I'm planning to use a TEC (thermo-electric cooler) in my computer and the math of these things are throwing me off.
The simplest and most inexpensive approach I was thinking of was using a small 30watt TEC on the north-bridge of my motherboard using the stock heat sink. I did some analysis but I am confused as to how much heat is coming out. The TEC device I was planning on is by a company in New Jersy called Melcor, here's the link for the device.
These are the important stats:
Imax=3.9 amps (max current draw at maximum voltage)
Qmax=33.4 watts (at maximum current and voltage)
Note: according to Melcor, Qmax is the maximum heat absorption, the cold side.
Using the equation V=IR,
Resistance of the device at Vmax/Imax is just under 4 ohms.
Assuming that R is constant with temp, this TEC draws out a bit over 3 amps @ 12.38 volts from my PSU, which is 40 watts of electricity to run.
I've set up an efficiency rating, that ratio of Q/(IV) should show how much Qmax is expected per total power used. For this device that efficiency is 0.55611 @ 25*C. If Melcor intends Qmax to be the coldside, that implies a 55.6% efficiency...is that realistic? That would mean that 44.4% of power going in comes out as heat. It is hard for me to accept that which is why I feel I may have made a mistake.
As of my north-bridge, I'm not worried. The power levels are small and the stock aluminum heat-sink can easily pump out 22 watts of heat, if I think Qmax is the heat expelled to run the device.