1-1

ME 306 Fluid Mechanics II

Part 1

Irrotational Flow

These presentations are prepared by

Dr. Cneyt Sert

Department of Mechanical Engineering

Middle East Technical University

Ankara, Turkey

csert@metu.edu.tr

Please ask for permission before using them. You are NOT allowed to modify them.

1-2

Inviscid Flow

Shearing stress in a fluid motion is related to viscosity and velocity gradient.

All real fluids have a certain nonzero viscosity.

But there may be regions in a flow field where change in velocity and therefore shear stress is negligibly small.

These regions are usually away from the solid walls (outside boundary layers).

In these regions flow may be treated to be inviscid (frictionless).

Reference: Munson, Young, Okiiishi

1-3

Inviscid Flow (contd)

For an inviscid flow Navier-Stokes equation reduces to Eulers equation.

Further it is possible to use Bernoullis equation along a streamline.

However, considering a flow field to be totally inviscid may result in unrealistic results.

For example, a body immersed in a uniform, inviscid flow will not experience any drag force, which contraditcs with experiments (dAlamberts paradox).

An inviscid flow disregards the no slip boundary condition and therefore no boundary layer can develop over the body.

Also the separation and the wake behind the body can not be predicted by considering the whole flow field to be inviscid.

Boundary layer separation over a body is the result of

viscous action.

1-4

Inviscid Flow and Rotation

For an inviscid flow there are no shearing forces.

Only pressure and body forces act on a fluid element, neither of which can cause the element to rotate.

Although there are other ways to make a flow rotational (such as going through a shock wave in compressible flow), viscous action is by far the most common one.

Therefore in an inviscid flow, fluid elements originating from an irrotational region will not posses any rotation.

In an inviscid flow a fluid element that originates from an irrotational flow region will remain irrotational.

Irrotational, uniform

upstream flow

1-5

Irrotational Flow

In this chapter well study flow fields that are irrotational everywhere.

This automatically tells that these flow fields are also inviscid everywhere.

Although a completely irrotational flow field is not realistic, studying irrotational flows is useful because

real world flows contain large irrotational regions away from solid surfaces.

they have very simple mathematics and can be studied analytically.

they show us the importance of boundary layers, viscous forces and related physics.

they provide very valuable insight such as the generation of lift force by a wing.

1-6

Vorticity

In ME 305 we studied vorticity to be a kinematic property of a flow field, which is related to the angular velocity (rotationality) of fluid particles.

= 2 =

=

+

+

=1

+

+1

1

For an irrotational flow angular velocity,vorticity and curl of velocity are all zero.

Cartesian :

Cylindrical :

Vorticity Angular velocity Curl of velocity

1-7

Vorticity (contd)

Exercise : In the boundary layer that develops next to a solid surface, flow becomes rotational. Find a relation between the amount of vorticity at the wall () and the magnitude of wall shear stress ().

Uniform flow approaching the solid wall

()

Slowing down due to no slip BC, i.e. viscous action

Boundary layer Edge of the

boundary layer

Flat plate

1-8

Circulation ()

Circulation is the line integral of a velocity field around a closed curve and it is closely related to the rotationality of the flow.

=

[2/]

Differential vector along the path of

integration Closed path of integration

For the 2D planar flow in the plane shown above

= +

= + = ( + )

Closed curve

1-9

Relation Between Circulation and Vorticity

Using Stokes theorem (studied in ME 210) line integral of circulation can be transformed into a surface integral.

=

=

=

Closed curve

For the 2D flow in the plane

=

( = )

=

=

Unit normal vector of (not )

1-10

Circulation and Vorticity

Important : For Stokes theorem to be applicable

should be defined properly and should be continuously differentiable over the area .

Area should be simply connected (should not contain any holes).

