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Author Topic: What factors affect the voltage produced by a Kelvin's Dropper?  (Read 8044 times)

Offline Laura_Kelly

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What different variables are likely to affect the voltage produced by a Kelvin's Dropper? And what effect will it have (i.e. is the voltage increased or reduced and how)?


 

Offline Soul Surfer

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Look up  http://en.wikipedia.org/wiki/Kelvin_water_dropper fo more information

This is an electrostatic inductive generator using positive feedback processes similar to those that happen in thunderstorms to generate lightning

The main process affecting the output is probably the number of dtrops and the speed of their passage through the system.
 

Offline rosy

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See:
http://www.thenakedscientists.com/HTML/content/kitchenscience/garage-science/exp/electricity-from-water-kelvin-water-drop-generator/
too.. (you probably already have, but just incase)
Playing with that one, the voltage tends to drop off due to (charged) water droplets landing on the (oppositely charged) rings...
 

Offline Laura_Kelly

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On those lines, because I can't easily get hold of a voltmeter going up to the maximum voltage that is suggested will occur due to the positive feedback loop, does anyone know an equation that can tell me the sparkover voltage (the voltage required to get a spark through the air) at different differences so I can then measure the voltage that is produced? That way I don't have to pay for an expensive voltmeter and risk the leaking of charge through the connection with the voltmeter as well.
 

Offline Laura_Kelly

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Also, does anyone think that these would also be factor affecting voltage?

1. The temperature of the water used
2. The length of the connecting wires
3. The type of water used (saline, deionised etc...)

The aim of the experiment I must do is to produce the maximum voltage possible by investigating the various parameters that limit it, and then prevent my findings in a physics fight. As I am the team leader for this experiment, I have to put together the hypothesis and method prior to our group constructing it. Are there any recommendations from those who have built one as to the best ways to help maximise voltage?
 

Offline daveshorts

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The limiting factor is definitely the rate at which you loose charge vs the rate at which you generate it.

The temperature could have an effect as water vapour could easily carry charge (I think easier than dry air) so this will tend to short out the machine.

Another major issue is sharp edges. A sharp edge will have a large electric field (= rate of change of voltage with distance) so smooth curves will help you.

Insulating things may help as well. Good luck
 

Offline syhprum

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A common or garden voltmeter reading up to 1kv normally has an input resistance of 10 Meg ohms you can extend this range with a string of 10 megohm resistors and test it on TV CRT that operates at about 28kv.
If the source resistance of what you are trying to measure is very high use it to charge up a capacitor to increase the power available (the TV CRT could well serve as this capacitor or make one with a jam jar)
 

Offline Bored chemist

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On those lines, because I can't easily get hold of a voltmeter going up to the maximum voltage that is suggested will occur due to the positive feedback loop, does anyone know an equation that can tell me the sparkover voltage (the voltage required to get a spark through the air) at different differences so I can then measure the voltage that is produced? That way I don't have to pay for an expensive voltmeter and risk the leaking of charge through the connection with the voltmeter as well.
An ordinary voltmeter will load the output of one of these generators down so far as to be meaningless.
Here are some tables of spark lengths
http://home.earthlink.net/~jimlux/hv/spherev.htm


BTW, my guess is thet the voltage will practically be limited by the insulation and corona losses.
However in theory I think the limit is the energy of the falling drop measured in electron volts.
That will turn out to be pretty big.
Also the current should be the (mean) number of electrons per drop times the charge on an electron times the number of drops per second
 

Offline syhprum

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A 10 MΩ voltmeter and a string of 99 10MΩ resistors should be able to measure the voltage to which the capacitor has charged a 1nf capacitor and 1000MΩ gives a time constant of 1 second which would enable one to calculate the the initial voltage with the aid of a stop watch.
have present day researches lost the art of improvisation ?
 

Offline daveshorts

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Be very careful with high voltages and capacitors. With a static machine like this it only doesn't kill you because it doesn't store enough charge to do so, it has plenty of voltage. A capacitor can store lots of charge and make them lethal if you don't do your sums right...
 

Offline Bored chemist

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What factors affect the voltage produced by a Kelvin's Dropper?
« Reply #10 on: 07/10/2009 19:27:05 »
A 10 MΩ voltmeter and a string of 99 10MΩ resistors should be able to measure the voltage to which the capacitor has charged a 1nf capacitor and 1000MΩ gives a time constant of 1 second which would enable one to calculate the the initial voltage with the aid of a stop watch.
have present day researches lost the art of improvisation ?
OK, that's fine if you have a 1 nF cap that will actually work at this sort of voltage- call it 10 KV or more. Most people don't have any UHT caps about the place so they are likely to use the intrinsic capacitance of the equipment. Clearly that's going to vary with the set up but lets see if we can get some sort of idea what the capacitance might be.
The capacitance of an isolated sphere (I know the joke about the racehorse already thanks) has a capacitance of 4pi e0 R
e0 is about 9E-12
so for 4 pi eo r to be 1E-9 you need r =about 9 metres.

I'm all for improvisation- but my bedroom's not big enough for that experiment.

Here's another estimate of the capacitance- it's the capacitance of a few bits of wire. Now, for some wire (like the stuff used for TV antenna cables) I can look up the capacitance. It's typically about 50pF per metre and I doubt thre's a whole metre of wire in the setup we might use (I also doubt that the wire is neatly surrounded by an insulated electrode giving a nice high capacitance).

So we might have a few tens of pF of capacitance.
If you spend a dull afternoon soldering a hundred resistors together (and I have done that BTW) then you will have about 10mSec or less to measure the voltage before it dissipates .

Or, like I said, you can use a spark gap.
 

Offline syhprum

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What factors affect the voltage produced by a Kelvin's Dropper?
« Reply #11 on: 07/10/2009 22:11:38 »
Don't use a voltmeter use a oscilloscope, then 10 m/s will be no problem, a 1nF capacitor is easily constructed from a glass jar and metal foil, they were good enough for Hertz and Marconi.
 

Offline Bored chemist

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What factors affect the voltage produced by a Kelvin's Dropper?
« Reply #12 on: 08/10/2009 19:05:13 »
10m/s is about 23MPH isn't it?
a 'scope will cope with 10ms quite easilly- even one based on a PC sound card should work. I mention that because, for many people, it's the only 'scope they have access to.
Spark gaps were also good enough for Hertz and Marconi
 

Offline Laura_Kelly

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What factors affect the voltage produced by a Kelvin's Dropper?
« Reply #13 on: 09/10/2009 04:25:46 »
Thanks for those ideas, will get onto building this stuff and let you know how it goes.
 

Offline Geezer

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What factors affect the voltage produced by a Kelvin's Dropper?
« Reply #14 on: 09/10/2009 05:51:44 »
You can turn your PC into a scope with, for example, an Optascope. I bought one a few years ago because I thought it might be useful, but I got sidetracked into other things, so I never used it in earnest, but it should work for your application. You will probably have to create a potential divider as those above have suggested, but the scope should let you record the voltage decay rate.
« Last Edit: 09/10/2009 05:53:23 by Geezer »
 

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What factors affect the voltage produced by a Kelvin's Dropper?
« Reply #14 on: 09/10/2009 05:51:44 »

 

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