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mike915
07-31-2010, 11:19 PM
which is what warms the cell, volts or amps??
So... which one of this cells will get hotter and which one will produce more HHO??

1. (-NNNNN+NNNNN-NNNNN+NNNNN-) At 12VDC 40 Amps 2V per gap

2. (-NNNNN+NNNNN-NNNNN+NNNNN-) At 24VDC 20 Amps 4V per gap

And for this...

1. (-59N+) At 120VDC 4 Amps 2V per gap

2. (-5N+5N-5N+5N-5N+5N-5N+5N-5N+5N-) At 120 VDC 4 Amps 20V per gap

And why??

Because I thought that the amps were the ones making the cell warmer and not the volts.

myoldyourgold
08-01-2010, 07:58 PM
Continue reading the forum use the search feature. Heat is caused by much more than just volts and amps. You need to understand it all. This takes a lot of study. Most of it has been discussed before.

sokeway
11-28-2010, 08:12 AM
which is what warms the cell, volts or amps??

So... which one of this cells will get hotter and which one will produce more HHO??

1. (-NNNNN+NNNNN-NNNNN+NNNNN-) At 12VDC 40 Amps 2V per gap

This one will produce more HHO


2. (-NNNNN+NNNNN-NNNNN+NNNNN-) At 24VDC 20 Amps 4V per gap

This one will get hotter.

To produce heat, current and voltage have a relationship.
But the voltage is high, the contribution of efficiency is not great, mainly produces hot.
So more waste voltage

kcarring
11-30-2010, 11:55 PM
I find this days newbies are quick to build some BIG cell to produce a geat deal of HHO, only to find out the size engine they are trying to feed is smaller than a 5.0 liter. You can produce enough from one cell configured +NNNNN- (3" X 11") plates and not worry about getting too hot. Then if you need more, let's say 2.5 lpm to feed a 5.0 liter engine you can add a second cell wired seperately. In the end with both cell running you can feed a 7.0 engine with them by adjusting the amp draw from your mixture or PWM if you are running the 28% solution. going big this days is a waste of time and your material. find a way to make your cell productive with less amps not big and draw lots of amps and create heat.

I agree totally. My engine is a 6.2l - and i reconfigured a -NNNN+NNNN-NNNN+NNNN- to incorporate 5N.

While it was an improvement, i'm not so convinced i wouldnt have been better to cut it down to 6N, and just two stacks. Turns out my diesel seemingly, anyway - only like "so much" HHO, and my personal number gets less, and less. With my current config, I have to use the pwm at 100hz, and or dial the constant current down. in my last dry cell it was huge, but leaky and also a voltage hog utilizing nearly 3v per cell. it put out the HHO 9and maybe steam?!) like no tomorrow but sucked up almost 40 amps and my truck ran like ****, due to power loss (electrical) to key components like pumps. Alternator strain too, no doubt, probably lucky i didnt destroy it hahaha i think eventually I will settle with an even less stacks and shoot for 1.8v-2.0v per cell.

Though i still feel I got a good deal, and will not, "knock the vendor" - i was sold on watching a video of it putting out nearly 4l/min. @ 41 amps. Retarded. I figure, no matter how much or little you feed to a motor (HHO) - if that amount does not equal the maximum amount you can utilize to fully combust the given fuel, during a stroke (at the lowest cost measured in amps) - u haven't dialed it in yet.

Farrahday
12-11-2010, 05:09 PM
which is what warms the cell, volts or amps??

So... which one of this cells will get hotter and which one will produce more HHO??

1. (-NNNNN+NNNNN-NNNNN+NNNNN-) At 12VDC 40 Amps 2V per gap

This one will produce more HHO


2. (-NNNNN+NNNNN-NNNNN+NNNNN-) At 24VDC 20 Amps 4V per gap

This one will get hotter.

To produce heat, current and voltage have a relationship.
But the voltage is high, the contribution of efficiency is not great, mainly produces hot.
So more waste voltage

There are some disturbing inaccuracies in this post, and I'm quite surprised that no one has picked up on them and pointed them out. :confused:

Heat produced by a cell, that is any cell or electrolyser, is a measure of the power dissipated. And as P = V x I, then examples 1 and 2 both dissipate 480 Watts, and hence will both get as hot as each other. However as the gases produced are proportional to the current (and not the voltage), all things being equal, example 1 is a more efficient electrolyser and will produce twice as much gases as example 2.

Roland Jacques
12-12-2010, 07:00 PM
Heat produced by a cell, that is any cell or electrolyser, is a measure of the power dissipated. And as P = V x I, then examples 1 and 2 both dissipate 480 Watts, and hence will both get as hot as each other. However as the gases produced are proportional to the current (and not the voltage), all things being equal, example 1 is a more efficient electrolyser and will produce twice as much gases as example 2.

