Navier Stokes Equation Derivation and Inertial Forces

[Red_CCF]

Hi

I was reading Introduction to Fluid Mechanics by Nakayama and Boucher and I got lost in their derivation of the Navier Stokes Theorem.

They basically started out with a differential of fluid with dimensions dx, dy, and b. Then they say that the force acting on it F = (F_x, F_y) is F_x = dx*dy*b*density*du/dt and F_y = dx*dy*b*density*dv/dt where u is the velocity in x direction and v the velocity in y direction. Then the author says that F is the "inertial force which is the mass times the acceleration".

I don't get this statement, because I thought that inertial force only existed in non-inertial frames (which they used as such in earlier chapters) and they didn't specify that they were doing the derivation in non-inertial frame (all the stuff I said above was basically the first four sentence/equations in the derivation).

If someone could help me I would really appreciate it.

Thanks

 

[Waldheri]

Not sure if I understand exactly what you mean, but if a volume element is size b*dx*dy (= dV), then mass m = rho*dV. Acceleration is nothing but the change in velocity, so ax = du/dx. A force Fx = m*ax = b*dx*dy*du/dx.

Perhaps it is good to keep in mind that the Navier-Stokes equation is nothing but a specific form of the conservation of momentum. If you know the Reynolds transport theorem, it is probably easier to just write out the integral of momentum over a domain, and set the time-derivative of that integral to zero (no change in momentum: momentum is conserved). This is basically what you see on the Wikipedia page.

 

[Red_CCF]

Not sure if I understand exactly what you mean, but if a volume element is size b*dx*dy (= dV), then mass m = rho*dV. Acceleration is nothing but the change in velocity, so ax = du/dx. A force Fx = m*ax = b*dx*dy*du/dx.

Hi

Thanks for the response. What I'm confused about is basically the author's use of the term inertial force (which he defined as m*a) to describe the force on the element. There was no mention or indication that the derivation was done in a non-inertial frame.

Thanks.

 

[aeroguy77]

The Reynolds transport theorem is the derivative of a blob of fluid taken along a path which is moving with a certain velocity. I can present an analogy. If a plane is flying over a country side, that plane is said to be in a inertial frame with respect to an observer on the ground. The blob of fluid in Reynolds transport theorem is analogous to an observer on the wing of the plane. It is sometimes called a material derivative. Hope that helps.

 

[Red_CCF]

The Reynolds transport theorem is the derivative of a blob of fluid taken along a path which is moving with a certain velocity. I can present an analogy. If a plane is flying over a country side, that plane is said to be in a inertial frame with respect to an observer on the ground. The blob of fluid in Reynolds transport theorem is analogous to an observer on the wing of the plane. It is sometimes called a material derivative. Hope that helps.

Hi, thanks for the response

So it's typical that the navier stokes equations be derived from an inertial frame and would it be correct if I assume the authors is using inertial forces in the same context as just a normal force (since they defined it as m*a)?

Thanks

 

[aeroguy77]

The Navier Stokes equations is the application of Newton's second law to a fluid. Basically, it is the conservation of momentum. The NS equations can be written for an inertial reference frame:

rho*(dv/dt+v*grad(v)) = -grad(P)+grad(T)+f

the terms on the LHS represent the inertial and non-inertial forces on the fluid blob (the second term is non-linear, also note that a solution of the NS equations is a velocity field). The RHS breaks up the inertial and non-inertial forces into a pressure gradient, a stress tensor and any additional body forces. The NS equations can also be written as a material derivative. It takes the form:

rho*Dv/Dt = -grad(P)+grad(T)+f

the term Dv/Dt represents acceleration and combines time dependent and convective effects. Convective acceleration (the term v*grad(v)) is a spatial effect. This can be diverging or converging nozzles, the Coriolis effect of the earth, etc. etc.

I hope this helps!

 

[Red_CCF]

The Navier Stokes equations is the application of Newton's second law to a fluid. Basically, it is the conservation of momentum. The NS equations can be written for an inertial reference frame:

rho*(dv/dt+v*grad(v)) = -grad(P)+grad(T)+f

the terms on the LHS represent the inertial and non-inertial forces on the fluid blob (the second term is non-linear, also note that a solution of the NS equations is a velocity field). The RHS breaks up the inertial and non-inertial forces into a pressure gradient, a stress tensor and any additional body forces.

