Helicopter air projection velocity problem

1. Jun 17, 2013

FBayer

So, I got a challenge question from my physics teacher but despite attacking it from different angles I’m not quite sure I’ve gotten the correct solution, in fact most of it was just shady assumptions which I’m not sure were supposed to be made, so I’d like to ask whether this is anywhere near correct and, if it isn’t, be provided with some pointers as to how to do this.

A helicopter hovers in still air well above the ground. If the helicopter has a mass of 4000 kg, and rotor blades of diameter 10 m, what is the velocity of air projected downwards by the blades? (g = 9.8 ms-1, density of air = 1.3 kgm-3)

So, first thing I modelled the hovering helicopter as a cylinder. With blade diameter 10 (= radius of 5), the covered area is 25π, or roughly 78.5 m².

Now, we know that F = ṁv where ṁ is flow rate and v is velocity, as proven by the units: kgs-1 × ms-1 = kgms-2 = N.

The flow rate can be described as m/t, and as m = ρV where ρ is density and V is volume, flow rate ṁ = ρV/t.

We can substitute this into our earlier equation while rearranging, such that:

$\vec{v} = \frac{\vec{F}}{\dot{m}} = \frac{\vec{F}}{\frac{\rho V}{t}} = \frac{\vec{F}t}{\rho V}$

And here’s where it gets shady. I don’t know how to get a volume if the height isn’t given, and the same deal goes for time, so I just assume they're all 1.

So, 4000 kg × 9.8 ms-2 = 39200 kgms-2, multiplied by 1 second is 39200 kgms-1.

Area, ~78.5m², multiplied by 1 m = 78.5m³.

Now:

$\frac{39200\: kg\, m\, s^{-1}}{78.5\: m^{3} \times 1.3\: kg\, m^{-3}} = \frac{39200\: kg\, m\, s^{-1}}{102.05\: kg} = 384\: ms^{-1}$

However, I made the assumptions that I can just use the area for the volume and the force for what is essentially momentum/impulse (Ft), which leads me to believe I probably did it wrong...so, how's it supposed to be done?

2. Jun 17, 2013

consciousness

You don't know the height, time but there is a simple relation between them in terms of a variable we know.

3. Jun 17, 2013

FBayer

Is there? Let's look at the units.

m = kg
g = ms-2
r = m
ρ = kgm-3

I don’t see any second unit involving time here, so there’s no way to get to ms-1, which is what we want – velocity.

4. Jun 17, 2013

Filip Larsen

You shouldn't have time t in your rate equations. But before you continue further you may want to search your references for "helicopter rotor wake" diagrams to learn that the control volume is not a cylinder and that you need a few more equations to arrive at a realistic result.

In simple rotor models it is usually assumed that the speed of the air mass far upstream (above) the rotor is effectivly zero, which together with conservation of mass, energy and momentum across different sections of the control surface allows one to derive the rotor thrust as a function of induced air speed at the rotor. I know this is probably something of a mouthful, but see what you can make of it and then return here if you have specific questions.

5. Jun 18, 2013

FBayer

Hmm, I've tried finding some materials on that...are you sure that level of complication is necessary?

Because this ("Evaluation of the Induced Velocity Field of an Idealised Rotor") talks about things like Taylor series, vector fields and vorticity, which are all things we haven't even come close to doing.

I'm doing A2 Physics (= US Grade 12), and this looks to me like undergraduate-level engineering, right? Now this question was set for Wednesday on last Wednesday along with a few other, easier questions (which I could do), and I can't imagine he'd set us something where we have to learn about an undergrad engineering topic over 7 days. Might he perhaps have missed out some variable or phrased something wrong?

6. Jun 18, 2013

consciousness

In

$\vec{v} = \frac{\vec{F}}{\dot{m}} = \frac{\vec{F}}{\frac{\rho V}{t}} = \frac{\vec{F}t}{\rho V}$

$Volume=Area * Height$

$Height=Velocity*Time$

Time gets cancelled and we can find the velocity. I think this is what you were supposed to do.

