CFD Online Logo CFD Online URL
www.cfd-online.com
[Sponsors]
Home > Forums > CFX

displacemente pump as a momentum source in CFX

Register Blogs Members List Search Today's Posts Mark Forums Read

Reply
 
LinkBack Thread Tools Display Modes
Old   December 26, 2014, 11:09
Default displacemente pump as a momentum source in CFX
  #1
New Member
 
ivan
Join Date: Dec 2014
Posts: 14
Rep Power: 4
iloc86 is on a distinguished road
Sponsored Links
im new in CFD simulation in CFX. Im currently simulating a simple pipe system with one displacemente pump. i have read that with a "source" i can model the effect of a pump. i have issues with introducing the gneral momentum value. the system is simple

(in= 570 psi)=====PUMP====(out=2000 psi)

i have done some runs (giving values to momentum) and have a flow to the output. in a simple way, how can i calculate the value of momentum or do a function to simulate the pump. i wanna a flow of 100 bpd to the output. from the pump i have a Q vs P table.

thanks for advance
iloc86 is offline   Reply With Quote
Sponsored Links

Old   December 27, 2014, 05:10
Default
  #2
Super Moderator
 
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 13,735
Rep Power: 106
ghorrocks is a jewel in the roughghorrocks is a jewel in the roughghorrocks is a jewel in the rough
Put a pressure "sensor" at the inlet and outlet of the pump. This could be a probe point or a plane to do averaging over. You then have the pressure difference across the pump. Use the pump curve to give the flow velocity which would occur for this pressure difference and apply that flow velocity as a momentum source term (using the defined value approach - let me know if you are not aware of this approach).
ghorrocks is offline   Reply With Quote

Old   December 27, 2014, 11:40
Default
  #3
New Member
 
ivan
Join Date: Dec 2014
Posts: 14
Rep Power: 4
iloc86 is on a distinguished road
I really appreciate your response. I saw that the momentum is entered as c(v2-v1) but i havent used this kind of approach. i have read too that i have to define a user function. i dont know if its necesary to use the p vs q of the pump cuz i already know that i have to keep a 100 bpd of flow and the diferencial pressure its going to be 2000 psi - 570 psi. i dont know if iam taking this in the wrong way :S thanks for advance
iloc86 is offline   Reply With Quote

Old   December 27, 2014, 13:05
Default
  #4
New Member
 
ivan
Join Date: Dec 2014
Posts: 14
Rep Power: 4
iloc86 is on a distinguished road
i have enter this in momentum source

(Velocity u - (0.0001840130728 [m^3 s-1] /area()@in) )

but i dont know what to do next
Attached Images
File Type: jpg sa.jpg (15.3 KB, 19 views)
iloc86 is offline   Reply With Quote

Old   December 29, 2014, 12:44
Default
  #5
New Member
 
ivan
Join Date: Dec 2014
Posts: 14
Rep Power: 4
iloc86 is on a distinguished road
im trying this in momentum source:

(massFlow()@in*(Velocity u - (0.0001840130728 [m^3 s^-1] /area()@in) ))/volume()@Subdomain 1
iloc86 is offline   Reply With Quote

Old   January 2, 2015, 17:11
Default
  #6
New Member
 
ivan
Join Date: Dec 2014
Posts: 14
Rep Power: 4
iloc86 is on a distinguished road
i cant solve this problem =(, i really dont know how to implement C(v2-v1) momentum source in this model. Any ideas?
iloc86 is offline   Reply With Quote

Old   January 3, 2015, 05:46
Default
  #7
Super Moderator
 
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 13,735
Rep Power: 106
ghorrocks is a jewel in the roughghorrocks is a jewel in the roughghorrocks is a jewel in the rough
You seem to have made yourself very, very confused.

