Inertial Force in Fluid Mechanics

  • #1
person123
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According to one explanation, the left hand acceleration terms of Navier Stokes equations are the called the inertial terms. If you were to balance forces on the fluid particle, they would have to be equal and opposite to the forces on the right hand side (pressure gradient, viscous, and body). To me it seems like you're in the non-inertial reference frame of the fluid particle and treating the inertial force as a fictitious force in this reference frame. Is this at all accurate?

Another explanation I thought I heard is that the inertial force is due to dynamic pressure only. Is this at all accurate (is it an approximation of the previous description)?

On a slight tangent, when finding Reynold's number, the inertial term is ##\rho V^2##. If all fluid particles of a dense fluid were moving at a high but constant velocity, this term should be very large. However, the acceleration is 0, and so shouldn't the inertial force be 0?
 
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  • #2
Arjan82
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I have a little bit of trouble interpreting your question, but here is a shot.

According to one explanation, the left hand acceleration terms of Navier Stokes equations are the called the inertial terms. If you were to balance forces on the fluid particle, they would have to be equal and opposite to the forces on the right hand side (pressure gradient, viscous, and body). To me it seems like you're in the non-inertial reference frame of the fluid particle and treating the inertial force as a fictitious force in this reference frame. Is this at all accurate?

Navier stokes states indeed that the mass times acceleration of a fluid parcel is equal to the gradient of the (static) pressure, the viscous forces and the body forces. This is very accurate indeed.

Whether you choose to solve these Navier-Stokes equation in a inertial or non-intertial frame of reference is a choice which is unrelated, both is possible in principle. Either choice can be solved accurately in principle. But the Navier-Stokes equations look different if you describe them in a non-inertial reference frame, for which case you indeed need to add these fictitious forces due the acceleration of the reference frame. These fictitious forces are however unrelated to the real forces that affect the motion of the parcel.

Does this answer your question?

Another explanation I thought I heard is that the inertial force is due to dynamic pressure only. Is this at all accurate (is it an approximation of the previous description)?

The inertial force is a fictitious force equal and opposite to the force applied to a fluid parcel to change its velocity. The dynamic pressure is just the kinetic energy per unit volume. I don't see how the inertial force can be 'due to' a dynamic pressure; a flow parcel only reacts to a static pressure gradient (or a viscous or body force...).

So, in short, I'm struggling with the question a bit.

On a slight tangent, when finding Reynold's number, the inertial term is ρρV2. If all fluid particles of a dense fluid were moving at a high but constant velocity, this term should be very large. However, the acceleration is 0, and so shouldn't the inertial force be 0?

Yes, if you have a dense fluid moving at a high constant speed, there are no inertial forces. But that's not the point of the Reynolds number. This number gives the ratio of the inertial forces to the viscous forces to characterize a flow.

A high Reynolds number means that inertial forces are more important in the dynamics of the flow than viscous ones if you were to perturb the flow (there should be an asterisk attached to this remark). This would mean that you can neglect viscous forces and describe the problem with potential flow theory (more asterisks to be attached here). This would mean your problem is easier to solve, you don't need to solve the full Navier-Stokes equations. On the other side, for Reynolds numbers much lower than one, the inertial forces are not important anymore and you can just use the Stokes equations to solve your problem.

In this case the ##\rho V^2## term is just a measure of the inertial forces to be expected.
 
  • #3
person123
307
45
Whether you choose to solve these Navier-Stokes equation in a inertial or non-intertial frame of reference is a choice which is unrelated, both is possible in principle. Either choice can be solved accurately in principle. But the Navier-Stokes equations look different if you describe them in a non-inertial reference frame, for which case you indeed need to add these fictitious forces due the acceleration of the reference frame. These fictitious forces are however unrelated to the real forces that affect the motion of the parcel.

Does this answer your question?
Yes, it essentially does, although I'm not fully clear on the non-inertial reference frames. I was getting at the point that the inertial force is a fictitious force; I heard that it wasn't but you wrote that it was (maybe it's somewhat ambiguous). I was trying to say that it's not an inertial force in the same way the coriolis force is for example, which appears in a rotating frame of reference. Instead it would simply appear if you were in the frame of reference of an accelerating fluid particle; in your frame of reference there would be a force equal and opposite to the force applied on the parcel.

The inertial force is a fictitious force equal and opposite to the force applied to a fluid parcel to change its velocity. The dynamic pressure is just the kinetic energy per unit volume. I don't see how the inertial force can be 'due to' a dynamic pressure; a flow parcel only reacts to a static pressure gradient (or a viscous or body force...).

So, in short, I'm struggling with the question a bit.
I'm not entirely sure; I think this was what my professor told me, but there may have been some confusion over that.

A high Reynolds number means that inertial forces are more important in the dynamics of the flow than viscous ones if you were to perturb the flow (there should be an asterisk attached to this remark). This would mean that you can neglect viscous forces and describe the problem with potential flow theory (more asterisks to be attached here). This would mean your problem is easier to solve, you don't need to solve the full Navier-Stokes equations. On the other side, for Reynolds numbers much lower than one, the inertial forces are not important anymore and you can just use the Stokes equations to solve your problem.
Thank you, this clarifies it. So in the case I gave, if the fluid were to be perturbed, inertial forces would likely be more significant than viscous forces; the characteristic of the flow in general is one in which inertial forces dominate.
 
