bhaazee Posted March 12, 2012 Posted March 12, 2012 I would like to find the engine power/ engine size (in cc) required to lift a helicopter that weights 25 to 30 Kg. Can someone please provide me some calculation? Although the question is vague with not much information I would really appreciate some basic equations to calculation the required engine capacity/power.<br style="font-family: Verdana, Tahoma, Calibri, Geneva; font-size: 13px; background-color: rgb(227, 227, 227); "><br style="font-family: Verdana, Tahoma, Calibri, Geneva; font-size: 13px; background-color: rgb(227, 227, 227); ">Regards
ewmon Posted March 12, 2012 Posted March 12, 2012 (edited) The answer is not so simple. Power requirements involve the area of the "disk" that is swept by the lifting rotor (called the "swept disk"). The smaller the area, the higher the power requirement; and the larger the area, the lower the power requirement (due to something called "disk loading"). It's why the Harrier is such a gas guzzler when performing like a helicopter, and the Osprey is more economical. It's also why have such huge rotors to lift one human, whereas using a (where the jet engine compressor/turbine is the "rotor") requires lots of power. There's also something called "ground effect" as well as other factors. If you google helicopter aerodynamics or something like helicopter power requirements, you'll find lots of good introductory and advanced material. Edited March 12, 2012 by ewmon
Enthalpy Posted May 13, 2012 Posted May 13, 2012 It takes much power just to keep a helicopter in the air, even without losses here: Lifting force is (worse than) F=ñ*V Power consumption is (worse than) P=ñ*0.5V^2 where I misuse ñ for the mass throughput in kg/s and V is the down speed after acceleration by the rotor (supposed zero upstream, or subtract the upstream V and 0.5V^2) Because of V and V^2, a small V and a big ñ (through a big rotor area) need less power for the same force, as said ewmon. This holds for horizontal propulsion as well, but here air speed before the propeller limits the exaggeration: once the relative speed increase is small, the game is over. Though, the trend is toward bigger propellers to be efficient: just compare recent airliners with the B-707. Unducted fans and others are attempts to make propellers larger than a turbofan. Turboprops are preferred on slower aircraft for being wider. And two-flow turbofans result equally from this logic. The same reason tells that magneto-hydrodynamic propulsion (MHD), which prefers very fast flows, is inefficient at slow boats.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now