# Helicopter downwash

1. Dec 23, 2005

### anders

Hello!

We all know that helicopters can hover, stand still in the air. As they do this, a powerful downwash, down-directed wind, results. This is very obvious when the helicopter is near the ground, since the downwash often kicks up a dust cloud.

My question is this: How strong will this downwash be?

I'm looking for an approximate formula, a function of the following parameters:

* Lift (vertical force needed to keep helicopter hovering)
* Density and Viscosity of the air
* Rotor diameter (diameter of the helicopters 'propellor')

(Have I missed any important parameter?)

My guess is that the downwash will:
- Increase with higher lift
- Increase if viscosity decreases.
- Increase with smaller rotor-diameters

best regards,
/anders

2. Dec 23, 2005

### Clausius2

Hi anders!, what do you mean with "how strong"?. Sorry, you should translate it into a physically measurable variable.

The most important variable of the downwhashing jet produced by the blades is its momentum (or kinetic energy), because it is the multiplication of mass flow $$\dot m$$of air downwhased and its velocity $$U$$ what will give the lift force.

Of course, the more $$\dot m$$, the more lift. The more diameter, the more $$\dot m$$, and then the more lift. Viscosity is not relevant in this kind of flows. Usually Reynolds number is so high that viscous effects are felt only at very small scales compared with the rotor diameter, so don't think of it when thinking of a real helicopter.

You may be able of working out some dependence with the Dimensional Analysis. Your problem may depend upon:
$$L$$ lift force (N)
$$\rho$$ density (Kg/m^3)
$$\mu$$ viscosity (Kg/ms)
$$D$$ rotor diameter (m)
$$\omega$$ Angular rotor speed (rad/s)
$$W$$ needed power for running the rotor (Watts)
$$U$$ relative velocity of the downwashing jet to the helicopter

I think you can write 4 adimensional parameters with this information. Write it down here and we can discuss the dependence upon each parameter.

See you.

3. Dec 23, 2005

### FredGarvin

I echo Clausius' question as to what you mean by "strong." I can tell you from perssonal experience that downwash can have enough force to knock you and small buildings right on your butt.

While I agree completely that viscosity is a factor, I would venture that it has minimal impact. The main driving factors will be:

- Angular velocity of the rotors
- Local ambient temp and pressure

All of these I think, except blade area, are covered by what Clasuius posted.

4. Dec 24, 2005

### anders

Hello!
First - thanks for taking the time to even begin to answer my question!

I agree that 'how strong will the downwash be' is not very clear. What I mean is this:

What will the mass flow through the rotor arc be (kg/s)?

Classius: You say that the lift equals the product of the mass flow and the velocity of the air. Which velocity are we talking about here? I think it has to be the velocity _added_ to the column of air moving downwards through the rotor blades, right?

This will probably be a useful relation for when the helicopter has just started to hover. However, almost immediately, the air above the helicopter will start to be sucked down towards the helicopter rotor. So, the air-velocity near the helicopter will be nonzero.

Following a typical air molecule, passing through a hovering helicopter rotor, I think it will go something like this:

1: Pass through the rotor, going downwards.
2: Start curving out to the side.
3: Swept upwards, outside the rotor arc.
4: Swept sideways in over the rotor again.
5: Repeat from 1.

I hear regular airplanes suffer from something called 'induced drag', which is especially pronounced at low speeds. A helicopter is standing still, so it should have very high induced drag - right?

Let's just for the sake of argument make the assumption that the air above the rotor gets a downward directed speed of 10 m/s. The helicopter still manages to hover, by sucking at this air, and effectively expelling it downwards at a speed of say 15m/s. Lift will in this case be massflow * 5m/s. Am I on the right track here?

I did a dimensional analysis, and found that the following formula is dimensionally correct:

mass flow = lift^0.5 * density^0.5 * rotor_diameter

If I recall correctly, dimensional analysis isn't guaranteed to give correct answers. I'm not sure I trust the formula above.

What do you guys think?

5. Dec 24, 2005

### FredGarvin

Induced drag is drag created due to the generation of lift. Helicopters are not more or less prone to it. Induced drag will vary with the load. Just because the airframe is not translating doesn't mean the rotors are not moving.

6. Dec 25, 2005

### anders

Ok, I agree that the rotors are moving.

