Asked by Artur M. G. Lourenço
on 25 May 2013

Hello Guys,

I have a simple problem. You remember the mechanics of fluids? To calculate the velocity distribution in a circular tube (actual fluid) use the equation "u" and then to further develop the known Hagen-Poiseuille equation. If we consider the tube without inclination have this equation:

u = (-N 2 - R 2) / 4 * mi

if I assign values to 'r' and 'mi', we have a paraboloid of revolution that describes the velocity distribution of the fluid in the tube. How can I make this chart in matlab?

See the example:

a = [-50:50]; u = -((a.^2-(0.001^2))/(4*1.485)); plot(u,a)

or

syms x ezplot(-((a^2-0.001^2)/(4*1.485)))

I put an fig in attach

https://docs.google.com/file/d/0B9JmyPHF_ZMPTFVmOWw3bk9FR1E/edit?usp=sharing

Thank you in advance for all the help!

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Answer by Youssef Khmou
on 26 May 2013

Edited by Youssef Khmou
on 26 May 2013

Accepted answer

hi,

i think your method works for this type of problems, try :

N=40; % Discretization Vmax=20; % 20m/s xc=0; yc=0; zc=0; R=0.5; % radius of the tube [x,y,z]=ellipsoid(xc,yc,zc,R,R,Vmax,N); z(z<0)=0; % trick to truncate the unwanted elements figure, surf(x,y,z), shading interp xlabel('X axis (m)'); ylabel(' Y axis (m)'); zlabel(' Velocity (m/s)'); title(' Velocity profile');

Artur M. G. Lourenço
on 27 May 2013

Answer by Youssef Khmou
on 25 May 2013

hi here is an example before staring to answer the problem :

the veolcity is defined as :

V(r)= Vmax*(1-r²/R²), R is the radius of the tube :

R=.50 ; %radius in meters: r=linspace(-R,R,30); % varying radius Vmax=20 ; % suppose that the maximum velocity of fluid is 20 m/s

V=Vmax*(1-r.^2/R^2); figure, bar(r,V); figure, plot(V,r); xlabel(' Velocity'),ylabel(' varying radius')

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Youssef Khmou
on 25 May 2013

ok try this way first :

R=.50 ; %radius in meters: r=linspace(-R,R,30); % varying radius Vmax=20 ; % suppose that the maximum velocity of fluid is 20 m/s V=Vmax*(1-r.^2/R^2); VV=sqrt(V'*V); figure, surf(r,r,VV), shading interp,

Artur M. G. Lourenço
on 26 May 2013

almost! Missing only the base should be circular. I am trying here to change that!

Artur M. G. Lourenço
on 26 May 2013

see this, my graph looks like a half of cylinder

[x, y, z] = ellipsoid(0,0,0,5.9,3.25,3.25,30); surfl(x, y, z) colormap copper axis equal

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