# Pressure constant B.C. in fractional step method

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 April 10, 2012, 03:31 Pressure constant B.C. in fractional step method #1 Member   Amin Shariat KHah Join Date: Apr 2011 Location: Shiraz Posts: 86 Rep Power: 8 Sponsored Links Hi everyOne As we know pressure in fractional step method is not real pressure and it's pressure-like variable as we see in every reference. and beside this for this pressure every where B.C. is dp/dn=0 . with this point how we can use pressure constant B.C. in this method? It's like the other method for example SIMPLE-like methods? dp/dn=0 doesn't make any problem? thanks for your attention

April 10, 2012, 06:52
#2
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Filippo Maria Denaro
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Quote:
 Originally Posted by amin144 Hi everyOne As we know pressure in fractional step method is not real pressure and it's pressure-like variable as we see in every reference. and beside this for this pressure every where B.C. is dp/dn=0 . with this point how we can use pressure constant B.C. in this method? It's like the other method for example SIMPLE-like methods? dp/dn=0 doesn't make any problem? thanks for your attention
for example in the pressure-free method (Moin & Kim) a constant pressure outlet gives you (one-dimension x, second order discretization)

( (dp/dx)i+1/2 - (dp/dx)i-1/2 )/dx = (1/dt) ( (u*)i+1/2 - (u*)i-1/2 )/dx (1)

with the general condition provided by the Hodge decomposition

(dp/dx)i+1/2 = (1/dt) ( (u*)i+1/2 - (un+1)i+1/2 )

Generally, you physically know and prescribe the velocity un+1 at the boundaries and rewrite the pressure equation as:

( (dp/dx)i-1/2 )/dx = (1/dt) ( (un+1)i+1/2 - (u*)i-1/2 )/dx

Now, you have to use Eq.(1) and (2) in the same congruent way but prescribing the constant pressure (that is you have to prescribe something in the LHS of Eq.(2)). Be carefull in the (un+1)

Last edited by FMDenaro; April 10, 2012 at 15:34.

April 10, 2012, 11:08
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Amin Shariat KHah
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 Originally Posted by FMDenaro for example in the pressure-free method (Moin & Kim) a constant pressure outlet gives you (one-dimension x, second order discretization) ( (dp/dx)i+1/2 - (dp/dx)i-1/2 )/dx = (1/dt) ( (u*)i+1/2 - (u*)i-1/2 )/dx (1) with the general condition provided by the Hodge decomposition (dp/dx)i+1/2 = (1/dt) ( (u*)i+1/2 - (un+1)i+1/2 ) Generally, you physically know and prescribe the velocity un+1 at the boundaries and rewrite the pressure equation as: ( (dp/dx)i-1/2 )/dx = (1/dt) ( (un+1)i+1/2 - (u*)i-1/2 )/dx Now, you have to use Eq.(1) and (2) in the same congruent way but prescribing the constant pressure (that is you have to prescribe something in the LHS of Eq.(2)). Be carefull in the (un+1)
Hi again dear Filippo
thanks very much
You are very professional in this method

understanding your explanations and equations is hard for me
Can you explain me easy like Pierc CFD book and in the form of this book?
you are professional and your explanations is professional too

Last edited by amin144; April 10, 2012 at 16:29.

April 10, 2012, 11:18
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Filippo Maria Denaro
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Quote:
 Originally Posted by amin144 Hi again dear Filippo thanks very much You are very professional in this method understanding your explanations and equations is hard for me Can you explain me easy like Pierc CFD book and in the form of this book? you are professional and your explanations is professional too

what is the issue you have not understood?

 April 10, 2012, 13:47 #5 Member   Amin Shariat KHah Join Date: Apr 2011 Location: Shiraz Posts: 86 Rep Power: 8 Thanks again very very much dear Filippo If I be right your text mean that we know P(i+1) (from pressure constant boundary) and (u*)i+1/2 so we derive p(i) from eq(1) and after we can derive u(n+1) from eq(2) ? If I'm wrong I'll appreciate you if you explain me what should I do step by step. I have another question why you vanished (dp/dx)i-1/2 in eq(3) if this is for outlet? In outlet we have (dp/dx)i+1/2=0 . am I right?