But the situation might be different from that of airplanes. Each rotor is flying in the wake of the rotor before it. Couldn't this be quite significant? Each rotor is flying in air that is already moving downwards (set into motion by the rotor before it, and the rotors before that).

What do you guys think of the following view of the problem?

Assumptions:
1: Our helicopter is far enough from the ground or other objects, that we may consider the air it is in to be infinite in all directions.

2: The helicopter will set the air in motion, but only the air that is close to it. (For each V, there will be a distance D so that the local air-velocity is less than V.)

3: The helicopter rotor gives the air around the helicopter kinetic energy at a rate identical to its engine output power.

4: The velocity vector field around the helicopter, has the approximate shape of a doublet (I'm not sure about this, but it seems plausible). A doublet is the vector field attained at the limit when a source and a sink are moved infinitesimally close to each other, while their strength is brought towards infinity.

5: The air will have a certain velocity V1 directly beneath the helicopter, close to the rotor. This velocity will be directed downwards.

6: There is a steady-state solution (one with constant, finite air-velocities, apart from the pulsations probable from the rotation of the rotor).

7: Because of 3 and 4, the kinetic energy added to the air is dissipated somewhere in the system. I propose that this loss is entirely made up of viscous losses in the air. If air viscosity was 0, the helicopter could not hover.

8: Assume that reynolds numbers are small enough for air to be incompressible.

A possible way to calculate the needed engine power, might be the following:

9: Calculate the viscous losses P1 in the air, for the vector field, as a function of V1. Needed engine power P2=lift*V1. Calculate the value V1 for which P1=P2. The mass-flow is now rotor-area*density*V1.

I didn't want to write too much about my ideas for solving the problem in the first post - I didn't want to lead anyone astray with my dubious theories :-) .

If anyone managed to follow my reasoning above, I'm very curious to hear what you think!

7. Dec 26, 2005

### FredGarvin

The idea that one rotor goes into the wash of another is a valid concern. You have two things happening: You have the induced flow from the rotor forcing the air through it and you have the rotational or circulating flow as well. However, as the following link shows you, the big power consumption comes from blade tip vorticies (thanks to the Air Cav guys I never thought I would say that!):

http://www.cavalrypilot.com/aerodynamics/hovering.html [Broken]

Speaking of that, I think you should also stipulate a couple of other boundary conditions.

You should also throw in the caveat that the aircraft does not have any forward motion and or there is no wind. Either of those would give rise to a difference in relative velocities somewhere in the rotor which would involve more complications, i.e. retreating blade stall and the such.

Your assumption #1 is called hovering "out of ground effect." Actually, in ground effect usually helps a helicopter to hover because less power is required (usually).

I think your assumptions listed are pretty sound. I may have a couple of issues with #3, but I think if you go with that your model won't be too far off.

Last edited by a moderator: May 2, 2017
8. Dec 26, 2005

### Clausius2

I meant it is of the same order, not just the same. In order to calculate the lift force you only need to make an integral momentum balance over the rotor, calculating RELATIVE speeds.

You are claiming here with words the solenoidality of the vorticity field once the rotor has started to downwash. Nice.

Induced drag is a concept of unsteady flow. The induced drag is an additional force one needs to exert to a body in order to put it in motion in a fluid at rest. The power employed in counteracting the induced drag is the power needed for counteracting the inertia of the displaced fluid. Once the steady regimen is reached, there is only normal (steady) drag.

Sorry, this is not a suitable relation. You didn't dimensional analysis. You only obtained a very strange parameter (there were 4) and you equaled it to a constant (what is false).

9. Dec 26, 2005

### Clausius2

I don't know what you mean with that.

It could be.
The helicopter may set the air in neighborhood of distance $$D$$ (rotor diameter) around the helicopter. (this is the correct expression)

That's generally false, due to dissipation of mechanical energy into heat.

Sounds good, but you are not taking into account the azimuthal component of $$V$$. I mean, the rotor is exerting a torque into the fluid. The downwash jet has a swirl.

This is right. The azimuthal component of the incoming stream is negligible due to the small role played by viscosity.

Sure, there is a time averaged solution. Again, the flow through the rotor is periodically unsteady.

