Unsteady Aerodynamics: Panel Method (Morino's Method) Flashcards

1
Q

Why are panel methods used?

A

The Theodorsen model is very powerful, but there are limitations related to the incompressibility of the flow and the hypothesis of large aspect ratio in case of 3D wings. Therefore, panel methods are needed in order to take into account the effects of 3D and compressibility.

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2
Q

What can Morinos method be used for?

A

1) Unsteady flow
2) Compressible flows
3) Subsonic/Supersonic flow (even if it’s not the best for supersonic cases)
4) Any geometry

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3
Q

What is the procedure to formulate Morinos method?

A

1) Starting from the divergence theorem, we can reach Green’s identity.

2) Considering to take as function G the fundamental solution to Laplce equation (ie the computation of the potential that results from having a singulatrity in the point x0), so the spatial Dirac delta. This is a function that goes to zero anywhere except at x0, where it goes to infinity. The problem is linear and can be seen as the superimposition of the combination of peturbations. The solution to this problem is known and computable, so G can be found

3) Subtituting the fundamental solution to the Laplace problem, it is easy to verify that Sφ grad(G)^2 = E(x0)φ(x0), where E(x0) is equal to 0 if x0 is external from V, 1 if x0 is inside the boundary, 1/2 if x0 is on the boundary.

4) Using this expression in the Green identity, we can find an expression for E(x0)φ(x0). This expression can be used to compute the potential φ when the solution to the fundamental problem G is known. The function E(x0) comes from the definition of the Dirac delta and it’s integral.

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4
Q

What are the BC for the incompressible case?

A

1) Irrotational flow
2) Normal velocity of the airfoil is equal with the velocity of the flow at that direction
3) Normal velocity of the wake is equal to zero (the wake is aligned with the local flow field)

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5
Q

Dsicuss about the analysis of the wake for the steady case.

A

The wake, by definition, is a surface of constant pressure, because if the jump of pressure over the wake was different from zero, the the wake will move from where it was because the jump of pressure will cause a movement of the wake itself. So the jump of pressure is a constant that we set to zero.

Therefore using Bernoullis theorem and all derivatives wrt are null, we can find that the jump of potential on the wake along the direction of the flow x is null. The jump of potential on the wake at the TE is equal to the circulation at that span station y, as prescribed by the Kutta condition.

This means that the jump of potential that exists at the trailing edge is transported to infinity. So, there could be a difference in the y direction, maybe cause the airfoil at a certain span is producing a different lift than at another span, but in the x direction it is constant.

Using this in Morino’s expression, we obtain an integral expression that can be used to compute the potnential on every point in the flow field when we know the potential on B, or to compute the potential on B itself. We obtain an expression that has sources and doublets on the airfoil, and the jump of potential in the wake.

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6
Q

What is the discretization panel method?

A

1) The surface is divided into N=Nx x Ny panels and the wake into Ny semi infinite stripes
2) Define the type of approximation for the potential φ over each panel. For 0-th order, the potential is constant on each panel, while for the 1st order, the potential is linear on each panel, different values of potential on each node.
3. Decide how and where the BC must be imposed on each panel

For the 0-th order panel, the best collocation point (point where the BC is applied) is at the center of the panel.

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7
Q

Discuss how the discretization and loads are obtained for the incompressible steady case.

A

The integral expression is discretized and transformed to a sum expression. The integral of the wake depends on a value hk, which take values 0, 1, -1 if the panel is not a TE panel, is a TE upper panel, is a TE lower panel correspondingly.

It is not necesasry to have potential values associated with the wake panels. The jump of potential of the wake panel is equal to the different between the potential of the corresponding TE panels (imposition of Kutta condition).

We can write everything in matrix form using Yφ = Z ∂φ/∂n = Uinf Z ∂η/∂x (steady case) (Y = matrix of coeffiecients, phi is vector of potentials and Z is the integral)

The coefficient of pressure is equal with -2/Uinf * ∂φ/∂x. We approximate the continuous potential function with a shape function Νφ times the discretized potential φ. The derivative wrt x of this function is equal to Pφ, where P is the derivative of the shape function wrt x. Then using Yφ = Z∂φ/∂n = UooZ∂η/∂x and setting as: A = YZ^-1 and α = ∂η/∂x, Cp = -2PA^-1*α.

