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Math 575-Lecture 12 1 Failure of Ideal Fluid; Vanishing Viscosity

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Math 575-Lecture 12 1 Failure of Ideal Fluid; Vanishing Viscosity Math 575-Lecture 12 In this lecture, we investigate why the ideal fluid is not suitable sometimes; try to explain why the negative circulation appears in the airfoil and introduce the vortical wake to resolve the D'Alembert paradox. 1 Failure of ideal fluid; Vanishing viscosity 1.1 Drawbacks of ideal fluids We have seen that for any finite body in ideal harmonic flow, there is no drag on the body by the D'Alembert paradox. Suppose we have an ideal fluid. The airfoil begins to accelerate from velocity zero. Initially, there is no circulation around the body. Later on, there should be vorticity as well by Kelvin's theorem if the fluid is ideal. This means that the airfoil can not fly. This is, however, clearly not the case. Further, consider a flapping wing. In the ideal fluid, the force on the body is zero and there is no thrust. All these situations imply that the ideal fluid assumption is not good. This is the difference between the ideal fluid and `almost potential flow'. 1.2 vanishing viscosity To understand this, we will start with a viscous flow and send the viscosity to zero. We will have then the limit of vanishing viscosity. In the limit, the fluid is ideal in most regions and however, in some regions the fluid can not be regarded as ideal. Instead, there could be vorticity introduced. It is this feature that will help to resolve these issues. The viscous stress, as we shall see, is µ(ru + ruT ) for Newtonian fluid. If we send µ ! 0, in most regions where the fluid does not change rapidly, the viscous stress disappears. However, near the boundary of the solid, this is not the case. On the boundary, the fluid particle attaches on the surface of the body. There is a small transition region. The boundary condition u · n = 0 should be understood at the outer layer of the transition region. Inside this region, the velocity changes rapidly and its width is related to µ. This region is called the boundary layer. As µ ! 0, the region is shrinking and therefore the velocity gradient is becoming bigger. Hence, the viscous stress will not vanish near the boundary. Also, if there is a corner on the surface of the body, the velocity gradient is large there also. The significance is that there is vorticity inside this boundary layer. When the vorticity is confined in the layer, then, we have the bound vorticity. If the body is moving rapidly, then, then vorticity is left behind the trailing edge. This process is called the vortex shedding. 1 2 Vortex Shedding and the starting vortex Consider the motion of airfoil. When the foil moves, a start vortex is released into the fluid. This vortex is called the starting vortex. As a result, we will have a `roll-up' core of vortex. This core has positive circulation. By Kelvin's theorem, the circulation on a closed curve enclosing this start vortex and the body is zero. Hence, there must be a negative circulation developed around the body and consequently, a negative bound vorticity is in the body (to be precise, inside the bounday layer). After the foil reaches its steady velocity and the circulation is the value needed to satisfy the Kutta-Joukowski condition, the release of the vortex is stopped. The lift is then generated. (P159 of A-book) 3 Vortex wake and drag Let a viscous fluid passing through a circular cylinder. If the velocity is big enough, there is a turbulent region behind the cylinder. This tubulent region will not vanish if we send the viscosity to zero µ ! 0. The body is shedding un-steady vorticity into this region, called the vortical wake. The vorticity released from the trailing is alternating between a positive and negative value. The energy is dissipated into the wake and hence the drag appears. This then resolves the D'Alembert paradox. The vortical wake happens if the body is doing unsteady motion such as in the flapping wing case. The birds need to flap to generate thrust and overcome this drag. This vortical wake can usually be modeled as some curves merging from the body. Outside these curves, the fluid can be regarded as ideal. 