Unsteady Two-Dimensional Thin Airfoil Theory

23 Pages·2011·0.8 MB·English

Unsteady Two-Dimensional Thin Airfoil Theory
1 General Formulation
Consider a thin airfoil of infinite span and chord length c. The airfoil may have a small
motion about its mean position. Let the x axis be aligned with the airfoil mean position and
centered at its midchord O. The flow upstream of the airfoil may have small nonuniformities
but its mean velocity U is uniform and in the x-direction. The velocity field can be written
as
V(x,y,t) = U+u (x,y,t)+u(x,y,t), (1)
∞
where u (x,y,t) represents the velocity of the imposed upstream nonuniformities, and
∞
u(x,y,t)standsfortheperturbationvelocitygeneratedbythepresenceoftheairfoil. Theas-
sumptionofsmallairfoilmotionandsmallupstreamnonuniformitiesimplythat|u (x,y,t)|/U <<
∞
1 and |u(x,y,t)|/U << 1.
• Normalize length with respect to c/2
• Normalize velocity with respect to the freestream U
• Determine a velocity field (cid:126)u(x,y,t) such that
∇×(cid:126)u = ∇·(cid:126)u = 0 (2)
• Boundary Conditions: (cid:126)u = {u,v}
1. y = 0, −1 < x < +1:
(a) v = specified
(b) v = v
+ −
2. y = 0, x > +1:
(a) pressure is continuous, p = p
+ −
(b) normal velocity is continuous, v = v
+ −
1
3. |(cid:126)u| → 0 as (x2 +y2)2 → ∞
From the Euler Equations,
(cid:195) (cid:33)
∂ ∂ 1∂p
±
+ u = − . (3)
±
∂t ∂x ρ ∂x
Using condition 2,
(cid:195) (cid:33)
∂ ∂
+ ∆u = 0, (4)
∂t ∂x
where ∆u = u −u . Therefore,
+ −
∆u = f(x−t), (5)
and (∆u) = f(1−t), is the velocity jump at the trailing.
T.E.
1
1.1 Formulation in terms of sectionally analytic functions
(cid:90)
1 ∞ ∆u−i∆v
u−iv = dx(cid:48) (6)
2πi x(cid:48) −z
−1
or
(cid:90) (cid:90)
1 1 ∆u 1 ∞ f(x−t)
u−iv = dx(cid:48) + dx(cid:48) (7)
2πi x(cid:48) −z 2πi x(cid:48) −z
−1 1
As z → x±0, −1 < x < +1,
(cid:90) (cid:90)
1 1 1 ∆u 1 ∞ f(x(cid:48) −t)
(u−iv) = ± ∆u+ c dx(cid:48) + dx(cid:48) (8)
± 2 2πi x(cid:48) −x 2πi x(cid:48) −x
−1 1
Taking the imaginary parts of both sides,
(cid:90) (cid:90)
1 1 ∆u 1 ∞ f(x(cid:48) −t)
v = c dx(cid:48) + dx(cid:48) (9)
± 2π x(cid:48) −x 2π x(cid:48) −x
−1 1
The real parts give u = ±1∆u, hence, u = −u = 1∆u.
± 2 + − 2
Ifwespecifythedependenceontime f(x−t)willbeknown. Therefore, wehavethefollowing
integral equation for the unknown ∆u:
(cid:90) (cid:90)
1 1 ∆u 1 ∞ f(x(cid:48) −t)
c dx(cid:48) = 2v − dx(cid:48) (10)
π x(cid:48) −x π x(cid:48) −x
−1 1
The solution to this singular integral equation is not unique. However, if we apply the Kutta
condition, i.e., the solution must be finite at the trailing edge, we get
(cid:115)  (cid:113) (cid:195) (cid:33) 
1 1−x (cid:90) 1 1+x(cid:48) 1 (cid:90) ∞ f(x(cid:48)(cid:48) −t)
∆u = − c 1−x(cid:48) 2v − dx(cid:48)(cid:48) dx(cid:48). (11)
π 1+x x(cid:48) −x π x(cid:48)(cid:48) −x(cid:48)
−1 1
The double integral can be reduced to a single integral, and we have
(cid:115)  (cid:115) (cid:113) 
1 1−x (cid:90) 1 1+x(cid:48) 2v (cid:90) ∞ x(cid:48)(cid:48)+1f(x(cid:48)(cid:48) −t)
∆u = − c dx(cid:48) − x(cid:48)(cid:48)−1 dx(cid:48)(cid:48). (12)
π 1+x 1−x(cid:48)x(cid:48) −x x(cid:48)(cid:48) −x
−1 1
The pressure jump, ∆p = p −p , is given by integrating Equation 3.
