Lower Dirichlet boundary condition for heat equation PDE

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Marten Meerburg 2023-5-31

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评论： Marten Meerburg 2023-6-5

Dear community,

I am currently working on modeling a packed bed thermal storage system for a school project. To do this I am solving two PDE's which represent the fluid and solid temperature respectively. The problem I am encountering is that I want to limit the lower temperature to 100 degrees (currently 0 degrees as can be seen in the figure). I've been playing around with the boundary conditions function but I can't get it to work, does anyone have an idea on how to get this done? Any help is much appreciated.

clear all;close all;clc
L = 15;
t_max = 8000;
steps = 1000;
x = linspace(0,L,steps);
t = linspace(0,t_max,steps);
m = 0;
sol = pdepe(m,@heatpde,@heatic,@heatbc,x,t);
Ta = sol(:,:,1);
Tb = sol(:,:,2);
figure(1)
plot(x,Ta(:,steps/10),'LineWidth',1.5)
hold on 
plot(x,Tb(:,steps/10),'LineWidth',1.5)
plot(x,Ta(:,steps/2),'LineWidth',1.5)
plot(x,Tb(:,steps/2),'LineWidth',1.5)
plot(x,Ta(:,round(steps/1.01)),'LineWidth',1.5)
plot(x,Tb(:,round(steps/1.01)),'LineWidth',1.5)
title('Temperature over length')
xlabel('Tank Length (m)')
ylabel('Temperature (degrees)')
function [c,f,s] = heatpde(x,t,u,dudx)
crho_s = 2.1e6; crho_l = 600; m = 150; cp_s = 920; cp_l = 1046; A = 160; h = 4; e = 0.35; rho_l = 0.61;
c = [1;
     1];
f = [0;
     0];
s = [-(m/(A*e*crho_l))*dudx(1) - (h*(u(1)-u(2))/(crho_l*e));
     (h*(u(1)-u(2)))/(crho_l*e)];
end
function u0 = heatic(x)
u0 = [500;
      500];
end
function [pl,ql,pr,qr] = heatbc(xl,ul,xr,ur,t)
pl = [ul(1);ul(2)];
ql = [0;0];
pr = zeros(2,1);
qr = ones(2,1);
end

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Answer 1

Torsten 2023-5-31

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编辑：Torsten 2023-5-31

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Your equations are not suited to be solved with pdepe.

Both contain no second derivatives in space which is necessary for an equation to be of type parabolic-elliptic (pdepe) - the type of equation that pdepe solves. The first one is a hyperbolic transport equation, the second is a simple ordinary differential equation that has no spatial derivatives at all.

For the first equation, you have to supply one boundary condition at x=0; setting a boundary condition at x=L is wrong for this type of equation. The second equation does not need boundary conditions at all.

If you neverthess want to try to use "pdepe", you will have to change pl. At the moment, your boundary condition setting

pl = [ul(1);ul(2)];

sets both temperatures (especially the inflow temperature of the fluid) to 0.

Further I doubt that the source term for the heat transfered between liquid and solid is

(h*(u(1)-u(2))/(crho_l*e)

for both liquid and solid because of the different heat capacities and densities of the two substances.

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Marten Meerburg 2023-5-31

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Hello Torsen,

First of all thanks for your quick responce. I'm still in the process of exactly understanding the thing I'm modeliing so your awnser is of great help. To make the equation/model more complete I added the diffusivity term (second derivative) in the equations, to hopefully better fullfill the purpose of the PDE solver. The resulting graph still reasonably represents the one I initially posted (don't mind the heat capacity, conductivity terms etc. these are just some preliminary values to tet how thing work out).

function [c,f,s] = heatpde(x,t,u,dudx)
crho_s = 2.1e6; crho_l = 600; m = 150; cp_s = 920; cp_l = 1046; A = 160; h = 4; e = 0.35; rho_l = 0.61; k_l = 0.05; k_s = 1.7;
c = [1;
     1];
f = [(k_l/crho_l)*dudx(1);
     (k_s/crho_s)*dudx(2)];
s = [-(m/(A*e*crho_l))*dudx(1) - (h*(u(1)-u(2))/(crho_l*e));
     (h*(u(1)-u(2)))/(crho_l*e)];
end

''For the first equation, you have to supply one boundary condition at x=0; setting a boundary condition at x=L is wrong for this type of equation. The second equation does not need boundary conditions at all.''

I thought about this and this is indeed seems logical. The thing Im trying to model should have an operating range between 100-500 degrees. Is it even possible then to set a lower temperature limit with the PDE solver?

Furthermore I was under the impression that the following line:

pl = [ul(1);ul(2)];

Sets the temperatures of both the fluid and solid at x=0 to 500 degrees, as defined in:

function u0 = heatic(x)
u0 = [500;
      500];
end

Am I wrong in this as you said: ''this boundary condition setting sets both temperatures (especially the inflow temperature of the fluid) to 0.''?

You were right on the source term, as of yet I did not implemented the different liquid and solid heat capacities and densities but is certainly something I will dive into. I was especially focussed on getting the temperature range correct.

Hope you could help me out or come with other suggestions, they are very much appreciated.

Kind regards, Marten

Torsten 2023-5-31

在 MATLAB Online 中打开

Is it even possible then to set a lower temperature limit with the PDE solver?

You can't set limits on the solution from pdepe. The solution results from the equation, the initial and the boundary conditions- no way to influence it.

Furthermore I was under the impression that the following line:

pl = [ul(1);ul(2)];

Sets the temperatures of both the fluid and solid at x=0 to 500 degrees, as defined in:

function u0 = heatic(x)
u0 = [500;
    500];
end

Am I wrong in this as you said: ''this boundary condition setting sets both temperatures (especially the inflow temperature of the fluid) to 0.''?

The boundary values for u(1) and u(2) at x=0 result from pl(1)+ql(1)*f(1) = 0, pl(2)+ql(2)*f(2) = 0. If you set pl(1) = ul(1), pl(2) = ul(2), ql(1) = 0 and ql(2) = 0, you set ul(1) + 0*f(1) = ul(1) = 0, ul(2) + 0*f(2) = ul(2) = 0, thus both temperatures to 0.

Marten Meerburg 2023-6-5

Thanks Torsten for anwswering my question. It really helped a lot.

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Lower Dirichlet boundary condition for heat equation PDE

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