Downwash v Lift?

[CyclicRick]

Heres one for the Aerodynamic experts among you:

How do you calculate how much downwash ie air pressure and how much is lift that keeps the helicopter in the air?
Now we all know that lift is 1/3 pressure(underneath) and 2/3 low pressure (on the top!) so you must have a third force involvement, because when you have pressure underneath pushing up and at the same time pressure pushing down, they must influence each other to some degree.
Was I asleep in Aerodynamic theory? Am I asking a stupid question?

[Heliport]

Split from the 'Helicopter Cowboy' thread.
Worth a discussion of it's own.

[CRAN]

Cyclicrick,

This is definitely not a stupid question in fact the real answer to your question is far far far too complicated for this forum, for those who are interested a good place to pick up the trail to the real answer would be some of the industry bibles such as:

Principles of Helicopter Aerodynamics - Leishman
Rotary Wing Aerodynamics Vol I & II - Stephniewski & Keys
Helicopter Theory - Johnson

....none of these references will expain the current state-of-the-art, but they do point you in the direction of it! (After that your into technical papers i'm afraid!)

The simplest model of helicopter aerodynamics is the momentum theory model that uses an actuator disk representation of the rotor. It is from this analysis that the 1/3 and 2/3 figures arrise.

Momentum theory assumes that the rotor is a disk made up of an infinite numbers of blades that do not suffer from tip losses. The flow is deemed to be inviscid and incompressible. Also this actuator disk is capable of supporting a pressure difference only.

The velocity of the air a small distance above and a small distance below the disk are the same. This must be the case in order that the law of mass conservation is upheld. These characteristics give rise to the conditions that the flow very far above the rotor is still and that the rotor is bound by a definite 'streamtube' Within this streamtube we can apply Bernoulli.

Bernoulli's equation can be applied which states that the total pressure along a streamline is constant. In order for Bernoulli to be true with zero velocity above the rotor and a finite velocity below the rotor and twice this velocity in the far wake a pressure difference results across the disk. This analysis is outlined clearly in Leishmans book.

The important thing about momentum theory is that it produces the theoretical ideal performance as it only considers ideal induced losses - i.e. the kinetic energy imparted to the wake by the rotor. All other losses are neglected. Therefore momentum theory is used in the calulation of figure of merit for a real rotor as it is the theoretical optimum to which the real rotor is compaired.

In answer to your initial question, 'how do you calculate the downwash', well using momentum theory the induced velocity at the disk in the hover:

vi = [(Mass x 9.81) / (2 x Density of air x Disk Area)]^0.5

The ^0.5 means take the square root of the contents of the square brackets.

The velocity of the downwash in the far wake is w=2 x vi

This theory is an idealisation and doesn't take into account things like tip losses, the highly important effects of vorticity in the wake, wake swirl, non-uniform inflow etc etc etc......this is a long list!

With regards the lift that the rotor produces...

Once you have the pressure difference accross the rotor all you need to do is multiply this by the Area of the disk to get the lift.

Nice.

Hope this goes some way to answering your question, though this is clearly a brief overview of the theory. Leishman does it really well so for the proper answer look it up!

Hope this helps
CRAN
:cool:

[Roofus]

I gotta go lie down...............:p

[sierra-papa]

Cyclic Rick,
the vectors of the low pressure area on top of the disc and the high pressure area below the disc both point upward. They don't cancel each other but make your total lift vector together.

Cran,
nice of you take precious time to answer questions on this very very very simple peoples forum. (Just kidding because of the opening in your post!)
I figure all you wrote pertains to a stabilized hover in a theoretical no wind situation with a constant AOA?? Now there is no hover without extensive vortices production. How can the mentioned downwash formula work without taking in account that the outbord section of the disc is busy with the production of vortices rather then lift?? Should the area covered with vortecies production simply be subtracted from the total disc area?

sp

[Arraitch]

Oh dear...this gives me a headache like the Bernouilli mumbo-jumbo rammed down my throat at university. Interesting equations and all, but turn that wing upside down (an aeroplane flying inverted) and try to use those equations to make it fly.

