i think we as a group really should try determining how much fuel we need, then design our electrolysers, and related power equipment to match, instead of building a box, plugging it into the wall, and trying to run an engine untill it runs out of fuel. Think of it like building a braking system for a new vehicle- you're REALLY gonna want to figure out how much pressure you need to apply to stop that tub before you go careening down a mountainside! ok, maybe thats overkill, but you get my point! lol
Thanks all!
Statistics: Posted by Veritech — Thu May 25, 2006 11:05 pm
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There is so much confusion about ratio's required to use in an IC engine. I was hoping you would comment about it. But I see now that it would depend on what type of hydrogen you are making - something I hadn't really paid attention to until now.
For example, an electrolyzer like your 60 plate welder design using 6 amps or less and without resonance, in your opinion what is the most likely volume ratio of air/hydroxy in a low pressure series multiplate cell required to run an engine? (I mean by hydroxy all of the types of hydrogen plus oxygen produced in a low pessure cell)
regards,
Statistics: Posted by zendog — Thu May 25, 2006 10:53 pm
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If you switch a car to methanol, the calories per unit of fuel is much less, but you need to richen the mixture up to prevent detonation. Methanol requires about 2.5 times the amount of liquid fuel to run the same engine, however will be less likely to knock and because there is more fuel, will make more power.
Different fuels require different amounts to stave off the knock. And tuning any engine is where you are riding the fine line between knock and maximum power per bang. Sometimes this wastes fuel and sometimes you can burn up a lot. Methanol is a wasteful fuel, hence dragsters shoot spectacular flames out the exhaust, wheras, LPgas powered engines will burn the fuel much more efficiently.
It just so happens that with petrol, that around 14:1 oxygen:fuel ratio is the ideal amount to achieve efficient burning, power and minimal knock.
The old fashioned backyard way of tuning a car engine was to adjust things till it pinged (knocked) and back it off a smidge on the timing and add a bit more fuel. That was how it was done before o2 sensors and dynos were common.
I know this still doesn't answer the question but it does open up new questions like:
What is the octane rating of hydrogen?
Will it knock?
How can we measure the air:fuel ratio or is there another way?
If a conventional ICE requires 1% to 3% hydrogen to each unit of air drawn in, then if we're working with a conventional 2 litre engine, each cylinder will require 5 to 15ml of hydrogen per piston firing.
3000 revolutions is 6000 piston firings, so 6000 X 5ml is 30 litres of hydrogen gas per minute. Multiplied by the volumetric efficiency, so possibly knock off 1/3rd of this figure for most inefficient engines.
A turbo engine will guzzle double the amount for every 14psi of boost you run....
If hydrogen expands 1800 times it's size, you will need to convert approximately 17ml of water every minute to run the 2 litre engine in my example.
17ml of water replaces 100ml of petrol.
Now I know what all the fuss is about, these figures are sounding pretty good to me.
Statistics: Posted by biggeorge — Wed May 24, 2006 5:46 pm
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it just doesnt make sense that an engine can run on a litre of gasoline for quite a while, but that same engine will only run for a minute on a litre of hydroxy- it jsut dont make sense.
at some point, i did take into consideration volumetric effiecency by using the fuel (gasoline & air) demands of a real working ICE- just trying to figure out the gasoline to hydroxy ratio.
what i'm wondering is that say hydrogen is 2X as powerful as gasoline- then to convert an ICE to hydroxy would result in an air/fuel mix of about 28:1 right? now how far can hydroxy be diluted with air before it wont ignite?
thanks for the reply!
Statistics: Posted by Veritech — Wed May 24, 2006 1:53 pm
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The o2 sensors measure the amount of oxygen being dumped out the exhaust to give you an indication of how much fuel was used at the moment before it was measured.
So 14.7:1 is 14.7 parts oxygen to 1 part fuel.
Hydrogen is a completely different ball game.
Also, the calculations in the original post seem to be the amount of air being consumed by the engine, not taking into account Volumetric Efficiency and so forth.
For an engine that does 1km in 60 seconds doing 3000rpm (average kind of cruise speed) and uses 100ml of fuel to do it, According to my calculations the engine would be consuming 0.033ml of petrol per engine revolution.
Your talking about miniscule amounts of fuel per rev.
Normal fuel injectors can spray around 300cc per minute and on cruise will only be open for 2-3ms per engine stroke. It's a tiny amount!!
Statistics: Posted by biggeorge — Wed May 24, 2006 1:02 am
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2.85 liters per revolution X 6000 rpm=17100 liters of air fuel per min X .04 =684 liters per min. of h2.
TRY THAT
Statistics: Posted by bigj8550 — Fri Apr 28, 2006 1:29 pm
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Statistics: Posted by Veritech — Fri Apr 28, 2006 12:24 am
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or
5.7 liters X 6000rpm X 0.04%=1368 liters of H2?
how about this- to make things more complicated! lol lets say my car with this carbeurated V8 uses 1 gallons per 20 miles- at 60mph (approx 2250rpm) thats a mile a minute= 1/20th of a gallon(231cubic inches) of fuel per minute= 11.55 cubic inches of gasoline per minute at 2250 rpm.
As per Big Jeremy's suggestion, lets triple this to get to approx. 6000rpm- that would be 34.65 cubic inches of gasoline per minute at 6750rpm. Note that this doesnt include air. this is strictly gasoline.
Now, not taking into consideration the difference in power from gas vs/ hydrogen- we'd only need to make 34.65 inches of hydroxy per minute. And if water expands about 1800 times when it converts to hydroxy- we're talking converting 0.01925 cubic inches of water per minute to power my raggedy ol' worn out 5.7L V8 at 6750rpm right??? and this doesn't even take into consideration Bob's 96/4% hydrogen/air mix. But since the numbers are based on real world numbers- these calculations should automatically take into consideration the volumetric effiency of the standard engine, correct?
2 quick questions on that Bob, is that 4% of total air/fuel- or is that 4% of the fuel only part of the air fuel ratio? i mean...
5.7 X 4% =0.228L of H2
or
5.225/0.475 (12:1) X 4% =0.019L of H2
also, is that 4% of pure H2, or is that 4% hydroxy?
Thanks a ton Guys!!keep it coming!
Dana
Statistics: Posted by Veritech — Fri Apr 28, 2006 12:19 am
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