Start here your topics on the water injector & ambient airgasprocessor system
Started by unknown · · 132 posts · last reply 4 November 2011
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#1 · date not recorded
Start here your topics on the water injector with ambient airgasprocessor system.
Steve -
#2 · date not recorded
thought i would share a interesting perspective i found today.. the attached pic is a snapshot right from stans pdf... page 154...
stan states that a 50 hp 1600cc needs 7.4 micro liters to maintain a speed of 65 mph
then he states a 325 hp diesel would require 48.1 to maintain the same average of 65mph...
what i find interesting is that the water amount needed is proportional to hp between the 2 engines (diesel and gasoline)
325 requires 48.1
50 requires 7.4
325/50= 6.5 .... so the diesel is 6.5 times more hp then then buggys gas engine...
7.4 ml x 6.5 times more hp = 48.1
so lets pretend a motor runs 2000 rpm at 65 mph... and its a 6 cylinder.. each cylinder receives 1 injection cycle per 2 cycles (revolutions) one of them is used to push out combusted gases and the other is compressing the gas right berfore ignition...
at 2000rpm requires a simple equation
divide rpm by 2 since only one of the revolutions is used for injection.. and thats your frequency of injection...
2000 / 2= 1000 ......
6 cyclinders x 7.4ml = 44.4ml
7.4 every 2 revolution means 2000rpm / 2 = 1000 injection cycles
1000 injection cycles x 6 injection (each cylinder) = 6000 x 7.4ml per cycle = 44,400ml aka .0444 liters @ 2000rpm
.0444 liters per minute x 60 min= 2.664 liters a hour
attachment_5344 pg 154 pdf under mode of operability.jpg
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#3 · date not recorded
Find out how much gasoline is used per injection cycle in a normal engine, and compare the "energy content" ... it should be the same. -
#4 ·
scuse me outlawstc but i think you calculation is a bit wrong
he said that the dune buggy consumed 44ml per minute so would consume about 2,6 liters of water per hour. 65mh 3000 rpm
However rpm doesn't matter as he stated the consumption for minute... thats where you was wrong in your calculations
and
he said 0,000007 liters per injection or 7 ul -
#5 ·
seb.. have u read the clip at the bottom of my post... it is right out of stans wfc memo on page 154 -
#6 ·
exactly
check what i said -
#7 ·
0,000007.4 liters * 6000 injections in 1 minute * 60minutes =2,6 liters per hour maintaining 3000 rpm he clearly stated this -
#8 ·
about the energy content he say that in one gallon of water you have 1,66 lb of hydrogen and in one gallon of gasoline you have 0,66 if you divide 1,66/0,66= 2,666 considering the contaminants he said 10 % nitrogen present on water should give you the famous 2,5 times more powerful
this is a comparison by weight -
#9 ·
your right i made a mistake... i will go over it tomorrow and fix it my brains tired.. have a nasty lil sinus cold and been reading to much today.. thanks for posting the by weight comparison.. -
#10 ·
ok
we could also state 2,6 liters of water *1847 gas expansion = 4800 liters of gas in one hour
at 3000rpm about 80 liters of gas per minute or 1,3 liters of gas per second
injections per second 3000/2*4/60 = 100 counting all injectors
so for every injection of gas in the case you would need 1,3 liter / 100 = 0,013 liters of oxygen hydrogen per injection
i estimated that each injector would works thus 25 times for second so if we have 4 cycles the time you would have to inject this gas inside the engine would be like 1/100 of a second or 0,01 second or 10 milliseconds remember at 3000 rpm if you double rpm the time to inject will be halved. -
#11 ·
to calculate maximum time for injection,devide 120,000 by the rpm,and you get milli seconds.
eg. 120,000/6000 rpm = 20 ms.
I use to do these numbers all the time when I was tuning pulse width for programable computers in cars.
Don -
#12 ·
Nice dynodon
But what is this number? And where it comes from ? Can you explain to us ? thank you
I have an explanation for the difference between your number and mine
I considered the four cycles (1°down intake) (2°up compression) (3°down explosion) and up for exhaust and theorized you should only inject the gas during the intake
at 3000 rpm you have 1500 complete cycles in a minute, witch = 25 complete cycles per second, if you consider a complete cycle 4 semi-cycles you might find why i found this number
25 complete cycles take 1 sec to occur right so 1sec / (100 quarters of cycle) or (25 * 4 semi cycles)
thats how i estimated 10ms absolute maximum injection time for 3000rpm
Actually I think the timing for the injection should be fix at about 4 ms and that the amount of hydrogen per injection should remain the same, the car should accelerate and de-accelerate by only controlling the amount of air that comes in and or also the exhaust gas re-entering the combustion chamber... the hydrogen will be generated on demand and maintained under min working pressure. If you think about when you mix more exhaust the motor will lower its rpm automatically lowering the consumption of hydrogen as its consume is dictated by number of injections . Probably meyer regulated the max rpm mixing with the air and the minimum adding also exhaust gases. What do you think?
HAPPY NEW YEAR!!!