Farrah you have some interesting post and you seem very knowledgeable in many areas. I for one am glad you are here to help folks sort this stuff out.

My experience (be it limited on this subject) shows that the fewer neutral (or floating ) plates the hotter the cell gets with the same watts. This is a Very excepted concept in the Hydroxy community. (if it is wrong it is good to sort it out)

I have never tested for heat specifically to confirm this, but i did notice this to be consistent with a few tests general performance test that I have done and seen through the years.

My thought is Power dissipated could be ETHER Browns Gas OR heat (= steam in many cases). It seems to be more a by-product of higher voltage per gap than plate area. I could be wrong about this I have never really tested for this personally.

What about this example 12 volts system both pulling 10 amps or 120 watts and same plate area.
Will these make the same heat or HHO?

+nn-nn+ (4 volt per gap)

or

+nnnnn- (2 volts per gap)


What roll do you think plate area plays in this?

Farrahday
12-13-2010, 09:43 AM
What about this example 12 volts system both pulling 10 amps or 120 watts and same plate area.
Will these make the same heat or HHO?

+nn-nn+ (4 volt per gap)

or

+nnnnn- (2 volts per gap)


What roll do you think plate area plays in this?

There's definitely some confusion and misconceptions in this area Roland. I x V = P. So it does not matter how the cell is configured, ultimately if the combination of I x V = the same P, then both cells will dissipate the same Power. The fact that some cells might not get as hot as others will come down to the design and the amount of electrolyte and plates therein. Obviously it takes longer to heat 2 litre of water than it does 1 litre from any given power, and a unit with more floating plates to absorb heat will seem cooler.

As for your example, 4 volts is too much across each cell as this will immediately double the power dissipated when compare to the generally accepted optimum of 2 volts per cell. That said, it's not quite as simple as that, for the following reasons.

Take Ohm's Law V/I = R, where R is the resistance of the cell. If we reduce the voltage across a cell, without also reducing the resistance of the cell, then the current through the cell will reduce proportionally - and so will the gas output.

For example:

4 volts/10 amps = 0.4 ohms resistance per cell
2 volts/10 amps = 0.2 ohms resistance per cell

So in this example, 4 volts/10 amps = 0.4 ohms, if we now reduced the voltage across the cell from 4 volts to the optimum 2 volts, the current draw will drop to just 5 amps. (V/R = I)

From what I can tell, some people seem to forget the relationship between voltage, current and resistance and assume some quite odd things.

Plate area plays its part in that the greater the plate surface in contact with the electrolyte, the more area there is for charges to exchange and hence the resistance of the cell is lowered.


What about this example 12 volts system both pulling 10 amps or 120 watts and same plate area.
Will these make the same heat or HHO?

+nn-nn+ (4 volt per gap)

or

+nnnnn- (2 volts per gap)



Both these configurations dissipate the same 120W power, and the total resistance of both configurations is 1.2 ohms, but the 4 volt configuration is made up of two series configurations put in parallel, and each series leg only draws 5 amps. Therefore only 5 amps is passing through all the plates compared to 10 amps of the 2 volt series configuration. As the gas output is directly proportional to the current passing through the system, then the 4 volt configuration will produce half the gas of the 2 volt configuration.

Incidentally, putting two of these series (+nnnnn- (2 volts per gap)) configurations in parallel would give an overall resistance of 0.6 ohms and accordingly draw 20 amps of current through the cell.

The key thing is to have enough floating plates to allow a healthy 2 volt across each individual cell. And, the cells must then be evenly spaced as otherwise you might, for example get one cell seeing 2.5 volts while its neighbour only sees 1.5 volts (which might not be high enough under all conditions to sustain electrolysis).

The larger the surface area of the plates the lower the cell resistance, and the surface area can be vastly increased by cross hatching with an abrasive pad, paper or cloth. And of course, the smaller the space between the plates the lower the resistance of the cell.

Two things that are easily made optimum are the 2 volts across each cell and the cell spacing. As far as the spacing is concerned, you can get plates spaced too closely wherey the evolving gas simply cannot escape and starts to halt the electrolysis process. 1/8th of an inch or 3mm is generally accepted as being the closest practical spacing.

The plate surface area will tend to be limited to the availability of suitable ss and indeed practical size restrictions of the unit you intend to fabricate. Once you have the plates cross-hatched, the appropriate number of plates to give around 2 volts per cell and the plate spacing set to optimum, then it is then simply a matter of adding water and doping the cell with a suitable electrolytic compound until the desired current is drawn.

Current makes gas, not voltage, but to increase the current you either have to up the cell voltage or decrease the cell resistance. Increasing the cell resistance by any number of the methods mentioned above is the way to go. Sure the power dissipation will rise accordingly, but it will be efficent use of the power.