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2.10 PANEL METHOD
The technique in Sections 2.8 and 2.9 is useful for a variety of simply-shaped bodies, but complicated configurations require a numerical scheme. The panel method provides a quick way of computing such flow fields. To introduce the ideas, consider the flow past a closed body in the panel method, the effect of the body is then represented by a large number, N say, of line vortices, distributed over the surface of the body. The strengths Γ 1 , ..., Γ N of the vortices are essentially determined from the requirement that the normal velocity is zero at N 'target' positions around the surface of the body. It is often convenient to take these target positions to be mid-way between the vortices. Once the Γ n 's are known, the velocity and pressure throughout the flow field can be readily calculated.
U^ n
U
Aerofoil in a uniform stream
Target positions at which u. n = 0
Aerofoil replaced by N line vortices
Γ N
To see why this is a reasonable approach, recall that the tangential fluid velocity increases from zero on the body surface to Us , the inviscid flow value, across a boundary layer in which the vorticity is nonzero. Since the boundary layer is very thin, this vorticity can be considered as concentrated at the body surface. The body can therefore be replaced by a thin sheet of vorticity, with zero velocity inside and the inviscid tangential fluid velocity Us outside. The vorticity/unit length, γ( l ), varies with arc length l.
l
γ( l ) dl
α (^) U ( x v( l ), y v( l ) )
( x , y )
A length dl of surface has a vortex of strength γ( l ) dl associated with it. The corresponding
complex potential is − i γ( l ) dl ln z ( − z v ( l )) / 2 π, where z v ( l ) = x v ( l ) + iy v( l )describes the
position of the vortex element at arc length l from the trailing edge. Superimposing the effects of the uniform flow and the vorticity distribution gives
F ( z ) = Ue^ −^ i^ α^ z − 2 i π γ( ) l ln z ( − z v ( ) l ) dl (2.37)
γ( l ) can be determined from the condition that the normal velocity is zero everywhere on the
surface. Equation (2.37) then gives the complex potential of the flow around the body.
γ( l ) has a simple physical meaning. To determine it apply Stokes theorem,
C^ u dl ·^^ = ^ ω · dS
to a curve C of short length dl , which straddles the boundary layer.
Us C body 0 surface dl
Then ω · dS =γ ( l )dl , while u is zero on one side of C and has the inviscid streamwise
tangential value U , on the other, i.e. u dl · = U sdl , with the upper sign on the upper body
surface and the lower sign on the lower surface. Hence
γ ( l ) = Us (2.38)
Equation (47) is an exact description of the inviscid flow field: the complex potential is an analytic function of z and leads to velocities that satisfy the boundary condition. The panel method approximates this vorticity distribution. The body surface is first covered by N small plates called panels. In two-dimensions these 'panels' are just straight lines. The
Equation (2.41) and the N −1 independent conditions that u. n = 0 = 0 at N −1 target positions are
sufficient to determine the Γ m , m =1,..., N. The complex potential is then given by equation (2.40). The streamlines can be calculated from the Imaginary part of the complex potential in the usual way. The fluid velocity follows from differentiation of the complex potential and the pressure throughout the flow field can then be calculated from Bernoulli's equation.
As N is increased, the panel method gives an increasingly good approximation to the flow. Figure 11 shows streamlines calculated using the panel method for the flow around a Joukowski aerofoil using about 35 panels. Streamlines for the same aerofoil, calculated using the 'function of a function method' are also shown, but are indistinguishable except very close to the aerofoil, where the discrete nature of the panel approximation becomes clear. For most of the flowfield it is clearly an excellent approximation.
Figure 11 Streamlines for flow around a Joukowski aerofoil calculated by the panel method ( N = 35 ).
The circulation in the panel method is 1
N m m =
Γ and so the method gives
1
lift force/unit length
N m m
ρ U
=
= −^ Γ
In Section 2.9, equation (2.36), we determined an exact, analytical form for the lift on a Joukowski aerofoil and it is constructive to compare the estimate in (2.42) with the exact form:
N, number of panels
Estimated lift Lift
The panel method gives a good estimate of the lift, with surprisingly few panels.
Although we have described the method through the example of flow past a Joukowski aerofoil. It applies equally well to a whole range of geometries, including internal flows.
An alternative method for obtaining a numerical solution is to approximate the partial differential equation ∇^2 ψ = 0 by a system of finite difference equations (as taught in 3A3 Numerical
Analysis course). These equations relate the unknowns ψ i , j at nodes throughout the ( x , y ) plane,
and then have to be solved to determine the solution. This approach is not particularly convenient for an unbounded region of flow, which would contain many such nodes! In
contrast, instead of being distributed throughout the flow like ψ i , j , the unknowns Γ n in the panel
method are confined to the surface. There are significantly less of them, and the method works well even for an unbounded flow.
Summary of Section 2. A rigid, two-dimensional surface can be replaced by a distribution of line vortices. The strength of the vortices is determined from the discretized form of the boundary conditions of zero normal velocity on the rigid surface, together with, where appropriate, a condition on the circulation.
If the aerofoil is lifting , so that its circulation Γ is nonzero , then far away the effect of the
aerofoil is like that of a line vortex of strength Γ. If, however, the circulation is zero , then we need to look at the next term in (2.42). The effect of the aerofoil is then like that of a doublet.
