Thermo electric engineers help please?

Question

If power of the T.E.Module is doubled and the volume is K would it hep me to duoble up the speed of cooling.

Answer 1

As any Electrical Engineer will tell you the resistive or "Joule heat" created is proportional to the square of the current applied (I2 R). This is NOT the case with thermoelectric modules.

The heat created is actually proportional to the current (amperes x volts) because of the flow of current is working in two directions (the Peltier effect). Therefore, the total heat ejected by the module is the sum of the current times the voltage plus the heat being pumped through the cold side.

To understand the capabilities of a thermoelectric module, and related assembly, it is necessary to understand what TE module specifications represent and their implications. The four standard specifications for a module are:

1.) The heat pumping capacity or Qmax in watts

2.) The maximum achievable difference in temperature between the hot and cold sides of the module known as the Delta Tmax or DTmax, usually represented in degrees Celsius

3.) The maximum (optimal) input current in amps or Imax

4.) The maximum input voltage or Vmax when the current input is optimal (Imax).

As a practical matter it is only possible reach either heat pumping capacity in watts or to obtain the maximum temperature differential in degrees.

In other words, the DTmax is the maximum temperature difference between the hot and cold side of the module when optimal power is applied and there is no heat load (Q=0).

As a thermal load Q is added, the difference in temperature between the two surfaces will decrease until the heat pumping capacity or Qmax value is achieved and there is no net cooling (DT=0).

Since your application will likely require net cooling of an object with a thermal mass, the actual heat pumped or Q will be less than Qmax and the actual difference in temperature will be less than the DTmax

Answer 2

You will surely need more cooling. I am not sure exactly what you mean by double the "speed" of cooling. You will definitely need to remove twice the BTUs of heat as you had been removing before. In order to keep the whole thing functioning efficiently, however, I would prefer to see a larger increase in heat dissapation, more like three times, four, or five times that of your original design. The cooler the mechanics run, the longer they will perform at optimum specs.

Consider forced air cooling, or encasement in water cooled jackets with a tight thermomechanical bonding.