Fluid Dynamics 21MAT23C3 Past Paper PDF

Fluid Dynamics M.Sc. Mathematics (Semester – III) MAHARSHI DAYANAND UNIVERSITY, ROHTAK (A State University established under Haryana Act No. XXV of 1975) 'A+' Grade University Accredited by NAAC Author Dr Poonam Redhu Assistant Professor Department of Mathema...

Fluid Dynamics M.Sc. Mathematics (Semester – III) MAHARSHI DAYANAND UNIVERSITY, ROHTAK (A State University established under Haryana Act No. XXV of 1975) 'A+' Grade University Accredited by NAAC Author Dr Poonam Redhu Assistant Professor Department of Mathematics M.D. University, Rohtak – 124001.  Paper Code: 21MAT23C3 Fluid Dynamics M. Marks = 100 Term End Examination = 80 Assignment = 20 Time = 3 Hours Course Outcomes Students would be able to: CO1 Be familiar with continuum model of fluid flow and classify fluid/flows based on physical properties of a fluid/flow along with Eularian and Lagrangian descriptions of fluid motion. CO2 Derive and solve equation of continuity, equations of motion, vorticity equation, equation of the moving boundary surface, pressure equation and equation of impulsive action for a moving inviscid fluid. CO3 Calculate velocity fields and forces on bodies for simple steady and unsteady flow including those derived from potentials. CO4 Understand the concepts of velocity potential, stream function and complex potential, and their use in solving two-dimensional flow problems applying complex-variable techniques. CO5 Represent mathematically the potentials of source, sink and doublets in two-dimensions as well as three-dimensions, and study their images in impermeable surfaces.  Section – I Kinematics - Velocity at a point of a fluid, Eulerian and Lagrangian methods, Streamlines, path lines and streak lines, Velocity potential, Irrotational and rotational motions, Vorticity and circulation, Equation of continuity, Boundary surfaces, Acceleration at a point of a fluid, Components of acceleration in cylindrical and spherical polar co-ordiantes. Section – II Pressure at a point of a moving fluid, Euler equation of motion, Equations of motion in cylindrical and spherical polar co-ordinates, Bernoulli equation, Impulsive motion, Kelvin circulation theorem, Vorticity equation, Energy equation for incompressible flow, Kinetic energy of irrotational flow, Kelvin minimum energy theorem, Kinetic energy of infinite fluid. Uniqueness theorems. Section – III Axially symmetric flows, Liquid streaming part a fixed sphere, Motion of a sphere through a liquid at rest at infinity, Equation of motion of a sphere, Kinetic energy generated by impulsive motion, Motion of two concentric spheres, Three-dimensional sources, sinks and doublets, Images of sources, sinks and doublets in rigid impermeable infinite plane and in impermeable spherical surface. Section-IV Two dimensional motion; Use of cylindrical polar co-ordinates, Stream function, Axisymmetric flow, Stoke stream function, Stoke stream function of basic flows, Irrotational motion in two-dimensions, Complex velocity potential, Milne-Thomson circle theorem, Two-dimensional sources, sinks, doublets and their images, Blasius theorem. Note:- The question paper of each course will consist of five Sections. Each of the sections I to IV will contain two questions and the students shall be asked to attempt one question from each. Section-V shall be compulsory and will contain eight short answer type questions without any internal choice covering the entire syllabus. Books Recommended: 1. W.H. Besaint and A.S. Ramasey, A Treatise on Hydromechanics, Part II, CBS Publishers, Delhi, 1988. 2. F. Chorlton, Text Book of Fluid Dynamics, C.B.S. Publishers, Delhi, 1985 3. O’Neill, M.E. and Chorlton, F., Ideal and Incompressible Fluid Dynamics, Ellis Horwood Limited, 1986. 4. R.K. Rathy, An Introduction to Fluid Dynamics, Oxford and IBH Publishing Company, New Delhi, 1976. 5. G.K. Batchelor, An Introduction to Fluid Mechanics, Foundation Books, New Delhi, 1994.  Contents Unit Title of Chapter Page No. 1 Eulerian and Lagrangian Descriptions of Fluid Motion 1-66 and Conservation of Mass 2 Pressure and Equation of Impulsive Action 67-126 3 Axially Symmetric Flow and Three-dimensional Sources, 127-169 Sinks and Doublets 4 Two-dimensional Motion and Stoke’s Stream Function 170-220 Fluid Dynamics 21MAT23C3 UNIT- 1 Eulerian and Lagrangian Descriptions of Fluid and Conservation of Mass Content 1.1 Objective 1.2 Elementary Concepts 1.3 Introduction 1.4 Kinematics - Lagrangian and Eulerian Methods 1.5 Streamlines, Pathlines and Streaklines 1.6 Velocity Potential 1.7 Vorticity 1.8 Flow and Circulation 1.9 Equation of Continuity 1.10 Boundary Surface 1.11 Acceleration at a Point of a Fluid 1.12 Check Your Progress 1.13 Summary of the Unit 1.14 Keyword 1.15 Self-Assessment Test 1.16 Suggested Reading MDU Rohtak 1 Fluid Dynamics 21MAT23C3 1.1 Objective In this unit reader will learn about Eulerian and Lagrangian method, Conservation of Mass, Flow lines, Velocity Potential, Rotational and Irrotational Motion, Boundary Surface and Acceleration. 1.2 Elementary Concept Vectors are used to express physical quantities that are direction dependent, such as force, displacement, and velocity. Other physical quantities, such as stress and strain, require more than one direction to fully describe them, and these are referred to as tensors. Some useful results on vectors are as follows:   (i) Let q  ui  vj  wk  (u, v,w) be a non-zero vector, then q  q  u 2  v2  w 2   q u v w    A unit vector q in the direction of q is given by q   i  j  k , where q q q q u v w   cos   l,  cos   m,  cos   n are called the d.c.’s of q ;  ,  ,  denote the angles which q q q  q makes with the coordinate axes. Clearly, l2  m 2  n 2  1.   (ii) If a and b are two non-zero vectors, then    a.b  ab cos  , a  b  absin  n     where  is the acute angle between a and b , n being the unit vector normal to the plane of a and b.       Here a and b , n form a right handed screw system and thus a  b   b  a       Note that a. b =0 and a  b =0 are respectively the conditions of perpendicularly and parallelism.          (iii) If a , b and c are nonzero vectors, then a. b  c  a b c  is called scalar triple product.             Note that a b c    b c a  &  b a c    a b c  MDU Rohtak 2 Fluid Dynamics 21MAT23C3    Similarly, the vector quantity a. b  c is defined as the vector triple product which can be represented        as a. b  c = (a.c)b  (a.b)c      (iv)   i  j k , where  is a scalar function, is called gradient of the scalar  x y z    and   i  j  k is a vector (operator). x y z The total differential of  (x, y, z) is    d  dx  dy  dz x y z          i  j k . dxi  dyj  dzk    x y z  In vector form, this gives     d   dx   dy   dz and d  .dr where r  xi  yj  zk , dr  idx  jdy  kdz, x y z         i j k x y z   u v w  (v) divq  .q    , q  (u, v, w) x y z   If .q  0 , then q is said to be solenoidal vector.  