How to determine streamlines for two-dimensional steady flow of a fluid?

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Homework Statement



The two-dimensional steady flow of a fluid with density ρ is given by

v=K(-yex + xey) / (x2 + y2) where K is a constant
(a) Can this flow corespond to the flow of anincompressible fluid?
(b) Determine and sketch the streamlines of this flow.
(c) Determine and sketch the aceleration of this field
(d) Determine the pressure difference between two points that are a distance r1 and r2 away from the origin. (r2 > r1) ?

Homework Equations



dx/u=dy/w


The Attempt at a Solution



I see that there is x^2+y^2, this makes me think that there is a circle based streamline flow?? I do not understand how I can visualize this in my head, such an equation. How can I know whether its incompresssible? The acceleration should be the derivative of those streamlines I guess.. Can anybody help me out with these questions PLEASE?
 

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  • #2
There is a condition involving the divergence of velocity...
 
  • #3
You're correct. The streamlines are circles. Under those circumstances, what is the acceleration of a particle traveling around the circumference of a circle?
 
  • #4
The acceleration is a=v^2/r ?
 
  • #5
Incompressibilty implies the divergence of velocity is zero...
 
  • #6
hvthvt said:
Incompressibilty implies the divergence of velocity is zero...

Good. Can you check that?
 
  • #7
(-ex + x)/ (x^2 + y^2) + (-ex + x)/(x^2 + y^2) does NOT equal zero
So its not an incompressible fluid? What to do with the derivative to x and to y of (x^2 + y^2)
 
  • #8
I am not sure what you did, but that does not look like you computed divergence. Do you understand what it is?
 
  • #9
Not really.. I am pretty stuck at this moment to be honest. Can you help me out?
 
  • #10
I need to check whether ∇.v = 0
So whether d/dx(formula) + d/dy(formula) = 0 right? But I'm having difficulty with deriving the x^2+y^2 part
 
  • #11
No, it is not d/dx(formula) + d/dy(formula).

Divergence applies to a vector, which can be written as two compoents $$

f(x, y) \hat \imath + g(x, y) \hat \jmath , $$ where ##f(x, y)## is the x-component of the vector, and ##g(x, y)## is the y-component of the vector. The divergence then is $$

\frac {\partial f} {\partial x} + \frac {\partial g} {\partial y}. $$

This is what you need to compute and check whether it is zero.
 
  • #12
Oooh really, thanks! I understand that it can be written as two components: but what is the x component, f(x,y) in this formula?
Is that
-yex / (x2 + y2) ??
 
  • #13
hvthvt said:
The acceleration is a=v^2/r ?
Yes. Do you know how to get v^2 when v is a vector expressed in component form?
 
  • #14
hvthvt said:
Oooh really, thanks! I understand that it can be written as two components: but what is the x component, f(x,y) in this formula?
Is that
-yex / (x2 + y2) ??

Yes it is. But without the ##e_x## part, it is there to denote that it is the x component.
 
  • #15
I found:
∂f/∂x= (2xye - y^3 - x^2y)/(x^2+y^2)^2
∂g/∂x=(-2xye+x^3 + xy^2)/(x^2 + y^2)^2

Computing ∂f∂x+∂g∂y = (x^3 - y^3 + xy^2 - x^2y)/(x^2+y^2)^2

This is not zero. So the fluid is imcompressible. Is this correct?
 
  • #16
Ooooh, without the e_x part, I get it! Then it gets much easier!
df/dx = 2xy/(x^2 + y^2)^2
dg/dx=-2xy(x^2+y^2)^2
df/dx-dg/dx = 0 so the flow corresponds to the flow of an incompressible fluid! right?
 
  • #17
Your final result is correct, but you there is a typo: it is not df/dx - dg/dy, but df/dx + dg/dy. If it were df/dx - dg/dy, you would not have obtained zero.

So, the flow is incompressible.
 
  • #18
Yes, sorry, that was my mistake. I typed it incorrectly, but I understand it right now. Thank you.
So now I am asked in (b) to determine and sketch the streamlines of the flow.
How should I determine them?
First of all, the flow is incompressible, so what kind of information does this supply?? Does that influence the way in which the streamlines flow?

How to determine the streamlines?
dx/vx = dy/vy --> dx/dy=vx/vy ??
 
  • #19
I think you have already determined - or, at least, guessed - that the streamlines are circular.

So what you can do is this. Choose some radius. Select a few points on a circle of that radius. Compute the velocity vector at those points. Draw the resultant vector as arrows on paper. Repeat.
 
  • #20
I figured out that 1/2(x^2+y^2) = constant. Does this have any significance for answering (b)?
 
  • #21
How can I compute the velocity vector at a specific point? Should i, for example, if i want to know what is happening in point (2,2)
Fill it in in the original equation?
f(2,2)i = -0.25
g(2,2)j= 0.25
 
  • #22
Yes, that is correct. All you need to do is to depict the velocity (-0.25, 0.25) at (2, 2) graphically.
 
  • #23
Thank you. I know this is very simple, but I haven't done this before. How to depict the velocity at (2,2)?
Does this mean that the arrow at (2,2) goes -0,25 to the left and 0.25 up ?? How can I visualize this?
 
  • #24
Yes, that would do it. Or you could scale the length of the arrow down or up. Just be consistent everywhere.
 
  • #25
hvthvt said:
Yes, sorry, that was my mistake. I typed it incorrectly, but I understand it right now. Thank you.
So now I am asked in (b) to determine and sketch the streamlines of the flow.
How should I determine them?
First of all, the flow is incompressible, so what kind of information does this supply?? Does that influence the way in which the streamlines flow?

How to determine the streamlines?
dx/vx = dy/vy --> dx/dy=vx/vy ??
Maybe we should be asking "what are your equations for vx and vy in this problem?" Please state your answers to this question. If you can answer this question, then you should be able to integrate the differential equation to get how x and y vary mutually.

Chet
 

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