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

[R.I.D.E.]

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

[TerryG]

Originally Posted by R.I.D.E.

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

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

[R.I.D.E.]

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

[R.I.D.E.]

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

[R.I.D.E.]

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

[hoke]

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

[itjstagame]

Originally Posted by TerryG

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

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: http://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.

[dtvg]

Originally Posted by JohnNeiferd

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

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.

[dtvg]

Originally Posted by TerryG

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

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.

[dtvg]

Originally Posted by itjstagame

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.

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.