2  2  2 2 Also, .   2    ,  is Laplacian operator. x 2 y 2 z 2 i j k      (vi) Curlq    q  x y z u v w  w v    u w    v u   i     j  k    y z   z x   x y  MDU Rohtak 3 Fluid Dynamics 21MAT23C3 (vii) (a) Gradient of a scalar is a vector. (b) Divergence of a scalar and curl of a scalar are meaningless. (c) Divergence of a vector is a scalar and curl of a vector is a vector.  (viii) curlgrad  0, divcurlq  0       Curlcurlq  graddivq   2 q i.e.,  2 q  graddivq  curlcurlq (ix) Gauss’s Divergence Theorem    (a)  q.dS   divqdV S V    (b)  n  qdS   curlqdV S V (x) Green’s Theorem   (a)  . dV   .dS    2 dV  .dS   2.dV V S V S V     (b)  ( 2  2 ).dV      dS V V  n n       (xi) Stoke’s Theorem  q.dr   curlq.dS   curlq.ndS C S S (xii) Orthogonal Curvilinear Co-Ordinates: Let there be three orthogonal families of surfaces f1 (x, y, z)   , f 2 (x, y, z)   ,f 3 (x, y, z)   (1) where x, y, z are cartesian co-ordinates of a point P(x, y, z) in space. The surfaces α=constant, β=constant, γ=constant (2) form an orthogonal system in which every pair of surfaces is an orthogonal system. The values  ,  ,  are called orthogonal curvilinear co-ordinates. From equations in (1), we can get x  x( ,  ,  ), y  y( ,  ,  ), z  z( ,  ,  ) MDU Rohtak 4 Fluid Dynamics 21MAT23C3 The surfaces (2) are called co-ordinate surfaces.  Let r be the position vector of the point P  x, y, z    i.e., r  xi  yj  zk  r( ,  ,  )  r A tangent vector to the  -curve (  =constant,  =constant) at P is. A unit tangent vector is   r   2 2 2    r  r  x   y   z  e1   or  h1 e1 where h1         r             Similarly, e2 , e3 are unit vectors along  -curve and  -curve, respectively such that   r  r  h 2 e2 ,  h 3 e3       r r r Further, dr  d  d  d  h1 e1d  h 2 e2 d  h 3 e3d      Therefore, (ds) 2  dr.dr  h12d 2  h 2 2 d 2  h 32d 2 where h1d , h 2d , h 3d are arc lengths along  ,  ,  - curves. In orthogonal curvilinear co-ordinates, we have the following results.  1  1  1   (a) grad   , ,   h 1  h 2  h 3     1      (b) If q  (q1 , q 2 , q3 ) , then divq    (h 2 h 3q1 )   (h1h 3q 2 )   (h1h 2q 3 )  h1 h 2 h 3     (c) If curlq    (1 ,  2 , 3 ), then 1     1    (h 3q 3 )   (h 2q 2 )  h 2h 3   MDU Rohtak 5 Fluid Dynamics 21MAT23C3 1    2   (h1q1 )  (h 3q3 )  h 3h1     1     3   (h 2q 2 )  (h1q1 )  h 2 h1     (d) 1    h 2 h 3     h1h 3     h 2 h1     2         h 1h 2 h 3    h1     h 2     h 3    (xiii) The Cartesian co-ordinate system (x, y, z) is the simplest of all orthogonal co-ordinate systems. In many problems involving vector field theory, it is convenient to work with other two most common orthogonal co-ordinates i.e. cylindrical polar co-ordinates and spherical polar co-ordinates denoted respectively by (r,  , z) and (r,  , ). (a) Cylindrical Polar Co-ordinates (r,  , z) Here  , ,     r, , z  , related to cartesian co-ordinates by x  r cos , y  r sin , z  z (r  0, 0    2,   z  ) Scalar factors are h1  1, h 2  r, h 3  1 and volume element is dxdydz  rdrddz. (b) Spherical Polar Co-ordinates (r, ,  ) Here  , ,     r, ,   , related to cartesian co-ordinates by x  r sin  cos , y  r sin  sin , z  r cos  (r  0, 0    , 0    2) Scalar factors are h1 = 1, h2 = r, h3= r sin  and volume element is dxdydz  r 2 sin drdd MDU Rohtak 6 Fluid Dynamics 21MAT23C3 1.3. Introduction 1.3.1 Fluid Dynamics The branch of science that deals with fluids in motion is known as fluid dynamics. A substance that may flow is referred to as a fluid; otherwise, it is referred to as a solid. There are two types of fluids: (i) incompressible liquids (those whose volumes do not change when the pressure changes) and (ii) compressible gases (those whose volumes do change when the pressure changes). The science of moving incompressible fluids is often referred to as hydrodynamics. However, there are no sharp distinctions between the three states of matter, i.e. solid, liquid and gases. In reality, there is not much of a difference between the three states of matter. Matter is thought to be made up of molecules in random relative motion under the influence of intermolecular forces in a microscopic view of fluids. The distance between molecules in solids is small, and this spacing persists even under intense molecular forces. The spacing between molecules is larger in liquids, resulting in a less well-ordered arrangement of molecules and weaker intermolecular forces; nevertheless, intermolecular forces are much weaker in gases, and molecules appear in disordered arrays with very high intermolecular spacing. If we imagine that our microscope, which has observed the molecular structure of matter, has a variable focal length, we could change our observation of matter from a fine detailed microscopic viewpoint to a longer range macroscopic viewpoint, in which the gaps between the molecules would not be visible and the matter would appear to be continuously distributed. We will use this macroscopic view of fluids throughout our research, in which physical quantities related with fluids inside a particular volume V are assumed to be distributed continuously and uniformly within a sufficiently small volume V. The Continuum hypothesis is the name given to this discovery. It means that at any location of a fluid, we can specify a different velocity, pressure, density, and so on. Moreover, for a continuous or MDU Rohtak 7 Fluid Dynamics 21MAT23C3 ideal fluid we can define a fluid particle as the fluid contained within an infinitesimal volume whose size is so small that it may be regarded as a geometrical point. 1.3.2 Stresses A fluid element is subjected to two types of forces. Body force is one of them, and surface force is the other. The surface force is proportional to the mass of the body on which it acts, whereas the body force is proportional to the  mass of the body on which it operates. Suppose F is the surface force acting on an elementary surface area dS at a  point P of the surface S. Let F1 and F2 be resolved parts of F in the directions of the tangent and normal at P. The normal force per unit area is called the normal stress and is also called pressure. The tangential force per unit area is called the shearing stress. 