I built 1 in my shed @ home. Difference was i filled a Boxing bag with water. Dripped it through ring's ontop ov each other with 1 flow. Bucket was negative looped with positive ring to batterree.
With a little tricky design you can beat 10 kV by about a factor of three. If you used curved supports for the rings and let the generator charge a Leyden jar instead, you can get at least 30 kV (based on spark length). The very limited power available from the low head and flow of the water means this isn't a viable source of generation for electrical energy as we usually think of it -- but there's no reason one of these couldn't be used to power a TEA laser (the UV beam from which can then excite a dye laser to produce a visible beam). I've seen one work off a Wimshurst machine; this has a little lower voltage, but it only takes about 5 kV to make the laser cavity discharge in a TEA laser, so this ought to work -- and as a bonus, the laser discharge won't completely draw down the Kelvin generator's charge, so it won't take long to start building charge separation again after each pulse.
If your container has 1 liter of water, and the height is 1 meter. The flow rate at 1 liter/hour. That is 1 kg/3600 seconds. At 9.8 (Joules/kg)/meter. Watts_maximum = MassFlowRate*GravitationalAcceleration*Height = (1 kg/3600 seconds) * ((9.8 Joules/kg)/meter) * (1 meter) Watts_maximum is about 2.72 milliWatts Watts = Volts*Amperes = (10,000 Volts) * Current Current_maximum = (2.72E-3 Watts/10000 Volts) = 2.72 E-7 Amperes = 272 nanoAmperes ( 2.72E-7 Coulombs/second) = (1 kg/3600 second)/(Constant_kg_per_Coulomb) Constant_kg_per_Coulomb = (1 kg/3600 second)/( 2.72E-7 Coulombs/second) = 1021.24183 (Kilograms/Coulomb) FaradaysConstant = AvogadrosNumber*ElectronCharge = (6.02214076E23 particles/mole)*(1.602176634E-19 Coulombs/Electron) FaradaysConstant = 96485.3321233 Coulombs/Mole FaradaysConstant = 96,485,332.1233 KiloCoulombs/KiloMole MolecularWeightWater = 18.01528 Kilograms/KiloMole PH water is about 1E-7 moles Hydrogen ions per liter (1000 grams/18.01528 grams/mole) = 55.5084351 Moles Water 1E-7 moles/55.5084351 moles = 1.801528E-9 Hydrogen ions per neutral molecules of H2O. But breaking up the droplets causes frictional charge separation, and that can be measured by measuring the current in your experiment with a nanoAmpere meter of some sort. But could it be that the charge is just the hydrogen ions already there but getting separated? CHECK ME. I get tired and it is hard to remember all the constants. Use CoData and a spreadsheet or Jupyter notebook. There are about 20,000 colleges and universities and most are on the Internet. And 100s of thousands of high schools. I could do these calculations in middle school. And NONE of those many groups are working globally on making sure the thousands of duplicate presentations is consistent. Get serious please. Richard Collins, The Internet Foundation
Did you measure the pH of the water in both canisters? They should be different right? Also what is the reason the voltage does not increase above 10-12 kV?
It's a point. pH is the log of the concentration of hydrogen ions, so the few free hydrogen ions in solution would be strongly directed to one container only. I don't know much about how this would work in a system with an overall static charge though. @@VoidHalo
This seems to have a high voltage low current. But how would analogue voltmeter detect this? How would an analogue voltmeter detect the voltage but higher currents? An analogue voltmeter determines the emf by how much current is passing in its coils?
see Pyramid Code documentary. I think evaporating water below would have negative ions attracted UP thru the structure toward the apex to interface with the sky. If the granite is also emitting ions (slightly radioactive), that helps it go as well.
I live in a place with many severe wind storms in I was thinking that I could use fences that sort of criss-crossed double fence lines as a wind block.
psst. now, after the pails have been charged, suspend a conductive disc from a firm linkage, aka pendulum, which should oscillate between the two pails. thus, you have created a clock which was originally a method of measuring the length of time the device remained operational for proof of perpetual motion. the device was marketed as a power-drain instead to consumers for profit by the electrical-industrial complex. have the fun.
I built 1 in my shed @ home. Difference was i filled a Boxing bag with water. Dripped it through ring's ontop ov each other with 1 flow. Bucket was negative looped with positive ring to batterree.
With a little tricky design you can beat 10 kV by about a factor of three. If you used curved supports for the rings and let the generator charge a Leyden jar instead, you can get at least 30 kV (based on spark length).
