Reference

and https://www.nature.com/articles/s41586-020-1985-6 . What components would be needed to store this triboelectric generation, which from the PDF is about 150v @ 275-300 microamps per cell if sufficient cells used to generate 750 watts of power (5A)? eg: supercapacitor-voltage regulator-battery. I don't know if those would be the correct components for such a charge circuit, just an example. Please be as specific as possible. Battery bank can be configured in multiples of 12v, ie 12v, 24v, 36v, 48v etc.... if that could simplify things. Would like to be able to use inexpensive off the shelf components. For high school science lab experiments.

 

What the heck is "new" about this? It looks like a pretty classic Kelvin generator that has been a stable of home-made experiments in static electricity for over a century. The video doesn't provide enough information about the tweaks they made or give any indication of the relative improvement in efficiency, so there's not much to go on.

If you are doing something for high school lab experiments, why do you need 750 W? Why do you need anything more than enough to demonstrate the principle and that the concept actually works?

Even if you could get all 300 µA from each cell without loss, you'd need almost 17,000 cells to get to your 5 A. So, again, what is it that you are envisioning doing in a high school lab experiment that could justify that?

 

The whole problem with your idea is your thinking that static electricity is the same as current electricity, but it's not. After the staic is discharged there is nothing left to continue powering what ever your trying to power.

 

By my math his math for the power in one droplet stream is off by 24x.

1 LED = 2 V (Vf) x 5 mA = 10 mW
100 LEDs = 1 W
1 W / 140 V = 7.14 mA

7.14 mA / 300 uA = 23.8x

ak

 

What the heck is "new" about this? It looks like a pretty classic Kelvin generator that has been a stable of home-made experiments in static electricity for over a century. The video doesn't provide enough information about the tweaks they made or give any indication of the relative improvement in efficiency, so there's not much to go on.

If you are doing something for high school lab experiments, why do you need 750 W? Why do you need anything more than enough to demonstrate the principle and that the concept actually works?

Even if you could get all 300 µA from each cell without loss, you'd need almost 17,000 cells to get to your 5 A. So, again, what is it that you are envisioning doing in a high school lab experiment that could justify that?

The video is indeed inadequate which is why the PDF of the actual paper was included. Though there are some similarities it is not a kelvin generator, there are no secondary rings. The 750w is to demonstrate that it is possible to do something practical with the energy as opposed to simply demonstrating a theoretical possibility and if the paper is read that is the energy available in per m^2 which is a not unreasonable goal to achieve. So instead of raising "it can't be done" blockades how about indicating the circuitry asked for, please.

 

The whole problem with your idea is your thinking that static electricity is the same as current electricity, but it's not. After the staic is discharged there is nothing left to continue powering what ever your trying to power.

If it will power the lights the charge isn't static and therefore can be collected. The question being asked is how?

 

The principle here seems more like a deactivated battery than a static electricity energy device. The possible amount of KE extracted as electrical energy seems pretty small.

https://www.nature.com/articles/s41378-021-00269-8

A related issue to DEG energy generation is the manufacturing of DEG devices and the pre-charge needed before they can operate at full efficiency. In this paper, we restricted the discussion to the steady-state regime because we believe that the technology is too immature to discuss this point. For example, Xu et al. generator requires 16,000 impinging droplets1 (about 2.4 J) and Wu et al. charged their substrates for 15 min using homogeneous electrowetting-assisted charge injection (h-EWCI)2,4,13 which requires 1.5 mJ of electrostatic energy in theory, but nearly 81 kJ in practice due to the 90 W consumption of the voltage amplifier used in their study. We are hopeful that suitable material choice and optimized industrial setup will dramatically reduce these pre-charge energies.

 

The video is indeed inadequate which is why the PDF of the actual paper was included. Though there are some similarities it is not a kelvin generator, there are no secondary rings. The 750w is to demonstrate that it is possible to do something practical with the energy as opposed to simply demonstrating a theoretical possibility and if the paper is read that is the energy available in per m^2 which is a not unreasonable goal to achieve. So instead of raising "it can't be done" blockades how about indicating the circuitry asked for, please.

Also remember that is per drop, how many drops per second per m^2 fall in a rainstorm?