BTU value of Hydroxy

[Laesa]

Hi All

I'm new here and was wondering if anyone could give me the btu value of Hydroxy?

[Laesa]

302.375 btu/mole

[AlaskaStar]

[Laesa]

[Laesa]

[AlaskaStar]

The answer to your question isn't defined and will take some thought, because I neither agree nor disagree with the approach to the answer, which is of course not a silver bullet solution to the problem.

I live North of Anchorage, in a sprawling community known to me as home, but to th rest of earth it is known as a place called "Wasilla" Same place that Sarah Palin is from. Her dad was one of my teachers in school, I know Sarah personally, and I drive past her house nearly every day. The 'official' population of Wasilla is small, although the residents in the "Wasilla Area" numbers are huge. The main stretch of road to get to my place sees at LEAST 23,000 vehicles PER DAY, and as much as 79,000 vehicles per day. Many of these are the same vehicle going round trip 4 times per day. Some are making the trip 10 times per day. I know this as I was on the Construction crew rehabilitating that road. The paver did the WORST job in the history of earth. They only re-paved it FIVE TIMES in the same summer!! So with having to jam up that much traffic....yeah, I saw more middle fingers than a few...

I know one lady who does deliveries and her vehicle makes about 30 trips per day out Knik Goose Bay Road. (KGB Road) KGB Road is only 27 miles long, and ends at the Pt Mckenzie Airfield. Turn off and go for another 20 or so miles and you will arrive at the Pt Mckenzie Minimal Security Prison Farm, and the Pt McKenzie Port.

So Wasilla is a bit bigger than the media posts about....and our state isn't as small as they portray it to be. I can drive for 3 hours and not see a single car on the road when headed out some places. If we place Alaska on top of the remainder of the USA, we would consume more than 1/2 the land mass. If we cut Alaska in half, Texas would be the 3rd largest state. Kinda small now isn't it? The media hasn't a clue.

I will respond with a more detailed explanation of my reasons in the next few days, keeping it short, sweet, and not so complex that it confuses the entire forum population.

AlaskaStar

[FlashBang]

[Laesa]

[AlaskaStar]

There are many variables unanswered to your question, but I will nonetheless attempt to answer it.

HHV system of calorific determination of energy contents is based on what could be (in layman's terms...) considered the highest possible point that an exacting amount of fuel combusted (not useful in my research lines) with the proper oxidizer content will make. Kinda like a slingshot:

The distance of stretch will be the same each time on the rubber band, but the size and weight of the rock to be flung is the variable. Other variables are the age of the rubber and the remaining amounts of elasticity in the rubber when stretched, but we will omit those variables for the sake of argument and to keep it simple. Now the direction of travel is exactly vertical in relation to the surface of the earth, and perpendicular to the 'level plane' at that spot on this sphere (yeah get level on a beach ball will ya!) we call Earth. So we intend to sling a rock straight UP. There is NO WIND, and gravity is constant (representing the 25 degrees C). The only variables to contend with now are the following: Weight and shape. Let's assume that SHAPE will be the required oxidizer because the shape is what determines the aerodynamic drag on the rock. The weight will be the number of hydrogen atoms and carbon atoms represented in this layman's term of a HHV value system of measurement.

So the rock is launched after much hashing over details and the person whom is holding the slingshot has now currently BLUE FINGERS that are in severe pain! The rock travels upward X distance before returning to earth. This X Distance answer is and can be repeatedly made using the exact same rock in the exact same scenario. This X Distance is determined entirely by the drag (oxidizer requirement) and the weight (Hydrogen and Carbon Content).

So after much ado, we now only achieve a system of measuring the farthest reaches of the fuel in a set environment of conditions.

(Mixing Chemistry and Physics into laymans terms can be fun isn't it?)

Now the LHV is a tricky part....

We now COMBINE the HHV to get us PAST a specified X Distance with the vertical rock launch and have replaced the poor human with a machine that doesn't tire from the long winded talk whilst holding a rock in a slingshot pulled taut for extended periods of time causing sore muscles and a bad attitude toward the person talking (likely resulting with the rocks being launched at guest speaker or property of, like car windows....).

(I am trying to be funny here too...)

So with the LHV system not only do we get PAST a SPECIFIED POINT by keeping the constants of the HHV system, but now, we add a twist, like changing the rules in the middle of the game. Here's the twist:

So we put the cutoff point at 15 feet elevation. We ASSUME using the LHV process that anything below 15 feet is UNUSABLE ENERGY (I now refer you to the Aerothermal Engine....and note that I am using temperatures between -100 and + 100 F to run it) because this 15 foot mark is referenced to represent the 150 degree mark. They ASSUME that ANYTHING below 150 degrees C is NOT USABLE ENERGY in terms of relative heat either through enthalpy or entropy, and so our cutoff is 15 feet above the level plane of the ground.

