Can anyone help me?

The one I want to see could not appear.

function [u, Re] = Q2i(Q)
d = 5
u =Q*0.000001/(pi*(d*0.001)^2/4)
end

This my function script.

clear all;clc;close

P = [0.62 0.94 1.13 1.72 2.34 2.7 2.98 4.05 4.54 5.32 5.98 6.12 6.73 7.78 8.95 9.91 10.25 11.3 12.44]
Q = [3.87 4.49 5.26 5.43 6.02 6.44 6.87 7.55 7.73 8.42 8.75 10.32 10.79 11.32 14.13 15.66 16.06 17.98 19.11]

d = 5
density = 789
viscosity = 0.0012318

u = Q2i(Q)
Re = (density * u * (d*0.001)) / viscosity

a = num2str(Re) 
for k = length(Q)
if  2300 > a
    fl = 8./a
else 
    ft = 0.0396 * a.^-0.25 
end
end

This is my working script.

The thing is that 'fl' did not show in the working space. May I know what mistakes I have made?

Answer 1

[Simon Chan]

Varaible a becomes a character array after you use function num2str, so it is comparing a 'character' instead of a number.

On the other hand, index k is not used inside the for loop.

Actually in your case, for loop is not requried and just use the following:

fl = Re(2300>Re);
ft = 0.0396 * Re(Re>=2300).^-0.25; 

[John D'Errico]

+1

[Sein Lim]

May I know why you use that code?

[Simon Chan]

Sorry for typo error. Variable fl should be calculated as follows:

fl = 8./Re(2300>Re)

The comparisons (2300>Re) and (Re>=2300) calculates the logical indexes which satisy their condition and hence for-loop is not required.

[Torsten]

function main
  P = [0.62 0.94 1.13 1.72 2.34 2.7 2.98 4.05 4.54 5.32 5.98 6.12 6.73 7.78 8.95 9.91 10.25 11.3 12.44];
  Q = [3.87 4.49 5.26 5.43 6.02 6.44 6.87 7.55 7.73 8.42 8.75 10.32 10.79 11.32 14.13 15.66 16.06 17.98 19.11];
  d = 5;
  density = 789;
  viscosity = 0.0012318;
  u = Q2i(Q,d)
  Re = (density * u * (d*0.001)) / viscosity
  f = zeros(size(Q));
  for i = 1:numel(Re)
    if  2300 > Re(i)
      f(i) = 8/Re(i)
    else 
      f(i) = 0.0396 * Re(i)^-0.25 
    end
  end
end

function u = Q2i(Q,d)
  u = Q*0.000001/(pi*(d*0.001)^2/4)
end

[Sein Lim]

Thank you everyone.

[Torsten]

@Simon Chan

The comparisons (2300>Re) and (Re>=2300) calculates the logical indexes which satisy their condition and hence for-loop is not required.

I doubt that the OP really wanted to cut the correspondence to the Re array by splitting the flow resistance parameter in an array for laminar regime (fl) and an array for turbulent regime (ft). Maybe there is an indexing solution also for the case where the results (dependent on the flow regime) are written in one array f which in length corresponds to Re, but - to be honest - I still find it more transparent with an if statement within a for-loop. I'm aware that convinced MATLABers will disagree.

[Simon Chan]

Thanks for your comment.

The following code uses NaN as the result when the data is not valid to the specific group.

idx.fl = 2300>Re;                           % Index for laminar regime (fl)
fl = NaN(1,length(Q));                      % Initialize fl
ft = NaN(1,length(Q));                      % Initialize ft
fl(idx.fl) = 8./Re(idx.fl);                 % fl stores its valid data (belongs to idx.fl)
ft(~idx.fl) = 0.0396 *Re(~idx.fl).^-0.25;   % ft stores its valid data (belongs to ~idx.fl)

[Torsten]

And for one array only:

idx.fl = 2300>Re;                          % Index for laminar regime (fl)
f = zeros(1,length(Q));                    % Initialize f
f(idx.fl) = 8./Re(idx.fl);                 % f stores laminar flow resistance parameters (belongs to idx.fl)
f(~idx.fl) = 0.0396 *Re(~idx.fl).^-0.25;   % f stores turbulent flow resistance parameters (belongs to ~idx.fl)

Yes, it's nice.

Answer 2

[Image Analyst]

Because you have this:

if  2300 > a
    fl = 8./a
else 
    ft = 0.0396 * a.^-0.25 
end

and because you never initialized fl before the loop, and because fl never showed up, that means you never entered the top part of your if block, meaning that "a" was never less than 2300.