Need some help with my math, how much HHO needed?
 

First off, I know its not practical at all to run an engine on solely HHO made with a common hydrogen generator. But that?s besides the point, I need some help with the math here.


Can someone help me figure out what I did wrong with these calculations and what steps I need to add in?

A Geo Metro 1.0L engine running at 3200rpms (estimated at 65MPH) and at 60% throttle would require 67.8042CFM (cubic feet per minute) of the air fuel ratio.

(1000cc X 3200rpms) / 2.54 / 2.54 / 2.54 / 12 / 12 /12 = 113.0069cfm X 60% = 67.8042

We?ll assume that we are running a stoichiometric ratio at 14.7:1. That means there is a total of 15.7 parts (14.7 + 1 = 15.7). We?d need 63.4854 of air and 4.3187CFM of fuel.

(67.8042 / 15.7) X 14.7 = 63.4854CFM of air.
(67.8042 / 15.7) X 1 = 4.3187CFM of fuel.

An important key to remember is that Hydrogen Generators split the water, H2O. This means it not only generates hydrogen but also oxygen. There are 2 hydrogen molecules and 1 oxygen molecule, which gives us a total of 3 molecules. (2 + 1 = 3). So, in order to get 4.3187CFM of fuel (Hydrogen), we?ll need more than 4.3187CFM of HHO because part of it is oxygen. So we?d need 6.4781cfm of HHO to create 4.3187cfm of Hydrogen.

(4.3187cfm of HHO / 2) X 3 = 6.4781

With 6.4781cfm of HHO, we?d have?

(6.4781 / 3) X 2 = 4.3187cfm of Hydrogen
(6.4781 / 3) X 1 = 2.1594cfm of Oxygen

Therefore, in order for a 1.0L engine to run on HHO we?d need to inject 6.4761CFM of HHO and the engine would have to suck in 61.3260CFM of air.

However, when talking about engines and hydrogen generators, liters are more commonly used them CFM. So, we have to inject 183.3827 liters per minute of HHO to run an engine on pure HHO.

Using www.ask.com 6.4761CFM is 183.3827 liters per minute.
Using www.ask.com 61.3260CFM is 1736.5590 liters per minute.


I checked my math and it seems to work out, but I must be missing something. Because if I was trying to figure out how much gasoline an engine needed, wouldn?t I use the same method? But 183.3827 liters (48.4446 gallons) of fuel per minute is definitely not correct.

Can someone help me figure out what I did wrong with these calculations and what steps I need to add in?

The volume of gas you produce for a gallon of gas is very different then putting HHO as a gas into your intake. As the gas travels through the intake it changes from a liquid to a gas state, while your HHO is already in a gaseous state.
I am interested in the method you are using to produce your HHO gas?

I'll try to find some of the obvious mistakes. First, the engine only sucks in air/fuel during the intake stroke, so your 3200 rpm = 1600 intake strokes/min. Change the 3200 to 1600. Next, throttle position has very very little effect on the amount of air/fuel being sucked in, 3200rpm uses approximately the same amount of air/fuel whether at 20% throttle or 100% throttle, so take out that calculation. Next, 14.7:1 ratio is by weight, not volume, which is what you're doing. So to find the amount of fuel you're using, you first have to find the weight of the air your engine is pumping. That makes the rest of your calculations moot, so when you get to this point, re-post and I'll try to "help" more. Good luck.

I haven't been able to construct a HHO generator that puts out resonable amounts of HHO within taking in too many amps. I made one large one that takes around 30 amps, might work good vehilces with large alternators, but not small cars. I've just been using the classic steel plates, baking soda, and water. No measurements or anything, I just mix some baking soda, water, and run some electricity through it.

nowhhs, as for the calculations, I'll give them another try by converting it to weight. Not quite sure how to do that for sure, especially in dealing with HHO in its gaseous state. ill see what i can do though

Speaking for myself, thats why I don't believe in HHO, you can't make enough of it to make any difference at all. If you do, using tons of electrical power, the current draw at the alternator drags down your engine more than the HHO benefits you. You can't win. I don't want to get into an argument or "discussion" with anyone about it, thats just my 2 cents. YMMV

I know what you mean, I don't believe in HHO either. However, if anyone was skilled enough to create a small version of a heat exchanger, where the exhaust heat boiled some water and the steam turned a turbine attached to an alternator, then you'd be able to turn that wasted exhaust heat into additional fuel, HHO.

