Navier Stoke Problem

1. Jul 8, 2012

iloc86

hi every body, im doing an analysis of a paper about a tool used in oil well drilling, they use 3D navier stoke equation and by lubrication aproximation the get this

-(1/r) ∂p/∂θ + k (∂( 1/r ∂(rw)/∂r )/∂r )=0

its necesary to get w so i use integral but i obtain an irreal i used mathematica wolfram. but in the paper they obtain

w= 1/2k ∂p/∂θ (ln r - 1/2) + C1 1/2 r + C2/r

somebody could confirm this

2. Jul 8, 2012

iloc86

sorry imaginary

3. Jul 9, 2012

JJacquelin

-(1/r) ∂p/∂θ + k (∂( 1/r ∂(rw)/∂r )/∂r )=0
∂( 1/r ∂(rw)/∂r )/∂r = (1/k) (∂p/∂θ) (1/r)
1/r ∂(rw)/∂r = (1/k) (∂p/∂θ) ln(r) + C1
∂(rw)/∂r = (1/k) (∂p/∂θ) r ln(r) + C1 r
(rw) = (1/k) (∂p/∂θ) [(r²/2) ln(r)-r²/4] + C1 r²/2 + C2
w = (1/2k) (∂p/∂θ) [r ln(r)-(r/2)] + C1 r/2 + C2/r

4. Jul 9, 2012

jackmell

I have issues with this. In particular, you have:

$$\int \left(\frac{1}{k} \frac{\partial p}{\partial \theta} \frac{1}{r}\right)dr=\int \partial\left(\frac{1}{r}\frac{\partial}{\partial r}(rw)\right)$$

I do not see how you can integrate the left side with respect to r and get:

$$\int \left(\frac{1}{k} \frac{\partial p}{\partial \theta} \frac{1}{r}\right)dr=\frac{1}{k}\frac{\partial p}{\partial \theta} \ln(r)+c$$

You're assuming $\frac{\partial p}{\partial \theta}$ is not a function of r and I do not think you can assume this. Also, wouldn't the constant of integration be an arbitrary function of theta in this case?

Last edited: Jul 9, 2012
5. Jul 9, 2012

iloc86

Hi every body thanks for the reply Jacqueline u obtain the same as the paper And jack in the doc they assume p doesnt change in r But as You say a c depent in theta What do u think?