Solving a PDE using PDE Toolbox
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Tony Stianchie
am 6 Mai 2023
Kommentiert: Tony Stianchie
am 6 Mai 2023
This problem invoves an inner pipe with a constant wall temp, an annular space with conduction only, and an insulated outer pipe (no heat transfer)
I'm trying to solve the following Partial Differental Equation as my Governing Equation:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1376394/image.png)
Initial Condition:
@ t = 0s, T = 573 K (uniformly distributed)
Boundary Conditions:
@ r = ri, T = 673 K (constant)
@ r = ro, dT/dr = 0 (insulated end)
This is my first time using pdepe, any help would be appreciated.
I'd like to be able to plot (Temperature vs time) for varying points between the inner and outer radii.
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Torsten
am 6 Mai 2023
Bearbeitet: Torsten
am 6 Mai 2023
C.ri = 0.0125;
C.ro = 0.0375;
C.alpha = 1.905*10^-5;
C.Ti = 673;
C.T0 = 573;
r = linspace(C.ri,C.ro,50);
t = linspace(0,2000,2001);
m = 1;
eqn = @(r,t,T,dudx)heatcondPDE(r,t,T,dudx,C);
ic = @(r)heatcondPDE_IC(r,C);
bc = @(rl,Tl,rr,Tr,t)heatcondPDE_BC(rl,Tl,rr,Tr,t,C);
sol = pdepe(m,eqn,ic,bc,r,t);
T = sol(:,:,1);
plot(r,[T(1,:);T(5,:);T(10,:);T(20,:);T(30,:);T(40,:);T(80,:)])
grid on
function T0 = heatcondPDE_IC(r,C)
if r == C.ri
T0 = C.Ti;
else
T0 = C.T0;
end
end
function [c, f, s] = heatcondPDE(r,t,T,dudx,C)
c = 1;
f = C.alpha *dudx;
s = 0;
end
function [pl,ql,pr,qr] = heatcondPDE_BC(rl,Tl,rr,Tr,t,C)
pl = Tl - C.Ti;
ql = 0;
pr = 0;
qr = 1;
end
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