1.3.3 Viscosity The internal friction between the fluid's particles provides resistance to deformation. Tangential and shearing forces are the source of friction (stresses). Fluids with this attribute are known as viscous or real fluids, while those without it are known as inviscid, ideal, or perfect fluids. Although all fluids are real, viscous effects are often ignored when the rates of variation in fluid velocity with distance are small. The body force per unit area at every point of surface of an ideal fluid acts along the normal to the surface at that point, as defined by the definitions of body force and shearing stress. As a result, perfect fluid has no shearing stress. Thus, we conclude that viscosity of a fluid is that property by virtue of which itis able to offer resistance to shearing stress. It is a kind of molecular frictional resistance. 1.3.4 Velocity of Fluid at a Point MDU Rohtak 8 Fluid Dynamics 21MAT23C3  Assume that at time t , a fluid particle at the point P having position vector r and at time t   t the   same particle has reached at point Q having position vector r   r.        (r   r)  r  r dr The particle velocity q at point P is q  Lt  Lt  , where the limit is assumed to  t 0 t  t 0  t dt  exist uniquely. Clearly q is in general dependent on  both r and t , so we may write      q  q(r, t)  q(x, y, z, t) , r  xi  yj  zk  Suppose, q  ui  vj  wk and since   dr dx dy q  i  j  dz k dt dt dt dt dx dy dz Therefore, u  ,v  ,w  dt dt dt  Remarks: (i) A point where q  0 , is called a stagnation point. (ii) When the velocity at each point is independent of time, i.e. the flow pattern is the same at each instant, the motion is termed as steady; otherwise, it is termed as unsteady. 1.3.5 Flux across any surface   The flux i.e. the rate of flow across any surface S is defined by the integral   (q.n)dS S , where  is the  density, q is the velocity of the fluid and n is the outward unit normal at any point of S. Also, we have Flux = density × normal velocity × area of the surface. 1.4 Eulerian and Lagrangian Methods (Local and Total range of Change) We have two methods for studying the general problem of fluid dynamics. 1.4.1 Eulerian Method MDU Rohtak 9 Fluid Dynamics 21MAT23C3 In this method, we fix a point in the space occupied by the fluid and observation is made of whatever changes of velocity, density pressure etc, take place at that point. i.e. point is fixed and fluid particles are allowed to pass through it. If P(x, y, z) is the point under reference, then x, y, z do not dependent  y, upon the time parameter t, therefore x,  z do not exist (dot denotes derivative w.r.t. time t). Suppose f (x, y, z, t) be a scalar function associated with some property of the fluid (e.g. its density)   i.e. f (x, y, z, t)  f (r, t) , where r  xi  yj  zk is the position vector of the point P, then   f f (r, t   t)  f (r, t)  Lt (1) t  t 0 t f Here, is called local time rate of change. t 1.4.2 Lagrangian Method In this approach, observations are made at each point and instant, i.e., any particle of the fluid is chosen and its particular motion is observed and studied throughout its course. Let a fluid particle be initially at the point (a, b, c). After lapse of time t, let the same fluid particle be at (x, y, z). It is obvious that x, y, z are functions of t. However, after the motion is allowed, the particles that had different positions at the beginning possess different positions. As a result, the final position's coordinates (x, y, z) are affected by (a, b, c). Thus x  f1 (a, b, c, t), y  f 2 (a, b, c, t), z  f3 (a, b, c, t). For this case, if f (x, y, z, t) be scalar function associated with the fluid, then    df f (r   r, t   t)  f (r, t) dt Lt  (2) t 0 t  y, where x,  z exist. df Here is called an individual time rate or total rate or particle rate of change. dt MDU Rohtak 10 Fluid Dynamics 21MAT23C3 Now we established the relation between these two time rates (1) and (2). We have f  f (x, y, z, t) df f dx f dy f dz f Therefore,     dt x dt y dt z dt t df  f  f  f    dx  dy  dz   f  f  i j  k . i  j  k   = f.q  dt  x y z   dt dt dt  t t  dx dx dx  where q  i  j  k  (u, v, w) dt dt dt df f  Thus   q.f dt t Remarks: (i) The relation df f    q.f dt t df        q.  f dt  t  d        q.  dt  t  d The operator is called Lagrangian operator or material derivative, i.e. time rate of change in dt Lagrangian view. Sometimes, it is called ‘differentiation following the fluid’.  (ii) Similarly, for a vector function F(x, y, z, t) associated with some property of the fluid, we can show that   dF   F       q. F  dt  t  Hence the relation (3) holds for both scalar and vector functions associated with the moving fluid. (iii) The Eulerian method is sometimes also called the flux method. MDU Rohtak 11 Fluid Dynamics 21MAT23C3 (iv) Both Lagrangian and Eulerian methods were used by Euler for studying fluid dynamics. (v) Lagrangian method resembles very much with the dynamics of aparticle (vi) The two methods are essentially equivalent, but depending upon the problem, one has to judge whether Lagrangian method is more useful or the Eulerian. Some important types of flows: (i ) Laminar(streamline) and Turbulent Flows: Laminar flow is defined as a flow in which each fluid particle traces a distinct curve and the curves traced by any two individual fluid particles do not intersect. The flow is said to be turbulent if each fluid particle does not trace out a specific curve and the curves traced out by fluid particles intersect. The fluid particles in such flows move in a zig-zag pattern. (ii) Uniform and Non–Uniform Flows: Uniform flow is defined as a flow in which the fluid particles have similar velocities at each section of the channel or pipe. The flow is said to be non-uniform if the fluid particles have different velocities in each portion of the channel or pipe. For channel flows, such flows are specified. (iii) Steady and Unsteady Flows: A steady flow is one in which the properties and conditions associated with the fluid's motion are independent of time, therefore the flow pattern remains constant. The flow is considered to be unsteady if the properties and conditions associated with the fluid's motion change with time, causing the flow pattern to change. (iv) Rotational and Irrotational Flows: A rotational flow is one in which the fluid particles continue to revolve around their own axes during motion, whereas an irrotational flow is one in which the fluid particles do not rotate about their own axes during motion. (v) Barotropic Flow: The flow is considered to be barotropic when the fluid pressure is merely a function of density. These fluids are known to as barotropic fluids. MDU Rohtak 12 Fluid Dynamics 21MAT23C3 Remarks: A fluids is said to be isotropic w.r.t. some property such as pressure, density, etc. if that property is the same in all directions at a point. If such property is not to be anisotropic w.r.t. that property. 