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Steady flow of an incompressible perfectly conducting fluid past a thin airfoil

Boundary Value Problems volume 2013, Article number: 231 (2013) | Download Citation

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Abstract

We consider the linearized Euler and Maxwell equations and Ohm’s law. We calculate the fundamental matrix and give integral representations for the velocity, magnetic induction and pressure. We use the boundary (slip) condition to obtain an integral equation for the jump of the pressure. We give some graphic representations of the velocity and magnetic induction for the case of the flat plate.

MSC:76X05, 45E05, 35E05.

Introduction

In papers dedicated to the motion of a wing in an electro-conductive fluid, the lift, drag and moment coefficients were calculated. Recent technological advances claim also for the study of the velocity and electromagnetic fields. We mention two examples: the plasma actuators for aircraft flow control (see [1]) and concealing aircrafts from radar using interaction between ionized gas and electromagnetic radiation. In the present paper, we study the steady two-dimensional flow of an ideal perfectly conducting incompressible fluid around a thin insulating airfoil. We consider the linearized partial differential equations of magnetohydrodynamics (consisting of Euler’s and Maxwell’s equations and Ohm’s law) and calculate the corresponding fundamental matrix. In order to obtain the integral representations for the velocity, the magnetic induction and the pressure fields (which represent the original result of this work), we perform the convolution of the components of the fundamental matrix with the simple layer distributions determined by the jump of the functions we are looking for. We notice that every integral representation has an elliptic as well as a hyperbolic part, this last one being determined by the presence of simple waves bounded by straight characteristics (weak shocks). From the integral representation of the velocity and the boundary conditions (linearized slipping condition and the continuity of the magnetic induction), we rediscover the singular integral equation for the jump of the pressure across the airfoil. We consider the particular case of the flat plate for which the solution of the integral equation is known. Then we perform some numerical integrations to calculate the velocity and the magnetic induction at the points of a two-dimensional grid in order to provide graphic representations for the velocity and magnetic induction fields.

Fundamental matrix of the linearized equations for the two-dimensional incompressible flow of perfectly conducting fluids

Let v, b and p designate the nondimensional perturbations of the velocity, magnetic induction and pressure, respectively, determined by the presence of a thin insulating airfoil whose equation is

y= h ± (x),x∈[0,1],| h ± (x)|≪1,| h ± ′ (x)|≪1.
(1)

At infinity, we assume that the unperturbed motion is uniform and parallel to the Ox-axis and that there exists a homogeneous magnetic field whose nondimensional expression is

B 0 =( α x , α y ), α x =cos(α), α y =sin(α),α∈(−π/2,π/2).
(2)

As it is shown in [2], Section 5.2, v=( v x , v y ), b=( b x , b y ) and p satisfy the following system of linear partial differential equations obtained by means of the small perturbations technique:

∂ v x ∂ x + ∂ p ∂ x + α y A ( ∂ b y ∂ x − ∂ b x ∂ y ) = 0 , ∂ v y ∂ x + ∂ p ∂ y − α x A ( ∂ b y ∂ x − ∂ b x ∂ y ) = 0 , ∂ v x ∂ x + ∂ v y ∂ y = 0 , ∂ b x ∂ x + ∂ b y ∂ y = 0 , b y + α y v x − α x v y = 0
(3)

with A= 1 Rh being Alfvén’s number (Rh is the magnetic pressure number). We introduce like in [3] the fundamental matrix of linear system (3)

( v x ( 1 ) v y ( 1 ) b x ( 1 ) b y ( 1 ) p ( 1 ) v x ( 2 ) v y ( 2 ) b x ( 2 ) b y ( 2 ) p ( 2 ) v x ( 3 ) v y ( 3 ) b x ( 3 ) b y ( 3 ) p ( 3 ) v x ( 4 ) v y ( 4 ) b x ( 4 ) b y ( 4 ) p ( 4 ) ) ,
(4)

whose components are the fundamental solutions of the systems

∂ v x ( j ) ∂ x + ∂ p ( j ) ∂ x + α y A ( ∂ b y ( j ) ∂ x − ∂ b x ( j ) ∂ y ) = δ j 1 δ ( x , y ) , ∂ v y ( j ) ∂ x + ∂ p ( j ) ∂ y − α x A ( ∂ b y ( j ) ∂ x − ∂ b x ( j ) ∂ y ) = δ j 2 δ ( x , y ) , ∂ v x ( j ) ∂ x + ∂ v y ( j ) ∂ y = δ j 3 δ ( x , y ) , ∂ b x ( j ) ∂ x + ∂ b y ( j ) ∂ y = δ j 4 δ ( x , y ) , b y ( j ) + α y v x ( j ) − α x v y ( j ) = 0 ,
(5)

where δ(x,y) is Dirac’s distribution and δ j i ={ 1 , j = i , 0 , j ≠ i .