I was just trying to do the math to see how much HHO it would be necessary to produce if someone did make this type of system. I know some automakers are working on a "turbine alternator" where the flow of exhaust gases turns the alternator, but I figured one that uses heat would be a little more efficient since it wouldn't generate any additional backpressure.

Yeah, your throttle calculations are a bit off. There is a huge difference in manifold vacuum from idle to 50% but after 50% there isn't much difference to 100% but that depends entirely on your rpm. On the dyno, my car will make 75ftlbs at 2000 rpm at 25% throttle and 100 at 50% and 115 at 100% but 25% drops off to almost no power(35 ft lbs) by 3250 when 100% just finished its first peak to 130 ft lbs.

johnneifred,

that's just water injection though. if you boil water, you get steam not hydrogen and oxygen. when the steam cools down it recondenses.

there have been benefits to water injections especially in turbo engines if for nothing else, to keep the turbo cool

*edit* my bad, I re-read your post and you said to create steam to turn an alternator. sorry man

Thanks for the edit BEEF, I was just about to try to re-explain what I meant, but I you've figured it out.

As for my calculations, I'm not sure how soon I'll get to them. I've been talking to Erik about how to lean out my '87 Civic and right now I'm waiting on a vacuum gauge I ordered from Ebay. Once I get the vacuum gauge I can compare reading at WOT (should be zero) and at half-throttle and such, and that should help me figure out how to get my calculations straight.

You calculate the total volume of HHO as fuel, when only 12.5% of HHO has any BTU energy. That is the Hydrogen. Oxygen contains no BTUs of energy.

You need to multiply your HHO volume by a factor of 8 to get your calculated fuel volume up to the level you have stated.

If you want to directly compare the liquid volume of Hydrogen to Gasoline you need to reduce your Hydrogen to a liquid volume which is about 1/860th of its gaseous volume.

Even then its mass is much less than Gasoline, not sure of the weight ratio, but I would bet Gasoline is 10 times heavier for the same liquid volume.

regards
gary

A gallon of gas contains 120,000 BTU of heat energy.

A cubic foot of pure hydrogen contains 325 BTU.

28 liters per cubic foot.

HHO is only 12.5% by weight hydrogen (the only part that is fuel).

1 liter of PURE (no oxygen) hydrogen contains 325/28 = 11.6 BTU of energy.

1 liter of HHO contains 11.6/8 = 1.45 BTU of potential heat energy.

You would need 120,000/1.45 = 82,759 liters of HHO to replace each gallon of gasoline, to supply the exact same amount of BTU of heat energy.


regards
gary

I would like to know where that number came from, that doesn't make any sense. What is comprising the other 87.2% of the HHO production. If your getting two hydrogen atoms per one oxygen atom during electrolysis then I don't see how you could have more oxygen than hydrogen. So what else is making up the 87.2% of the browns gas.

Also are you specifically referring to a baking soda electrolyte, would not other electrolytes like KOH be different in their percentages?

Terry

Check your periodic table of the elements.

Atomic weight of hydrogen is just over 1.
Oxygen is 16.

Two hydrogen molecules weight 2/16th (or 1/8th) the atomic weight of one oxygen molecule.

regards
gary

Actually I goofed slightly.

The total atomic weight of one molecule of water is 18, of which 2/18ths is hydrogen, so the quoted 12.5% is actually only 11.1% which means even less hydrogen fuel.

My VX uses one gallon of gasoline per hour at 55 MPH, at about 2000 RPM.
That is 120,000 (2000X60) revolutions of the engine per hour.

That equals 1 BTU of fuel energy per revolution, or 1/2 BTU per combustion pulse per cylinder. 33.3 revolutions per second, requiring 33.3 BTU per second of fuel energy.

Every liter of your HHO contains less than my quoted 1.45 BTU of hydrogen fuel energy. You would need over 20 liters per second to run the same engine.

Simple physics, which I learned in 1968, last year in high school.

regards
gary

Pop flew a B17 in WW2. Last mission was D-Day 6-6-44. First mission was Christmas eve 1943. They used water injection when they ran the engines at War emergency power rating to keep the engines from blowing up.

I am not so sure about water injection being an advantage as far as efficiency is concerned unless you are talking about a highly boosted engine that would have reliability issues without water injection, like those in the B17 which was turbo supercharged for performance at altitude.

The heat necessary to vaporize the water is heat that would normally create greater expansion of the combusted mixture. In the B17 it was not there to create more power, it was there to keep the engine from disintegrating.

regards
gary

I Used A Scan Gage Ii To Monitor My Miledge And Added One Quart Bottles. With 5 Bottles I'm Getting Close To 30mpg.

