[Bored chemist]

Hydrogen-Oxygen chain of combustion


ho.jpg (13.59 kB . 342x239 - viewed 786 times)

So, it's not just physics you don't understand; you are also clueless about chemistry.

[guest39538]

Hydrogen-Oxygen chain of combustion


ho.jpg (13.59 kB . 342x239 - viewed 786 times)

So, it's not just physics you don't understand; you are also clueless about chemistry.

You have no thinking ability Mr C , I had to redraw it for elsewhere as they had the same obtuse problem .


h1.jpg (93.71 kB . 524x323 - viewed 479 times)ΔH = ΔF³ = Δp where p is pressure


You are all obtuse seriously , just soapy water and strange smells ….

[Bored chemist]

It's not the drawing, as such, that's the problem.
The issue is the nonsense
For example, there seem to be 5 hydrogens and one oxygen in the first picture.
As I said, you must be pretty poor at chemistry to think that's' the right ratio to make water.

And in both pictures you have scribbled random characters on them with no indication of what they are meant to mean.

It would be less of a problem, but we actually know what the details of this reaction are, so we know exactly how wrong you are.

[Bored chemist]

You keep failing
In particular, you have repeatedly failed to address this

What are the units of the quantities involved?

[guest39538]

It's not the drawing, as such, that's the problem.
The issue is the nonsense
For example, there seem to be 5 hydrogens and one oxygen in the first picture.
As I said, you must be pretty poor at chemistry to think that's' the right ratio to make water.

And in both pictures you have scribbled random characters on them with no indication of what they are meant to mean.

It would be less of a problem, but we actually know what the details of this reaction are, so we know exactly how wrong you are.

No there is lots of hydrogen and oxygen in the picture but you can't see it , the one's you can see are conceptual Mr C so you can see it .

[guest39538]

You keep failing
In particular, you have repeatedly failed to address this

What are the units of the quantities involved?

Ask a mathematician , I'm a theoretical physicist , not a mathematician , science can put in the values , I'm not going to do all the work for them , lazy I tell ya .

[Bored chemist]

No there is lots of hydrogen and oxygen in the picture but you can't see it , the one's you can see are conceptual Mr C so you can see it .

So, you have provided a picture of hydrogen and oxygen- in which the hydrogen and oxygen are invisible.

Did you think that might be helpful somehow?

[Bored chemist]

You keep failing
In particular, you have repeatedly failed to address this

What are the units of the quantities involved?

Ask a mathematician , I'm a theoretical physicist , not a mathematician , science can put in the values , I'm not going to do all the work for them , lazy I tell ya .

No.
Explaining what the terms in the equation and what the units should be is part of your job.
If you learned how to do it, you would see that what you posted can't be true.

[guest39538]

No there is lots of hydrogen and oxygen in the picture but you can't see it , the one's you can see are conceptual Mr C so you can see it .

So, you have provided a picture of hydrogen and oxygen- in which the hydrogen and oxygen are invisible.

Did you think that might be helpful somehow?

Oh my word , you are so obtuse , have you ever tried to draw something that's invisible to the eye ? 

The diagram demonstrates what's happening within a volume of hydrogen-oxygen when you ignite it .  The ''orange'' hydrogen is the ignition point, the surrounding volume if forced to big crunch at a very fast speed , this then pressures the ignition point to cause the reaction chain .  Learn some physics Mr C please .

[Bored chemist]

Since you have made up some stuff that's wrong; then drawn it badly and written random symbols on it, it's not me that needs to learn physics.

[guest39538]

Since you have made up some stuff that's wrong; then drawn it badly and written random symbols on it, it's not me that needs to learn physics.

https://en.wikipedia.org/wiki/Illusory_superiority

In the field of social psychology, illusory superiority is a condition of cognitive bias wherein a person overestimates their own qualities and abilities, in relation to the same qualities and abilities of other persons. Illusory superiority is one of many positive illusions, relating to the self, that are evident in the study of intelligence, the effective performance of tasks and tests, and the possession of desirable personal characteristics and personality traits


I just got banned elsewhere for telling the mod a few home truths about his illusory superiority , something most people in physics with an education suffer from .

The process I've explained involves gravity Mr C , the ignition point has greater mass/energy therefore attracts the lesser energy state surrounding volume of hydrogen and air .  It's how gravity works Mr C and I have had enough of science and their Dunning-Kruger obviousness .  They are really clueless  and obtuse , I can't believe how naïve they are .  Yesterday elsewhere I destroyed the Hubble and the cosmological model of the big bang using their own theory .
Now I'm sick and tired of been treated by science as some fool , they have it wrong and are naïve , not me .  I provide the objective evidence and correct semantics unbiased and undeluded .

You haven't a clue how the universe works , seriously .

I admit I'm crap at maths and I admit I don't always explain too well , I don't suffer from it at all .

[guest39538]

My ability is my envision of physical process , I do know my chit because I can bloody well see it .


str.jpg (23.2 kB . 524x323 - viewed 461 times)

My 7D complete model is objectively a superior model to the present model because it has no shortfalls and conforms to present physics , explaining almost everything in the process .

ΔV = Δ   


added -

A 1kg mass divided by V 10cm^3 = 1 / 10000 = 0.0001ρ

So 1kg + U 1kg =2kg  so 2kg / 10000 = 0.0002ρ


so 2kg / 20000 should equal 0.0001p  and it does . 0.0001p

See it works ...

[Bored chemist]

The process I've explained involves gravity Mr C , the ignition point has greater mass/energy therefore attracts the lesser energy state surrounding volume of hydrogen and air .

Flames work in zero gravity.

You haven't a clue how the universe works , seriously .

I know how one small bit of the universe works in at least one important way.
I know that gravity can no possibly be involved in the combustion of hydrogen, because it still works in free fall.

So I know that you are wrong; not because of any conceit on my part, but because that's what the evidence says.

So,  Dunning-Kruger it is; but not the way round you were thinking.

explaining almost everything in the process .

It has yet to explain anything.

[Bored chemist]

because I can bloody well see it .

There are a number of words for people who see things that are not real.
One of the more polite ones is "deluded".

[guest39538]

Flames work in zero gravity.

It's not zero gravity , you're are not accounting for the flames mass.

[Bored chemist]

Flames work in zero gravity.

It's not zero gravity , you're are not accounting for the flames mass.

Are there any more  straws you want to clutch at?
The gravitational forces involved in a hydrogen flame are tiny.

[guest39538]

Flames work in zero gravity.

It's not zero gravity , you're are not accounting for the flames mass.

Are there any more  straws you want to clutch at?
The gravitational forces involved in a hydrogen flame are tiny.

Relative to itself they are massive

[Bored chemist]

Relative to itself they are massive

That doesn't actually mean anything.
The forces are still tiny.

[guest39538]

∈

Relative to itself they are massive

That doesn't actually mean anything.
The forces are still tiny.

No , when you ignite the hydrogen you create a strong force at that point . E=UF³

=0.0001j per point ?  Based on 10cm³


Hmm, I get 90000J per point so 0.0001 kg must equal 90000J?

= = = ==900j per point = 0.000001kg per point

Scratches head …..

added- So the given point of ignition would have kE=0.000944076141J  so the total output is 944.076141J  and an increase of 44.076141j  of energy .

Added- Sorry I might have done that wrong , the result may be 8496685270J  of energy I'll have to rethink .

[Bored chemist]

Scratches head

That's because none of what you posted makes sense.