Last edited: Jun 18, 2013
7. Jun 18, 2013

rcgldr

It's a function of the induced air speed at the rotor, but at the rotor the air speed is about 1/2 the "exit" speed of the air as most of the work done by the rotor is an increase in pressure with little increase in speed of the incoming induced wash. Wiki article for propellers, which is similar if the propeller is in a static situation instead of moving forwards:

http://www.grc.nasa.gov/WWW/K-12/airplane/propanl.html

8. Jun 18, 2013

Filip Larsen

Since the problem is posed as homework I thought it prudent not to serve the answer on a platter. Anyone with a disposition for reading the "answer" can also find a relevant wikipedia page that derive the relationship between rotor thrust and induced air speed. Btw, I read the boldfaced problem in the first post as asking for a relationship with induced air speed, i.e. the air speed just below the rotor.

9. Jun 18, 2013

FBayer

Right, okay, I figured it out today:

First reformulate F as change in momentum (assuming v is constant)

$F = \frac{\mathrm{d} p}{\mathrm{d} t} = \frac{\mathrm{d} m\vec{v}}{\mathrm{d} t} = \frac{\mathrm{d} m}{\mathrm{d} t}\vec{v}$

Substitute available variables into dm/dt.

$\frac{\mathrm{d} m}{\mathrm{d} t} = \frac{\mathrm{d} \rho V}{\mathrm{d} t} = \frac{\mathrm{d} V}{\mathrm{d} t}\rho$

Recognise the following equivalence (where x1, x2, x3 are side lengths of a cube).

$\frac{\mathrm{d} V}{\mathrm{d} t} \equiv \frac{x_{1}x_{2}x_{3}}{t} \equiv x_{1}x_{2} \frac{\vec{x}_{3}}{t} \equiv A\frac{\vec{x}_{3}}{t} \equiv A\vec{v}$

Expand A.

$A = \pi r^{2}$

Therefore,

$\frac{\mathrm{d} V}{\mathrm{d} t} \rho = A \vec{v} \rho = \pi r^{2} \vec{v} \rho$

Recall that (dV/dt)*ρ = (dm/dt) and p = mv. Thus,

$F = \frac{\mathrm{d} p}{\mathrm{d} t} = \frac{\mathrm{d} m}{\mathrm{d} t} \vec{v} = \left (\frac{\mathrm{d} V}{\mathrm{d} t} \rho \right ) \vec{v} = (\pi r^{2} \vec{v} \rho) \vec{v} = (\pi r^{2} \vec{v}\, ^{2} \rho)$

Rearrange to make v the subject.

$\vec{v} = \sqrt{\frac{F}{\pi r^{2} \rho}}$

Substitute F = mg.

$\vec{v} = \sqrt{\frac{mg}{\pi r^{2} \rho}}$

Now this is entirely comprised of known variables:

$\vec{v} = \sqrt{\frac{mg}{\pi r^{2} \rho}}$

And the numerical solution:

$\vec{v} = \sqrt{\frac{4000 \, \mathrm{kg}\times 9.8 \, \mathrm{ms}^{-2}}{\pi \times (5\, \mathrm{m})^{2} \times 1.3 \, \mathrm{kg}\, \mathrm{m}^{-3}}} \approx \sqrt{\frac{39200 \, \mathrm{kg} \, \mathrm{ms}^{-2}}{78.5 \, \mathrm{m}^{2} \times 1.3 \, \mathrm{kg}\, \mathrm{m}^{-3}}} \approx \sqrt{\frac{39200 \, \mathrm{kg} \, \mathrm{ms}^{-2}}{102.1 \, \mathrm{kg}\, \mathrm{m}^{-1}}} \approx \sqrt{384 \, \mathrm{m}^{2}\mathrm{s}^{-2}} \approx 19.6 \, \mathrm{ms}^{-1}$

Done.