Vortex Motion fluid dynamics by MD raisinghania

11.1A. Introduction.

It is known that all possible motions of an ideal liquid can be subdivided into two classes; vortex free irrotational or potential flows, whose characteristics can be derived from a velocity potential ( , , , ), x y z t and vortex or rotational motions for which this is not the case. Rotational motions differ from potential flows in that, as the name applies, all particles of the fluid or at least part of them rotate about an axis which moves with the fluid. Potential flow, on the other hand, is irrotational by definition. So far we paid attention almost entirely to cases involving irrotational motion only. In the present chapter we wish to discuss the theory of rotational or vortex motion.

11.1B. Vorticity, vorticity components (or components of spin).

If q be the velocity vector of a fluid particle, then the vector quantity, Ω (= curl q ), is called the vorticity vector or simply the vorticity and is a measure of the angular velocity of an infinitesimal element. Let Ω _xi _y j _zk so that (_x, _y, ) are the vorticity_z components or the components of the spin. Then, if q = ui + vj + wk.

  x w/   y v/ z   y u/  z w x/   z v/  x u/y,

If x,y, are all zero, the motion is irrotational and the velocity function  existsz,

and if x,y,  are not all zero, the motion is rotational.z

In the case of two-dimensional motion, we know that w = 0 and u and v are functions of x and y only and hence for two-dimensional case,

_{ x 0,}   and _y 0      z ( v/ x) ( u/y)

It follows that in two-dimensional motion there can be at the most only one component of spin and its axis is perpendicular to the plane of the motion.

11.1C. Vortex line. [Kanpur 1999]

A vortex line is a curve drawn in the fluid such that the tangent to it at every point is in the direction of the vorticity vector. The differential equations of the vortex lines are

.

x y z

dx dy dz

 

  

In two dimensional motion, since axis of rotation at every point is perpendicular to this plane of motion and hence the vortex lines must be all parallel (because all vortex line will be perpendicular to the plane of motion).

11.1

11.1D. Vortex tube and vortex filament (or vortex). [Kanpur 2001] The vertex lines drawn through each point of a closed curve enclose a tubular space in the fluid which is called the vortex tube.

A vortex tube of infinitesimal cross section is known as a vortex filament or simply a vortex.

11.2. Helmholtz’s vorticity theorems. Properties of vortex tube.

(1) The product of the cross section and vorticity (or angular velocity) at any point on a vortex filament is constant along the filament and for all time when the body forces are conservative and the pressure is a single-valued function of density only

Let  be the vorticity vector and let ω be the angular velocity vector. Then we have Curlq and 2 . ...(1) LetS1,S2 be two sections of a vortex tube and let n1, n2 be the unit normals to these

sections drawn outwards from the fluid between them. Again, suppose S be the curved surface of the vortex tube and

S

 = total surface area of element =     S1 S S2

V

 = total volume which S contains. Then

. 0,

      



Sn dS



V dV by Gauss divergence theorem ...(2)

Since   curlq0. Hence (2) gives 1 2 0 S dS S dS S dS         



n 



n 



n  _...(3)

Since  is tangential to the curved surface of the vortex tube,n  0 at each point ofS. Hence (3) reduces to 1 2 S dS S dS      



n 



n  _or 1 2 S d S d      



 S



 S _...(4) or 1 2 2 2 S d S d      



 S



 S _...(5)

Thus, to the first order of approximation, (4) and (5) give

 1 S1  2 S2 and  1 S1  2 S2. ...(6)

Equation (6) shows that S or 2 is constant over every section SS  of the vortex tube. Its value is known as the strength of the vortex tube. A vortex tube whose strength is unity is called a unit vortex tube.

(2) Vortex lines and tubes cannot originate or terminate at internal points in a fluid. Let S be any closed surface containing a volume V. Then we have

0,

S d  S  dS  V  dV 



 S



n 





which shows that the total strength of vortex tubes emerging from S must be equal to that entering S. Hence vortex lines and tubes cannot begin or end at any point within the liquid. They must either form closed curves or have their extremities on the boundary of the liquid.

(3) Vortex lines move along with the liquid (i.e. they are composed of the same elements of the liquid) provided that body forces are conservative and the pressure is a single-valued function of density.

Let C be a closed circuit of liquid particles and let S be an open surface with C as rim. Then the circulation  is constant in the moving circuit C by Kelvin’s circulation theorem (refer Art. 6.4). Thus we have curl C d S d S d  

_

q r 

_

q S 

_

 S so that S d



 S = constant ...(7) since  is constant. Thus for a surface S moving with the fluid (7) holds. The L.H.S. of (7) represents the total strength of vortex tubes passing through S. This shows that the vortex tubes move with the fluid. By takingS 0, it follows that the vortex lines move with the liquid. 11.3. Illustrative solved examples.

Ex. 1. If _{u(ax by ) /(x}2_y2_{) v(ay bx ) /(x}2_y2_{) w = 0, investigate the nature of}

motion of the liquid. Also show that

(i) the velocity potential is 



( / 2) log (a  x2y2)btan ( / ) ,1 y x



(ii) the pressure at any point (x, y) is given by

2 2 2 2 1 const. 2 p a b x y      Sol. Given u(ax by ) /(x2y2) v(ay bx ) /(x2y2), w = 0 ...(1) From (1), 2 2 2 2 2 2 2 2 2 2 ( ) 2 ( ) 2 ( ) ( ) u a x y x ax by ay ax bxy x x y x y            and 2 2 2 2 2 2 2 2 2 2 ( ) 2 ( ) 2 . ( ) ( ) a x y y ay bx ax ay bxy y x y x y            v

We see thatu/  x v/ y 0 and hence the equation of continuity is satisfied by (1). Therefore (1) represents a possible motion. Moreover (1) represents a two-dimensional motion and hence vorticity components are given by

 x 0,  y 0,   z ( v/  x) ( u/y) ...(2) From (1), 2 2 2 2 2 2 2 2 2 2 ( ) 2 ( ) 2 ( ) ( )              u b x y y ax by bx by ayx y x y x y and 2 2 2 2 2 2 2 2 2 2 ( ) 2 ( ) 2 ( ) ( ) b x y x ay bx bx by axy y x y x y             v

so from (1),  z 0. Thus,      x y z 0, showing that the motion is irrotational.

