SOLUTION FOR THE BOUNDARY LAYER ON A FLAT PLATE. Consider the following scenario. A steady potential flow has constant velocity U in the x direction. An infinitely thin flat plate is placed into this flow so that the plate is parallel to the potential flow (0 angle of incidence ).
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A STEADY RECTILINEAR POTENTIAL FLOW HAS ZERO PRESSURE GRADIENT EVERYWHERE
A steady, rectilinear potential flow in the x direction is described by the relations
According to Bernoulli’s equation for potential flows, the dynamic pressure of the potential flow ppd is related to the velocity field as
Between the above two equations, then, for this flow
For the case of a steady, laminar boundary layer on a flat plate at 0 angle of incidence, with vanishing imposed pressure gradient, the boundary layer equations and boundary conditions become (see Slide 15 of BoundaryLayerApprox.ppt with dppds/dx = 0)
Tangential and normal velocities vanish at boundary: tangential velocity = free stream velocity far from plate
Until now we have not given a precise definition for boundary layer thickness. Here we use to denote nominal boundary thickness, which is defined to be the value of y at which u = 0.99 U, i.e.
The choice 0.99 is arbitrary; we could have chosen 0.98 or 0.995 or whatever we find reasonable.
Consider a plate of length L. Based on the estimate of Slide 11 of BoundaryLayerApprox.ppt, we can estimate as
where C is a constant. By the same arguments, the nominal boundary thickness up to any point x L on the plate should be given as
One triangle is similar to another triangle if it can be mapped onto the other triangle by means of a uniform stretching.
The red triangles are similar to the blue triangle.
The red triangles are not similar to the blue triangle.
Perhaps the same idea can be applied to the solution of our problem:
Suppose the solution has the property that when u/U is plotted against y/ (where (x) is the previously-defined nominal boundary layer thickness) a universal function is obtained, with no further dependence on x. Such a solution is called a similarity solution. To see why, consider the sketches below. Note that by definition u/U = 0 at y/ = 0 and u/U = 0.99 at y/ = 1, no matter what the value of x. Similarity is satisfied if a plot of u/U versus y/ defines exactly the same function regardless of the value of x.
Similarity not satisfied
So for a solution obeying similarity in the velocity profile we must have
where g1 is a universal function, independent of x (position along the plate). Since we have reason to believe that
where C is a constant (Slide 5), we can rewrite any such similarity form as
Note that is a dimensionless variable.
If you are wondering about the constant C, note the following. If y is a function of x alone, e.g. y = f1(x) = x2 + ex, then y is a function of p = 3x alone, i.e. y = f(p) = (p/3)2 + e(p/3).
Maybe, maybe not, you never know until you try. The problem is:
This problem can be reduced with the streamfunction (u = /y, v = - /x) to:
Note that the stream function satisfies continuity identically. We are not using a potential function here because boundary layer flows are not potential flows.
We want our streamfunction to give us a velocity u = /y satisfying the similarity form
So we could start off by guessing
where f is another similarity function.
But this does not work. Using the prime to denote ordinary differentiation with respect to , if = f() then
So if we assume
then we obtain
This form does not satisfy the condition that u/U should be a function of alone. If F is a function of alone then its first derivative F’() is also a function of alone, but note the extra (and unwanted) functionality in x via the term (Ux)-1/2!
So our first try failed because of the term (Ux)-1/2.
Let’s not give up! Instead, let’s learn from our mistakes!
This time we assume
Now remembering that x and y are independent of each other and recalling the evaluation of /y of Slide 10,
Thus we have found a form of that satisfies similarity in velocity!
But this does not mean that we are done. We have to solve for the function F().
Our goal is to reduce the partial differential equation for and boundary conditions on , i.e.
to an ordinary differential equation for and boundary conditions on f(), where
To do this we will need the following forms:
The next steps involve a lot of hard number crunching. To evaluate the terms in the equation below,
we need to know /y, 2/y2, 3/y3, /x and 2/yx, where
Now we have already worked out /y; from Slide 12:
we now work out the two remaining derivatives:
Similarity works! It has cleaned up the mess into a simple (albeit nonlinear) ordinary differential equation!
From Slide 9, the boundary conditions are
But we already showed that
Now noting that = 0 when y = 0, the boundary conditions reduce to
Thus we have three boundary conditions for the 3rd-order differential equation
There are a number of ways in which the problem
can be solved. It is beyond the scope of this course to illustrate numerical methods for doing this. A plot of the solution is given below.
To access the numbers, double-click on the Excel spreadsheet below. Recall that
By interpolating on the table, it is seen that u/U = F’ = 0.99 when
Recall that the nominal boundary thickness is defined such that u = 0.99 U when y = . Since u = 0.99 U when = 4.91 and = y[U/(x)]1/2, it follows that the relation for nominal boundary layer thickness is
In this way the constant C of Slide 5 is evaluated.
Let the flat plate have length L and width b out of the page:
The shear stress o (drag force per unit area) acting on one side of the plate is given as
Since the flow is assumed to be uniform out of the page, the total drag force FD acting on (one side of) the plate is given as
The term u/y = 2/y2 is given from (the top of) Slide 17 as
The shear stress o(x) on the flat plate is then given as
But from the table of Slide 20, f’’(0) = 0.332, so that boundary shear stress is given as
Thus the boundary shear stress varies as x-1/2. A sample case is illustrated on the next slide for the case U = 10 m/s, = 1x10-6 m2/s, L = 10 m and = 1000 kg/m3 (water).
Sample distribution of shear stress o(x) on a flat plate:
U = 0.04 m/s
L = 0.1 m
= 1.5x10-5 m2/s
= 1.2 kg/m3
Note that o = at x = 0.
Does this mean that the drag force FD is also infinite?
No it does not: the drag force converges to a finite value!
And here is our drag law for a flat plate!
We can express this same relation in dimensionless terms. Defining a diimensionless drag coefficient cD as
it follows that
For the values of U, L, and of the last slide, and the value b = 0.05 m, it is found that ReL = 267, cD = 0.0407 and FD = 3.90x10-7 Pa.
is plotted below. Notice that the plot is carried only over the range 30 ReL 300. Within this range 1/ReL is sufficiently small to justify the boundary layer approximations. For ReL > about 300, however, the boundary layer is no longer laminar, and the effect of turbulence must be included.
The solution presented here is the Blasius-Prandtl solution for a boundary layer on a flat plate. More details can be found in:
Schlichting, H., 1968, Boundary Layer Theory, McGraw Hill, New York, 748 p.