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Inlet Boundary velocity

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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?

2 Replies Last Post 2015年6月29日 GMT-4 09:55
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago 2013年2月6日 GMT-5 00:59
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
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

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Posted: 9 years ago 2015年6月29日 GMT-4 09:55
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
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

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