report hydrofoil

7/28/2019 Report Hydrofoil

1/23

Vortex Lattice Method for Hydrofoils

Gerrie Thiart, Mechanical Engineering,

University of Stellenbosch

Problem Group:

Eyaya Eneyew, AIMS, Cape TownAlistair Fitt, University of Southampton, UK

Neville Fowkes, UWA, Perth, Australia

David Mason, WITS, Johannesburg

Eric Newby, WITS, Johannesburg

Prosper Ngabonziza AIMS, Cape Town

15th January 2010
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2/23

Basic Problem

How to deal with difficulties in implementing vortex latticemethod

Validity of model

What happens as the hydrofoil moves near to the water

surface How to deal with some nearly-divergent integrals that occur

How to deal with some highly oscillatory integrals that may

occur

NOTE: we do not know the details of how Gerrie currently deals

with all these difficulties - so if you already know all this - sorry!
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3/23

Description of Hydrofoil

Hydrofoil is essentially a wing placed underneath a boat The hydrofoil lifts the boat out of the water

The drag is reduced (no hull drag), allowing the boat to

travel faster
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4/23

Vortex Lattice Method

Historically, one of the first CFD methods

Pioneered as long ago as late 1940s at ARC (Aeronautical

Research Council - founded 1909, UK)

A completely INVISCID IRROTATIONALINCOMPRESSIBLE method

- but in spite of this doesallow the prediction of lift and drag

Based on Prandtl lifting line theory

Often used in the early rough stages of aircraft design
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5/23

Vortex Lattice Method (VLM) - General Theory

Assume irrotational flow of an incompressible inviscid fluid

so that fluid velocity q= where is velocity potential We need to solve the Laplace equation 2 = 0 with

suitable BCs

VLM essentially relies on the following observation:

For flow past a body of length L with free stream velocity Uex,we have

= Ux +L

0 f(t) log

(x t)2

+ y2

dt +L

0 g(t) arctany t

x

dt

where

f(t): distribution of sources providing displacement but not liftg(t): distribution of vortices providing lift but not displacement.
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6/23

VLM - Basic

Basic idea of VLM is to use vortices only (zero-width hydrofoil) -

discretize wing into a sequence of panels, each panel

associated with a vortex whose strength ij is to be determined
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7/23

VLM - general 3-D wings with dihedral and camber

Coordinates for 3-D hydrofoil with dihedral and camber


8/23

VLM - Theory and Boundary Conditions

When vortex strengths ij are known, lift and drag follow easily

Trailing vortex contributions cancel since F = Ux

To set up equations to determine ij:

Must impose boundary condition of no flow normal to

hydrofoil

Here we have a free surface (water surface) z = (x, y, t)

Need to satisfy boundary conditions

z = t + xx + yy at z = (x, y, t)1

2(2x +

2y +

2z) + g = 0 at z = (x, y, t)

0 as z


9/23

VLM - Linear Theory This theory is nonlinear

z = t + xx + yy at z = (x, y, t)

12

(2x + 2y +

2z) + g = 0 at z = (x, y, t)

0 as z

No hope of a nice simple solution for the fully nonlinear

problem.

The only hope is to linearize by observing that if the hydrofoil

only creates small waves on the free surface, then

Ux + O()

so the boundary conditions become

z = t + Ux at z = 0

1

2U2 + g = 0 at z = 0

0 as z
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10/23

Three types of problem integral

Though in principle the method is straightforward, it has beenobserved that there can be problems in actually carrying out

the required computations.

We will identify three general circumstances where the

integrations have to be carried out carefully and may causeproblems.

Where do all the troublesome integrals come from?

- to set up the equations to determine the vortex strengths ij,we need to know the velocity components induced by each

vortex on the system. After much simplification, this boils down

to calculating and then summing a whole lot of integrals like:
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11/23

Integrals required for ij equations

u = ij

2

/2

/2

Im[f(, , , )[J(, r, r, r) J(, l, l, l)] cos d

v = ij2

/2/2

Im[f(, , , )[J(, r, r, r) J(, l, l, l)] sin d

w =ij

2

/2

/2

Re[f(, , , )[J(, r, r, r) J(, l, l, l)]d

where

J(, , , ) = 2iH()exp ((z+ ) + i)

+1

1

(z+ ) + i)

0

etdt

t + ((z+ ) + i))

f(, , , ) = sec + tan

cos + sin +

=g

U2

sec2 , = (x ) cos + (y ) sin
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12/23

Problem integral 1:

The first difficulties that we shall discuss come from the function

f(, , , ) = sec + tan

cos + sin + .

Assume that for simplicity = 0 (flat hydrofoil). Then (forexample)

u

/2/2

Jr Jlsin( )

d

wheretan =

.

