Incompressible Flow Over Airfoils

Transcription

1 Chapter 7 Incompressible Flow Over Airfoils Aerodynamics of wings: -D sectional characteristics of the airfoil; Finite wing characteristics (How to relate -D characteristics to 3-D characteristics) How to obtain -D characteristics? (a)experimental methods (NACA airfoils) (b)analytical methods (c)numerical methods Airfoil characteristics: () C l α () C d α (3) C m,c/4 α Characteristics of C l α curve: (a)α l= (b)a = dc l dα (c)c lmax 6

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3 7.. VORTEX FILAMENT At low to moderate angles of attack C l α curve is linear. The flow moves slowly over the airfoil and is attached over most of the surface. At high angles of attack, the flow tends to separate from the top surface. C l,max occurs prior to stall C l,max is dependent on R e = ρvc µ C m,c/4 is independent of R e except for large α C d is dependent on R e The linear portion of the C l α curve is independent of R e and can be predicted using analytical methods. Theoretical/analytical methods to evaluate -D characteristics: Recall from potential flow that a spinning cylinder produces lift L = ρv Γ. clockwise): ψ = Γ lnr For a -D Vortex (spinning φ = Γ θ v θ = Γ r v r = 7. Vortex Filament Consider -D/point vortices of same strength duplicated in every plane parallel to the z-x plane along the y-axis from to. The flow is -D and is irrotational everywhere except the y-axis. y-axis is the straight vortex filament and may be defined as a line. Definition: A vortex filament is a straight or curved line in a fluid which coincides with the axis of rotation of successive fluid elements. Helmholt s vortex theorems:. The strength of a vortex filament is constant along its length. Proof: A vortex filament induces a velocity field that is irrotational at every point excluding the filament. Enclose a vortex filament with a sheath from which a slit has been removed. The vorticity at every point on the surface=. Evaluate the circulation for the sheath. Circulation = C V d s = S ( V ) d A. Sheath is irrotational. Thus V =, C V d s = or V d s = b V d s + d V d s + a V d s + V d s = a b c d However, a V d s + V d s = b d Copyright c 5 6

4 7.. VORTEX FILAMENT as it constitutes the integral across the slit. Thus, b d V d s + V d s = a c b a d V d s = c V d s = d V d s = Γ. A vortex filament cannot end in a fluid; it must extend to the boundaries of the fluid or form a closed path. 3. In the absence of rotational external flow, a fluid that is irrotational remains irrotational. 4. In the absence of rotational extenal force, if the circulation around a path enclosing a definite group of particles is initially zero, it will remain zero. 5. In the absence of rotational extenal force, the circulation around a path that encloses a tagged group of elements is invariant. Copyright c 5 63

5 7.. VORTEX FILAMENT Vortex Sheet or vortex surface An infinite number of straight vortex filaments placed side by side form a vortex sheet. Each vortex filament has an infinitesimal strength γ(s). ds v θ v r r θ vortex sheet γ(s) is the strength of vortex sheet per unit length along s. v θ = Γ r for -D (point vortex). A small portion of the vortex sheet of strength γds induces an infinitesimally small velocity dv at a field point P (r, θ). So, v θ vortexfilament = γds r... dv P = γds r. Circulation Γ around a point vortex is equal to the strength of the vortex. Similarly, the circulation around the vortex sheet is the sum of the strengths of the elemental vortices. Therefore, the circulation Γ for a finite length from point a to point b on the vortex sheet is given by: Γ = b a γ(s)ds Across a vortex sheet, there is a discontinuous change in the tangential component of velocity and the normal component of velocity is preserved. z u w n w s u x Copyright c 5 64