+ −
(cid:90)
x
∆p = −(∆u−∆u )− ∆u˙ dx(cid:48) (13)
T.E.
1
The lift L(cid:48) is given by
(cid:90) (cid:90) (cid:90) (cid:90)
+1 +1 +1 x
L(cid:48) = − ∆pdx = ∆udx−2(∆u) + ∆u˙ dx(cid:48)dx. (14)
T.E.
−1 −1 −1 1
Integrating the last integral by parts gives
(cid:90) (cid:90)
+1 +1
L(cid:48) = ∆udx−2(∆u) − (1+x)∆u˙ dx. (15)
T.E.
−1 −1
2
Evaluation of the integrals is done by substitution of Equation 12 into Equation 15.
(cid:115) (cid:115) 
(cid:90) (cid:90) (cid:90)
+1 +1 1+x(cid:48) ∞ x(cid:48)(cid:48) +1
∆udx = −2 v(x(cid:48),t)dx(cid:48) + f(x(cid:48)(cid:48) −t) −1dx(cid:48)(cid:48) (16)
1−x(cid:48) x(cid:48)(cid:48) −1
−1 −1 1
(cid:115)  (cid:115) 
(cid:90) (cid:90) (cid:90)
+1 +1 1+x(cid:48) ∞ x(cid:48)(cid:48) +1
(1+x)∆u˙dx = −2 x(cid:48) v˙dx(cid:48)+ f˙(x(cid:48)(cid:48)−t)x(cid:48)(cid:48) −(x(cid:48)(cid:48) +1)dx(cid:48)(cid:48) (17)
1−x(cid:48) x(cid:48)(cid:48) −1
−1 −1 1
1.2 Application of Kelvin’s Theorem
Kelvin’stheoremstatesthatthecirculationaroundacircuitmovingwiththefluidisconstant
for a barotropic fluid with no viscosity and body forces.
(cid:90) (cid:90) (cid:90)
∞ +1 ∞
0 = ∆udx = ∆udx+ f(x(cid:48)(cid:48) −t)dx(cid:48)(cid:48) (18)
−1 −1 1
Substituting ∆u from Equation 12 into (18) and using (16)
(cid:115) (cid:115)
(cid:90) (cid:90)
+1 1+x(cid:48) ∞ x(cid:48)(cid:48) +1
−2 vdx(cid:48) + f(x(cid:48)(cid:48) −t) dx(cid:48)(cid:48) = 0. (19)
1−x(cid:48) x(cid:48)(cid:48) −1
−1 1
This equation will determine the intensity of the vortex shedding in the wake.
1.3 The Lift Formula
Substituting (16 and 17) into (15), gives
(cid:115)
(cid:90)
+1 1+x(cid:48)
L(cid:48) = −2 vdx
1−x(cid:48)
−1
(cid:115)
(cid:90)
+1 1+x(cid:48)
+2 x(cid:48) v˙ dx(cid:48)
1−x(cid:48)
−1
 (cid:115)  
(cid:90)∞ x(cid:48)(cid:48) +1 
+ f(x(cid:48)(cid:48) −t) −1dx(cid:48)(cid:48) −2(∆u)
 x(cid:48)(cid:48) −1 T.E.
