When to add body forces, and when not?

lWiNdY · November 6, 2023, 10:45

Typically, the effect of a wind turbine on a flow (especially wake dynamics) is modeled by an external force in the momentum equation where the turbine is an actuator disk i.e.

$\rho\frac{\delta\vec{V}}{\delta t} + \rho(\vec{V}\cdot \nabla)\vec{V}= -\nabla p + \mu\nabla^2\vec{V} + \vec{f}$ where the wind turbine's effect is put in to the last term. Suppose a turbine is tilted, then it can be equivalent to a lift and drag term. Equivalently, there is a pressure and suction side on the actuator disk.

However, when solving for a wing or airfoil, it seems that the body forces are neglected. Think for example of the potential flow solutions. Still, in the solution there is a lift and drag as is in the wind turbine case. Again, there is a pressure and suction side.

Why is the lift of the airfoil not added to the momentum equation in the case of an airfoil?

I am aware that this is a rather odd question. Especially because the airfoil's surface sets up a pressure distribution which is seen as lift. Therefore, let me rephrase it to

'Why is the turbine modeled as an external force rather than surface distribution as with an airfoil/wing? '

Best,
L.

LuckyTran · November 6, 2023, 10:59

The keyword is external body force.

Lift and drag forces arise from integrating the forces acting on the wing, these forces arising from the flow inside the computational domain: namely, the pressure acting over the wing. The lift and drag force are not being neglected, they're already accounted for.

Draw your overall control volume (the fluid equivalent of a free body diagram) and then determine whether the forces you want to model are already present inside the domain or truly come from some other dark magic (i.e. an external body force).

An actuator disk is modeled as a external force because.... the computational domain is just an empty volume with no features inside it. The entire turbine is neglected. Hence, the influence of the turbine is instead modeled by (the actuator disk model) a magical fictitious force due to the turbine being fictitious. There is no turbine surface to compute the lift/drag force on the actuator disk, the actuator disk doesn't exist.

aerosayan · November 6, 2023, 11:05

In turbomachinery, the reference frame of motion is on the rotating part, and since the rotating part moves extremely fast, the body/coriolis/centrifugal forces have to be taken into account, because they make up a significant portion of the forces acting on the fluid.

I don't know for wind turbines, but it's true for compressors, impellers, etc.

Even for rotating helicopter blades, the body forces are important, because the different forces cause flow along the length of the blade from root to tip, or in the opposite direction, so, since these spanwise flows can change the lift/drag characteristic of the rotor, thus they're significant enough for the final result.

But for simpler bodies like aerofoils, majority of the forces are due to simple linear motion of the aerofoil through the air, and due to chordwise flow, so the other body forces are not as significant enough.

lWiNdY · November 6, 2023, 11:13

"An actuator disk is modeled as a external force because.... the computational domain is just an empty volume with no features inside it. The entire turbine is neglected. Hence, the influence of the turbine is instead modeled by (the actuator disk model) a magical fictitious force due to the turbine being fictitious. There is no turbine surface to compute the lift/drag force on the actuator disk, the actuator disk doesn't exist."

Allright, clear. Thus, when one would model the actuator disk with a finite thickness and the same properties for example by placing a porous frisbee in the domain, then one does not need to add those forces any more?

LuckyTran · November 6, 2023, 11:20

Quote:

Originally Posted by lWiNdY

Allright, clear. Thus, when one would model the actuator disk with a finite thickness and the same properties for example by placing a porous frisbee in the domain, then one does not need to add those forces any more?

I understand why you're struggling with this, I do. An actuator disk with finite thickness or a porous frisbee is modeled by the external force. The problem is you are using some software that highlights the actuator disk on your screen in pink cells and so you think it is present in the model. But it is a virtual model.

When you generate the grid... do you mesh around an actuator disk or do you end up with elementary cells: cubes, tets, etc? Do you mesh the pores in the frisbee or is it just plain cells? You have to actually mesh the features to reveal these natural forces.

The thickness of the actuator disk is adjusted by increasing the region where the external body force is applied. The porous zone adds a pressure gradient proportional to more models. Neither of these cases, actually has an actuator disk, or a porous frisbee in the domain. These models are made possible via implementing them as external body forces. Due to their popularity, if you say you have an actuator disk model, or a porous model, then everyone knows you are added external forces to the momentum equation. The question is how to know if you have an external body force if you have a jetpack-cat.

