How Does Gravity Affect Convection Between Two Plates?

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Homework Statement



From my lecture notes, here are the equations for convection between two plates. I have derived equations 9.6, 9.7 and 9.8. But for 9.4 there's a problem when gravity becomes involved.

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Homework Equations



Navier stokes: ## \rho \frac{D \vec u}{D t} = -\nabla p + \mu \nabla^2 \vec u + \vec F ##

The Attempt at a Solution


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However, I was reading through Tritton's book on flows where he detailed the derivation:

Starting from the navier-stokes equation:

[tex]\rho \frac{D \vec u}{D t} = -\nabla p + \mu \nabla^2 \vec u + \vec F[/tex]

where ##\vec F## represents contribution of other forces (such as gravity).

They then begin to define ##\vec F##:

By letting density vary, we have ##\rho = \rho_0 + \Delta \rho##. Gravitational acceleration can be defined through a potential: ##\vec g = -\nabla \phi = -\nabla gz##. Thus,

[tex]\vec F = -(\rho_0 + \Delta \rho)\nabla \phi = -\nabla(\rho_0 \phi) + \Delta \rho \vec g[/tex]

Introducing ##P = p + \rho_0 \phi##, navier stokes becomes:

[tex]\rho_0 \frac{D\vec u}{D t} = -\nabla P + \mu \nabla^2 \vec u + \Delta \rho \vec g[/tex]
 
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Chestermiller said:
You haven't told us what your problem is.

Chet

Equations 9.5 from the lecture notes and eqn from the book doesn't match
 
Chestermiller said:
What it is about them that you feel doesn't match?

Chet

Substituting ##P## inside and changing ##\nabla## to ##\frac{\partial}{\partial z}##, it gives:

[tex]\rho_0 \frac{D\vec u}{D t} = -\nabla P + \mu \nabla^2 \vec u + \Delta \rho \vec g[/tex]
[tex]\rho_0 \frac{D\vec u}{D t} = - \frac{\partial}{\partial z}(p + rho_0 \phi) + \mu \nabla^2 \vec u + \Delta \rho \vec g[/tex]
[tex]\rho_0 \frac{D\vec u}{D t} = - \frac{\partial}{\partial z}(p - rho_0 z \vec g) + \mu \nabla^2 \vec u + \Delta \rho \vec g[/tex]
[tex]\rho_0 \frac{D\vec u}{D t} = - \frac{\partial p}{\partial z} + \rho_0 \vec g + \mu \nabla^2 \vec u + (\rho - \rho_0) \vec g[/tex]
[tex]\rho_0 \frac{D\vec u}{D t} = - \frac{\partial p}{\partial z} + \mu \nabla^2 \vec u + \rho \vec g[/tex]
[tex]\frac{D\vec u}{D t} = - \frac{1}{\rho_0} \frac{\partial p}{\partial z} + \frac{1}{\rho_0}\mu \nabla^2 \vec u + \frac{\rho}{\rho_0} \vec g[/tex]
 
Chestermiller said:
It appears that the p's in Eqns. 9 are what you are calling P. The ρ0g has apparently been absorbed into the pressure term.

Chet

I don't think that's right, as applying the same equation in the horizontal direction (w) gives eqn 9.4. The small ##p## in eqn 9.4 should not include ##\rho_0\phi##.
 
Chestermiller said:
The derivative of ##\rho_0g## is zero in the horizontal direction.

Ah that's true. Quite annoying when the lecture notes don't specify the derivation, but this makes sense! Thanks a lot.
 
Got the answer, thanks a lot!
 
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