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Inlet Boundary velocity
Posted 2013年2月5日 GMT-5 21:30 Version 3.5a 2 Replies
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Inlet boundary velocity profile = 4*s*(1-s)*Umax
What is actually represent this equation? What is s?
What is the different between normal velocity and Umax?
What is actually represent this equation? What is s?
What is the different between normal velocity and Umax?
2 Replies Last Post 2015年6月29日 GMT-4 09:55
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Posted:
1 decade ago
Hi
if you check the doc, and the Forum you will see that "s" is a COMSOL internal variable taking the value 0 to 1 along a 2D boundary. so if you select an edge in 2D (=a boundary) and write a velocity 4*s*(1-s)*Umax you get an parabolic profile with u=0 at s=0 and s=1 (the walls intersection points) and a velocity Umax at s=0.5 the middle of the edge/boundary
This is a good starting velocity profile for FCD for a pipe or any fluid flowing between two (no-slip) walls
--
Good luck
Ivar
if you check the doc, and the Forum you will see that "s" is a COMSOL internal variable taking the value 0 to 1 along a 2D boundary. so if you select an edge in 2D (=a boundary) and write a velocity 4*s*(1-s)*Umax you get an parabolic profile with u=0 at s=0 and s=1 (the walls intersection points) and a velocity Umax at s=0.5 the middle of the edge/boundary
This is a good starting velocity profile for FCD for a pipe or any fluid flowing between two (no-slip) walls
--
Good luck
Ivar
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Posted:
9 years ago
Dear Ivar
First question:
In examples I encountered
4*s*(1-s)*Umax
versus
6*s*(1-s)*Umax
As the average of s*(1-s) is 1/6, I assume 6*s*(1-s) is the best choice as it gives you U_max at s=0.5?
Or why do you propose 4*s*(1-s)?
Second question:
Why use 6*s*(1-s)*Umax
and not (-4*(s-0.5)^2+1)*Umax?
I admit, your expression is shorter, however, the velocity profile is parabolic, so of the form a*x^2 (and not a*x^2+b*x).
Third question:
I have an inlet with a symmetry plane. I select the inlet boundary to give it a velocity profile that is half of 6*s*(1-s)*Umax . But 's' always goes up to 1. Do I use 6*s/2*(1-s/2)*Umax to get half of the profile?
Thx
First question:
In examples I encountered
4*s*(1-s)*Umax
versus
6*s*(1-s)*Umax
As the average of s*(1-s) is 1/6, I assume 6*s*(1-s) is the best choice as it gives you U_max at s=0.5?
Or why do you propose 4*s*(1-s)?
Second question:
Why use 6*s*(1-s)*Umax
and not (-4*(s-0.5)^2+1)*Umax?
I admit, your expression is shorter, however, the velocity profile is parabolic, so of the form a*x^2 (and not a*x^2+b*x).
Third question:
I have an inlet with a symmetry plane. I select the inlet boundary to give it a velocity profile that is half of 6*s*(1-s)*Umax . But 's' always goes up to 1. Do I use 6*s/2*(1-s/2)*Umax to get half of the profile?
Thx