This code performs heat transfer.

2023 4 25
1 个回答
更新时间：2023 4 28
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Tf=100; % Final Temp
Tint=1000; % Initial Temp
Tinf=20; % Water Temp
h=3000; % W/m^2*K
rho=5550; %kg/m^3
dx=0.005; %m
L=.25/2;%m
num=(L/dx);
T=zeros(num,1);
T(:,1)=Tint;
Cp=600;
k=27.5;
Alpha=(k/(rho*Cp));
Bi=((h*dx)/k);
dt=(dx^2)/(2*Alpha*(1+Bi));
Fo=((Alpha*dt)/(dx^2));
i=1;
while T(num,i)>Tf
for j=1:num
if j==1
T(j,i+1)=(2*Fo(j)*Bi(j))*Tinf+(2*Fo(j))*T(j+1,i)+(1-(2*Fo(j)*Bi(j))-(2*Fo(j)))*T(j,i);
elseif j==num
T(j,i+1)=2*Fo(j)*T((j-1),i)+(1-(2*Fo(j)))*T(j,i);
else
T(j,i+1)=(Fo(j))*(T((j-1),i)+T((j+1),i))+(1-(2*Fo(j)))*T(j,i);
end
if (20<=(T(j,i)))&&((T(j,i))<600)
Cp(j)=425+(.773)*(T(j,i))-.00169*((T(j,i))^2)+.00000222*((T(j,i))^3);
elseif (600<=(T(j,i)))&&(T(j,i))<(735)
Cp(j)=666+(13002/(738-(T(j,i))));
elseif (735<=(T(j,i)))&&(T(j,i))<(900)
Cp(j)=545+(17820/((T(j,i))-731));
elseif (900<=(T(j,i)))&&(T(j,i))<=(1000)
Cp(j)=600;
end
if(20<=(T(j,i)))&&((T(j,i))<800)
k(j)=54-.0333*(T(j,i));
elseif (800<=(T(j,i)))&&(T(j,i))<=(1000)
k(j)=27.3;
end
Alpha=(k(j)/(rho*Cp(j)));
Bi(j)=((h*dx)/k(j));
Fo(j)=((Alpha*dt)/(dx^2));
end
i=i+1;
end
Index exceeds the number of array elements. Index must not exceed 1.

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Torsten
Torsten 2023-4-25
编辑：Torsten 2023-4-25
Sorry, but your code is full of errors. Maybe it makes more sense if you explain what you are trying to do.
M Miller
M Miller 2023-4-26
This is part of the code to find how long it takes a slab of steele to cool from 1200° C to 200° C by using water jets at 40° C
Torsten
Torsten 2023-4-26
Seems you try to solve the transient heat conduction equation in 1d.
MATLAB's "pdepe" is the correct tool to do this.

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回答（1 个）

Cris LaPierre
Cris LaPierre 2023-4-25
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Your code is written expecting Fo and Bi to be a vectors, but they are scalars, so only ever has 1 element. You get this error any time you try to index using a value greater than 1.
The easiest solution based on the information we have is to not index these 2 variables.
Tf=200; % Final Temp
Tint=1200; % Initial Temp
Tinf=40; % Water Temp
h=4000; % W/m^2*K
rho=7850; %kg/m^3
dx=0.005; %m
L=.5/2;%m
num=(L/dx)+1;
T=zeros(num,1);
T(:,1)=Tint;
Cp=650; % initial Cp
k=27.3; % Initial k
% used to solve for dt
Alpha=(k/(rho*Cp));
Bi=((h*dx)/k);
dt=(dx^2)/(2*Alpha*(1+Bi));
Fo=((Alpha*dt)/(dx^2));
i=1;
while T(num,i)>Tf
for j=1:num
if j==1
T(j,i+1)=(2*Fo*Bi)*Tinf+(2*Fo)*T(j+1,i)+(1-(2*Fo*Bi)-(2*Fo))*T(j,i);
elseif j==num
T(j,i+1)=2*Fo*T((j-1),i)+(1-(2*Fo))*T(j,i);
else
T(j,i+1)=(Fo)*(T((j-1),i)+T((j+1),i))+(1-(2*Fo))*T(j,i);
end
% Cp and k need to be defined each time it finds a temperature since they are both temperature dependent
if (20<=(T(j,i)))&&((T(j,i))<600)
Cp(j)=425+(.773)*(T(j,i))-.00169*((T(j,i))^2)+.00000222*((T(j,i))^3);
elseif (600<=(T(j,i)))&&(T(j,i))<(735)
Cp(j)=666+(13002/(738-(T(j,i))));
elseif (735<=(T(j,i)))&&(T(j,i))<(900)
Cp(j)=545+(17820/((T(j,i))-731));
elseif (900<=(T(j,i)))&&(T(j,i))<=(1200)
Cp(j)=650;
end
if(20<=(T(j,i)))&&((T(j,i))<800)
k(j)=54-.0333*(T(j,i));
elseif (800<=(T(j,i)))&&(T(j,i))<=(1200)
k(j)=27.3;
end
%These values are recalculated each time as well based on the Cp and k
Alpha=(k(j)/(rho*Cp(j)));
Bi=((h*dx)/k(j));
Fo=((Alpha*dt)/(dx^2));
end
i=i+1;
end
ttot=(i-1)*dt
ttot = 826.5942

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M Miller
M Miller
2023-4-25

编辑：

M Miller
M Miller
2023-4-28

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