Course Number: 804538

1. Course Number: 804538 Advance Fluid Mechanics Faculty Name Prof. A. A. Saati

2. Lecture 3 contd. Chapter: 2 Inviscid Irrotational Flows Ref. ADVANCED FLUID MECHANICS By W. P. Graebel

3. 2.1 Inviscid Flows • Finding solutions with Navier-Stokes equations that were introduce in Chapter 1 is a challenge. • Most 19th century fluid mechanics was concerned with study of inviscid flows • There was no clear understanding of the effectof Reynoldnumber on the flow, and the study of turbulence was left mostly untouched • At a time of great excitement in the field of fluid mechanics, electricity, and magnetism, and thermodynamics

4. 2.1 Inviscid Flows • As a result of the study: • Scientists discovered a great similarity in their fields • Fields share many terms such as source, sink, potential, and current, among others • In this Chapter, the electrical, and magnetic analogies are set aside, but keep in mind a minor change in terminology

5. 2.1 Inviscid Flows • If viscosity terms are omitted (due to high ) in Navier-Stokes equations • This is called the Euler equation.

6. 2.2 IrrotationalFlows and Velocity Potential • From the circulation theorem • If mass density is constant and viscous effects con be neglected • The integral in the right hand side is an exact differential, so the integral around the path vanishes. • persistence of irrotationality: for flow with no upstream circulation, as the flow moves downstream, it must continue to be vorticity-free ( irrotational)

7. 2.2 IrrotationalFlows and Velocity Potential • By the definition irrotational flow • Since for irrotational flow for any C, it follows that the integral must be an exact differential, • This allows us to write velocity field written as the gradient of scalar • Where is called the velocity potential • Note: the surfaces of constant are locally orthogonal to the velocity vector.

8. 2.2 IrrotationalFlows and Velocity Potential • The continuity equation for an irrotationalincompressible flow is • This equation, called the Laplace equation, is used to determine for a given flow situation. • By this the flow in completely determined from irrotationalityand continuity • What then of the dynamics of the flow? • Since Euler’s equation can be Written in the form.

9. 2.2 IrrotationalFlows and Velocity Potential • For irrotational flows and g can be written as • Euler’s equation can be Written in the form. • Integrate this equation gives Bernoulli equation for irrotational flows • Where is constant or function of time

10. 2.2 IrrotationalFlows and Velocity Potential • Note that for an irrotationalincompressible flow can be written as • eq. 1.11.7 • Where and are the values of velocity potential at the start and end points of the traverse of C

11. 2.2.1 Intersection of Velocity Potential Lines and Streamlines in Two Dimensions • Surfaces of constant velocity potential and constant stream functionintersect one another throughout the flow • Since by definition the velocity is always normal to a constant potential surface, • It follows that the constant stream surfaces are generally perpendicular to constant potential surfaces • There are exception to this: neat stagnation point

12. 2.2.1 Intersection of Velocity Potential Lines and Streamlines in Two Dimensions • To see the relationship between and lines • and • The slop of and is given by:

13. 2.2.1 Intersection of Velocity Potential Lines and Streamlines in Two Dimensions • Multiplying the two slopes gives • This leads to the conclusion that constant and lines are orthogonal exceptat places where the velocity is ether zero ( stagnation ) or infinite ( singularities )

14. 2.2.1 Intersection of Velocity Potential Lines and Streamlines in Two Dimensions • Taylor series expansion gives • Since the first derivatives vanish at point of stagnation point, then Taylor series expansion to second-order gives • Solving this quadratic equation • at stagnation point • A similar expiration can be found involving

15. 2.2.2 Basic Two-Dimensional Irrotational Flows • In two-dimensional flows the basic solutions must satisfy the equations • Also, the equation • The incompressible continuity equation for irrotational flows corresponding to while , and • this must be satisfied the irrotational condition two-dimensional incompressible flows

16. 2.2.2 Basic Two-Dimensional Irrotational Flows • Uniform stream • A uniform stream is flow whose velocity is the same at every point • The velocity components are • Integrating the equation, we • find that • A similar expiration can be • found for the stream function

17. 2.2.2 Basic Two-Dimensional Irrotational Flows • Line source or sink • The velocity potential • Where mis called the strengthif mis positive ( line source ) and if mis negative ( line sink) • It extends from in the z direction. • The velocity components are given by

18. 2.2.2 Basic Two-Dimensional Irrotational Flows • Line source or sink • Differentiating the velocity components gives • Therefore, continuity is satisfied everywhereexceptpossibly at the point

19. 2.2.2 Basic Two-Dimensional Irrotational Flows • Line source or sink • A similar expiration can be found involving the stream function for a source is seen to be

20. 2.2.2 Basic Two-Dimensional Irrotational Flows • Line doublet (dipole) • Consider a source and sink pair of equal strength and let them approach one another • The pair then has combined potential • The potential of a line doublet • Where B gives the strength & direction

21. 2.2.2 Basic Two-Dimensional Irrotational Flows • Line doublet (dipole) • Constant line are similar family of circles, but 90 degrees with respect to the circles. • The stream function for a line doublet is

22. 2.2.2 Basic Two-Dimensional Irrotational Flows • Line doublet (dipole) • For constant values of this is the equation of a circle whose radiuscentered at • For constant values of this is the equation of a circle whose radiuscentered at • The velocity field associated with doublet is

23. 2.2.2 Basic Two-Dimensional Irrotational Flows • Line vortex • The vortex is a “reverse analog” of the source • It has concentrated vorticity rather than concentrated discharge • And also because the constant line are radial line while the line of constant are concentric circles • The velocity potential is

