Excellent work, brother. Howerver let me tell you, that you complicate yourself with the calculations, and this gives you lower efficiency (partly due to loses in apparatus).
The useful energy contained in 1 liter of HHO can be easily calculated if you know some chemistry. HHO is a stechiometric mixture of H2 and O2 (dismiss comments about monoatomic H and O in HHO, thy have no justification).
So the proportion is the same as in water: 2 volumes of H2 every one volume of O2.
NOTE that this proportion is in volume, not in weigth: Oxygen weigths 18 times more than Hydrogen, so the weigth of an amount of HHO will be mostly due to its content of O2.
So, in 1 Liter (volume) of HHO you have 2/3 Liter of H2 (biatomic gas) and 1/3 Liter of O2 (biatomic gas). No more, no less, nor anything else.
Now, the useful energy of a fuel is expressed by its "Lower Heating Value" or LHV (see Wikipedia). LHV of H2 (biatomic gas) is measured to be (in "Normal Conditions": 25ºC and 1 atm of pressure) 2.974 KWh/m3 or, what is the same, 2.974 Wh/Liter (mult & divide by 1000).
Excellent video, well presented. Just a small bit of advice! Cut that lens cap off your camera, the constant tapping is driving me crazy!!! Also, a ceiling mounted camera arm may be just what you need!! Keep posting!
I'd just like to say, an Excellent video that was very informative and very accurate! Couldn't have done it better myself, even on a good day!
As for hydrogen usage;
Personally I think and from my experiments, the answer doesn't lie with electrolysis, no matter how good the system works, it will never be close to a high effeciency, 80% yeild for 100% power input.
Hydrogen itself is known to be approx 1/10th the energy of gasoline/etc.
You will need the proper mixture of AIR with HHO.Problem is there is not enough AIR.H2 is lighter than O2.So there is only H2 in the sysem,and very little O2 or very little AIR. So to be certain,
(1)one should have 5 times more AIR than HHO;
(2)ignite where the level of H2 in the top (as it will flow to the top),is suppose to touch the lower positioned AIR, OR create enough spark depth,to reach that point.
Now there is enough mixture for all H2 to burn.First lift weight 5 times the volume HHO
The Problem with your theory is the gas is HHO and seems to be homogenious it doesn't separate in layers of H2 H and O2 or O what ever the configuration. seem to think its pure H2 not so.
Ive been thinking about this video since you posted it i keep wondering if the distance traveled was limied due to vacuum created as the cylinder moved up. i would like to see you do the same experament with air/vapor gasoline ratio 14 to 1 (125 ml) (supposed ratio gasoline burns to air)[way easier said then done]
and dont forget the implosion effect that happens right after the initial explosion. as you turn the higher volume hho back into water as steam. i'd think that would limit the jump of the plunger.
wait, so you only got 1.8% effenciency from the work the hydrogen did? It still even seemed to do a lot of work... Just imagine what it would be 60-90%effecient! Very interesting, good job.
Possibly you could pull say 100 ml air in then 100 ml HHO this would give a 50/50 mix then make other tests and adjust 60/40 70/30 Est. This would produce a power chart that would provide very useful info. A pressure compression gage at the bottom of your cyl would also add useful info as to how the test works at varying compression ratios. An auto fuel air mix and HHO mixture would be REALLY GREAT INFO!! But we all are Dreamers thats why we do this. Thanks for the hard work
Most engines run on a fuel to air mixture. And from what I have read even HHO has a more efficient burn cycle when in a mixture. I'm not sure how you would produce the mix with your rig but it would be very interesting to see the different power strokes at different mixtures.