Some components of the fundamental matrix

We shall use the Fourier transform

F[f]( ξ 1 , ξ 2 )= ∬ R 2 exp [ i ( x ξ 1 + y ξ 2 ) ] f(x,y)dxdy.
(6)

Taking into account that

F [ ∂ f ∂ x ] ( ξ 1 , ξ 2 )=−i ξ 1 F[f]( ξ 1 , ξ 2 ),F [ ∂ f ∂ y ] ( ξ 1 , ξ 2 )=−i ξ 2 F[f]( ξ 1 , ξ 2 ),
(7)
F[δ]=1,
(8)

and using the notation F[f]= f ˜ , we obtain from (5)

− i ξ 1 v ˜ x ( j ) − i ξ 1 p ˜ ( j ) + i α y ξ 2 A 2 b ˜ x ( j ) − i α y ξ 1 A 2 b ˜ y ( j ) = δ j 1 , − i ξ 1 v ˜ y ( j ) − i ξ 2 p ˜ ( j ) − i α x ξ 2 A 2 b ˜ x ( j ) + i α x ξ 1 A 2 b ˜ y ( j ) = δ j 2 , − i ξ 1 v ˜ x ( j ) − i ξ 2 v ˜ y ( j ) = δ j 3 , − i ξ 1 b ˜ x ( j ) − i ξ 2 b ˜ y ( j ) = δ j 4 , b ˜ y ( j ) + α y v ˜ x ( j ) − α x v ˜ y ( j ) = 0 .
(9)

Solving system (9), we get for j=2

v ˜ x ( 2 ) = i ξ 1 2 ξ 2 ( ξ 1 2 + ξ 2 2 ) ( ( ξ 1 α x + ξ 2 α y A ) 2 − ξ 1 2 ) , v ˜ y ( 2 ) = − i ξ 1 3 ( ξ 1 2 + ξ 2 2 ) ( ( ξ 1 α x + ξ 2 α y A ) 2 − ξ 1 2 ) , b ˜ x ( 2 ) = i α x ξ 1 2 ξ 2 + i α y ξ 1 ξ 2 2 ( ξ 1 2 + ξ 2 2 ) ( ( ξ 1 α x + ξ 2 α y A ) 2 − ξ 1 2 ) , p ˜ ( 2 ) = − i ξ 1 2 ξ 2 ( ξ 1 2 + ξ 2 2 ) ( ( ξ 1 α x + ξ 2 α y A ) 2 − ξ 1 2 ) + i α y A 2 ξ 1 α x + ξ 2 α y ( ξ 1 α x + ξ 2 α y A ) 2 − ξ 1 2 .
(10)

For (l,k)∈{(3,0);(2,1);(1,2)}, we have

ξ 1 l ξ 2 k ( ξ 1 2 + ξ 2 2 ) ( ( ξ 1 α x + ξ 2 α y A ) 2 − ξ 1 2 ) = a l k ξ 1 + b l k ξ 2 ξ 1 2 + ξ 2 2 + c l k ξ 1 + d l k ξ 2 ( ξ 1 α x + ξ 2 α y A ) 2 − ξ 1 2
(11)