A far better understanding of moving air to make something fly is on this site:


http://http://jefraskin.com/forjef/jefweb-compiled/published/coanda_effect.html

makes very interesting reading.

[Nick Lappos]

Arraitch,
The nice thing about what Cran has posted is that it is not Bernouilli, but rather that staid Englishman, Newton. Simply put, who gives a damn about pressure, the rotor tosses air down, and in the process, picks the aircraft up. If the rotor were a machine gun shooting large pellets of lead downward, we could calculate how much lead, how fast, to create the lift we need. That is all the momentum equation Cran posted does.

The great confusion created by Bernouilli, pressures and curved airfoils gives me a headache, and I have several thousand hours of experimental test flying!

Remember this: The rotor lifts the aircraft up because it throws air down. Small rotor throws less air, so the air has to be going faster. Big rotor throws more air, so the air can go downward slower. Downwash is the name we give that air stream that lifts us. If you are doing 100 knots, or at a steady hover, you have to transfer the momentum to that air to stay up. Same downwash, at any speed.

Tip vortexes and other losses cost energy, and they add velocity that is in outward directions, so they don't help lift you, they are a measure of the extra (lost) energy that the engine must produce.

[CRAN]

Sierra-Papa,

If this is a very very very simple persons forum then I must be a very very very simple person too - cos I like it here! :)

The reason why I say that it is a really complecated area is because it really is. If I try to explain Blade Element Theory, Vortex Theory, Potential Theory, or the numerical solution of the Reynolds Averaged Navier Stokes equations then 1) that little blue list at the top of the forum would soon deminish and 2) you'd be in for a very long post! :D

The momentum stuff I mentioned is strictly only applicable to the hover in the form I gave, though it can be readily extended to forward flight. (Which is the same as hovering into wind - your other point.)

All of the limitations that you pulled up are correct - these are the limitations of momentum theory, the progressively more rigorous rotor models I listed above take these factors into account.

With regards the state of the rotor - momentum theory only allows you to specify the disk area (i.e. Radius and root cutout) no other real rotor parameters are considered. Aerofoil characteristics, AoA, twist, taper, blah blah - all ignored. But don't get me wrong its still useful for many things.....

Last thing.....yeah, one of the effects of the tip vortex is to create a locally high inflow which reduces the local angle of attack, for a simple case you might assume the outer 3% of the blade produces no lift but DOES produce drag. (Simple fix, used in a blade element type calc.) Again put complex tips on the blades and it all gets a bit hairer and we have to think a bit harder.

Rock On
CRAN
:cool:

[CyclicRick]

Thanks all for your time but...yee gods!

I'm sure this, in detail, is a VERY complicated subject especially for my little brain.

I understand completely the vector diagramm of a blade in the hover IGE/OGE and climbing and decending, that in itself is fairly simple BUT Nick said that a helicopter hovers because it pushes air down.
Can't understand that one, a blade produces not only downward pressure but also lift does it not. A normal aerofoil use bernoulli's principle of pressure, 1/3 pressure underneath the lifting surface and 2/3 suction (for want of a better word), so a blade must produce the same and in addition downward thrust which a normal aerofoil does not, I assume. So we do in fact have three factors.
Am I still missing the point?

[Dave Jackson]

CyclicRick

This should help. Airfoils and Airflow (http://www.monmouth.com/~jsd/how/htm/airfoils.html)

Dave J.

[[email protected]]

Cyclic, if you have looked at the web page Dave Jackson points you to you will probably now understand that your 1/3 and 2/3 concept and Nick's blowing air downwards are in fact the same. the 1/3 underneath the aerofoil produces pressure upwards and in doing so is deflected downwards by the blade - the 2/3 above the aerofoil is accelerated over the top (thus reducing it's pressure and sucking the blade upward) but continues to follow the shape of the blade and is spat off the trailing edge which is pointing downward. The overall effect is to deflect all the air downwards (giving the downwash) whilst producing a force (lift) upwards. When the next blade comes round a fraction of a second later the air it meets is already going downwards, and by the time the blades of your rotor have passed round a few times the air will be going straight down through the rotor which has induced the airflow through it (the big green Induced Flow arrow on the vector diagram).

