# Trying to build an 11x9 matrix from another 11x9 matrix and other elements

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Michael Costa on 2 Nov 2020
Commented: Mathieu NOE on 3 Nov 2020
I am writing a program that compares different heat exchangers in order to select the best one for the application. I have successfully built an 11x9 array for the overall heat transfer coefficient (U), but now I am trying to build another 11x9 matrix for the log mean temperature difference (delTlm), and I am getting errors or answers that don't make sense. For example, when I did get an output for the delTlm matrix, every row had all values as the same. Here is the code that I have so far.
A = [2.4; 4.3; 7.4; 4.3; 7.4; 9.1; 2.4; 4.3; 7.4;].*0.0929; %m^2
%From problem 1, the exit temperature curves significantly flatten out
%after an NTU of 3 (returns diminish quickly). I will run the program from
%2.5 NTU to 3.5 NTU
NTU = (2.5:0.1:3.5).';
for i = 1:length(NTU)
%Calculating the effectiveness using the HE_effectiveness program
eff(i) = effectiveness(NTU(i), c, 'One Shell Pass');
%Finding the actual heat transfer from Qmax and the effectiveness
Q = (eff.*Qmax); %kW
%Finding the overall transfer coefficient
U = (NTU.*Cmin)./A.'; %kW/m^2*deg C
%Finding the log mean temperature difference
deltaTlm = Q./U.*A.'
end
Michael Costa on 2 Nov 2020
As I said in the question, I have successfully built the U matrix (thanks to another guy on here)...so you shouldn't have to worry about the c variable or the effectiveness function. Do you see either one of those in the calculation of the matrix that I am asking about? Maybe you have misunderstood my question.

Mathieu NOE on 2 Nov 2020
hello
there were only a few minor indexing issues in your main code.
also I recommend not using i or j as index var names as they are reserved for imaginary / complex numbers
so , here main code fixed :
%Specific heat capacity of water at 60 deg C
C_c = 4.185*m4; %kJ/s*deg C
%The brine is modelled as pure water.
C_h = 4.185*m13; %kJ/s*deg C
%Solution
%---------
%Finding Cmin, Cmax, Qmax, and c
Cmin = min(C_c, C_h); %kJ/s*deg C
Cmax = max(C_c, C_h); %kJ/s*deg C
Qmax = Cmin*(T13 - T4); %kW
c = Cmin/Cmax;
A = [2.4; 4.3; 7.4; 4.3; 7.4; 9.1; 2.4; 4.3; 7.4;].*0.0929; %m^2
%From problem 1, the exit temperature curves significantly flatten out
%after an NTU of 3 (returns diminish quickly). I will run the program from
%2.5 NTU to 3.5 NTU
NTU = (2.5:0.1:3.5)';
for ci = 1:length(NTU)
%Calculating the effectiveness using the HE_effectiveness program
eff(ci) = effectiveness(NTU(ci), c, 'One Shell Pass');
%Finding the actual heat transfer from Qmax and the effectiveness
Q = (eff(ci).*Qmax); %kW
%Finding the overall transfer coefficient
U = (NTU(ci).*Cmin)./A.'; %kW/m^2*deg C
%Finding the log mean temperature difference
deltaTlm(ci,:) = Q./U.*A';
end
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Mathieu NOE on 3 Nov 2020