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1 TAME synthesis problem Tert-Amyl Methyl Ether (TAME) is an ...

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Algorithm to solve the <strong>problem</strong><br />

We have a non-linear system of Ordinary Differential Equations (ODE) to solve:<br />

⎧ 1B<br />

⎪ − ℜ1B<br />

= 0<br />

⎪<br />

2B<br />

⎪ − ℜ2<br />

B = 0<br />

⎪<br />

⎪ M<br />

⎨ − ℜ M = 0<br />

⎪<br />

⎪<br />

dfT<br />

− ℜT<br />

= 0<br />

⎪ dX<br />

3<br />

⎪ dT b<br />

⎪ − ∑<br />

⎩ dX Cp j=<br />

dX<br />

df<br />

dX<br />

df<br />

dX<br />

df<br />

ρ τ<br />

ε<br />

ρ<br />

1<br />

R ( ∆H<br />

r ) + NTU ( T −T<br />

)<br />

j<br />

j<br />

w<br />

= 0<br />

The program we developed uses subroutine DDASSL (Bren<strong>an</strong> et al., 1989) to solve th<strong>is</strong> system. Th<strong>is</strong><br />

code solves a system of differential/algebraic equations of the form delta(t, y, yprime)=0, with<br />

delta(i) = yprime(i) – y(i), using the backward differentiation formulas of orders one trough five. t <strong>is</strong><br />

the current value of the independent variable (in our case t = X), y <strong>is</strong> the array that contains the solution<br />

components at t (in our case we have: y(i) = f(i), i=1,4 <strong>an</strong>d y(5) = T) <strong>an</strong>d yprime <strong>is</strong> the array that<br />

contains the derivatives of the solution components at t.<br />

The program solves the system from t to tout <strong>an</strong>d it <strong>is</strong> easy to continue the solution to get results at<br />

additional tout. In our case, we are going to get results at different values of X, between 0 <strong>an</strong>d 1.<br />

Th<strong>is</strong> <strong>problem</strong> <strong>is</strong> rather complex because most of the other variables depend on yi: the kinetic, adsorption<br />

<strong>an</strong>d thermodynamic equilibrium “const<strong>an</strong>ts” depend on T (y5), the reaction rate <strong>an</strong>d the components rate<br />

of formation depend on yi (fi (i=1,4) <strong>an</strong>d T), as the activity coefficients.<br />

In Figure 2 <strong>is</strong> the algorithm to solve our <strong>problem</strong>.<br />

(1)<br />

10

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