April 10, 2012, 15:34
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Filippo Maria Denaro
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Quote:
 Originally Posted by amin144 Thanks again very very much dear Filippo If I be right your text mean that we know P(i+1) (from pressure constant boundary) and (u*)i+1/2 so we derive p(i) from eq(1) and after we can derive u(n+1) from eq(2) ? If I'm wrong I'll appreciate you if you explain me what should I do step by step. I have another question why you vanished (dp/dx)i-1/2 in eq(3) if this is for outlet? In outlet we have (dp/dx)i+1/2=0 . am I right?

Sorry, I was wrong in copying and pasting the LHS of (3), it writes

( (dp/dx)i-1/2 )/dx = (1/dt) ( (un+1)i+1/2 - (u*)i-1/2 )/dx

... the Eq.(2) prescribes a relation between the normal component of the pressure gradient, the normal component u* and the normal component un+1. The mathematical problem requires only 1 boundary condition to be well posed, therefore generally one prescribe the physical velocity on the boundary. If you know un+1, then you just substitute the (2) into (1) and get (3), this way you does not prescribe dp/dx=0 on the boundary.
In your case you want to prescribe (p)i+1 with a Dirichlet boundary condition, therefore you can not prescribe the physical velocity. You must use some extrapolation from the interior

 April 10, 2012, 16:57 #7 Member   Amin Shariat KHah Join Date: Apr 2011 Location: Shiraz Posts: 86 Rep Power: 8 1) If we put eq(2) in eq(3) we will have: ( (dp/dx)i-1/2 )/dx = (1/dt) ( (un+1)i+1/2 -2* (u*)i-1/2 )/dx it's a little different from eq(3) 2) why we see in refrences says that dp/dn=0 on boundaries? 3) so if I will have a interpolation to derive u(n+1) on velocity, where the pressure diffrence between inlet and outlet effect to drive the fluid? only in solving poisson equation for pressure and using value of pressure on boundaries for pressure ? and this work is sufficient to apply boundary condition?

April 10, 2012, 17:21
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Filippo Maria Denaro
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Quote:
 Originally Posted by amin144 1) If we put eq(2) in eq(3) we will have: ( (dp/dx)i-1/2 )/dx = (1/dt) ( (un+1)i+1/2 -2* (u*)i-1/2 )/dx it's a little different from eq(3) 2) why we see in refrences says that dp/dn=0 on boundaries? 3) so if I will have a interpolation to derive u(n+1) on velocity, where the pressure diffrence between inlet and outlet effect to drive the fluid? only in solving poisson equation for pressure and using value of pressure on boundaries for pressure ? and this work is sufficient to apply boundary condition?

1) put (dp/dx)i+1/2 = (1/dt) ( (u*)i+1/2 - (un+1)i+1/2 ) into the LHS of

( (dp/dx)i+1/2 - (dp/dx)i-1/2 )/dx = (1/dt) ( (u*)i+1/2 - (u*)i-1/2 )/dx (1)

and will get

( (dp/dx)i-1/2 )/dx = (1/dt) ( (un+1)i+1/2 - (u*)i-1/2 )/dx

2) dp/dn = 0 is sometimes referenced but it is not really exact... however, the Neumann homogeneous condition applied on Eq.(2) will drive to u* = un+1 and the equivalence with Eq.(3) exists. For this reason you read in literature this condition. But when you compute the normal pressure gradient on the boundary with dp/dn=0, or dp/dn = (1/dt) (u* -un+1), the solutions are different

3) need to considere the extrapolation also in the momentum quantity. You must verify at the end of the steps that Div Vn+1 = 0 also on the cell at outlet

 April 10, 2012, 17:43 #9 Member   Amin Shariat KHah Join Date: Apr 2011 Location: Shiraz Posts: 86 Rep Power: 8 1) you are right like always. I guess I,m being crazy slowly 2) so this B.C. (dp/dn=0) make unexact results? what should do? 3) satisfying Div(Vn+1)=0 need only more itteration or any other yhing?

 April 10, 2012, 17:47 #10 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 3,416 Rep Power: 39 dp/dn = 0 is an approximation congruent to a boundary layer approximation ... but can lead to some problems, that is largely discussed in literature. the divergence-free constraint is satisfied at machine accuracy in the Exact Projecton Method and at the level of the local truncation error in the Approximate Projection Method. Generally, EPM is for second order accurate discretization on staggered grids, APM is on non-staggered. amin144 likes this.

 April 10, 2012, 17:58 #11 Member   Amin Shariat KHah Join Date: Apr 2011 Location: Shiraz Posts: 86 Rep Power: 8 Thanks your perfect answers

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