Part (i) We have, d   ( / x dx)   ( / y dy) = – udx – vdy

2 2 2 2 2 2 2 2

2 2

2

ax by ay bx a xdx ydy xdy ydy

dx dy b x y x y x y x y          _  _ _   _        

Thus,   ₂ {log ( 2 2)} _tan1 _

 

a y

d d x y b d

x Integrating,   



( / 2) log (a  x2y2)btan ( / ) .1 y x



Part (ii). Let q be the fluid velocity. Then, pressure is given by _{p/ q}2_{/ 2 const. or p/ const.– (q}2_{/ 2) ...(3)} But 2 2 2 2 2 2 2 2 ( ) ( ) , ( ) ax by ay bx q u x y        v using (1) or 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ( ) ( ) ( ) a x y b x y a b q x y x y         ...(4)

Using (4), (3) gives the required result.

Ex. 2. Prove that the necessary and sufficient condition that the vortex lines may be at right angles to the stream lines are

u, v, w, = (      / x, / y, / z), where µ,  are functions of x, y, z, t.

[G.N.D.U. Amritsar 1998; I.A.S. 2005; Agra 2001, 03, 05] OR

Find the necessary and sufficient condition that vortex lines may be at right angles to the

streamlines. [G.N.D.U. Amritsar 2002; Kanpur 1999]

Sol. Streamlines are given by dx/u = dy/v = dz/w ...(1) and vortex lines are given by dx/ _x dy/ _y dz/_z ...(2) (1) and (2) will be at right angles, if u _x v   _y w _z 0 ...(3) But   x w/   y v/ z,   y u/  z w x /      z v/ x u/y ...(4)

Using (4), (3) may be re-written as

(u w /    y v/ z) v(u/  z w/ x) w(   v/ x u/y)0,

which is the necessary and sufficient condition in order that udx + vdy + wdy may be a perfect differential. So we may write

udx dy wdz     d __xdx__ydy_zdz

 

v

 u   ( / x), v   ( / y) and w   ( / z). Ex. 3. In an incompressible fluid the vorticity at every point is constant in magnitude and direction; show that the components of velocity u, v, w are solutions of Laplace’s equation.

[Agra 2012; I.A.S. 2004; Meerut 1999; Rajasthan 2000, 05] Sol. Let _x, _y,_zbe the components of vorticity  so that       ( 2x 2y 2 1/ 2z ) and direction cosines of its direction are  x/ ,  y/ ,  z/ .

 and its direction being given, it follows that x,y,z are all constants. Moreover,,

/

  x w    y v/ z,   y u/  z w x/ ,   z v/  x u/y ...(1)

Differentiating second equation in (1) w.r.t. ‘z’ and third equation in (1) w.r.t ‘y’ and then subtracting, we get 2 2 2 2 2 2 0 u w u x z y x z y               v _or 2 2 2 2 0 u u w x y z z y          _  _        v ...(2) The equation of continuity is

/ / 0

u x y w z

Using (3), (2) reduces to _2_u/x2_{ }2_{u y/} 2_{ }2_u/z2 _0,

showing that u satisfies Laplace’s equation. Similarly we can show that v and w also satisfy Laplace’s equation.

Ex. 4. Assuming that in an infinite unbounded mass of incompressible fluid, the circulation in any closed circuit is independent of time, show that the angular velocity of any element of the fluid moving rotationally varies as the length of the element measured in the direction of the axis

of rotation. (Guwahati 2000; 03; Roorkee 2000)

Sol. Let  be the angular velocity of the element ds and let  be its area of cross-section. Then circulation in any closed circuit surrounding this element is 2 Let the element ds form. a part of the vortex filament s so that the circulation is constant all along this element. Furthermore, since circulation is assumed to be independent of time, we get

2 constant. ...(1)

Again, since the liquid is incompressible dsconstant. ...(2) Dividing (1) by (2), we get /dsconstant,which proves the required result.

Ex. 5. It udx + vdy + wdz = d    where ,   are functions of x, y, z, t, prove that thed vortex lines at any time are the lines of inter- section of the surfaces  = constant and  = constant (Agra 2012) Sol. Given udxvdy wdz     d d

   _ _ _ _  _ _ _ _    udx dy wdz dx dy dz dt dx dy dz dt x y z t x y z t v , , and 0 u dx x dy y w z z t t             _        _{ }       _      v ...(1)

Hence the components of spin x, y,zare given by

2     _  _  _  _,           x w y z y z z z y y v using (1) 2 2 2 2 y z y z y z z y z y z y y z z y                                           or 2 x  _{ }/_/  _{ /}/ y z y z Similarly, 2 / / / /           y z x z x and / / 2 / /           z x y x y  / / / 2 / / / 0. / / /                         _{  }   x y z  _{     } x y z x y z x y z x y z             _x( / x) _y( / y) _z( / z) 0. ...(2) Similarly, we have            _x( / x) _y( / y) _z( / z) 0. ...(3) Equations (2) and (3) show that the vortex lines at any time are the lines of intersection of the surfaces  = constant and  = constant.