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8
Q

Discuss how the discretization and loads are obtained for the incompressible unsteady case.

A

Since the case is unsteady, the full Bernoulli theorem has to be considered, and at the border conditions at the trailing edge, where the jump of potential is equal to the one at the TE.

The jump of pressure of the wake is equal to zero, so using the full bernoulli eq we have:
-2/Uinf^2 * (Uinf ∂Δφ/∂x + ∂Δφ/∂t) = 0. Transforming the equation in frequency domain, we obtain a solution of the jump of potential which depends on the circulation and a the position of the panel. The jump of potential along the wake panels is not constant anymore: it is equal to the product of the jump of potential at the trailing edge, multiplied by a delay operator that express the fact that the vorticity travels in the wake at the asymptotic speed.

Now for the discretization, it is necessary to divide the wake into individual panels also along the length x direction due to the delay term. If the distance between panels is larger than the size of the panel, the delay could be considered cosntant and moved out of the integrals.

Assigning a reduced frequency k, it is possible to compute the loads generated by the different body displacements and so a tabulated version of the aerodynamic transfer matrix.

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8
Q

How is the problem formulated in the case of compressible flow: steady flow?

A

In these cases, we apply the PRANDTL-GLAUERT transformation to transform the equation into the original, incompressible format.

The original is: ∇^2φ = Minf^2 * ∂^2φ/∂x^2. Setting as x’ = x/β, y’ = y and z’ = z, with β^2 = 1-Minf^2, we obtain: ∇^2φ’ = 0. Then, we calculate the distance r’ using the new coordinate system, and we obtain a new expression for the discretized problem.

The compressible case is like an incompressible case with an expanded coordinate in the x direction.

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9
Q

How is the problem formulated in the case of compressible flow: unsteady supersonic flow? What is the explanation for τ (tau). What is the difference between subsonic and supersonic flow?

A

Using the euation for the linearized potential :∇^2φ -1/α^2 ∂^2φ/∂t^2 = 0. We look for the ufndamental solution of this modified Laplace equation and transform it in the frequency domain, from which we can obtain a general solution for the fundamental solution G, which depends on the distance and τ, where τ is the time required by the peturbation to travel to x0, the point where it started, to x.

Given x and x0, τ is the amount of time we have to go back in x0 to understand the intensity of the peturbation that reached x in the current instant of time (αinf * τ = r, so τ = sqrt(r^2/ainf^2)). If we are at a certain position x and we generate at the initial instant in time a peturbation, we will feel the peturbation only after time tau. Before that instant of time, the wave that has been generated will not have reached x. Before we assumed that the peturbation travelled at an infinite speed, but here it travels at a finite speed.

If the peturbation sourc is moving forward with speed Uinf in the direction x. If the fluid is moving at constand speed, after the τ, x0 will move to a different place so the delay will be different in the different directions. Solving the new equation, we obtain two solutions of τ, where a negative delay doesnt make sense.

For subsonic flows (Minf <1), the negative tau is discarded and we substitute tau in the fundamental solution to the compressible equation of motion in frequency domain.

For supersonic flows, (Minf>1), there are different possibilities, as β^2 becomes negative. There are tow possibilities.
If rβ >= 0 , the point is within the Mach cone and it is reached by two waves that started at different times from different positions. (τ1,τ2>0)
If rβ <0 The point is outside the Mach cone and it is not reached by any peturbation (τ1,τ2<0)
This means that there is a portion of space where there are two time delays and one portion where we have no time delay possible. In the supersonic case, the source of emission is travelling faster than the speed of sound, whole the information is travelling at a certain speed. After τ seconds, the information emitted has reached some points, but the point that emitted is outside the range, because it is travelling faster.

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10
Q

What are the domains of dependance in supersonic flow panelization?