The easy way is to just plot the mesh. Can you see the jetpack-cat in the mesh?

lWiNdY · November 6, 2023, 11:26

Quote:

Originally Posted by LuckyTran

I understand why you're struggling with this, I do. An actuator disk with finite thickness or a porous frisbee is modeled by the external force. The problem is you are using some software that highlights the actuator disk on your screen in pink cells and so you think it is present in the model. But it is a virtual model.

Well, I think we must consider the goal of the simulation. That is in this case the 'far wake' of a wind turbine. In that case, it does not necessarily matter how the pressure is distributed exactly. It is just to find out what happens to a flow when a force is imposed on it. That is where actuator disk shows up.

Now, you do not really have to use the actuator disk. One can also mesh a poreous frisbee as if it were a wing. Even more so, one could create a full 3D windturbine where one resolves earth's boundary layer AND blade's boundary layer. In both cases, you do not need to add the external force, do you?

LuckyTran · November 6, 2023, 15:05

Btw a porous frisbee will not generate any thrust because the porous media model has only a pressure drop term.

You have to understand that the actuator disk model and porous media model are just fancy convoluted implementations for a source term in the momentum equation that someone else has coded up for you. You are asking why you need no external body force if you use an actuator disk... It is a paradoxical question, the actuator disk model is the external body force. And so is a porous frisbee. Consider that you do not use an actuator disk model or a porous model and write the source term yourself as a field function.

Take a step back and consider what your mesh looks like when you use an actuator disk or porous model. You have just basic shape (i.e. a box or a cylinder). Can you distinguish this mesh from a homogeneous flow in an ambient environment? The actuator disk and porous media are used to model the influence of these bodies that are not in your grid because the simulation needs some way of knowing that it is not a homogeneous ambient flow. When I grid the entire planetary surface with a wind turbine I include all the influences in my simulation and therefore I don't need external forces. If I mesh a box, delete the entire planet, delete the turbine, then I need to substitute the effects of the wind turbine that is not there and therefore I need to add them as a source/sink in my equation in order to model their effect. That is to say, if you plot just a mesh, it will be blatantly obvious that there is not a wind turbine in your simulation, and therefore you need extra sources/sinks to model it.

It is clear that there is no one way to model any problem and I am not asking you why you are not doing DNS of the entire universe dating back to the Big Bang. You model your problems however you like. My response is strictly explaining why and why not external forces miraculously appear. Again, consider the problem in classical mechanics of free body diagrams. Internal forces become exposed as external forces depending on what you define as your body. The same happens for control volumes.

lWiNdY · November 7, 2023, 03:16

Quote:

Originally Posted by LuckyTran

Btw a porous frisbee will not generate any thrust because the porous media model has only a pressure drop term.

I am not talking about the poreous media model, rather I am talking about meshing a porous disk as if it were an airfoil with a slat and flap. As long as this 'body' generates thrust i.e. drag in axial flow direction, and lift in the case of a tilted rotor, representing a wind turbine; it would be fine and I do not need any porous media model or external force on an actuator disk.

November 6, 2023, 10:45	When to add body forces, and when not?	#1
lWiNdY New Member L. Join Date: Jun 2022 Posts: 12 Rep Power: 4	Typically, the effect of a wind turbine on a flow (especially wake dynamics) is modeled by an external force in the momentum equation where the turbine is an actuator disk i.e. $\rho\frac{\delta\vec{V}}{\delta t} + \rho(\vec{V}\cdot \nabla)\vec{V}= -\nabla p + \mu\nabla^2\vec{V} + \vec{f}$ where the wind turbine's effect is put in to the last term. Suppose a turbine is tilted, then it can be equivalent to a lift and drag term. Equivalently, there is a pressure and suction side on the actuator disk. However, when solving for a wing or airfoil, it seems that the body forces are neglected. Think for example of the potential flow solutions. Still, in the solution there is a lift and drag as is in the wind turbine case. Again, there is a pressure and suction side. Why is the lift of the airfoil not added to the momentum equation in the case of an airfoil? I am aware that this is a rather odd question. Especially because the airfoil's surface sets up a pressure distribution which is seen as lift. Therefore, let me rephrase it to 'Why is the turbine modeled as an external force rather than surface distribution as with an airfoil/wing? ' Best, L.