EXERCISE 11 (A)

1. Each particle of a mass of liquid is revolving uniformly about a fixed axis with the angular velocity varying as the nth power of the distance from the axis. Show that the motion is irrotational only if n + 2 = 0.

If a very small spherical portion of the liquid be suddenly solidified, prove that it will begin to rotate about a diameter with an angular velocity (n + 2)/2 of that with which it was revolving about the fixed axis.

11.4. Rectilinear vortices.

Vortex lines being straight and parallel, all vortex tubes are cylindrical, with generators perpendicular to the plane of motion. Such vortices are known as rectilinear vortices.

Derivation of velocity potential, stream function, velocity components and complex potential due to a rectilinear vortex filament.

Consider a rectilinear vortex filament with its axis parallel to the axis of z. The motion being similar in all planes parallel to xy-plane, we have no velocity along the axis i.e. w = 0. Moreover u and v are independent of z i.e.

u/ z 0 and v/ z 0 ...(1) If x, y,z be the vorticity components, then

 x 0,  y 0 and      z v/ x u/y ...(2)

Now the equations of lines of flow are

dx/u = dy/v i.e. vdx – udy = 0. ...(3) The equation of continuity is u/  x v/ y 0

so that     v/ y ( u) /x ...(4)

Equation (4) shows that vdx udy– must be perfect differential, d (say). Thus, we have vdx udy   ( / x dx)   ( / y dy)

so that u    and ( / y) v  / x. ...(5) Then the lines of flow are given by d  0 i.e.  = const. Hence  is the stream function. Using (5), (2) gives     _z 2 / x2   2 / y2 ...(6) Thus the stream function  satisfies

_{  }2 _/ 2_{   }2 _/ 2 _{ ,}

z

x y on the vortex filament ...(7A)

= 0, outside the filament ...(7B) Let P (r,  ) be any point outside the vortex filament. Since the motion outside the vortex is irrotational, the velocity potential  exists such that

1 .

r r

 

 

  ...(8)

Moreover, outside vortex filament,  satisfies the equation (re-writing (7B) in polar

co-ordinates) 2 2 2 2 2 1 1 0. r r r r            ...(9)

There being symmetry about the origin,  must be independent of  and so (9) reduces to

2 2 1 0     d d r dr dr or 1 0   _     d d r r dr dr

Integrating, (r d/dr)c c being an an arbitrary constant, ...(10)

Integrating (10),  clog .r ...(11)

Since  is independent of , (8)  d/dr–(1/ ) (r  / ) ...(11) Now (10) and (11) give c 1

r r

  

 so that    c . ...(12) If w(    be the complex potential outside the filament, then we havei )

w   c iclogric(logr i  ) iclog (rei)iclog .z Let k be the circulation in the circuit embracing the vortex. Then, we have  2 2 0 0 1 2 , _{ }   _ _        





k rd c d c r by (12)  ck/ 2 . Hence, we have

  ( / 2 ) ,k    ( / 2 ) logk  r and w( / 2 ) log ,ik  z ...(13) Here k is called the strength of the vortex.

If there be a rectilinear vortex of strength k at z₀ (= x₀ + iy₀), then

0

( / 2 ) log ( )

w ik  z z ...(14)

We now determine velocity components due to a rectilinear vortex of strength k at A₀ (z₀). Let P (x, y) be any point in the fluid. Then, if r₀ be the distance between A₀ (z₀) and P (z),

r02(x x 0)2(y y 0)2 and  ( / 2 ) logk   r0  0 0 0 0 2 0 2 0 0 2 r k y y k y y u y r y r r r                   ...(15) and 0 0 0 0 2 0 2 0 0 2               r k x x k x x x r x r r r v _...(16) Hence, 0 2 2 2 2 0 0 2 0 {( ) ( ) } , 2 2 k k q u x x y y r r          v ...(17)

which gives velocity at P (x, y).

Remark 1. Some authors defineKk/ 2 as the strength of the vortex. So, they take

0 0 0 0 2 2 0 0

, log , log , log ( )

, ,              _    _  K K r w iK z w iK z z y y x x K u K q r r v = K r ...(18)

However, we shall not use these results in the present discussion unless otherwise stated. Remark 2. The case of several rectilinear vortices.

Let there be a number of vortices of strengths k₁, k₂, k₃, … situated at z₁, z₂, z₃, … Then the complex potential is given by

3

1 2

1 2 3

log ( ) log ( ) log ( )

2 2 2 ik ik ik w z z  z z  z z      ₂ nlog ( n) i k z z   



...(19)

Here vortices of strengths k₁, k₂, k_3, … are situated at (x₁, y₁), (x₂, y₂), (x₃, y₃), … Hence, using (15) and (16), the velocity components u and v due to these vortices at any point outside there filamentsare given by

2 1 2 n n n y y u k r    



and 2 1 2 n n n x x k r   



v _...(20) where r = (x – x_n2 _n)2 _{+ (y – y} n)2, n = 1, 2, 3,… ...(21)

Result (20) may also be deduced from ₂ n

n k dw i u iv dz z z        

Let k_m be the strength of a vortex situated at (x_m, y_m). Then we omit the term containing k_m while finding the velocity of that vortex. Thus the motion of the m th vortex situated at (x_m, y_m) is given by 2 1 , 2 m n m n mn y y x k r    



 _and 1 ₂ 2 m n m n mn x x y k r   



 _...(22) where m n and rmn2 = (xm – xn)2 + (ym – yn)2. ...(23) Using (22), we have 1 ₂ 0, 2 m n m m m n mn m m n y y k x k k r     





 

_...(24)

since m, n can be interchanged and the denominator is positive.