6 7.. VORTEX FILAMENT As n, we get Γ = Γ = v d l Box v d l = [w n u s w n + u s] γ s = (u u ) s + (w w ) n γ s = (u u ) s or γ = (u u ) γ = (u u ) states that the local jump in tangential velocity across the vortex sheet is equal to the local sheet strength. Kutta Condition: For a given airfoil at a given angle of attack, the value of Γ around the airfoil is such that the flow leaves the trailing edge smoothly. The Kutta condition tells us how to find Γ; it is based on experimental observation. A body with finite angle TE in relative motion through a fluid will create about itself a circulation of sufficient strength to hold the rear stagnation point at the TE. If the TE has a zero angle, the Kutta Condition requires that the velocity of fluid leaving upper and lower surfaces at the TE be equal and non-zero. A body with a finite TE angle will have crossing streamlines at the TE unless the TE is a stagnation point. The Kutta Condition eliminates the crossing streamlines. Consider the TE as a vortex sheet: γ(t E) = V u V l If the TE has a finite angle V u = V l = because TE is a stagnation point. Or γ(t E) =. If the TE is a cusp (zero angle), V u = V l and hence γ(t E) = V u V l =. Thus, Kutta Condition expressed mathematically in terms of vortex strength is γ(t E) =. 7.. Bound Vortex and Starting Vortex The question might arise: Does a real airfoil flying in a real fluid give rise to a circulation about itself? The answer is yes. When a wing section with a sharp T.E is put into motion, the fluid has a tendency to go around the sharp T.E from the lower to the upper surface. As the airfoil moves along vortices are shed from the T.E which form a vortex sheet. Copyright c 5 65

7 7.. VORTEX FILAMENT Helmholtz s theorem: If Γ = originally in a flow it remains zero. Kelvin s theorem: Circulation around a closed curve formed by a set of continuous fluid elements remains constant as DΓ the fluid elements move through the flow, Dt =. Substantial derivative gives the time rate of change following a given fluid element. The circulation about the airfoil is replaced by a vortex of equal strength. This is the bound vortex since it remains bound to the airfoil. A true vortex remains attaced to the same fluid particles and moves with the general flow. Thus, as far as resultant forces are concerned, a bound vortex of proper strength in a uniform flow is equivalent to a body with circulation in a uniform flow. Both the theorems are satisfied by the starting vortex and bound vortex system. In the beginning, Γ = when the flow is started within the contour C. When the flow over the airfoil is developed, Γ within C is still zero which includes the starting vortex Γ 3 and the bound vortex Γ 4 which are equal and opposite. Copyright c 5 66

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9 7.. FUNDAMENTAL EQUATION OF THIN AIRFOIL THEORY 7. Fundamental Equation of Thin Airfoil Theory Principle: Mean camber line is a streamline of the flow. Velocity induced by a -D vortex is V = v θ ê θ = Γ r êθ where Γ is the strength of the -D vortex. Similarly the velocity induced by the vortex sheet of infinitesimal length ds is given by d v P = γ(s)ds r êθ To force the mean camber line to be a streamline, the sum of all velocity components normal to the mcl must be equal to zero. Consider the flow induced by an elemental vortex sheet ds at a point P on the vortex sheet. It is perpendicular to the line connecting the center of ds to the point P given by d v P = γ(s)ds r êθ Thus dw P the velocity normal to the mcl is: dw P γ(s) cos βds = dv P cos β = r where β is the angle made by dv p to the normal at P, and r is the distance from the center of ds to the point P. The induced velocity due to the vortex sheet representing the entire mcl is given by; w P (s) = T E γ(s) cos β ds r Now determine the component of the freestream velocity normal to the mcl. Copyright c 5 68

10 7.3. FLAT PLATE AT AN ANG OF ATTACK V,n = V sin(α + ɛ) where α is the angle of attack and ɛ is the angle made by the tangent at point P to the x-axis. The slope of the tangent line at point P is given by: or dz = tan( ɛ) = tan ɛ dx ɛ = tan ( dz dx ) V,n = V (sin α + tan ( dz dx )) In order that the mcl is a streamline. w P (s) + V,n =, or T E γ(s) cos β ds + V (sin α + tan ( dz r dx )) = within thin airfoil theory approximation s x, ds dx, cos β = and r (x x), where x varies from to c, and x refers to the point P. After changing these variables and making the small angle approximation for sin and tan, and upon rearrangement we get: c γ(x) x x dx = V (α dz dx ) 7.3 Flat Plate at an Angle of Attack The following analysis is an exact solution to the flat plate or an approximate solution to the symmetric airfoil. The mean camber line becomes the chord and hence: dz dx = c γ(x) (x x) dx = V α In order to facilitate analytic solution, we do a variable transformation such that: x = c ( cos θ) x = c ( cos θ ) θ = at and θ = at TE and θ increases in CW, dx = c sin θdθ γ(θ) c sin θ c [( cos θ ) ( cos θ)] dθ = V α Copyright c 5 69

