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Author: Subject: house 40ampers phase limitation ?
avi66
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[*] posted on 27-1-2011 at 13:38
house 40ampers phase limitation ?


i want to run a big cell on 50 amper current, but i can't because my house electricity phase design to deliver max 40amper total, my question is, if i use 2 cell inline using 12 volt, with 25 amper current, i will double the yield ?!?! i will get a yield equal to 50 amper cell by using 2 inline 25 amper cells ?!?! or my phase will jump, because 2 cells on 25 amper current will rise the total phase current to 50 amps?
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Xenoid
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[*] posted on 27-1-2011 at 16:14


In simple terms, neglecting AC and losses.

So your house has 40 Amp MAINS current. That's 40 Amps at say (whatever) 230 Volts.

Power = Voltage x Current = 230 x 40 = 9200 Watts = 9.2 KW

Methinks you could run quite a few cells with 9.2 KW

Your cell is say 50 Amps (Hah!.. do you realise how hot it will get) at say 4 Volts, that's 50 x 4 = 200 Watts = .2 KW

You could therefore run 46 cells.

I hope you are using a transformer or some sort of low voltage power supply, please don't connect your cell directly to the mains :(

Edit: Yes, two separate cells in series, operating at twice the voltage and half the current will be the same as one larger cell. The power consumed will be the same!

[Edited on 28-1-2011 by Xenoid]
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[*] posted on 27-1-2011 at 16:39


That's not electricity! :D
I have three phase with each phase rated at 150 Amps.
So 230 x 150 x 3 = 103500 Watts = 103.5 kW
But I am working in light industrial premises so normal household rules do not apply.
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avi66
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[*] posted on 27-1-2011 at 18:40


so in 1 phase i can use up to 9200 Watts, divide it to 5 volts =1840A, you want to tell me that in 1 free house phase i can run a cell with 1840A ?
and if i use the 12 volts output its will be:
V*I=W
12*X=9200(ONE PHASE)
X=9200/12 (766A)
Then i will be able to use up to 766A at one phase ?
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Xenoid
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[*] posted on 27-1-2011 at 18:51


Quote: Originally posted by avi66  

Then i will be able to use up to 766A at one phase ?


Yeeeee....ss...s! In very simple terms, neglecting losses, etc. But you will need VERY thick cables to carry 766 Amps :o

What sort of a power supply are you planning on using for this 50 Amp cell. You'll need thick (welder, jump start) type cables for 50 Amps.

By the way, are we talking about a chlorate cell here!
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[*] posted on 27-1-2011 at 21:03


Idiotic configuration to so completely unbalance power pulling that many amps off of one phase. Use a 3 phase transformer, output as Wye, float neutral and tap 2 phases, feed your 208 Vac single phase to low voltage high current transformer.

http://www.federalpacific.com/university/transbasics/chapter...

A few more ideas:

http://www.equitech.com/faq/3phase.html

http://cr4.globalspec.com/thread/27383/Three-Phase-to-Single...


[Edited on 1-28-2011 by IrC]




"Science is the belief in the ignorance of the experts" Richard Feynman
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Xenoid
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[*] posted on 27-1-2011 at 21:55


Quote: Originally posted by IrC  
Idiotic configuration .....


Well of course! You don't seriously think it's going to happen do you :o
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avi66
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[*] posted on 28-1-2011 at 03:27


i don't need 766 amper for my cell, i just need 60 A/H maximum, i got mmo anode with the size 5x0.3x36cm(30 in sol) which have surface area of 300c^2 multiple by 200 ma per square cm is 60000 ampers .... i will use probebly 50 ampers with 3 computer power supply 5 volts output.
its just hard to believe that my home can deliver at each phase at list 1500 ampers ... multiple by 3 is 4500 ampers ... i can open a chlorate factory !!!! :D
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hissingnoise
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[*] posted on 28-1-2011 at 03:51


Quote:
... i can open a chlorate factory !!!!

Given the apparent state of your knowledge of electricity, you also have a good chance of electrocuting yourself!
Your time, it seems, would be better spent reading up on all things electrical . . .

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[*] posted on 28-1-2011 at 04:15


I whas just joking man .... just talking theoretically, Thanks for help peoples.
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[*] posted on 28-1-2011 at 09:01


I'm guessing you were also planning to just hook up those three power supplies in parallel to your cells, right?

Seriously, take hissingnoise's advice and spend a bit of time so you understand the basics at least.




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[*] posted on 28-1-2011 at 09:55


if i have a a barrier which capable of standing 50 ampers current , and 3 dc power supply which capable Carrie 20 ampers at 5 volts each one, i will be able to start all of 3 power supply and then open the barrier, so all they power supply will be able to give 50 ampers hours together ?!
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[*] posted on 28-1-2011 at 11:36




Hello Folks,

50 Amps is not a bad size of cell to start off with (shakes head). Suppose it's a case of, reach for the stars and you will (probably) get to the tree tops :D

You can try paralling your three computer power supplies by connecting the three outputs (red, +5V) to the Anode and having three Cathodes. Each Cathode will need to be connected to a seperate black wire(s) from each (computer) power supply. That's the only way I know of combining (in parallel) computer power supplies for this type of application. Manipulate the spacings to get each supply to supply approx. 17 Amps of current.
You will have a bang if you connect the supplies in parallel to one Anode and one Cathode IMO.

You could simply cut your MMO and make three 20 Amp Anodes and run smaller (more sensible size for a beginner) Cells.

See here for some wise words of wisdom!!
http://oxidizing.110mb.com/chlorate/powersup.html

@phlogiston Have you used that signature for long in your career?????? Just wondering. It takes me back........
I believe he once said>
I aim ALL my rockets at the moon.
Sometimes they happen to land on London.

Dann2



[Edited on 28-1-2011 by dann2]
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[*] posted on 28-1-2011 at 14:51


Avi6, from your last post I suspect you believe that you need a 50 ampere supply to reach 50 Ah. If so, then you will probably enjoy learning that you can reach the same number of Ah's with any supply, it will just take longer. With a 50A supply, it takes one hour to reach 50Ah. With 25A, it takes 2 hours. With a more reasonable 2 A and a tiny anode, it will still only take a day. With that 776A supply you previously considered it will take 3 minutes and 52 seconds :)

Dann2, yes, c'est moi. It's been a while... lovely to see you never left this alone for all this time. Can you imagine how I enjoyed reading the immense progress that has been made here with respect to homebrewing anodes when I discovered this forum!

Re the .sig:I recently found out it actually came from a brilliant song by the comedian Tom Lehrer (http://www.youtube.com/watch?v=kTKn1aSOyOs). I guess I've got to change it now.




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