SOLUTION FOR THE BOUNDARY LAYER ON A FLAT PLATE

1
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).
• Viscosity should retard the flow, thus creating a
  boundary layer on either side of the plate. Here
  only the boundary layer on one side of the plate
  is considered. The flow is assumed to be
  laminar.
• Boundary layer theory allows us to calculate the
  drag on the plate!

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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 Accordin
g to Bernoullis 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
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BOUNDARY LAYER EQUATIONS FOR A FLAT PLATE
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
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NOMINAL BOUNDARY LAYER THICKNESS
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.
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STREAMWISE VARIATION OF BOUNDARY LAYER THICKNESS
Consider a plate of length L. Based on the
estimate of Slide 11 of BoundaryLayerApprox.ppt,
we can estimate ? as or thus 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
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SIMILARITY
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
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SIMILARITY SOLUTION
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 satisfied
Similarity not satisfied
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SIMILARITY SOLUTION contd.
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).
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BUT DOES THE PROBLEM ADMIT A SIMILARITY SOLUTION?
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.
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SOLUTION BY THE METHOD OF GUESSING
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 But
so that
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SOLUTION BY THE METHOD OF GUESSING contd.
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
(U?x)-1/2! So our first try failed because of
the term (U?x)-1/2. Lets not give up! Instead,
lets learn from our mistakes!
not OK
OK
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ANOTHER TRY
This time we assume Now remembering that x and
y are independent of each other and recalling the
evaluation of ??/?y of Slide 10, or
thus 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(?).
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REDUCTION FROM PARTIAL TO ORDINARY DIFFERENTIAL
EQUATION
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
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REDUCTION contd.
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?/?y?x, where Now we
have already worked out ??/?y from Slide
12 Thus
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REDUCTION contd.
Again using we now work out the two remaining
derivatives
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REDUCTION contd.
Summarizing,
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REDUCTION contd.
Now substituting into yields or
thus Similarity works! It has cleaned up the
mess into a simple (albeit nonlinear) ordinary
differential equation!
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BOUNDARY CONDITIONS
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
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SOLUTION
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.
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SOLUTION contd.
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 ? 4.91.
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NOMINAL BOUNDARY LAYER THICKNESS
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 ? yU/(?x)1/2, it
follows that the relation for nominal boundary
layer thickness is Or In this way the
constant C of Slide 5 is evaluated.
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DRAG FORCE ON THE FLAT PLATE
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
b
L
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DRAG FORCE ON THE FLAT PLATE contd.
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).
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DRAG FORCE ON THE FLAT PLATE contd.
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 (air)
Note that ?o ? at x 0. Does this mean that
the drag force FD is also infinite?
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DRAG FORCE ON THE FLAT PLATE contd.
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.
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DRAG FORCE ON THE FLAT PLATE contd.
The relation 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 gt about 300, however,
the boundary layer is no longer laminar, and the
effect of turbulence must be included.
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REFERENCE
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.