Ecuaciones diferenciales

Function xdot=vdv_ode(t,x);

%Resolver dos ecuaciones diferenciales que modelan la reacción tipo Van de Vusse en un reactor Batch isotérmico así como ver la variación de A y B en el reactor

[t,x]=ode45(vdv_ode, [0 5],x0)

% integrar de t=0 a t=5 minutos, con las condiciones iniciales de ca0=x0(1) y cb0=x0(2), y x0 es un vector columna

Ca=x(1);

Cb=x(2);

Los parámetros son:

K10=1287000000000

K20=1287000000000

K30=9043000000

E1=9758.3

E2=9758.3

E3=8560

H1=4.2

H2=-11.0

H3=-41.85

CP=3.01

RO=0.9342

Ca0=5.10

Ki=k10*exp((E1/(T+273.15)))

K1=K10* exp((E1/(T+273.15)))

K1=K20* exp((E2/(T+273.15)))

K1=K30* exp((E3/(T+273.15)))

T=25

Subplot(2,1,1),plot(t,x(:,1),’k’), xlabel(´t’),ylabel(´ca´)

Subplot(2,1,2),plot(t,x(:,2),’k’), xlabel(´t’),ylabel(´cb´)

function xdot = vdv_ode(t,x);

%

% Solves the two differential equations modeling the van de vusse reaction

% scheme in an isothermal CSTR. The states are the concentration of A and

% B in the reactor.

%

% [t,x] = ode45(vdv_ode,[0 5],x0)

% integrates from t = 0 to t = 5 min, with initial conditions

% ca0 = x0(1) and cb0 = x0(2), and x0 is a column vector

% 16 Jan 99

% b.w. bequette

%

% since the states are passed to this routine in the x vector,

% convert to natural notation

ca = x(1);

cb = x(2);

% the parameters are:4

k1 = 5/6; % rate constant for A_B (min^-1)

k2 = 5/3; % rate constant for B_C (min^-1)

k3 = 1/6; % rate constant for 2A_D (mol/(l min))

% the input values are:

% dilution rate (min^-1)

caf = 10; % mol/l

% the modeling equations are:

dcadt =-k1*ca-k3*ca*ca;

dcbdt = k1*ca-k2*cb;

% now, create the column vector of state derivatives

xdot = [dcadt;dcbdt];

% end of file

>>[t,x] = ode45('vdv_ode',[0 10],x0);

>> subplot(2,1,1),plot(t,x(:,1),'k'), xlabel('t'), ylabel('ca')

>> subplot(2,1,2),plot(t,x(:,2),'k'), xlabel('t'), ylabel('cb')

Function xdot=vdv_ode(t,x);

%Resolver dos ecuaciones diferenciales que modelan la reacción tipo Van de Vusse en un reactor Batch isotérmico así como ver la variación de A y B en el reactor

[t,x]=ode45(vdv_ode, [0 5],x0)

% integrar de t=0 a t=5 minutos, con las condiciones iniciales de ca0=x0(1) y cb0=x0(2), y x0 es un vector columna

Ca=x(1);

Cb=x(2);

Los parámetros son:

K10=1287000000000

K20=1287000000000

K30=9043000000

E1=9758.3

E2=9758.3

E3=8560

H1=4.2

H2=-11.0

H3=-41.85

CP=3.01

RO=0.9342

Ca0=5.10

Ki=k10*exp((E1/(T+273.15)))

K1=K10* exp((E1/(T+273.15)))

K1=K20* exp((E2/(T+273.15)))

K1=K30* exp((E3/(T+273.15)))

T=25

Subplot(2,1,1),plot(t,x(:,1),’k’), xlabel(´t’),ylabel(´ca´)

Subplot(2,1,2),plot(t,x(:,2),’k’), xlabel(´t’),ylabel(´cb´)

function xdot = vdv_ode(t,x);

%

% Solves the two differential equations modeling the van de vusse reaction

% scheme in an isothermal CSTR. The states are the concentration of A and

% B in the reactor.

%

% [t,x] = ode45(vdv_ode,[0 5],x0)

% integrates from t = 0 to t = 5 min, with initial conditions

% ca0 = x0(1) and cb0 = x0(2), and x0 is a column vector

% 16 Jan 99

% b.w. bequette

%

% since the states are passed to this routine in the x vector,

% convert to natural notation

ca = x(1);

cb = x(2);

% the parameters are:4

k1 = 5/6; % rate constant for A_B (min^-1)

k2 = 5/3; % rate constant for B_C (min^-1)

k3 = 1/6; % rate constant for 2A_D (mol/(l min))

% the input values are:

% dilution rate (min^-1)

caf = 10; % mol/l

% the modeling equations are:

dcadt =-k1*ca-k3*ca*ca;

dcbdt = k1*ca-k2*cb;

% now, create the column vector of state derivatives

xdot = [dcadt;dcbdt];

% end of file

>>[t,x] = ode45('vdv_ode',[0 10],x0);

>> subplot(2,1,1),plot(t,x(:,1),'k'), xlabel('t'), ylabel('ca')

>> subplot(2,1,2),plot(t,x(:,2),'k'), xlabel('t'), ylabel('cb')

H+

:

Reacción 3:

Nota Importante: La expresión de la constante de equilibrio para la reacción de

formación del precipitado de hidróxido férrico (Reacción 1) es:





       

(1)

3

3

1

Keq

...