Similarly, m m 0

m

k y 



 ...(25)

Since k_m is independent of t, integration of (24) and (25) yield



k x m m const. and



k y m m const. ...(26)

Also, x 



k xm m

/



km and y



k ym m

/



km. ...(27)

Using (26), (27) show that ,x y are constants. Hence if k₁, k₂, k₃, … be supposed to be the masses situated at z₁, z₂, z₃, …, then their centre of gravity is fixed throughout the motion. This point is known as the centre of vortices. Thus if there be several vortices, they move in such a manner that their centre is stationary. If



k _m 0, the centre of vortices is at infinity or else indeterminate

Remark 3. Single vortex in the field of several vortices.

‘‘To show that a single rectilinear vortex in an unlimited mass of liquid remains stationary, and when such a vortex is in the presence of other vortices it has tendency to move of itself but its motion through the liquid is entirely due to the vortices caused by the other vortices.”

Proof. The value of stream function at any point inside of a circular vortex tube is given by 2 2 1 2 d d r dr dr      or 1 d rd 2 r dr dr   _{ }     or 2 d d r dr dr dr         

Integrating it twice,  (1/ 2) r2c₁logr c c c ₂, ,_{1 2} being arbitrary constants ...(1)  Velocity at right angles to the radius vector  d /dr  r c r/ ...(2) Since the velocity at the origin is finite, c must be zero. Then (2) gives

(d/dr)r0 0,

showing that the velocity at the origin due to a single vortex must vanish. It follows that a vortex filament (vortex) induces no velocity at its centre. Thus, if a vortex is in the presence of other vortices it has no tendency to move of itself but its motion through the liquid will be caused by the other vortices.

11.5. Two vortex filaments.

Case I. When the filaments are in the same sense. Let us consider two rectilinear vortices of strengths k₁ and k₂ at A₁ (z = z₁) and A₂ (z = z₂). Then complex potential due to stationary system is

w(ik1/ 2 ) log ( z z 1) ( ik2/ 2 ) log ( z z 2) ...(1)

However, vortices situated at A₁ and A₂ would start

moving due to the presence of each other. Let u₁, v₁ be the components of the velocity q₁ of A₁ which is due to A₂ alone. Then, we have

1 1 2 1 1 1 1 2 1 1 2 _{z z} 2 ik dw ik u i z z dz _ z z     _  _{ }           v ...(2) 1 1 1 2 2 1 2 1 2 | | 2 | | 2 ( ) k k q u i z z A A        v _...(3) Similarly, 2 2 1 2 1 1 2      ik u i z z v ...(4) and q2 k1/ 2 ( A A1 2) ...(5) From (2) and (4), (u1iv1) /k2 (u2iv2) /k1 i.e. k₁ (u₁ – iv₁) + k₂ (u₂ – iv₂) = 0 or (k₁u₁ + k₂u₂) – i (k₁v₁ + k₂v₂) = 0 so that k₁u₁ + k₂u₂= 0 and k₁v₁ + k₂v₂= 0. ...(6)

Sincek1k20, (6) shows that a point G, the centroid of masses k1, k2 at z1 and z2, moving

with velocities (u₁, v₁), (u₂, v₂) is at rest. Hence the line A₁A₂ rotates about G. Since G is C. G. of k₁ and k₂, we have k₁ A1G = k2 A2G or 1 2 2 1 1 2 2 1 1 2 1 2 A G A G A G A G A A k k k k k k       so that 1 2 1 2 1 2 , k A G A A k k   and 1 2 1 2 1 2 k A G A A k k   ...(7) Re-writing (3), we have 1 2 1 2 1 2 ₂ 1 1 2 1 2 , 2 ( ) k A A k k q A G k k A A        ...(8) where 1 2 ₂ 1 2 2 ( ) k k A A     and 2 1 2 1 1 2 .   k A A A G k k ...(9)

Thus angular velocity of A₁ is  about G. Similarly, we may show that the angular velocity of A₂ is  about G. Hence the line A₁A₂ revolves about G with uniform angular velocity .

Remark. As a particular case, let k₁= k₂= k and A₁A₂= 2a. Then, we have q1k/ 4 a q2 k/ 4 and a  k/ 4a2

and the stream function is given by

 ( / 2 ) logk   r1( / 2 ) logk   r2 ( / 2 ) log (k   r r1 2)

where r₁= A₁P, r₂= A₂P and P is any point in the fluid. The streamlines are given by const,

  i.e. r₁ r₂= const, which are Cassini’s ovals.

A1 k1 r1 G k2 A2 r2 P r( , )

Case II. When the filaments are in the opposite sense.

Let k₁ and k₂ be of opposite signs. Then G will not lie in between A₁ and A₂. However, if k₁ > k₂, then G will lie on A₁A₂ produced and if k₂ > k₁, it will lie on A₂ A₁ produced. As before, it can be shown that the line A₁ A₂ revolves about G with uniform angular velocity . 11.6A. Vortex pair.

