Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 1-by-2. don't know how to fix this

K = 1;% Assuming all minor losses to currently be 1
g = 32.2;
NL = 2;%Number of loops
NJ = [6;5]; % The total number of lines for the given problem/network
IW = 50000;%Number of iterations
epsln = 0.00085; % Roughness factor for Cast Iron Pipes
% Pipe Length Matrix [ft] 
ZL(1,1) = 600; ZL(1,2) = 500; ZL(1,3) = 200; ZL(1,4) = 400; ZL(1,5) = 400; ZL(1,6) = 500;
ZL(2,1) = 450; ZL(2,2) = 500; ZL(2,3) = 300; ZL(2,4) = 250; ZL(2,5) = 500;
% Pipe Diameter Matrix [ft] 
D(1,1) = 8/12; D(1,2) = 16/12; D(1,3) = 6/12; D(1,4) = 6/12; D(1,5) = 6/12; D(1,6) = 8/12;
D(2,1) = 6/12; D(2,2) = 6/12; D(2,3) = 6/12; D(2,4) = 8/12;  D(2,5) = 8/12;
% Flow rate Matrix (cfs
Q(1,1) = 0.31175; Q(1,2) = 0.31175; Q(1,3) = 0.31175; Q(1,4) = 0.22265; Q(1,5) = 0.13355; Q(1,6) = 0.1559;
Q(2,1) = 0.1559; Q(2,2) = 0.114; Q(2,3) = 0.0695; Q(2,4) = 0.28945;  Q(2,5) = 0.1559;
% Common lines
ID(1,1) = 0; ID(1,2) = 0; ID(1,3) = 0; ID(1,4) = 0; ID(1,5) = 0; ID(1,6) = 2;
ID(2,1) = 0.; ID(2,2) = 0; ID(2,3) = 0; ID(2,4) = 0;  ID(2,5) = 1;
for IT = 1:IW
    for I = 1:NL 
    for J = 1:NJ(I)
        ReD(I,J) = 4*abs(Q(I,J))/(pi*D(I,J)*v);
         % Calculate Friction factor and DF/DQ
        if ReD < 2000
            Fr(I,J) = 64/ReD;
            DF_DQ(I,J) = -64/(ReD(I,J)*Q(I,J));
        else
            H = (abs(ReD(I,J)))^0.9;
             A = epsln/(3.7*D(I,J)); B = 5.74/(ReD(I,J).^0.9); C = 5.74/H;
             Fr(I,J) = 1.325/((log(A + B)).^2);
             DF_DQ(I,J) = (13.69*(A + C)^-1)/(H*Q(I,J)*((log(A + C))^3));
        end
        alpha = 8*ZL(I,J)/(pi^2*g*D(I,J)^5); beta(I,J) = (8*K)/(pi^2*g*(D(I,J)^4));  %#ok<SAGROW>
        if Q(I,J) < 0
                 Hf(I,J) = -(alpha*Q(I,J)^2*Fr(I,J))+ beta*Q(I,J)^2; 
                 DHDQ(I,J) = -(2*alpha*Fr(I,J)*Q(I,J) + alpha*(Q(I,J)^2)*DF_DQ(I,J));   
             else
           

Getting Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 1-by-2.for thhis code

    145; Hf(I,J) = alpha*abs((Q(I,J).^2)*Fr(I,J)) + abs(beta*Q(I,J)^2);   
                DHDQ(I,J) = (2*alpha*Fr(I,J)*Q(I,J) + alpha*(Q(I,J)^2)*DF_DQ(I,J));
        end
    end
    end
 % Define the correction factor for each loop
    for I = 1:NL
        sum_Hf(I) = 0; sum_DHDQ(I) = 0;  
        for J = 1:NJ(I)
            sum_Hf(I) = sum_Hf(I) + Hf(I,J);
            sum_DHDQ(I) = sum_DHDQ(I) + DHDQ(I,J); 
        end
        DeltaQ(I) = -sum_Hf(I)/(sum_DHDQ(I)); % % Correction Factor for each loop
    end
    for I = 1:NL
        for J = 1:NJ(I)
            if ID(I,J) == 0 % For Uncommon Lines With No Shared Loops
                Q_N(I,J) = Q(I,J) + DeltaQ(I);  
            else % For Common Lines With Shared Loops
                g = ID(I,J); 
                Q_N(I,J) = Q(I,J) + DeltaQ(I) - DeltaQ(g);   
            end
        end
    end
    % Convergence Test
    for I = 1:NL
        for J = 1:NJ(I)
            Q_D = abs(Q_N(I,J) - Q(I,J));
        end
    end
    if Q_D > EPSLN
        for I = 1:NL
            for J = 1:NJ(I)
                Q(I,J) = Q_N(I,J); % Do another iteration until it passes the convergence test
            end
        end
    end
    if Q_D < EPSLN
        break; % It passes the convergence test
    end
end
fprintf('The new flow rates (ft^3/s) for each pipe are \n');
fprintf('----------------------------------------------\n');
fprintf('   Line:      1   |   2   |    3   |   4    \n');
fprintf('   Loop 1:  %1.3f | %1.3f | %1.3f | %1.3f | %1.3f %1.3f \n',Q_N(1,1),Q_N(1,2),Q_N(1,3),Q_N(1,4),Q_N(1,5),Q_N(1,6));
fprintf('   Loop 2:  %1.3f | %1.3f | %1.3f | %1.3f | %1.3f %1.3f \n',Q_N(2,1),Q_N(2,2),Q_N(2,3),Q_N(2,4),Q_N(2,5),Q_N(2,6));
fprintf('----------------------------------------------\n');

[Daniel Sode]

Line 145 which I've labeled is the line in question

Answer 1

[Walter Roberson]

alpha = 8*ZL(I,J)/(pi^2*g*D(I,J)^5); beta(I,J) = (8*K)/(pi^2*g*(D(I,J)^4));  %#ok<SAGROW>

The second part of that line is growing beta as the loop indices increase.

Hf(I,J) = -(alpha*Q(I,J)^2*Fr(I,J))+ beta*Q(I,J)^2; 

That code uses all of beta. As soon as beta grew to two elements, the right hand side becomes two elements, but the left hand side only names one location to store into.