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How is induced velocity in far wake region and actuator disk related in climb condition?(a) w=√v(b) w=v^2(c) w=2v(d) w=3vI had been asked this question in semester exam.This interesting question is from Momentum Theory in section Helicopter Dynamics of Aerodynamics

Answer»

Correct answer is (c) w=2v

Easy explanation: The momentum equation for the ROTOR in climb condition is given by:

T=\(\dot {m}\)(V+w)-\(\dot {m}\)V=\(\dot {m}\)w

And the energy equation for climb condition rotor is given by:

T(V+v)=\(\frac {1}{2}\dot {m}\)(V+w)^2–\(\frac {1}{2}\dot {m}\)(V)^2=\(\frac {1}{2}\dot {m}\)w(w+2V)

On rearranging and eliminating T/\(\dot {m}\) term, we get w=2v which is similar to the condition for hovering helicopter. Thus, the induced velocity in the wake region is TWICE that of the velocity at the rotor DISK.


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