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Jack A Lopez
Jack A Lopez

Best Answer 5 years ago

Well, there is a simple model for heat pumps, based on conservation of energy.

Qhot = Qcold + W

where Qcold is heat moved out of the cold reservoir. W is work done on the system. Qhot is heat dumped into the hot reservoir, and due to conservation of energy, it turns out that, Qhot = Qcold + W.

For refrigerators, the goal is large Qcold in return for small W, so the figure of merit is,

COP = Qcold/W

where COP is well known acronym for "coefficient of performance". It is a measure of how good the fridge is at cooling.

For a heat pump as heater, the goal is large Qhot in return for small W, so the figure of merit is actually COP+1, since:

Qhot/W = (Qcold + W)/W = COP + 1

The people who make Peltier modules, publish their numbers for COP, and the exact number is variable, depending on the temperature difference across the module.

Search for "typical coefficient of performance for peltier modules"


returns a bunch of stuff. This page,


is kind of interesting, because it has a calculator to play with, and it does a good job of giving me an approximate number for COP for a (typical?) Peltier device. E.g. if I enter: 25 C for cold side, and 50 C for hot side, it says Peltier can give me COP=0.596 (and COP+1 = 1.596)

Also I found a blurb on this Wikipedia page,


"...the coefficient of performance of current commercial thermoelectric refrigerators ranges from 0.3 to
0.6, one-sixth the value of traditional vapor-compression refrigerators"

So if we go with Qhot/W =1.6, that is 1.6 times better than an ordinary (Qhot/W =1), cheap, electric heater.

Is that good enough for your thing? Does it justify the expense of using a Peltier module? I dunno. That's up to you, I guess.

BTW, I am just using that link to coolchips.gi for their calculator. I have no idea if the technology they're promising is real or not, or if they're just some sort of scam artists. I suspect the latter, but I don't know for sure.


Reply 2 months ago

I had a few questions regarding a Peltier module.
1.Can a one Peltier module both boil and cool water at the same time.( one side for heating the water and the other side for cooling water)
2.Can a politer run of a solar panel if so how( what is needed)
3.If this is possible could you please tell me how and what is needed to boil water and cool water at the same time with one Peltier module all while being solar powered.

Jack A Lopez
Jack A Lopez

Reply 2 months ago

There is probably a better way to make hot and cold water from sunlight, than by using Peltier modules and photovoltaic (solar) panels.


That article mentioned the "ISAAC Solar Icemaker" and I think that one is worth looking at, erm, searching for...


Regarding question 1, it is difficult to get a single Peltier module to pump heat across a temperature difference (deltaT) of more than about 50 C.

There are graphs published for Peltier modules, showing heat flow (in watts) as a function of deltaT, and also voltage across the module (in volts) as a function of deltaT, and there is a set of these curves for different amounts of current (in amperes) through the module.

I mean really, it is like there are two independent variables: deltaT (in degrees C) and current (in amperes).

It will make more sense if you see the graph.

This page,


has one, under the heading, "Design Using Function Diagrams." Also that page gives some explanation about how to use this diagram.

Also I noticed, a search for images of "peltier data sheet"


several results that are pictures of the same diagram.

Anyway, it is graphs like these, that are the basis for my assertion that deltaT greater than about 50 C, is challenging. 50 C is roughly the difference between boiling water (100 C) and hot water (50 C).

Regarding question 2, it is possible to drive a Peltier module with a solar (photovoltaic) panel, but I do not think it is exactly practical. I mean, you will discover what I mean when you calculate the amounts of electric power (in watts), the product of voltage (in volts) and current (in amperes), and then go shopping for a PV panel big enough to supply that power.

Regarding question 3, I am not going to design a system for you. I am going to let you do the math yourself. I can kind of guess at the results though. That is: large amounts of expensive, solar PV, electricity will be needed to produce small amounts of hot water, and even smaller amounts of cold water.


5 years ago

1. All caps is seen as shouting - please don't the bold isn't making that better.

2. Raise it to what from 50 deg C?

3. What the 60 watts is desn't matter in theory but in practice you will have a lot of losses.

2.26 x 10 to the 6 joules per Kg. - this is to raise the water to boiling point.

you should be able to work out the rest. Sounds like homework.


Answer 5 years ago

Not only does it sound like homework again, the profile pic looks familiar too.