Blasiu's Theorem And Its Applications

© OCT 2019 | IRE Journals | Volume 3 Issue 4 | ISSN: 2456-8880

Blasiu’s Theorem and Its Applications NWAE OO HTIKE1, NGWE KYAR SU KHAING2, KAY THI WIN3 1, 2, 3 Department of Mathematics,TU(Yamethin), Myanmar Abstract- The basic concepts of fluid motion are expressed Laurent’s Theorem, Residue Theorem, Circle theorem are presented-Blasiu’s theorem and it’s application are also discussed. A source outside a cylinder with and without circulation around the cylinder is described. A doublet outside a cylinder is studied.

Indexed Terms- Moment, Laurent. Residue, Circle Theorem.

I.

INTRODUCTION

When a long cylinder is placed with its generators perpendicular to the incident stream of a moving fluid containing hydrodynamic singularities, such as sources, sinks it experience forces tending to produce translation and rotation of the cylinder. These effects are calculated using the following theorem due to Blasius. The cylinder can be of any general section.

In practice it can be an aerofoil. We confine attention to the case when the fluid is incompressible.

LAURENT’S THEOREM

II.

Let f(z) be analytic in a domain containing two concentric circles C1 and C2 with center Z0 and the annulus between them. Then f(z) can be represented by the Laurent series 

f(z) =

a



n (z  z 0 )

n



n 0

 z  z n 1

bn

0

n

(1)

bn 

  z

1 2i

*

 z0



n 1

 

f z* dz* (2)

c

Takencounter clockwise around any simple closed path C that lies in the annulus and encircles the inner circle.

III.

RESIDUE THEOREM

Let f(z) be analytic inside a simple closed path C and on C, except for finitely many singular points z1, z2, ..., zk inside C. Then the integral of f(z) taken counterclockwise around C equals 2i times the sum of the residues of f(z) at z1, z2,…, zk.



f  z dz  2i

k

 j1

c

IV.

Res f  z  (3) zz j

CIRCLE THEOREM

Suppose there is irrotational two-dimensional motion of an incompressible inviscid fluid, having no rigid boundaries, in the z-plane given by a complex potential f(z), where the singularities of f(z) are at a generator distance than a from the origin. If a circular cylinder

whose

cross-section

C

is

z  a is

introduced into the fluid and held fixed, then the complex potential becomes   a2 W  f (z)  f   z 

  (4)  V. DOUBLET

= a 0  a1  z  z0   a 2  z  z0   ...

2



b1 b2   ... z  z 0 (z  z 0 ) 2

Consisting of nonnegative and negative powers. The coefficients of this Laurent series are given by the integrals an 

1 2i

 z  c

*

  z 

f z*

n 1

dz,

0

Figure 1.Doublet

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© OCT 2019 | IRE Journals | Volume 3 Issue 4 | ISSN: 2456-8880 Suppose there is a source of strength m at the point A, z0  aei , and a sink of equal and opposite strength at the origin, Then OA  a, z0  aei

For the system of sources,



W  m log z-m log z  aei







  m[ log z  aei  log z] z  aei z

  m log

 aei    m log 1 (5) z   Using logarithmic series, we get W

Figure 2. Thrust on a cylinder

 aei a 2 e 2i a 3e3i  =  m       2z 2 3z3  z  i

=

mae z



2 2i

ma e 2z

2



3 3i

ma e

3z3

 ...

ei ae2i ua 2 e3i    ... (6) z 2z 2 3z3 Taking a  0 and m  , so that  = constant we Then W =

The above figure shows the section C of the cylinder in the plane XOY. PP’ is an arc elements of C of length s . Then if p denotes the pressure at P,. the force on unit length of the section s is p s normal to C. If  is the inclination to OX of the tangent at P to C, then x and y components of this force are

pyand  px, where x  s cos  and y = ssin .



Hence X =  pdy, Y=

get the complex potential for a doublet at the origin z

ei = 0 as w = . z VI.

c

So that Y+ iX =

 pdx. c

 p  dx  idy . c

BLASIU”S THEOREM (THRUST ON A CYLINDER)

An incompressible fluid moves steadily and irrationally under no external forces parallel to the zpane past a fixed cylinder whose section in that plane is bounded by a closed curve C. The complex potential for the flow is w. Then the action of the fluid pressure on the cylinder is equivalent to a force per unit length having components (X,Y) and a couple per unit length of moment M, where Y+ iX = 

P 2

 c

2

 dw   dz  dz;  

 P  M = Re   2

c

density.

