I'm trying to code these equations. This is my attempt. I would like to plot Cf vs x/L, but x/L as I have it shows me an error..the same with delta/x, not sure why? any help will be greatly appreciated. Thanks much.
Re = 0:500:1000;
x/L = 0:0.2:1;
Cf = 0.664*(Re)^-0.5;
delta/x = 5.0*(Re)^-0.5;

1 Comment

Les Beckham
Les Beckham on 3 Feb 2023
Edited: Les Beckham on 3 Feb 2023
What are x, L, and delta?
Also, you can't put an expression like x/L on the left hand side of an assignment statement, only a variable name (optionally indexed).

Sign in to comment.

 Accepted Answer

Here is the corrected code:
Re = 0:50:1000;
xL = linspace(0,1, numel(Re));
Cf = 0.664*(Re).^(-0.5);
deltaX = 5.0*(Re).^(-0.5);
plot(Cf, xL, 'ro', 'markerfacecolor', 'y')
xlabel('C_f')
ylabel('x_L')
grid on

4 Comments

Thanks much for your help. Just a quick question, I coded this simple equation. I got a plot. It shows a straight line, buyt I think it should be some kind of curve, is it right? or is there is something wrong? Thanks.
Ue = @(x)2*(1-x);
x = (0:0.1:1);
plot(Ue(x),x)
grid on
syms x
Ue(x) = 2*(1-x)
Ue(x) = 
diff(Ue,x)
ans(x) = 
The derivative of the function is constant; therefore the function must be a straight line, not a curve.
Hello I would like to plot, q, h and Re in the same graph, no sure about it, any help will be appreciated. Thanks
1 = 30;
T2 = 40;
v = 0.5;
x1 = 0.01;
x2 = 0.05;
Pr = 5.18;
nu = 7.708;
k = 0.6184;
d = 995;
Re = v*(x2-x1)/nu
h = 0.3314*(Re.^(0.5))*(Pr.^(0.3))*(k/x2 - x1)
q = h*(x2 - x1)*1*(T2-T1)
plot (h, q, Re)
Your h, q, and Re are all scalars, so plotting them on the same graph is not going to be interesting.
T1 = 30;
T2 = 40;
v = 0.5;
x1 = 0.01;
x2 = 0.05;
Pr = 5.18;
nu = 7.708;
k = 0.6184;
d = 995;
Re = v*(x2-x1)/nu
Re = 0.0026
h = 0.3314*(Re.^(0.5))*(Pr.^(0.3))*(k/x2 - x1)
h = 0.3417
q = h*(x2 - x1)*1*(T2-T1)
q = 0.1367
plot([h, q, Re],'r*')

Sign in to comment.

More Answers (0)

Categories

Find more on Mathematics in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!