Exercise : Veloicty field of a steady, incompressible flow is given by

= 2 + (0.52 2)

Calculate the circulation for the shown square path using

a) line integration

b) area integration

(1,1)

(1,0)

(0,1)

(0,0)

1-11

Circulation and Vorticity (contd)

Exercise : Consider a forced (rotational) vortex around the origin with a velocity field of = , where is a constant. Calculate the circulation for a circular path of radius with its center at the origin using

a) line integration

b) area integration

Exercise : Consider a free (irrotational) vortex around the origin with a velocity field of = /. Calculate the circulation for a circular path of radius with its center at the origin using

a) line integration

b) area integration (BE CAREFUL)

=

= /

1-12

Velocity Potential ()

For an irrotational flow : = 0

As studied in ME 210, curl of the gradient of any scalar function is zero.

= 0

Using these two equations we can define a velocity potential function as

= +

In an irrotational flow field, velocity vector can be expressed as the gradient of a

scalar function called the velocity potential.

Due to the possibility of defining a velocity potential function, irrotational flows are also called potential flows.

Some books use a minus sign so that decreases in the flow direction, similar to temperature decreasing in the heat flow direction. But we use plus in this course.

A scalar function called velocity potential

1-13

Velocity Potential (contd)

For a 2D flow in the plane : = =

, =

For a 2D flow in the plane : = =

, =

1

If the irrotational flow is also incompressible (In ME 306 well NOT study compressible potential flows)

Continuity Equation : = 0

= 0

2 = 0

For an incompressible and irrotational flow velocity potential satisfies the Laplaces equation.

2: Laplaces operator

1-14

Velocity Potential (contd)

2 = 0

Note that the relation between and is similar to that of and .

In the plane : 2

2+2

2= 0

In the plane : 1

+1

22

2= 0

In the plane In the plane

=

=

=

=

=

=

1

=1

=

Cauchy Riemann Equations

Streamfunction

1-15

Potential Flow Exercises

Exercise : Using Cauchy Riemann equations show that streamfunction also satisfies the Laplaces equation for incompressible, potential flows.

Exercise : Show that constant streamfunction lines (streamlines) are always perpendicular to constant velocity potential lines for incompressible, potential flows.

Exercise :Draw constant velocity potential lines of the following flow fields for which streamlines are shown. Constant velocity potential lines and streamlines drawn together form a flow net.

Flow near a corner Flow over a cylinder

1-16

Bernoullis Equation for Incompressible Potential Flow

Incompressible and Potential Flow

Solve Laplaces equation for the velocity potential.

Obtain velocity field using =

Obtain pressure distribution using Bernoullis equation

Exercise : Show that for an incompressible potential flow BE is valid between any two points of the flow field, not necessarily two point on the same streamline

1 2

3

1 and 2 : 1+12

2+ 1 =

2+22

2+ 2

1 and 3 : 1+12

2+ 1 =

3+32

2+ 3

1 + 2 = 3 , 1 + 2 = 3 , 1 + 2 = 3

To obtain complicated flow fields we can combine elementary ones such as

Uniform flow

Source/sink

Vortex

1-17

Superposition of Elementary Potential Flows

Laplaces equation is a linear PDE.

Superposition can be applied to both velocity potential and streamfunction.

Potential flow 1 Potential flow 2 Potential flow 3

+ =

1-18

1. Uniform Flow

Consider uniform flow in the plane parallel to the axis, in + direction.

= , = 0

Lets find the equation for velocity potential.

=

=

= + ()

=

0 =

= 0 =

Taking = 0 for simplicity potential function turns out to be

=

Constant lines correspond to constant lines, i.e. lines parallel to the axis.

Exercise : For this flow show that streamfunction equation is =

1-19

1. Uniform Flow (contd)

Constant and constant lines are shown below.

=0

=1

= 0

= 1

Exercise : Find the equations for and for uniform flow in a direction making an angle of with the + axis.

Consider the 2D flow emerging at the origin of the plane and going radially outward in all directions with a total flow rate per depth of .

Velocity components are

=

2 , = 0

decreases as increases in accordance to the continuity equation. 1-20

2. Line Source at the Origin

View from the top

Constant lines

Streamlines

: Source strength

Lets find the equation for velocity potential.

=

2=

=

2ln () + ()

=1

0 =

1

= 0 =

Taking = 0 for simplicity

=

2ln ()

Constant lines correspond to constant lines as shown in the previous slide.

Exercise : Show that for a line source streamfunction equation is =

2

Exercise : For a line source calculate the vorticity for a circular path around the origin with radius and also for a square path with one side being equal to .

