Solving Navier Stoke Problem with 3D Equation and Lubrication Approximation

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Discussion Overview

The discussion revolves around the application of the 3D Navier-Stokes equations and lubrication approximation in the context of oil well drilling. Participants are analyzing a specific equation derived from these principles and exploring the implications of their mathematical manipulations and assumptions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents an equation derived from the lubrication approximation and seeks confirmation of their results, specifically the expression for w.
  • Another participant provides a detailed derivation of the equation for w, but raises concerns about the integration process and assumptions regarding the dependence of pressure on r.
  • There is a discussion about the assumption that the pressure gradient ∂p/∂θ is independent of r, with one participant questioning the validity of this assumption.
  • A participant notes that the constant of integration could potentially depend on θ, suggesting that this aspect may not have been adequately addressed in the original paper.

Areas of Agreement / Disagreement

Participants express differing views on the assumptions made in the derivation, particularly regarding the dependence of pressure on r and the nature of the constants of integration. No consensus is reached on these points.

Contextual Notes

Limitations include the assumption that the pressure gradient does not vary with r, which some participants challenge. The implications of this assumption on the integration process and the resulting expressions for w remain unresolved.

iloc86
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hi every body, I am 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
 
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sorry imaginary
 
-(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
 
JJacquelin said:
-(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

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

[tex]\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)[/tex]

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

[tex]\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[/tex]

You're assuming [itex]\frac{\partial p}{\partial \theta}[/itex] 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:
Hi every body thanks for the reply Jacqueline u obtain the same as the paper And jack in the doc they assume p doesn't change in r But as You say a c depent in theta What do u think?
 

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