# displacemente pump as a momentum source in CFX

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 December 26, 2014, 10:09 displacemente pump as a momentum source in CFX #1 New Member   ivan Join Date: Dec 2014 Posts: 14 Rep Power: 9 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

 December 27, 2014, 04:10 #2 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,037 Rep Power: 134 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).

 December 27, 2014, 10:40 #3 New Member   ivan Join Date: Dec 2014 Posts: 14 Rep Power: 9 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

December 27, 2014, 12:05
#4
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ivan
Join Date: Dec 2014
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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
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 December 29, 2014, 11:44 #5 New Member   ivan Join Date: Dec 2014 Posts: 14 Rep Power: 9 im trying this in momentum source: (massFlow()@in*(Velocity u - (0.0001840130728 [m^3 s^-1] /area()@in) ))/volume()@Subdomain 1

 January 2, 2015, 16:11 #6 New Member   ivan Join Date: Dec 2014 Posts: 14 Rep Power: 9 i cant solve this problem =(, i really dont know how to implement C(v2-v1) momentum source in this model. Any ideas?

 January 3, 2015, 04:46 #7 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,037 Rep Power: 134 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.

 January 3, 2015, 12:44 #8 New Member   ivan Join Date: Dec 2014 Posts: 14 Rep Power: 9 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 =(

 January 3, 2015, 22:35 #9 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,037 Rep Power: 134 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

 January 5, 2015, 09:37 #10 New Member   ivan Join Date: Dec 2014 Posts: 14 Rep Power: 9 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

 January 5, 2015, 10:51 #11 New Member   ivan Join Date: Dec 2014 Posts: 14 Rep Power: 9 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

 January 5, 2015, 16:10 #12 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,037 Rep Power: 134 You need the momentum source term coefficient to assist with convergence. Read the documentation about it.

 Tags cfx, momentum, pumps, source

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