1.5 Streamline, Pathlines and Streaklines 1.5.1 Streamline It is a curve drawn in the fluid such that the direction of the tangent to it at any point coincides with the   direction of the fluid velocity vector q at that point. At any time t , let q  (u, v, w) be the velocity at each point P(x, y, z) of the fluid. The direction ratios of the tangent to the curve at P(x, y, z) are  dr  (dx, dy, dz).    Since the tangent and the velocity at P have the same direction, therefore q  dr  0 , i.e.  (ui  vj  wk)   (dxi  dyj  dzk)  0   (vdz  wdy)i  (wdx  udz)j  (udy  vdx)k  0  vdz  wdy  0  wdx  udz  udy  vdx dx dy dz    u v w These are the differential equations for the streamlines and the solution of these equations give streamlines.   Remarks: (i) It is noted that the streamlines are not well defined at the stagnation point where q  0. A streamline may divide into two branches at such a point. (ii) If we draw the MDU Rohtak 13 Fluid Dynamics 21MAT23C3 streamlines through every point of a closed curve in the fluid, we obtain a stream-Tube. A stream-tube of very small cross-section is called a stream filament. (iii) The component of velocity at right angle to the streamline is always zero, so there is no flow across the streamlines. Thus, if the boundary of stream tube is replaced by a rigid boundary, the flow is not affected. The principle of conservation of mass, then gives that the flux across any cross section of the stream tube should be the same. 1.5.2 Pathlines When the fluid motion is steady so that the pattern of flow does not vary with time, the paths of the fluid particles coincide with the streamlines. But in case of unsteady motion, the flow pattern varies with time and the paths of the particles do not coincide with the streamlines. However, the streamline through any point P does touch the pathline through P. Pathlines are the curves described by the fluid particles during their motion i.e. these are the paths of the particles. The differential equations for pathlines are  dr  q dt dx dy dz i.e.  u,  v,  w (1) dt dt dt where ( x, y , z ) are the cartesian co-ordinates of the fluid particle and not a fixed point of space. The equation of the pathline which passes through the point (x 0 , y 0 , z 0 ) , which is fixed in space, at time t = 0 say, is the solution of (1) which satisfy the initial condition that x  x 0 , y  y 0 , z  z 0 when t = 0. The solution gives a set of equations of the form x  x(x 0 , y0 , z 0 , t)   y  y(x 0 , y0 , z 0 , t)  (2) z  z(x 0 , y 0 , z 0 , t)  MDU Rohtak 14 Fluid Dynamics 21MAT23C3 which, as t takes all values greater than zero, will trace out the required pathline. Remarks: Streamlines give the motion of each particle at a given instant whereas pathlines give the motion of a given particle at each instant. We can make these observations by using a suspension of aluminum dust in the moving liquid. 1.5.3 Streaklines A streakline, in addition to streamlines and pathlines, is useful for observational purposes. This is the curve formed by all fluid particles that have interacted with a specific fixed point in space at some point in time. As a result, a streakline is the locus of various particles passing through a fixed point. The streakline is observed when a neutrally buoyant marker fluid is continuously injected into the flow at a fixed point of space from time τ =-∞. The marker fluid may be smoke if the main flow involves a gas such as air, or a dye such as potassium permanganate (KMnO4) if the main flow involves a liquid such as water. If the co-ordinates of a particle of marker fluid are (x, y, z) at time t and the particle coincided with the injection point (x 0 , y 0 , z 0 ) at some time t0, where τ≤ t , then the time-history (streakline) of this particle is obtained by solving the equations for a pathline, subject to the initial condition that x  x 0 , y  y0 , z  z 0 at t = τ. As τ takes all possible values in the angle -∞≤τ≤ t , the locations of all fluid particles on the streakline through (x0, y0, z0) are obtained. Thus, the equation of the streaklines at time t is given by x  x(x 0 , y0 , z 0 , t, )   y  y(x 0 , y0 , z 0 , t, )  (    t) (3) z  z(x 0 , y0 , z 0 , t, )  Remark: (i) For a steady flow, streaklines also coincide with streamlines and pathlines. (ii) Streamlines, pathlines and streaklines are termed as flowlines for a fluid. 1.6 Velocity Potential MDU Rohtak 15 Fluid Dynamics 21MAT23C3  Assume that q  ui  vj  wk is the velocity at any time t at each point P(x, y, z) of the fluid. Also assume that the expression udx  vdy  wdz is an exact differential, say d. Then, d = udx  vdy  wdz       i.e.   dx  dy  dz  dt   udx  vdy  wdz  x y z t  where    (x, y, z, t) is some scalar function, uniform throughout the entire field of flow.     Therefore, u  ,v  ,w  , 0 x y z t  But  0     (x, y, z) t         Hence q  ui  vj  wk    i  j k     x y z  where  is termed as the velocity potential and the flow of such type is called flow of potential kind.  In the above definition, the negative sign in q   is a convention and it ensures that flow takes place from higher to lower potentials. The level surfaces  (x, y, z, t) =constant, are called equipotential or equipotential surfaces. Theorem: At all points of the field of flow the equipotential (i.e. equipotential surfaces) are cut orthogonally by the streamlines.  Proof: If the fluid velocity at any time t be q  (u, v, w) , then the equations of streamlines are dx dy dz   (1) u v w   The surfaces given by q.dr  0 , i.e. udx  vdy  wdz  0 (2) MDU Rohtak 16 Fluid Dynamics 21MAT23C3 are such that the velocity is at right angles to the tangent planes. The curves (1) and the surfaces (2) cut each other orthogonally. Suppose that the expression on the left hand side of (2) is an exact differential say,  d , then d  udx  vdy  wdz (3) where  is velocity potential. The necessary and sufficient condition for the relations    u ,v   ,w   x y z   i.e. q   to hold is curlq  curl( )  0 (4) The solution of (2) i.e. d  0 is  (x, y, z) =constant (5) The surfaces (5) are called equipotentials. Thus the equipotentials are cut orthogonally by the streamlines.  Note: When curlq  0 , the flow is said to be irrotational or of potential kind, otherwise it is rotational.  