Navier-Stokes Problem: Solving for Pressure Gradient in a 2D Rectangular Cavity

  • Thread starter Thread starter member 428835
  • Start date Start date
  • Tags Tags
    Navier-stokes
Click For Summary
SUMMARY

The discussion focuses on solving the Navier-Stokes equations for a 2D rectangular cavity filled with liquid, where the top plate moves with velocity V_0. The participants derive the velocity profile and pressure gradient, concluding that the pressure gradient is significant and necessary for flow analysis. They emphasize the importance of boundary conditions and integral constraints in solving the equations, particularly in low Reynolds number scenarios. The conversation also touches on the use of stream functions to satisfy continuity equations and visualize flow patterns.

PREREQUISITES
  • Understanding of Navier-Stokes equations
  • Familiarity with boundary conditions in fluid dynamics
  • Knowledge of Reynolds number and its implications
  • Experience with stream functions in fluid flow analysis
NEXT STEPS
  • Study the derivation of the Navier-Stokes equations in 2D flow scenarios
  • Learn about the significance of Reynolds number in fluid dynamics
  • Explore the application of stream functions in solving fluid flow problems
  • Investigate numerical methods for solving the Navier-Stokes equations
USEFUL FOR

Fluid dynamics researchers, mechanical engineers, and students studying fluid mechanics who are interested in the mathematical modeling of flow in confined geometries.

  • #31
I'm having second thoughts on your equation for ##\omega##. Shouldn't it be a function of x times a function of y, not the sum of such functions?

cosh cos, cosh sin, sinh cos, sinh sin
 
  • Like
Likes   Reactions: member 428835
Engineering news on Phys.org
  • #32
Chestermiller said:
I'm having second thoughts on your equation for ##\omega##. Shouldn't it be a function of x times a function of y, not the sum of such functions?

cosh cos, cosh sin, sinh cos, sinh sin

Shoot, you're totally right. Then ##\nabla \cdot \psi = \omega## is not separable. I can't think of a way to analytically solve. How would you do this numerically? Finite difference ##\nabla^4 \psi=0##?
 
  • #33
joshmccraney said:
Shoot, you're totally right. Then ##\nabla \cdot \psi = \omega## is not separable. I can't think of a way to analytically solve. How would you do this numerically? Finite difference ##\nabla^4 \psi=0##?
No. Solve the equations involving both omega and psi. The only tricky part is numerically specifying the boundary conditions on omega. But I know how to do that.
 
  • #34
Yea, how do you do that?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 18 ·
Replies
18
Views
8K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
4K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 51 ·
2
Replies
51
Views
9K
  • · Replies 3 ·
Replies
3
Views
2K