And so Y + iX = c



  dx  idy   2  q c

2

 dx  idy 

c

=



 2

2

 q

dy   dx  ds  i ds  ds, since c  dx  idy   0   c



2

c

 2



But

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P 1 2  q  c,  2

1 c = constant we obtain, P = c  q 2 , 2 where q is the fluid velocity, p the pressure, and 

=

 d   z  w  d z  .. (7)  dz  



From Bernoulli’s equation

 q

 2

2

 cos   sin   ds

c

 q e

2  i

ds

c

dw   u  iv   qei dz

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q

2

e

2i

Y  iX  

 2

2

 dw   dz  and so    dw 

2

  dz  e

i

ds.

f  z    m log  z  f  (10)

By circle theorem, the complex potential with boundary z  a.

c

Since dz = dx +idy =(cos   i sin ) ds = ei ds, The moment about O of the force components where M = pxx + pyy, so that total moment about O is

py, px is M, M

 p  xdx  ydy  c

M

1 2   c  2 q   xdx  ydy   

 c

M 

 2

 q

2

 xdx  ydy  .

c

2

 dw  2 2i Now   dx  i dy   dz   x  i y  q e dz  

 a2   m log(z  f )  m log   f    z  a2   m log(z  f )  m log z  m log z   f   Neglecting constant term  mlog(f )].

Then differentiating with respect to z, we get m m m    dw zf z a2 z dz f Squaring both sides, we get 2

 q 2  x  i y   dx  i dy 

1 1 1 2  dw  2  2   dz   m [ 2 2   z  f  z  a2  z  f  z  z  f    2 2   ]   a2  a2   z  f   z   z  z   f  f   

 q 2  xdx  ydy  + iq 2  ydx  xdy 

If the above expression is put into partial fractions,

  x  i y  q 2 e2i ei ds  q 2  x  i y  ei ds

 q 2  x  i y   cos   i sin   ds

(8) Hence from Equation (7) and (8), we have   2  

M  Re 

2  dw    z dz  (9)   dz   

 c

Evaluation of the integrals for Y + iX and for M is effected using residue calculus.

VII.

w

 a2   f z  f    z 

A SOURCE OUTSIDE A CYLINDER

we obtain 2   z   f z

2 2  2    a a2   z  f   z   z  z   f  f    2f 2f 2f 2f 2 2  2 2 2 2 2 2  f  f  f  a  f  a2  a  a 2 zf z zf z a a z z f f 2

 dw  Therefore the sum of residues of   at z = 0  dz 

A. Without circulation around the cylinder We consider a source of strength m at a distance f centre of the cylinder z  a f  a . from the Then the complex potential due to the source without boundary is given by

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and z =

a2 is f

2f 2f 2f  2 m2   2  2  2 2 a a  f f a



2m2 a 2 f (f 2  a 2 )

Using the theorem of Blasius,

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© OCT 2019 | IRE Journals | Volume 3 Issue 4 | ISSN: 2456-8880   2 Y  iX

  ma 2  m    zc  a2  cz  z   c  

2

 dw    dz c  dz 



 2m2 a 2     2i  2 2  2  f (f  a ) 

Therefore Y  0 and X 

  ik   z    

Therefore

2m2 a 2 f (f 2  a 2 )

i

B. With circulation around the cylinder We shall consider here a source of strength m which is situated at the point z = c on the real axis outside the circular cylinder z  a and there is a circulation

2

m2 m2 (m  ik) 2 2m 2  dw       dz     z2   a2  z  c 2  a 2 2 (z  c) z   z     c c    2m(m  ik) 2m(m  ik)   (z  c)z a2 (z  )z c and hence 2

of a circulation of strength 2k around the cylinder.

m2 m2  dw     dz     z  c 2  a 2  z  c 

  