Here there is a contradiction. You said in 3 that the kinetic energy added to the fluid is EXACTLY the same than the power engine output. Now you are presenting dissipation here. Your statement about viscosity is trivially true, in part because no fluid in this universe has 0 viscosity. The hovering effect is caused by the downwashing jet, but you wouldn't obtain such jet if there were not possible the boundary layer being developed on the blades.

Reynolds number has NOTHING to do with incompressibility. In fact, the Reynolds number of this flow may be of the order of some millions. However, the flow may be regarded as incompressible because the Mach number associated is usually below compressibility limits.

A very rough and mistaken method. I advice you to take a look at the integral conservation laws of fluid mechanics.

10. Dec 27, 2005

### FredGarvin

Clausius,
You just reminded me with your comments on #8 and the Reynolds number. I didn't really think too much about that originally, but this situation could be complicated quite a bit because the Re is going to change drastically as one goes from the root to the blade tip. I wonder if it would be wise to start off with an average Re so as to not get too complicated.

11. Dec 27, 2005

### Clausius2

Anyway, if you take the average Re it would be very large. What I meant is Re has nothing to do with the assumption of incompressibility.
To clarify a little bit the things, I will show anders how to go about the dimensional analysis making some small change:
$$L$$ lift force (N)
$$\rho$$ density (Kg/m^3)
$$\mu$$ viscosity (Kg/ms)
$$D$$ rotor diameter (m)
$$\omega$$ Angular rotor speed (rad/s)
$$U$$ relative velocity of the incoming air stream to the rotor
$$a$$ characteristic sound speed
A wise combination would give:
$$\frac{L}{\rho \omega^2 D^4}=\Phi\left(\frac{\rho \omega D^2}{\mu}; \frac{\omega D}{a}; \frac{\omega D}{U}\right)$$

Where you can see inside $$\Phi$$ (from left to right): Reynolds Number, Mach Number, I don't remeber the name of the last parameter but it has to do with the blade angle of attack.

Actually, Re>>1 and $$M\sim 0.3$$, therefore

$$\frac{L}{\rho \omega^2 D^4}=\Phi\left(\frac{\omega D}{U}\right)$$

is the most suitable relation. The most important factors are in fact the rotor angular speed (which in fact turns out to be a measure of the mass air flow), the rotor diameter, and the internal aerodynamics of the blade (angle of attack and angle of incidence) given by the parameter inside $$\Phi$$.

Last edited: Dec 27, 2005
12. Dec 27, 2005

### FredGarvin

I totally agree with you on the incompressibility part. Your statement had reminded me of another sticking point in the model which is particular only to helicopters. The Re for a wing on a fixed wing aircraft should be pretty constant across the entire span. In the case of the helicopter, that is not true.

I am going to have to pull out my helicopter theory text. I haven't looked at that in a long time. I don't recognize that last dimensionless parameter.

13. Dec 28, 2005

### Clausius2

Hi Fred, you're right because the characteristic velocity varies from the root to the blade tip due to the solid rotation.

At least you are lucky, you have one. I have never studied at college nothing about helicopter dynamics. :rofl:

Think of the kinematic triangle resulting of the composition of the solid rotation $$\omega D$$ and the absolute incoming stream velocity$$U$$ (here I mean absolute as the relative speed of the stream to the rotor as a whole). The ratio of both things is proportional to the tangent of the incidence? angle. Variation of this angle (internal kinematics of the blades) would vary this parameter. And you may remember that the galilean composition of both velocities gives as the relative flow velocity to blade $$W$$. Moreover, $$W$$ is of capital importance because the variation of relative flow kinetic energy will give the variation of Static Pressure through rotor (proportional to power) and so will determine ultimately the blade aerodynamics.

Also that ratio may be interpreted as an Strouhal Number, because it is the ratio of the characteristic time of variation of the periodically unsteady flow ($$1/\omega$$) and the characteristic time of residence of a fluid particle inside the rotor ($$D/U$$).

I remember to have dealt with this parameter when learning wonderful things about how wind turbines work, and in my course of turbomachines.

Last edited: Dec 28, 2005
14. Dec 28, 2005

### anders

I don't understand what you mean by making an integral momentum balance, and what you mean by relative speed. The relative speed between the rotor and the air?
Solenoidality! Hehe, that was a new word for me. I suppose because of the similarity to the shape of the magnetic field around a solenoid?
Are you sure that induced drag is only a factor during acceleration? I thought it could be a factor even at constant velocity?
http://www.grc.nasa.gov/WWW/K-12/airplane/induced.html
Hehe, it seems I only _thought_ I knew what dimensional analysis was :-).