Hi

I was wondering, if the above equation is derived from an inertial frame, how can there exist non-inertial forces in the equation?

Thank you

 

[boneh3ad]

The Navier-Stokes equations are derived from a infinitesimal fluid element in a non-inertial frame typically called the Lagrangian frame. In other words, your frame is following the fluid element as it goes through some arbitrary motion (translation and/or deformation) in an implied inertial frame.

 

[aeroguy77]

The convective acceleration can be converted from a non-inertial FoR to an inertial one. If the analysis is to be done in an inertial FoR, than the force caused by convective acceleration would be replaced by an equivalent force as seen from that reference point. So to completely answer your question, non-inertial forces (or fictitious forces) cannot be present in a inertial FoR, by definition. However, as stated before, a fictitious force can be seen as an inertial force by a fixed observer. For example: if you have a steady flow through a nozzle, the flow is being accelerated w.r.t. position along the nozzle. The forces would be represented as inertial forces acting on the inlet and the outlet of the nozzle.

 

[Red_CCF]

The convective acceleration can be converted from a non-inertial FoR to an inertial one. If the analysis is to be done in an inertial FoR, than the force caused by convective acceleration would be replaced by an equivalent force as seen from that reference point. So to completely answer your question, non-inertial forces (or fictitious forces) cannot be present in a inertial FoR, by definition. However, as stated before, a fictitious force can be seen as an inertial force by a fixed observer. For example: if you have a steady flow through a nozzle, the flow is being accelerated w.r.t. position along the nozzle. The forces would be represented as inertial forces acting on the inlet and the outlet of the nozzle.

Hi, thanks for your response

I think my main confusion lies with the terminology and not the actual derivation, as the book I'm reading doesn't specify the FoR, I have to figure it out from its wording.

When you use the term "inertial force", do you mean a real force (in an inertial frame)? Is this common? Because I have always found the term to be used as a synonym to fictitious forces. Although the author of the book used inertial force as a synonym to fictitious forces (as I'm used to), he begins the N-S derivation by stating that the external forces acting on a fluid body is m*a, which he defined to be "inertial forces"; I personally this is a mis-print or he is changing his definition the term but I'm not 100% sure?

Thank you

 

[boneh3ad]

Hi, thanks for your response

I think my main confusion lies with the terminology and not the actual derivation, as the book I'm reading doesn't specify the FoR, I have to figure it out from its wording.

When you use the term "inertial force", do you mean a real force (in an inertial frame)? Is this common? Because I have always found the term to be used as a synonym to fictitious forces. Although the author of the book used inertial force as a synonym to fictitious forces (as I'm used to), he begins the N-S derivation by stating that the external forces acting on a fluid body is m*a, which he defined to be "inertial forces"; I personally this is a mis-print or he is changing his definition the term but I'm not 100% sure?

Thank you

Like I said, the Lagrangian frame in which you are deriving the Navier-Stokes equations is not inertial. That should solve your issue.

 

[aeroguy77]

Sorry about all the confusion. Let me try one more approach. If you are in a car and you slam on the brakes, your body tends to be pushed forward. When you analyze this situation from your point of view, you find that your motion forward is caused by no apparent force. This is called a fictitious force. At this point, since you are accelerating, you are in a non-inertial FoR. If a by stander was watching you, as your body flung forward, he would note that there is no force being applied, and that you simply continued forward due to Newton's simple law. As applied to N.S., a fluid blob experiences inertial forces due to the fluctuations (or small accelerations) caused in every direction by the law of inertia in a inertial FoR.

 

[Red_CCF]

Like I said, the Lagrangian frame in which you are deriving the Navier-Stokes equations is not inertial. That should solve your issue.

My apologies, I missed your post. However, the author stated at the beginning of the book that he felt a Eulerian frame is "more common and effective" than Lagrangian but I may be reading too much into this. Is there another book with a relatively simple derivation that you recommend?

As applied to N.S., a fluid blob experiences inertial forces due to the fluctuations (or small accelerations) caused in every direction by the law of inertia in a inertial FoR.

So, in a horizontal pipe (pressure drop occurs from left to right), if I was the blob of fluid, how would I "feel" or "see"?

But basically, the form of the N-S equation is the same in both types of frames, except in the non-inertial there is fictitious forces lumped into the body forces and the coordinates are moving with the blob?

Thanks