1
 (cid:115) 
(cid:90)∞
x(cid:48)(cid:48) +1
− f˙(x(cid:48)(cid:48) −t)x(cid:48)(cid:48) −(x(cid:48)(cid:48) +1)dx(cid:48)(cid:48) (20)
x(cid:48)(cid:48) −1
1
Noting that
(cid:90) (cid:90)
∞ ∞
(x(cid:48)(cid:48) +1)f˙(x(cid:48)(cid:48) −t)dx(cid:48)(cid:48) = f(x(cid:48)(cid:48) −t)dx(cid:48)(cid:48) −2f(1−t), (21)
1 1
and using ?? to show that
(cid:115) (cid:115) (cid:115)
(cid:90) (cid:90) (cid:90) (cid:90)
∞ x(cid:48)(cid:48) +1 ∞ x(cid:48)(cid:48) +1 ∞ f 1 1+x(cid:48)
fdx(cid:48)(cid:48) − x(cid:48)(cid:48) f˙dx(cid:48)(cid:48) = √ dx(cid:48)(cid:48) −2 dx(cid:48), (22)
1 x(cid:48)(cid:48) −1 1 x(cid:48)(cid:48) −1 1 x(cid:48)(cid:48)2 −1 1 1−x(cid:48)
3
gives the following expression for the lift,
(cid:115)
(cid:90) +1 1+x(cid:48) (cid:90) +1√ (cid:90) ∞ f
L(cid:48) = −2 vdx(cid:48) −2 1−x(cid:48)2v˙ dx(cid:48) + √ dx(cid:48)(cid:48). (23)
−1 1−x(cid:48) −1 1 x(cid:48)(cid:48)2 −1
The first term,
(cid:115)
(cid:90)
+1 1+x(cid:48)
L(cid:48) = −2 vdx(cid:48)1 (25)
q.s. 1−x(cid:48)
−1
represents the quasi-steady lift. It is the lift obtained if the velocity, v, is time-independent.
For example, if the airfoil is at angle of attack α to the upstream flow, then v = −α and
L(cid:48)q.s. = 2πα, which is the classical result for the lift coefficient of a thin airfoil. This is also
the case when the time variation of v is small, then v˙ ≈ 0 and the vortex shedding in the
wake is also negligible, f(x−t) ≈ 0.
The second term,
(cid:90) +1√
L(cid:48) = −2 1−x(cid:48)2v˙ dx(cid:48)2,. (27)
a.m.
−1
is proportional to the acceleration of the velocity v˙ and therefore represents the apparent
mass lift. This force dominates the response of the airfoil when v˙ depends strongly on time.
The last term,
(cid:90)
∞ f
L(cid:48) = √ dx(cid:48)(cid:48), (28)
w 1 x(cid:48)(cid:48)2 −1
1In dimensional form, we get,
(cid:90) (cid:115) (cid:90) (cid:114)
+c/2 c/2+x(cid:48) c x(cid:48)
L(cid:48) = −2ρ U vdx(cid:48) =−2ρ U vdx(cid:48). (25)
q.s. 0 c/2−x(cid:48) 0 c−x(cid:48)
−c/2 0
Note that
(cid:115)
(cid:90)
+c/2 c/2+x(cid:48) c
dx(cid:48) = π ,
c/2−x(cid:48) 2
−c/2
(cid:90) (cid:114)
c x(cid:48) c
dx(cid:48) = π .
c−x(cid:48) 2
0
2In dimensional form,
(cid:90) (cid:90)
c +1(cid:112) c(cid:112)
L(cid:48) = −2ρ ( )2 1−x(cid:48)2v˙dx(cid:48) =−2ρ x(cid:48)(c−x(cid:48))v˙dx(cid:48). (27)
a.m. 0 2 0
−1 0
Note that
(cid:90)
+1(cid:112) π
1−x(cid:48)2dx(cid:48) = ,
2
(cid:90) −1
c(cid:112) π
x(cid:48)(c−x(cid:48))dx(cid:48) = c2.
8
0
4
represents the wake shedding contribution to the lift. Thus the lift can be written as the
sum of three terms representing respectively, the quasi-steady lift, the apparent mass lift,
and the wake induced lift,
L(cid:48) = L(cid:48) +L(cid:48) +L(cid:48) . (29)
q.s. a.m. w
This result was first derived by von Karmen and Sears in 1938.
1.4 The Moment Formula
An elementary force dF produces a moment with respect to the origin O, dM/O = xdF.
