summation of aij & i, j dependent function in matlab

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BenK 2024-4-22

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https://ww2.mathworks.cn/matlabcentral/answers/2110196-summation-of-aij-i-j-dependent-function-in-matlab

评论： BenK 2024-4-22

采纳的回答： VBBV

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Hello, I am trying to create a code that can calculate the local resistance coefficient via formula above. the values of ai and aij are defined.

I have created a code using for loops, but getting a really high number.

I have defined example values for Re, f0, and f1 values below. Using the published graph, I should be getting around 1-1.1

ai and bi are values defined.

Please forgive me, it might be that I got matrix algebra incorrect, as I am thinking that might be the suspicion but not sure what I need to change at this point.

Thank you.

%
% initial contraction from flow, formula from diagram 4.10
%
%
%given test values
Re_0=60.46;
f0=1;
f1=2.7225;
%
ind=8;
acoeff=0;
ai=[-25.12458,18.5076,-170.4147,118.1949,-44.42141,9.09524,-.9244027,.03408265];
for i = 1:ind
acoeff=acoeff+ai(i)*log(Re_0)^i;
end
acoeff
acoeff = -1.6830e+03
bi=[1.07 1.22 2.933;.05 -.51668 .8333; 0 0 0];
iB=3;
jB=3;
count=0;
B_inner=0;
Bcoeff=0;
for i=1:iB
    for j=1:jB
        B_inner=B_inner+bi(i,j)*((f0/f1)^j);
    end
    Bcoeff=Bcoeff+B_inner*log(Re_0)^i;
end
Coeff_local=(acoeff*Bcoeff)*(1-f0/f1)
Coeff_local = -6.6410e+04

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采纳的回答

VBBV 2024-4-22

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2110196-summation-of-aij-i-j-dependent-function-in-matlab#answer_1445856

编辑：VBBV 2024-4-22

在 MATLAB Online 中打开

%
% initial contraction from flow, formula from diagram 4.10
%
%
%given test values
Re_0=60.46;
f0=1;
f1=2.7225;
%
ind=8;
acoeff=0;
ai=[-25.12458,118.5076,-170.4147,118.1949,-44.42141,9.09524,-.9244027,.03408265];
for i = 0:ind-1
acoeff=acoeff+ai(i+1)*log(Re_0)^i;
end
acoeff
acoeff = -0.0814
bi=[1.07 1.22 2.933;.05 -.51668 .8333; 0 0 0];
 
iB=3;
jB=3;
count=0;
B_inner=0;
Bcoeff=0;
for i=0:iB-1
    for j=0:jB-1
        B_inner=B_inner+bi(i+1,j+1)*((f0/f1)^j);
    end
    Bcoeff=Bcoeff+B_inner*log(Re_0)^i;
end
Coeff_local=(acoeff*Bcoeff)*(1-f0/f1)
Coeff_local = -2.1331

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VBBV 2024-4-22

编辑：VBBV 2024-4-22

@BenK that value is natural log to decimal log conversion factor. since acoeff and Bcoeff are computed separately involving log function it needs to be used twice

BenK 2024-4-22