A

Knowing the shape of the Mach cone for every panel, it is possible to plot two areas:
1) Domain of dependance
2) Domain of influence

For example, Panel A influences the panels within the Mach cone directed downward and is influenced only by the panels included in the Mach cone directed upward. If a panel is in both domains, we can refine the panel distribution or perform the integrals only on portions of the panels.

Also, by using zero order panels, we will end up having potential that is jumping from one point to another, which is a generator of peturbation. Indeed, by using zero-order panels, we will end up having a lot of shockwaves that will appear as a discontinuity of potential. For this rerason it is better to use first order panels to avoid discontinuiities for supersonic flows.

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11
Q

Discuss the indicial response: compressible vs incompressible flow.

A

Looking at the Wagner diagrams (result of a step applied at the boundary conditions) we can examine the lift and moments generated.

For the lift, the potential solution goes to infinity at the first instant of time because we have a jump due to added mass terms proportional to acceleration, which is infinite at the first instant of time, then slowly grows to reach the steady value. In a compressible flow, the information does not travel instantaneously, so the jump will be smaller and smaller and then goes to a steady value whish is different than the incompressible case, due to Prandtl-Glauert correction.

For the moment, also for the incompressible flow the moment will start from infinite, while the compressible will start from a finite value. After that instant in time, the incompressible moment will always go to zero, while the compressible goes to zero with a delay. The AC will move backward because the moment will have a different constant in time

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12
Q

What is the Lifting Surface?

A

The idea of the lifting surface is to obtain a formulation where given the aerodynamic loads developed in a certain lifting surface it is possible to derive the induced inflow (and so the change of AoA) on the surface. The advantage is that there is no need to discretize the wake. w/Uoo = int(K * ΔCP)), where K is a kernel function.

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13
Q

What is the Piston Theory?

A

When the Mach number is very large, the region of influence and dependance are very small, so it is reasonable to consider the variation of pressure on every point only function of the movement of that point (not influenced by others). Every panel generates a change of pressure that is locally generated and locally defined. Every panel works for itself. Every points functions as a piston.

If Minf»1 (Minf>2 or higher), it is possible to compute the pressure difference of each point given the movement of the point itself. The epxression is nonlinear and if w/ainf «1 then a linearization of 1st and 2nd order is possible.

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14
Q

What is a Recap of all the method described?

A

1) 2D flows
1.1. M = 0.
a) Theodorsen h, α, β (We obtain Ham(k))
b) Cicala: Any change of η (We obtain Ham(k))
c) Wagner + Duhamel integral (Time domain formulation)

1.2) M>= 0 Subsonic or supersonic (Ham(k))
a) Panel method
b) Lifting surface
c) Vortex Method (compute indicial response)

1.3) M»1
Piston theory

1.4) M ~ 1
CFD

2) 3D flows
a) Strip theory (combine 2D models)
b) Panel methods

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15
Q

What is the aeroelastic interface?

A

Typically we deal with two systems. The first is the structural system and the other is the aerodynamic system, which are developed using ncompatible misrepresentations. For instance, we can represent the structure of the wing using a beam. This must be connected with the dynamic model that is mainly represented through panels. The job of the aeroelastic interface is to cope with this problem.

With the beam assumption, we assume that the beam section is rigid, so we can compute the displacement of the airfoil using this rigid assumption (xf = xea + wxd).

The aeroelastic interface must have:
1) The ability to interface matching and non-matching grids
2) Ability to deal with irregular grids and local refinements
3) Correct representation of rigid movements
4) Computational efficiency
5) Independence from the numerical formulation used in the structure or aerodynamic field
6) Ability to manage extrapolation
7) Conservation of exchanged energy
8) Smoothness

The first step is to compute displacement of aerodynamic panels, by knowing the displacement of the structured loads. The second step is to do the opposite. So we take the pressures that are applied on the panel and revert them into loads that are applied on the structure: nodal and sectional loads.

There is a need to have some compatibility between the way we transfer kinematics from structure to aerodynamic grid and this compatibility comes from the conservation of virtual works.