11 7.4. -D LIFT COEFFICIENT FOR A THIN/SYMMETRICAL AIRFOIL Here we simply state a rigorous solution for γ(θ) as: γ(θ) sin θ (cos θ cos θ ) dθ = V α γ(θ) = αv + cosθ sin θ We can verify this solution by substitution as follows: γ(θ) sin θ (cos θ cos θ ) dθ = V α We now use the following result to evaluate the above integral. V α Thus, it satisfies the equation: ( + cos θ) (cos θ cos θ ) dθ = V α (cos nθ) (cos θ cos θ ) dθ = sin nθ sin θ ( + cos θ) (cos θ cos θ ) dθ dθ + cos θ cos θ cos θ dθ (7.) cos θ cos θ = V α ( + ) = V α (7.) γ(θ) sin θ (cos θ cos θ ) dθ = V α In addition, the solution for γ also satisfies the Kutta condition. When θ =, By using L Hospital rule, we get Thus it satifies the Kutta condition. γ() = V α γ() = V α sin cos = 7.4 -D lift coefficient for a thin/symmetrical airfoil Where s is along the mcl. By using thin airfoil approximation: L L ρv = ρ V Γ = ρv T E T E γ(x)dx = ρv γ(s)ds γ(x)dx Copyright c 5 7

12 7.4. -D LIFT COEFFICIENT FOR A THIN/SYMMETRICAL AIRFOIL Using the transformation x = c ( cosθ) Substituting the solution: L ρ V γ(x)dx = ρ V c γ(θ) sin θdθ L γ(θ) = αv ( + cos θ) sin θ ρ V c L αρ V c L V α( + cos θ)dθ cαρ V ( + cos θ)dθ dc l dα L α( ρ V )(c ) L αq S or C l = L q = α, and dc l S dα = = shows that lift curve is linearly proportional to the angle of attack Calculation of Moment Coefficient M = = = = ρ V = ρ V x(dl ) (7.3) x(ρ V dγ) (7.4) x(ρ V γ(x)dx) (7.5) = αρ V c 4 = αρ V c ( = q c α ) γ(x)xdx (7.6) αv ( + cos θ) sin θ ( ) c c ( cos θ) sin θdθ (7.7) ( cos θ)dθ (7.8) (7.9) (7.) Copyright c 5 7

13 7.5. THIN AIRFOIL THEORY FOR CAMBERED AIRFOIL C m, = M q Sc = M q c = α C l = α C m, = C l 4 M = M c/4 L c/4 C m, = C m,c/4 C l /4 C m, = C l /4 C m,c/4 is equal to zero for all values of α. C m,c/4 = c/4 is the aerodynamic center. Aerodynamic center is that point on an airfoil where moments are independant of angle of attack. 7.5 Thin Airfoil Theory for Cambered Airfoil γ(x)dx x x = V (α dz dx ) where dz dx is the slope of mcl at x. For symmetric airfoil, mcl is a straight line and hence dz dx = everywhere. On the other hand, for a cambered airfoil dz dx varies from point to point. As before, we do a variable transformation given by: Equation (A) becomes: x = c ( cos θ) dx = c sin θdθ (A) c γ(θ) sin θdθ (cos θ cos θ ) = V (α dz dx ) (B) has a solution of the form: Sub. solution in equation (B) γ(θ) = V [A ( + cos θ sin θ ) + A n sin nθ] n= A ( + cos θ) (cos θ cos θ ) dθ + n= A n sin nθ sin θ dz dθ = (α ) (cos θ cos θ ) dx x Using the integral cos nθ (cos θ cos θ ) dθ = sin nθ sin θ Copyright c 5 7