Or think of it another way - a cubic metre of air weighs 1.225Kg (according to ICAO) so a 5000kg helicopter will have to move 6125 cubic metres of air constantly through it's rotor disc to hover! As Nick said, the bigger the rotorarea, the easier it is to achieve and a smaller rotor will have to work harder (using more power) and produce a more concentrated and powerful downwash.

[Nick Lappos]

Crab,

You can't tell the cubic feet of air needed to hover from the weight of the helo. Need to know the disk size, too.

You have the momentum thing sort of right, but the real calculation is the mass of air times the speed change you make. A helicopter can be held up if it takes a small amount of air and imparts lots of speed to it, or a large amount of air with little speed. Its identical to a scuba diver's fin size. Big fins move a bunch of water, so the diver needs to pump his legs more slowly. Small fins take more strokes per minute to move the diver the same way.

We could hover a helicopter with a cannon shooting bullets downward, if the bullets weighed enough and the cannon shot enough of them per second. If the velocity of the bullets were high, fewer of them are needed to lift the helo.

[[email protected]]

Nick, good point, well made and right as usual! I was trying to oversimplify the concept of a mass of air being pushed down to keep a mass of helicopter up.

As an aside is it Bernoulli's theorem that gives the final velocity of the downwash as twice the velocity across the disc or have I got that wrong as well?

[CyclicRick]

Dave.J

Thanks for the web address, good one. I've saved it so I'll be taking a really good look at that later on, Thanks again.

I think I've got it now:cool:

[Nick Lappos]

Crab,

The outcome of the flow equations is the fact that the rotor smoothly accelerates the air, so that half the speed change takes place before the disk, and half afterward. This is similar to the way a wing affects the flow around it, since the flow realigns far ahead of the wing. This is true in subsonic flight for all flow disturbences (in supersonic flight, the wing hits the air before the pressure disturbence can get there).

[[email protected]]

Thanks Nick, is there a simple way of calculating the mass of air required through the disc given the AuM and the disc area? What processes do the designers go through to determine downwash velocity, disc size and power requirements given a design AuM for the aircraft.

[Q max]

I made this up myself .... but it looks plausible!

MV = FT (momentum equals impulse)

so
F=MVi/T or Vi M/T (Vi is final velocity, M/T is the mass arrival rate)

M/T = pAV (p is density Rho, A area of disc, and V is Vi/2 'cos that's the speed at the disc ie the 'average speed')


so
F = Vi M/T = Vi pA Vi/2 = 1/2 pAVi^2 = W (W is weight of heli)


So Vi is related to Weight and Area like this:


Vi=(2W/pA)^0.5


so if you Quadruple the Area you Halve the Vi ....!WOW!


(Since Area = pi r^2

Vi depends on 1/r ! ) (where r is the rotor radius)



I like the momentum approach - much more intuative
and I really like Nicks brand of Flying Fisiks (even if he doesn't believe the TR above the disc would try and roll the heli the 'other way')

But if you have a helicopter in EXTREME ground effect (ie a hovercraft) how do you use the momentum argument then?

[[email protected]]

Qmax, thanks, but I did ask for a simple method!!! One of my workmates who is trying for ETPS started with the momentum idea but got confused with the kinematics of it all (my brain had turned to mush long before he got stuck).

[Q max]

... I thort that was simple .... sorry when I look at it again it does look a bit inpenerable.

In English it means:

The momentum you give the air depends on how fast you make it go.
AND
The amount of air you do it, which Depends on this speed AND the area of the disc.


Can I teach at ETPS now?;)

inpenerable ?

[[email protected]]

Qmax, yes but only if you can manipulate your formulae to give the amount of air in cubic metres and the speed of the downwash if you know the disc area and the aircraft AUM.

[Aro]

I'll have a pop...

This is particularly easy to overcomplicate, so at the simplest and most uncomplicated level 'every action has an equal and opposite reaction'...

