The helicopter rotor can be idealised as a “momentum disk”. It imparts a uniform velocity (vi) to the airflow creating a change in momentum which will result in an upward thrust (T).
HOVER
For the hover case air is sucked in from all directions so the far-field inlet velocity is effectively zero, accelerated to vi at the disk and then propelled to a final slipstream velocity of vs below. If the area of the disk is A then the mass flow of air being accelerated over a small time step is Δm/Δt = ρ A vi
The change in momentum of the stream will be ΔM = Δm/Δt. vs
and this will be equal to the force produced, the thrust (T).T = ρ A vi . vs
The change in Energy per unit time of the stream will be ΔE/Δt = 1/2 Δm/Δt vs2
The work done per unit time on the air by the thrust isΔW/Δt = T . Δs/Δt. = T.vi
hence ΔW/Δt = T.vi = ρ A vi2.vs = ΔE/Δt =1/2 ρ A vi vs2
A vi2.vs =1/2 A vi vs2 giving vs = 2.vi
Hence thrust of the ideal rotor in hover is
T = 2 ρ A vi2
As the thrust is required to be equal the weight (W) of the vehicle, it is possible to use this equation to predict the induced flow required to be produced by the rotor,
The variation of pressure will drop below atmospheric above the disk, increas due to the energy input of the disk and then drop bad to atmospheric in the slipstream. Applying and Energy balance will again lead to v s = 2 v i so that, T = 2 A v i V c v
Power required from engine to produce this thrust from the induced velocity will be
P(hover induced velocity) = Phi = T.vi
CLIMB
For the case where the helicopter is climbing, the rotor will capture flow from a fixed area above and accelerate this to a final slipstream velocity. The
momentum balance will be slightly different as the incoming air has an initial momentum due to the helicopter climb speed and the rotor is just augmenting this. The mass flow rate through the disk will be,Δm/Δt = ρ A vi Vc
The thrust due to momentum change will be,T = Δm/Δt vsVc−Vc = ρ A viVc vs
There will be a variation of static pressure. Th pressure will drop below atmospheric above the disk, and increase due to the energy input of the
disk and then drop back to atmospheric in the slipstream.
Applying an Energy balance will again lead to the same result as for the hover case,
vs = 2 vi
so that,
T = 2 ρ A vi Vc vi
rearranging gives,
Solving this as a quadratic equation for vi gives,
or
If climb rate is small then it can be assumed that T = W so that required induced velocity can be predicted for varying climb rate.
Less induced velocity is required for climb as the momentum change applied to a higher initial velocity is more effective.
Power requirement is not reduced in a similar fashion. While engine power is still needed for thrust creation, there is additional power required to move the vehicle at a climb speed.
P(ideal climb total) = P(induced velocity for climb) + P(climb)
P(ideal climb total) = T . vi + T . Vc
The power required for climbing will thus be,
Assuming climb rates are small, so that thrust approximately equals weight, the variation of power required for different climb rates is shown in the following figure (weight of typical helicopter assumed).
Note : The above power estimates only cover the requirements for the rotor acting as a ideal momentum disk. This covers induced velocity and climb requirements only. In reality to produce the momentum change rotor blades must be pushed around within the disk, hence there will be an additional requirement of power to overcome the profile drag of the rotor blades. This will be covered analytically in a following section or may be predicted numerically using blade element theory.
DESCENT
For descent rates that are small and do not approach the magnitude of vi, the analysis can be carried out using a negative value of Vc in the above equations. However there are physical limits on this approach due to the changing flow pattern. As the capture area is now below, problems will arise as the rotor is now capturing its own wake an recirculating it to create momentum change. Eventually at a higher rate of descent the system will become closed and the rotor will simply recycle its own wake ('ring state'). In this condition there is no momentum change and hence no thrust. If decent is fast enough then flow pattern will fully reverse and rotor will 'windmill' or auto-rotate. The momentum change is now negative with the rotor be powered by the airflow. upward thrust will again be produced in this auto-rotation state.
a) Slowly descending rotor.
b) rotor descending in vortex ring state. c) fast descent in autorotation (windmill) state. Vd = - Vc
vi --direction reversed.