Why is the gradient decreasing in the Blasius equation?

[johnt447]

Blasius equation help :(

Hello so I'm working on the blasius solution for incompressible viscous flow. So far I have solved for f' and successfully plotted a graph of flow speed against [tex]\eta[/tex]. So from what I understand this is correct I got a gradient of velocity from 0 to the flow speed. Now i want to extend it to different values of x. Now this is my issue, the book I have has a equation for du/dy so my plan was find the gradient for each different value of x, hence since i know it always starts at 0 and ends at the flow speed i could find the points in between. Although when trying the equation out I found my gradient was in fact decreasing as x increases. As far as I understood the gradient (du/dy) should become larger as x increases. The equation I'm using is this
[tex]\frac{du}{dy}=V_{\infty}\sqrt{\frac{V_{\infty} }{\nu x}}f''(0)[/tex]

where [tex]V_{\infty}[/tex] is flow speed, [tex] \nu [/tex] is kinematic viscosity, u is speed and why y and x are the coordinates.

So what I did was use the equation with f''(0) bit included I wasn't sure if that has to change I thought since the gradient in a incompressible flow is constant.

I was thinking would it be ok (and not cheating :P) to just find the gradient manually say we know at the edge of a boundary layer u=10 and at the surface u=0 and hence we also know the value of y for each one we could find the gradient that way instead of using the above formula

 

[boneh3ad]

You need to be careful how you use the term gradient here. You seem to use it inconsistently and what seems to be incorrectly here.

At any rate, you are making this much harder than necessary. If you correctly solved for f'(eta), you have all you need. Since eta is a similarity variable incorporating both x and y an f' is nondimensional velocity, all you need to find the dimensional velocity profiles at a given x location is to plug in your x and sweep across y to get f' and just multiply that by the free stream velocity.

Keep in mind that you generally have to go to eta=10 to get outside the boundary layer, so as you move farther along in x, expect to have to go to larger y values. Physically, this is the boundary layer growing as it moves downstream. This means that te velocity gradient will actually get smaller as you move downstream.

 

[johnt447]

You need to be careful how you use the term gradient here. You seem to use it inconsistently and what seems to be incorrectly here.

At any rate, you are making this much harder than necessary. If you correctly solved for f'(eta), you have all you need. Since eta is a similarity variable incorporating both x and y an f' is nondimensional velocity, all you need to find the dimensional velocity profiles at a given x location is to plug in your x and sweep across y to get f' and just multiply that by the free stream velocity.

Keep in mind that you generally have to go to eta=10 to get outside the boundary layer, so as you move farther along in x, expect to have to go to larger y values. Physically, this is the boundary layer growing as it moves downstream. This means that te velocity gradient will actually get smaller as you move downstream.

Thanks for the reply but how do i go around finding a dimensional velocity profile? What do i plug in my x value to in order to get a dimensional velocity? (Also what is the te velocity gradient?)

Do you mean use [tex]\eta=y\sqrt{\frac{V_{\infty}}{\nu x}}[/tex] where i would just put in a value of x and go through all the values of y up to the boundray height finding a new eta each time? Thing is if i do this surely I can get any value of eta but my program can only find a finite value of eta

 

[boneh3ad]

Yes, you just plug your x location into eta and then sweep through the y direction. It will give you a range of etas to use. You will obviously still need to know viscosity and free stream velocity.

Doing this will give you f', which can be multiplied by the free stream to get the dimensional velocity.

And gradient means the multi-dimensional derivative of a function. It is often just used to refer to a single derivative as well. The way you were using it seemed off, but I may have just misinterpreted the way you were using it.

 

[johnt447]

Yes, you just plug your x location into eta and then sweep through the y direction. It will give you a range of etas to use. You will obviously still need to know viscosity and free stream velocity.

Doing this will give you f', which can be multiplied by the free stream to get the dimensional velocity.

And gradient means the multi-dimensional derivative of a function. It is often just used to refer to a single derivative as well. The way you were using it seemed off, but I may have just misinterpreted the way you were using it.

Thanks managed to get it sorted earlier today when i said gradient i wasn't meaning the grad function but like the gradient of how speed is changing with y similar to y/x is a gradient of a line

 

[boneh3ad]

That is because for a straight line, y/x is the derivative.