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wake state
Vortex
ring state
2
1
Experimental
induced velocity
Normal
working state
–4 –3 –2 –1 0 1 2
Windmill
brake state
v i
Rotor aerodynamics in axial flight 45
the flow upstream of the rotor is irrotational, there can be no swirl ahead of the rotor
disc. Behind the rotor disc there are two contributions to the swirl velocity: that due
to the bound circulation about the blades and that due to the spiral vortex lines
forming the slipstream, Fig. 2.9. Let the swirl angular velocity contributions be
denoted by ω b and ω t respectively. Ahead of the rotor we have ωb + ω t = 0 or
ω b = – ω t. On passing through the disc the contribution from the bound circulation
changes sign, while the contribution from the trailing vortices remains the same;
hence, behind the disc the total angular velocity is ω = –ωb + ω t = 2ω t. This value
remains constant in the slipstream because no extra circulation is added; we should
also expect the swirl velocity to remain constant from the fact that the bound vorticity
contribution diminishes as we go away from the disc but the trailing vortex contribution
steadily increases from its value at the disc to twice that value in the ultimate slipstream
since, for any point there, the vortex lines are doubly infinite.
To relate the swirl velocity to the thrust on the rotor disc, consider axes fixed in the
advancing and rotating rotor blade. With respect to these axes the flow is steady and,
since the flow is irrotational, except at the vortex lines leaving the blade, the constant
in Bernoulli’s equation must be the same everywhere. Let qz, qr, qψ be the velocity
components of the air relative to fixed axes situated in the rotor disc, and let the rotor
rotate with angular velocity Ω, Fig. 2.10. The velocity components relative to a given
point on the blade are qz, qr, qψ – Ωr. At a great distance ahead of the rotor, the
velocity components relative to the same point of the blade are qz = Vc, qr = 0,
qψ = – Ωr. Also, p = p∞. Bernoulli’s equation for the flow which applies everywhere, is
p V p q q q r z r ∞ + = + + + ( – ) 12
c
2 12
2 12
2 12
ρ ρ ρ ρ ψ Ω 2
Let p1 be the pressure just in front of the disc; since there is no swirl in front of the
disc qψ = 0, hence
p V p q q r z r ∞ + = + ( + + ) 12
c
2 12
ρ 1 ρ 2 2 2
Ω2 (2.19)
Just behind the disc the pressure will have jumped to p1 + Δp, the axial velocity
will be unchanged, and the radial velocity will have changed sign since we shall have
qr
qz
qψ
Ω
Fig. 2.9 Spiral vortices in axial flight Fig. 2.10 Velocity components relative to blade
r
46 Bramwell’s Helicopter Dynamics
passed through the vortex sheet leaving the blade. (For an infinite number of blades,
the radial velocity will be zero.) Thus, behind the disc we have
p V p p q q q q r + r z r ∞ + = + + ( + + – 2 ) 12
c
2 12
ρ 1 ρ 2 2 ψ2 ψ 2
Δ ΩΩ2 (2.20)
Subtracting eqn 2.19 from eqn 2.20 gives
Δ Ω p q r q = ( – ) 12
ρ ψ θ (2.21)
If we write qψ = ωr, eqn 2.21 can be expressed as
Δ Ω p r = ( – ) 12
ρω ω 2 (2.22)
The total head pressure just ahead of the disc and relative to axes fixed in the disc
is
H p q q z r = + ( + ) 12
2 2
1 ρ
Just behind the disc we have
H H p p q q q z r + = + + ( + + ) 12
Δ 1 Δ ρ 2 2 ψ2
giving
Δ Δ H p q = + 12
ρ ψ2
= + 12
Δp ρω 2r2 (2.23)
Thus, the change in total head pressure exceeds the jump in static pressure across
the disc by a term representing the kinetic energy of the swirl of the slipstream.
To get some idea of the swirl velocity, qψ = ωr, in a typical case, let us note that
eqn 2.22 can be expressed in terms of the disc loading as
Δp = wD = (Ω – r
12
ρω ω ) 2 (2.24)
and take the values wD = 250 N/m2, Ω = 25 rad/s and r = 6 m. For sea-level density
we find ω = 0.23 rad/s and qψ = 1.38 m/s. Since the induced velocity in hovering for
this disc loading has been found to be 10.2 m/s, the angle of flow relative to the rotor
axis is 7.8°. We also see that the second term in eqn 2.23 is only about 12
per cent
larger than Δp; this justifies the neglect of the swirl velocity in the earlier analysis.
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