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Grabbing two Roots and Assigning them to subsequent variables?

Posted 5 years ago
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So basically I have this assignment where I need to compute a Pxy Diagram of a species Mixture. I have to "guess" x1, x2, P (Pressure) such that all these calculations lead the last line in the code to equal 1.
Currently I am just copying and pasting the values of the roots into Zl and Zv. I know I know how to take one root i.e. the first by adding [[1]] but I do not know how to assign that to the appropriate variable. In this case I need the first root to equal Zl and the third root to Zv. This would save me like 30 minutes of copying and pasting. Here is the code. Its not a fun calculation.

bar = 10^5;
P = 1 bar;(*choose*)

x1 = 1;(*choose*)
x2 = 0;(*Choose*)

omega1 = 
 0.48 + 1.574 acc1 - 0.176 acc1^2;
a1 = 0.42748 (R^2 Tc1^2)/Pc1 (1 + omega1 (1 - Sqrt[Tred1]))^2

b1 = 0.08664 (R Tc1)/Pc1;

omega2 = 0.48 + 1.574 acc2 - 0.176 acc2^2;

a2 = 0.42748 (R^2 Tc2^2)/Pc2 (1 + omega2 (1 - Sqrt[Tred2]))^2;

b2 = 0.08664 (R Tc2)/Pc2;

a = x1^2 a1 + x2^2 a2 + 2.0 x1 x2 (1 - k12) Sqrt[a1 a2];

b = x1 b1 + x2 b2;

Ap = ( a P)/(R T)^2;

Bp = ( b P)/(R T);

**NRoots[Z^3 - Z^2 + (Ap - Bp - Bp^2) Z - Ap Bp == 0, Z]**

*How do get the first root equal Zl and third to equal Zv without just copying and pasting?*

**Zl = 0.005206170900992007;**

**Zv = 0.955658798860841**;

C1 = Ap/Bp (-b1/b + 2/a*(x1 a1 + x2 Sqrt[a1 a2] (1 - k12)));

C2 = Ap/Bp ((-b2/b) + (2/a) (x2 a2 + x1 Sqrt[a1 a2]*(1 - k12)));

phiL1 = Exp[b1/b (Zl - 1) - Log[(Zl - Bp)] - C1*Log[(Zl + Bp)/Zl]];

phiV1 = Exp[b1/b (Zv - 1) - Log[(Zv - Bp)] - C1*Log[(Zv + Bp)/Zv]];

phiL2 = Exp[b2/b (Zl - 1) - Log[(Zl - Bp)] - C2*Log[(Zl + Bp)/Zl]];

phiV2 = Exp[b2/b (Zv - 1) - Log[(Zv - Bp)] - C2*Log[(Zv + Bp)/Zv]];

K1 = phiL1/phiV1;

K2 = phiL2/phiV2;


**x1 K1 + x2 K2 (*Need to satisfy x1 K1+x2 K2\[Equal]1*)**
Posted 5 years ago

Consider this expression

{z1, z2, z3} = Last /@ (List @@ NRoots[(z + 2) (z - 3) (z - 4) == 0, z])

Here's an explanation of what's going on: The NRoots part returns an Or. We convert the Or to a list by applying List@@. That gives us a list of 3 equations. We take the right hand side of each equation using Last/@. Then, we set the values of z1, z2, z3 by assigning the list to {z1, z2, z3}, or whatever names we wish.

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