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of the rotor, reaching a value in the ultimate wake of Vc + v2, where the slipstream
radius is R2 and the pressure p2.
Since the slipstream velocity is higher than the undisturbed axial velocity Vc, it is
clear that the mass of fluid leaving the bottom end of the control volume exceeds that
entering at the top. There must therefore be some flow through the cylindrical sides
of the control surface. If this flux is denoted by Q , we have
Q = π(R – R )Vc + πR (Vc + 2 ) – πR Vc 1
2
2
2
2
2
1
v 2
= π R2 2
2v
Thus the total mass per unit time entering the control surface is
ρ πR1V ρπR
2
2
2
c + v2
and the total mass leaving the surface is
ρ π(R1 – R )Vc + ρπR(Vc + 2)
2
2
2
2
2 v
Since the flux entering the control surface consists of air having velocity Vc, the
momentum per unit time entering the surface is
ρ Vc(πRVc + πR 2) 1
2
2
2v
and the momentum per unit time leaving the surface is
ρ π (R1 R V ρπR V
2
2
2
c
2
2
2
c 2
– ) + ( + v )2
Hence, the rate of change of momentum in the axial direction is
ρ π (R1 R V ρπR V ρπRV ρπRV
2
2
2
c
2
2
2
c 2
2
1
2
c
2
2
2
– ) + ( + v) – – cv2
= ( + ) 2
2
ρπR Vc v2 v2
The total force in the axial direction acting on the control surface consists of the
rotor thrust plus the pressure forces on the ends of the cylinder. Equating this force
to the rate of change of momentum, we get
T + R p – (R – R)p – R p = R(V + ) 1
2
1
2
2
2
2
2
2 2
2
π ∞ π ∞ π ρπ c v2 v2
or T = R(V + ) + R p – p ) 2
2
c 2 2
2
ρπ v2 v π ( 2 ∞ (2.4)
Continuity of the flow requires that
ρ(Vc + vi)A = ρ(Vc + v2)πR2
2 (2.5)
Rotor aerodynamics in axial flight 39
so that eqn 2.4 can be written
T/A = Δp = ρ(Vc + vi)v2 + (p2 – p∞)(Vc + vi)/(Vc + v2) (2.6)
Applying Bernoulli’s equation to points upstream of the rotor gives
p V p V ∞ + = + ( + ) 12
c
2 12
c i
ρ ρ v 2 (2.7)
and for points downstream of the rotor
p p V p V + + ( + ) = + ( + ) 12
c i
2 12
c 2
Δ ρ v 2 ρ v2 (2.8)
Subtracting eqn 2.7 from eqn 2.8 gives
Δ p = p – p + (Vc + )
12
2 ∞ ρ v2 v2 (2.9)
and equating eqns 2.6 and 2.9 we have
ρ v2 vi v v v v
12
( – 2) = (p2 – p∞ )( 2 – i )/(Vc + 2 ) (2.10)
Let us assume, as for the classical actuator disc, that the pressure in the final wake
is the same as the ambient pressure, i.e. that p2 – p∞ = 0. Then, from eqn 2.10,
v i v
12
= 2
irrespective of the axial velocity Vc of the rotor. Thus the increment of velocity at the
rotor disc, which we usually refer to as the ‘induced’ velocity, is half the value in the
ultimate wake. Putting this relationship, and p2 = p∞, in eqn 2.6 we have
T = 2ρA(Vc + vi)vi (2.11)
from which the induced velocity may be calculated when the thrust is known. In
particular, in hovering flight, Vc = 0 and
vi = v0 = √(T/2ρA) (2.12)
in which v0 is termed the ‘thrust velocity’.
If wD is the disc loading, T/A, in N/m2, and ρ has the International Standard
Atmosphere (ISA) value corresponding to sea-level,
vi = v0 = 0.64√wD
A typical value of wD is 250 N/m2, giving an induced velocity (thrust velocity) of
10.2 m/s.
To calculate the power being supplied by the rotor, we must consider the rate at
which kinetic energy is being imparted to the air. The rate at which kinetic energy
enters the control surface is
12
c 2 c
(ρπR1V + ρπR V2
2
2
2v )
and the rate at which kinetic energy leaves the control surface is
12
c
3
[ρπ(R1 – R )V + ρπR (Vc + 2 ) ]
2
2
2
2
2 v 3
40 Bramwell’s Helicopter Dynamics
The power P delivered by the rotor is found to be
P = A(Vc + i )(Vc + ) + (p – p )A(V + )
12
ρ v v2 v2 2 ∞ c vi (2.13)
in which the first term on the right-hand side is total rate of change of kinetic energy,
the second term derives from rate of doing work by the pressures on the ends of the
　
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