My question has 2 parts:

  1. I just followed the derivation of Navier Stokes (for Control Volume CFD analysis) and was able to understand most parts. However, the book I use (by Versteeg) does not derive it in its entirety. He pulls a lot of results directly from Schlichtling and continues his analysis. I want to understand the derivation in its full form. Is there any resource other than Schlichtling (My library doesn't have it) for deriving the NV equations in their full form? I would prefer an online (free) PDF or similar. (I'm not sure my library would have many books on this including the ones discussed here)

  2. After the derivation, most books follow it up with :

$$\underbrace{\frac{\partial (\rho\phi) }{\partial t}}_{\text{Rate of increase of }\phi} +\underbrace{\text{div}(\rho\phi \vec u)}_{\text{Convective Term}} = \underbrace{\text{div}(\Gamma \text{ grad}(\phi))}_{\text{Diffusive Term}} + \underbrace{S_\phi}_{\text{Source Term}}$$ where $\phi$ is property per unit mass, $\vec u$ is velocity vector and $\Gamma$ is diffusive term (Like viscosity or thermal conductivity).

What I can't understand is the use of the terms "convective" and "diffusive"? What do they mean? What is the physical interpretation of these terms?

Their dictionary meanings seem to exacerbate the situation:

(convection) the transfer of heat through a fluid (liquid or gas) caused by molecular motion.

(diffusion) The spreading of something more widely or the intermingling of substances by the natural movement of their particles

  • $\begingroup$ Good question! Unfortunately I'm not especially familiar with this branch of physics so I can't answer it, but I'll try to draw some attention to it. $\endgroup$ – David Z Mar 5 '12 at 20:05
  • $\begingroup$ What is $p$..? Did you see this... ? $\endgroup$ – Vijay Murthy Mar 6 '12 at 10:08
  • $\begingroup$ Whoops. That was $\rho$, edited. Wikipedia is far from complete. I want something that is entirely complete from mass continuity to energy conversation. $\endgroup$ – user7950 Mar 6 '12 at 10:10
  • $\begingroup$ You want something online and also include energy equations. I cant think of anything right now. But this is by a master. Besides, the structure of the equations remains the same for energy as those for mass and momentum. $\endgroup$ – Vijay Murthy Mar 6 '12 at 10:18

I don't know a good answer to your first question (I'd be interested in a good text for that myself), but I can answer the second.

It's easier to explain if we temporarily imagine $\phi$ represents the concentration of some dye made up of little particles suspended in the fluid. The convective term (aka advective term) is transport of $\phi$ due to the fact that the fluid is moving: a single "particle" of $\phi$ will tend move around according to the velocity of the fluid around it. The diffusive term, on the other hand, represents the fact that the dye tends to spread out, regardless of the motion of the fluid, because each particle is undergoing Brownian motion. So if you were moving along at the same velocity as the fluid you would see a small spot of dye tend to become more and more blurred over time.

For quantities like energy and momentum the diffusion happens for a slightly different reason (transfer of the quantity between fluid molecules when they collide) but the principle is the same. The property is transported along with the fluid's bulk velocity (convective term) but also tends to spread out and become blurred of its own accord (diffusive term).

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"The terms on the left hand side of the momentum equations are called the convection terms of the equations. Convection is a physical process that occurs in a flow of gas in which some property is transported by the ordered motion of the flow. The terms on the right hand side of the momentum equations that are multiplied by the inverse Reynolds number are called the diffusion terms. Diffusion is a physical process that occurs in a flow of gas in which some property is transported by the random motion of the molecules of the gas. "


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  • 2
    $\begingroup$ It may be better to phrase it in your own words. As you state it, it doesn't explain more than the explanation in the original question. $\endgroup$ – Bernhard Oct 17 '12 at 6:23

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