14 7.5. THIN AIRFOIL THEORY FOR CAMBERED AIRFOIL The first term becomes: Using the integral The second term becomes: = = A A ( + cos θ) (cos θ cos θ ) dθ n= A (cos θ cos θ ) dθ + A cos θ (cos θ cos θ ) dθ sin nθ sin θ (cos θ cos θ ) dθ = cos nθ A n sin nθ sin θ (cos θ cos θ ) dθ = A n cos nθ n= Therefore Equation (B) becomes: A n= A n cos nθ = α dz dx x Upon rearrangement the slope at a point P on the mcl is given by: dz dx = (α A ) + A n cos nθ n= From Fourier serious Where, f(θ) = B + B = B n = n =,,..., B n cos nθ n= f(θ)dθ (α A ) = B = A n = B n f(θ) cos nθdθ ( dz dx )dθ Evaluation of Γ From thin airfoil theroy Γ = γ(x)dx = c γ(θ) sin θdθ γ(θ) = V [ A ( + cos θ) + A n sin nθ] sin θ n= Γ = c γ(θ) sin θdθ = c V A ( + cos θ)dθ + c = cv [A (θ + sin θ) + A n sin nθ sin θdθ] n= V n= A n sin nθ sin θdθ Copyright c 5 73

15 7.5. THIN AIRFOIL THEORY FOR CAMBERED AIRFOIL Using A n sin nθ sin θdθ n= = for n = = for n Γ = cv [A + A ] L = ρ V Γ = ρv c[a + A ] C l = L q s = L q c = [A + A ] C l is normalized by the α as seen by the chord connecting the and TE of the mcl. c is the chord connecting the and TE of the mcl. C l = A + A ( = α [ = α + as is the case for symmetric airfoil. Also, Determination of moment coefficient ( ) dz dx ( dz dx ) ( dθ + ) (cos θ )dθ dc l dα = C l = dc l dα (α α L=) = (α α L= )... α L= = ] (cos θ ) dz dx dθ M = ρ V xγ(x)dx C m, = M q sc = V c As before we do a variable transformation from x to θ. Thus, x = c ( cos θ) xγ(x)dx ( ) ) dz cos θdθ dx dx = c sin θdθ γ(θ) = V [ A ( + cos θ) + A n sin nθ] sin θ n= C m, = A ( cos θ)dθ A n ( cos θ) sin θ sin nθdθ n= Copyright c 5 74

16 7.5. THIN AIRFOIL THEORY FOR CAMBERED AIRFOIL Using the following definite integrals: cos θdθ = sin θdθ = sin θ sin nθdθ = for n = = for n =,..., sin θ cos θ sin nθdθ = for n = = 4 for n = = for n = 3,..., C m, = A dθ + A cos θdθ (A n sin θ sin nθdθ A n n= = A + A A + A 4 = A A + A 4 C l = A + A C m, = [ ] A + A A ) sin θ cos θ sin nθdθ [ Cl = 4 + ] 4 (A A ) M = M c 4 L c 4 C m, = c m, c 4 C l 4 = C m, c 4 = 4 (A A ) C m, = x cp C l c [ Cl 4 + ] 4 (A A ) x cp = [ c 4 + c (A A )] = c [c + (A A )] 4C l 4 C l Relationship between pressure on mcl and γ () () (3) dl = (p l p u ) cos η(ds ) L = L = T E T E (p l p u ) cos ηds (A) ρv γ(s)ds Equating (A) and (B) and setting cos η =, we get T E (p l p u )ds = T E (B) ρv γ(s)ds Copyright c 5 75

17 7.5. THIN AIRFOIL THEORY FOR CAMBERED AIRFOIL or Using Bernoullis equation: (p l p u ) = ρv γ(s) () From vortex sheet theory: From (), () and (3) p l + ρ(u l + w ) = p u + ρ(u u + w ) p l p u = ρ (u u + u l )(u u u l ) () u u u l = γ(s) (3) V = u u + u l i.e., within thin airfoil approximation, the average of top and bottom surface velocities at any point on the mcl is equal to the freestream velocity. c p,l c p,u = (p l p ) q (p u p ) q = p l p u q = ρ(u u + u l )(u u u l ) q = γ(s)(v ) v = γ(s) V Copyright c 5 76