November 6, 2023, 10:59		#2
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,747 Rep Power: 66	The keyword is external body force. Lift and drag forces arise from integrating the forces acting on the wing, these forces arising from the flow inside the computational domain: namely, the pressure acting over the wing. The lift and drag force are not being neglected, they're already accounted for. Draw your overall control volume (the fluid equivalent of a free body diagram) and then determine whether the forces you want to model are already present inside the domain or truly come from some other dark magic (i.e. an external body force). An actuator disk is modeled as a external force because.... the computational domain is just an empty volume with no features inside it. The entire turbine is neglected. Hence, the influence of the turbine is instead modeled by (the actuator disk model) a magical fictitious force due to the turbine being fictitious. There is no turbine surface to compute the lift/drag force on the actuator disk, the actuator disk doesn't exist. aerosayan likes this.

November 6, 2023, 11:05		#3
aerosayan Senior Member Sayan Bhattacharjee Join Date: Mar 2020 Posts: 495 Rep Power: 8	In turbomachinery, the reference frame of motion is on the rotating part, and since the rotating part moves extremely fast, the body/coriolis/centrifugal forces have to be taken into account, because they make up a significant portion of the forces acting on the fluid. I don't know for wind turbines, but it's true for compressors, impellers, etc. Even for rotating helicopter blades, the body forces are important, because the different forces cause flow along the length of the blade from root to tip, or in the opposite direction, so, since these spanwise flows can change the lift/drag characteristic of the rotor, thus they're *significant enough* for the final result. But for simpler bodies like aerofoils, majority of the forces are due to simple linear motion of the aerofoil through the air, and due to chordwise flow, so the other body forces are not as significant enough.

November 6, 2023, 11:13		#4
lWiNdY New Member L. Join Date: Jun 2022 Posts: 12 Rep Power: 4	"An actuator disk is modeled as a external force because.... the computational domain is just an empty volume with no features inside it. The entire turbine is neglected. Hence, the influence of the turbine is instead modeled by (the actuator disk model) a magical fictitious force due to the turbine being fictitious. There is no turbine surface to compute the lift/drag force on the actuator disk, the actuator disk doesn't exist." Allright, clear. Thus, when one would model the actuator disk with a finite thickness and the same properties for example by placing a porous frisbee in the domain, then one does not need to add those forces any more?

November 6, 2023, 15:05		#7
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,747 Rep Power: 66	Btw a porous frisbee will not generate any thrust because the porous media model has only a pressure drop term. You have to understand that the actuator disk model and porous media model are just fancy convoluted implementations for a source term in the momentum equation that someone else has coded up for you. You are asking why you need no external body force if you use an actuator disk... It is a paradoxical question, the actuator disk model is the external body force. And so is a porous frisbee. Consider that you do not use an actuator disk model or a porous model and write the source term yourself as a field function. Take a step back and consider what your mesh looks like when you use an actuator disk or porous model. You have just basic shape (i.e. a box or a cylinder). Can you distinguish this mesh from a homogeneous flow in an ambient environment? The actuator disk and porous media are used to model the influence of these bodies that are not in your grid because the simulation needs some way of knowing that it is not a homogeneous ambient flow. When I grid the entire planetary surface with a wind turbine I include all the influences in my simulation and therefore I don't need external forces. If I mesh a box, delete the entire planet, delete the turbine, then I need to substitute the effects of the wind turbine that is not there and therefore I need to add them as a source/sink in my equation in order to model their effect. That is to say, if you plot just a mesh, it will be blatantly obvious that there is not a wind turbine in your simulation, and therefore you need extra sources/sinks to model it. It is clear that there is no one way to model any problem and I am not asking you why you are not doing DNS of the entire universe dating back to the Big Bang. You model your problems however you like. My response is strictly explaining why and why not external forces miraculously appear. Again, consider the problem in classical mechanics of free body diagrams. Internal forces become exposed as external forces depending on what you define as your body. The same happens for control volumes.