0
\$\begingroup\$

Suppose I power a Peltier device to heat up/cool down some water. Then I later use the same Peltier device, in reverse, to convert the separated heat energy back into electricity.

I have read somewhere that a Peltier device can realistically achieve about 10% efficiency. Since the device is used twice, will I only recover about 1% of my input energy?

I am building a machine and considering the operation mentioned above. If only 1% would be recovered, I should explore other means; If so, then what heat-pumping/electricity-generating options should I consider if I am concerned about efficiency?

\$\endgroup\$
  • 2
    \$\begingroup\$ It's worse than that, because the temperatures of the two reservoirs will equalize to the environment over time. Peltier-Seebeck modules are not economical outside of some really unusual circumstances to begin with; this arrangement just won't do what you want. \$\endgroup\$ – Hearth May 21 at 15:07
  • 4
    \$\begingroup\$ You would be lucky to even get 1%. The Peltier module itself is a poor thermal insulator, so will conduct heat to the cold side without converting it to energy. \$\endgroup\$ – evildemonic May 21 at 15:14
1
\$\begingroup\$

Lets suppose water (in a perfectly insulated bath) is heated with a resistor (where there will be 100% electric energy to heat) and you put in 1 Joule of energy (1 Watt/second). With a peltier that is connected to something colder than the water, one could get 5 to 8% of the energy back, or around 0.05J.

But this is assuming all of the heat in the water will go back through the peltier, which is not possible in the real world. A good insulated water bath will be needed if you want to store energy in this way. It would be better to use a resistor to heat the water, if you use a peltier to do the heating, there could also be an efficiency loss through the heatsink or whatever other thing its connected to.

It may also be advantageous to heat 1 water bath and cool another, and insulate both. One problem with thermal storage is heat leakage as any heat stored will want to equilibrate with any temperature outside of the heat storage container. So it makes up for a poor long term energy storage solution.

\$\endgroup\$
0
\$\begingroup\$

You should be concerned about insulation between the hold and cold water sides.

Sharing the same water defeats the purpose of the temperature differential sharing both sides.

That would require isolated heatsinks on both sides to achieve 10% efficiency for each response.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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