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Re: simultaneous equations for chemical speciation
*To*: mathgroup at smc.vnet.net
*Subject*: [mg100222] Re: simultaneous equations for chemical speciation
*From*: dh <dh at metrohm.com>
*Date*: Thu, 28 May 2009 06:50:27 -0400 (EDT)
*References*: <gvk9v0$ono$1@smc.vnet.net>
Hi Jem,
First, the equation:
a=b+c+d
e=f+g+c+2d
j=mg/f
k=lm
n+m+b+c=g+l
do not add any information because they only introduce a new variable
that does not appear anywhere else. This leaves 3 equations:
h=c/(bg)
i=d/(cg)
hi=d/(bg^2)
Multiplying the first 2 gives the third. Therefore, we actually have 2
equations:
h=c/(bg)
i=d/(cg)
These do not have a general solution. Only under special circumstances
we get a solution. For this problem we have Reduce. Therefore, you get
possible solutions by:
Reduce[{h == c/(b*g), i == d/(c*g)}, {g}]
Daniel
Jem & Tracy wrote:
> Hi, I am a postgrad student and completely new to mathematica, I am just
> starting to realise the possibilities.
>
> I am trying to do a chemical speciation calculation.
> I have a set of simultaneous equations which represents the equilibria,
> mass balance, and charge balance in the system.
>
> I want to eliminate some of the variables, and then be able to solve for
> the remaining ones.
> My system of equations looks like this:
> a=b+c+d
> e=f+g+c+2d
> h=c/(bg)
> i=d/(cg)
> hi=d/(bg^2)
> j=mg/f
> k=lm
> n+m+b+c=g+l
>
> Basically, I can solve for the variable I want, but I am having trouble
> eliminating the ones I don't want (ie. I am getting an answer that is in
> terms of the wrong variables)
> Sofar my formula looks like:
>
> Solve[{a == b + c + d, e == f + g + c + 2 d, h == c/(b*g),
> i == d/(c*g), h*i == d/(b*g^2), j == m*g/f, k == l*m,
> n + m + b + c == g + l}, {g}]
>
> If I use a similar formula for the "Eliminate" function, I get error
> mesages.
>
> I have been in the help files for ages, and don't really understand how
> to do this.
> I would really appreciate some pointers!!
> Thanks
>
>
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