# Physics of an incompressible fluid

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March 21, 2015, 11:49
#61
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Filippo Maria Denaro
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 Originally Posted by Martin Hegedus So, you are saying that Euler equations do not boil down to the Bernoulli equation. I believe they do. But, unfortunately, I just don't have the time to prove it or dig up a reference. I'll have to leave it at this. And we both agree that the velocity profile can be found using potential theory, correct?

I depends on the hypothesis....Bernoulli requires strictly the flow to be homoentropic and density homogeneous whilst you can write the Euler equation for a case where the entropy changes both in time and space, provided that ds/dt + u ds/dx=0 (constant entropy along the path-line). Of course if ds/dx is no zero Bernouilli cannot be applied. Only if you assume also steady state then ds/dx must vanish too and you have the condition to use Bernouille

 March 21, 2015, 12:05 #62 Senior Member   Martin Hegedus Join Date: Feb 2011 Posts: 500 Rep Power: 19 Sorry, I truly need to end it here. Best of luck to everyone!!

 March 21, 2015, 12:10 #63 Senior Member   Martin Hegedus Join Date: Feb 2011 Posts: 500 Rep Power: 19 BTW, I do want to take the time to agree with you that if wants to couple dp/dt directly into the Euler equations for the 1-D example then it needs to be done through the drho/dt term. And this I mentioned earlier too. Otherwise one needs to calculate dp/dt directly after the Euler equations are solved. For the case of incompressible flow dp/dt is uncoupled from the Euler equations.

 March 21, 2015, 12:14 #64 Senior Member   Martin Hegedus Join Date: Feb 2011 Posts: 500 Rep Power: 19 Actually, I guess my statement is always true since dp/dt does not show up in the Navier stokes equations. Sorry, my focus is just not on this at the moment.

March 21, 2015, 12:26
#65
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Filippo Maria Denaro
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 Originally Posted by Martin Hegedus BTW, I do want to take the time to agree with you that if wants to couple dp/dt directly into the Euler equations for the 1-D example then it needs to be done through the drho/dt term. And this I mentioned earlier too. Otherwise one needs to calculate dp/dt directly after the Euler equations are solved. For the case of incompressible flow dp/dt is uncoupled from the Euler equations.

Ok, I'll try to resume if we are talking about the same issue...if we assume incompressible flow such that rho is constant both in space and time then du/dx must be zero from continuity (hence u=u(t)) and the momentum equation is reduced to

du/dt + (1/rho0) dp/dx = 0.

From the equation of state p = rho0*R*T, therefore you must have a term T=T(x) to produce a dp/dx.

Note that the energy equation written in terms of pressure still produces (du/dx=0)

dp/dt + u(t) dp/dx = 0 -> dT/dt + u(t) dT/dx = 0.

In conclusion, I see momentum and pressure equations still coupled...
So, I have doubt we are not talking of the same problemmm

 March 21, 2015, 13:52 #66 Senior Member   Martin Hegedus Join Date: Feb 2011 Posts: 500 Rep Power: 19 My point is that the equation for dp/dt is decoupled from the incompressible conservation of mass and momentum equations. However, dp/dx is not decoupled. First one must for those four equations and then solve for dp/dt. Or, at least that is one way to do it. There might be others.

 March 21, 2015, 13:57 #67 Senior Member   Martin Hegedus Join Date: Feb 2011 Posts: 500 Rep Power: 19 I guess another way to look at is that the pressure distribution sets itself up instantaneously for an incompressible fluid throughout the fluid. It's an elliptic equation rather than a hyperbolic one. It is only based on the boundary conditions.

 March 21, 2015, 14:05 #68 Senior Member   Martin Hegedus Join Date: Feb 2011 Posts: 500 Rep Power: 19 BTW, I'll admit that I can not grasp what the equation of state for an incompressible material is.