2



(m  ik) 2

 c c  2 2 2 2 c a a  c2  2m    zc a2 z  c

z2  1  1   c    2m(m  ik)   c  z c z    

 c c   2 2  2m(m  ik)  a 2  a  . z   a  z  c 

By Blasius’ theorem, we get Figure 3. With Circulation around the Cylinder

Y  iX  

 2

2

 dw    dz c  dz 



Then, by Circle theorem, we obtain

 a2  z   ik log   . a   z 

w  f (z)  f 



 a2  z  m log(z  c)  m log   c   ik log   . z a    Then

  m m  a2 dw    2   2 dz zc a  z  c  z

 ik    z a

 1  a 

2   a 2   dw  2i sum of residues of   at z = 0 and z = 2 c    dz 

 2m2 c 2m(m+ik) 2m(m+ik)c 2m(m+ik)c   -πρi  2 2 + + +  c a2 a2  a -c 

1  2mk   c c  c a

Y  iX  2m 2i  

c

2

 2  c  c 2  a 2  2mk  2m 2i    c c2  a 2  c  



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

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  

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

2m 2 a 2



c c2  a 2



i

Therefore we obtain Y =

2mk . c

2mk 2m 2 a 2 and X = . c c c2  a 2





If  is the angle made by the resultant thrust with the positive x axis, it is given by \

 = tan 1

Y X



 

 2mk c c  a  = tan 1   c 2m 2 a 2   = tan

1



 k c2  a 2   ma 2 

2

2

 

 

w=

  zf

z

w=

  a 2   2  z  f f f

 a2 f  z  f 

   1 z

a2 f

a 2 2   w=   f 2 , neglecting constant  zf f a z f

a 2 2 dw  Hence,   f 2 dz z  f a z f Squaring and expressing the result inpartial fractions,

  we obtain

VI. A DOUBLET OUTSIDE A CYCLINDER 2

Let a doublent of strength  be at a distance f from the

2 2

f 

due to the doublet with boundary z  a is given by

 a2  w = f z  f    z  w=

a4 f4

  dw    dz      z  f 4  a 2  z  f 

centre of the cylinder z  a.  f  a  .

Figure 4. Doublet Outside a Cylinder Then the complex potential due to the doublet at z = f is  f z  . zf By circle theorem, the complex potential

2

2

2

a2 f2

[

2f 2 2

 a2

z

a2 f

f

f3 2

 a2

zf

  f 3

  f 3

4

f2 2

 a2



2

 z  f 2

f2  a2



 a2  z  f 

  

2

  

2

2

By Boasiu’s theorem we obtain, 2

Y  iX = 

  dw  c  dz  dz 2

   Y  iX =   2i sum of residues of 2  

2 a2   dw   at z =   dz  f    

   z  f a2 f z

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© OCT 2019 | IRE Journals | Volume 3 Issue 4 | ISSN: 2456-8880  2 2 a 2 Y  iX =  i   2  f 

Y  iX =

4 2 a 2 f

f

2

 a2



3

 3  2f  2 f  a2 





3

     

i

Therefore Y=0 and X =

4 2 a 2 f

f

2

 a2



3

[4] M.E O’Neill, and F.Chorlton, “Ideal and Incompressible Fluid Dynamics”, John Wiley & Sons, New Yrok, 1986.

[5] J.W.Brown, R.V, Churchill, “Complex Variables and Applications”, 8th edition, Mc Graw-Hill Company, 2009.

(26)

VIII. CONCLUSION Finally, this paper is concluded that, Blasiu’s theorem stated that an incompressible fluid is moved by a steady irrotational fluid motion under no external force. Blasiu’s theorem is expressed by the action of the fluid pressure on the cylinder.

The Blasiu’s theorem gives a convenient formula for the force on a two- dimensional body in an incompressible potential flow field. The direct way to find the force on the body is to integrate the pressure forces over the surface.

ACKNOWLEDGEMENTS I want to thank Dr.Khin Maung Chin Pro Rector and Dr.Ngwe Kyar Su Khaing Assistant Professor of Technological University (Yamethin) for their permission, kindness and her urge wrote this research paper well.

I am also indebted to all the teachers at my department who permitted me and friends who gives me all kind of encouragement and necessary help to write this research paper.

REFERENCES [1] D.H Wilsom, “Hydrodynamics”, dword Amold (Publsiher) Ltd, London, 1966.

[2] F. Cholton,”Textbook of fluid dynamics”, D.Von Nostrand Co.Ltd, London, 1967.

[3] J.W... Brown, R...V. Churchill, “Complex Variables and Application,” 8 thedition, Mc GrawHill Company, 2009.

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