Say your chopper weighs 1300Kg
1m^3 of air weighs 1.3Kg
1000m^3 of air weighs 1300Kg

Gravity is trying to accelerate you downwards at 9.8ms^2
you have to accelerate 1000m^3 of air through the blades at 9.8ms^2 to counter it.

Or accelerate 2000m^3 at 4.9ms^2...etc

The pressure during hover (ideally) will be the weight of the chopper over the area of the disc.


Does that work?

[Q max]

Well ...
the mass per second (M/T)

M/T = Square root (2WpA) (Crab)

so if you quadruple the weight you only double the Mass per second

Intuitively this makes sense cos if you double the mass per time(independantly) then you'll ALSO have to be making it go twice as fast -

so you don't need to double the mass per time if you double your weight !

.... a very different answer to ARO - I am simplifying it - just not over-simplifying!

[p is density 1.3kg/m^3 W is weight in Newtons ie AUMass times g, (10Newtons/kg), A area], Formulae unverified

and Vi= Square root(2W/pA) (Crab)

[[email protected]]

Aro - you converted the mass of the helo to a force in newtons eg 1300kg x 9.8m/s = 12740 newtons mass x acceleration = force - therefore IMHO you should do the same to the air eg 1.3kg x 9.8 m/s = 12.7 newtons because the air is at rest before you push it through the disc and is still affected by the gravitational constant.

[heedm]

crab, he did.

[Aro]

Hi Q,

You said earlier...
"F = Vi M/T = Vi pA Vi/2 = 1/2 pAVi^2 = W (W is weight of heli)"

You've proved Force (mKgs^-2) equals Weight(or Mass?) (Kg)

I think it should read...

Fr = Vi M/T = Vi pA Vi/2 = 1/2 pAVi^2 = Fh

Where Fr is Force required and Fh is Force imparted by the heli.

Or...The force required to hold the heli is the same as the force the heli imparts.

Pressure is newtons per metre squared.

Say the disk is 10m^2 and the heli is 1300kg, gravity is 9.8m/s^2.

F=12740N
A=10m^2

Pressure=12740/10
=1274Pa

or the Force of the chopper over the area of the disk.

This demonstrates that a smaller disk will create a higher pressure for the same chopper, due to the increased work rate, and also concurs with Vi being dependent on the radius of the disk.


There is an issue with the following equations...

No1
M/T = pAV (p is density Rho, A area of disc, and V is Vi/2 'cos that's the speed at the disc ie the 'average speed')

and

No2
M/T = Square root (2WpA) (Crab)


Lets first find Vi with our (aribtrary) numbers

vi = [(Mass x 9.81) / (2 x Density of air x Disk Area)]^0.5
vi = [(1300 x 9.8) / (2 x 1.3 x 10)]^.05
vi = (12740 / 26) ^.5
vi = 22m/s

No 1
M/T = pAV
M/T = 1.3 x 10 x 11
M/T = 286Kg/s

No 2
M/T = Square root (2WpA)
M/T = (2x1300x1.3x10)^.5
M/T = 33800^.5
M/T = 183kg/s

If we say
No 3
M/T = sqrt(2FpA)
M/T = (2x12740x1.3x10)^.5
M/T = 576Kg/s

I'm inclined to go with the last one which is an equivalent of 443m^3/s. (For a 1300kg chopper with a 10m^2 disk).

To provide a force equal to that of the chopper, that air would need to be accelerated at...


1000/443=2.25

2.25 * 9.8 = 22m/s^2

Which would see the air at vi after 1 second.


My apologies for the misuse of 'weight' in the previous post.

[CyclicRick]

I wish I hadn't asked now.:confused:

[ShyTorque]

Doh! My brain hurts.

I often get asked how much downwash there is in the hover.

I normally use the following formula:

(Dese5tonsachopperup = 5tonsadatairdown = lotsawindsoplizdon'tweardemhatsorputupdoseumbrellasuntildewh oledamnlotisondegroundandstopped) squared.

