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Blasius Laminar - Foundations of Fluid Mechanics I - Lecture Notes, Study notes of Fluid Mechanics

This is the first course of a two-semester fluid mechanics sequence for graduate students in the thermal sciences. This course deals with solutions of these equations, both exact and approximate. Key points of this lecture are: Blasius Laminar, Derivative Vector, Runge-Kutta Solution, Bottom Portion, Root Function

Typology: Study notes

2012/2013

Uploaded on 10/03/2013

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Here the function Rkadapt is used, which is similar to rkfixed except it internally uses adaptable spacing
instead of fixed spacing (more accuracy where needed). It reports at fixed spacing however.
ηstart 0:= ηend 10:= num_steps 2000:= Z Y1guess( ) Rkadapt YBC Y1guess()ηstartend,num_steps,D,
()
:=
Now find the correct boundary condition, using the root function, and then re-set Y1guess to this best value:
best root Z Y1guess()
num_steps 1+3,1Y1guess,
()
:= best 0.332057=
Zfinal Rkadapt YBC best()ηstartend,num_steps,D,
()
:=
Middle portion of Zfinal (to find δ): Bottom portion of Zfinal (to check boundary conditions):
η Y1=f '' Y2=f ' Y3=f η Y1=f '' Y2=f ' Y3=f
Zfinal
1234
1999
2000
2001
9.99 99526·10
-9 1 8.269213
9.995 19427·10
-9 1 8.274213
10 42908·10
-9 1 8.279213
=
Now generate a plot of the similarity variables: n 1 num_steps..:=
0 0.5 1 1.5 2
0
1
2
3
4
5
6Blasius BL Similarity Solution
f'', f', and f
eta
Zfinaln1,
Zfinaln1,
Zfinaln1,
Zfinaln2,Zfinaln3,
,Zfinaln4,
,
Blasius flat plate boundary layer similarity solution by the Runge-Kutta method
The equation to solve is f''' + cff'' = 0, where prime denotes d/dη. Here, let c = 1/2. c 0.5:= ORIGIN 1:=
We will define a vector Y which contains three unknowns, Y1 = f'', Y2 = f', and Y3 = f. Vector YBC is Y at η = 0.
Note that YBC1 at η = 0 mus t be guessed in order to satisfy all boundary condit ions in t he problem.
Y1guess 1:= YBC2 0:= YBC3 0:=
YBC Y1guess()
Y1guess
YBC2
YBC3
:= Verify the vector: YBC Y1guess()
1
0
0
=
DηY,
()
cY3
Y1
Y1
Y2
:=
Define the derivative vector D which contains the first derivative with respect to η of each
variable in the Y vector. This derivative vector D is needed for the Runge-Kutta solution.
Now calculate the solution as η marches from ηstart to ηend, using Runge-Kutta. Here Z is the solution matrix,
where the first column is η, the second column is Y 1, the third column is Y2, and the fourth column is Y3.
Zfinal
1234
982
983
984
4.905 0.01855 0.989908 3.189153
4.91 0.018403 0.99 3.194103
4.915 0.018256 0.990092 3.199053
=
1
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Here the function Rkadapt is used, which is similar to rkfixed except it internally uses adaptable spacing instead of fixed spacing (more accuracy where needed). It reports at fixed spacing however.

ηstart := 0 ηend := 10 num_steps := 2000 Z Y1guess( ) :=Rkadapt YBC Y1guess(^ ( ) , ηstart, ηend, num_steps,D)

Now find the correct boundary condition, using the root function, and then re-set Y1guess to this best value:

best := root Z Y1guess( ( ) num_steps 1+ , 3 − 1 ,Y1guess) best =0.

Zfinal :=Rkadapt YBC best(^ ( ) , ηstart, ηend, num_steps,D)

Middle portion of Zfinal (to find δ): Bottom portion of Zfinal (to check boundary conditions): η Y 1 =f '' Y 2 =f ' Y 3 =f η Y 1 =f '' Y 2 =f ' Y 3 =f

Zfinal

1 2 3 4 1999 2000 2001

9.99 99526·10 -9^1 8. 9.995 19427·10 -9^1 8. 10 42908·10 -9^1 8.

Now generate a plot of the similarity variables: (^) n := 1 ..num_steps

0 0.5 1 1.5 2

0

1

2

3

4

5

6

Blasius BL Similarity Solution

f'', f', and f

eta

Zfinaln 1,

Zfinaln 1,

Zfinaln 1,

Zfinaln 2, , Zfinaln 3,,Zfinaln 4,

Blasius flat plate boundary layer similarity solution by the Runge-Kutta method

The equation to solve is f''' + cff'' = 0, where prime denotes d/dη. Here, let c = 1/2. (^) c := 0.5 ORIGIN := 1 We will define a vector Y which contains three unknowns, Y 1 = f'', Y 2 = f', and Y 3 = f. Vector YBC is Y at η = 0. Note that YBC 1 at η = 0 must be guessed in order to satisfy all boundary conditions in the problem.

Y1guess := 1 YBC2 := 0 YBC3 := 0

YBC Y1guess( )

Y1guess YBC YBC

:= Verify the vector:^ YBC Y1guess( )

D (^ η ,Y)

− c ⋅ Y 3 ⋅Y 1

Y 1

Y 2

Define the derivative vector D which contains the first derivative with respect to η of each variable in the Y vector. This derivative vector D is needed for the Runge-Kutta solution. Now calculate the solution as η marches from ηstart to ηend, using Runge-Kutta. Here Z is the solution matrix, where the first column is η, the second column is Y 1 , the third column is Y 2 , and the fourth column is Y 3.

Zfinal

1 2 3 4 982 983 984

4.905 0.01855 0.989908 3. 4.91 0.018403 0.99 3. 4.915 0.018256 0.990092 3.

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