Sciencemadness Discussion Board
Not logged in [Login ]
Go To Bottom

Printable Version  
Author: Subject: Problems with the Kamlet-Jacobs Equations
Elemental Phosphorus
Hazard to Others
***




Posts: 177
Registered: 11-11-2016
Location: Is everything
Member Is Offline

Mood: No Mood

[*] posted on 5-2-2022 at 17:46
Problems with the Kamlet-Jacobs Equations


I had recently read about the Kamlet-Jacobs equations, and thought that they were very interesting, it was claimed that they could predict within about a 5% error the detonation pressure and VoD of energetic materials (at least those that have a density of above 1.0 g/cc and are composed solely of C-H-N-O atoms.)

I am writing a paper for a chemistry class (a high school class, not college) I'm in about predicting behavior of real and theoretical energetic materials using the equations.

In order to get familiar with the equations I decided to calculate VoD for TNT and EGDN to see how accurate the equations were. However, I always got values that were far above the real VoDs for the explosives.

I am using these equations:


D = A x (1 + B x p) x sqrt(phi)

PCJ = K x p^-2 x phi

phi = N (sqrt( M x Q ))

Where:

D = detonation velocity in mm/us (km/s)
PCJ = Detonation pressure in kilobar
N = moles gas/gram of explosive
M = Mean molar mass of gaseous products
Q = Explosive detonation enthalpy per gram
p = Density in grams per cubic centimeter

A, B and K are computed constants.

A = 1.01

B = 1.30

K = 15.58

(I apologize in advance if the formatting gets messed up)

German Wikipedia has a page on it:
https://de.wikipedia.org/wiki/Kamlet-Jacobs-Gleichungen


As for my sample calculations:


For EGDN:

N = 0.033
M = 30.41
Q = 7400
p = 1.49

C2H4N2O6 -> 2CO2 + 2H2O + N2

That's 5 moles gas per mole EGDN. And the molar mass is 152.1g/mol

5/152.1 = 0.03287

The mean molar mass of the products would be ((18.02)2 + (44.01)2 + 28.01)/5 = 30.41g/mol

As for Q:

(-285.8kJ/mol)2 + (-393.5kJ/mol) + (0kJ/mol) = -1358.6kJ/mol

-1358kJ/mol - (-233kJ/mol) = -1125.6kJ/mol enthalpy of detonation

-285.8 being the enthalpy of formation of water, -393.5 that of carbon dioxide, and -233 being that of EGDN

-1125.6 / 152.1 = 7.4004 kJ/g or 7400.4 J/g

(the enthalpy of formation of EGDN was taken from NIST https://webbook.nist.gov/cgi/cbook.cgi?ID=C628966&Mask=2)

So the equation would seem to be:

phi = 0.033 ( sqrt(30.41 x 7400.4) ) = 15.65

D = 1.01 (1 + 1.3 x 1.49) x sqrt(15.65)
D = 11.74 km/s

Which is obviously way too high, that would be higher than for any known explosive.

For TNT:

N = 0.0253
M = 28.52
Q = 5883
p = 1.65

I'll spare all the other calculations, but note that since TNT is very oxygen-deficient, I calculated moles of gas and detonation enthalpy using the assumption that H2O and CO2 were produced, not CO gas, as the production of CO2 seems more favorable (at least at standard conditions, of course an explosion is not that, but still).

The equation is then


phi = 0.0253 ( sqrt(28.52 x 5883) ) = 10.36

D = 1.01 x (1 + 1.3 x 1.65) x sqrt(10.36)
D = 10.23 km/s

Which is also obviously too high, closer to that of octanitrocubane than TNT.

Interestingly enough, if the equation is amended to:

D = A x (B x p) x sqrt(phi)

D = 1.01 x (1.3 x 1.65) x sqrt(10.36)
D = 6.97 km/s

and

D = 1.01 (1.3 x 1.49) x sqrt(15.65) = 7.74 km/s

For TNT, that is pretty close to the literature value of 6900m/s, and for EGDN the literature says 7500m/s, also pretty close.

However, this doesn't make any sense. I must be calculating something wrong, but I've calculated several times and I get the same results.

So, any ideas? I could really use the help. Thanks!:P
View user's profile View All Posts By User
MineMan
International Hazard
*****




Posts: 889
Registered: 29-3-2015
Member Is Offline

Mood: No Mood

[*] posted on 5-2-2022 at 20:47


There is a paper by an Iranian prof. It’s a much better and simpler equation.
View user's profile View All Posts By User
Microtek
International Hazard
*****




Posts: 652
Registered: 23-9-2002
Member Is Offline

Mood: No Mood

[*] posted on 6-2-2022 at 01:07


You have to convert the heat of explosion to calories per gram (by dividing with 4.186). I realize that some sources claim otherwise, and you could transform the formulas to work with SI-units by adjusting the parameters, but that is the reason you get erroneous values. Also, you are using the heat of formation for water in the liquid state. It is gaseous until long after the detonation is over, so you should use -240.6 kJ/mol.



[Edited on 6-2-2022 by Microtek]
View user's profile View All Posts By User
Elemental Phosphorus
Hazard to Others
***




Posts: 177
Registered: 11-11-2016
Location: Is everything
Member Is Offline

Mood: No Mood

[*] posted on 7-2-2022 at 05:46


Quote: Originally posted by MineMan  
There is a paper by an Iranian prof. It’s a much better and simpler equation.


Would you mind linking it or telling me how to find it? I’m intrigued.


Also, thank you Microtek, that’s very helpful. Using kcal and -240.6 for the heat of formation of water, I got 7.08km/s for TNT, and 8.03 km/s for EGDN. Both a bit high, but not too far off.

That’s an error of 2.6% for TNT and 7.1% for EGDN, which is within the error.
View user's profile View All Posts By User
Microtek
International Hazard
*****




Posts: 652
Registered: 23-9-2002
Member Is Offline

Mood: No Mood

[*] posted on 8-2-2022 at 01:26


The Iranian scientist referred to is probably Mohammad Keshavarz, who has written a host of papers on different mathematical models of energetics. Just be aware that it is inherently problematic to extrapolate fitted empircal models to unknown substances. You can use the models to locate candidate energetics for further experimental studies though.
View user's profile View All Posts By User
MineMan
International Hazard
*****




Posts: 889
Registered: 29-3-2015
Member Is Offline

Mood: No Mood

[*] posted on 8-2-2022 at 23:12


Quote: Originally posted by Microtek  
The Iranian scientist referred to is probably Mohammad Keshavarz, who has written a host of papers on different mathematical models of energetics. Just be aware that it is inherently problematic to extrapolate fitted empircal models to unknown substances. You can use the models to locate candidate energetics for further experimental studies though.


I know of no other way unless you employ an EOS such as that of EXPLO 6. Modern EOS can handle chlorine and metal ions. But the equation from that prof is about the best you can get without EXPLO… it also works on non ideal explosives such as AN and AP mixtures. It can handle chlorine I remember and maybe some Al as well?
View user's profile View All Posts By User
Dornier 335A
Hazard to Others
***




Posts: 229
Registered: 10-5-2013
Location: Northern Europe
Member Is Offline

Mood: No Mood

[*] posted on 14-2-2022 at 09:06


My experience is that Keshavarz's equations are better than Kamlet-Jacobs, but slightly harder to use. I've had good results with both chlorine and aluminium as well!

[Edited on 2022-2-14 by Dornier 335A]




Some of my old videos are available here:
https://archive.org/details/@dornier335a
View user's profile Visit user's homepage View All Posts By User

  Go To Top