Peltier devices are not very efficient. One problem is that heat is conducted through the device. It's hard to conduct electricity without also conducting heat. As a result, a lot of the heat pumped from the cold side to the hot side leaks back.
Another problem is ohmic heating from the electric current. The amount of heat pumped by the Peltier is proportional to the current, whereas the ohmic heating is proportional to the square of the current. The greater the current the lower the efficiency. For this reason, if you drive the Peltier with rectified AC, it should be filtered to smooth DC for greatest efficiency.
....Single stage thermoelectric devices are capable of producing a "no load" temperature differential of approximately 67°C. Temperature differentials greater than this can be achieved by stacking one thermoelectric on top of another."...
..."The typical efficiency of TEGs is around 5–8%."...
It's a little more complicated. Just as a running motor acts as a generator to generate a back EMF, a TEC generates a reverse voltage driven by the temperature difference. So the voltage applied is the sum of the IR drop plus the reverse voltage.
The amount of heat removed from the cold side is proportional to current, but you have to deduct heat leaking through from the hot side (themal conductivity) and the heat added by ohmic losses (I squared R).
"The figure above represents a thermoelectric couple. It shows some terms used in the mathematical equation:
L = element height
A = cross-sectional area
Qc = heat load
Tc = cold-side temperature
Th = hot-side temperature
I = applied current
Additionally, there is the following:
S = Seebeck coefficient
R = electrical resistivity
K = thermal conductivity
V = voltage
N = number of couples
Here are the basic equations:
Qc = 2 * N* [S * I * Tc -1/2 * I^2 * R * L/A – K * A/L * (Th – Tc)]
V = 2 * N * [S * (Th -Tc) + I * R * L/A]
The first Qc term, S*I*Tc, is the peltier cooling effect. The second term,1/2*I^2*R*L/A, represents the Joule heating effect associated with passing an electrical current through a resistance. The Joule heat is distributed throughout the element, so 1/2 the heat goes towards the cold side, and 1/2 the heat goes towards the hot side. The last term, K*A/L*(Th-Tc), represents the Fourier effect in which heat conducts from a higher temperature to a lower temperature. So, the peltier cooling is reduced by the losses associated with electrical resistance and thermal conductance.
For the voltage, the first term, S*(Th-Tc) represents the Seebeck voltage. The second term, I*R*L/A represents the voltage related by Ohm’s law.
These equations are very simplified and are meant to show the basic idea behind the calculations that are involved. The actual differential equations do not have a closed-form solution because S, R, and K are temperature dependent. Unfortunately, assuming constant properties can lead to significant errors."
Peltier devices are much more efficient when the temperature differential is lower:-
I can't immediately find a graph which extends into the negative differential region, but, I doubt if it gets worse than the 40 Watts at 0 shown here for this device.
On the other hand I can't quite see why you need anything more than a "simple" heat exchanger for your situation.
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Use as many TEC1-12706 12V peltier modules as you need. Less than $1.50 each
Buy a (several) PC PSU(s) for the 12V supply. You just need to short the power good pin to ground to make the supply operate in the absence of a PC motherboard. Unless you're going for a huge system, this is easily the most cost effective way to buy 12V power supplies.
The "mechanical" design of the equipment will be much more challenging.
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