with

a 30 = − A 2 ( A 2 + 1 − 2 α x 2 ) 1 + 2 A 2 ( 1 − 2 α x 2 ) + A 4 , b 30 = − 2 A 2 α x α y 1 + 2 A 2 ( 1 − 2 α x 2 ) + A 4 , c 30 = α y 2 ( A 2 + 1 + 2 α x 2 ) 1 + 2 A 2 ( 1 − 2 α x 2 ) + A 4 , d 30 = 2 α x α y 3 1 + 2 A 2 ( 1 − 2 α x 2 ) + A 4 ,
(12)
a 21 = 2 A 2 α x α y 1 + 2 A 2 ( 1 − 2 α x 2 ) + A 4 , b 21 = − A 2 ( A 2 + 1 − 2 α x 2 ) 1 + 2 A 2 ( 1 − 2 α x 2 ) + A 4 , c 21 = 2 α x α y ( A 2 − α x 2 ) 1 + 2 A 2 ( 1 − 2 α x 2 ) + A 4 , d 21 = α y 2 ( A 2 + 1 − 2 α x 2 ) 1 + 2 A 2 ( 1 − 2 α x 2 ) + A 4 ,
(13)
a 12 = A 2 ( A 2 + 1 − 2 α x 2 ) 1 + 2 A 2 ( 1 − 2 α x 2 ) + A 4 , b 12 = 2 A 2 α x α y 1 + 2 A 2 ( 1 − 2 α x 2 ) + A 4 , c 12 = ( A 2 − α x 2 ) ( A 2 + 1 − 2 α x 2 ) 1 + 2 A 2 ( 1 − 2 α x 2 ) + A 4 , d 12 = − 2 α x α y 3 1 + 2 A 2 ( 1 − 2 α x 2 ) + A 4 .
(14)

Hence

v ˜ x ( 2 ) = i a 21 ξ 1 + i b 21 ξ 2 ξ 1 2 + ξ 2 2 + i c 21 ξ 1 + i d 21 ξ 2 ( ξ 1 α x + ξ 2 α y A ) 2 − ξ 1 2 , v ˜ y ( 2 ) = − i a 30 ξ 1 − i b 30 ξ 2 ξ 1 2 + ξ 2 2 + − i c 30 ξ 1 − i d 30 ξ 2 ( ξ 1 α x + ξ 2 α y A ) 2 − ξ 1 2 , b ˜ x ( 2 ) = α x v ˜ x ( 2 ) + α y v ˜ y ( 2 ) + i α y ξ 1 ( ξ 1 α x + ξ 2 α y A ) 2 − ξ 1 2 , p ˜ ( 2 ) = − v ˜ x ( 2 ) + i α y A 2 ξ 1 α x + ξ 2 α y ( ξ 1 α x + ξ 2 α y A ) 2 − ξ 1 2 .

Using the inverse Fourier transforms ([4], Appendix 1)

F − 1 [ ξ 1 ξ 1 2 + ξ 2 2 ] = − i 2 π ∂ ∂ x ln x 2 + y 2 , F − 1 [ ξ 2 ξ 1 2 + ξ 2 2 ] = − i 2 π ∂ ∂ y ln x 2 + y 2 , F − 1 [ ξ 1 ( ξ 1 α x + ξ 2 α y A ) 2 − ξ 1 2 ] = i A 2 α y ∂ ∂ x H ( x − α x α y y − A α y | y | ) , F − 1 [ ξ 2 ( ξ 1 α x + ξ 2 α y A ) 2 − ξ 1 2 ] = i A 2 α y ∂ ∂ y H ( x − α x α y y − A α y | y | )

(where H(x)={ 1 , x ≥ 0 , 0 , x < 0 is Heaviside’s function), we get

v x ( 2 ) = a 21 2 π x x 2 + y 2 + b 21 2 π y x 2 + y 2 − A c 21 2 α y ∂ ∂ x H ( x − α x α y y − A α y | y | ) − A d 21 2 α y ∂ ∂ y H ( x − α x α y y − A α y | y | ) ,
(15)
v y ( 2 ) = − a 30 2 π x x 2 + y 2 − b 30 2 π y x 2 + y 2 + A c 30 2 α y ∂ ∂ x H ( x − α x α y y − A α y | y | ) + A d 30 2 α y ∂ ∂ y H ( x − α x α y y − A α y | y | ) ,
(16)
b x ( 2 ) = α x v x ( 2 ) + α y v y ( 2 ) − A 2 ∂ ∂ x H ( x − α x α y y − A α y | y | ) ,
(17)
p x ( 2 ) = − v x ( 2 ) − α x 2 A ∂ ∂ x H ( x − α x α y y − A α y | y | ) − α y 2 A ∂ ∂ y H ( x − α x α y y − A α y | y | ) .
(18)

Integral representations

In thin airfoil theory, the linearized boundary conditions are usually imposed on the segment [−1,1] and the functions we are looking for are defined on R 2 ∖[−1,1]. Since v x , v y , b x , b y and p are integrable functions, they may be regarded as regular distributions. Taking into account the boundary conditions ([2], Chapter 5)

