Question

Because of their knowledge of physics and interest in transportation, you have got a summer job at the US company Sikorsky, they are interested in investigating the energy consumption of your new 3-rotor helicopter (Sikorsky S-97 Raider) The blade of a rotor can be considered as a thin rod. If each of the three blades of the helicopter rotor is 3.75 meters long and has a mass of 135 kg. a) Determine the energy needed to run the 2 rotors until the blades reach a speed of 50.0 rev / s. b) If regular gasoline were used to produce that energy (the energy resulting from question A). How many gallons of it would be needed? c) If the engine of a Sikorsky CH-53K helicopter provides a useful shaft power of 13.5x10^6 watt, how many revolutions per minute should its blades rotate to keep it in the air for an hour? ta tol- hef d) Explain a way to reduce the energy consumption of these new LES helicopters. aplain a way to reduce the

Answer 1

Given: 3-Rotor, Mass of blade M =135 kg, Length of the blade, L = 3.75m, $2\pi N = 2\pi \times 50= 314.16 rad/s$

Moment of inertia of one blade   $I = ML^{2}/3 = 135 \times (3.75)^{2}/3 = 632.81 kg-m^{2}$ .

So the moment of inertia of 3- blade( 1- rotor)   $= 632.81\times 3 = 1898.43 kg-m^{2}$ .

(a) The energy needed to run one rotor =$\frac{I\omega ^{2}}{2} =(1898.43\frac{314.16^{2}}{2}) = 9.36\times 10^{7}J$

To run Two rotors = $18.68 \times 10^{7}J$

(b) The lower calorific value of gasoline = 43.4 MJ/kg, 1Gallon = 2.8 kg/hr.

Fuel energy = $m_{f}\times C_{v}$ , $\frac{Fuel energy}{C_{v}} = \frac{18.68\times 10^{7}}{43.4\times 10^{6}} =4.3 kg/s =4.3\times 3600 kg/hr =15480 kg/hr = 15480/2.8 = 55285.57 Gallons$

(c) Power = $13.5 \times 10^{6}W$ , So work done = (power) (time) =$13.5 \times 10^{6} \times 3600=4.86\times 10^{10}J$

Work needed to run 3- rotor =$3\times 9.36\times 10^{7}= 28.08\times 10^{7}J$

The rotational energy of the rotors $\frac{I\omega ^{2}}{2}\times 3\times 3= 4.86\times 10^{10}J, \omega =4131rad/s . N = 39449.9 rev/min.$

(d) by the decrease in component friction losses. The energy can be saved. by shutting down 1 or 2 rotors when they are not needed