View previous topic :: View next topic 
Author 
Message 
slartibartfasz Veteran
Joined: 29 Oct 2002 Posts: 1462 Location: Vienna, Austria

Posted: Tue Jun 24, 2003 5:59 pm Post subject: physic/chemistry/math wizzards  solve this... 


in order to compute the viscosity of gas mixtures at various temperatures i need to determine the reduced dipol momentum of a substance. in namly formula i have to use* the dipol momentum of the substance in a very strange unit:
1 Debye = 10^18 * SQRT(10^13 J m^3)
this seems to be correct since i get sensible results when looking at the resulting unit and the computed values are within 1 or 2% of the measured values.
my problem is that i simply cannot see how the above unit conversion can be achieved. i've read through all definitions i could find, trashed more paper, than a whole forest can deliver and emptied several pens. but still ... how i'm sure its not really complicated it just has a vicious twist somwhere...i'm curious if and how fast this one gets solved here
links:
http://physics.nist.gov/Pubs/SP811/appenB8.html
http://www.bipm.fr/enus/welcome.html
http://physics.nist.gov/cuu/Units/
http://www.chemie.fuberlin.de/chemistry/general/si_en.html
^...power
SQRT()...square root
Debye...Unit of the dipol momentum [C*m]
J...Joule [kg m^2 s^2]
* formulas and units are out of the VDI Wärmeatlas (german)  the german standard book for everything concerning engineering and heat... _________________ To an engineer the glass is neither half full, nor half empty  it is just twice as big as it needs to be. 

Back to top 


far Guru
Joined: 10 Mar 2003 Posts: 394 Location: Stockholm, Sweden

Posted: Tue Jun 24, 2003 6:32 pm Post subject: 


sqrt(kg * m^5 * s^2) is not equal to A * s * m
It does not make sense. Perhaps the constants in the expression have units? _________________ The Porthole Portage Frontend 

Back to top 


slartibartfasz Veteran
Joined: 29 Oct 2002 Posts: 1462 Location: Vienna, Austria

Posted: Tue Jun 24, 2003 6:53 pm Post subject: 


far wrote:  sqrt(kg * m^5 * s^2) is not equal to A * s * m
It does not make sense. Perhaps the constants in the expression have units? 
seconded  it doesnt seem to make sense...
but if i look up a dipol momentum somewhere [D], mangle it through this and use the resulting value i get a correct solution in the numerical and the unit context. if i use sqrt(J m^3) as a unit in the formula of the reduced dipol momentum, i get an dimensionless figure, which looks sensible to me for a 'reduced' figure.
there has to be some connection...
i put part of my work on my server (pdf)  the formula in question is in the first line page 4  its not corrected yet, i'm sure there are some errors in the text. _________________ To an engineer the glass is neither half full, nor half empty  it is just twice as big as it needs to be. 

Back to top 


vers_iq Apprentice
Joined: 18 May 2002 Posts: 264

Posted: Tue Jun 24, 2003 7:07 pm Post subject: 


Firstly the formule u gave is not very vivid,
Is this somekind of proofing question?, looks like u have to use dipol equition which i dont really like to see the shape to find "viscosity" and by entering the variables and constants base on your gas mixture condition then all you have to do is twist them until the formula match
just a theory _________________ "You know the world is going crazy when the best rapper is a white guy, the best golfer is a black guy, the Swiss hold the America's cup, France accusing the US of arrogance and Germany doesn't want to go to war." 

Back to top 


far Guru
Joined: 10 Mar 2003 Posts: 394 Location: Stockholm, Sweden

Posted: Tue Jun 24, 2003 7:13 pm Post subject: 


slartibartfasz wrote:  i put part of my work on my server (pdf)  the formula in question is in the first line page 4  its not corrected yet, i'm sure there are some errors in the text. 
Is this what you are referring to?
Code:  were <foo> is the reduced dipol momentum ... 
It doesn't look like the one above.
It is "where <foo> is the reduced dipole moment," btw. _________________ The Porthole Portage Frontend 

Back to top 


slartibartfasz Veteran
Joined: 29 Oct 2002 Posts: 1462 Location: Vienna, Austria

Posted: Tue Jun 24, 2003 7:42 pm Post subject: 


far wrote:  Is this what you are referring to?
Code:  were <foo> is the reduced dipol momentum ... 
It doesn't look like the one above. 
yes  the thing above is the unit of mu to be used in <foo>  its only a small note in the book. its not in the text because i noticed it after i did the first test with a prog i wrote. it turned out that i have to use the dipol moment in the exact unit of my original post to get a correct result...
[EDIT:] if u take <foo> (the reduced dipol moment) and assume that the dipol moment (mu) has the dimension on the right side of the first equation in this thread u will see that <foo> becomes dimensionless as it should be for a 'reduced' value
Quote: 
It is "where <foo> is the reduced dipole moment," btw. 
thx  technical terms can be quite tricky...
@vers_iq:
yes it is a proofquestion. i know that the rest works and i actually get correct results for mixtures of 15 different gases and temperatures higher than 1000°C... what i dont know is WHY it is correct  i understand every single step involved save the unit of the dipol moment > see first post _________________ To an engineer the glass is neither half full, nor half empty  it is just twice as big as it needs to be. 

