Integral of a Matrix Issue

Question

uki71319 2022-11-8

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编辑： uki71319 2023-3-27

Dear all, i'm trying to integrate a matrix to get a radial average mean temperature along three axial positions; however the codes always return the same matrix vlaue, instead of a single T mean value along axial site..

Does anyone know how could i get a real mean temperature value at each axial dirctection?

r=[0,5,10,15]
T=[300,310,320,335;
    302,312,322,337;
    305,315,325,340;] % as matrix 3*4
R_t=15            % radius
T*r'
fun=@(r)T*r'
integral(fun,0,r_t,'ArrayValued',true)
T_mean=(2./(r_t).^2).*integral(fun,0,r_t,'ArrayValued',true)

The output:

T = 3×4    
   300   310   320   335
   302   312   322   337
   305   315   325   340
T*r':   
ans = 3×1    
        9775
        9835
        9925
integral(fun,0,r_t,'ArrayValued',true):
ans = 3×4    
1.0e+04 *
    3.3750    3.4875    3.6000    3.7687
    3.3975    3.5100    3.6225    3.7912
    3.4312    3.5437    3.6562    3.8250
   
T_mean=(2./(r_t).^2).*integral(fun,0,r_t,'ArrayValued',true):
T_mean = 3×4    
  300.0000  310.0000  320.0000  335.0000
  302.0000  312.0000  322.0000  337.0000
  305.0000  315.0000  325.0000  340.0000

Thanks in advnce!!!!

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Answer 1

Torsten 2022-11-8

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Why do you name the same radius R in the denominator R_T ? Both are equal to the cylinder radius (in your case 15, I guess).

r=[0,5,10,15];
T=[300,310,320,335;
    302,312,322,337;
    305,315,325,340];
Tr = r.*T;
T_mean_cyl = 2/r(end)^2 * trapz(r.',Tr.')
T_mean_cyl = 1×3
  322.7778  324.7778  327.7778
%T_mean_cart = 1/r(end) * trapz(r.',T.')

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uki71319 2022-11-16

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Dear @Torsten, Sir, I still have some questions regarding this issue. The following are the plots of the solution.

Then I further plotted the Temperature profile along the radius but only at z=1 as follows. (Assuming that z=[1 2 3])

z=[1,2,3]
r=[0,5,10,15]                         % radius
T_z_1=[300,310,320,335]               % Temp at z=1 along r dimension
Ttest=r.*T_z_1
trapz(r.',Ttest.')
T_mean_test=2/r(end)^2*trapz(r.',Ttest.')
plot(T_mean_test,'b.','MarkerSize',20)
hold on 
plot(r,T_z_1,'r.-','MarkerSize',20)
hold off
title('T vs r at z=1','FontSize',16)
xlabel('r [m]')
ylabel('T [K]')
legend('T\_mean\_test','T at z=1')

My questions are:

Why the Tr_mean is shown higher than Tr=10? (I was expecting Tr_mean lower than Tr_10......I know this is from the calculated formula, but could you please explain to me in detail? It will be super helpful! I wonder if it's becasue of the cross area integration through the radius? But how?)
For the second graph, T vs r at z=1, why the corresponing radius of Tr_mean is at r=1?
WIll it make the result a big difference if we choose to use integral function in this case?

Any advice is greatly appreiated!

Thanks in advance!

Torsten 2022-11-16

编辑：Torsten 2022-11-16

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1.

Because the area of the cylinder grows with r^2 and not with r and the high temperature between r=10 and r=15 is weighted much more than the low temperature between r=0 and r=10.

2.

Use

plot([0 15],[T_mean_test,T_mean_test],'b','MarkerSize',20)

instead of

plot(T_mean_test,'b.','MarkerSize',20)

3.

r=[0,5,10,15];
T=[300,310,320,335;
    302,312,322,337;
    305,315,325,340];
for i = 1:3
    fun = @(rq)2*pi*rq.*interp1(r,T(i,:),rq,'spline')/(pi*r(end)^2);
    T_mean_test(i) = integral(fun,r(1),r(end));
end
T_mean_test
T_mean_test = 1×3
  321.4815  323.4815  326.4815

Torsten 2022-11-16

Sir, for Question 1, you said that cylinder goes with r^2, do you mean "the r inside the integral will becomes (r^2 /2) afterwards"? Right?

You must interprete the mean temperature as an area-weighted average, and the areas grow quadratically with r. Thus high temperatures for big r values influence the mean temperature more than low temperatures for small r. This is different for a plate where the areas are equal throughout.

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