summation of aij & i, j dependent function in matlab

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

Answer 1

[VBBV]

%
% 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

[VBBV]

There is a type error in the ai coefficient

ai=[-25.12458,118.5076,-170.4147,118.1949,-44.42141,9.09524,-.9244027,.03408265];
            % this is 118.5076

Similarly check the values for bi , may be there is typo errors with values

[BenK]

Thank you for that! Ugh, these simple typos shouldn't be here.

Modifying that value 'improved', but didn't result in the correct answer- still there are negative values and off by a factor.

I am curious if I am actually performing the double summations correctly (I did check the values on the bij matrix), but the fact they are asking for a dot b makes me think A and B needs to be vectors, but that didn't make sense to me, unless if I am incorrectly performing the summation of aij matrices.

*edit- I uploaded both pages, showing all values.

[VBBV]

%
% 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)*log10(Re_0)^i;
end
acoeff/2.3

ans = 0.6863

bi=[1.07 1.22 2.9333;.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(j+1,i+1)=bi(i+1,j+1)*((f0/f1)^j);
    end
    Bcoeff(i+1)=sum(B_inner(:,i+1))*log10(Re_0)^i;
end

Coeff_local=(acoeff*sum(Bcoeff))*(1-f0/f1)/2.3

Coeff_local = 0.8099

[VBBV]

@BenK I guess you need to use log10 in place of log. From the look up table, for the given Re value of 60.46, and F0/F1 ratio of 0.367, i.e. for (1-f0/f1) ~0.6326 you need to get a value around 0.8099.

Re_0=60.46;
f0=1;
f1=2.7225;
f0/f1

ans = 0.3673

1-f0/f1

ans = 0.6327

[BenK]

Thank you! Looking at your modified code, I realized that I made a mistake by setting indices from 1 to ind vs 0 to ind-1 since the there are values calculated based on the power.

not sure about your lookup table - looking at the table range is from 1.4- to 1.5?

One question- is there a reason why you divided both the acoeff and bcoeff by 2.3? thanks for the continued help!

[VBBV]

I guess you also need to interpolate the values using the lookup table, since Re = 60.46

[VBBV]

@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]

Thank you for helping me with this issue. I will mark this as resolved