Back to top 


far Guru
Joined: 10 Mar 2003 Posts: 394 Location: Stockholm, Sweden

Posted: Tue Jun 24, 2003 8:10 pm Post subject: 


slartibartfasz wrote:  [EDIT:] if u take <foo> (the reduced dipol moment) and assume that the dipol moment (mu) has the dimension on the right side of the first equation in this thread u will see that <foo> becomes dimensionless as it should be for a 'reduced' value 
Here is <foo>: mu_R = mu^2 * p_c / (k * T_c)^2
Lets start with the denominator. I am assuming that k is Boltzmann's constant and that T_c is a temperature, then the denominator is J^2.
If mu^2 is J * m^3 then if mu_R is supposed to be dimensionless, p_c must be J / m^3 (energy density?)
Where does Debye and the electric moment come into this? Does it have anything at all to do with electromagnetism? _________________ The Porthole Portage Frontend 

Back to top 


slartibartfasz Veteran
Joined: 29 Oct 2002 Posts: 1462 Location: Vienna, Austria

Posted: Tue Jun 24, 2003 8:45 pm Post subject: 


far wrote: 
Here is <foo>: mu_R = mu^2 * p_c / (k * T_c)^2
Lets start with the denominator. I am assuming that k is Boltzmann's constant and that T_c is a temperature, then the denominator is J^2.
If mu^2 is J * m^3 then if mu_R is supposed to be dimensionless, p_c must be J / m^3 (energy density?)

k...boltzmann constant [J/K]
T_c...critical temperature [K]
p_c...critical pressure [bar]
assume mu^2.... [J m^3]
exactly  only p_c can be written as N/m^2 because of course 1J=N*m
Quote:  Where does Debye and the electric moment come into this? Does it have anything at all to do with electromagnetism? 
thats my question  if i would not have found the strange unit conversion from the first post, i would have inserted the dipol moment mu in Debye [C*m] which is clearly wrong. the formula only works if i use mu as [SQRT(J m^3)] and this is the step i dont understand.
the general influence of the dipol moment is clear  it influences the way gas molecules bump into each other > influences friction > influences viscosity. i just dont know why the equation in the first post is correct  and as u may have noticed i hate to work with math i dont understand _________________ To an engineer the glass is neither half full, nor half empty  it is just twice as big as it needs to be.
Last edited by slartibartfasz on Tue Jun 24, 2003 8:47 pm; edited 1 time in total 

Back to top 


Hypnos Advocate
Joined: 18 Jul 2002 Posts: 2878 Location: Omnipresent

Posted: Tue Jun 24, 2003 8:46 pm Post subject: 


Well, here's the dipole momentum per mol (Jackson Eq. 4.81):
Code:  p_mol=1/3*p_0^2*E/kT 
where p_0 is the permanent dipole moment of a single molecule in the fluid, E is applied electric field and T is the absolute temperature . Obviously, the units work out if you consider that the units of E is J/C/m in SI.
Hope that helps ... are you trying to derive the permanent dipole moment, which would be related to the local electrodynamics, and hence the viscosity? If you invert the formula above, it looks something like your original formula ... _________________ Personal overlay  Simple backup scheme 

Back to top 


slartibartfasz Veteran
Joined: 29 Oct 2002 Posts: 1462 Location: Vienna, Austria

Posted: Tue Jun 24, 2003 9:00 pm Post subject: 


Hypnos wrote:  Well, here's the dipole momentum per mol (Jackson Eq. 4.81):
Code:  p_mol=1/3*p_0^2*E/kT 
where p_0 is the permanent dipole moment of a single molecule in the fluid, E is applied electric field and T is the absolute temperature . Obviously, the units work out if you consider that the units of E is J/C/m in SI. 
this looks great  can u please give me the whole citation of the book (or enough that i can find it)... this the best hint i found so far  thx a lot _________________ To an engineer the glass is neither half full, nor half empty  it is just twice as big as it needs to be. 

Back to top 


Hypnos Advocate
Joined: 18 Jul 2002 Posts: 2878 Location: Omnipresent

Posted: Tue Jun 24, 2003 9:05 pm Post subject: 


slartibartfasz wrote:  this looks great  can u please give me the whole citation of the book (or enough that i can find it)... this the best hint i found so far  thx a lot 
I'm a silly physicist  I thought "Jackson Eq. 4.81" would be more than sufficient!
_Classical Electrodynamics_, Jackson, John David, 2nd edition, John Wiley & Sons, New York (1975).
Some notes:
* There is a newer, 3rd edition, which is partly in SI units (2nd edition is all in Gaussian, i.e. "God's units.")
* Jackson is a professor emeritus here, and he's somewhat intimidating. _________________ Personal overlay  Simple backup scheme 

Back to top 


slartibartfasz Veteran
Joined: 29 Oct 2002 Posts: 1462 Location: Vienna, Austria

Posted: Tue Jun 24, 2003 9:14 pm Post subject: 


Hypnos wrote:  I'm a silly physicist  I thought "Jackson Eq. 4.81" would be more than sufficient! 
hehe  i am a chemist  i dimly recall that i had to sign a paper  vowing never to learn anything a physicist might call standard electrodynamics *urg*  but i have to admit that i may be handy from time to time
thx again  i keep u updated if i figure it out completely  tomorrow... _________________ To an engineer the glass is neither half full, nor half empty  it is just twice as big as it needs to be. 

Back to top 


carambola5 Apprentice
Joined: 10 Jul 2002 Posts: 214 Location: Madtown, WI


Back to top 


slartibartfasz Veteran
Joined: 29 Oct 2002 Posts: 1462 Location: Vienna, Austria

Posted: Wed Jun 25, 2003 4:22 am Post subject: 


LOL
that was  of course  the first thing i tried _________________ To an engineer the glass is neither half full, nor half empty  it is just twice as big as it needs to be. 

Back to top 


