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March 21, 2015, 12:49 

#61  
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Filippo Maria Denaro
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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 pathline). 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, 13:05 

#62 
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Martin Hegedus
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Sorry, I truly need to end it here. Best of luck to everyone!!


March 21, 2015, 13:10 

#63 
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Martin Hegedus
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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 1D 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, 13:14 

#64 
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Martin Hegedus
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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, 13:26 

#65  
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Filippo Maria Denaro
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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, 14:52 

#66 
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Martin Hegedus
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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, 14:57 

#67 
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Martin Hegedus
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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, 15:05 

#68 
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Martin Hegedus
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BTW, I'll admit that I can not grasp what the equation of state for an incompressible material is.


March 21, 2015, 15:36 

#69 
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Martin Hegedus
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Or, another way to approach this, in the 1D 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, 15:37 

#70  
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Filippo Maria Denaro
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see at the end of http://www.cfdonline.com/Wiki/Incompressible_flow specific formulation for lowMach 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, 16:52 

#71 
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Martin Hegedus
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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, 05:20 

#72  
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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, 05:39 

#73 
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Filippo Maria Denaro
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...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 

March 22, 2015, 06:22 

#74  
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March 22, 2015, 06:28 

#75  
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Filippo Maria Denaro
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March 22, 2015, 22:55 

#76  
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Jonas T. Holdeman, Jr.
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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. 

March 23, 2015, 01:47 

#77 
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Martin Hegedus
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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, 05:53 

#78  
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Filippo Maria Denaro
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I think you used a stream functionvorticity formulation which is still a counterpart of the Hodge decomposition. Now tangential velocity is prescribed. However, the same problem can be solved in the velocitypressure 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 divergencefree acts as source term in the elliptic pressure equation 

March 23, 2015, 10:25 

#79  
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Jonas T. Holdeman, Jr.
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FMDenaro said: I think you used a stream functionvorticity formulation which is still a counterpart of the Hodge decomposition. I used a pressurefree 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 velocitypressure formulation ... My result is velocitypressure, but minus the pressure. Look at what you would have to do with usual vp 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 divergencefree 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 nonconservative and divergencefree. Let me repeat the argument for my finite element method. Divergencefree velocity components are strongly correlated by virtue of being divergencefree, 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 divergencefree, 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 divfree velocity components. I take the curl of this element to get a divfree 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 divfree 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 divfree fields. 

March 23, 2015, 10:36 

#80  
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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 10:43. Reason: Engrish.. ;) 

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actionatadistance, body force, incompressible fluid, physics 
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