Two vortex filaments of strengths k and – k form a vortex pair. Let us consider two rectilinear vortices of

strengths k and – k at A₁ (z = z₁) and A₂ (z = z₂). Then complex potential at any point P(x, y) due to stationary system is 1 2 log( ) log ( ) 2 2 ik ik w z z  z z   ...(1)

However, the vortices situated at A₁ and A₂ would start moving due to the presence of each other. Let u₁, v₁ be the components of the velocity q₁ of A₁ which is due to A₂ alone. Then,

1 1 1 1 1 2 1 1 2 _{z z} 2 ik dw ik u i z z dz _ z z     _  _{ }          v _...(2)  1 1 1 1 2 1 2 | | 2 | | 2 ( ) k k q u i z z A A        v _...(3) Similarly, 2 2 2 1 1 2 ik u i z z     v _...(4) and q2k/ 2 ( A A1 2) ...(5) Let q₁= q₂= q, (say) ...(6)

Thus the velocity q₁ of A₁ due to A₂ is q perpendicular to A₁ A₂. Similarly, the velocity q₂ of A₂ due to A₁ is q perpendicular to A₂ A₁ in the same sense as that of A₁. Hence the vortices situated at A₁ and A₂ move in the same direction perpendicular to A₁ A₂ with uniform velocity q. However, the line may move forward or backward according to the directions of rotation.

Let w    z = (x, y), zi , ₁= (x₁, y₁) and z₂= (x₂, y₂).  (1)     i ( / 2 ) log[(ik  x x 1)i y y(  1)] ( / 2 ) log[( ik  x x 2)i y y(  2)]

Equating imaginary parts of both sides, we get

2 2 2 2

1 1 2 2

( / 4 ) log[(k x x) (y y) ] ( / 4 ) log[(k x x ) (y y ) ]

          

or  ( / 4 ) (logk  r₁2logr₂2) ( / 4 ) log ( / ) k  r r_{1 2} 2( / 2 ) log( / )k  r r_{1 2} where r₁= A₁P and r₂= A₂P.

The streamlines are given by  const. i.e. r₁/r₂= const. which clearly form a system of coaxial circles having A₁ and A₂ as their limiting points. 11.6B. Vortex doublet or dipole.

A vortex pair consisting of two vortices of strengths k and – k at a distance s apart, where 0

s

  and k   in such a manner that    s ( / 2 )k  is finite, is called a vortex doublet of strength µ. To simplify out working, we take    where s 2   as 0  s 0.

Thus    2 ( / 2 )k   ( ) /k  ...(1) A1 r1 –k A2 r₂ P x, y( ) k

Consider two vortex filaments of strengths k and – k at

1( i )

A z e and A z2(   ei) so that A A   . Then A1 2 2 1 A2

will be the axis of the doublet, inclined at an angle  to the x-axis. The complex potential is given by

( / 2 )[log ( i ) log ( i )] w ik  z e  z e ₂ log 1 log 1 i i ik e e z z            _  _ _  _       2 2 3 3 2 2 3 3 2 3 2 3 2 i i i i i i ik e e e e e e z z z z z z                   _   _     2 , 2 i i i ik e ik e z re             as z = ree i

[To first order of approx. of small quantity  ] Thus, w _{  ( / )i r e}i()_{, by (1) ...(2)}

      i ( / )[cos (i r    ) isin (  )]

   ( / ) sin (r    and )    ( / ) cos (r   ) ...(3) As a particular case, let the doublet be at O and the axis along y-axis. Then putting  = /2 in (3), we have

  ( / ) cosr  and    ( / ) sinr  ...(4) Remark 1. If  Ua2, then (4) gives   (Ua2/ ) sin ,r 

which is the stream function for a circular cylinder of radius a moving with velocity U along the x-axis. Thus the motion due to a circular cylinder is the same as that due to a suitable vortex doublet placed at the centre with axis perpendicular to the direction of motion.

Remark 2. Some authors define v = 2 as the strength of the vortex doublet. Then (4) takes the form

 ( / 2 ) cosv r  and   ( / 2 )sin .v r  ...(5) 11.7. Motion of any vortex.

When there are any number of vortices in an infinite liquid, we can find the motion of any one of them. It depends not on itself but on others, hence to find the motion we have to subtract from the stream function of the system the term that corresponds to it.

Let there be a number of vortices of strengths k₁, k₂, k₃, … situated at z₁, z₂, z₃, … respectively, where z_n = x_n + iy_n.Then the complex potential of the system at any outside point is given by

( / 2 ) nlog ( n) w i 



k z z or    i ( / 2 )i 



knlog[(x x n)i y( yn)] or ₂ n 1₂log (



n)2 ( n)2



tan 1 n n y y i i k x x y y i x x         _     _ 



_  _  log (



)2 ( )2



4 n n n k x x y y      



 The stream function  at the vortex (x_m, y_m) is given by

log (



)2 ( )2



log (



)2 ( )2



4 4 n m n n m m k k x x y y x x y y            



A1 k  y X X y A2 –k O – 

If  be the stream function for the motion of the vortex (x_m, y_m), we have        _{ } y_{m } y _m and _m m x x           

by equating the components of velocity of the vortex (x_m, y_m).

