# Convection between two plates

Tags:
1. Nov 12, 2014

### unscientific

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. The attempt at a solution

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

Starting from the navier-stokes equation:

$$\rho \frac{D \vec u}{D t} = -\nabla p + \mu \nabla^2 \vec u + \vec F$$

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,

$$\vec F = -(\rho_0 + \Delta \rho)\nabla \phi = -\nabla(\rho_0 \phi) + \Delta \rho \vec g$$

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

$$\rho_0 \frac{D\vec u}{D t} = -\nabla P + \mu \nabla^2 \vec u + \Delta \rho \vec g$$

2. Nov 12, 2014

### Staff: Mentor

You haven't told us what your problem is.

Chet

3. Nov 13, 2014

### unscientific

Equations 9.5 from the lecture notes and eqn from the book doesn't match

4. Nov 13, 2014

### Staff: Mentor

What it is about them that you feel doesn't match?

Chet

5. Nov 13, 2014

### unscientific

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

$$\rho_0 \frac{D\vec u}{D t} = -\nabla P + \mu \nabla^2 \vec u + \Delta \rho \vec g$$
$$\rho_0 \frac{D\vec u}{D t} = - \frac{\partial}{\partial z}(p + rho_0 \phi) + \mu \nabla^2 \vec u + \Delta \rho \vec g$$
$$\rho_0 \frac{D\vec u}{D t} = - \frac{\partial}{\partial z}(p - rho_0 z \vec g) + \mu \nabla^2 \vec u + \Delta \rho \vec g$$
$$\rho_0 \frac{D\vec u}{D t} = - \frac{\partial p}{\partial z} + \rho_0 \vec g + \mu \nabla^2 \vec u + (\rho - \rho_0) \vec g$$
$$\rho_0 \frac{D\vec u}{D t} = - \frac{\partial p}{\partial z} + \mu \nabla^2 \vec u + \rho \vec g$$
$$\frac{D\vec u}{D t} = - \frac{1}{\rho_0} \frac{\partial p}{\partial z} + \frac{1}{\rho_0}\mu \nabla^2 \vec u + \frac{\rho}{\rho_0} \vec g$$

6. Nov 13, 2014

### Staff: Mentor

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

7. Nov 13, 2014

### unscientific

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 $\rho_0\phi$.

8. Nov 13, 2014

### Staff: Mentor

The derivative of $\rho_0g$ is zero in the horizontal direction.

9. Nov 13, 2014

### unscientific

Ah that's true. Quite annoying when the lecture notes don't specify the derivation, but this makes sense! Thanks alot.

10. Nov 15, 2014