theory and computation of vortex dominated flows: assignment

8/11/2019 Theory and Computation of Vortex Dominated Flows: Assignment

1/12

AS5400 - Theory and Computation ofVortex Dominated Flows

Indian Institute of Technology Madras

Page 1 of 12

Homework Assignment 4

By C R Rakesh, AE11B02617 Apr 2014

1 Question 1The way we got the complex potential for the Klein-Kaffeeloffel experiment is

as follows:

1. First we wrote the complex potential of a stationary and infinitely long

cylinder centred at the origin.

2. Using Joukowski transform, we transformed the circle (side view of the

long cylinder) to an ellipse. Even now the ellipse is stationary

3. Then we said that in the Klein-Kaffeeloffel experiment, the 2

dimensional plate (as is called by Saffman in page 95) is moving with

velocity U and not the fluid against a stationary plate. Hence, we used

the principle of Galilean invariance to transform the co-ordinate system

(since it is inertial) so that the ellipse now moves with velocity U and the

fluid far away from it is stationary

4.

Then, in order to model a two-dimensional plate, we modified the 2constants (a and b in our notation) so that a = b. This yielded a 2

dimensional plate and the appropriate substitution in the complex

potential we derived (up until step 3) yielded the results as obtained

by Klein-Kaffeeloffel.

We shall now verify if the complex potential (as prescribed by inspection,

Saffman page 95) does indeed model the Klein-Kaffeeloffel experiment. The

complex potential so defined is:

U z z (1)We assumed that the two-dimensional plate is at the y-axis and centred at the

origin and time t = 0. We wish to see if equation (1) gives the correct velocity

field and if the boundary conditions are satisfied. The velocity field can be

found out as follows:

1 (2)


2/12



Page 2 of 12

We know the velocity field at t = 0 for the two-dimensional plate to evaluated

in the regions immediately to the left and right of it. Hence, we have to apply

the following to equation (2):

0, ||< , (3)Applying equation (3) to equation (2), we have:

1 (4)

Hence, the velocity components are given by:

, |= |= (5)

The velocity field obtained from equation (5) is exactly the same as that

obtained in the class. Now, before we compute the vortex sheet strength, we

have to verify if the complex potential (equation (1)) also satisfies the

following boundary conditions as mentioned in Saffman:

, 0 ||< 0 (6)Substituting conditions in equation (3) to equation (1), we have:

(7)

From equation (7), we see that condition (6, i) is satisfied.

Now, to show condition (6, ii):

lim lim

lim 1 1 , 1


3/12



Page 3 of 12

lim 1 1 lim 1 1 2 8 lim 2 38 0

From equation (8), we see that condition(6, ii) is satisfied.

To find the vortex sheet strength, from equation (5), we have:

. 2 The general expression for vortex sheet strength is given by

(8)At t=0 and at x=0, we have:

2 (9)

The result obtained in equation (9) is the same as the one obtained in class

and equation (9) represents the strength of the vortex sheet.

2 Question 2For the rolling up of semi infinite vortex sheet, the following equations can

be used to describe the evolution of the vortex sheet:

33 ; 123 ; 14 ; (10)

2 0(11)


4/12



Page 4 of 12

3233 exp[ ] (12)

3(13)

In the above set of equations, we have the following variables:, , ,, , , ,, , and we are going to plot z as a function of time.Since our vortex sheet is made of up infinite vortex filaments, we can use

Kelvins circulation theorem and say: 0or that the value of is same for

all time. (Since we assumed incompressible and inviscid flow to arrive at theabove equations (10 to 13), we can safely say that Kelvins circulation theorem

will also hold).

Excluding z and t, we have 8 parameters and 5 equations (except equation 13).

Hence, we can choose and modify 3 parameters. We choose to modify , , .Initial condition: A vortex sheet in the x-axis in [0,100]unitsThe following cases were chosen for the simulations


5/12



Page 5 of 12

Case 1: 0, 4, /64: this is chosen as the base case. All the othercases will be compared with this case.

Figure 1 The Base case

It can be seen that for the above combination of the tweakable parameters,

the vortex sheet rolled to form a spiral at time t = 1 units.

For the further cases, note the following: Value of time t, Axis limits. These will

be used to explain what is observed.


6/12



Page 6 of 12

Case 2: 0, 4, /16

Figure 2 Epsilon is increased

When we increase the value of by 4 times, it is seen that the vortex sheetcurls up to a very less extent. The number of spiral loops seen (with respect to

the same axes window) is less compared to that observed in Figure 1. Hence,

we conclude that increasing

reduces the number density of spiral loops in

the xy plane after the same amount of time.

Hence, we can say that the parameter denotes the number density ofspirals and that the number density of spiral loops varies inversely with.


7/12



Page 7 of 12

Case 3: 0, 8, /64

Figure 3 -

is increased

Figure 4 - is increased and viewed at half time


8/12



Page 8 of 12

By comparing figure 4 and figure 1, we can see that they are very similar. But

it should be noted that figure 4 is for t = 0.5 units. Hence, we can say that

increasing

essentially increases the roll-up rate of the vortex sheet. Figure

3 shows the vortex sheet at time t = 1 units. It is seen that the vortex sheet

curls up further.

Hence, we conclude that increasing increases the speed of rolling upor speed of evolution of vortex sheet.

Case 4:

1, 4, /64

Figure 5 - is made positiveIt is seen from figure 5 that making positive makes the centre of the curledup vortex sheet move towards the positive y-axis. Hence, we conclude that


9/12



Page 9 of 12

the parameter denotes (to some extent) the position of the centre ofthe vortex sheet.

Case 5: 1, 4, /64

Figure 6 - is made negativeIt is seen from figure 5 that making negative makes the centre of the curledup vortex sheet move towards the negative y-axis (See the point (-1, -1.2)).

The simulation is done till time t = 0.5 because at time t = 1 units, the spiral

diverges very much and the positions of the individual vortex filaments are

too large to be stored by the computer, i.e., the positions become NaN.

The above statement also demonstrates another fact: the vortex sheet curls

up and expands as time progresses. It is essentially unstable, which is why

at higher time steps, we are not able to represent the position of the vortex

sheet (on a computer).


10/12


11/12



Page 11 of 12

In the above example, the key ingredient in proving condition (14) was

choosing a function f(n) such that

= is finite.(Note: we assume

that the other terms viz. A0, k, t0, t, U2, U1, are all finite)Hence, we can choose any function f(n) which makes the value of theimproper integral to be finite.

Some sample functions:

3.1 f(n) = A0.n.exp(-n2)

lim

lim 12

| 12

(17)

, exp sin 2

=

Consider t < t0:

|, | exp = |, | =

12

< 0 <

Hence, solution converges for t < t0. The solution diverges for t > t0because > 0 > .3.2 f(n) =

+The integral of f(n) is finite:

lim

1

lim tan 2


12/12



Page 12 of 12

By using the same argument as in section 3.1, we can prove condition (14).

3.3 f(n)=

+

The integral of f(n) is finite:

lim lim 1 lim tan | 4

By using the same argument as in section 3.1, we can prove condition (14).

Hence, the singularity character of Birkhoff-Rott equation is established.

theory and computation of vortex dominated flows: assignment

Documents