Yeah, he meant H2 to O2 is a 1:8 ratio by weight which works out to 1/9th by weight for 11.1% or so.

I've only seen one guy try to use just H2 and he was running a 750cc motorcycle. It's odd because in his setup he puposefully seperated the O2 and put it out to atmosphere and combusted using H2 and air. That doesn't make a lot of sense to me, to me you already have H2 and O2, so why use air? You could use a close system even if you actually produced enough. In my mind HHO at it's normal 1:8 ratio by weight is already designed to perfectly combust. So now all that's left is to cram however much you want into your engine. At 100% VE in a 1L that'd be 1L of displacement for every 2 revolutions (1 OTTO cycle), so if you had 100% VE and wanted to use your full displacement I would say 3200rpm = 1600 cycles = 1600L/m.

Of course if you seperate off and just use H2 and air then I have no idea. Clearly the only thing that 'burns' in air is O2 and I think that's 8% of air by volume. I don't know the % by weight of air, but if you had that you'd know the weight of H2 you'd need based on the 1:8 ratio and could go from there. You can't win though because whatever the H2 requirement would have to come from the 1600L/m HHO. The only advantage I can see is if you can keep it from preigniting then maybe you can run half the H2 and just get half the power and go to full power as you needed it.

Here's the best link I've found: https://www.free-energy-info.co.uk/Chapter10.pdf
Starting on page 78 is Bob Boyce's electrolyzer which supposedly produces 1200% more than it should by drawing 'energy from the environment'. If you have any hope of creating 1600L/m you might as well use a 'free energy' device ;P. I gotta admit I'm not electrician but I don't see how his big field coil in aluminum can do that. The idea of AC pulsing makes good sense though because it keeps the gas from sticking to the plates.

Efficiency
 

My work with HHO generators shows me that lower voltages per cell produce HHO more efficiently. Efficiency is most often measured in MMW milliliters per minute per watt. Basically how much energy was consumed to produce an amount of HHO in a period of time.

The equation is MMW=60,000/(watts*seconds per Liter) or
(60 seconds per minute * 1000 milliliters in a liter )/(Volts*Amps*Seconds per Liter)
Your production rate in seconds per Liter is 60,000/(Watts * MMW)


Actual value may vary some for your design and electrolyte, but my experience shows me:

1 if i drop 13.8 volts across 1 cell (2 plates) i produce about 0.97 milliliters per minute per watt.

2. if a have 2 cells and drop 13.8 volts across the whole thing, diving the voltage in 1/2 for each cell by electrically connecting them in series with each other, i get a MMW around 1.8 milliliters per minute per watt

3. if a have 4 cells and drop 13.8 volts across the whole thing, diving the voltage in 1/4 for each cell by electrically connecting them in series with each other, i get a MMW around 3.4 milliliters per minute per watt.

Hooking up cells in series increases the total resistance, lowers the watts consumed, and lowers overall production. There are 3 ways to compensate.
1 increase surface area of your plates in each cell to lower resistance.
2 increase the concentration of your electrolyte to lower resistance.
3 decrease plate spacing to lower resistance.

Lowering resistance draws more amps, increasing watts consumed, while producing more hydrogen with your new designs that have higher MMW
While each solution has certain practical cost and physical limitations, it is best to combine all 3 solutions.

In the end having a higher efficiency means less heat production, while using more of the energy provided for HHO production.

Atomic weight of oxygen is about 16 (8 neutrons 8 protons) while atomic weight of hydrogen is about 1 (1 proton). Each molecule of water has about 10 protons and 8 neutrons or 18 nuclei.
Based on this
2 Hydrogen protons /total 18 nuclei =11.11% by weight. Sorry i didn't calculate isotopes, electrons, or small weight differences of protons and neutrons into total weight.

