View Full Version : Calculating Rotor Downwash
09-19-2004, 07:20 PM
How do you calculate the rotor downwash speed during hover, assuming a 7 in. chord rotor, a rotor diameter of 24 feet, all up weight of 650 lbs, rotor rpm of 500, with 0 wind speed ?
09-19-2004, 07:45 PM
Using the formula below, which is found in helicopter texts, I get 17.4 ft/sec.
Apparently the downwash can be estimated without knowing the blade chord or rpm. It is mainly a function of disk loading and air density.
09-20-2004, 03:15 AM
Al hit the nail on the head.
What are you calculating?
09-20-2004, 03:35 AM
al, your equation is similar to this one
lift = Cz * 1/2 *R * V2 * S
cz depends on airfoil, i guess its the 0.002378
V wind speed
S surface (main disc - central disc)
but this equation (i gave to dean first then, deleted it) is not accurrate as it is used to calculate the lift for WINDMILLS, assuming a wind normal to the disc.
like christian said in other post, another method should be more accurate
09-23-2004, 09:50 AM
Victor, your and Al's formulas are derived from momentum theory; Force = Mass x Acceleration.
The factor; 0.002378 is air density, slugs/ft³ (weight of air, lb. per ft³ divided by acceleration of gravity)
Momentum theory assumes uniform flow through the rotor disc and ignores the influence of ground effect.
Power required is down wash velocity x gross weight divided by 550. The answer will be optimistic by 10% or so and to that must be added rotor blade profile drag power.
09-23-2004, 10:10 AM
c BEATY, thats right, i think thats a similar calculus.
ok for air density (i m dumb considering a Cx of 0.002 ;))
why 550 ? can you explain or give a link ?
i have various formula and the only method i consider is : perform them all and design a little above of all.. (hope my english is understandable)
09-23-2004, 11:38 AM
James Watt discovered that Welsh mine ponies could move coal carts at a rate of 550 ft/lb/sec over the course of a working day. A drawbar pull of 100 lb., moving at the rate of 5.5 fps (3.75 mph) = 1 HP.
Therefore, one Imperial HP = 550ft/lb/sec.
Velocity (fps) x force (lb.) divided by 550 = HP.
*That would be 76 kg/meters/sec but you round off to 75 kg/m/sec.
During the prime of the Citroen 2 CV, I saw horses pulling sugar beet wagons in the south of France that would equal 3 Welsh mine ponies.
The 2 CV, by the way, probably was 15-20 hp. The 2 CV came from the tax collector's formula.
09-23-2004, 12:10 PM
LOL okay C beaty, i m in Metric system ;)
10-10-2004, 05:38 PM
Way cool thread, and you are right on, C. Beaty, except that the helicopter is actually far less efficient than the 10% you estimate (the rotor itself loses 20%, the main transmission 2%, the downwash on the fuselage 3 to 5%, the tail rotor about 3 to 5%). For most helicopters, the entire system is only about 65 to 75% efficient, so take the momentum theory horse power value and increase by about 40% to get at the real number.
10-10-2004, 06:45 PM
Nick, I was only discussing that portion of rotor power used in producing a momentum change in the air mass (induced power).
The ~10% figure results from slipstream rotation and non uniform velocity over the rotor disc.
I agree that blade profile power, fuselage negative lift and transmission losses can substantially increase hovering power. Profile power alone can amount to as much as 40% of induced power, depending on blade loading and resultant tip speed.
10-10-2004, 07:10 PM
Welcome aboard, Nick! :)
Although, I don't visit PPrune as often, I usually seek out your posts there.
For those of you who don't know of him, Nick is the S-92 Program Manager (http://www.sikorsky.com/details/0,3036,CLI1_DIV69_ETI890,00.html) at Sikorsky Aircraft.
10-10-2004, 11:39 PM
With all this loss of power it's obviously that the tail rotor has to go. ;)
For decades, Western aerodynamic texts have mathematically shown that the tail rotor wastes 8 - 10% of the power. Recently, Kamov has stated in an article entitled Aerodynamic Features of Coaxial Configuration Helicopter (http://www.kamov.ru/market/news/petr11.htm) that the tail rotor wastes 10-12% of total power. In addition, the latest aerodynamic text (year 2000), by Leishman (University of Maryland) supports Kamov's position.
If a craft's gross weight to empty weight is 2:1, then the above might result in a difference in payload of up to 24%.
10-11-2004, 12:31 AM
Welcome Nick, looking forward to your input in the forum.
10-11-2004, 04:56 AM
hello dave, christian, nick i v not be introcuded but pleased any way .. did i miss something ? coax or not coax? hoax ?
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