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(unsteady aerofoil)
ψ = 90°
160°
180°
200°
210°
220°
270°
250°
240°
360°
330°
300°
CN
Fig. 6.48 Variation of pitching moment and normal force coefficients with azimuth angle
Rotor aerodynamics in forward flight 233
6.5 The boundary layer on a rotating blade
The first attempt at calculating the boundary layer on a rotating blade was made by
Fogarty39 in 1951 for the case of hovering flight. Fogarty made use of a theorem of
Sears40 which states that if φ1(x, z) is the potential for plane steady flow past the
cylinder in a parallel stream at unit speed, then the potential φ for flow about the
cylinder when rotating at angular velocity Ω is
φ = Ωy[φ1(x, z) – x] (6.31)
It was also shown that the velocity components u1, v1, w1, relative to the blade are,
Fig. 6.49,
u1 = Ωy ∂φ1/∂x
v1 = Ω[φ1(x, z) – 2x]
w1 = Ωy ∂φ1/∂z
Thus the velocity components u1 and w1 in the plane of the cylinder are the same
as those of steady plane flow about the cylinder in a stream of velocity Ωy. The
spanwise component v1, which is independent of the radial distance, can be found
directly from the plane flow potential φ1. These relationships establish the appropriate
boundary conditions outside the boundary layer.
The equations used are the familiar Navier–Stokes equations referred to co-ordinates
rotating with the blade, Fig. 6.49. If the usual boundary layer approximations are
made, and certain terms are neglected on account of the high aspect ratio of the blade,
we have for the boundary layer equations
u u
x
w u
z
x
p
x
u
z
+ – = – 1 +
2
2
∂
∂
∂
∂
∂
∂
∂
∂
Ω2
ρ ν
u
x
w
z
u y
p
y z
+ + 2 – 2 = – 1 +
2
2
∂
∂
∂
∂
∂
∂
∂∂
v v Ω Ω ρ ν v
∂p/∂z = 0
together with the continuity equation
∂u/∂x + ∂w/∂z = 0
To find the pressure gradients ∂p/∂x and ∂p/∂y we use Bernoulli’s equation for the
external potential flow, i.e.
Ω
z
w
u
y
v
0
x
Fig. 6.49 Co-ordinates of rotating blade
234 Bramwell’s Helicopter Dynamics
p u x y / + + ) = ( + ) + constant 12
1
2
1
2 12
ρ ( v Ω2 2 2
Then, neglecting the small terms v1∂v1/∂x and v1∂v1/∂y, we have approximately
1 = 2 –
ρ 1
∂
∂
∂
∂
p
x
x u
u
x
Ω 1
1 = 2 –
ρ 1
∂
∂
∂
∂
p
y
y u
u
y
Ω 1
so that, on substituting for the pressure gradients, the boundary layer equations are
u u
x
w u
z
u
u
x
u
z
+ = 1 +
∂
∂
∂
∂
∂
∂
∂
∂
1
2
2 ν (6.32)
u
x
w
z
u u
u
y z
+ + 2 = 1 +
∂
∂
∂
∂
∂
∂
∂∂
v v Ω 1 v
2
2 ν (6.33)
∂ u/∂ x + ∂w/∂ z = 0
together with the boundary conditions
u = v = w = 0 for z = 0
u → u1 = Ωy ∂φ1/∂x, v → v1 = Ω[φ1(x, z) – 2x]
w → w1 = Ωy ∂φ1/∂z, for z → ∞
We notice that eqn 6.32 is the same as the boundary layer equation for twodimensional
plane flow and may therefore be solved by any of the known methods.
Having solved this equation for u and w, we can obtain v from the solution of
eqn 6.33. This has been done by Fogarty for the case of a flat plate and the section
defined by z = kx(1 – x2). The results for the flat plate are shown in Fig. 6.50.
The importance of Fogarty’s results is that the spanwise flow, or ‘centrifugal pumping’,
is usually very small. As an example, let us take a point at a radial distance of 6 m
from the hub. Since the chord will be about 0.5 m at most, let us take x = 0.3 m.
5
4
3
2
1
0 0.2 0.4 0.6 0.8 1.0
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Bramwell’s Helicopter Dynamics(117)