This exists only as a Cauchy Principal Value (CPV) integral and

MUST be evaluated by suitable methods.
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13/23

Problem integral 1:

Standard numerical integration methods (e.g. Gaussian

quadrature) will not, in general, cope with CPV integrals.To give some concrete illustrative examples, lets take the case

where

We consider a flat non-tapered Hydrofoil

We use a one by one grid to calculate the integrals


14/23

The term involving the first (non-integral) part of J looks like

- showing a typical Cauchy-type singularity behaviour where

cos + sin = 0 - i.e. at = /4 (note also the influence ofthe Heaviside step function - which causes no problems at all).
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15/23

Numerical evaluation of CPV integrals

Normal tactic is to approximate the integrand using a spectralmethod (e.g. splines) and use exact integration (approximate

the function UNDER the integral sign and do all integrals in

closed form) to ensure that the collocation points do not bump

into the singularity as the mesh is refined.

EXAMPLE: use trapezium rule to evaluate

1

0

t4

t 13

dt =49

108+

1

81log2 0.4622610763

using a regular mesh of N intervals that does not bump into the

point t = 13 .

P bl i l
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16/23

Problem integral 1:

Numerical results - integral value vs N - the middle points are

the right answer (spectral method) the upper and lower points

are N = 1 mod 3 and N = 2 mod 3!

P bl i l 2


17/23

Problem integral 2:Another severe difficulty arises from the integral term appearing

in the function J. The third term in J(,,, )) is of the form

0

et

t + ((z+ ) + i)dt

which in the simple case of a 1 1 grid on a flat, non-tapered1 1 hydrofoil becomes

0 sec2

0

et

t + 0 sec2

2z + 12 i(cos + sin )dt

So there will be a term in u of the form2

2

sec2

cos + sin Im

0

et

t + 0 sec2

2z+ 12

i(cos + sin )dt

d.

(Note that the sec2 term in the outer integral is saved by the

one in the inner integral as /2).

P bl i t l 2
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18/23

Problem integral 2:multiplying by the c.c., we get

2

2

sec2

cos + sin

0

et 12 0 sec2 (cos + sin )(t + 2z0 sec2 )2 + 14

20 sec

4 (cos + sin )2 dt d

Though the troublesome cos + sin terms cancel, thedenominator in the t-integral is 0 exactly when = 4 andt = 4z0. Since z < 0 this will inevitably happen, giving a

hypersingular integral and severe numerical problems.

Suggestion: Do not use the Giesing & Smith simplification of J,

but instead use

J = iexp((z+)+i)+ 1

0

exp((z+)+)d

- this avoids hypersingular integrals, replacing them with a CPV

integral (together with the one discussed previously - which are

NOT simultaneously singular).

P bl i t l 3
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19/23

Problem integral 3:Finally, the first term in J(,,, ) is of the form

2iH()exp ((z + ) + i)

The exponential part of this term becomes highly oscillatoryunder the right conditions, which correspond to the hydrofoilbeing moved close to the surface of the water. As an illustrativeexample, for a flat hydrofoil, the exponential can be written as

E = exp0 sec2 (2z+ i) .Since z < 0, when is within a distance of /2, we have

E exp

0

2 + O(4)(2|z|+ i)

The complex term in the exponential causes high frequencyoscillations as 0, but the real term damps them out.For the oscillation to be sufficiently damped we require (by

numerical experiment - more could be done with more time)

2|z|

10

Allo able distance from fl id s rface
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20/23

Allowable distance from fluid surface:

Now consider a hydrofoil (span 2l, chord ) with the centre ofthe leading edge at (x, y) = (0, 0).

Divide the hydrofoil into a 2M N grid

Task: determine the largest possible value of

= (x ) cos + (y ) sin

this will allow us to determine a bound for |z|

- thereby telling us how close to the surface we can bring

the hydrofoil.

Obervation: max() ocurs when calculating the value inducedat the top corner control point by the vortex segment starting at

the origin.

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21/23


This maximal value of was found to be given by

=

2N

cos +

ll

2m

sin

The maximum value of is

max =

2N2

+

ll

2m2

So we have

|z| 120

c c2N2

+

l l2M2

Placing the hydrofoil any closer to the surface then this causes

the integrand to become highly oscillatory corresponding to the

breakdown of the linearised boundary condition.

Effect of placing the hydrofoil close to the surface
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22/23

Effect of placing the hydrofoil close to the surface

Final Conclusions
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23/23

Final Conclusions

You MUST use a suitable method to calculate the CPV

integrals - spectral methods are highly recommended

Do not use the Giesing & Smith simplification of J, but

instead use the less convenient-looking original version.

This has CPV integrals, but no hypersingular integrals Use the criterion that we have found to determine how

close to the surface the hydrofoil may be placed

Dont complain that things go wrong when z is small and

negative - at this point the theory cannot be linearised andthe model is no longer valid.
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