Suppose there is a single vortex k at (x₁, y₁) in front of a fixed wall taken as y = 0. We have to introduce the image – k at (x₁, – y₁) and the stream function of the system is

 ( / 4 ) log (k 



x x 1)2(y y 1)2



( / 4 ) log (k 



x x 1)2(y y 1)2



    ( / 4 ) log (k 



x x 1)2(yy1)2



 – 1 1 1 2 2 1 1 1 1 , 2( ) 4 ( ) ( ) _{ }        _{ }                 x x y y y y k y y x x y y 1 2 1 1 4 4 4 4 y k k y y      and 1 1 1 2 2 , 1 1 1 1 2( ) 0 4 ( ) ( )        _  _{ }  _              x x y y x x k x x x x y y    ( / 4 ) logk  y1

Hence the path of the vortex is the streamline for the vortex, i.e. y₁ = constant. 11.8. Kirchhoff vortex theorem. General system of vortex filament. (Agra 2011)

If ( , ),r 1 1 ( ,r 2 2), …, ( ,r n n) be the polar coordinates at any time t of a system of

rectilinear vortices of strengths k₁, k₂…, k_n then 1 , n p p p k x A  



1 , n p p p k y B  



2 1 , n p p P k r C  



2 1 , n p p p P k r D   





where A, B, C, D are constants and   p d p/ .dt



Proof. The complex potential w due to n vortex filaments of strengths k_p at the points

(cos sin ) p p p p p p z x iy r  i  is given by 1 log ( ). 2 n p p P ik w z z    



Hence, the velocity at any point of the fluid, not occupied by any vortex is given by

1 . 2 ( ) n p p P ik dw u i dz _ z z       



v

Since the velocity (u_p, v_p) of vortex k_p is produced by the remaining other vortices (because any particular vortex cannot move solely on its own account), we have

log ( ) 2 p _p p q p p q q p z z _{z z} dw d ik u i z z dz _ dz _          _{ }         _



_ v 2 ( ) q p q q p ik z z     



...(1)

Multiplying (1) by k_p and summing up from p = 1 to p = n, we obtain

1 1 ( ) 0, 2 ( ) n p q p p p p q p p q p ik k k u i z z         



v

 

...(2)

the double summation on R.H.S is zero because the terms cancel in pairs, for example ik_p k_q/(z_p – z_q) cancels ik_p k_q/(z_q – z_p) and there are no terms in k etc.2p

Equating real and imaginary parts, (2) gives



k u _{p p} 0 and



kp pv 0 or k_p dxp 0 dt 



and k_p dyp 0. dt 



...(3) Integrating (3),



k x_{p p} A and



k y_{p p} B, where A and B are constants of integration.

Again, multiplying (1) by k_pz_p and summing from p = 1 to p = n, we obtain 1 1 ( ) 2 ( ) n n p q p p p p p p q p p q p ik k z k z u i z z        



v

 

or 1 1 ( ) ( ) 2 n n p q p p p p p p p q p p q p k k z i k x iy u i z z         



v

 

or 1 ( ) ( ) , 2 n p p p p p p p p p p q p i k x u y i x y u k k          _



v v



_...(4)

because in the double summation on R.H.S., the sum of pairs of terms such as k_p k_q z_p/(z_p – z_q) and k_p k_q z_q/(z_q – z_p) reduces to k_p k_q and there are no terms in 2_.

p

k Equating real and imaginary parts, (4) gives



k_p(x u_{p p}y_{p p}v )0 ...(5) and ( – ) 1 constant , 2 p p p p p p q k x y u  k k  D 



v



say ...(6) Re-writing (5), we have k_p 2x_pdxp 2y_p dyp 0 dt dt        



or ₍ 2 2_{) 0} p p p d k x y dt  



or 2 0 p p dr k dt 



2 2 2 p p p r x y       Integrating, 2 _, p p k r C



where C is constant of integration.

From (6), p p p p p dy dx k x y D dt dt        



...(7)

But, y_p/x _p tan_{p } xprpcospandyp rpsinp_ Differentiating both sides of the above equation w.r.t. ‘t’, we get

₂ sec2 p p p p p p p dy dx x y _d dt dt dt x  _   or _p p _p p 2_psec2 _p p _p2 p dy dx x y x r dt  dt        cos p p p x r        (7) can be re-written as



k rp p2 p D.

11.8A. Illustrative solved examples

Ex. 1. Verify that the stream function  and velocity potential  of a two-dimensional vortex flow satisfies the Laplace equation.

Sol. We know that the stream function  and velocity potential  of a two-dimensional vortex flow are given by

  (k/ 2 ) ...(1) and  ( / 2 ) logk  r ...(2) From (1),   / r 0,   2 / r20,    2 / 2 0. so that 2 2 2 2 2 2 1 1 0. r r r r               ..(3) Form (2), , 2 k r r     2 2 ₂ 2, k r r       2 2 0     so that 2 2 2 2 2 2 1 1 0. r r r r               ...(4)

The equations (3) and (4) show that  and  satisfy Laplace equation.

Ex. 2. (a) When an infinite liquid contains two parallel, equal and opposite rectilinear vortices at a distance 2a, prove that the streamlines relative to the vortex are given by the equation,

2 2 2 2 ( ) log , ( ) x y a y c a x y a      

the origin being the middle point of the join, which is taken for axis of y. [Kanpur 2009] (b) Show that for a vortex pair the relative streamlines are given by k{(y/2a) + log (r₁/r₂)} = constant, where 2a is the distance between the vortices and r₁, r₂ are the distances of any point from them.

