Wow? Intake to HHO ratio

[Philldpapill]

Has anyone ever done a very simple calculation? I was thinking about this today and I'm amazed...

Let's say we have a fantastic 10LPM cell on our vehicle. Let's also make an assumption that this is a piddly little 2.0 liter four stroke engine. If your car is just idling at 500rpm's... That means it is sucking in 2.0 liters, 250 times a second(once every other revolution). That's 500 liters of air per minute.

In other words, the amount of HHO going into the air intake only constitutes 2% of the total volume of air. Wow. On top of that, 33% is just oxygen. The atmospher contains roughly 21% oxygen.

Now, granted, even this is a VERY liberal estimate of how much HHO is being pumped into an engine this size. Even half of the 10LPM would be great for an engine this size. So, in reality, it's more like 1% of the intake is HHO... Or am I missing something big here? If this is the case, it seems WAY more efficient to produce bottled H2 and feed it in while driving...

[Roland Jacques]

Take it to the next step. Most cars are seeing results at 0.5 liter per 1.0 litter of engine displacement. So a more common number for the cell would be 1LPM cell. Then add that HHO is only 2/3rds H2 (by molecule). By volume it may be closer to 1/3 because O2 molecule is larger than H2 ( I'm not sure how much larger O2 molecule is over a H2 molecule so I'm guessing)

So less than 0.2% of H2 may be a more accurate number.

[redrat100]

Has anyone ever done a very simple calculation? I was thinking about this today and I'm amazed...

Yes.

Solving for engine airflow rate, where:
3456 = volumetric constant
CFM = cubic feet per minute
CID = cubic inch displacement
RPM = revs per minute
VE = volumetric efficiency, assumed to be 75% (not to be confused with thermal efficiency which is around 25% at best)

Using your figures (500 rpm seems a bit low for idle... my 2L engine idles at 950 rpm):
CFM = (CID x RPM x VE)/3456
CFM = (122.047 {2 liters} x 500 x .75)/3456
CFM = 13.243 = 375 LPM
Your 2L engine pumps 375 lpm of fuel air mixture throught the combustion chambers at 500 idle rpm. So, your idle HHO percentage, at 10lpm, is really 2.67%.

Now for highway speeds. Substituting 3000 rpm in the above equation gives 79.46 CFM = 2250 LPM. Your 10 lpm HHO ratio freeway speeds is closer to .44%.

It seems the consensus here that 1 lpm HHO for every liter of engine displacement is what is required for measurable mileage increases. So the ratio is even lower. The trick is building a cell small enough, with a high enough output, with low power consumption.

Or, you could stuff a H2 gas cylinder into your trunk as you suggest but I would not want to be anywhere near the car when it goes off.

[Roland Jacques]

Yes.

Solving for engine airflow rate, where:
3456 = volumetric constant
CFM = cubic feet per minute
CID = cubic inch displacement
RPM = revs per minute
VE = volumetric efficiency, assumed to be 75% (not to be confused with thermal efficiency which is around 25% at best)

Using your figures (500 rpm seems a bit low for idle... my 2L engine idles at 950 rpm):
CFM = (CID x RPM x VE)/3456
CFM = (122.047 {2 liters} x 500 x .75)/3456
CFM = 13.243 = 375 LPM
Your 2L engine pumps 375 lpm of fuel air mixture throught the combustion chambers at 500 idle rpm. So, your idle HHO percentage, at 10lpm, is really 2.67%.

Now for highway speeds. Substituting 3000 rpm in the above equation gives 79.46 CFM = 2250 LPM. Your 10 lpm HHO ratio freeway speeds is closer to .44%.

It seems the consensus here that 1 lpm HHO for every liter of engine displacement is what is required for measurable mileage increases. So the ratio is even lower. The trick is building a cell small enough, with a high enough output, with low power consumption.

Or, you could stuff a H2 gas cylinder into your trunk as you suggest but I would not want to be anywhere near the car when it goes off.

is 375 lpm correct?
Or would it really be half that 188 LPM since only 50% of your RPMs intake strokes.

Do you know how to determine By volume how much H2 is in 1 liter of HHO?

[Philldpapill]

Roland, for every 2 molecules of H20 that are electrolyzed, you get 2 molecules of H2, and 1 of O2. I've done an expirement that seperates the gases during electrolysis(I'll show you if you want - it's all PVC pipe). The volume that the H2 took up was exactly twice the volume of the O2(give or take <1%). This makes sense because you have exactly twice as many H2 molecules floating around in there.

So, in a nutshell, for every liter of HHO, you have 2/3L of H2, and 1/3L of O2

[Stevo]

Wouldn't VE be more like ~15-20% at idle? Just a guess, but here's a good VE thread: http://www.physicsforums.com/archive...p/t-65238.html

[Roland Jacques]

Roland, for every 2 molecules of H20 that are electrolyzed, you get 2 molecules of H2, and 1 of O2. I've done an expirement that seperates the gases during electrolysis(I'll show you if you want - it's all PVC pipe). The volume that the H2 took up was exactly twice the volume of the O2(give or take <1%). This makes sense because you have exactly twice as many H2 molecules floating around in there.

So, in a nutshell, for every liter of HHO, you have 2/3L of H2, and 1/3L of O2

Ok, I'll go with your experiment numbers. I just seem to remember somthing to the effect that the Oxygen molecules was much larger than Hydrogen, like 2-3 times maybe not.

So is your idea to boost with H2 from a compressed bottle? How cost effective is that?
Assuming we go with the standard 0.5 to 1.0 LPM (x.66= 0.33 to 0.66 LPM) per Liter of engine displacment numbers.

[redrat100]

Do you know how to determine By volume how much H2 is in 1 liter of HHO?

It is always a 2 to 1 ratio, 2 Hydrogen and 1 Oxygen. Molecule size has nothing to do with it. The equation looks like this:

2H2O (liquid) + vdc --> 2H2 (gas) + O2 (gas)

I liter of HHO = 2/3 liter H2 and 1/3 Liter O2. Always.

[Philldpapill]

redrat100 is right for 99% of the cases... Unless you've got serious oxidation issues with your cells like if your using a really crappy plate material that oxidizes easily. If that's the case, then you'll have a lot more H2 than O2 because the O2 is tide up in the oxidized material(like iron oxide).

[Roland Jacques]

It is always a 2 to 1 ratio, 2 Hydrogen and 1 Oxygen. Molecule size has nothing to do with it. The equation looks like this:

2H2O (liquid) + vdc --> 2H2 (gas) + O2 (gas)

I liter of HHO = 2/3 liter H2 and 1/3 Liter O2. Always.

I would like some clarification if you can.

1. Are you saying that H2 molecules are the same size as O2 molecules?
2. Or are you saying it does not matter if one is bigger than the other?

I dont know much about chemistry so i don't know for sure "size" of this molecules. But i was just thinking the size of the molecules would matter. If H2 molecules were the size of grapes and the O2 was the size of watermelons 100 grapes would displace far less space than 50 watermelons. What am i missing?