2 3.1 Drag in 2D ideal fluid For the 2D ideal fluid, people usually use the so-called free streamlines to model the vortical wake. The vortices are then concentrated on these curves. Inside the region between these free streamlines, the fluid is dead. The velocity is discontinuous across these two curves but pressure is continuous. The model then resolves the D'Alembert paradox in 2D. For more discussions, see Section 5.4 in Childress. 3.2 3D wing: Combining Kutta-Joukowski lift theory and the vortical wake Consider a relatively real airfoil. In the middle part of the airfoil, the flow is close to the 2D ideal flow and there is negative circulation. The circulation in the middle causes lift by Kutta-Joukowski lift theorem. This negative circulation corresponds to a vortex tube. The airfoil actually falls into a vortex tube. By Kelvin's theorem, the vortex tube can not end in the fluid. Hence, the vortex tube then turn and extends out from the trailing edge of the wing. As a result, the vortex is shed into the fluid and inside this tube, a vortical wake is formed. It is this vortical wake that causes the drag on the wing. For more discussion, see Section 5.5 in Childress, the lifting line theory. 4 Karman Vortex Street In this section, we look at a model for the vortical wake in 2D, called the Karman vortex street. 3 The wake is modeled by an array of point vortices, located at b a b (na; ); (na + ; − ): 2 2 2 b b The point vortices at y = 2 have negative strength −γ while the vortices at y = − 2 have positive strength. If we observe the street far away from the body, the field can be viewed as a doubly infinite array. We know the complex potential introduced by a point vortex with strength Γ is given iΓ by − 2π log(z − z0). Consider the lower array: the complex potential is N iγ X a ib fl(z) = − lim log(z − − na + ) = 2π N!1 2 2 n=−N N N iγ a ib Y a ib iγ a ib Y 1 a ib − lim log(z− + ) ((z− + )2−n2a2) = − lim log(z− + ) (1− (z− + )2) 2π N!1 2 2 2 2 2π N!1 2 2 n2a2 2 2 n=1 n=1 Note that log(−1) = iπ and log(n2a2) are just some constants which do not have physical sense. Using the series expansion z2 z2 sin(z) = z(1 − )(1 − ) ::: π2 22π2 we find iγ π a ib f = − log sin( (z − + )) l 2π a 2 2 Similarly, we find that the complex potential generated by the upper array is iγ π ib f = log sin( (z − )) u 2π a 2 The vortex wake is generated by a uniform flow and hence the total complex potential is ! iγ sin( π (z − ib ) f(z) = Uz + f + f = Uz + log a 2 l r 2π π a ib sin( a (z − 2 + 2 ) 4.1 velocity of the point vortex As we have seen, the velocity of the point vortex is given by the fields not including the b one contributed by itself. Hence, the velocity for the point vortex at (x; y) = (0; 2 ) could 4 be computed as: df(z) iγ 1 γ πa W = lim [ − ] = U − tanh( ): (4.1) ib dz 2π ib 2a b z! 2 z − 2 If the vortex is shed with a time frequency f, then the distance is a = W=f = (U − V )=f: Usually, V U and hence, the a ≈ U=f. For a cylinder with diameter D, one has approximately f ≈ 0:2U=D, and thus a ≈ 5D. 4.2 Drag induced by the Karman vortex street As the body sheds point vortices, the vortex on one side generates a pressure field that induces a force on the body, denoted by D1. This force can be computed by the Blasius theorem. On the other side, the vortices carry momentum into the fluid. By the conservation of momentum, there is an averaged drag D2 on the body. Hence, the total drag is given by FD = D1 + D2: To compute D2, we compute the momentum carried by the two adjacent point vortices. The complex velocity field is given by iγ iγ w(z) = − a 2π(x + iy − ib=2) 2π(x + iy − 2 + ib=2) The complex momentum is therefore Z 1 Z 1 m1 + im2 = ρ dy w(z)dx −∞ −∞ For the x integral, 1 x + iy − ib=2 1 iγ log a j−∞ 2π x + iy − 2 + ib=2 If jyj < b=2, the integral is 2πi while jyj > b=2, the integral is zero. Hence, m = −ργb Since such two adjacent vortices correspond to a period of the shedding, and the period is given by T = a=(U − V ). Then, on average, the force is equal to momentum shed per unit time or m U − V D = − = ρbγ 2 T a 5 where the negative sign appears because we compute the loss of momentum of the body. To compute D1, the idea is to apply the theorem of Balsius and we have I iρ df 2 D1 − iFy = ( ) dz: 2 Cb dz However, this is not easy to evaluate. Instead, one is going to pick a big rectangle contour such that one edge AB is in the middle of the street so that the field it sees can be approx- imately by the doubly infinite array.
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