Sol. Part (a) Let there be two rectilinear vortices of strengths k and – k at A₁ (z = 0 + ia) and A₂ (z = 0 – ia) respectively. Thus A₁ A₂ = 2a, origin being the middle point of A₁ A₂ and y-axis being taken along A₁ A₂ as shown in figure. Here we have a vortex pair and hence (by Art. 11.6) the vortex pair will move with a uniform velocity k/(2A A or / 41 2) k  perpendicular toa

the line A₁ A₂ (i.e. along the x-axis). To determine the streamlines relative to the vortices, we must impose a velocity on the given system equal and opposite to the velocity / 4k a of motion of the vortex pair. Accordingly, we add a termkz/ 4a to the complex potential of the vortex pair.. Note that

_dzd ₄kz_a _4k_a,

 

and hence the term added is justified. So, for the case under consideration, the complex potential is given by

w( / 2 ) log (ik  z ia ) ( / 2 ) log ( ik  z ia ) ( kz/ 4a) or    i ( / 2 ) log{ik  x i y a (  } ( / 2 ) log { ik  x i y a (  )}k x iy(  ) / 4xa

Equating the imaginary parts, we have

2 2 2 2 log[ ( ) ] log [ ( ) ] 4 4 4 k k ky x y a x y a a            A1 P(x, y) a a O A2(0, – ) a –k Y X k Y (0, )a

or 2 2 2 2 ( ) log . 4 ( ) k x y a y a x y a  _{ }       _   _ ...(1)

Hence the required relative streamlines are given by  const., that is,

2 2 2 2 ( ) log . ( ) x y a y c a x y a      

Part (b). As in part (a), do upto (1). Let r₁= A₁P and r₂= A₂P so that 2 2 2

1 ( )

r x  y a

and 2 2 2

2 ( ) .

r x  y a Putting these in (1) of part (a), we obtain

2 1 1 2 2 2 log log 4 2 2 r r k y k y a r a r       _  _ _  _     ...(2)

Hence the relative streamlines are given by  const., i.e. k{(y/2a) + log (r₁/r₂)} = const. Ex. 3. An infinite liquid contains two parallel, equal and opposite rectilinear vortex filaments at a distance 2a. Show that the paths of the fluid particles relative to the vortices can be represented

by the equation 2 2 2 2 2 cos cos log const. 2 cos r a ar r a r a ar          [Kanpur 1997]

Sol. To obtain the desired result, modify solution of example 2(a) as follows : Let the vortex pair lie along x-axis in place of y-axis. Then interchanging x and y, we obtain

2 2 2 2 ( ) log 4 ( )          _   _ k y x a x a y x a or 2 2 2 2 2 2 2 log . 4 2        _  _ _    _ k x y a ax x a x y a ax ...(1)

Let x r cos , yrsin . Then, in polar coordinates, (1) takes the form 2 2 2 2 2 cos cos log . 4 2 cos        _  _ _    _ k r a ar r a r a ar ...(2)

Hence the relative streamlines are given by  const.,

i.e. 2 2 2 2 2 cos cos log const. 2 cos r a ar r a r a ar         

Ex. 4. If two vortices are of the same strength and the spin is the same in both, show that the relative stream lines are given by log (r4 + a4 – 2a2r2 cos 2  ) – (r2/ 2a) = const.,  being measured from the join of vortices, the origin being its middle point, 2a being the distance between the vortices.

Sol. Let there be two vortices each of strength k at C (a, 0) and D (– a, 0) and let O, the middle point of C D be taken as the origin as shown in figure. Then, by case I of Art. 11.5, line DC revolves about O with uniform angular velocity  given by

2 2 2 1 2 2 2 ( ) 2 (2 ) 4 k k k k A A a a         ...(1)

Let P(r,  ) be any point in the fluid. Then, CP = r₁= (r2 + a2 – 2ar cos  )1/2 and DP = r₂= (r2 _{+ a}2 _{+ 2ar cos  )}1/2

D k O  k X C Y P r2 r₁ r

Corresponding to the given vortices at C and D, the stream function  is given by

1/ 2

2 2 2 2

1 2 1 2

log log log ( ) log ( 2 cos ) ( 2 cos )

2 2 2 2 k k k k r r r r  r a ar r a ar       _       _     2 2 2 2 2 2 4 4 2 2 2

log[( ) 4 cos ] log[ 2 (2 cos 1)]

4 4

k k

r a a r r a a r

        

 

Thus, _{ ( / 4 ) log (k   r}4_a4_{2a r}2 2_{cos 2 ) ...(2)}

Since the vortices are moving with angular velocity  about O, there would be linear velocity due to vortex system. To determine the streamlines relative to the vortices, we must impose velocity  on the given system so as to reduce to the whole system to rest. If r  be the stream function to produce the desired velocity, then we have

₂ 4 kr r r a          , by (1) so that 2 2. 8 kr a      ...(3)

Hence for the case under consideration, the stream function      is given by( )   ( / 4 ) log (k  _ r4a42a r2 2cos 2 ) r2/ 2a2_. ...(4) Hence the required streamlines are given by  const.,

i.e. log (r4 + a4 – 2a2r2 cos 2  ) – (r2/2a2) = const.

Ex. 5. Three parallel rectilinear vortices of the same strength k and in the same sense meet any plane perpendicular to them in an equilateral triangle of side a. Show that the vortices all move round the same cylinder with uniform speed in time _(42 2_{a )/3k . (Kanpur 2011)}

Sol. Let r be the radius of the circumcircle of the equilateral triangle ABC. Let O be the circumcentre. From figure, r = OB = (a/2) sec 30° = / 3.a There are three vortices of strength K at A, B, C which are situated at the points 2m i/ 3_,

m

z re  m = 1, 2, 3. Then the complex potential of the vortices at B, C, A, is given by

2 /3 4 /3 6 /3

( / 2 ) [log ( – i ) log ( i ) log ( i )]

w ik   z re   z re   z re  ( /2 ) log (ik   z3r3) Then the velocity induced at _zre6 / 3i _{r, by others is given by}

1 1 3 3 _{2 2} 2 log ( ) ( ) 2 2 2 d ik ik ik z r u i z r z r dz z zr r       _    _         v Thus, 1 1 1 _{2 2} 2 | | 2 _{z r} 2 k z r k q u i r z zr r        _{ }        v

circumference of the circumcircle The required time

velocity at z r   2 2 2 2 / 3 4 4 , /2 3 3 3 a ra a a as r k r k k        

Ex. 6. If n rectilinear vortices of the same strength k are symmetrically arranged as generators of a circular cylinder of radius a in an infinite liquid, prove that the vortices will move round the cylinder uniformly in time 82 2a / (n1) ,k and find the velocity of any part of the liquid.