Have a look at section 1.3.2.2.2 General Momentum Source of the documentation (Modeling Guide) for how to implement this type of momentum source.
ghorrocks is offline   Reply With Quote

Old   January 3, 2015, 13:44
Default
  #8
New Member
 
ivan
Join Date: Dec 2014
Posts: 14
Rep Power: 4
iloc86 is on a distinguished road
thanks for your reply. yes i read that. As far as i know i can model the pump effect like a momentum source, in CFX its a force/volume ok? thats why the unit are Newton/volume or kg m s^-2 m^-3. the way to do that it by the equation of Momentum Source s= -C(v - vspec) in x y z components. where v is the variable of velocity, vspec is the one i specify (velocity that is function of the dp of the pump, in this case i want to keep it at 100 bpd that is an average velocity of 0.2 m/s in my case) and c as far i know is the coefficient of momentum that is the mass flow/volume , so i understand it as the mass flow that goes trought the pump and the volume of the pump giving units as kg s^-1 m^-3. but CFX modeling says that C is just a large number with units kg s^-1 m^-3 and that its good to define a coefficient to help convergence, so as far i cant understand i define it like

sx=-100000 kg s^-1 m^-3 *(u - 0.2 m s^-1)
sy= 0 cuz i dont want source in that direction
sz= 0 cuz i dont want source in that direction

coefficient= massflow enters the pump /volume of the pump

whats your opinion, i messed all? ahaha sorry =(
iloc86 is offline   Reply With Quote

Old   January 3, 2015, 23:35
Default
  #9
Super Moderator
 
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 13,735
Rep Power: 106
ghorrocks is a jewel in the roughghorrocks is a jewel in the roughghorrocks is a jewel in the rough
You are close, but you missed the point of the coefficient. The equations are meant to be:

sx=-C *(u - 0.2 m s^-1)
sy= 0 cuz i dont want source in that direction
sz= 0 cuz i dont want source in that direction

C=100000 kg s^-1 m^-3

Momentum source term coefficient = -C
ghorrocks is offline   Reply With Quote

Old   January 5, 2015, 10:37
Default
  #10
New Member
 
ivan
Join Date: Dec 2014
Posts: 14
Rep Power: 4
iloc86 is on a distinguished road
thanks!!. So if i use the Mom Source Coeff. i define too like in the source entry? Is there a way to calculate that coeffcient intead of just "give" a large number?
thanks for advance
iloc86 is offline   Reply With Quote

Old   January 5, 2015, 11:51
Default
  #11
New Member
 
ivan
Join Date: Dec 2014
Posts: 14
Rep Power: 4
iloc86 is on a distinguished road
im trying this:

sx=-100000000 [kg m^-3 s^-1]*(Velocity u - 0.2 [m s^-1])
sy= 0 cuz i dont want source in that direction
sz= 0 cuz i dont want source in that direction

C=-1e+08

it take some time to convergence. im running in steady state, what do you think to run it in transient? i have seen i have trouble to convergence when the fluid start to go to the 2000 psi output as i increase C. it would be nice if i foud a way to calculate that number :S
iloc86 is offline   Reply With Quote

Old   January 5, 2015, 17:10
Default
  #12
Super Moderator
 
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 13,735
Rep Power: 106
ghorrocks is a jewel in the roughghorrocks is a jewel in the roughghorrocks is a jewel in the rough
You need the momentum source term coefficient to assist with convergence. Read the documentation about it.
ghorrocks is offline   Reply With Quote

Reply

Tags
cfx, momentum, pumps, source

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are On
Pingbacks are On
Refbacks are On


Similar Threads
Thread Thread Starter Forum Replies Last Post
SparceImage v1.7.x Issue on MAC OS X rcarmi OpenFOAM Installation on Windows, Mac and other Unsupported Platforms 4 August 14, 2014 06:42
Problem compiling a custom Lagrangian library brbbhatti OpenFOAM Programming & Development 2 July 7, 2014 11:32
OpenFOAM without MPI kokizzu OpenFOAM Installation 4 May 26, 2014 09:17
Trouble compiling utilities using source-built OpenFOAM Artur OpenFOAM Programming & Development 14 October 29, 2013 11:59
Error bulding swak4Foam sfigato OpenFOAM Installation 18 August 22, 2013 12:41

Sponsored Links


All times are GMT -4. The time now is 10:09.