Distribution graph velocity : how to make simple paraboloid of revolution?

4 ビュー (過去 30 日間)
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
Thank you in advance for all the help!

採用された回答

Youssef  Khmou
Youssef Khmou 2013 年 5 月 26 日
編集済み: Youssef Khmou 2013 年 5 月 26 日
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');
  1 件のコメント
Artur M. G. Lourenço
Artur M. G. Lourenço 2013 年 5 月 27 日
O/ Thank you so much! From here I think I can move forward only. Just leave it on the lack horinzontal and insert my data. Thank you again.

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その他の回答 (3 件)

Artur M. G. Lourenço
Artur M. G. Lourenço 2013 年 5 月 25 日
sorry 'u' is:
u = (-a^2 - r^2) / 4 * mi

Youssef  Khmou
Youssef Khmou 2013 年 5 月 25 日
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')
  6 件のコメント
Artur M. G. Lourenço
Artur M. G. Lourenço 2013 年 5 月 26 日
almost! Missing only the base should be circular. I am trying here to change that!
Artur M. G. Lourenço
Artur M. G. Lourenço 2013 年 5 月 26 日
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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Artur M. G. Lourenço
Artur M. G. Lourenço 2013 年 5 月 26 日
i can´t do it

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