Application of partial derivatives

Click For Summary
SUMMARY

The discussion focuses on applying partial derivatives to determine the steady flow around a long circular cylinder of unit radius in a fluid. The user seeks assistance in solving the problem, which involves calculating the fluid speed at the cylinder's surface, expressed as 2A|SinØ|. The solution involves defining the cylinder's equation as x² + y² = 1 and deriving the gradient vector from the function f(x,y) = x² + y² - 1, resulting in the gradient vector <2x, 2y>, which corresponds to the uniform velocity vector of the fluid flow.

PREREQUISITES
  • Understanding of partial derivatives and gradient vectors
  • Familiarity with fluid dynamics concepts
  • Knowledge of cylindrical coordinates and equations
  • Basic calculus skills
NEXT STEPS
  • Study the application of partial derivatives in fluid dynamics
  • Learn about the Navier-Stokes equations for fluid flow analysis
  • Explore cylindrical coordinate systems in mathematical modeling
  • Investigate the concept of boundary layers in fluid mechanics
USEFUL FOR

Students and professionals in mathematics, physics, and engineering fields, particularly those focusing on fluid dynamics and mathematical modeling of flow around objects.

vips
Messages
2
Reaction score
0
sorry folks i don't even have an idea to this question`s solution so i hope u people may like to help me. i`m stuck to it since last week nd i hope its from partial derivative... please suggest me a book or a hint or the solution.

Let a long circular cylinder of unit radius be placed in a large body of fluid flowing with uniform velocity, the axis of cylinder being perpendicular to the direction of flow. determine the steady flow.also show that the speed of the fluid at points on the cylinder surface is 2A|SinØ|.
 
Physics news on Phys.org
I would say your formula you are given is that the radius of the cylinder is say x2 + y2 = 1 (as it is a unit radius), and allow your function f(x,y) = x2 +y2 -1 = some constant C (as your Z value can be any value you want it to be as the cylinder extends in either direction of Z). So given this, you can solve for your gradient (which is the vector given by the partial derivatives). So partial f respect to x = 2x, partial f respect to y = 2y, so the gradient is given by the vector <2x,2y>. This vector would be equivalent to your uniform velocity vector (flow rate). After that I'm not really sure how to determine the exact velocity though.. that may start you?
 
i`ll try though i have only 1 day left to solve it... thank you for the reply...
 

Similar threads

Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
0
Views
1K
Replies
5
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K