Channel forcing (pressure gradient)
 

Hello everyone!

My question of the day: If you have a finite volume code, and you want to run a channel flow case, how do you drive the flow?

I am slightly confused, because even if I do not impose any pressure gradient, the flow will just keep on moving due to the periodic boundary conditions (no mass or momentum is being destroyed, so it has to keep on moving).

Basically, does the following procedure work?

1. compute the average flow rate
2. compute the delta flow rate = actual - target
3. define dpdx = delta flow rate/channel area
4. add dpdx as a source term in the momentum equation and rho*u*dpdx as a source term in the energy equation.

Is that ok? I am confused because I would have thought that my simulation would converge towards the pressure gradient required to drive the flow. Instead, following the previous procedure, the simulation will converge to the correct flow rate when dpdx goes to zero. Does that make sense?

Thanks!

Joachim

usually, two ways are possible: imposing a constant pressure gradient or imposing a flow rate.
If you work with pressure gradient then a constant forcing term is included in the momentum equation and the pressure solver adjusts the periodical part of the pressure to ensure the continuity constraint

Thanks!
It is possible to impose a flow rate without using a pressure gradient? How could I do that?

Dear Joachim,

the flow in the periodic settings MUST have a source term to be sustained. Otherwise the viscosity will eventually kill your motion. In all the cases this source term is a pressure gradient (the non-periodic part). As stated by Filippo, you have two choices:

- fixed pressure gradient -> the mass flow rate is just an output of your computation

- fixed mass flow rate -> you iterate on the source term (pressure gradient) until the resulting mass flow rate reaches your target value. As this is a control problem, there is no unique way to iterate on the source term and you have to consider additional information to make a proper choice.

Viscosity is only acting as a transport phenomena in the equations. Shouldn't it only move the momentum around instead of dissipating it?

In my case, the flow is compressible, is that fine if I simply add the source term without further consideration?

Thanks!

Don't you expect the effect of viscosity to be dissipative? Thus, if you initialize the flow in the periodic channel with a parabolic laminar profile and no source term, do you expect the fluid to keep moving without any energy input?

You should check the kinetic energy equation, it will show you this is not actually the case. For compressible flows there are additional considerations on the energy equation ,where an additional source term is needed.

Check this:

http://onlinelibrary.wiley.com/doi/1...O;2-6/abstract

Well, if you consider the case of an incompressible laminar flow, then yes, I would expect the flow to magically goes on forever I guess. :D

without any forcing, for compressible flows you would find a vanishing velocity field, the kinetic energy will be converted in internal energy. Density can eventually have a final gradient but pressure is homogeneous

...and laminar or not is the same ;)

agreed. I only meant that a viscous incompressible flow (my scenario) theoretically does not require any pressure gradient to drive its motion. Correct?

well, for example a wall-free constant velocity field is a solution with zero pressure gradient, but in a channel, to substain the motion you need a pressure gap ...

I understand that this is the case in nature.
For a finite volume solver however, I do not really see how momentum could dissapear without any dissipative term in the equation...The momentum and mass are only being convected and transported throughout the domain...

why FV should change the physics? I don't understand... maybe you are solving Euler equations? in viscous flows, the momentum equation has the convective, diffusive and pressure fluxes. The diffusive flux does dissipate the velocity gradient by means of molecular viscosity. Without a difference in the pressure, the acceleration will diminish until the rest

Well, viscosity acts as a transport term in the momentum equation. Basically, you can use Green's theorem to show that a transport term only redistribute the variable. Over the entire domain, the total amount of momentum should stay the same. If you compare with the TKE equation for example, you still have the transport of TKE due to viscosity, but you also have the dissipation term (also due to viscosity), with explains the decay of TKE. Maybe I missed something...

but the kinetic energy does not conserve... it has a production term that takes into account for the dissipation of energy in internal energy.

If you integrate the momentum equation over the boundaries of the channel, the total convective flux vanishes, the pressure flux vanishes (if you consider only the periodical field) and the diffusive flux has a contribute due to the integral along the walls. To substain the transport of momentum you need a forcing term

For compressible flow I agree. However, if the flow is incompressible, the continuity/momentum and energy equations are decoupled. It follows that you can compute your velocity profile independently of the temperature/internal energy field.

Yes, the decoupling from energy is true fro incompressible, but that does not mean that your kinetic energy is conserved....
you can solve continuity and momentum alone but the evolution of the kinetic energy remains imposed by the velocity field obtained from the equation system and you can always write the kinetic energy to see that exists a production term that dissipate the v^2/2 you obtained from the continuity/momentum.

For example, the continuity equation (divergence-free velocity) must be resolved very well as the residual in the continuity acts (implicitly) as source term in the kinetic energy and can false the energy level (or also driving to a numerical instability)

Consider again, for the sake of simplicity, the steady, incompressible, laminar case for the channel flow. What are the momentum equations and how the terms are balanced in them? You will notice that only the x-momentum equation is non-null and the dp/dx balances the diffusion term mu*d2u/dy2. If you neglect the pressure gradient, the only possible outcome in a fixed channel is a no flow condition. You can't mix diffusion and dissipation, one is a transport mechanism, the other is a net energy effect. The momentum is not an effective measure for the energy of the flow.

If you let the flow in the channel evolve without pressure gradient, than you have to consider the unsteady term too, which will balance your momentum equation to the no flow condition.