Cheaper, gas or diesel???

[FordGuy100]

Lets all pay attention here.

Work = Force x distance. Force = mass x acceleration (gravitational acceleration). Work is measured in Joules. 1055.1 Joules/1 BTU

So.

Work = mass x gravitational acceleration x distance

Mass of a 1 ton truck? Lets say 7000 pounds. We need it in Kg. 2.205 pounds/kilogram. 7000/2.205 = 3174.6Kg, lets round it to 3100Kg. Gravitational acceleration (force holding objects to earth) is 9.8m/s^s

Distance? You said 306 miles. We need meters. 1609.344meters/mile. 306 x 1609.344 = 492459.3m

Now we have our equation

Work = mass x gravity x distance:
3100Kg x 9.8 x 492459.3m = 1.496091353 x 10^10 Joules (14960913530 joules). 1055.1 joules per BTU

14960913530j / 1055.1 = 14,179,616.66 BTU's. So without taking into ANY wind resistance, or tire resistance, mechanical resistance...in a frictionless world it would take that much energy, here measured in BTU's, to move that truck 306 miles.
Now, you said you used 2.4 gallons of diesel. 1 gallon of diesel has 139000 BTU's of energy. 2.4x 139000 = 333,600 BTU's of energy in that diesel. Subtract from the number above and you are claiming to have burned (14179616.66 - 333600) 13,846,016.66 BTU's of Hydrogen.

1 gallon of liquid hydrogen contains 33,700 BTU's. So 13,846,016.66 / 337000 = 41 gallons of liquid hydrogen.

 

[rjjp]

Justin forgot to mention that the math assumes 100% efficiency.

 

[FordGuy100]

But...there are 420.6grams of hydrogen in 1 gallon of water vs. 264 grams of hydrogen in 1 gallon of lidquid hydrogen. 420.6 / 264 = 1.59x more H2 in water than in liquid hydrogen.

41 gallons liquid hydrogen / 1.59 = 25.78 gallons of water consumed.

So to travel that distance negating any resistance whatsoever, using only 2.4 gallons of diesel, would require the consumption of 25.x gallons of water.

25.78 x 420.6 grams = 10845.66 grams of hydrogen. 1 mole of H2 gas is 2 grams of hydrogen. 10845.66/2 = 5422.8 moles of hydrogen. 1 mole of gas = 22.4L in gaseous form. 5422.8 x 22.4 = 121471.39L of gas. Now, you said you system was in the 5-6L/minute range, so lets take the high end of 6 liters per minute. 121471.69/6liters per minute =20245.23 minutes worth of gas at the rate you said. 20245.23 / 60 minutes per hour = 337.42 Hours.

337.42 Hours? What does that mean Justin. That means to get the amount of BTU's that it would require to make the truck go that distance, with the amount his HHO kit produced, it would take his truck 337.42 Hours to go from A to B over the 306 miles. Or, at an average speed of (306/337.42) .9mph. Yes .9mph. Any faster than that, and his HHO kit would not be able to sustain the amount of BTU's required to move the truck along with the diesel...AKA physically impossible.

Since in real life you would have averaged lets say 45mph over those 306 miles that means it would have taken you 6.8 hours of driving, or 408 minutes. 408 minutes at 6 liters per minute would mean a total amount of Hydrogen consumed was 2448L. Compare that to what it would really require BTU wise (121471.69L) and you can see the math is so far off that it is NOT POSSIBLE.

The only way it would have happened is if he idl'ed along at .9mph THEORETICALLY. But even then idling along requires something along the lines of at least 1 gallon per hour for your engine to run. Because even an old cummins at idle the pump will inject a certain amount of fuel no matter what is being put into the intake.

 

[FordGuy100]

In conclusion, it didnt happen.













Time for bed.

 

[FordGuy100]

Also, if you look at the very first part, you can see how a much smaller vehicle (in weight) requires much less energy to move a distance. Thats why the nightmoose is able to get such good fuel mileage.

 

[PwrSmoke]

I think I just had a tech geek ******!

 

[wmoguy]

Wowsa FordGuy100, your a smart dude! My head hurt just trying to follow all that math.

#goingbacktomylowtechmathworldnow

 

[wmoguy]

I will pay you $500 to take a couple days off work, and drive down here...then let me drive a truck equiped with that for 300 miles.

It is....literally...impossible. HHO Hydrogen is extremly low in BTU's. BTU's is a form of Energy. It takes a certain amount of energy to move a certain amount of weight a certain distance.

! I had a similar proposition, but maybe more lucrative since he's only 1hr South of me in Denver..

My Proposition:

Design your 100mpg system for my pickup. The savings I see in decreased fuel costs, I will split that fuel savings with you 50/50 and send you a monthly royalty payment. I drive about 25k miles a year on this pickup so it's a fairly lucrative financial offer. After 1 year of use, if the system pans out even 50% of what you claim, I'll finance the entire business venture of this HHO system and we both can retire very very wealthy.

 

[FordGuy100]

Dont get me wrong, I'm all for HHO systems, and 5-6lpm is actually a very good designed system. But the fact of the matter is that the only reason you would see an increase in fuel mileage with HHO is that it will help burn the diesel a little better. HHO will not be able to replace diesel as the main fuel source. Thats because your engine is having to turn an alternator, then all that current is sent through the HHO system, hydrogen and oxygen is realised (Browns gas) and then directed into the intake. The claim that the energy output of HHO gas is greater vs the draw on the alternator defies physics. Nothing is 100% effecient. If it were, we would have perpetual motion machines, and we wouldnt have to rely on fossil fuels ever again.

5-10% increase with HHO I can wrap my mind around. What is being claimed is just.....yeah.

Quote from another forum:

"And you didn't know Cummins can manage 127mpg??? They also deliver babies, go to Mars, and cure AIDS. DUH "

 

[93fordturbo]

May have went far with the response to the 127 mpg. Ok, so it was outlandish maybe. Leave it at that. I appreciate all the responses. Very good ideas. If I had a steady supply of, I would run that. But finding it is a problem for me. I guess we will have to see.

 

[chvycmnslvr68]

Well all that math is very impressive ... but the fact of the matter is that the most fuel efficient diesel engine built to date only burns at best 55% of the diesel fuel the rest goes out the exhaust system ...The 93 dodge engine that I am running is not any where near that efficiency ... so lets say maybe around 40% .... then the hydrogen/oxygen mix makes it 100% efficient ... plus the extra power that is generated by the h2 fuel induction ... it is very possible... I was getting 33 MPG just by having the double O/D and dropping my engine RPM down to 950 @ 75 mph .... I had my timing retarded about 3 degrees to allow me to have more torque at a lower RPM and maintain turbo boost ... If you take the 40% engine efficiency and jump it up to 100% that alone gives you around 100MPG then add in the extra power from the h2 gas and there you are .. I did do it and all the math in the world isnt going to change that

 

[jim x 3]

Well all that math is very impressive ... If you take the 40% engine efficiency and jump it up to 100% that alone gives you around 100MPG then add in the extra power from the h2 gas and there you are .. I did do it and all the math in the world isnt going to change that

chvy, Yes, it is. Because you haven't shown cause and effect and you are mistaken. Show us your variable controlled dyno runs measuring BRAKE SPECIFIC FUEL CONSUMPTION.

On the other hand, impressive as Fordguys numbers are, he makes a gross error in determining the amount of work required to drive 306 miles. Instead he has calculated (and I actually didn't check his arithmetic) the amount of work required to LIFT the truck 306 miles. But his basic premise is correct that the energy needed is not available. Kudos to him for his effort to approach this issue analytically.

BS to chvy. And we have been here before.

Regards,