side. As soon as you start to suck calories out of either side, this differential temperature starts to drop. Worst case is attaching a heat sink on both sides, and immersing the heat sinks in separate large buckets of water at equal temperature. Differential temperature now equals 0K, at least initially.

Another analogy is shoving a 100 watt soldering iron into a bucket of water. While the 100 watt iron will easily melt solder at 200C, the same iron will barely raise the temperature of the bucket of water unless you want to leave it in the bucket for several hours.

The hot version of your analogy is the fan on the warm side. Heat sinks are specified in C/W rise. If about 100 watts is going into heating the warm side, a 20C differential will require a: 20C / 100W = 0.2C/W heat sink which is rather large. You can do better with forced air cooling of a smaller heat sink, but it has to be calculated.

Bottom line is that I can't calculate the final temperature of your system without knowing more about what you have and what you're doing.

Drivel: I saw a fan powered by a Peltier/Seebeck module using the heat from a wood burning stove at a friends. I immediately decided to throw together something similar for my wood burner. Unfortunately, I neglected to calculate the heat gradient across the heat sink and ended up melting the solder in the Peltier/Seebeck junctions. Oops. Different operating temperatures also require different solder types (Pg 2):

--
Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
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Reply to
Jeff Liebermann

Grin... I've seen a module fail because it overheated itself and melted the solder holding one of the elements in place. (Each element in a peltier is a little hunk of n or p semiconductor. I should look (I-V wise) at some of the pieces.)

George H.

Reply to
George Herold

Hot (or cold) reading...

Ferrotec "Thermoelectric Reference Guide"

"Everything you've wanted to know about TEC's

"Performance Evaluation of a Thermoelectric Refrigerator"

"Improvement In The COP Of Thermoelectric Cooler"

--
Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
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Reply to
Jeff Liebermann

Mine had some help from my wood burner that normally runs at about 150

- 300C surface temperature: I've never tried to melt solder on the stove top. Yet another experiment to try.

I think you'll be disappointed. It's just bulk resistance as in a straight line. It will look something like the Volts-vs-Amps graph on the lower right: Note that the above model is rated to 200C max and might survive overheating. The others are rated at 110 or 125C max. The better modules are made with Bismuth Telluride. Also see Fig 6 at: "Performance Evaluation of a Thermoelectric Refrigerator"

Odd but useful collection of relevant material properties: and an interesting page on TIM (thermal interface materials).

--
Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
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Reply to
Jeff Liebermann

Ooops! That is why they mount them a bit further up. Worth getting one if you have a wood burner where warm air gets trapped in a pre-existing large chimney from an old fireplace. It makes quite a difference.

I tried making one using a high temperature Peltier and a "solar" powered toy motor but it didn't last. Motor didn't like being on its side much or having to do some actual work.

One thing I would like to do although have so far failed is to run a modern high efficiency LED from the waste heat of a candle flame. That is why I had the high temperature Peltier in the first place. There was an ordinary one above it and for simplicity a slab of ice on top.

Even with a slab of the coldest ice I could get and the base hot enough to melt lead solder I couldn't get enough power out to be worthwhile.

I wanted to avoid using any power miser electronics to couple the output to the LED but it seemed pretty hopeless even to drive a red led.

--
Regards, 
Martin Brown
Reply to
Martin Brown

I was thinking of heating up one side and seeing if I could measure the voltage difference..... (probably boring as you say...)

George H.

Reply to
George Herold

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