Index exceeds the number of array elements, must not exceed 1, in a nested for loop

In my code, I am analyzing a heat concentric heat exchanger and am solving for h, the local convection coefficients at points along the length of the of tube. In my nested for loop, indexed by j, I am attempting to update the length 0, dDh, for each iteration of the Nsselt number Nud. However, I keep getting the error that the index exceeds the number of array elements. there are 10 distance steps and 10 mass flow rates to analyze, so I am confused on why the sizes are a probelm. Any help is appreciated.

```clear

clc

% givens for the air

ma = 15; %kg/s

Tai = 1200; % deg-C

cpa = 1.207; % kJ/kg-K @ 1200C

ODa = 0.05; % m

IDa = 0.04; % m

% givens for the feedwater

mw = 10; %kg/s

Twi = 250; % deg-C

Two = 500; % deg-C

cpw = 4.860; % kJ/kg-K @ 250C

ODw = 0.01; %m

IDw = 0.008; %m

L = 5; %m, length of HEX bundle

Cc = mw*cpw;

Ch = ma*cpa;

A = [Cc, Ch];

Cmin = min(A);

Cmax = max(A);

Cr = Cmin/Cmax;

Tao = Tai - ((Cmax/Cmin)*(Two-Twi)); %air outlet temperature in deg-C

dT1 = Tai-Two;

dT2 = Tao - Twi;

LMTD = (dT2 - dT1)/log(dT2/dT1); %log mean temperature difference for the HEX

e = (Cc*(Two-Twi))/(Cmin*(Tai-Twi)); %HEX effectiveness

NTU = (1/(Cr-1))*log((e-1)/((e*Cr)-1)); % Net Transfer Units

Aht = pi*ODw*L; % heat transfer area (ft^2), or the outside surface area of the feedwater tube

U = (NTU*Cmin)/Aht; % overall heat transfer coefficient

%%% PART B

mw2 = [0.01:0.01:0.1]; %water mass flow rate in kg/s

rho_w = 798.6; %lg/m^3

Ac = 0.25*pi*IDw^2; %cross-sectional area of the inside of the feedwater tube

Dh = (4*Ac)/(4*pi*IDw); % hydraulic diameter for the feedwater tube

muw = 0.0001063; % dynamic visocsity of the water @ 250C (N*s/m^2)

Uw = zeros(1,length(mw2));

Red = zeros(1,length(mw2));

Nud = zeros(1,length(mw2));

k = 8; %W/m-K. thermal conductivity of water @250-C, guess beased on lower figures

step = Dh/9;

dDh = [0:step:Dh]; %size of the hydraulic diameter

h = zeros(length(mw2),length(dDh));

for i = 1:10

Uw(i) = mw2(i)/(rho_w*Ac); %velocity of the feedwater for the varying flow rates

Red(i) = (rho_w*Uw(i)*Dh)/muw;

Pr = (muw*cpw)/k;

if Red <= 5000

Nud = 4.36;

elseif Red > 5000

Nud(i) = 0.023*(Red(i)^0.8)*(Pr^0.4);

end

for j = 1:10

h(i,j) = (Nud(i)*k)/dDh(j);

end

end```

Answer 1

[Torsten]

Use

clear
clc
% givens for the air 
ma = 15; %kg/s
Tai = 1200; % deg-C
cpa = 1.207; % kJ/kg-K @ 1200C
ODa = 0.05; % m
IDa = 0.04; % m
% givens for the feedwater
mw = 10; %kg/s
Twi = 250; % deg-C
Two = 500; % deg-C
cpw = 4.860; % kJ/kg-K @ 250C
ODw = 0.01; %m
IDw = 0.008; %m
L = 5; %m, length of HEX bundle
Cc = mw*cpw;
Ch = ma*cpa;

A = [Cc, Ch];
Cmin = min(A);
Cmax = max(A);
Cr = Cmin/Cmax;
Tao = Tai - ((Cmax/Cmin)*(Two-Twi)); %air outlet temperature in deg-C
dT1 = Tai-Two;
dT2 = Tao - Twi;
LMTD = (dT2 - dT1)/log(dT2/dT1); %log mean temperature difference for the HEX
e = (Cc*(Two-Twi))/(Cmin*(Tai-Twi)); %HEX effectiveness
NTU = (1/(Cr-1))*log((e-1)/((e*Cr)-1)); % Net Transfer Units
Aht = pi*ODw*L; % heat transfer area (ft^2), or the outside surface area of the feedwater tube
U = (NTU*Cmin)/Aht; % overall heat transfer coefficient 
%%% PART B
mw2 = [0.01:0.01:0.1]; %water mass flow rate in kg/s
rho_w = 798.6; %lg/m^3
Ac = 0.25*pi*IDw^2; %cross-sectional area of the inside of the feedwater tube
Dh = (4*Ac)/(4*pi*IDw); % hydraulic diameter for the feedwater tube
muw = 0.0001063; % dynamic visocsity of the water @ 250C (N*s/m^2)
k = 8; %W/m-K. thermal conductivity of water @250-C, guess beased on lower figures
step = Dh/9;
dDh = [0:step:Dh]; %size of the hydraulic diameter
h = zeros(length(mw2),length(dDh));
for i = 1:10
    Uw = mw2(i)/(rho_w*Ac); %velocity of the feedwater for the varying flow rates
    Red = (rho_w*Uw*Dh)/muw;
    Pr = (muw*cpw)/k;
    if Red <= 5000
        Nud = 4.36;
    elseif Red > 5000
        Nud = 0.023*(Red^0.8)*(Pr^0.4);
    end
    for j = 1:10
        h(i,j) = (Nud*k)/dDh(j);
    end
end

or change at least this part

if Red <= 5000
    Nud = 4.36;
elseif Red > 5000
    Nud(i) = 0.023*(Red(i)^0.8)*(Pr^0.4);
end

to

if Red(i) <= 5000
    Nud(i) = 4.36;
elseif Red(i) > 5000
    Nud(i) = 0.023*(Red(i)^0.8)*(Pr^0.4);
end

[Matthew Palermo]

This worked! Thanks, I was really struggling to see what was going on