8 views (last 30 days)

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.

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

Luca
on 18 Sep 2018

Stephen Cobeldick
on 19 Sep 2018

On paper I drew a blue box with the operations to be performed.

Walter Roberson
on 19 Sep 2018

When you solve() a polynomial system with rational coefficients, solve returns all of the solutions (in an undefined order). When you solve a system with software floating point, solve may assume that you only want one solution, and which of the solutions it chooses is undefined and small changes in the values or starting point can have it pick a quite different solution.

Also, when software floating point is used, the meaning of of raising a value to a power is defined by

A^B = exp(B*log(A))

and for negative A that can end up creating a complex valued solution. For example (-1.0)^(1/3) is always a complex root of unity, never -1. Therefore when you use software floating point you can end up calculating something different.

When you use software floating point, approximations are made that can lead to quite different solutions. Consider

1/((10^40+1)-(10^40))

Algebraically this is 1/1 but

1/((10.^40+1)-(10.^40))

Will lose precision on the addition, causing this to evaluate as 1/0

If I knew exactly what input values you had used in your screen snapshots then I could trace to figure out the exact situation for you, but I was not able to figure out what mol inputs you used.

Stephen Cobeldick
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

Walter Roberson
on 19 Sep 2018

Luca
on 20 Sep 2018

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

Walter Roberson
on 20 Sep 2018

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