Question about amp vs. volts

The wires heat up because they are too thin for the amount of current being used. The amount of H2 is proportional to current, all other factors being equal (which they are generally not).

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Dirk

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Measure the current, and use that measurement to determine the wire size you need. There's got to be a wire size vs. current table on the web -- I just use the one in the ARRL handbook. If the current is high enough (above 10 amps) then you could just use the AWG ratings for house wire.

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Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

Another thought:

I'm not too familiar with the physics of electrolysis, but I imagine that the voltage drop due to the electrolysis itself is pretty constant, so you cell current will be limited largely by the resistance of your external circuitry. If this is the case you'll want to (a) find out what the magic voltage is and don't use much more than that for efficiency, and (b) use something to limit the current to your cell -- in your case I'd suggest a whopping big power resistor, which you can make on the cheap from a bit of heater wire.

Don't burn your house down -- it makes for embarrassing explanations.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

to Jim's Engineering page to the wiretable excel spreadsheet

Jim

Ah, time for a back-of-the-envelope calculation...presuming you put in two electrons per hydrogen molecule and your nominal one liter per minute is at STP, that should be about 2electrons/molecule * 1mole/

22.4liters * 1 liter/minute * 6.022e23 molecules/mole * 1.602e-19 coulombs/electron * 1minute/60seconds = 143 coulombs/second = 143 amps (assuming I didn't make some dumb mistake).

I'm not surprised that your wires are getting hot, if you're using something like 12AWG or smaller.

Have you calculated the efficiency of the system? That is, how does the energy available from burning the generated hydrogen compare with the energy you put in to generate that hydrogen? Depending on your reason for generating hydrogen gas, you may or may not care about that...

Ohms law!

V/I = R

P = V^2/R = I^2*R

Either cut down V, which cuts down I or decrease R!!!!

If you need that much current(as you essetnially cut I in half and got about half the amount of hydrogen) or cut R

Its much easier to cut I if you can.

You can use several smaller generators instead of one large one.

Essentially you can run 4 generators at half the current/voltage for the same "heat price" at one but get twice as much hydrogen.

e.g., run 2 smaller generators at half the voltage, get 1/2 the "heat"(1/4 per generator) and get the same amount of hydrogen. (assumes current and resistance do not change)

Decreasing R will decreasing the heat dissipated (P = I^2*R) but you must decrease V too!!! (else I increases!!)

So, say you have V = 12V, R = 1ohm

This gives

I = 12A and P = 144W

Now if you decrease R by 1/2 so R = 0.5ohm

then

I = 24A and P = 288W. (V is still 12V)

But if you make V = 6V the

I = 12A and P = 72W!

So same current in first one but half the voltage and half the resistance will cut the power in half! (and you'll still get the same amount of hydrogen!!)

Point being that you have control of V and R but I and P depend on them. You can play around with V and R to get what you want.

You can also solve for I and P such that

R = P/I^2 V = P/I

So if you want 36W dissipation and 12A current then

R = 1/4Ohm V = 3V

(note same current as all the other examples but lower power dissipation(heat in the wires(and larger wires)))

hehe, he'll figure that out the hard way when he gets his electric bill. 30A continously for a month is about 150$ at 0.06$/kWH(which now days is unheard of).

If he runs that stuff for a month his bill will be up in the thousands. I wonder how much hydrogen he can buy with that? ;) (I bet more than he generated!! ;)

He's probably a Democrat and doesn't understand that there's no hydrogen tooth fairy ;-)

...Jim Thompson

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Oh, goody. More back of the envelope stuff. $150/$0.06/kWH =

2500kWH. Assuming 30*24 hours in a month, that's just under 3.5kW. The OP mentioned 12V, I believe, for the high rate I figured, and assuming 90% efficiency getting from the 3.5kW power line voltage, that's about 260 amps available. That would be about 2 liters per minute, or 86400 liters per month. At 22.4 liters/mole and 2 grams per mole, that works out to about 7.7 kilograms of hydrogen. I gather from a quick Google search that hydrogen goes for around $1 to $2 per kilogram, so at the outside, that's about $15 worth of hydrogen. How does the energy available from burning 7.7 kilograms of hydrogen compare with the energy available in a hundred or so kilos of gasoline that you could buy for $150?

I suspect the OP is just the Troll alter-ego of some regular poster. That doesn't make the b.o.t.e. stuff any less fun though.

oops, I used 110V... forgot that it was 12V. So I'm off by a factor of about

1. (in any case my conclusion is still right ;)

You need to do the following:

Measure the current for each case. Theoretically, the quantity of hydrogen produced is proportional to the current. The electrolysis cell presents a certain V-I characteristic, probably not linear like a resistor. But the higher the voltage, the higher the current once you have overcome the potential needed to electrolyze H2O.

Other variables will be the composition of either electrode and the type and concentration of electrolytes or other additives you add to the solution.

Your goal should be to produce as close to the theoretically ideal amount of gas for a given current with the lowest voltage possible. And without corroding the electrodes (or blowing up your house).

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Paul Hovnanian     mailto:Paul@Hovnanian.com
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Tom, He is probably measuring 993ml of GAS being produced, not hydrogen. Since he is putting in so much more than the overvoltage, he is wasting most of his power in generating steam and water vapor. He is proabably also just producing Browns gas, so we wont' have to worry too much in replying, as he should have killed himself before the week is out...

-- Charlie Edmondson Edmondson Engineering Inc