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#13 · date not recorded
hmmmm i always figured injection time changes when revolution time changes (rpm) and the changes are made proportionately with rpm.. having a predetermined variable being horsepower.
seb check my math above once more please.. pretty sure i have it right... so it is 7.4 ml per cylinder for 50 hp motor amounting to 2.6 liters per hour at 65mph and 2000rpm? and a average hydrocarbon fuel gets around 6.5 liters a hour at 65mph and 2000rpm due to the fact that there is 2.5 more power in water from having a higher quantity of hydrogen per 1:1 by volume. -
#14 · date not recorded
sebosfato,that number is 120k ms. (milliseconds)
1 second = 1000 ms
1 minute = 60 seconds
60 seconds x 1000 = 60,000 ms
It takes two revolutions to complete one cycle,so that means two 2 prm = 120 seconds
120 seconds x 1000 ms = 120,000 ms
Don -
#15 · date not recorded
hmmmm i always figured injection time changes when revolution time changes (rpm) and the changes are made proportionately with rpm.. having a predetermined variable being horsepower.
seb check my math above once more please.. pretty sure i have it right... so it is 7.4 ml per cylinder for 50 hp motor amounting to 2.6 liters per hour at 65mph and 2000rpm? and a average hydrocarbon fuel gets around 6.5 liters a hour at 65mph and 2000rpm due to the fact that there is 2.5 more power in water from having a higher quantity of hydrogen per 1:1 by volume.
Now you corrected it right? Attention because you still confusing micro liters with milliliters. But now the results seems right! I don't understand why you talk about 2000 rpm? Do you mean for 6 cylinders? still not clear...
however
stan said 7.4 micro liters per injection cycle for a 50hp 4 cylinders running at 3000rpm and 65mph
I believe he used the mixing of the gas to control the motor speed i mean he used more or less exhaust recirculation accordingly with the rpm desired. I say this because he talked about the speed of the explosion of the hydrogen witch would be to great alone. So to reduce the speed or the rpm (MEYER TRANSLATION) he substituted the air in for exhaust gas. why? Because is much easier (KISS) to make a pulse of fix width that changes only in frequency (rpm) than is to create a pulse that changes frequency and also width right ?
If the pressure of the cell is fix and the quenching circuit works the way i proposed it becomes obvious that he stated 7.4ul per injection cycle because for every injection cycle you will inject the same amount of hydrogen. The amount of gas injected will be controlled by the rpm since as you add exhaust gas the RPM will change REDUCE automatically.
He probably had a sensor on the valves that every time the intake valve open the solenoid opens for 4ms injecting the gas as i said 13,3ml of hydrogen oxygen gas per injection...
than on the intake he probably used a kind of three way switch that has the air input and also the exhaust return... he than just regulated the speed rotating the switch... Isn't it very simple?
All is needed here now is to calculate the pressure needed to be maintained on the cell for having this amount of gas passing inside the quenching circuit in this predetermined time at a velocity greater than as he stated 350cm/second.
here is a link with a calculator i'm studying :
http://www.pipeflowcalculations.com/airflow/index.htm
pressure in the cell is free so you can use it so no need for gas pump.
Right you also understood well the 2,5 times more powerful comparison is (by) volume but or also more probably or precisely (BY) weight...
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#16 · date not recorded
well said seb.. i understand your perspective on injection time now.. makes sense... -
#17 · date not recorded
sebosfato,your getting Stans systems mixed up.The injection of 7.4ul is water through the water injectors,and not the gas injectors with the quenching tubes.Two different systems.Those injection numbers are for the water injection system,that injects water into the injector at the spark plug hole and it explodes coming out of the injector directly into the cylinder.
He's not talking about injecting hydroxy gasses into the injectors.
Don -
#18 ·
i know but its the same. however even working with gas meyer used solenoids and injectors to run his buggy. -
#19 ·
hello,
some mental gymnastics on the air processor, my thoughts:
How to go about making one knowing it will work? well surely stan had a way of calculating what was needed. Instead of just attempting to copy the picture, i am wondering a few things.
For the air intake diameter used on the engine cc size for the volume of air to be passed through it, for the voltage intensifier coil windings to be calculated, do we have any ballpark idea of what may have been the intended voltage across the positive and negative voltage zones in the air processor and possibly any clue on the shape of the s/s electrodes? (an air capacitor with a moving medium)
ie is there a requirement for a particular length of electrode to provide a contact area for a minimum length of time to obtain the intended effect.?
For the electrodes, is there any preferred or given distance between the electrodes with relation to the voltage zone and voltage used.
in the air processor, for the voltage intensifier coils seen in the pictures (on site here somewhere) is there any know way to figure out how many windings are required - also noticed the air passes through the centre of the voltage intensifier.
At the opposite end, how many windings would be needed to be an effective pickup coil for the electron extractor..
The air processor seen in the pictures may work on the 1.6l buggy - but not all engines are that size. so im posting these thoughts to ponder - if the above items could be worked out somehow, could there be a way of calculating a way to make, say, a larger diameter air processor for a larger engine.
cheers -
#20 ·
and googling the dielectric value of aire comes up
The dielectric strength of air is approximately 3 kV/mm. Its exact value varies with the shape and size of the electrodes and increases with the pressure of the air.
so question to ponder is - what is the voltage required to perform the ionisation function. -
#21 · date not recorded
I don't think that you understand the usage of the air gas processor.It does not process all of the air going into the engine.It only does a small amount.This processed air is injected under the throttlebody along with exhaust gas recirculation.Also the idle control air is added in there as well to control the engines idle spped.So you have ionized air,exhaust gasses and metered air,all going in under the throttlebody,in the vacuum area.
The air gas processor.tubes are the same size as the resonant cell.3 inches long.To ionize the air several thousand volts will be needed.
Don -
#22 · date not recorded
Don, did the gas processor use the same VIC as the 3" wfc tubes? -
#23 · date not recorded
I didn't see anything hooked to the processor.I can only assume that the vic coil would be very similar.
Don -
#24 · date not recorded
in the Hydrogen Gas gun, do you know where the primary is? we see two coils at the bottom and one at the top. Any guess on the wire gage? -
#25 · date not recorded
I don't know anything about that unit.Couldn't tell which wires went where.Can't help out any.
I don't think that it was ever used.
Don