0
\$\begingroup\$

As a part of experiment I also want the one side of this TEC(Peltier module) to be held at 50°C(where the room temperature is 25°C) and need to size a heatsink this time for the cold side.

The hot side to be monitored includes only a taped small wire element along with a thermistor for feedback to a TEC controller, and nothing else. And in this case, the heatsink will act as a heat reservoir. But as far as I understand, this time to the cold side I have to attach a heatsink otherwise the cold side will struggle. (For temperature regulation, I can use TEC controller which can both regulate for heating and cooling application.)

Here, its about to add heat to the system rather than remove heat. In cooling, we add Qc and electrical power to find total heat. But its a bit confusing in this case.

So in this case, what should be the way of thinking in steps to come up with a reasonable heatsink for the cold side? An example using the charts of the TEC would be great.

\$\endgroup\$
3
  • \$\begingroup\$ What is your Qh? If it is just the bare hot face of the TEC then I doubt you'll need anything unless you want to prevent condensation on the cold face. \$\endgroup\$
    – vir
    Commented Dec 7 at 19:02
  • \$\begingroup\$ Running water at 50 °C over the part you want to keep at 50 °C should work well. \$\endgroup\$
    – Andrew Morton
    Commented Dec 7 at 20:48
  • \$\begingroup\$ it is just the bare hot face of the TEC \$\endgroup\$
    – user16307
    Commented Dec 7 at 22:33

1 Answer 1

Reset to default
0
\$\begingroup\$

If you want to heat the hot side to 50 C, then you need do nothing with the cold side, except perhaps seal it from condensation. As heat is removed from the cold side, its temperature will drop, the ΔT increases, and at high enough ΔT, Qc will drop to zero. At this point, as much heat is getting to the cold face by conduction through the TEC from its internal resistive heating as is being pumped away by Peltier action. Basically, the TEC has become an expensive, fragile, resistive heater.

If you do add some heat to the cold side, this will appear at the hot side in addition to the resistive heating.

\$\endgroup\$
11
  • \$\begingroup\$ I see it makes very sense now. One more question whats the matter with condensation? What could happen without heatsink on cold side I mean if we dont prevent condensation on the cold face O TEC. What is negative about condensation in this context? \$\endgroup\$
    – user16307
    Commented Dec 7 at 22:41
  • \$\begingroup\$ It's bad because water and electronics usually don't mix. Yours doesn't appear to be sealed so condensation could work its way into the module and cause corrosion or other problems. If the condensation freezes then your Qc could drop since the ice will insulate the cold face. \$\endgroup\$
    – vir
    Commented Dec 7 at 22:56
  • \$\begingroup\$ Neil_UK says in heating mode the ΔT will start to increase and at high enough ΔT Qc eventually will got to zero. Its I think because the TEC controller will supply such current and at that current and ΔT the Oc will be zero. Only heating contribution will be from current and internal resistance of the Peltier. So if not sealed are you saying some moisture or water can get into the Peltier module? Lets say I prevent condensation but use no heatsink, would effect the 50 C regulation? \$\endgroup\$
    – user16307
    Commented Dec 7 at 23:14
  • \$\begingroup\$ Can we say that these charts here peltiermodules.com/peltier.datasheet/TEC1-12706.pdf useless in case of using it as heater? \$\endgroup\$
    – user16307
    Commented Dec 7 at 23:22
  • \$\begingroup\$ @user16307 if you are already using it as a cooler, and sometimes want to heat, then it's excellent as a heater. It's at least 100% efficient, slightly more so if you allow heat into the cold side where it will add a little to the output. If you only ever want to heat, then you are far better off with a resistor, alli-clad bolt-down ones are ideal for heating surfaces. \$\endgroup\$
    – Neil_UK
    Commented Dec 8 at 16:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.