CEE 262A H YDRODYNAMICS

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CEE 262A H YDRODYNAMICS. Lecture 16 Boundary layers: Scaling and Blasius. Boundary layers over a solid surface (classical aerodynamics). Imagine a flow that has a velocity parallel to the surface far from the surface that varies as U E (s) where s is the distance along the surface.

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### CEE 262A HYDRODYNAMICS

Lecture 16

Boundary layers: Scaling and Blasius

Imagine a flow that has a velocity parallel to the surface far from the surface that varies as UE(s) where s is the distance along the surface.

UE(s)

s

How does the boundary layer thickness, drag etc, depend on the flow and its variation?

We start with the 2D, steady Navier Stokes equations (written in our usual Cartesian coordinates) for a homogeneous fluid. If the curvature of the surface is small (and hence the flow is steady):

where x1 is the distance parallel to the surface and x3 is the distance normal to the surface. Note that we have used the perturbation pressure (hydrostatic pressure is removed).

Suppose that the freestream velocity ~ U and the length scale for any flow variations is L, then a scaling for the momentum equations would be:

where the importance of viscous stresses is measured by the Reynolds number

I.e. if Re >> 1, viscous stresses can be neglected, at least in the outer (potential) flow.

Define inner- and outer-layer variables:

Outer layer:

u, w, x, z, p

Weak viscous stresses.

Inner layer:

u+, w+, x+, z+, p+

Strong viscous stresses.

In order to make viscous stresses important in the inner layer, we want (assume all quantities are dimensionless)

Dimensionally, this implies that the vertical scale for the inner layer is d:

(z dimensional here)

we don't stretch x. So we have

Thus, continuity requires that

Thus, the inner-layer momentum equations are:

These could also have been obtained via nondimensionalizationwith

If we substitute these into the governing equations and look at the lowest-order terms:

• Typical of the governing equations for “thin shear layers”, these show that:
• Streamwise derivatives << flow normal derivatives
• Pressure in the inner layer is the same as outside, i.e. p0+ is independent of z+

Which give to lowest order the inviscid flow problem

Outer layer:

Inner layer:

We can complete the problem specification by requiring that the following asymptotic matching hold:

(outer limit of inner variable = inner limit of outer variable)

Since the inner vertical momentum equation implies

Then the matching condition requires

However, from the outer x-momentum equation, we know that at z=0

Thus, the pressure gradient inside the boundary layer is the same as

outside and

For the velocity

Thus, the pressure gradient in the boundary layer and the streamwise velocity that must be matched to are determined by the outer, potential flow, which is what determines UE(x).

Note that we require

The problem of boundary layers then becomes one in which

the pressure field is known and we only need to solve for

the developing velocity field:

Eliminate p0+

For boundary layers on solid surfaces, we do the following:

• Find inviscid (potential) flow
• Use the inviscid flow to solve the lowest layer boundary layer equations and find wall stress, boundary layer growth
• If desired look at next terms in expansion (higher order boundary layer theory)

This approach does not allow us to describe separated flow regions

Van Dyke “Album of Fluid Motion”

Uniform upstream flow

d0(x)

x = 0

In general, the pdes that describe bls are nonlinear and hence tough.

Like Stokes' first problem, there is no imposed vertical length scale. Can we find a similarity solution?

UE

d0(x)

x(t)

Consider moving at speed UE and observing the growth of the

boundary layer in this moving frame that moves according to

x(t)=UEt. After some time t the frame has moved a distance

t=x/UE from the leading edge at x=0. In the moving frame, by

analogy to Stokes' first problem, the boundary layer grows roughly

as d0(t)=(nt)1/2. Substitution gives

Recall that

Then convert partials wrt x and z+ to derivatives wrth (let x+=x):

To convert the pdes to an ode using the similarity variable, we look at the momentum equation in terms of the streamfunction

Let’s choose the streamfunction as follows

Why this form?

The derivatives are given by

So substituting into the momentum equation we get

Which is subject to the conditions that (noting that u0+=f ')

(no-slip)

(no-flux)

(match outer solution)

Unfortunately, this is not analytically integrable and so must be done numerically. This requires a “shooting” method – start at the wall and integrate out. Since we don’t have f’’, we must guess f’’ and see if the f’ tends to 1 far from the wall. The value at the wall is f''(0)=0.332.

The result looks like this:

u0+

Schlichting “Boundary Layer Theory”

z+

We found that the curvature of the velocity at the wall is determined by the pressure gradient

Favorable

C

B

A

Thus, if the pressure gradient < 0, the boundary layer will be thin, whereas if pressure gradient > 0, the boundary layer will be thick. The first case is known as a favorable pressure gradient, whereas the second case is known as an adverse pressure gradient. This latter case leads to separation, e.g. for flow around a cylinder

BL

Separation

vortical wake

Using the Blasius solution, we can define the boundary layer thickness more accurately, for example if we define the boundary layer by the point at which the velocity is 99% of the freestream velocity:

(note that this works for Rex< 105)

If we calculate the (dimensional) vorticity

Thus

As expected, most of the vorticity is contained within the BL. In terms of vorticity (dimensional) we have

diffusion

If we hadn’t accounted for u3 or the variation with height of u1, we would have repeated Stokes first problem with t replaced by x1/U .

What about the flux of vorticity from the wall?

Vorticity in the flow is introduced ONLY at the leading edge at x=0.

The dimensional wall stress can be calculated with

Thus, the skin friction coefficient cfdepends inversely on Rex

why?

"young" BL: mdu1/dx3=t1

"old" BL: mdu1/dx3=t2<t1