 March 21, 2015, 14:36 #69 Senior Member   Martin Hegedus Join Date: Feb 2011 Posts: 500 Rep Power: 19 Or, another way to approach this, in the 1-D case for example, is to take the time derivative of the momentum equation. So, for example, if the acceleration was constant then d(dp/dt)/dx = 0.

March 21, 2015, 14:37
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Filippo Maria Denaro
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 Originally Posted by Martin Hegedus BTW, I'll admit that I can not grasp what the equation of state for an incompressible material is.
Again in the general perspective of the incompressible (and homoenthalpic) flow model, no state equation exists...

see at the end of http://www.cfd-online.com/Wiki/Incompressible_flow

specific formulation for low-Mach flow (e.g. combustion), slightly modify the set of equations introducing a state equation for the base pressure.

In the Euler equation I wrote before, if one set rho= constant, T= constant, it would be no acceleration using the state equation since dp/dx=0, the only solution is u=constant ...

 March 21, 2015, 15:52 #71 Senior Member   Martin Hegedus Join Date: Feb 2011 Posts: 500 Rep Power: 19 I totally agree with you. And I did state earlier that whatever density one chooses must be compatible with the conservation of energy and the state equation. But I do hope that Jonas is stating something more interesting than everything is constant. Sort of boring. That's why I'm confused about this post. Which brings me to the next point, Jonas has basically put very little effort into explaining himself so it's probably best to move on. If he wants to state/analyze something that he has very little understanding of, he should go for it. It may provide me, along with others, free entertainment down the road which pays for the time taken discussing this.

March 22, 2015, 04:20
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 Originally Posted by momentumwaves To get flow moving & sustain it, without a forcing function, would see it eventually come to a halt. Very often, in simulations, we overlook how the inlet velocity is applied. Most often, the load is applied at the instant of simulation startup, & held there - in that sense it acts like Heaviside's step function at time t=0.
I think that a velocity inlet condition usually comes with a faulty initial condition (in which case it acts as a Heaviside's step function at t=0). If I have a Diriclet condition on the velocity I like to incorporate the velocity boundary condition in the pressure boundary condition instead and set the velocity boundary condition to a Neumann one. Tomato tomato, but it feels more 'logical' to apply a driving force to a flow rather than prescribing a velocity.

On a side note: I have seen many solutions, using velocity inlet, which ignores physics (flow is usually pressure driven and friction is a function of velocity).

 March 22, 2015, 04:39 #73 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,764 Rep Power: 71 ...why? I consider inlet velocity or Neumann pressur BC.s equivalent...Consider the decomposition (from Euler or NS is not a matter): vn+1 + Grad phi = v* with the constraint that Div vn+1 =0 Therefore Div Grad phi = Div v* and the elliptic equation admits a solution (apart a function of time) if the BC.s d phi/dn = n.( v*-vn+1) is prescribed. You see that the compatibility condition is satisfied and is totally equivalent to solve Div Grad phi = Div v_mod* d phi/dn = 0 In conclusion, you can prescribe a velocity inlet profile by means of the normal velocity to the boundary. As a consequence, the Neumann BC.s are implied anon_h likes this.

March 22, 2015, 05:22
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 Originally Posted by FMDenaro ...why? I consider inlet velocity or Neumann pressur BC.s equivalent...Consider the decomposition (from Euler or NS is not a matter): vn+1 + Grad phi = v* with the constraint that Div vn+1 =0 Therefore Div Grad phi = Div v* and the elliptic equation admits a solution (apart a function of time) if the BC.s d phi/dn = n.( v*-vn+1) is prescribed. You see that the compatibility condition is satisfied and is totally equivalent to solve Div Grad phi = Div v_mod* d phi/dn = 0 In conclusion, you can prescribe a velocity inlet profile by means of the normal velocity to the boundary. As a consequence, the Neumann BC.s are implied
Yes I agree (that's why I said tomatos tomatos ), it is just my personal preference. Also it boils down to what type of discretization you have performed as consistency in the discretization method usually is more important than prescribing a boundary condition based on PDE analysis alone.