1-21

2. Line Source (contd)

Consider a line source that is located NOT at the origin, but at a point A as shown below

Equations for and change slightly as

=

2ln 1

=

21

or equivalently using and coordinates

=

2ln 2 + 2

=

2arctan

To study a line sink for which the flow is radially inward towards a point, we simply use a negative value.

1-22

2. Line Source (contd)

1

1 Source

A

For an irrotational vortex located at the origin of the plane, origin is a singular point.

The flow field is irrotational except the origin. Circulation around any path that does not enclose the origin is zero.

We consider that all the circulation is squeezed into the origin.

Velocity components are

=

, = 0

In slide 1-11 we showed that for a path enclosing the origin, circulation is

= 2

which is known as the strength of the vortex.

Direction of the vortex is determined as > 0 : CCW rotating vortex

< 0 : CW rotating vortex 1-23

3. Irrotational Vortex

Using the given velocity components and the Cauchy-Riemann equations one can determine

=

2 , =

2ln ()

1-24

3. Irrotational Vortex (contd)

Constant lines

Streamlines

Exercise : Elementary components of a potential flow of water ( = 1000 /3) is shown below. Find the velocity at point A and find the pressure at A if the pressure at infinity is 100 kPa.

1-25

Exercises for Elementary Potential Flows

= 7.542

= 0.8

= 0.6

= 3

= 36.81

Sink () Source ()

A

Exercise : For the previous problem determine the equations of velocity potential and streamfunction by superimposing elementary flows. Find the velocity at point A by differentiating both and equation.

Consider a source of strength at the origin and a uniform flow in + direction.

1-26

Superposition of a Source and Uniform Flow (Flow Past a Half Body)

= + = +

2ln ()

= + = +

2

= +

2cos () +sin ()

= +

2cos () +

2sin ()

where

= 2 + 2 , = arctan

or

= cos() , = sin()

2

+

2cos () +sin ()

1-27

At point velocity is zero : + = 0

2 = 0 =

2

Flow Past a Half Body (contd)

We expect to have a stagnation point with zero velocity on the negative part of the axis.

Source ()

, ,

= ?

1-28

Flow Past a Half Body (contd)

Streamline passing through point is called the stagnation streamline.

Value of the streamfunction of this stagnation streamline is

= +

2 = 0 +

2 =

2

Equation of the stagnation streamline is +

2 =

2

From slide 1-25

Stagnation streamline

= sin = arctan

1-29

Flow Past a Half Body (contd)

Flow outside the stagnation streamline resembles a flow over an body with a blunt nose.

Equation of the half body is given by the equation of the stagnation streamline.

Movie Flow Over Half Body

1-30

Flow Past a Half Body (contd)

Exercise : For the flow of water ( = 1000 /3) with = 5 / and = 102

a) Determine the equation of the upper part of the half body and plot it.

b) Find the thickness of the body at = 0. Find the maximum thickness of the half body as .

c) Calculate the pressure distribution on the body for /4 < < and plot vs . Take the pressure away from the origin to be = 100 .

d) Calculate the fluid speed on the body for /4 < < and plot vs .

Superposition of

a source of strength at = ,

a sink of strength at = and

uniform flow of magnitude

in + direction.

= + + = +

2ln 1

2ln (2)

= + + = +

21

22

= +

21

22

1-31

A Source and a Sink in Uniform Flow (Flow Past a Rankine oval)

A 1 2 1

2

1-32

Flow Past a Rankine oval (contd)

Exercise : For = 1000 /3, = 5 / , = 102 and = 0.5

a) Determine the equation of the upper part of the half body and plot it.

b) Calculate the pressure distribution on the body for 0 < < and plot vs . Take the pressure away from the origin to be = 100 .

c) Calculate the fluid speed on the body for 0 < < and plot vs .

1-33

Superposition Exercises

Exercise : We want to study the potential flow over the following bodies. Which elementary flows need to be superimposed to get the desired shape?

Which other interesting shapes can you obtain by combining elementary flows?