Note that this analysis only applies to a streamlined body, because we have assumed that the wake is so thin that the flow outside the boundary layers is irrotational. A bluff body has a wide wake. The effect of such a body is best modelled by a source as in Section 2.2.3. The source correctly accounts for the displacement of the streamlines of the irrotational flow by the wide wake.
Summary of Section 2. Far away from a 2D body in a uniform flow: a bluff body has the same effect on the flow as a line source a lifting, streamlined body has the same effect on the flow as a line vortex, with the same circulation the far-field effects of other 2D bodies decay more rapidly with distance, like the field of a doublet.
Question 15 A large airplane weighs 400 tonnes, has a wing span of 100 m and is flying at 250 m/s through air of density 0.3 kg /m^3. The engines are mounted under and ahead of the wing, as sketched in the figure. The centre of the intake inlet plane C is 4 m ahead and 2 m below the wing centre of lift L. Assuming the lift to be distributed uniformly across the span of the wing, estimate the "droop" angle at which the engine inlet must be set such that the flow is parallel to the centre line at point C.
Droop Angle
C
L
Flight
2.12 IMAGES
In the previous sections, we have been considering bodies in unbounded flows. However, many practical problems involve a nearby rigid boundary. For example, while an aircraft at altitude may be modelled by an aerofoil in an unbounded expanse of fluid, a more appropriate model for an aircraft taking off or landing is an aerofoil near a rigid, plane surface. Often the effect of simple rigid boundaries can be derived from basic unbounded potential flow solutions by the method of images.
2.12.1 Plane walls
Consider a line source above a rigid wall. The boundary condition requires zero normal velocity on the wall and this can be achieved by placing an identical image source the same distance below the wall.
y
line source m (^) is equivalent to m
m
By symmetry, the two sources have a plane surface streamline midway between them and this can be taken to be the wall, i.e. the line source and its image produce the same fluid flow in y ≥ 0 as the line source and the rigid wall. A vortex near a wall requires an image vortex the same distance below but of opposite sign:
line vortex (^) is equivalent to Γ
Similarly, the effect of the ground on a nearby aerofoil can be replaced by an image aerofoil.
Aerofoil is equivalent to
Question 17 Set up an image system for the flow field produced by the sink shown in the figure. Sketch the variation of velocity along the wall y = 0, x ≥ 0. Where on this wall might separation occur? (Detailed calculations are not required.)
2 a
a
m
y
x
Question 18 The trailing vortices left by an aircraft are modelled by two infinitely long line vortices of equal strength Γ but opposite sign, a distance 2 a apart. Calculate the velocity of the vortex pair when they are far above the ground.
Show that as the vortices approach the ground, y = 0, they follow the trajectories
2 2 12 x = ay / y − a. Sketch these trajectories.
Question 19 A source of strength m is placed on the centre-line of a wind-tunnel of height 4 a. Determine the fluid velocity far upstream and far downstream of the source.
A half-body in a wind tunnel is modelled by this line source and uniform stream of velocity α.
For the particular dimensions shown in the figure, use continuity to determine the fluid velocity
far downstream. Hence show that m = 4 aU ∞ and α =3 2 U ∞
Determine the image system for the line source and calculate the position of the nose of the body.
Note that
(^1) coth n^4 4
s s in a a a
=∞
= ^
2.12.2 Cylindrical walls
As an elementary example, consider a line source of strength m , at a distance d from the centre of a rigid cylinder of radius a ( a < d ).
a
− m^ +m
a^2 / d
+m
d
The image point of d is at a^2 / d. Place an image line source of strength m there, together with an equal but opposite line source at the origin. We need to check that this image system satisfies the boundary condition that the cylinder surface is a streamline.
The complex potential for this source and its images is
( ) ( )
2 2 2 2 F z m^ ln z d m^ ln z a^ mln z
π π d π
= − + ^ − −
and we want to show that ψ = Im F is constant on the cylinder surface.
For a position z on the surface of the cylinder, we can write z = aei^ θ^ and then
( )
( )( 2 ) 2
i i i i
m^ ae^ d^ ae^ a^ / d F ae ln ae
θ θ θ
The factor ae ( i^ θ^ −^ a^2 /^ d ) aei^ θ^ can be rewritten as^1 −^ ae
− i θ (^) / d = − (^) ( ae − i θ (^) − d )/ d. This leads to
( ) ( )( ) 2
i i i
i
F ae mln ae d ae d d m ln ae d d
θ θ θ
θ
recalling that the product of a complex number with its complex conjugate is just its magnitude squared.
Using ln (−1) = i π , we finally obtain
( )
2 2 2
i F aei m^ ln ae^ d im d
θ θ
Summary of Section 2.
Source Image(s)
Type Position Type Position
Line source, strength m
distance d above rigid plane surface
line source, strength + m
distance d below plane surface
distance d > a from centre of cylinder
line source, strength + m line source, strength − m
distance a^2 / d from centre of cylinder at centre of cylinder
Line vortex, strength Γ
distance d above rigid plane surface
line vortex, strength −Γ
distance d below plane surface
distance d > a from centre of cylinder
line vortex strength −Γ line vortex strength Γ
distance a^2 / d from centre of cylinder at centre of cylinder