We now launch Rock A. Rock A went from an elevation of 5 feet (starting point) to a total height of 160 feet and back to 5 feet (ending point must be same as starting point) . So Rock A has an HHV of 155.
Now.... Rock A when applied to the LHV end of the deal, is rated at an LHV of 145. This is because we cutoff at 15 feet from origination point to ending point. We cannot get something from nothing per say, so this arbitrary pick make it work out this way.

Rock B undergoes the same brutal abuse and consistently gives an HHV score of 200, but only receives an LHV score of 185.

Rock C same thing, but gets 105 HHV and 90 LHV.

So we have the HHV system in which the starting point and ending point are the SAME (which is good science) but ignores the Enthalpy and Entropy latent heat LOSSES of the material of the canister for testing (which is again another variable that SHOULD be calibrated to adjust for).

The LHV system the starting and ending points are different (bad science in my opinion only confusing, and based entirely on bad logic), but ignores the Latent heat, but accepts absorbed heat losses (partial picture here....) of the containment test vessel!!!

Both are a sour deal and will only confuse a person in the end. Too many variables, too much assumption, and an arbitrary pick of ASSUMING that any amount of heat that is below 150 Degrees C is USELESS and cannot be used in any possible or useful manner is not taken lightly by the poor guy who is suffering from hypothermia!

So If I had to pick a process, the HHV comes close, BUT I would have to calculate out ALL the variables to make it TRUELY a scientific method to be utilized for any REAL DATA COMPARISONS, so even the HHV system isn't proper. Let's do it right, and get the ENTIRE PICTURE.

AlaskaStar

[AlaskaStar]

Another note:

The 150 Degree arbitrary pick of LHV is real, and they do state that any Delta Potential Differentials BELOW the 150 C degrees is useless and will not work.

This is in conjunction with Boyle's Law of the Ideal Gas.

Now My Aerothermal engine (same engine for the X-Rig topic here in the forum) is being built to show the following:

A LOW DELTA T CAN BE USED TO GENERATE USEFUL AMOUNTS OF POWER.

REVERSING THE 95% INTERNAL LOSSES AND 5% TRANSFER OF USEFUL ENERGY TO 95% TRANSFER OF ENERGY AND ONLY 5% INTERNAL LOSSES CAN BE DONE.

REALIZATION THAT SMALL AMOUNTS OF ENERGY USED TO THE FULLEST EXTENT CAN BRING ABOUT HUGE IMPLICATIONS FOR THE INDUSTRIAL AGE AS WELL AS SCIENCE.

Those are the things I hope to prove with my engine. I am not looking for fame/ fortune, but I to want to have those that are educated enough to understand be able to understand it in case a miscreant decides I am a 'danger' to the oiled economy (the grease that makes the wheel turn and the money burn....) and removes this 'danger', at least the technology is not lost forever.

AlaskaStar

[Laesa]

AlaskaStar

Thanks so much for that analogy (and yes, you did a great job at making me laugh) The real issue at hand, is that within an industrial environment (power generation) in particular Btu content is king, so my thought was to attempt to bring each side of the equation (i.e. energy used versus energy created) to the same denominator for a more accurate method of determining efficiency of the various types of production of HHO.

In my research with Stan Meyers, Mr. Boyce and Mr. Lawton I have examined all of their written and video evidence, however I have not been able to find anywhere, were any of them stated the method they were using to determine the efficiency of the cells they had created.

Mr. Boyce seems to be the only one that has publicly stated the production rates with his cell (and you pf course, with your chemical process) which has been of invaluable help.

If all of them are in fact using Faraday's law, why then is there so much controversy on the method used?

I am like you in the fact, that fame and fortune are not my goals, but rather the salvation of a country in dire need.

Again thanks very much for taking the time to answer

[AlaskaStar]

Well the way to think of it in terms of power generation is like this:

How many BTU was required to 'make' the fuel?

How many BTU was harnessed in utilizing the fuel?

Subtract the BTU requirement from the 'making' of the fuel from the BTU harnessed in using the fuel, and you will now be left with a 'net' energy content in terms of BTU.

So if you use a system that consumes 5 KW of power to make the fuel, and you only get 4.6 KW output from the fuel, the it is easy to say you have a NEGATIVE 400 watts of power in BTU content from losses in both systems. You can overcome this.

If you have a system that uses the fuel 2x more efficiently then the means to make the fuel, then you are only going to have a net of say (using the last scenario) 4.2 KW energy content that is 'usable' by standards of excess energy to be used.

So both ends of the equation need looked at under heavy scrutiny to determine WHERE and HOW the efficiencies of both systems can be made and if they are cost effecive using that set of hardware, or if a completely different set of hardware needs utilized. (turbine verses piston engine? Flywheel for kinetic storage? etc...etc...)

So depending of where and how you go about accomplishing the task, you may be looking into several thousands of dollars worth of hardware and tools.

Look at some of the stuff I make. Hardware is not cheap, and neither are the tools to make the parts from scratch.

Are you prepared to take on the so called 'impossible'?


AlaskaStar

[Laesa]

[AlaskaStar]

[Laesa]