[b](x)=0, v y (x,±0)= h ± ′ (x),x∈[−1,1],
(19)

we obtain the following linearized system for the distributions v x , v y , b x , b y and p:

∂ v x ∂ x + ∂ p ∂ x + α y A ( ∂ b y ∂ x − ∂ b x ∂ y ) = 0 = f 1 , ∂ v y ∂ x + ∂ p ∂ y − α x A ( ∂ b y ∂ x − ∂ b x ∂ y ) = [ p ] δ [ − 1 , 1 ] = f 2 , ∂ v x ∂ x + ∂ v y ∂ y = [ h ′ ] δ [ − 1 , 1 ] = f 3 , ∂ b x ∂ x + ∂ b y ∂ y = 0 = f 4 , b y + α y v x − α x v y = 0 ,
(20)

where [b](x) and [p](x) represent the jumps of b and p over the segment [−1,1] and [p] δ [ − 1 , 1 ] , [ h ′ ] δ [ − 1 , 1 ] are simple layer distributions. We may easily verify that

v x = ∑ j = 1 4 v x ( j ) ⋆ f j , v y = ∑ j = 1 4 v y ( j ) ⋆ f j , b x = ∑ j = 1 4 b x ( j ) ⋆ f j , p = ∑ j = 1 4 p ( j ) ⋆ f j ,
(21)

where ⋆ stands for the convolution product. We shall consider, for the sake of simplicity, the case of zero thickness wing, i.e., h + (x)= h − (x). Hence f 1 = f 3 = f 4 =0 and

v x = v x ( 2 ) ⋆ [ p ] δ [ − 1 , 1 ] , v y = v y ( 2 ) ⋆ [ p ] δ [ − 1 , 1 ] , b x = b x ( 2 ) ⋆ [ p ] δ [ − 1 , 1 ] , p = p ( 2 ) ⋆ [ p ] δ [ − 1 , 1 ] .

We calculate the convolutions and obtain the following representation for v x :

v x ( x , y ) = a 21 2 π ∫ − 1 1 [ p ] ( ξ ) x − ξ ( x − ξ ) 2 + y 2 d ξ + b 21 2 π ∫ − 1 1 [ p ] ( ξ ) y ( x − ξ ) 2 + y 2 d ξ − A c 21 2 α y ∂ ∂ x ∫ − 1 1 [ p ] ( ξ ) H ( x − ξ − α x α y y − A α y | y | ) d ξ − A d 21 2 α y ∂ ∂ y ∫ − 1 1 [ p ( ξ ) ] H ( x − ξ − α x α y y − A α y | y | ) d ξ = a 21 2 π ∫ − 1 1 [ p ] ( ξ ) x − ξ ( x − ξ ) 2 + y 2 d ξ + b 21 2 π ∫ − 1 1 [ p ] ( ξ ) y ( x − ξ ) 2 + y 2 d ξ − A 2 α y 2 ( c 21 α y − d 21 α x − d 21 A y | y | ) × [ p ] ( x − α x α y y − A α y | y | ) × H ( 1 − | x − α x α y y − A α y | y | | ) .
(22)

Similarly, we get

v y ( x , y ) = − a 30 2 π ∫ − 1 1 [ p ] ( ξ ) x − ξ ( x − ξ ) 2 + y 2 d ξ − b 30 2 π ∫ − 1 1 [ p ] ( ξ ) y ( x − ξ ) 2 + y 2 d ξ + A 2 α y 2 ( c 30 α y − d 30 α x − d 30 A y | y | ) × [ p ] ( x − α x α y y − A α y | y | ) × H ( 1 − | x − α x α y y − A α y | y | | ) ,
(23)
b x ( x , y ) = α x v x ( x , y ) + α y v y ( x , y ) − A 2 [ p ] ( x − α x α y y − A α y | y | ) × H ( 1 − | x − α x α y y − A α y | y | | ) ,
(24)
p ( x , y ) = − v x ( x , y ) + y 2 | y | [ p ] ( x − α x α y y − A α y | y | ) × H ( 1 − | x − α x α y y − A α y | y | | ) .
(25)