I am a simple man ... I just like watching the darn piston lift off ...
that is a pretty darn cool rig you got there ... and if it helps you to enhance our general understanding of how this fuel acts, that's even better ...
but I still get a kick outta watching that thing dance ... LOL ...
one thing, given the time of the burn is less than a second then the joule convertion is not quite right. you are taking work done in x time, but your not figuring the time component into the 9.5inches at 25lbs, in what time? I don't think you can do it that way, and be right, and if you adjust for the time, I think your eff. will go up alot, also how much contraction force is applied? that will change the eff. too. You will proly have to do two separate tests to get that info. invert piston
and lift a weight with the contraction, and measure the time. then combine the data from the two experienments to get a complete cycle energy output, I think if you do this with acuracy will will be surprized by the results. energy is currently only measured on one side of the equation, the other side is ignored, because from a linear approach it would be a negative energy, but with a simple rachet device all motion would go the same direction and therefore be additive.
actually thats what you need to do, use a bidirectional rachet using cables, with the weight after the rachet, so as the piston goes up it lifts a weight and as it contracts it continues to lift the weight further. and measure the time, that would be a complete data set for the cycle.
of the amount of energy. It is not. It takes the same amount of energy to lift a mass through a distance regardless of the time. Thats the definition of work Mass through a distance.
Now Horse power is an energy rate not a fixed amount of energy. If I wanted to figure out the potential energy rate That is different. HHO Use rate -VS- burn rate and developed HP would require a much more sophisticated mechanism and lots of instrumentation.
I think if you check you math and units you'll find something missing, even using the energy content of hydrogen as the baseline energy, not consumed watts for hho production, the energy should be 890 joules, 125ml=82.5ml of hydrogen. do a wiki for energy density, it gives for gaseous H at rm temp. There has to be an accounting for the known energy of H, and 26joules of mechanical work just aint right, there isn't much enegy lost to heat and sound in your system to account for it.
And I am pretty sure that the time factor will come into play, I know what your saying AG but the test data doesn't agree, and thats normally a good indicator of something.
If bubbler was empty,you might be right,but since H2 is lighter than O2;gas is drawn from bottom so hardly any H2 is drawn in cilinder.H2 is in top of bubbler!Idea is fine, yet execution entirely wrong.If done right,so drawn from top(longer hose),it is a different story.Next thing is:enough air,by lifting first weight up some factor 5 to 125 ml.Better protect everything first and take less:50 ml.Otherwise BANG!Good health to our new mr.Nobel.He deserves it.Science should have done it long ago.
actualy friend time is important in determining power ,as the work is the same the time reqiured to do that work in a given time is called power,so you may whant to look into physics to get e better perspective of what energy is,you are write about work ,but the caliqual use of the word is not going to cut it ,work = force x distance ,,,power = work/time
example one newton will exellerate a mass i 1 kg 1 meter in 1 second this is one joul of energy , ill finnish
what i meen is when you measure the energy on your unit you have to no how fast that wiegt is going if you dont beleave me i under stand but trust me you need to no the time to calculate the jouls friend
Excellent work, have you an update?
MrZobow 1 year ago
Do you think it's realisable to make a car owrking with this kind of system?
atb1248 1 year ago
the amount of pressure from the gas will also affect your test, correct?
911truthseekers 2 years ago
mind your tether dude, eff that is annoying
chasbot 3 years ago
nice demonstration ..
jdcmusicman 3 years ago
Excellent work, brother. Howerver let me tell you, that you complicate yourself with the calculations, and this gives you lower efficiency (partly due to loses in apparatus).
The useful energy contained in 1 liter of HHO can be easily calculated if you know some chemistry. HHO is a stechiometric mixture of H2 and O2 (dismiss comments about monoatomic H and O in HHO, thy have no justification).
So the proportion is the same as in water: 2 volumes of H2 every one volume of O2.
juanpazago 3 years ago
NOTE that this proportion is in volume, not in weigth: Oxygen weigths 18 times more than Hydrogen, so the weigth of an amount of HHO will be mostly due to its content of O2.
