How to save the results of all iterations?

I have an equation with a summation term in it, so i used for loop to evaluate the summation term. however, i only get the final result of the for loop. I have tried pre-allocatation but still did not work. I am not sure if the mistake is in the for loop or somewhere else in my code. could you please take a look and advice my how to tweek my code? Thank you

R=1.1/100; %m
k=1; %W/m.K
rho=1100; %kg.m^3
Cp=4180; %J/kg.K
h=2700; %W.m^2.K
alpha=k/(rho*Cp);
tmax=5000;
Tb=120;
Ti=5;
T=zeros(tmax,1);
for t=1:tmax;
    Y=zeros(20,1);
    for n=1:20;
    X=(-1).^n.*exp(-alpha.*n.^2.*pi^2.*t/R^2);
    end 
    Y=sum(X)
    T=Tb+2*(Tb-Ti).*Y;
end 
plot(t,T);
hold on 

Answer 1

[David Fletcher]

Something like this?

R=1.1/100; %m
    k=1; %W/m.K
    rho=1100; %kg.m^3
    Cp=4180; %J/kg.K
    h=2700; %W.m^2.K
    alpha=k/(rho*Cp);
    tmax=5000;
    Tb=120;
    Ti=5;
    T=zeros(tmax,1);
    for t=1:tmax;
        Y=zeros(20,1);
        for n=1:20;
        X(n)=(-1).^n.*exp(-alpha.*n.^2.*pi^2.*t/R^2);
        end 
        Y=sum(X)
        T(t)=Tb+2*(Tb-Ti).*Y;
    end 
    plot(1:tmax,T);
    hold on 

[Alaa H]

Thank you for your response! I have tried that and still get only one final value for T when i should be getting 5000 values, any other guess?

[David Fletcher]

You sure? When I run my amended version I get 5000 values for T

[Alaa H]

it totally worked now!! I forgot to change the T to T(t) the 1st time I ran the code for some reason, but now I ran it again with all the modifications you made and it worked just fine. Thank you so much!!!

Answer 2

[Walter Roberson]

T=zeros(tmax,1);

Okay, good so far

for t=1:tmax;

Each time you iterate the for loop, t will be a scalar containing one of those values. You declared T to be tmax rows, same as the maximum value for t, so you are on track.

T=Tb+2*(Tb-Ti).*Y;

Each iteration of the for loop, you are overwriting all of T. So after the for loop finishes, T is going to contain a single value that is the one calculated in the last iteration of the loop.

    Y=zeros(20,1);
    for n=1:20;

That is setting up to write into 20 different rows for Y, which is fine so far.

    X=(-1).^n.*exp(-alpha.*n.^2.*pi^2.*t/R^2);
    end

Each iteration of the n loop, you overwrite all of X, and you do not write into Y at all.

     Y=sum(X)

but X is a scalar... and what is the point of having initialized Y if you are going to overwrite it with sum(X) ?

plot(t,T);

As described above, at each iteration of the for t loop, t will be a scalar. At the end of the loop, t will have the last value assigned to it, which would be tmax, and will be a scalar.

As explored above, you are overwriting T with a scalar for each iteration of the loop, so T is going to end up a scalar.

Hint:

for K = 1 : 5
   x(K) = K.^2;
end

Also

X=(-1).^n.*exp(-alpha.*n.^2.*pi^2.*t/R^2);

with n = 20 and t = 5000 the exp() term will be exp(-35479) approximately, which is far far too small to bother with.

You can demonstrate that with those coefficients and the maximum possible n, that there is no point doing t > 100 -- that beyond that you would underflow to 0.

Are you certain you want to proceed at integer t up to 5000? It would be more common to have a factor to convert that to a shorter time scale, such as t/1000 for bound of 5 seconds. Time 5000 seconds for an exponential process is not going to have anything useful left of the equation by 5000.

[Alaa H]

Thank you for your respnse. I get now that I have to write X and T as functions of n and t respectivly. you are right about the n and t values, they were just initial guesses since i did not know the approximate time at which T conveges to Tb eventually, I guess there are some equations i can use to get an estimates of the time but i just went with trial and error process. Thank you for taking the time to review the code!