The expression for the total moment is
(cid:90)
+1
M/O = − x∆pdx. (30)
−1
Substituting ∆p from 13 gives
(cid:90) (cid:90) (cid:90)
+1 x
M/O = x∆udx+ −1+1x ∆u˙dx(cid:48)dx. (31)
−1 1
Integrating the last term by parts,
(cid:90) (cid:90)
+1 1 +1
M/O = x∆udx+ (1−x2)∆u˙ dx. (32)
2
−1 −1
After considerable reduction, we obtain
(cid:90) +1√ (cid:90) +1 √ 1 (cid:90) ∞ f
M/O = 2 1−x2vdx− x 1−x2v˙ dx− √ dx. (33)
−1 −1 2 1 x2 −1
As for the lift formula, the moment is the sum of three terms representing the quasi-steady
moment, the apparent mass moment and the wake induced moment.
1.5 Harmonic Time Dependence
Equations (23 and 58)determine the lift and moment of an airfoil in terms of the velocity v
and the distribution of the vortex shedding in the wake f(x−t). The velocity v is specified
but f(x−t) must be determined from the time history of the motion. In many instances the
airfoil is undergoing periodic oscillations or subject to periodic gust excitations. For these
cases the velocity v is a periodic function of time and it is possible to consider the airfoil
response to a single harmonic component v = v˜e−iωt. In the general case it is also convenient
to consider the Fourier transform
(cid:90)
+∞
v(x,t) = v˜(x,ω)e−iωtdω, (34)
−∞
and to calculate the response to every Fourier component v˜e−iωt. This approach has the
advantage of immediately determining the expression for the wake vortex distribution
f(x−t) = c ei(ω(x−t)), (35)
o
5
where c is constant which can be readily determined from the Kelvin theorem condition by
o
substituting (35) into (19),
(cid:82) (cid:113)
2 +1 1+x vdx
coe−iωt = (cid:82) −(cid:113)1 1−x . (36)
∞ x+1 eiωxdx
1 x−1
The numerator in (36) is −L(cid:48) and the integral in the denominator can be expressed in
q.s.
terms of Hankel functions, giving
−L(cid:48)
c e−iωt = q.s. , (37)
o iπH (ω)
2 +
where H (ω) = H(1)(ω) + iH(1)(ω). The wake-induced lift, L(cid:48) , can also be evaluated by
+ 0 1 w
substituting (35) into (28),
H(1)(ω)
L(cid:48) = −L 0 . (38)
w q.s. H (ω)
+
Substituting (38) into (23) gives
L(cid:48) = L(cid:48) C(ω)+L(cid:48) , (39)
q.s. a.m.
where
iH(1)(ω)
C(ω) = 1 (40)
H (ω)
+
is the complex conjugate of the Theodorsen function3. It can be readily shown that as
ω → 0,C(ω) → 1, and as ω → ∞,C(ω) → 0.5.
Similarly, the expression for the moment becomes
(cid:90) +1√ (cid:90) +1 √ 1
M/O = 2 1−x2vdx− x 1−x2v˙ dx+ L(cid:48) (1−C(ω)) (41)
2 q.s.
−1 −1
In what follows we apply the general formulas derived for the lift and moment to special
cases. We first consider the cases of translatory and rotational oscillations as models for
bending and torsional oscillations. We then study the case for a harmonic transverse gust
impinging on the airfoil. Closed form analytical expressions for the lift and moments will be
derived for these fundamental cases. The results will also be used to study the fundamental
problems of a step change in the angle of attack and that of a sharp edged gust.
1.5.1 Translatory Oscillations
For a translatory oscillation the airfoil normal velocity v is independent of x and has the
form v = ve−iωt, where v, the magnitude of the oscillation, is constant. Using this expression
for v in the various lift and moment formulas gives
3Theodorsen considered a time dependence of the form exp(iωt).
6
Figure 1: Vector diagram of the real and imaginary parts of the Theodorsen function as the
reduced frequency increases from 0 to ∞.
L(cid:48) = −πρcUv, (42)
q.s.
π
L(cid:48) = − ρc2v˙, (43)
a.m. 4
L(cid:48) = −L(cid:48) (1−C(ω)), (44)
w q.s.
π
L(cid:48) = L(cid:48) C(ω)− ρc2v˙ (45)
q.s. 4
c
M /O = − L(cid:48) , (46)
q.s. 4 q.s
M /O = 0, (47)
a.m.
c
M /O = − L(cid:48) , (48)
w 4 w
c
M/O = − L(cid:48) C(ω) (49)
4 q.s.