For irrrotational flow, q  . Example: The velocity potential of a two dimensional flow is   cxy. Find the streamlines. Solution: The streamlines are given by dx dy dz   u v w  where q  (u, v, w)  For an irrotational motion, we have curlq  0  curl(  )  i.e. q   , where  is the velocity potential. From here, MDU Rohtak 17 Fluid Dynamics 21MAT23C3      (u, v, w)    , ,   (cy, cx, 0)  x y z  i.e. u  cy, v  cx, w  0 Therefore, streamlines are dx dy dz   cy cx 0 i.e. xdx  ydy  0, dz  0 i.e. x 2  y 2  a 2 , z  K which are rectangular hyperbola in the plane parallel to xy-plane. Example: If the speed of fluid is everywhere the same, show that the streamlines are straight. Solution: The streamlines are given by the differential equation dx dy dz   u v w where u, v, w are constants. The solutions are vx  uy  constant, vz  wy  constant The intersections of these planes are certanity straight lines. Hence the result. Example: Find the streamlines and pathlines of the particles for the two-dimensional velocity field x u , v  y, w  0 1 t Solution: For streamlines, the differential equations are dx dy dz   u v w Therefore, dx dy dz (1  t)   x y 0 MDU Rohtak 18 Fluid Dynamics 21MAT23C3 Here t =constant = t0 (at given instant), therefore the solutions are (1  t 0 ) log x  log y  c1 , z  c2 Or log x (1 t 0 )  log y  log a, z  c 2 Or x (1 t 0 )  ay, z  c 2 which are the required streamlines. For pathlines, we have dx dy dz  u,  v,  w dt dt dt dx x dy dz Therefore,  ,  y,  0 dt 1  t dt dt dx dt dy   ,  dt, dz  0  log.x  log(1  t)  log a, log y  t  log b, z  c x 1 t y  x  a(1  t), y  be t , z  c x a a  y  be ;z  c which are the required pathlines. Note: In case of pathlines, t must be eliminated since these give the motion at each instant (i.e. independent of t). Example: Obtain the equations of the streamlines, pathlines and streaklines which pass through (l, l, x t  y 0) at t = 0 for the two dimensional flow u  1   , v  , w  0. t0  t0  t0 Solution: We define the dimensionless co-ordinates X, Y, Z and time T by writing x y z t dx dy dz dt X , Y  , Z  , T  such that dX  , dY  , dZ  , dT  and l l l t0 l l l t0 Xl Yl u (1  T), v  , w  0 t0 t0 Streamlines are given by MDU Rohtak 19 Fluid Dynamics 21MAT23C3 dx dy dz   u v w t ldX t ldY dZ  0  0  Xl(1  T) Yl 0 dX dY dZ    X(1  T) Y 0 Integrating these, we get Z=constant= C1 (1) and log X  (1  T ) log Y  log C2 , where C2 is constant  X  C2Y (1T ) (2) As variable X, Y, Z and T are independent and C1 , C2 are constants, (1) & (2) given the family of streamlines at all times t  t 0 T. In particular, x  1  y, z  0, t  0  C1  0, C 2  1 And we get streamline as Y = X, i.e., y = x and z = o dX dY dZ Pathlines are given by  X(1  T),  Y, 0 dT dT dT Now, X, Y, Z are the dimensionless co-ordinates of a fluid particle and are functions of T. dX  T2  Therefore,  (1  T)dT  log X   T    log K1 X  2  2  X  K1eT  T /2 (3) dY dY Y  dT  log Y  T  log K 2 dT Y  Y  K 2eT (4) dZ  0  Z  cons tan t  K 3 (5) These are the parametric equations of pathlines. The pathline through P(1,1,0) i.e. X=1=Y,Z=0,T=0 is obtained when K1  K 2  1, K 3  0 MDU Rohtak 20 Fluid Dynamics 21MAT23C3 T2 T Xe 2 , Y  eT , Z  0 Elimination of T gives,  T  T  T 1   log Y  1 T  1   1   2  2    2 T Xe  e   2 Y Y ,Z  0 The pathline which passes through X=Y=1, Z=0 when T=  is given by  1 1  X  exp. T  T 2     2  ,  2 2  Y  exp(T   ), Z  0 These are the parametric equations of the streakiness true for all values of T. At T=0 the equation given as  1  X  exp.     2  ,  2  Y  exp(  ), Z  0 Eliminating  , we have,   log Y    log Y Therefore,           1   lo g Y   1  1    2   1   2  2   X  e   e   2  Y  Y ,Z  0 Article: To obtain the differential equations for streamlines in cylindrical and spherical co- ordinates. Solution: We know that the streamlines are obtained from the differential equations.    q  dr  0 (1)   where q is the velocity vector and r is the position vector of a liquid particle. If the motion is irrotational, then MDU Rohtak 21 Fluid Dynamics 21MAT23C3  q   Therefore, the differential equation (1) become     dr  0 (2) (i) In cylindrical co-ordinates (r,  , z) , we have    1    dr  (dr, rd , dz) and   grad   , ,   r r  z  Thus, the different equation (2) become   1      , ,   (dr, rd , dz)  0  r r  z  dr rd dz    (3)  / r 1/ r. /   / z (ii) In spherical co-ordinates (r,  , ) , we have  dr  (dr, rd , r sin  d )   1  1   and   grad   , ,   r r  r sin      1  1    The differential equation (2) become  , ,   (dr, rd , r sin  d )  0  r r  r sin    dr rd r sin  d   (4)  / r 1/ r. /  1  /  r sin  Equations (3) and (4) are the required differential equations. A Example: Show that if the velocity potential of an irrotational fluid motion is    cos  , where r2 (r,  , ) are the spherical co-ordinates of any point, the lines of flow lie on the surface r  k sin 2  , k being a constant. MDU Rohtak 22 Fluid Dynamics 21MAT23C3 Solution: The differential equations for lines of flow (streamlines) are dr rd r sin  d    / r 1/ r. /  1  /  r sin  From first two members, we have dr rd  2A 1 A   3 cos     sin   r r  r2  dr 2rd dr cos     2 d cos  sin  r sin   log r  2 log sin   log k  r  k sin 2  Hence the result. Note: In cartesian co-ordinates, the above velocity potential can be written as y   A(x 2  y 2  z 2 )3/ 2 z.tan 1   , where x  r sin  cos , y  r sin  sin , z  r cos  are spherical x   polar substitutions. Also, the streamlines r  k sin 2  can be written as r 3  kr 2 sin 2   (x 2  y 2  z 2 )3/ 2  k(x 2  y 2 )  x 2  y 2  z 2  k 2/3 ( x 2  y 2 ) 2/3 which are the streamlines in cartesian co-ordinates. Example: At the point in an incompressible fluid having spherical polar co-ordinates (r,  , ) , the velocity components are (2Mr 3 cos  , Mr 2 sin  , 0) where M is a constant. Show that velocity is of potential kind. Find the velocity potential and the equations of streamlines.  Solution: Here dr  drr  rd  r sin  d  q  2Mr 3 cos  r  Mr 2 sin  Then, MDU Rohtak 23 Fluid Dynamics 21MAT23C3 r r r sin   1 curlq     2 r sin  r   3 2 2Mr cos  Mr sin  0  1  curlq  r.0  r.0  r sin  (2Mr 3 sin   2Mr 3 sin  )   0 2 r sin    Therefore, the flow is potential kind. Now, using the relation    1   1    q      r     r r  r sin    From here,     2Mr 3 cos  ,   Mr 2 sin  , 0 r   Therefore,    d  dr  d  d r    (2Mr 3 cos  )dr  ( Mr 2 sin  )d  d(Mr 2 cos  ) Integrating, we get   Mr 2 cos  , which is the required velocity potential. The streamlines are given by dr rd r sin  d    / r 1/ r. /   1  /  r sin  dr rd r sin  d 3  3  2Mr cos  Mr sin  0 From the last term,  = constant. From the first two terms, we get MDU Rohtak 24 Fluid Dynamics 21MAT23C3 dr 2 cos   d  2 cot  d r sin  Integrating, we get log r  log sin 2   constant  r  A sin 2  , = constant The equation  = constant shows that the streamlines lie in planes which pass through the axis of symmetry   0. 1.7 Vorticity 1.7.1 Vorticity   If q  (u , v, w) be the velocity vector of a fluid particle, then the vector  defined by      curlq    q is called the vortex vector or vorticity and its components are (1 ,  2 , 3 ) given by w v u w v u 1   ,2   , 3   y z z x x y In other words, we can say the fluid motion is said to be rotational if     curlq  0    If   curlq  0 , then the fluid motion is said to be irrotational or of potential kind and then q  . 