Power Required Losses

Losses which occur between the torquemeter and the rotors must be made up by the engine and are additive to the power required by the rotors. Sources of these losses include:

• Main rotor transmission

• Tail rotor gearboxes

• Engine nose gearboxes

• Transmission-mounted accessories such as generators and hydraulic pumps

• Cooling fans driven from the drive system

Gearbox and transmission losses are produced by friction between the gear teeth and in the bearings and by aerodynamic drag, or "windage.” The losses are a function of the size of the gearbox as well as the power being transmitted at any given time. For preliminary design purposes, the losses in gearboxes—including power to run their lubrication systems—can be estimated from the following equation:

•Power loss per stage = iC[Design max. power + Actual power], h. p.

where К = 0.0025 for spur or bevel gears and К = 0.00375 for planetary gears.

The example helicopter will be assumed to have two engine nose gearboxes with one stage of bevel gears, each designed for 2,000 h. p. The main rotor transmission has two spur gears stages and one planetary gear stage, with a design capacity of 4,000 h. p. The tail rotor drive system contains two gear boxes with one stage of bevel gears each and a design capability of 750 h. p. The total transmission losses are thus:

h-p-trans. = 0-0025 (4,000 + Eng. power) + 0.00875 (4,000 + Main rotor power)

+ 0.0050 (750 + Tail rotor power)

Assuming that the engine power is the sum of the main and tail rotor powers, the losses become:

h-p-trans. = 49 + .0112 main rotor power + .0075 tail rotor power

Losses due to generators and hydraulic pumps are a function of the load on them during any given flight condition. Assuming typical efficiencies:

Load in watts

Generator loss =■-—} ,h. p.

Hydraulic pump loss = (P^P^re psi) (Flow rate, gpm)

7 r F (.80)(1,714) r

For performance calculations on the example helicopter, it will be assumed that the generator produces a constant 2,200 watts, which results in a loss of 4 h. p. Hydraulic pumps deliver most of their flow during maneuvers, but even during steady flight they are pumping some fluid. For the example helicopter a minimum flow rate of 1.3 gpm with a 3,000 psi system will be assumed. This gives a hydraulic pump loss of 3 h. p. No separate shaft-driven cooling fans are in the configuration, but, of course, electrically or hydraulically driven fans are already accounted for.

Since the rotor performance is a function of the atmospheric density ratio, the calculations of total engine power will be simplified if it is assumed that both the transmission and accessory losses are also proportional to the density ratio. (This is an assumption more weighted toward convenience than accuracy, but valid, enough for power estimating.) Using this assumption for the example heli­copter:

h-p-iraii[5] [6].cc. = ~ (56 + .0112 h. p.,, + .0075 h. p.T) Po