The integral equation for the jump of the pressure

Using the Plemelj formulas and linearized boundary conditions (19), we get from (23) the integral equation

h ′ ( x ) = v y ( x , ± 0 ) = − a 30 2 π p . v . ∫ − 1 1 [ p ( ξ ) ] x − ξ d ξ + A ( c 30 α y − d 30 α x ) 2 α y 2 [ p ( x ) ] , x ∈ ( − 1 , 1 ) ,
(26)

which is equivalent to

−β[p](x)+ k π p . v . ∫ − 1 1 [ p ] ( ξ ) ξ − x dξ=χ h ′ (x),
(27)

where k= A 2 +1−2 α x 2 , β= α y (1+ A 2 ) A − 1 and χ=2(4 α x 2 + A − 2 − A 2 ). As it is shown in [3, 5] and [2], Chapter 5, the solution of equation (27) is

[ p ] ( x ) = − β χ β 2 + k 2 h ′ ( x ) − k χ π ( β 2 + k 2 ) ( 1 − x 1 + x ) θ p . v . ∫ − 1 1 ( 1 + ξ 1 − ξ ) θ h ′ ( ξ ) ξ − x d ξ + 2 C sin θ π ( 1 − x ) 1 − θ ( 1 + x ) θ , tan θ π = k β , 0 ≤ θ < 1 .
(28)

We may take C=0 if we impose Kutta-Joukovsky’s condition. Other choices of the constant C were considered by Stewartson [6].

The flat plate

In this case, h(x)=−εx, h ′ (x)=−ε. Taking into account that

p . v . ∫ − 1 1 ( 1 + ξ 1 − ξ ) θ 1 ξ − x dξ= π sin θ π { 1 − ( 1 + x 1 − x ) θ cos θ π } ,

we get from (28)

[p](x)= ε χ β 2 + k 2 ( 1 − x 1 + x ) θ .

The lift is ([2], 5.2.6)

L=− ∫ − 1 1 [p](x)dx=4επ ( A 2 − 1 ) 2 + 4 A 2 α y 2 A 2 ( A 2 − 1 + 2 α y 2 ) θ.

In order to obtain graphic representations for the velocity (Figure 1), magnetic induction (Figure 2) and lift (Figure 3), we use the quadrature formulas

I 1 ( x , y ) = ∫ − 1 1 ( 1 − ξ 1 + ξ ) θ x − ξ ( x − ξ ) 2 + y 2 = ∑ k = 1 N W k x − t k ( x − t k ) 2 + y 2 , I 2 = ∫ − 1 1 ( 1 − ξ 1 + ξ ) θ y ( x − ξ ) 2 + y 2 = ∑ k = 1 N W k y ( x − t k ) 2 + y 2 ,

where

W k =− 4 Γ ( N + θ + 1 ) Γ ( N − θ + 1 ) [ ( N + 1 ) ! ] 2 P N − 1 ( θ + 1 , − θ + 1 ) ( t k ) P N + 1 ( θ , − θ ) ( t k ) .
Figure 1
figure1

Velocity field. The velocity is represented by arrows in the grid points.

Full size image
Figure 2
figure2

Magnetic induction field. The magnetic induction is represented by arrows in the grid points.

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Figure 3
figure3

Lift coefficient versus Alfvén’s number. The lift coefficient is calculated for various values of the angle α.

Full size image

P N − 1 ( θ + 1 , − θ + 1 ) ( t k ), P N + 1 ( θ , − θ ) ( t k ) are the Jacobi polynomials and t k , k=1,…,N, are the roots of P N ( θ , − θ ) ( t k ). We considered ε=0.2.

Conclusions

We have calculated the velocity, the pressure and the magnetic induction for the steady flow of an incompressible perfectly conducting fluid past a thin airfoil. The integral representations of the velocity, pressure and magnetic induction contain respectively an elliptic part and a hyperbolic one. Some calculations were performed for the flat plate. In the graphic representation of the velocity and magnetic induction from Figures 1 and 2, we may observe the simple waves which are determined by the influence of the magnetic field. In Figure 3 we represent the lift coefficient against Alfvén’s number A. We notice that the lift coefficient increases when A decreases, i.e., when the value of the magnetic induction at infinity increases.

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  1. Department of Mathematics, University of Bucharest, Academiei 14, Bucharest, Romania

    • Adrian Carabineanu

  2. Institute of Mathematical Statistics and Applied Mathematics, Romanian Academy, Calea 13 Septembrie 13, Bucharest, Romania

    • Adrian Carabineanu

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Keywords

  • linearized system
  • fundamental matrix
  • integral representation