March 22, 2015, 05:28
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Filippo Maria Denaro
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 Originally Posted by Simbelmynė Yes I agree (that's why I said tomatos tomatos ), it is just my personal preference. Also it boils down to what type of discretization you have performed as consistency in the discretization method usually is more important than prescribing a boundary condition based on PDE analysis alone.
yes, depending on the type of discretization for the pressure eqaution we can get an exact or approximate projection method, but that does not modify the physics implied by the PDE.

March 22, 2015, 21:55
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Jonas T. Holdeman, Jr.
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Quote:
 Originally Posted by Simbelmynė ...., but it feels more 'logical' to apply a driving force to a flow rather than prescribing a velocity. On a side note: I have seen many solutions, using velocity inlet, which ignores physics (flow is usually pressure driven and friction is a function of velocity).
I just got back tonight from my mathematics conference where the title for my talk was "On the Physics of Incompressible Fluids". Thanks for your help in getting ready guys. I haven't read posts since I left, but quick browsing turned up Simbelmynė's post. Not all problems can be expressed in terms of the pressure. Here is a problem I worked out about 8 years ago.

If I go down to Home Depot and look at ceiling fans for sale, I see they have data on blade diameter and flow rates (xxx CFM), but not pressure. This little 2D problem represents a ceiling fan in a small room with a box on the floor. The BCs are psi=0, u=0 on walls and other bounding surfaces, and delta psi at tips of blades to produce a specified flow rate. I show the streamlines in the attached figure. No attempt is made to model the blades.
Attached Images
 FanFlow.jpg (32.1 KB, 16 views)

 March 23, 2015, 00:47 #77 Senior Member   Martin Hegedus Join Date: Feb 2011 Posts: 500 Rep Power: 19 However, what a fan actually does is accelerate the flow and does not necessarily provide a given CFM. So what Simbelmynė's says is correct.

March 23, 2015, 04:53
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Filippo Maria Denaro
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 Originally Posted by Jonas Holdeman I just got back tonight from my mathematics conference where the title for my talk was "On the Physics of Incompressible Fluids". Thanks for your help in getting ready guys. I haven't read posts since I left, but quick browsing turned up Simbelmynė's post. Not all problems can be expressed in terms of the pressure. Here is a problem I worked out about 8 years ago. If I go down to Home Depot and look at ceiling fans for sale, I see they have data on blade diameter and flow rates (xxx CFM), but not pressure. This little 2D problem represents a ceiling fan in a small room with a box on the floor. The BCs are psi=0, u=0 on walls and other bounding surfaces, and delta psi at tips of blades to produce a specified flow rate. I show the streamlines in the attached figure. No attempt is made to model the blades.

I think you used a stream function-vorticity formulation which is still a counterpart of the Hodge decomposition. Now tangential velocity is prescribed.
However, the same problem can be solved in the velocity-pressure formulation, you fix the normal velocity component that provides the correct flow rate.
Just an observation, however... in you room the closed domain does not permit flow to go in/out from walls. Therefore, the fan must be introduced as an external force in the momentum equation to prevent continuity be unsatisfied. This force, if is not divergence-free acts as source term in the elliptic pressure equation

March 23, 2015, 09:25
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Jonas T. Holdeman, Jr.
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 Originally Posted by Martin Hegedus However, what a fan actually does is accelerate the flow and does not necessarily provide a given CFM. So what Simbelmynė's says is correct.
Fan, blower, pump ... whatever you call them, they all do the same thing. As Simbelmynė might say - tomatoes, tomatoes.

FMDenaro said: I think you used a stream function-vorticity formulation which is still a counterpart of the Hodge decomposition.

I used a pressure-free velocity equation, no stream function or vorticity in the formulation. Stream function appears in the boundary conditions. I used a simple mesh, structured except for the cutout for the box.