It normally helps for passengers and suffices on a practical basis for pilots of simple brain like wot mine is. :rolleyes:

[Q max]

Aro this is forensic Math !
- But your Fisiks is phukedup (not that mine isn't a bit also)


"You've proved Force (mKgs^-2) equals Weight(or Mass?) (Kg) "

No, no, no, ... yes:

Weight is a force! The OTHER expression for gravity is helpful ie. g= 9.8 N/kg
Weight isn't in Kg! Weight is Mass x g which is N (Newtons) so mKgs^-2 = N = (g x Kg)

So Fr = Fh = F = W = Mass x 9.8(N/Kg or ms^-2)

So 'your' "No 2" is "No 3" !!



Is this you trying to quote me?:
"vi = [(Mass x 9.81) / (2 x Density of air x Disk Area)]^0.5 "

if so it was:
Vi=(2W/pA)^0.5 (The 2 is on the top)

So
Vi = [(2x1300x9.8)/(1.3x10)]^0.5
= [25480/13]^.5 = 44 m/s Not 22 m/s



No 1
M/T = pAV
M/T = 1.3 x 10 x 22 (you've got 11 where I've got 22 BUT....)
M/T = 286Kg/s !!!!surprisingly same as your answer!!!
!!!you got there by making a mistake!!!


No 2
M/T = Square root (2WpA) or W=F so "No 2" is "No 3"
M/T = (2x1300x9.8x1.3x10)^.5
= 331240^.5
M/T = 575 Kg/s

But MY mistake this should have read:
M/T = Square root (0.5xWpA)

So No 2 which is the same as No 3
Gave an answer which was Square root(2/0.5) Too big so should have been:
M/T = 575/2 = 287.5 Kg/s which is the same as No 1 (rounding errors)


So No 1 = No 2 = No 3 !!!!!!


But to summaries (very clumsily) it's almost exactly what Nick Lappos elegantly said:

the 287.5 Kg/s is NOT "1300kg of air being accelerated cos thats the mass of the helicopter"
like most people wrongly assert.


Aro
so 287.5Kg/s is equivalent to 221m^3/s

Are you saying:

1000m^3, which is NOT the amount of air this helicopter is acting on per second,

is 1000/221 = 4.52 times the amount which it IS acting on per second

and so if it were acting on 221m^/s instead of 1000 m^3/s it would have to accelerate it at g x 4.52 = 9.8 x 4.52 = 44m/s^2 which happens to be the correct flow speed?

?

If so I think you are wrong (and right Post Script).

I think you are effectively calculating the factor by which you are wrong and multiplying your wrong answer by the correcting factor here.

effectively you are saying if you took a mass of air equal to the mass of the helicopter and multiplied g by the factor that that mass of air was incorrect by you'd get the correct flow speed...


Yea yea yea - I've got it! sorry I thought you were answering the question ... you're right in as far as you go (except for the basic errors ( factor of 2) which we both made.

What a waste of time - but didn't you find some of the relationships ineresting tho?

Like:

"Vi depends on 1/r ! "
and
"if you Quadruple the Area you Halve the Vi "

[Nick Lappos]

The easy way to look at the physics is to recall that the force is proportional to the time rate of change of the momentum (the real wording of Newton's second law). Don't use the acceleration, use the velocity change on a mass flow (a delta V on a mass/sec)

If the air has a mass M (the density times the volume) and you give it a change in velocity of V, and it is flowing through the disk at that velocity, then the force or lift generated is MV/T.

Use mass in slugs (a unit of 32.2 pounds), assume a lift of 20,000 pounds, and that air has a density of .002378 slugs/cubic foot. Assume a downwash velocity of about 35 knots (60 ft/sec)

20000=60 x M/T

The mass flow for the air = M/T = 333 slugs per second=10,000 lb/sec, which is about 140,000 cubic feet of air per second. If the column of air is 53 feet in diameter ( or 2200 square feet, the size of the disk) it would have a velocity of 63 ft/sec (which is almost exactly what we assumed).
So, to keep a 20,000 pound helo hovering, if the rotor has a 53 foot diameter, the mass that flows per second is 10,000 pounds of air, and it has a final velocity of 60 ft/sec.

In other words, if the helicopter had a disk loading of 9 pounds per square foot, it has to throw half its weight in air EVERY SECOND to stay up, and it throws that air at 60 feet per second downward.