Using simpleFoam with water

nico765 · October 22, 2007, 15:04

Hello,

Is there a way to setup the fluid density in simpleFoam?

I would like to use water, so far i have set up my cases to run at similar reynolds number using high velocities (~100m/s) in my cases. Not sure if this is the best way, since the cases are more unstable when running at higher velocities (same y+ values).

Nico

hjasak · October 22, 2007, 15:13

Fluid density is constant: that is why we use the kinematic viscosity. Therefore, divide your dynamic viscosity (in Pascal seconds) bu fluid density and specify that in the constant/transportProperties. Be careful to put back the density if you need wall forces or similar from the pressure (for consistency, p is kinematic pressure).

Enjoy,

Hrv

mkraposhin · October 23, 2007, 05:52

Dear Hrvoje Jasak! Please, answer a stupid qustion: how can i convert relative pressure from simpleFoam to normal (total Pressure (kg*m/s^2)??

Many thanks for advice!

hjasak · October 23, 2007, 06:02

Requires a Napoleonic answer

Multiply by the density

Jasak

mkraposhin · October 23, 2007, 06:11

But it is negative!!! for example, pressure in simpleFoam ranges from -1 to 1 - what does it means?

mkraposhin · October 23, 2007, 06:24

And, another question (i hope, i'm not too importunate, and sorry for bad English!)

This question arises when i increase mesh density in cavity tutorial by 2 (both in x and y dimensions) - pressure range increases by 4...

i think, it arises from Bernoulli eqn
i mean, p/rho + w^2/2 + g*z = const...

specific volume (= 1/rho) decreases by 4, so pressure range increasses 4 to satisfy equation. isn't it?

hjasak · October 23, 2007, 06:24

As you know, in incompressible flows, the pressure level does not matter: the flow is driven by the pressure gradient. Therefore, you can add any offset to the pressure field that you like - remember the pressure on the boundary being e.g. zero or the pressure in the reference point usually set to zero as well.

Therefore, if you know the absolute pressure in any point of the domain, just shift the complete pressure field for this number, maybe adding 101325 Pascal if it makes you feel better.

None of this actually matters when you are calculating the forces unless you've got vacuum outside.

Hrv

mkraposhin · October 23, 2007, 06:39

Many Thanks, Hrvoje Jasak, for Your advice!
Many Thanks!

I'll give it a try.

October 22, 2007, 15:04	Hello, Is there a way to se	#1
nico765 Member nicolas Join Date: Mar 2009 Location: Glasgow Posts: 42 Rep Power: 17	Hello, Is there a way to setup the fluid density in simpleFoam? I would like to use water, so far i have set up my cases to run at similar reynolds number using high velocities (~100m/s) in my cases. Not sure if this is the best way, since the cases are more unstable when running at higher velocities (same y+ values). Nico

October 22, 2007, 15:13	Fluid density is constant: tha	#2
hjasak Senior Member Hrvoje Jasak Join Date: Mar 2009 Location: London, England Posts: 1,907 Rep Power: 33	Fluid density is constant: that is why we use the kinematic viscosity. Therefore, divide your dynamic viscosity (in Pascal seconds) bu fluid density and specify that in the constant/transportProperties. Be careful to put back the density if you need wall forces or similar from the pressure (for consistency, p is kinematic pressure). Enjoy, Hrv __________________ Hrvoje Jasak Providing commercial FOAM/OpenFOAM and CFD Consulting: http://wikki.co.uk

October 23, 2007, 05:52	Dear Hrvoje Jasak! Please, ans	#3
mkraposhin Senior Member Matvey Kraposhin Join Date: Mar 2009 Location: Moscow, Russian Federation Posts: 355 Rep Power: 21	Dear Hrvoje Jasak! Please, answer a stupid qustion: how can i convert relative pressure from simpleFoam to normal (total Pressure (kg*m/s^2)?? Many thanks for advice! __________________ MDPI Fluids (Q2) special issue for OSS software: https://www.mdpi.com/journal/fluids/..._modelling_OSS GitHub: https://github.com/unicfdlab Linkedin: https://linkedin.com/in/matvey-kraposhin-413869163 RG: https://www.researchgate.net/profile/Matvey_Kraposhin

October 23, 2007, 06:02	Requires a Napoleonic answer h	#4
hjasak Senior Member Hrvoje Jasak Join Date: Mar 2009 Location: London, England Posts: 1,907 Rep Power: 33	Requires a Napoleonic answer Multiply by the density Jasak __________________ Hrvoje Jasak Providing commercial FOAM/OpenFOAM and CFD Consulting: http://wikki.co.uk

October 23, 2007, 06:11	But it is negative!!! for exam	#5
mkraposhin Senior Member Matvey Kraposhin Join Date: Mar 2009 Location: Moscow, Russian Federation Posts: 355 Rep Power: 21	But it is negative!!! for example, pressure in simpleFoam ranges from -1 to 1 - what does it means? __________________ MDPI Fluids (Q2) special issue for OSS software: https://www.mdpi.com/journal/fluids/..._modelling_OSS GitHub: https://github.com/unicfdlab Linkedin: https://linkedin.com/in/matvey-kraposhin-413869163 RG: https://www.researchgate.net/profile/Matvey_Kraposhin

October 23, 2007, 06:24	And, another question (i hope,	#6
mkraposhin Senior Member Matvey Kraposhin Join Date: Mar 2009 Location: Moscow, Russian Federation Posts: 355 Rep Power: 21	And, another question (i hope, i'm not too importunate, and sorry for bad English!) This question arises when i increase mesh density in cavity tutorial by 2 (both in x and y dimensions) - pressure range increases by 4... i think, it arises from Bernoulli eqn i mean, p/rho + w^2/2 + g*z = const... specific volume (= 1/rho) decreases by 4, so pressure range increasses 4 to satisfy equation. isn't it? __________________ MDPI Fluids (Q2) special issue for OSS software: https://www.mdpi.com/journal/fluids/..._modelling_OSS GitHub: https://github.com/unicfdlab Linkedin: https://linkedin.com/in/matvey-kraposhin-413869163 RG: https://www.researchgate.net/profile/Matvey_Kraposhin

October 23, 2007, 06:24	As you know, in incompressible	#7
hjasak Senior Member Hrvoje Jasak Join Date: Mar 2009 Location: London, England Posts: 1,907 Rep Power: 33	As you know, in incompressible flows, the pressure level does not matter: the flow is driven by the pressure gradient. Therefore, you can add any offset to the pressure field that you like - remember the pressure on the boundary being e.g. zero or the pressure in the reference point usually set to zero as well. Therefore, if you know the absolute pressure in any point of the domain, just shift the complete pressure field for this number, maybe adding 101325 Pascal if it makes you feel better. None of this actually matters when you are calculating the forces unless you've got vacuum outside. Hrv __________________ Hrvoje Jasak Providing commercial FOAM/OpenFOAM and CFD Consulting: http://wikki.co.uk

October 23, 2007, 06:39	Many Thanks, Hrvoje Jasak, for	#8
mkraposhin Senior Member Matvey Kraposhin Join Date: Mar 2009 Location: Moscow, Russian Federation Posts: 355 Rep Power: 21	Many Thanks, Hrvoje Jasak, for Your advice! Many Thanks! I'll give it a try. __________________ MDPI Fluids (Q2) special issue for OSS software: https://www.mdpi.com/journal/fluids/..._modelling_OSS GitHub: https://github.com/unicfdlab Linkedin: https://linkedin.com/in/matvey-kraposhin-413869163 RG: https://www.researchgate.net/profile/Matvey_Kraposhin

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