# Question about amp vs. volts

Discussion in 'Electronic Design' started by MeSoDumm, Aug 5, 2008.

1. ### MeSoDummGuest

I've made my Hydrogen generator, but I am having an issue with the
wires heating up really bad!! What do I need to do?? I have hooked
it up to 12 volts and it puts out 993ml per minute, when i hook it up
to 6 volts it cuts it to 422ml per minute. What should the ohms
resistance be? or does that matter? Thx for all the help!!

2. ### Dirk Bruere at NeoPaxGuest

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).

--
Dirk

http://www.transcendence.me.uk/ - Transcendence UK
http://www.theconsensus.org/ - A UK political party
http://www.onetribe.me.uk/wordpress/?cat=5 - Our podcasts on weird stuff

3. ### Tom BruhnsGuest

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...

4. ### Jon SlaughterGuest

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)))

5. ### Jon SlaughterGuest

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!!

6. ### Tom BruhnsGuest

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.

7. ### Jon SlaughterGuest

oops, I used 110V... forgot that it was 12V. So I'm off by a factor of about
10. (in any case my conclusion is still right

8. ### Charlie E.Guest

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...