[Kanpur 2005, 09; Rohilkhand 2000; Nagpur 2003, 05; Himanchal 1999, 2003; Kurukshetra 1999] 30° 30° 30_° M N B A C a/2 a/2 O L

Sol. Let A₁ A₂ A₃ be the circle of radius a. Suppose that n rectilinear vortices each of strength k be situated at points 2 im n/ _,

m

z ae m = 0, 1, 2,…, n – 1 of the circle. Then the complex potential due to these n vortices is given by

1 1

2 / 2 /

0 0

log ( ) log ( ) log ( ).

2 2 2 n n im n im n n n m m ik ik ik w z ae z ae z a             









[using a well known result of algebra] Now the fluid velocity q at any point out of all the n vortices is given by

1 1 2 2 n n n n n n dw ik nz kn z q dz z a z a         

Again the velocity induced at A₁ (z = a), by others is given by the complex potential

log ( ) log ( ) log

2 2 2 n n n n ik ik ik z a w z a z a z a             w ( / 2 ) log (ik  zn1zn2azan2an1) so that 2 2 2 1 2 2 1 ( 1) ( 2) 2                  n n n n n n n dw ik n z n z a a dz z z a za a  ₂ ( 1) ( 2) 2 1 ₄( 1)           _{ }   _{ }  z a dw ik n n ik n dz na a [By algebra (n 1) (n2) ... 2 1 {(    n1) / 2} {( n 1) 1]n n( 1) / 2 or 1 1 ( 1) , 4 z a dw ik n u i dz _ a       _{ }      v

so that u₁= 0 and v1k n( 1) / 4a. If qr and q be the radial and transverse velocity components

of the velocity at z = a, then we have q_r = 0 and q_ = k(n – 1)/4a. Due to symmetry of the problem, it follows that each vortex moves with the same transverse velocity (k n1) / 4 . Hencea

the required time T is given by

2 2 2 8 . ( 1) / 4 ( 1) a a T k n a n k       

Ex. 6 (b). Prove that if n rectangular vortices of equal strength k are symmetrically arranged as generators of a right cicular cylinder of radius a and infinite length in an incompressible liquid, then the two dimensional motion of the liquid is given by w( / 2 ) log (ik  znan), the origin of co-ordinates being the centre of the cross-section of the cylinder. Show that the vortices move round the cylinder with speed (n1) / 4 .k a [G.N.D.U. Amritsar, 1999, 2001]

Sol. Proceed as in Ex. 6 (a)

Ex. 7. If ( , ), ( ,r1 1 r2 2), be polar coordinates at time t of a system of rectilinear vortices,

of strength k₁, k₂, .... prove that kr2 const. and 2

1 2

(1/ 2 ) .

kr k k

    

[Agra 2003, 10; Allahabad 2000; Jadavpur 2003, 04; Kanpur 2001; Mumbai 1999; Rohilkhand 2000] O 2 /n 2 /n 2 /n A3( )z3 A2 A1 ( )z2 ( )z1 An(z )n y X

Sol. Refer remark 2 of Art. 11.4. We have 2 2 1 2 , 1 2 m n m n mn n m n m n mn n y y x k r x x y k r     _  _   _    _





  mn _{... (1)} where r_mn2 (x_mx_n)2(y_my_n) .2 ... (2)  ( ) 1 ( )₂ ( ) 2 m m n m m n m m m m m n m mn m m n x y y y x x k x x y y k k r       



 

 

, using (1) 1 ₂ 0, 2 n m m n m n mn m n x y x y k k r    



 

since m and n can be interchanged.

 _m ( _m2 _m2) 0 m d k x y dt  



or 2 0. m m m dr k dt 



 Integrating, m m2 const., m k r 



i.e.



kr 2 const. Also,



_m( _{m m}  _{m m} ) m k x y y x 1 ( )₂ ( ) 2 m m n m m n m n mn m n x x x y y y k k r     

 

, using (1) 2 2 2

Since interchanging , , we get

( ) ( ) ( ) 1 2 ( ) ( ) ( ) , using(2) m m n m m n n n m m n n n m m n m n mn m n m n x x x y y y x x x k k y y y x x y y r    _{    }     _{ }              

 

 2 1 , 2 m m m m n m n d k r k k dt   



 

as _{xy xy r}2_ i.e. 2 1 2 1 . 2 kr _{  } _ k k   







Ex. 8. Two parallel rectilinear vortices of strengths k₁ and k₂(k_l > k₂) are at a distance 2a apart in an infinite mass of liquid. If the vortices intersect a plane perpendicular to their length at points A and B, show that the point on AB at a distance b from the mid-point on the same side of mid-point as the vortex of strength k₁, is always occupied by the same fluid element if .

3 2 2 3

1 2 1 2

(k k ) / (k k ) ( b 5a b) / (ab 3 )a .

Sol. In what follows, we shall need results (7) and (9) of Art. 11.8.

If G be the centre of vertex, then 2 2

1 2 1 2 2 k ak AG AB k k k k     Similarly, BG(2ak1) / (k1k2)

Let O be the mid-point of AB. Let OC = b.

 2 1 2 1 2 1 2 2 – – – ak k k OG OA AG a a k k k k      B O G Cb A 2a k2 k1