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Blasius Secondary Flow - Foundations of Fluid Mechanics II - Lecture Notes, Study notes of Fluid Mechanics

M. J. P. Rohilkhand University Fluid Mechanics

This is the second course of a two-semester fluid mechanics sequence for graduate students in the thermal sciences. This course includes topics like fully turbulent flows, turbulent boundary layers and free shear flows, turbulence modeling, laminar boundary layers including axisymmetric and 3-D boundary layers. Key points in this lecture are: Blasius Secondary Flow, Boundary Layer Similarity Solution, Guessed Boundary Conditions, Known Boundary Conditions, Runge-Kutta Solution

Typology: Study notes

2012/2013

Uploaded on 10/03/2013

aanila 🇮🇳

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Z

123456

1

2

3

4

5

6

7

8

9

10

0 0.33206 0 0 1.08597 0

5·10

-3 0.33206 1.66028·10

-3 4.15071·10

-6 1.08097 5.41737·10

-3

0.01 0.33206 3.32057·10

-3 1.66028·10

-5 1.07597 0.01081

0.015 0.33206 4.98085·10

-3 3.73564·10

-5 1.07097 0.01618

0.02 0.33206 6.64114·10

-3 6.64114·10

-5 1.06597 0.02152

0.025 0.33206 8.30142·10

-3 1.03768·10

-4 1.06098 0.02684

0.03 0.33206 9.96171·10

-3 1.49426·10

-4 1.05598 0.03213

0.035 0.33206 0.01162 2.03385·10

-4 1.05098 0.0374

0.04 0.33206 0.01328 2.65646·10

-4 1.04598 0.04264

0.045 0.33206 0.01494 3.36208·10

-4 1.04098 0.04786

=

Top portion of Z:

Z Rkadapt Y ηstart,ηend,num_steps,D,

()

:=num_steps 2000:=ηend 10:=ηstart 0:=

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.

DηY,

()

c−Y3

⋅Y1

⋅

Y1

Y2

1−Y2

()

2

+Y2Y5

⋅+ 1

2Y3

⋅Y4

⋅−

Y4

⎡

⎢

⎣

⎤

⎥

⎦

:=

Now calculate the solution as η marches from ηstart to ηend. Here Z

is the solution matrix, where the first column is η, the second column

is Y1, the third column is Y2, the fourth column is Y3, the fifth

column is Y4, and the last column is Y 5.

Now 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.

Y

0.33206

0

1.08597

0

⎛

⎜

⎝

⎞

⎟

⎠

=

Verify the vector:

Y41.085973166:=Y10.332057:=Y50:=Y30:=Y20:=

Guessed boundary conditions:Known boundary conditions:

ORIGIN 1:=

First define a vector Y which contains five unknowns, Y1 = f'', Y2 = f', Y3 = f, Y4 = h', and Y5 = h.

Since two of these are at infinity, f''(0) and h'(0) need to be guessed until the boundary conditions at infinity are

satisfied.

The boundary conditions are f'(0)=1, f(0)=1, f'(∞)=1, h(0)=0, and h(∞)=0.

c 0.5:=

Here, let c = 1/2, following Kundu's book.

The equations to solve are f''' + cff'' = 0, where prime denotes d/dη, and h'' + 0.5fh' - f'h + 1 - (f')2 = 0.

Blasius-secondary flow flat plate boundary layer similarity solution

1

docsity.com

Partial preview of the text

Download Blasius Secondary Flow - Foundations of Fluid Mechanics II - Lecture Notes and more Study notes Fluid Mechanics in PDF only on Docsity!

Z

5·10 -3^ 0.33206 1.66028·10 -3^ 4.15071·10 -6^ 1.08097 5.41737·10 -

0.01 0.33206 3.32057·10 -3^ 1.66028·10 -5^ 1.07597 0.

0.015 0.33206 4.98085·10 -3^ 3.73564·10 -5^ 1.07097 0.

0.02 0.33206 6.64114·10 -3^ 6.64114·10 -5^ 1.06597 0.

0.025 0.33206 8.30142·10 -3^ 1.03768·10 -4^ 1.06098 0.

0.03 0.33206 9.96171·10 -3^ 1.49426·10 -4^ 1.05598 0.

0.035 0.33206 0.01162 2.03385·10 -4^ 1.05098 0.

0.04 0.33206 0.01328 2.65646·10 -4^ 1.04598 0.

0.045 0.33206 0.01494 3.36208·10 -4^ 1.04098 0.

Top portion of Z:

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

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.

D (^ η ,Y)

− c ⋅ Y 3 ⋅Y 1

Y 1

Y 2

− 1 + ( Y 2 )^2 + Y 2 ⋅Y 5 1

− ⋅ Y 3 ⋅Y 4

Y 4

Now calculate the solution as η marches from ηstart to ηend. Here Z

is the solution matrix, where the first column is η, the second column

is Y 1 , the third column is Y 2 , the fourth column is Y 3 , the fifth

column is Y 4 , and the last column is Y 5.

Now 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.

Y

Verify the vector:

Y 2 := 0 Y 3 := 0 Y 5 := 0 Y 1 :=0.332057 Y 4 :=1.

Known boundary conditions: Guessed boundary conditions:

ORIGIN := 1

First define a vector Y which contains five unknowns, Y 1 = f'', Y 2 = f', Y 3 = f, Y 4 = h', and Y 5 = h.

Since two of these are at infinity, f''(0) and h'(0) need to be guessed until the boundary conditions at infinity are

satisfied.

The boundary conditions are f'(0)=1, f(0)=1, f'(∞)=1, h(0)=0, and h(∞)=0.

