Fluid mechanics: intuition about the 'convective' term

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Hi,
I am doing an introductory course on fluid mechanics, and I'd like some intuition (that's why I'm posting on the engineering forums and not on the math forums, even if I'm studying for a degree in math) about the concept of the total derivative and, particularly, its convective component.

The total derivative of a vector field is defined as:
[tex]\frac {D \bf u}{Dt}= \frac {\partial u} {\partial t} + ({\bf u} \cdot \nabla){\bf u}[/tex]where the first term is the 'local' acceleration and the second the 'convective' acceleration.

I have no problem (I think) with the purely symbolic manipulation on the second term,
[tex]({\bf u} \cdot \nabla){\bf u} = u_1 \frac {\partial u_1}{\partial x} + u_2 \frac {\partial u_2}{\partial y} + u_3 \frac {\partial u_3}{\partial z}[/tex]where [itex]{\bf u} = u_1 {\bf i} + u_2 {\bf j} + u_3 {\bf k}[/itex] is the velocity vector. I just need to wrap my head around about the physical interpretation of it. The explanation of it being the 'acceleration following the motion' is a bit too vague, and I was wishing for some intuition about it.

A similar operation appears also applied to scalar fields, as in the continuity equation for (possibly compressible) flows,
[tex]\frac {\partial \rho} {\partial t} + ({\bf \rho} \cdot \nabla){\bf u} = 0[/tex]where [itex]\rho[/itex] is the density field, and which reduces to [itex]\mbox{div } {\bf u} = 0[/itex] if the flow is incompressible (constant [itex]\rho[/itex]). I believe I have a half-decent understanding of what 'divergence' means (in this context, if no sources or sinks are in the region, a compression*** on x has to be compensated by a decompression*** on y, in a two-dimensional example, just as a consequence of conservation of mass), and I hoped for a similar intuition for the 'convective' term, in both the scalar and vector field contexts.

Thanks a million if you have read so far! :)

*** P.S.: sorry, 'compression' and 'decompression' when talking about divergence is a poor choice of words; I meant 'change in speed' rather than 'change in density', since I was referring at that point to the continuity equation for an incompressible flow.
 
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Answers and Replies

  • #2
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Apologies for the terribly misquoted equation at the end. I'm trying to edit the above post, but apparently I can't after a while.

Please ignore the second half of the above post, from "A similar operation..." onwards.

Also, a boldface is probably missing in the first equation. Doh!

I got a very satisfactory answer from another person, but any additional insight is most welcome.
 
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  • #3
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Hi,
I am doing an introductory course on fluid mechanics, and I'd like some intuition (that's why I'm posting on the engineering forums and not on the math forums, even if I'm studying for a degree in math) about the concept of the total derivative and, particularly, its convective component.

The total derivative of a vector field is defined as:
[tex]\frac {D \bf u}{Dt}= \frac {\partial u} {\partial t} + ({\bf u} \cdot \nabla){\bf u}[/tex]where the first term is the 'local' acceleration and the second the 'convective' acceleration.

I have no problem (I think) with the purely symbolic manipulation on the second term,
[tex]({\bf u} \cdot \nabla){\bf u} = u_1 \frac {\partial u_1}{\partial x} + u_2 \frac {\partial u_2}{\partial y} + u_3 \frac {\partial u_3}{\partial z}[/tex]where [itex]{\bf u} = u_1 {\bf i} + u_2 {\bf j} + u_3 {\bf k}[/itex] is the velocity vector.
This is not correct. Try again.

Consider the case where the flow is steady state. So the acceleration of each fluid particle is given by:

$$\vec{a}=\frac{du_x}{dt}\vec{i}_x+\frac{du_y}{dt}\vec{i}_y+\frac{du_z}{dt}\vec{i}_z$$

But, by the chain rule
$$\frac{du_x}{dt}=\frac{\partial u_x}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u_x}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial u_x}{\partial z}\frac{\partial z}{\partial t}$$

Chet
 
  • #4
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Hi, Chet,
thanks for the input - I suspect I had in mind a scalar field all along and, when it comes to vector fields, I need to move back a few squares and review my vector calculus.

Thanks again!
 
  • #5
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Hi dodo,

The advective derivative, [itex]\vec u \cdot \nabla [/itex], is a tricky beast.

Higher level classes often use several 1-D modes to shed insight into dynamics modeled by this operator. If you're curious I highly suggest googling these models. There is a wealth of information out there. Also these models are often used in introductory numerical methods classes, and I bet you can find some simulations of these models to help visualize the dynamics.

In 1-D the advective derivative is simply [itex]u_x \frac{\partial}{\partial x} [/itex], where [itex]u_x [/itex] is the flow in the x direction.

The first model is the 1st order wave equation: [itex] \frac{\partial f}{\partial t}+c\frac{\partial f}{\partial x}=0[/itex], where [itex]c[/itex] is a constant flow velocity, and [itex] f[/itex] could be a number of quantities like density, temperature, or flow in the y direction. When you solve this equation you find that your initial profile is just shifted in time at a velocity [itex]c[/itex].

The second and third models and the inviscid Burgers equation: [itex] \frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=0[/itex], and the viscous Burger's equation: [itex] \frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=\nu \frac{\partial^2 u}{\partial x^2} [/itex]. The difference between these two models is that the viscous one includes dissipation.
 
  • #6
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Thanks for the advice, Wolfman, I'll give simpler 1-D models a look.
 

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