PDEPE solver failure (spatial discretization) ; partial differential/derivative temperature modelling
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Attempting to model temperature profile of an adsorption column based on concentration profile since this determines the heat of adsorption.
This is the function:
function[c,f,s] = pdefunEB(x,t,u,dudx)
global qmax_phenol K_L_phenol kf De u_s Rep Dp epsilon rho_liq cp_AC cp_feed H_ads h alpha
C = u(1) ;
q = u(2) ;
Tf = u(3) ;
dCdx = dudx(1) ;
dqdx = dudx(2) ;
dTdx = dudx(3) ;
Ts = 298.15 ; % adsorbent temp Kelvin
qstar = ((qmax_phenol)*(K_L_phenol)*C)/(1+(K_L_phenol)*C) ;
dqdt = kf*(qstar-q) ;
c = ones(3,1) ;
f = [De*dCdx;
0;
0];
s = [-u_s*dCdx-(1-epsilon)/epsilon*rho_liq*dqdt;
dqdt;
-u_s*dTdx+(((h*alpha*(Tf-Ts)-(H_ads/cp_AC))+(H_ads/cp_feed))*(dqdt*((1-epsilon)/epsilon)))] ;
end
%% initial condition
function u0 = icfunEB(x)
u0 = [0;
0;
298.15] ;
end
%% boundary condition
function [p1,q1,pr,qr] = bcfunEB(x1,u1,xr,ur,t)
global C_phenol
C0 = C_phenol ;
p1 = [u1(1)-C0;0;-298.15] ;
q1 = [0;1;0] ;
pr = [0;0;0] ;
qr = [1;1;1] ;
end
this is where the error appears (Error using pdepe Spatial discretization has failed. Discretization supports only parabolic and elliptic equations, with flux term involving spatial derivative)
mEB = 0 ;
sol1 = pdepe(mEB,@pdefunEB,@icfunEB,@bcfunEB,x,t) ;
CEB = sol1(:,:,1) ;
qEB = sol1(:,:,2) ;
TEB = sol1(:,:,3) ;
figure(2)
T0 = 298.15 ;
Cp1 = C1(:,end)./C01 ;
plot(t,TEB,'b-','LineWidth',2)
grid on
title('Column Temperature')
xlabel('time in minutes')
ylabel('temp in kelvin')
Heres a picture of my mathematical working before input into matlab:
0 commentaires
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Torsten
le 20 Avr 2024
Modifié(e) : Torsten
le 20 Avr 2024
Please include the constants defined as globals so that we can test your code.
One obvious error is the boundary condition for T in the left boundary point. Use
p1 = [u1(1)-C0;0;u1(3)-298.15] ;
instead of
p1 = [u1(1)-C0;0;-298.15] ;
Another problem can be that f = 0 for equations (2) and (3). "pdepe" - as the name indicates - is designed for parabolic-elliptic partial differential equations (thus for equations with second-order spatial derivatives present). Your second equation is a simple ODE, your third equation is hyperbolic in nature.
4 commentaires
Torsten
le 20 Avr 2024
Modifié(e) : Torsten
le 20 Avr 2024
As you can see above, the temperature explodes at the right endpoint (see above). Check the possible reasons (especially your sourceterm -u_s*dTdx+(((h*alpha*(Tf-Ts)-(H_ads/cp_AC))+(H_ads/cp_feed))*(dqdt*((1-epsilon)/epsilon))) )
h*alpha = 1.5e9 !!!
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