1.7.2 Vortex Line It is a curve in the fluid such that the tangent at any point on the curve has the direction of the vorticity  vector .    The differential equations of vortex lines are given by   dr  0 i.e. MDU Rohtak 25 Fluid Dynamics 21MAT23C3 dx dy dz   1 2 3  where   (1 , 2 , 3 ). 1.7.3 Vortex Tube The locus of vortex line drawn at each point of a closed curve forms a vortex tube. A vortex tube with small cross section is called a vortex filament. Let S be the surface and V be the volume of a vortex tube, then by Guass divergence thorem,     n̂. dS   . dV   .(  q)dV  0 (1) S V V  Here, n̂.  S is called the strength of the vortex tube. A vortex tube with a unit strength is called a unit vortex tube. Equation (1) shows that the total strength of a vortex tube emerging from S is equal to that entering S. This means that vortex lines and tubes cannot originate or terminate at internal points in a fluid. They can only form closed curves or terminate on boundaries. However, in case of vorticity, it is not necessary to assume incompressibility of the fluid. 1.8 Flow and Circulation 1.8.1 Flow Let A and B be two points in the fluid.    A Then  q.dr  0 is called the flow along any path from A to B B  B  B If motion is irrotational then q   and flow   .dr    d   (A)   (B) A A 1.8.2 Circulation MDU Rohtak 26 Fluid Dynamics 21MAT23C3 It is the flow round a closed curve. If C be the closed curve in a moving fluid, then circulation  about      C is given by    q.dr   n.curlqdS   dS   n. C S S   If the motion irrotational, then q   and thus,     .dr    d  (A)   (A)  0, where A is C C any point on the curve C. This shows that for an irrotational motion, circulation is zero. 1.8.3 Theorem The necessary and sufficient condition such that the vortexlines are at right angles to the streamlines, is      (u, v, w)    , ,   x y z   i.e. q   , where  and  are functions of x, y, z and t. Proof: Necessary condition: We know that the differential equation    q.dr  0 is integrable if  Q R     ... ..  0  pdx  Qdy  Rdz  0 is integrable if  z y    q.curlq  0 (exactness condition)    i.e. q.  0,   curlq This shows that the streamlines are at right angles to the vortex lines. Thus the streamlines and vortex   lines are at right angles to each other if the differential equation q.dr  0 is integrable. However, it is   certainty not an exact equation, because its exactness implies that q   , hence curlq  0 i.e. vortex lines cease to exist.   Hence, there exists an integrating factor  (function of x, y, z, t) such that  1 q.dr  0 is integrable. If this differential equation is integrable, then we can write    1 q.dr  d , where  is a scalar function of x, y, z, t MDU Rohtak 27 Fluid Dynamics 21MAT23C3       1 q.dr  .dr  d  .dr   q   .   Sufficient condition: let us take q       1 q  Then, curlq  curl (  )          (     )             Therefore,    q.  (   ).q  .(  q)    .(  1 q  q)  0 This shows that the directions of streamlines and vortex lines are at right angles to each other. 1.9. Equation of Continuity 1.9.1 Equation of Continuity by Euler’s Method (Equation of conservation of Mass) When the region of a fluid has no inlets (sources) or outlets (sinks) through which fluid can enter or leave the region, than the amount of fluid within the region is conserved in accordance with the principle of conservation of mass. We formulate this principle mathematically by means of the so-called equation of continuity. Let us consider a closed geometrical surface S fixed in space, which is drawn in a region through which fluid flows. It is supposed that S does not include points which are either sources or sinks. Let V be the volume enclosed by S and let n̂ denotes the unit vector normal to an element  S of S drawn outwards. MDU Rohtak 28 Fluid Dynamics 21MAT23C3  Let q be the fluid velocity and  be the fluid density. First, we consider the mass of fluid which leaves V by flowing across an element  S of S in time  t. This quantity is exactly that which is contained in a small cylinder of cross-section   t.  S of length (q.n)    t. S Thus the mass of the fluid is = density  Volume=  (q.n) Hence the rate at which fluid leaves V by flowing across the   S element  S is  (q.n). Summing over all such elements, we obtain the rate of flow of fluid coming out of V across the entire surface S. Hence, the rate at which mass flows out of the region V is       S  (  q).n S   .FdV by Gauss divergence   (q.n) S  S  S F.ndS  V    div(  q)dV (1) V Now, the mass M of the fluid possessed by the volume V of the fluid is M    dV , where    (x, y, z, t) with (x, y, z) the cartesian co-ordinates of a general point of V, a V fixed region of space. Since the space co-ordinates are independent of time t, therefore the rate of decrease of mass within V is dM d         dV     dV (2) | V does not change w.r.t. time dt dt  V  V t But the considered region is free from source and sink i.e. the mass is neither created nor destroyed, so by law of conservation of mass , i.e. mass is neither created nor destroyed and thus from equations (1) and (2), we get MDU Rohtak 29 Fluid Dynamics 21MAT23C3    div(  q)dV   t dV V V         div(  q)  dV  0 V t  Since V is arbitrary, we conclude that at any point of the fluid which is neither a source nor a sink,      div(  q)  0, i.e.  .(  q)  0 (3) t t Equation (3) is known as equation of continuity of fluid and holds at all points of the fluid free from sources and sinks. Corollary (1) We know that    div(  q)   divq  q.(grad )    Therefore (3) takes the form  .(q)  (q.)   0 (4) t D Corollary (2) We know that the differential operator is given by Dt D     (q.) Dt t D  Therefore, from (4), we obtain the equation of continuity as   (.q)  0 Dt D  i.e.   divq  0 (5) Dt 1 D  Corollary (3) Equation (5) can be written as  divq  0  Dt D   (log  )  divq  0 (6) Dt  Corollary (4) When the motion of fluid is steady, then  0 and thus the equation of continuity (3) t  becomes div(  q)  0 (7) MDU Rohtak 30 Fluid Dynamics 21MAT23C3 | Here  is not a function of time i.e.  =  (x, y, z) D Corollary (5) When the fluid is incompressible, then  =constant and thus =0 Dt The equation of continuity becomes  div(q)  0 (8) which is same for homogeneous and incompressible fluid. Corollary (6) If in addition to homogeneity and incompressibility, the flow is of potential kind such  that q   , then the equation of continuity becomes single word div( )  0  .( )  0   2  0 (9) which is known as the Laplace equation. 1.9.2 Equation of Continuity in Cartesian Co-ordinates Let (x, y, z) be the rectangular cartesian co-ordinates.  