FMDenaro said: However, the same problem can be solved in the velocity-pressure formulation ...

My result is velocity-pressure, but minus the pressure. Look at what you would have to do with usual v-p methods. You would have to cut the mesh along a line joining the blade tips (an "O" type mesh?). You would apply periodic BC for the velocity on the cut to assure continuity of velocity. You would have to put in a pressure jump across the cut while otherwise assuring that the pressure continues. You would have to assume a pressure jump, calculate flow volume, adjust the pressure jump, calculate flow volume, iterating until you satisfied the specified flow.

FMDenaro said: in you room the closed domain does not permit flow to go in/out from walls. Therefore, the fan must be introduced as an external force in the momentum equation to prevent continuity be unsatisfied. This force, if is not divergence-free acts as source term in the elliptic pressure equation.

The driving force (fan) was introduced as a boundary condition on the stream function, so of course it is non-conservative and divergence-free.

Let me repeat the argument for my finite element method. Divergence-free velocity components are strongly correlated by virtue of being divergence-free, and if you try to correlate them after the fact with a pressure, you are unnecessarily spitting into the wind. If a vector field is divergence-free, then there necessarily exists a stream function (or velocity potential in 3D (tomatoes, tomatoes?)) such that the field is the curl of the SF. They are inseparable! I choose a scalar Hermite stream function where the derivative DOFs are the div-free velocity components. I take the curl of this element to get a div-free velocity element. If the SF element is continuous, then Un will vanish on element boundaries not containing the node on which the shape function is defined, assuring continuity of the normal component of the velocity across element boundaries. This element has the primitive variables u & v as DOFs, but we have replaced a scalar pressure with a scalar stream function.

Consider a patch containing all the elements sharing an internal node. The vector field on the patch is exactly div-free because it is the curl of a SF. The normal components of the vector field vanish on the boundaries of the patch. These are exactly the conditions for orthogonality with respect to vector functions which are the gradient of a scalar potential. Used as test functions, they are projectors for div-free fields.

March 23, 2015, 09:36
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 Originally Posted by Jonas Holdeman I just got back tonight from my mathematics conference where the title for my talk was "On the Physics of Incompressible Fluids". Thanks for your help in getting ready guys. I haven't read posts since I left, but quick browsing turned up Simbelmynė's post. Not all problems can be expressed in terms of the pressure. Here is a problem I worked out about 8 years ago. If I go down to Home Depot and look at ceiling fans for sale, I see they have data on blade diameter and flow rates (xxx CFM), but not pressure. This little 2D problem represents a ceiling fan in a small room with a box on the floor. The BCs are psi=0, u=0 on walls and other bounding surfaces, and delta psi at tips of blades to produce a specified flow rate. I show the streamlines in the attached figure. No attempt is made to model the blades.
Thanks for your input, I hope that the conference was successful. =)

With regards to the pressure I was talking about, I think I might have been a bit confusing. I basically stated two things. First is that I like to express the flow in terms of pressure - usually by modifying the pressure boundary condition to account for the velocity needed. My second point about the velocity inlet sometimes being problematic is very well illustrated by your case. If a fan company has given you a flow rate for a ceiling fan it is very very likely that they have calculated this flow rate based on the fan working against no external pressure (such as friction). If you go to home depot and obtain a channel fan (don't know if that's the proper English word - anyway a fan that you use for ventilation) then it is more likely that it has a fan curve specification.

Another way of thinking about it is to ask yourself where the extra energy will come from if you have a fan that is producing a constant flow rate regardless of the system friction.

Finally, if your comment is just about the validity of using a velocity boundary condition then I have no argument - velocity boundary condition works well in most cases

Last edited by Simbelmynė; March 23, 2015 at 09:43. Reason: Engrish.. ;)

 Tags action-at-a-distance, body force, incompressible fluid, physics