Here, let c = 1/2, following Kundu's book. c :=0.

The equations to solve are f''' + cff'' = 0, where prime denotes d/dη, and h'' + 0.5fh' - f'h + 1 - (f') 2 = 0.

Blasius-secondary flow flat plate boundary layer similarity solution

1 docsity.com

Bottom portion of Z (to verify BCs):

Z

9.95 1.03781·10 -8^1 8.22921 -2.12027·10 -9^ 1.43271·10 -

9.955 1.01667·10 -8^1 8.23421 -1.69195·10 -9^ 1.4327·10 -

9.96 9.95948·10 -9^1 8.23921 -1.27186·10 -9^ 1.4327·10 -

9.965 9.75637·10 -9^1 8.24421 8.59864·10 -10^ 1.43269·10 -

9.97 9.55729·10 -9^1 8.24921 -4.558·10 -10^ 1.43269·10 -

9.975 9.36215·10 -9^1 8.25421 5.95158·10 -11^ 1.43269·10 -

9.98 9.17088·10 -9^1 8.25921 3.29137·10 -10^ 1.43269·10 -

9.985 8.9834·10 -9^1 8.26421 7.10305·10 -10^ 1.43269·10 -

9.99 8.79965·10 -9^1 8.26921 1.08413·10 -9^ 1.43269·10 -

9.995 8.61955·10 -9^1 8.27421 1.45076·10 -9^ 1.4327·10 -

10 8.44303·10 -9^1 8.27921 1.81033·10 -9^ 1.43271·10 -

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

Blasius BL with Secondary Flow

eta

f'', f', and f

Zn 2,

Zn 3,

Zn 4,

Zn 5,

Zn 6,

Zn 1,

(^2) docsity.com

Blasius Secondary Flow - Foundations of Fluid Mechanics II - Lecture Notes, Study notes of Fluid Mechanics

Related documents

Partial preview of the text

Download Blasius Secondary Flow - Foundations of Fluid Mechanics II - Lecture Notes and more Study notes Fluid Mechanics in PDF only on Docsity!

Z

5·10 -3^ 0.33206 1.66028·10 -3^ 4.15071·10 -6^ 1.08097 5.41737·10 -

0.01 0.33206 3.32057·10 -3^ 1.66028·10 -5^ 1.07597 0.

0.015 0.33206 4.98085·10 -3^ 3.73564·10 -5^ 1.07097 0.

0.02 0.33206 6.64114·10 -3^ 6.64114·10 -5^ 1.06597 0.

0.025 0.33206 8.30142·10 -3^ 1.03768·10 -4^ 1.06098 0.

0.03 0.33206 9.96171·10 -3^ 1.49426·10 -4^ 1.05598 0.

0.035 0.33206 0.01162 2.03385·10 -4^ 1.05098 0.

0.04 0.33206 0.01328 2.65646·10 -4^ 1.04598 0.

0.045 0.33206 0.01494 3.36208·10 -4^ 1.04098 0.

Top portion of Z:

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

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.

D (^ η ,Y)

− c ⋅ Y 3 ⋅Y 1

Y 1

Y 2

− 1 + ( Y 2 )^2 + Y 2 ⋅Y 5 1

− ⋅ Y 3 ⋅Y 4

Y 4

Now calculate the solution as η marches from ηstart to ηend. Here Z

is the solution matrix, where the first column is η, the second column

is Y 1 , the third column is Y 2 , the fourth column is Y 3 , the fifth

column is Y 4 , and the last column is Y 5.

Now 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.

Y

Verify the vector:

Y 2 := 0 Y 3 := 0 Y 5 := 0 Y 1 :=0.332057 Y 4 :=1.

Known boundary conditions: Guessed boundary conditions:

ORIGIN := 1

First define a vector Y which contains five unknowns, Y 1 = f'', Y 2 = f', Y 3 = f, Y 4 = h', and Y 5 = h.

Since two of these are at infinity, f''(0) and h'(0) need to be guessed until the boundary conditions at infinity are

satisfied.

The boundary conditions are f'(0)=1, f(0)=1, f'(∞)=1, h(0)=0, and h(∞)=0.

Here, let c = 1/2, following Kundu's book. c :=0.

The equations to solve are f''' + cff'' = 0, where prime denotes d/dη, and h'' + 0.5fh' - f'h + 1 - (f') 2 = 0.

Blasius-secondary flow flat plate boundary layer similarity solution

1 docsity.com

Bottom portion of Z (to verify BCs):

Z

9.95 1.03781·10 -8^1 8.22921 -2.12027·10 -9^ 1.43271·10 -

9.955 1.01667·10 -8^1 8.23421 -1.69195·10 -9^ 1.4327·10 -

9.96 9.95948·10 -9^1 8.23921 -1.27186·10 -9^ 1.4327·10 -

9.965 9.75637·10 -9^1 8.24421 8.59864·10 -10^ 1.43269·10 -

9.97 9.55729·10 -9^1 8.24921 -4.558·10 -10^ 1.43269·10 -

9.975 9.36215·10 -9^1 8.25421 5.95158·10 -11^ 1.43269·10 -

9.98 9.17088·10 -9^1 8.25921 3.29137·10 -10^ 1.43269·10 -

9.985 8.9834·10 -9^1 8.26421 7.10305·10 -10^ 1.43269·10 -

9.99 8.79965·10 -9^1 8.26921 1.08413·10 -9^ 1.43269·10 -

9.995 8.61955·10 -9^1 8.27421 1.45076·10 -9^ 1.4327·10 -

10 8.44303·10 -9^1 8.27921 1.81033·10 -9^ 1.43271·10 -

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

Blasius BL with Secondary Flow