SHP vs Delivered to the Blades
 

What's the rough number of the actual percentage of HP consumption from the engine accessories and transmission on a small turbine helicopter?
Couldn't find any info on that..

Depends on many things, especially the TR thrust required, Soave. In a hover OGE, the losses might be: TR 4 to 5%, Transmission losses 2 1/2%, accessory loads 1%. Total 8% or so. In cruise, the TR is perhaps 2% and the rest that same.
Anyone flying a Bell with mast torque vs engine torque needles want to provide some more precise data?

I'm in no way an expert but I really surprised by your figure of only 5% for TR thrust in hover. I would have thought it much more than that for 2 reasons, one being the big difference in max total torque in the hover vs high speed flight (unloaded TR) for the later Super Puma family helis. And the second being the ease of adding or subtracting 5% torque by a relatively small TR pedal input in the hover.

Edit: oh wait a minute, did you mean 5% lost in the drive train as opposed to 5% used by the entire TR thing?

Thanks, that's what I had in mind it would be (around 10%).

I am no expert either, but I would think for a medium twin that number is higher too, but for a small single, 5% is a very probable number for me.

Should you do a max rate spot turn to the left with a Puma the torque doesn't drop that much. A tail rotor has a long motion arm to the rotor hub so not a lot of thrust is required to hold the fuselage straight.

I think Mr. Lappos is being kind to transmission designers with his estimate of 2-1/2% losses within the MRGB of a small turbine helo. With some light turbine helos the main gearbox efficiency situation gets a bit complicated.

Consider the example of the OH-58D with the 650shp M250-C30 engine. The engine has a speed reduction gearbox that drops the power turbine speed from >30krpm down to ~6krpm at the main gearbox input. So the main gearbox only needs to provide a speed reduction ratio of ~17.4:1. The small input/output reduction ratio and low input speed allows good efficiency from this main gearbox design, and under optimum operating conditions the 97.5% efficiency number provided by NickLappos would be achievable. But since some of the speed reduction is performed by the engine's gearbox, it does not accurately reflect the full drivetrain losses between the power turbine and main rotor. If all the speed reduction was performed within the main gearbox, the peak gearbox efficiency would be something like 96%.

Another important factor to consider with main rotor gearboxes mated to turboshaft engines with high speed power turbines, is the effect of speed/load on gearbox efficiency numbers. There are various types of losses produced by the gears and bearings, such as friction, viscous, windage, etc. Friction losses tend to be more load dependent, while viscous and windage losses tend to be more speed dependent. I've attached a graph that illustrates just how rapidly gearbox "efficiency" drops off under conditions of constant speed and reduced torque. This is a great illustration of the effect of speed dependent losses on gearbox efficiency, and why you will usually obtain best gearbox efficiency at max load for a given speed.

Flat Pitch Torque as an Indicator
 

As a quick reference one can calculate the power required to drive the complete system based on the flat pitch torque. As examples – the B206L L4 takeoff limit is 100% Q 450 ESHP. Typical flat pitch Q is 15% or 67.7 ESHP. Similarly the H-53E takeoff limit is 137% or about 13000 ESHP. The H-53E flat pitch torque on all three engines is approximately 20% each or about 1900 ESHP just to drive the system.

That sounds a lot more like the numbers I expected as opposed to the few % mentioned earlier. Cars have similarly high drivetrain losses.

But of this 15-20% a big part comes from drag. If you run it at 100% Rpm without main- and tail blades you get a much lower value on torque.

Right. That brings us back to the original question, if an engine is rated 450 HP on its shaft what HP would actually get to the blades?

Using the info so far here gathered: 96,5% would be available for MR and TR.

know this isn't an answer to your question but you also need to account for the gearbox limit.


A helicopter may have 3 engines capable of delivered a total of 4500kW, but the gearbox may have a limit of maximum 3500kW.