The very limited power available from the low head and flow of the water means this isn't a viable source of generation for electrical energy as we usually think of it -- but there's no reason one of these couldn't be used to power a TEA laser (the UV beam from which can then excite a dye laser to produce a visible beam). I've seen one work off a Wimshurst machine; this has a little lower voltage, but it only takes about 5 kV to make the laser cavity discharge in a TEA laser, so this ought to work -- and as a bonus, the laser discharge won't completely draw down the Kelvin generator's charge, so it won't take long to start building charge separation again after each pulse.
Yes, but the most important question. If I put my tongue on the contacts can I (un)Safely discharge it? 😂😂
If your container has 1 liter of water, and the height is 1 meter. The flow rate at 1 liter/hour. That is 1 kg/3600 seconds. At 9.8 (Joules/kg)/meter.
Watts_maximum = MassFlowRate*GravitationalAcceleration*Height = (1 kg/3600 seconds) * ((9.8 Joules/kg)/meter) * (1 meter)
Watts_maximum is about 2.72 milliWatts
Watts = Volts*Amperes = (10,000 Volts) * Current
Current_maximum = (2.72E-3 Watts/10000 Volts) = 2.72 E-7 Amperes = 272 nanoAmperes
( 2.72E-7 Coulombs/second) = (1 kg/3600 second)/(Constant_kg_per_Coulomb)
Constant_kg_per_Coulomb = (1 kg/3600 second)/( 2.72E-7 Coulombs/second) = 1021.24183 (Kilograms/Coulomb)
FaradaysConstant = AvogadrosNumber*ElectronCharge = (6.02214076E23 particles/mole)*(1.602176634E-19 Coulombs/Electron)
FaradaysConstant = 96485.3321233 Coulombs/Mole
FaradaysConstant = 96,485,332.1233 KiloCoulombs/KiloMole
MolecularWeightWater = 18.01528 Kilograms/KiloMole
PH water is about 1E-7 moles Hydrogen ions per liter (1000 grams/18.01528 grams/mole) = 55.5084351 Moles Water
1E-7 moles/55.5084351 moles = 1.801528E-9 Hydrogen ions per neutral molecules of H2O.
But breaking up the droplets causes frictional charge separation, and that can be measured by measuring the current in your experiment with a nanoAmpere meter of some sort. But could it be that the charge is just the hydrogen ions already there but getting separated?
CHECK ME. I get tired and it is hard to remember all the constants. Use CoData and a spreadsheet or Jupyter notebook. There are about 20,000 colleges and universities and most are on the Internet. And 100s of thousands of high schools. I could do these calculations in middle school. And NONE of those many groups are working globally on making sure the thousands of duplicate presentations is consistent. Get serious please.
Richard Collins, The Internet Foundation
Very interesting 👍
Amazing.
Sorcery! I Tell Ya! Soorrccceerrryy! 😂
Did you measure the pH of the water in both canisters? They should be different right?
Also what is the reason the voltage does not increase above 10-12 kV?
Why would the PH be different?
It's a point. pH is the log of the concentration of hydrogen ions, so the few free hydrogen ions in solution would be strongly directed to one container only. I don't know much about how this would work in a system with an overall static charge though. @@VoidHalo
Yeah, I'm sure bro sent the water in for a genetic pregnancy test too. What kind of question is that hotdog boy? 😂 Jk
@@6022We're not building nuclear reactors here.
Thanks! Love it!
This seems to have a high voltage low current. But how would analogue voltmeter detect this? How would an analogue voltmeter detect the voltage but higher currents? An analogue voltmeter determines the emf by how much current is passing in its coils?
Kelvin Generator+RAM pump= unlimited Energy 🙏
💀
very good setup. i like it. now, work on your microphone setup.
All you need now is to use it to charge up a leyden jar.
But how much current ie power dose it have an infinite internal resistant if so no power 😊
What if the water isn't de ionized? Just wondering because they theorized the Pyramids utilized this method...
see Pyramid Code documentary. I think evaporating water below would have negative ions attracted UP thru the structure toward the apex to interface with the sky. If the granite is also emitting ions (slightly radioactive), that helps it go as well.
another good video on RUclips is ruclips.net/video/KMAtkjy_YK4/видео.html
can the water be return to the top so to never run out?
Ram pumps
I live in a place with many severe wind storms in I was thinking that I could use fences that sort of criss-crossed double fence lines as a wind block.
Dude
Good luck
psst. now, after the pails have been charged, suspend a conductive disc from a firm linkage, aka pendulum, which should oscillate between the two pails. thus, you have created a clock which was originally a method of measuring the length of time the device remained operational for proof of perpetual motion. the device was marketed as a power-drain instead to consumers for profit by the electrical-industrial complex. have the fun.
I tried making one these when I was in Junior High School in 1969