Let q  ui  vj  wk (1)       and  i j k (2) x y z   Then, the equation of continuity  div(  q)  0 can be written as t   (  u)  (  v)  (  w)    0 (3) t x y z      u v w  u v w    0 (4) t x y z  x y z  which is the required equation of continuity in Cartesian co-ordinates.  Corollary: (1) If the fluid motion is steady, then  0 and the equation (3) becomes t MDU Rohtak 31 Fluid Dynamics 21MAT23C3  (  u)  (  v)  (  w)   0 (5) x y z Corollary: (2) If the fluid is incompressible, then  =constant and the equation of continuity is   .q  0 u v w i.e.   0 (6) x y z Corollary: (3) If the fluid is incompressible and of potential kind, then equation of continuity is  2  0  2  2  2  i.e.    0 , where q  . x 2 y 2 z 2 1.9.3 Equation of Continuity in Orthogonal Curvilinear Co-ordinates Let (u1 , u 2 , u 3 ) be the orthogonal curvilinear co-ordinates and e 1 , e 2 , e 3 be the unit vectors tangent to the co-ordinate curves.  Let q  q1 e1  q 2 e 2  q 3 e 3 (1) The general equation of continuity is    div(  q)  0 (2) t We know from vector calculus that for any vector point  function f  (f1 , f 2 , f3 ) ,  1      .f   (h 2 h 3f1 )  (h1h 3f 2 )  (h 2 h1f 3 )  (3) h1h 2 h 3  u1 u 2 u 3  where h1 , h 2 , h 3 are scalars. Using (3), the equation of continuity (2) becomes MDU Rohtak 32 Fluid Dynamics 21MAT23C3  1        (h 2 h 3  q1 )  (h1h 3  q 2 )  (h 2 h1 q 3 )   0 (4) t h1h 2 h 3  u1 u 2 u 3  Corollary: (1) When motion of fluid is steady, then equation (4) becomes    (h 2 h 3  q1 )  (h1h 3  q 2 )  (h 2 h1  q 3 ) =0 (5) u1 u 2 u 3 Corollary (2) When the fluid is incompressible, the equation of continuity is (  =constant)    (h 2 h 3q1 )  (h1h 3q 2 )  (h 2 h1q 3 )  0 (6) u1 u 2 u 3 Corollary: (3) When fluid is incompressible and irrotational, then  =constant  1  1  1   q      , ,   and the equation of continuity becomes  h1 u1 h 2 u 2 h 3 u 3    h 2 h 3     h1h 3     h 2 h1        0 (7) u1  h1 u1  u 2  h 2 u 2  u 3  h 3 u 3  Now, we shall write equation (4) in cylindrical and spherical polar co-ordinates. 1.9.4 Equation of Continuity in Cylindrical Co-ordinates (r,  , z) Here,  q   q r , q ,q  & h1  1, h 2  r, h 3  1 The equation of continuity becomes  1        (r q r )  (  q )  (r q z )   0 t r  r  z   1  1    (r q r )  (  q )  (  q z )  0 (8) t r r r  z Corollary: (1) When the fluid motion is steady, then equation (8) becomes    (r q r )  (  q )  r (  q z )  0 (9) r  z MDU Rohtak 33 Fluid Dynamics 21MAT23C3 Corollary: (2) For incompressible fluid, equation of continuity is    (rq r )  (q )  r (q z )  0 (10) r  z Corollary: (3) When the fluid is incompressible and is of potential kind, then equation (8) takes the form       (r )  ( )  (r )  0 (11) r r  r z z  where q   ;  is expressed in cylindrical co-ordinates. 1.9.5 Equation of Continuity in Spherical Co-ordinates (r,  , )  q   q r ,q , q z  & h1  1, h 2  r, h 3  r sin  The equation of continuity becomes  1  2     2 t r sin   r (r sin  q r )   (r sin  q )   (r q )   0    1   2      2 t r sin  sin  r (r  q r )  r  (sin  q )  r  (  q )   0 (12)   Corollary: (1) For steady case, equation (12) becomes  2   sin  (r  qr )  r (sin  q )  r (  q )  0 (13) r   Corollary: (2) For incompressible fluid, we have  2   sin  (r q r )  r (sin  q )  r (q )  0 (14) r   Corollary: (3) When the fluid is incompressible and is of potential kind, then equation (8) takes the form  2     1  (r sin  )  (sin  ) ( )0 (15) r r    sin   MDU Rohtak 34 Fluid Dynamics 21MAT23C3  where q   ;  is expressed in spherical co-ordinates. 1.9.6 Symmetrical forms of Motion and Equation of Continuity for Them We have the following three types of symmetry which are special cases of cylindrical and spherical polar co-ordinates. (i) Cylindrical Symmetry: In this type of symmetry, with suitable choice of cylindrical polar co- ordinates (r, θ, z), every physical quantity is independent of both θ and z so that       0 & q  q(r, t)  z For this case, the equation of continuity in cylindrical co-ordinates, reduces to  1   (  q1r)  0 (1) r r r If the flow is steady, then equation (1) becomes  (  q1r )  0   q1r  cons tan t  F(t) (say) r Further, if the fluid is incompressible then q1r  constant  G(t). (ii) Spherical Symmetry: In this case, the motion of fluid is symmetrical about the centre and thus with the choice of spherical polar co-ordinates (r, θ,  ), every physical quantity is independent of     both θ & . So that   0 & q  q(r, t)   The equation of continuity in spherical co-ordinates, reduce to  1   2. (  q1r 2 )  0 (2) r r r  For steady motion, it becomes (  q1r 2 )  0   q1r 2  cons tan t  F(t) (say) r And for incompressible fluid, it has the form q1r 2  cons tan t  G(t) (say). MDU Rohtak 35 Fluid Dynamics 21MAT23C3 (iii) Axial Symmetry :- (a) In cylindrical co-ordinates (r, θ, z), axial symmetry  Means that every physical quantity is independent of θ i.e  0 and thus the equation of continuity   1     becomes   (  q r r)  (  q r)  0   0 r r  r z  (b) In spherical co-ordinates (r, θ,  ), axial symmetry means that every physical quantity is  independent of  i.e.  0 and the equation of continuity, for this case, reduce to   1  1   2. (qr r2 )  (  q sin  )  0 r r r r sin   Example: If σ(s) is the cross-sectional area of a stream filament, prove that the equation of   continuity is (  )  (  q)  0 , where  s is an element of arc of the filament and q is the fluid t s speed. Solution: Let P and Q be the points on the end sections of the stream filament. The rate of flow of fluid out of volume of filament is  (  q)Q  (  q) P  (  q) P  s (1) s where we have retained the terms upto first order only, since  s is infinitesimally small. Now, the fluid speed is along the normal to the cross-section. At time t, the mass within the segment of filament is  s and its rate of increase is   (  ) s  (  s )  0 (2) t t Using law of conservation of mass, we have from (1) and (2) MDU Rohtak 36 Fluid Dynamics 21MAT23C3   (  ) s  (  q s)  0 | Total rate=0 t s   i.e. (  )  (  q)  0 (3) t s which is the required equation at any point P of the filament.  