Originally Posted by kennedyford
Corey you math is incorrect for the application I posted.
1. Hydrogen can be produces with as little as a 1.5 volt battery. A 9 volt battery makes quite a bit of hydrogen with very lil amperage.
2. I can remove 1 light bulb and produce hydrogen to inject into the diesel air system without compromising the electrical system.
3. The truck along with every car on the road is making hydrogen all the time. Why do you think batteries bubble when charging or your not supposed to smoke around them and filtered water is needed to refill the batteries.
4. 1/3x1/2x1/2 is the same as .3x.5x.5= .75 That is 3/4 !
5. Where do you identify a UNIT and how do you come up with 8 percent number?
6. 1/3 wasted as heat? Hydrogen burns colder then fuel and warmer then alcohol. How can 1/3 be wasted on fuel and the other 1/3 trying to overcome friction?
Making statements that have no factual proof or in this case, have no bearing on my posting is slightly foolish. Its as if you dont want anyone to think outside the box or succeed. Im not attacking you, just your math. Provide a study on h2 injection in a diesel engine and ill read it with respect.
Actually his math is correct, yours is wrong KennedyFord.
1/3 x 1/2 x 1/2 is indeed 1/12 - in decimals that would be
.33 x .50 x .50 = .0825 or 8-1/4 percent - which is where he got that number
The numbers he came up with are generally accepted approximate numbers for the efficiency of an internal combusion engine.
Approximately 1/3 of the energy of the fuel burned is lost as heat - out the exhaust and into the metal of the engine block to be carried away by the coolant and dissapated into the air by the radiator. Best case thermal efficiency is around 70% (30% loss).
Approximately 1/3 of the energy of the fuel burned is used to overcome friction losses related to the mechanical components of the engine. This includes everything from the friction drag of the rings on the cylinder walls, to the friction of the lifters on the cam, to the friction of the bearings on the journals, to the friction of the coolant against the water pump blades, to the friction of the cooling can against the air. Again, best case scenario is about 70%-75% efficiency or a 25%-30% loss.
Those two factors compound to give the basic efficiency of an internal combution engine as around 45% (2/3 x 2/3 = 4/9 = 44.9%). Some engines are a bit more efficient than this, but not by a whole lot. Best case is about 55% (.7 x .75 = 52.5%).
His 50% efficiency number for the alternator (amount of energy required to turn the alternator vs. the electrical energy output of the alternator) isn't too far off the mark either. The range is anywhere from 45%-65%. I'm not so sure about the 50% efficiency number he's assuming for the electrolyte cell. It sounds like this is the one unknown factor and that your assertions about the low voltage and ampreage requirements required to produce the H2 are where you disagree with him.
Where his equation breaks down is that he is using the numbers that represent the losses - not the efficiency. Instead of 1/3 x 1/3 x 1/2 - which are the 3 loss factors, the equation should be the product of the 3 efficiencies. The efficiencies are calculated as 100% minus the loss. If the losses are 1/3, 1/3, and 1/2 then the efficiencies would be 2/3, 2/3, and 1/2. Compounding those numbers gives us 2/3 x 2/3 x 1/2 = 4/18 = 22%.
Since the engine is providing the mechanical power to turn the alternator to generate the electrical power to supply the electrolyte cell, then the efficiencies (or losses) do indeed all compound. If we assume BEST case scenario we have .7 (thermal efficiency) x .75 (mechanical efficiency) x .65 (alternator efficiency) = 34% total efficiency for producing the electrical power going into the electrolyte cell.
IF his assumption of 50% efficiency for the electrolyte cell is correct, then multiply that by the 22% (assumed efficiencies) or the 34% (best case efficiency) then we get 11% to 17% total efficiency for the output of the electolyte cell. Better, but still not great. If the home built electrolyte cell is 75% efficient - which I think is probably a real stretch, but we'll assume to be best case - then your overall efficiency is 75% of either 22% (assumed) or 75% of 34% (best case).
That gives us a range of 11% (using average assumed efficiencies) to 25.5% (using absolute best case efficiencies).