So, in 1 Liter (volume) of HHO you have 2/3 Liter of H2 (biatomic gas) and 1/3 Liter of O2 (biatomic gas). No more, no less, nor anything else.
juanpazago 3 years ago
Now, the useful energy of a fuel is expressed by its "Lower Heating Value" or LHV (see Wikipedia). LHV of H2 (biatomic gas) is measured to be (in "Normal Conditions": 25ºC and 1 atm of pressure) 2.974 KWh/m3 or, what is the same, 2.974 Wh/Liter (mult & divide by 1000).
juanpazago 3 years ago
Finally, 1 Liter of HHO will contain 2/3 of the energy of 1 Liter of H2. This is, 1.9827 Wh/Liter.
If your equipment produces, say, 2 Liters/minute of HHO, you have 3.9654 Wh/minute, or 238 Wh in 1 hour (120 Liters).
Don´t confuse Watts (Power=Energy/time) with Watts-Hour (Energy). 1 Wh = 3600 Joules.
juanpazago 3 years ago
wikipedia is full of sht have the time
i would be ware of any of its info any jackass can write down info on wiki its not safe to use
NOBOX7 2 years ago
Excellent video, well presented. Just a small bit of advice! Cut that lens cap off your camera, the constant tapping is driving me crazy!!! Also, a ceiling mounted camera arm may be just what you need!! Keep posting!
tex8596 3 years ago
Only if one could produce hydrogen with a "over-unity" effect, could it be an alternative.
Just my thoughts.
The answer to the world's energy/environmental issues is something else...
hvlabsdotcom 3 years ago
Hello AllgoodAutomation,
I'd just like to say, an Excellent video that was very informative and very accurate! Couldn't have done it better myself, even on a good day!
As for hydrogen usage;
Personally I think and from my experiments, the answer doesn't lie with electrolysis, no matter how good the system works, it will never be close to a high effeciency, 80% yeild for 100% power input.
Hydrogen itself is known to be approx 1/10th the energy of gasoline/etc.
hvlabsdotcom 3 years ago
You will need the proper mixture of AIR with HHO.Problem is there is not enough AIR.H2 is lighter than O2.So there is only H2 in the sysem,and very little O2 or very little AIR. So to be certain,
(1)one should have 5 times more AIR than HHO;
(2)ignite where the level of H2 in the top (as it will flow to the top),is suppose to touch the lower positioned AIR, OR create enough spark depth,to reach that point.
Now there is enough mixture for all H2 to burn.First lift weight 5 times the volume HHO
lagswing 3 years ago
The Problem with your theory is the gas is HHO and seems to be homogenious it doesn't separate in layers of H2 H and O2 or O what ever the configuration. seem to think its pure H2 not so.
AllgoodAutomation 3 years ago
Ive been thinking about this video since you posted it i keep wondering if the distance traveled was limied due to vacuum created as the cylinder moved up. i would like to see you do the same experament with air/vapor gasoline ratio 14 to 1 (125 ml) (supposed ratio gasoline burns to air)[way easier said then done]
g0a23657 3 years ago
Good Suggestion.
Currently working on a Large SID CELL Arrangement
AllgoodAutomation 3 years ago
and dont forget the implosion effect that happens right after the initial explosion. as you turn the higher volume hho back into water as steam. i'd think that would limit the jump of the plunger.
strapped9 3 years ago
Excellent!
sidyoung 3 years ago
wait, so you only got 1.8% effenciency from the work the hydrogen did? It still even seemed to do a lot of work... Just imagine what it would be 60-90%effecient! Very interesting, good job.
BasementBen 3 years ago
actuaky 40 to 60 psi will do what we seen here,hho is not as strong as i once thought ,it all bark and no bite
NOBOX7 2 years ago
Possibly you could pull say 100 ml air in then 100 ml HHO this would give a 50/50 mix then make other tests and adjust 60/40 70/30 Est. This would produce a power chart that would provide very useful info. A pressure compression gage at the bottom of your cyl would also add useful info as to how the test works at varying compression ratios. An auto fuel air mix and HHO mixture would be REALLY GREAT INFO!! But we all are Dreamers thats why we do this. Thanks for the hard work
Hydroginist 3 years ago
very good Ideas and yes this is where I am headed.