15. Dec 28, 2005

### anders

Do you think that is significant? Does helicopter downwash air heat up?
Ah, good point. Perhaps one could introduce some kind of dimensionless parameter to model this?
Agreed, there is a contradiction. What I meant was that the air would first be set into motion by the rotor, then be blown away far from the helicopter, and that the dissipation into heat would be a slow process, most of it taking place far from the rotor. I meant that I thought that the dissipiation into heat would take place in a very large (I'm guessing hundred of meters) volume of air around the helicopter. In this way, all the power of the engines could be dissipated into heat, without any large increase in air temperature. But I'm not sure about this. Perhaps there really is a significant amount of temperature increase of the air in the immediate vicinity of the rotors?
Ok, off topic, but why could you not? Sure,you would have a very inefficient rotor, but molecules impacting on the underside of the rotor would be kicked downwards, right?
Oops, my ignorance of aerodynamics became obvious. So if compressibility should enter the model is more a function of mach-number than of reynoldsnumber. I didn't know that.

16. Dec 28, 2005

### anders

What is $$\Phi$$ in this equation?

17. Dec 28, 2005

### Clausius2

I mean relative speed as the speed viewed from a reference frame attached to the rotor root. Otherwise the flow is totally unsteady.

Sure. There is some analytical similarity: $$\nabla \cdot \overline{B}=0$$ and $$\nabla \cdot \overline{\omega}=0$$ for the magnetic field and the vorticity field in an incompressible fluid. Wonderful, isn't it?

Here you win and I win also. You're right, there is another "induced drag" due to the vortex sheets generated in a wing, and also there is an "induced drag" due to acceleration. Both of them are real, but I didn't know what you meant.

18. Dec 28, 2005

### Clausius2

No, it is not too much significant. But it exists.
Sorry, I don't know what you mean.

Yes and no. There is some loss of mechanical energy just on the blades and the boundary layer. Such layer is a source of irreversibility, and based on the First Principle $$Tds=cdT$$ it does heat up the air (not too much by the way). The downwashing jet kinetic energy is dissipated far away from the rotor by viscosity (it does not mean it heats up the air).

I don't know what you mean there. For sure there is no fluid with null viscosity, and even assuming potential flow one must assume also some special boundary conditions regarding circulation in order to work properly.

ahahaha! I don't know!. Dimensional analysis does not give you the Functional dependence explicitly. $$\Phi$$ could be measured experimentally. But don't you realise of the power of dimensional analysis which has aided us eliminating two parameters and giving only a single dependence?

Nice to talk to you.

19. Dec 28, 2005

### anders

Off topic, but:
I'm probably just going to show I have a very vague grasp of what viscosity is, but wouldn't an incredibly thin gas behave like a fluid with very very low (near 0) viscosity? After all, if there's hardly any interaction between particles at all, there shouldn't be much viscosity? Perhaps that wouldn't be what is commonly referd to as a "fluid" at all though...
And it would still be possible to hover under such conditions. Just imagine a very very light helicopter, with a large rotor, with broad blades, with a 45 degree angle of attack. The air/gas molecules would bounce off the under-side of the rotor blades, in "elastic collisions" (don't know english term).

20. Dec 29, 2005

### Clausius2

To be sincere, it is an smart comment. You are referring to a Rarefied Gas, with very low density and viscosity. I am not an expert in this stuff, but I doubt pretty much it would generate lift. My view of the fundamentals of lift generation in an helicopter is the next: The rotor is comunicating momentum (via viscosity, otherwise the flow would slip) to fluid. The momentum balance says that such momentum increasing must be balanced with an increasing of pressure. Therefore the bulk of fluid behind the rotor is dynamically pressurized, comunicating such extra dynamic pressure to the rotor and the helicopter structure. In a rarified gas this wouldn't be possible, because of the dispersal of molecules. You wouldn't have the transmission of pressure waves as I have described, and the motion induced by the rotor to the fluid would seem more random. Even talking in an engineering point of view, it is not realizable at all (you would obtain little-null lift for a very heavy helicopter).