If your storing, why store only hydrogen and not oxyhydrogen
3 reasons ...
1 Weight. Why carry the extra weight around, 88.88% of oxyhydrogen is weight of oxygen.
2 Space. About 1/3 of the volume of oxyhydrogen is oxygen.
3 Safety. oxyhydrogen is more dangerous in that is doesn't need air to burn, where as hydrogen by itself needs oxygen from the air to burn.

okay, lost of posts on this topic, and I'm going to ignore most of them and reply mostly to the start.

dtvg,
First of all I'd like to show you the math, but with a different approach.

from your gas log I think it is fare to say that on a good day on the highway you can get about 57 MPG. so lets round up and make the assumption that your car uses 1 gallon of gas to go 60 miles at 60mph therefor running for 60 min on one gal of gas.
gas had about 125,000 BTU per gal. so 125,000/1= 125,000 BTU per hour

Okay now the HHO side of things.
And the amount of HHO we get is directly related to the electrical energy put in to it, so lets take a look at the necessary electrical system

125,000 btu * .293= 36,635W
Okay lets say you have a really good HHO set up with only about 10% loss. so you would need 1.1 X 36,635 =~40,000W
okay so at 14 V 40,000/14= 2857A
okay so a DF8D car battery (really big one) holds 200 Amp Hours (when drained slowly) and has a weight of 150lbs.
So if drained in an hour lets say it will give 180 Ah (generous)
2857A/180Ah = 204 batteries at 150 lb each
that's 30,600 lbs of batteries in a 1700 lb car, just to go for one hour.

or you could just go with an electric swap...

HHO is a waist of electricity, because ICE's are huge energy holes that give very little back.

That said small amounts of HHO can improve FE by helping gasoline combustion. Small amounts of propane can do the same thing.

If you do what to know how much HHO that is by volume find out what 36,635W is equal to in HHO and dived by 60 go get a min.

Re: Need some help with my math, how much HHO needed?
 

Here's a few things to consider when talking about running an ICE on hydroxy gas (the combined gas output from water electrolysis) vs. gasoline:

Gasoline doesn't burn as a liquid - why use a liquid BTU/volume comparison? If comparing liquid gasoline to hydroxy, you'd have to compare the BTU value of a gallon of gasoline to the BTU value of the water to be electrolyzed.

(By the way, a cubic meter of water contains more hydrogen than a cubic meter of liquid H2: 111kg vs. 71kg. See the pdf link below.)

To make an accurate comparison, you need to compare hydroxy's energy value w a gasoline VAPOR MIX value - the gaseous mix your engine actually burns. According to the link below, the BTU value of a gas vapor mix is approximately 95 BTU/ft3 at 1atm.
ref: https://cta.ornl.gov/bedb/biofuels/et...-No2Diesel.xls

By comparison, pure H2 has a value of 270 BTU/ft3 at 1atm.
ref: https://www1.eere.energy.gov/hydrogen...fs/fcm01r0.pdf

Both of those BTU values will rise when the gasses are compressed in the cylinder - to what extent I don't know.

Nor do I know what the BTU value of hydroxy is, but doesn't adding oxygen to an acetylene torch create a hotter burn? Not my area of expertise. You'd also have to consider the resulting expansion - does compressing hydroxy have the same expansion potential as a compressed gas vapor mix? Maybe you'd want some additional water vapor in the cylinder to expand into steam to help push the piston down. You'd also want to adjust your timing forward, since the hydroxy will burn faster.

Still, it seems like with the above numbers there exists the opportunity of replacing gasoline vapors with a hydroxy mix. The problem is how to create enough of it on demand, and perhaps how to deal with water vapor getting mixed in with your oil.

E=mc2 means that there is a theoretical 85 billion BTUs of 'energy' contained in a cubic centimeter of water. (https://en.wikipedia.org/wiki/Mass%E2...gy_equivalence) Energetic potential surrounds us - the question is not "where does the energy come from?", but rather, "how efficiently can we release it?"

Re: Need some help with my math, how much HHO needed?
 

Liquid measures are standard when talking about fuel, and the BTU's convert as gasoline has the same energy to mass ratio in any from.
The BTU rating of water is zero. There is no energy that can be extracted from it at this stage (at least not by any means we are talking about). The energy needs to be put into it with electricity to make a useful fuel.
Comparing liquid to gas is comparable in this situation, comparing gasoline to water is like comparing a bag of tomatoes to a seed with out light, water or soil.


So one gal. is equal to about 1300 cubic feet of vapor mix. This don't change the amount of energy in it. But with the Figure of 270Btu's per foot of H that would mean, on the previous example, that we would need about 7.7cubic feet per min. of hydrogen per min.
nope, just a decrease in size, same energy.
With acetylene the O2 makes it hotter that air mix because there is less room temp air going into it. This makes a more compact flame with the same energy, which is there for much hotter.
Your right on with the timing thing.
Never going to beat a electric motor with an ICE powered with electricity...
Well that's not really a topic related to HHO. With HHO you are just putting energy in and then getting some of it back. This isn't cold fusion or anything.