Deduction: For steady incompressible flow, (  ) =0 and equation (3) reduce to t   (  q)  0  ( q)  0   q  constant s s which shows that for steady incompressible flow product of velocity and cross-section of stream filament is constant. This result means that the volume of fluid acrossing every section per unit time is dis tan ce volume constant   q  c   c c t t Example: A mass of fluid in such a way that each particle describe a circle in one plane about a fluid axis, show that the equation of continuity is   ( )  (  )  0 t  where  is the angular velocity of a particle whose azimuthal angle is  at time t. Solution: Here, the motion is in plane and the particles describe a circle,   z = constant, r =constant  0 (1) z r i.e. there is only rotation. Now, the equation of continuity in cylindrical co-ordinates ( r ,  , z ) MDU Rohtak 37 Fluid Dynamics 21MAT23C3  1  1    (r q r )  (  q )  (  q z )  0 (2) t r r r  z Using (1) in (2), we get  1   (  q )  0 t r   1   (  r )  0 where q  q  r t r      (  )  0 t  Hence the result. Example: A mass of fluid is in motion so that the lines of motion lie on the surface of co-axial  1   cylinders, show that the equation of continuity is  (  v )  (  v z )  0 where v , vz the t r  z velocities are perpendicular and parallel to z. Solution: The equation of continuity in cylindrical co-ordinates ( r ,  , z ) is  1  1     (  rv r )  (  v )  (  v z )  0 , where q  (v r , v , v z ) t r r r  z Since the lines of motion (path lines) lie on the surface of cylinder, therefore the component of velocity in direction of dr is zero i.e. vr  0  1   Thus, the equation of continuity in the present case reduces to  (  v )  (  v z )  0 t r  z Hence the result. Example: The particles of a fluid move symmetrically in space with regard to a fixed centre, prove that     2 the equation of continuity is u . (r u)  0 , where u is the velocity at distance r. t r r 2 r MDU Rohtak 38 Fluid Dynamics 21MAT23C3 Solution: Firstly, derive the equation of continuity in spherical polar co-ordinates. The present case is the case of spherical symmetry as the motion is symmetrical w.r.t. a fixed centre.  Equation of continuity is  1     2. (  r 2q r )  0, 0 t r r    1    2. (  r 2 u)  0, q r  u t r r  1  2 1    2..(r u)  2  (r 2 u)  0 t r r r r     2  u  (r u)  0 t r r 2 r Hence the result. Example: If the lines of motion are curves on the surfaces of cones having their vertices at the origin and the axis of z for common axis, prove that the equation of continuity is   q r 2  q r cos ec    . (  q )  0 t r r r  Solution: The equation of continuity in spherical polar co-ordinates is  1   2      2 t r sin  sin  r (r  q1 )  r  (sin  q 2 )  r  (  q 3 )   0   In the present case, the lines of motion lie on the surfaces of cones, i.e.  =constant, so the velocity component in the  -direction is zero. i.e. q  0. The equation of continuity becomes  1  2 1    2 (r  q r )  (  q )  0 t r r r sin   MDU Rohtak 39 Fluid Dynamics 21MAT23C3  1  2   cos ec    2 r (  q r )   q r.2r   (  q )  0 t r  r  r    2  q r cos ec    (qr )   (  q )  0 t r r r  Hence the result. Example: Show that polar form of equation of continuity for a two dimensional incompressible fluid is  (ru) v  cos      0. If u   , then find v and the magnitude of the velocity q , where q =(u, v). r  r2 Solution: The equation of continuity in polar co-ordinates ( r ,  ) in two dimensional (z=0) is  1  1   (  rq r )  (  q )  0 t r r r       In the present case,  =constant and q =(u, v), so (ru)  (v)  0 r r r   cos  Now, u   , so we get r2   cos    cos  v v  cos  ( 2.r)  (v)  0  2  0  r r  r   r2  sin  v r2    q  q  u 2  v2  2 r Example: Liquid flows through a pipe whose surface is the surface of revolution of the curve x2 yak about the x-axis ( a  x  a ). If the liquid enters at the end x  a of the pipe with a velocity V, show that the time taken by a liquid particle to transverse the entire length of the pipe from 2a 2 1 x  a to x  a is 2 (1  k  k 2 ) , where k is assumed to be so small that the flow remains V(1  k) 3 5 appreciably one-dimensional throughout. MDU Rohtak 40 Fluid Dynamics 21MAT23C3 x2 Solution: Here, y  a  k a kx 2 2 Therefore, Area of the section distant x from 0 is   y 2   (a  ) a Area at x=-a is   (a  ka)2   a 2 (1  k)2 Applying the equation of continuity at the two sections, i.e. expressing equal rates of volumetric flow, kx 2 2 dx we get  a 2 (1  k) 2 V   (a  ) x where x  the velocity at the section distant x from 0 is. Thus, a dt 1 kx 2 2 we have dt  (1  2 ) dx (1  k)2 V a So, the required time of flow is a 1 kx 2 2 T  2 (1  ) dx 0 (1  k) 2 V a2 a 2 k2x4 kx 2  (1  4  2 2 )dx (1  k) 2 V 0 a a 2a  2 1 2  2 1  k  k  (1  k) V  3 5  Hence the result.  Remark: If the velocity vector q  (u, v, w) is kinematically possible for an incompressible fluid  u v w motion, then the equation of continuity must be satisfied. i.e. .q  0    0. x y z  c 2 (x j  yi)  Example: Test whether the motion specified by q  , c being constant, is a possible motion x 2  y2 for an incompressible fluid. If so determine the equations of the streamlines. Also test whether the motion is of potential kind and if so determine the velocity potential and equipotential surfaces. MDU Rohtak 41 Fluid Dynamics 21MAT23C3   y x  Solution: We have q  c 2  2 2 , 2 2 , 0   (u, v, w) x y x y      y    x     2xy 2xy  Therefore, .q  c 2   2 2    2 2   q  c2  2 2  2 2  0  x  x  y  y  x  y    x  y x  y  Thus the equation of continuity for an incompressible fluid is satisfied and so such a motion is possible. The equation of streamlines are dx dy dz   u v w dx dy dz  2 2 2  2 2 2  c y / (x  y ) c x / (x  y ) 0  xdx  ydy  0, dz  0  x 2  y 2  constant, z=constant (1) which are the circles whose centers lie on the z-axis and whose planes are perpendicular to this axis. Further, we have i j k     q  x y z c 2 y c2 x 0 x 2  y2 x 2  y2     x    y   curlq  c2   2 2    2 2  k  x  x  y  y  x  y    y2  x 2 x 2  y2    c2  2 2 2  2 2 2 k0  (x  y ) (x  y )   Thus, the flow is irrotational, i.e. of potential kind, so we can find  ( x, y, z ) such that q    c 2 y  c 2 x   u  2 ,   v  , 0 x x  y 2 y x 2  y 2 z MDU Rohtak 42 Fluid Dynamics 21MAT23C3 The last equation shows that    (x, y). The first equation is  1 c2 x  ,   (x, y)  c 2 tan 1 ( )  f (y) 2 (2) x y x y 1   y  c 2 x  2  f (y)  f (y)  0 y x  y 2 Using equation (2) We have, f(y) = constant (3) Since the constant potential gives rise to no velocity, so we may choose the constant in (3) to be zero x and therefore from equation (2), we get  (x, y)  tan 1   (4) y Now the equipotential surfaces are given by x   cons tan t   cont.  y  c1x (5) y From equation (1) and (5), we observe that the streamlines are accordingly orthogonal to the equipotentials.  Example: Show that a fluid of a constant density can have a velocity q given by   2xyz (x 2  y 2 )z y  q 2 2 2 , 2 2 2 , 2 2   (x  y ) (x  y ) (x  y )  Find the vorticity vector. Solution: We have u 3x 2  y 2 v y 2  3x 2 w  2yz. 2 ,  2yz. , 0 x (x  y 2 )3 y (x 2  y 2 )3 z   u v w  divq  .q    0 x y z MDU Rohtak 43 Fluid Dynamics 21MAT23C3  Thus, q is a velocity vector of a possible fluid of constant density (incompressible). Also, u v 3y 2  x 2 3y 2  x 2   2xz. 2  2xz. 2  0, y x (x  y 2 )3 (x  y 2 )3 v w x 2  y2 x 2  y2   2   0, z y (x  y2 ) 2 (x 2  y 2 ) 2 w u 2xy 2xy   2  2 0 x z (x  y ) (x  y 2 )2 2 2    v w   w u   u v    curlq       j     k  0 i    z y   x z   y x   Vorticity vector