# Why getting some strange fractions? The code is simple and correct,maybe.

20 views (last 30 days)
Luca on 17 Sep 2018
Commented: Walter Roberson on 20 Sep 2018
why these fraction?
nuove_pp(reag+i) -> ..and.. -> nuove_pp ->
nuove_pp = 1×4
14.6794 9.7862 44.0381 39.1449
nuove_pp
reag
prod
cs
prodotti_numeratore=1;
for i=1:prod
prodotti_numeratore = prodotti_numeratore.*((nuove_pp(reag+i)+cs(reag+i)*x)^cs(reag+i)) ;
end
reagenti_denominatore=1;
for i=1:reag
reagenti_denominatore = reagenti_denominatore*((nuove_pp(i)-cs(i)*x)^cs(i)) ;
end
Q=prodotti_numeratore /reagenti_denominatore
equ= Q == K_P;
sol=double(solve(equ,x,'Real',true))

Stephan on 17 Sep 2018
Edited: Stephan on 17 Sep 2018
Hi,
if you calculate symbolic equation, the calculated values (like factors for example) in symbolic equations are represented exactly, which means that they look like this.
If you want to have double representation use:
double(a)
If you create function handles from your symbolic equation the values will be converted to double automatically.
A way for more precision than double is vpa like Star Strider suggested in his answer. If you want to solve symbolic equations numeric have a look at vpasolve.
Best regards
Stephan

Star Strider on 17 Sep 2018
It looks as though you are using the Symbolic Math Toolbox.
To get the fractions as thier decimal equivalents, use the vpa (link) function.

Luca on 18 Sep 2018
thank you very much for your availability. The main question is: why is not simply inserted the value relative to pp_nuove (3)? Matlab generates a useless and inexplicably large fraction.
prodotti_numeratore=1;
for i=1:prod
prodotti_numeratore = prodotti_numeratore.*((nuove_pp(reag+i)+cs(reag+i)*x)^cs(reag+i)) ;
end
Stephen23 on 20 Sep 2018
That's all.
nuove_pp = 1×4
14.6794 9.7862 44.0381 39.1449
reag=2;
prod=2;
cs=[1 1 1 1];
prodotti_numeratore=1;
for i=1:prod
prodotti_numeratore = prodotti_numeratore.*((nuove_pp(reag+i)+cs(reag+i)*x)^cs(reag+i)) ;
end
reagenti_denominatore=1;
for i=1:reag
reagenti_denominatore = reagenti_denominatore*((nuove_pp(i)-cs(i)*x)^cs(i)) ;
end
Q=prodotti_numeratore /reagenti_denominatore
equ= Q == K_P;

Luca on 18 Sep 2018
Thank you, heartily. But why does the matlab result differ from the paper one? The implemented code is as simple as the exercise.
Walter Roberson on 19 Sep 2018
What input values should I be using? It looks like I should use 1 1 1 1 for the regents, but what should I enter for the mol values?

Luca on 20 Sep 2018
Mol values?
do not need mol values, but only the partial pressures I have already indicated. Why do you ask me for the mol?
(Excuse me for my bad English)
Walter Roberson on 20 Sep 2018
Your code prompts four times for "Inserisci il coefficiente stechiometrico del reagente". This values get displayed directly on your sample output diagrams as "cs" just above "Q". It looks like the answers to each of those should be 1 .
Your code then prompts two times for "Assegnazione moli di reagente" and two times for "Assegnazione moli di prodotto". The English translation of that is "Assignment of moles of reagent", and "Assignment of moles of product" which appears to be questions about mole values that would be responded to with mol quantities, rather than a question about partial pressures. These values get displayed in your code as moli_reag and moli_prod a full page above the portion of the code that you have posted images of.
I need you to pick one particular set of "Inserisci il coefficiente stechiometrico del reagente" values and one particular set of "Assegnazione moli di reagente" and "Assegnazione moli di prodotto" values and tell me what they are. Do not assume that I can read them off of the diagrams: your diagrams are cut off.
This is not hard: just tell me which 8 input values are to be used.

Luca on 20 Sep 2018
Leave default values. The range could be from 0.1 to 4 with 0.1 step.
moli_reag(reag)=0;
for i=1:reag
prompt = 'Inserisci le moli di reagente:';
title = 'Assegnazione moli di reagente';
dims = [1 55];
defaultinput = {'0.40'};
moli_reagx= inputdlg(prompt,title,dims,defaultinput);
moli_reag(i)=str2double(moli_reagx);
end
moli_reag
PS thank you for taking the time.
Walter Roberson on 20 Sep 2018
Sorry, I do not seem to be able to assist you on this matter; the communications problems seem to be too large to overcome.

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