24. 2.2.2 Basic Two-Dimensional Irrotational Flows • Line vortex • The velocity components • The velocity decreases inversely with the distance form • Also • Laplace’s equation is seen to be satisfied • - vortex clockwise, + counterclockwise

25. 2.2.2 Basic Two-Dimensional Irrotational Flows • Line vortex • Since is multivalued, we anticipate that there is circulation associated with vortex flows. • Note: multivalued: in traversing a path around the point changes by • Checking this : calculating the circulation around 2 by 2 • The result show that it depends only on the path encircles

26. 2.2.2 Basic Two-Dimensional Irrotational Flows • Line vortex • Note: any path not enclosing the vortex has zero • circulation, • Therefor, a vortex has concentrated vorticitybut not concentrated mass discharge. • The stream function for a vortex is

27. 2.2.2 Basic Two-Dimensional Irrotational Flows • Line vortex • Summaryof Velocity potential and stream function for irrotational flows (Table 2.2.1 pp57)

29. 2.2.4 Basic Three-Dimensional Irrotational Flows • Except for vortex, all our Two-dimensional irrotationalflows havethree-dimensional counterparts • An example of three-dimensional vortex is smoke ring • Mathematical representation of such a three-dimensional phenomenon is more complicated • Here, we list these counterparts. • Uniform stream

30. 2.2.4 Basic Three-Dimensional Irrotational Flows • Uniform stream • The velocity potential for uniform stream is • The velocity field • Surface of constant are planes perpendicular to U • When U has only a component in z direction

31. 2.2.4 Basic Three-Dimensional Irrotational Flows • Point source or sink (point monopole) • The velocity potential for a point source of strength m at • is • Here, m is the volume discharge from the source • The continuity satisfied everywhere except at • If m is positive, If m is negative • Irrotationality is satisfied everywhere. • Surface of constant are concentric spheres center at • The velocity is directed along the radius and dies out • The velocity is given by • When the source lies • on the z-axis

32. 2.2.4 Basic Three-Dimensional Irrotational Flows • Point doublet (point diploe) • The velocity potential for a doublet • The velocity components is given by • When the doublet lies on the z-axis, and Bpoint parallel to the z-axis

33. END OF LECTURE 3

34. 2.2.5 Superposition and the Method of Images • For simple cases, the solution for flow past a given body shape can be obtained by analogy between potential flow and geometrical optics. • Line source near a plan wall • Suppose we have a source of strength m a distance b from a plane wall. • The plane wall can be viewed as a mirror. • It sees an image source of the same strength at distanceb behind the mirror. • A mirror interchanges right and left. • This does not affect the source, but will affect signs of vortex and doublet.

35. 2.2.5 Superposition and the Method of Images • Line source near a plan wall • The potential for the two-dimensional case is, with the x-axis acting as a wall. • With velocity component • Note: that vanished on y=0

36. 2.2.5 Superposition and the Method of Images • Line source near a plan wall • This solution can be used as a simple model of physical problem such as pollution sources neat a straight coastline. • Note that if the source is in the corner, both walls act as mirror, and there will be three image source plus the original source.

37. 2.2.5 Superposition and the Method of Images • Line source near a plan wall • Example 2.2.1 • Point source near walls • The strength m is located at the point (1,2,3). • There is a wall at x = 0 and at y = 0 • What is the velocity potential for this flow? • Solution • Form section 2.2.4 the given source has velocity potential • The first wall image: strength m at (-1,2,3)

38. 2.2.5 Superposition and the Method of Images • Line source near a plan wall • Example 2.2.1 • Solution (contd.) • When the second wall is add, then both the original and the first image at (-1,2,3) will give the combined velocity potential • Strength m at (1,-2,3) and (-1,-2,3) • The total velocity potential with strength m at (1,2,3), (-1,2,3) (1,-2,3), and (-1,-2,3) • Note: If there where third wall at z-axis, eight sources would be needed

39. 2.2.6 Vortices Near Walls • The source near the wall is that of a vortex near a wall, • The image will have a circulation in the reverse direction of the original • The velocity potential • The velocity components

40. 2.2.6 Vortices Near Walls • The velocity at (0,b) due to the vortex is • This is called the induced velocity • The vortex at (0,b) will travel with the induced velocity, carrying its image with it • The stream function • With constant given by • Where C is a constant related to

41. 2.2.6 Vortices Near Walls • The streamline are circles of radius at (0,d) • Where • Conceder a vortex pair near a wall • as shown in the Figure • The total velocity potential • The stream function

42. 2.2.6 Vortices Near Walls • The velocity components

43. 2.2.6 Vortices Near Walls • The inducedvelocity at (b,a) is • Therefore, the equation of motion for this vortex are • Similar equations hold for the other three vortices • The path the vortices travel can be found by solving the above equations, which is given by

44. Read and study Example2.2.1, 2.2.2 & 2.2.3

45. END OF LECTURE 4

46. 2.2.7 Rankine Half-Boby • The source located at the origin in a uniform stream • The velocity potential • The stream function • In the three-dimensional flow

47. 2.2.7 Rankine Half-Boby • The velocity components • There is stagnation point at r = 0 • and • The stream function for z positive • The stream function for z negative • The streamlines goes from the source to minus infinity • At stagnation it splits goes along the curve • Where

48. 2.2.7 Rankine Half-Boby • The two-dimensional velocity components • The stagnation point located at . On for x positive and for x negative and start at

49. 2.2.8 RankineOvel • A closed ovel shape is obtained by putting an aligned source and sink pair in a uniform stream • The 2-D velocity potential • The 2-D stream function • The 3-D velocity potential and stream function

50. 2.2.8 RankineOvel • The stagnation points are therefore at • The streamline that makes up the body is given by • Notice that along , except in the range where • The equation giving the body shape is • The maximum height of the body will be at x = 0