0 Members and 1 Guest are viewing this topic.

Quote from: Kryptid on 02/04/2019 22:39:22Quote from: Thebox on 02/04/2019 22:36:35Well in theory yes Then that is what I'm trying to do. If "u" is equal to the net electric charge in a hydrogen molecule, then "u" has to be zero.Well yes if we're being honest , can we call this , the idol moment ?

Quote from: Thebox on 02/04/2019 22:36:35Well in theory yes Then that is what I'm trying to do. If "u" is equal to the net electric charge in a hydrogen molecule, then "u" has to be zero.

Well in theory yes

Quote from: Thebox on 02/04/2019 22:41:52Well yes if we're being honest , can we call this , the idol moment ? Okay then, so what would "E" represent in the equation in normal science language?

Well yes if we're being honest , can we call this , the idol moment ?

Difficult to answer but I think in your terms we can use hf³

Quote from: Thebox on 02/04/2019 22:48:43Difficult to answer but I think in your terms we can use hf³I assume "h" means Planck's constant. Does the "f" mean frequency? What would frequency represent for a hydrogen molecule? The frequency of the chemical bond?

The frequency of incoming hf I'm referring to , I'm not sure about your chemical bond but I imagine the frequency speeds up some what when ''heated'' to combustion ?

Quote from: Thebox on 02/04/2019 22:41:52Quote from: Kryptid on 02/04/2019 22:39:22Quote from: Thebox on 02/04/2019 22:36:35Well in theory yes Then that is what I'm trying to do. If "u" is equal to the net electric charge in a hydrogen molecule, then "u" has to be zero.Well yes if we're being honest , can we call this , the idol moment ? What Billy Idol? 😆

Quote from: Thebox on 02/04/2019 22:53:57The frequency of incoming hf I'm referring to , I'm not sure about your chemical bond but I imagine the frequency speeds up some what when ''heated'' to combustion ? It sounds like your equation assumes ignition by photons. Which means we can use any kind of photon frequency we want here, I guess?

Yes I think so , obvious the shorter the wave length , the quicker combustion is achieved .

Quote from: Thebox on 02/04/2019 22:58:02Yes I think so , obvious the shorter the wave length , the quicker combustion is achieved .Now we're making some progress. I assume that "V" is simply the volume of the hydrogen gas. So now on to "t", which is, I'm guessing, the time taken to burn the hydrogen. Depending on how the system is set up, you could burn hydrogen as fast or as slow as you want to. So can any number be put in for "t"? How about ten seconds?

Ten seconds will do , consider also there is the volume of the hydrogen and volume external of the hydrogen unless your talking about a compression chamber and a hydrogen fuel engine .

Quote from: Thebox on 02/04/2019 23:07:45Ten seconds will do , consider also there is the volume of the hydrogen and volume external of the hydrogen unless your talking about a compression chamber and a hydrogen fuel engine .As a gas, hydrogen will fill whatever container it's in. So those volumes can be taken as equal if you'd like.

Arr , ok, I see what you're on about , pressure equalling density I presume ? Obviously density also plays a part , there will be more energy per mm³ of hydrogen if pressured compared to less pressured .

Quote from: Thebox on 02/04/2019 23:12:32Arr , ok, I see what you're on about , pressure equalling density I presume ? Obviously density also plays a part , there will be more energy per mm³ of hydrogen if pressured compared to less pressured . If we are assuming normal atmospheric pressure, then knowing that we have one gram of hydrogen will already give us the volume.Unless you have anything else to add, I think I have all I need to do the math. Should I start?

Yea you can try , I'll be interested to see what you come up with .

Quote from: Thebox on 02/04/2019 23:16:37Yea you can try , I'll be interested to see what you come up with .I can't directly copy the equation text as you've written it (is it an image file?), but this should be an equivalent expression: (uE^{3})/(V/t)"E" equals "hf^{3}", which is the Planck constant multiplied by photon frequency. I'll use a very powerful photon in the form of a gamma ray (10^{19} hertz). 10^{19} cubed is 10^{57}. So "E" equals (6.62607015×10^{−34} joule-seconds) x (10^{57} hertz). Multiply that out and you get 6.62607015 x 10^{23}. One gram of hydrogen is equal to 0.49611 moles of hydrogen, which occupies 11.1129 liters of volume (0.0111129 cubic meters). So now I have the needed values:Combustion energy = (uE^{3})/(V/t)= ((0)*((6.62607015 x 10^{23})^{3}))/(0.0111129/10)= ((0)*((2.909163 x 10^{71})))/(0.00111129)= (0)/(0.00111129)= 0 joulesOkay, so we've run into a problem. By making "u" equal to zero, the equation predicts that no energy is released at all by hydrogen burning. However, there is a way around this. Maybe instead of net charge you meant gross charge instead? That way, instead of subtracting the charge on the electrons from the charge on the protons and getting zero, you add them together and get a finite number (6.4087064 x 10[/sup]-19[/sup] coulombs of charge, to be precise). Here's what happens if I do that instead:Combustion energy = (uE^{3})/(V/t)= ((6.4087064 x 10^{-19})*((6.62607015 x 10^{23})^{3}))/(0.0111129/10)= ((6.4087064 x 10^{-19})*((2.909163 x 10^{71})))/(0.00111129)= (1.864397 x 10^{53})/(0.00111129)= 1.677687 x 10^{56} joulesHow about that? Does that look good?

Wow , yes it looks good , I have no idea what it means lol , that's why I have a simple equation . Does it work ?

I have a simple equation . Does it work ?

Quote from: Thebox on 03/04/2019 01:08:13Wow , yes it looks good , I have no idea what it means lol , that's why I have a simple equation . Does it work ? The enthalpy of combustion of hydrogen is 141,580 joules per gram according to https://www.engineeringtoolbox.com/standard-heat-of-combustion-energy-content-d_1987.html141,580 joules per gram is nowhere remotely the same as 1.677687 x 10^{56} joules per gram.

Quote from: Thebox on 03/04/2019 01:08:13I have a simple equation . Does it work ? No, of course not.That's what we have said all along.Your calculation is nonsense.