1-34

Doublet

Superposition of

a source of strength at the orgin (moved from axis to the origin),

a sink of strength at the origin (moved from + axis to the origin) ,

Consider the limiting case of the source and sink of Slide 1-31 approaching to the origin. Skipping the details we can get

=

2cos , =

2sin

where is the strength of the doublet, which is related to .

Velocity field is given by

=

=

22cos

=1

=

22sin

Constant lines

Streamlines

1-35

A Doublet in Uniform Flow (Flow Past a Cylinder)

Superposition of

a doublet of strength at the origin with its axis aligned with the axis.

uniform flow of magnitude in + direction.

= + = +

2cos ()

= + =

2sin ()

1-36

Flow Past a Cylinder (contd)

Exercise : For the superposition of a doublet and uniform flow

Find the location of the stagnation points.

Determine the equation of the stagnation streamlines.

Find the pressure distribution over the cylinder.

Calculate the lift and drag forces due to pressure.

As seen from the above exercise potential flow theory predicts ZERO DRAG

FORCE on the cylinder.

Actually this is the case for any closed body, irrespective of its shape.

This result is not physical and it is known as dAlembert paradox (1752).

In a real viscous flow, shear stresses inside the boundary layer will cause a frictional drag force. Also viscous action will cause separation & full pressure recovery ould NOT be possible.

1-37

Flow Past a Rotating Cylinder

Superposition of

a doublet of strength at the origin

CCW rotating irrotational vortex of strength at the origin

uniform flow of magnitude

This will result in

= + + = +

2cos +

2

= + + =

2sin

2ln

1-38

Flow Past a Rotating Cylinder (contd)

Exercise : For the flow shown above, obtain the velocity components in the cylindrical coordinate system. Show that pressure distribution over a rotating cyliner of radius is

= +2

21 4 sin2

2

sin

2

2

which results in zero drag force and a downward lift force of

=

1-39

Magnus Effect (Kutta-Joukowski Theorem)

A rotating cylinder in a uniform flow will have a net lift force on it.

Direction of the lift force depends on the direction of and .

TOP Low velocity

High pressure

BOTTOM High velocity Low pressure

1-40

Magnus Effect (contd)

Exercise : Anton Flettner built a series of rotor ships in 1920s that are propelled by rotating cylinders driven by electric motors. Consider a version of Flettners ship a single cylinder of height 10 m and diameter of 2 m, rotating at 600 rpm. The ship is sailing with a speed of 5 km/h as shown below. Wind speed is 20 km/h. Considering = 100 and = 1.2 /

3 calculate

a) the thrust on the ship (component of the lift force in the direction of sailing)

b) velocity and pressure at point A.

= 5 / A

= 20 /

http://www.wikipedia.org

Reference: www.rexresearch.com/flettner/flettner.htm

1-41

Magnus Effect (contd)

Exercise : Magnus effect acts not only on cylinders but also on other rotating bodies such as spheres. It can be used to explain how a spinning ball moves in a curved trajectory. A football player wants to make a penalty kick as sketched below. Will a CW or a CCW spin do the trick? Explian using slide 1-39.

Exercise : Watch the following movies

http://www.youtube.com/watch?v=2pQga7jxAyc (Enercon's rotor ship. Audio in German) http://www.youtube.com/watch?v=23f1jvGUWJs (Magnus force acting on a cylinder) http://www.youtube.com/watch?v=wb5tc_nnMUw (Curved free kick from Roberto Carlos)

?

1-42

Kutta Condition (Lift on an Airfoil)

Magnus effect applies not only to cylinders but any closed shape.

Consider the flow over a slender body with a sharp trailing edge, such as an airfoil.

An airfoil is designed to generate high lift force.

There are two stagnation points, 1 and 2.

Experiments show that the streamlines leave the trailing edge smoothly as shown above, known as the Kutta condition.

1 2

The amount of vortex necessary () can be used to calculate the lift force generated on the airfoil.

=

If we add the correct amount of CW vortex to this flow field we can bring point 2 down to the trailing edge and obtain the correct streamline pattern of the previous slide.

1-43

Lift on an Airfoil (contd)

Potential flow theory will predict an unphysical location for point 2.

It is impossible for streamlines to make such a sharp turn at the trailing edge.

1

2

+

=