  • #4
Arjan82
411
305
Yes, it essentially does, although I'm not fully clear on the non-inertial reference frames. I was getting at the point that the inertial force is a fictitious force; I heard that it wasn't but you wrote that it was (maybe it's somewhat ambiguous). I was trying to say that it's not an inertial force in the same way the coriolis force is for example, which appears in a rotating frame of reference. Instead it would simply appear if you were in the frame of reference of an accelerating fluid particle; in your frame of reference there would be a force equal and opposite to the force applied on the parcel.

The forces in an non-inertial reference frame are fictitious in a different way than the inertial forces are fictitious. The first forces effectively cancel the acceleration due to the non-inertial reference frame such that the acceleration of the reference frame is deleted from your dynamics again. It is more of an administrative thing you could say.

There is an argument to be made that inertial forces are not really fictitious. It is the resistance of mass to movement, and you do actually feel a countering force when trying to accelerate a mass.
 
  • #5
person123
307
45
I think that makes sense now. Thank you very much for your responses.

The inertial force is a fictitious force equal and opposite to the force applied to a fluid parcel to change its velocity. The dynamic pressure is just the kinetic energy per unit volume. I don't see how the inertial force can be 'due to' a dynamic pressure; a flow parcel only reacts to a static pressure gradient (or a viscous or body force...).

So, in short, I'm struggling with the question a bit.
I want to come back to this because I think I might know what was meant by inertial force being from dynamic pressure. The rate of change of momentum due to an outflux of flow is ##\rho A v^2##. This could be considered a dynamic pressure in that the velocity of the fluid induces pressure. However, the rate of change of momentum ##\frac{dp}{dt}## is the inertial force. Does that make sense at all?
 
  • #6
pasmith
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Summary:: I've both heard that the inertial force is the force just from dynamic pressure and read that it's simply the term due to fluid acceleration, which I would imagine would incorporate all forces.

According to one explanation, the left hand acceleration terms of Navier Stokes equations are the called the inertial terms. If you were to balance forces on the fluid particle, they would have to be equal and opposite to the forces on the right hand side (pressure gradient, viscous, and body). To me it seems like you're in the non-inertial reference frame of the fluid particle and treating the inertial force as a fictitious force in this reference frame. Is this at all accurate?

The reference frame is generally the frame in which the spatial coordinates are defined. But there are two choices for relating those coordinates to the flow field, and partial differentiation with respect to time means different things in each of them.

In the Eulerian description the flow field is [itex]\mathbf(\mathbf{x},t)[/itex] and the spatial coordinates [itex]\mathbf{x}[/itex] are fixed in space and fluid parcels move past them. Partial differentiation with respect to time is differentiation at a fixed point in space, and so [itex]\left(\dfrac{\partial \mathbf{u}}{\partial t}\right)_{\mathbf{x}}[/itex] doesn't give he acceleration of a particular fluid parcel.

In the Lagrangian description, the flow field is [itex]\mathbf{v}(\mathbf{X},t)[/itex] and the spatial coordinates [itex]\mathbf{X}[/itex] labels the particular fluid particle which was initially at that position. Thus [itex]\left(\dfrac{\partial \mathbf{v}}{\partial t}\right)_{\mathbf{X}}[/itex] does give the acceleration of a fluid parcel, and if the frame of reference is inertial then there are no "ficticious forces". But partial differentiation with respect to time doesn't tell you about a fixed point in space, because again the fluid parcel has moved.

The two descriptions are related by noting that the fluid parcel [itex]\mathbf{X}[/itex] is at time [itex]t[/itex] at position [tex]
\mathbf{x} = \mathbf{X} + \int_0^t \mathbf{v}(\mathbf{X},t)\,dt[/tex] and by the chain rule [tex]
\left(\frac{\partial}{\partial t}\right)_{\mathbf{X}} = \left(\frac{\partial}{\partial t}\right)_{\mathbf{x}}
+ \mathbf{u} \cdot \nabla[/tex] which is where the "inertial force" in the Eulerian description comes from.

Alternatively, in the Eulerian description you can consider conservation of momentum within a fixed volume. This momentum changes both due to forces acting on the volume and due to fluid parcels moving into or out of the volume. Now the flux of the [itex]i[/itex] component of mementum is [itex]\rho u_i \mathbf{u}[/itex] and taking the divergence and simplifying using the mass conservation equation yields [tex]
\frac{\partial}{\partial t}(\rho u_i) + \nabla \cdot (\rho u_i \mathbf{u}) = \rho \left(\frac{\partial u_i}{\partial t} + \mathbf{u} \cdot \nabla u_i\right).[/tex]

Does that make the [itex]\mathbf{u} \cdot \nabla \mathbf{u}[/itex] term a ficticious force? Depending on how you derive it it's either a component of acceleration or a term representing momentum flux, and is present in inertial frames.
 

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