These results show that the center of pressure for both the the quasi-steady lift and the
wake-induced lift is located at the quarter chord, and that the center of pressure for the
apparent mass lift is located at the origin.
7
1.5.2 Rotational Oscillations
Foranairfoilinrotationaloscillation, theairfoilsurfaceis y−αx = 0, whereα = αexp(−iωt)
istheairfoilanglewiththex-axis, andα,theangularmagnitudeoftheoscillation, isconstant.
The expression for the normal velocity is obtained by taking D /Dt(y −α)x = 0, yielding
0
v = α˙x+αU. It is seen that the velocity is the sum of two terms. The second term, αU,
is independent of x and represents a translatory oscillation whose lift and moment are given
by (42 - 49). Thus, we need only to calculate the expressions for the lift and moment in
response to the first term α˙x by substituting this expression for v in the various lift and
moment formulas. This gives
Figure 2: Vector diagram of the real and imaginary parts of the lift coefficient of an airfoil
in translatory oscillation.
π
L(cid:48) = − ρc2Uα˙, (50)
q.s. 4
L(cid:48) = 0, (51)
a.m.
L(cid:48) = −L(cid:48) (1−C(ω)), (52)
w q.s.
8
L(cid:48) = L(cid:48) C(ω) (53)
q.s.
M /O = 0, (54)
q.s.
π
M /O = − ρc4α¨, (55)
a.m.
128
c
M /O = − L(cid:48) , (56)
w 4 w
c π
M/O = − L(cid:48) (1−C(ω))− ρc4α¨. (57)
4 q.s. 128
Figure 3: Vector diagram of the real and imaginary parts of the lift coefficient of an airfoil
in rotational oscillation.
1.5.3 Translatory and Rotational Oscillations
For an airfoil in translatory and rotational oscillation hinged at the point O(cid:48), the airfoil
surface is y − α(x − x ) = 0, v = α˙(x − x ) + αU. This is equivalent to the sum of a
O(cid:48) O(cid:48)
translatory oscillation, αU − α˙x , whose lift and moment are given by (42 - 49), and a
O(cid:48)
rotaional oscillation about the origin whose lift and moment are given by (50 - 57). Note
that the expression for the moment with respect to O(cid:48) is
9
M/O(cid:48) = M/O−x L(cid:48). (58)
O(cid:48)
1.5.4 Step Change in Angle of Attack
This problem examines the time evolution of the lift of an airfoil as the angle of attack α is
suddenly increased from zero to α , i.e.,
o
(cid:189)
0, t < 0,
α =
α , t > 0.
o
This can be written in terms of the unit step function
y +u(t)α x = 0, (59)
o
where
(cid:189)
0, t < 0,
u(t) =
1, t > 0.
The normal velocity is then given by
v = −u(t)α U −δ(t)α x, (60)
o o
where δ(t) is the unit impulse or Dirac function and u˙(t) = δ(t). Define the unit step and
Dirac functions as
(cid:90)
1 1 +∞ eiωt
u(t) = + c dω, (61)
2 2πi ω
−∞
(cid:90)
1 +∞
δ(t) = eiωtdω. (62)
2π
−∞
or,
(cid:90)
1 1 +∞ sinωt
u(t) = + dω, (63)
2 π ω
0
(cid:90)
1 +∞
δ(t) = cosωtdω. (64)
π
0
As in previous results, we calculate the lift in two stages. We first consider the lift produced
by the translatory part -u(t)α U. The constant term in (63) gives a steady lift of πρcU2α /2.
o o
The circulatory part of the lift produced by (α Ue−iωt)/(πω) is
o
ρcU2α C(ω)
L(cid:48) = − o e−iωt (65)
C (ω)
.
Hence, the lift produced by −(α Usinωt)/(πω) is the imaginary part of (65). The total
o
circulatory lift can be obtained by integration,
(cid:34) (cid:35)
(cid:179) (cid:180) 1 (cid:90) ∞ Im([e−iωtC(ω)]
L(cid:48) = πρcU2α − dω. (66)
C o 2 ω
0
10