AllgoodAutomation 3 years ago
You have a great start great info!!
Most engines run on a fuel to air mixture. And from what I have read even HHO has a more efficient burn cycle when in a mixture. I'm not sure how you would produce the mix with your rig but it would be very interesting to see the different power strokes at different mixtures.
Hydroginist 3 years ago
I am a simple man ... I just like watching the darn piston lift off ...
that is a pretty darn cool rig you got there ... and if it helps you to enhance our general understanding of how this fuel acts, that's even better ...
but I still get a kick outta watching that thing dance ... LOL ...
SmartScarecrow 3 years ago
Ditto!
Nice Work Allgood, I wanna see more things being moved by HHO alone!
MadStu1978 3 years ago
one thing, given the time of the burn is less than a second then the joule convertion is not quite right. you are taking work done in x time, but your not figuring the time component into the 9.5inches at 25lbs, in what time? I don't think you can do it that way, and be right, and if you adjust for the time, I think your eff. will go up alot, also how much contraction force is applied? that will change the eff. too. You will proly have to do two separate tests to get that info. invert piston
mielectric1 3 years ago
and lift a weight with the contraction, and measure the time. then combine the data from the two experienments to get a complete cycle energy output, I think if you do this with acuracy will will be surprized by the results. energy is currently only measured on one side of the equation, the other side is ignored, because from a linear approach it would be a negative energy, but with a simple rachet device all motion would go the same direction and therefore be additive.
mielectric1 3 years ago
actually thats what you need to do, use a bidirectional rachet using cables, with the weight after the rachet, so as the piston goes up it lifts a weight and as it contracts it continues to lift the weight further. and measure the time, that would be a complete data set for the cycle.
mielectric1 3 years ago
It is a mis conception that time is a component.
of the amount of energy. It is not. It takes the same amount of energy to lift a mass through a distance regardless of the time. Thats the definition of work Mass through a distance.
Now Horse power is an energy rate not a fixed amount of energy. If I wanted to figure out the potential energy rate That is different. HHO Use rate -VS- burn rate and developed HP would require a much more sophisticated mechanism and lots of instrumentation.
AllgoodAutomation 3 years ago
I think if you check you math and units you'll find something missing, even using the energy content of hydrogen as the baseline energy, not consumed watts for hho production, the energy should be 890 joules, 125ml=82.5ml of hydrogen. do a wiki for energy density, it gives for gaseous H at rm temp. There has to be an accounting for the known energy of H, and 26joules of mechanical work just aint right, there isn't much enegy lost to heat and sound in your system to account for it.
mielectric1 3 years ago
And I am pretty sure that the time factor will come into play, I know what your saying AG but the test data doesn't agree, and thats normally a good indicator of something.
mielectric1 3 years ago
If bubbler was empty,you might be right,but since H2 is lighter than O2;gas is drawn from bottom so hardly any H2 is drawn in cilinder.H2 is in top of bubbler!Idea is fine, yet execution entirely wrong.If done right,so drawn from top(longer hose),it is a different story.Next thing is:enough air,by lifting first weight up some factor 5 to 125 ml.Better protect everything first and take less:50 ml.Otherwise BANG!Good health to our new mr.Nobel.He deserves it.Science should have done it long ago.
lagswing 3 years ago
actualy friend time is important in determining power ,as the work is the same the time reqiured to do that work in a given time is called power,so you may whant to look into physics to get e better perspective of what energy is,you are write about work ,but the caliqual use of the word is not going to cut it ,work = force x distance ,,,power = work/time
example one newton will exellerate a mass i 1 kg 1 meter in 1 second this is one joul of energy , ill finnish
NOBOX7 2 years ago
what i meen is when you measure the energy on your unit you have to no how fast that wiegt is going if you dont beleave me i under stand but trust me you need to no the time to calculate the jouls friend
also gravity is effecting all of this
NOBOX7 2 years ago
cool design dude.
jesslessthemess 3 years ago