Force & Energy Changes of Air by Helicopter

[phantomvommand]

Homework Statement

I am trying to find the power generated by a helicopter blade spinning, sweeping out a circular area A each time, and causing the air below to flow uniformly at speed v. The density of air is $\rho$

Relevant Equations

Power = Fv
Power = Energy/time

Solution 1:
Force on air = Force on helicopter.
$F = \frac {dp} {dt} = v \frac {dm} {dt} = v^2 \rho A$
$P = Fv = \rho A v^3$

Solution 2:
Power from helicopter converted to rate of kinetic energy increase of air
$P = \frac {dE} {dt} = \frac {1} {2} v^2 \frac {dm} {dt} = \frac 1 2 \rho A v^3$

I am almost certain solution 1 is the correct solution. This appears similar to the sand falling on conveyor belt problem. The energy loss in the 'sand' problem comes from the fact that sand particles have to be sped up to reach speed v. In this case, is there energy loss due to air friction between air particles? And so energy is lost by speeding air particles up to speed v. This is quite hard to imagine, given how little friction between air particles there should be.

Thanks for all the help.

 

[andrewkirk]

Solution 1 can't be correct. When the blade first strikes an air particle, on average the particle's downward speed is zero, not v as the power formula in solution 1 suggests.

Consider the short period of contact between the blade and an air particle with zero initial vertical velocity.
Fv is zero immediately after the beginning of that period and is $\rho Av^3$ (summed over all particles in the blades' sweep) just before the end. So we'd expect the average power to be halfway between those. That gives us solution 2.

 

[haruspex]

Solution 1 can't be correct.

Looks ok to me. Just momentum conservation.

 

[phantomvommand]

Looks ok to me. Just momentum conservation.

Then what is wrong with what @andrewkirk said?
The v in P =Fv is actually dx/dt, which will give you 1/2 of final velocity v, so I am now inclined to go with solution 2.
And also solution 2? Thanks for the help :)

 

[haruspex]

Then what is wrong with what @andrewkirk said?
The v in P =Fv is actually dx/dt, which will give you 1/2 of final velocity v, so I am now inclined to go with solution 2.
And also solution 2? Thanks for the help :)

Well, it could be right. Momentum conservation only gets us as far as the force being $\rho Av^2$.
But I do not see the air as being smoothly accelerated. Each tranche is given a large impulse in a short time. No doubt the blade profile helps to smooth it a bit, but a lot of energy is going to end up as oscillations - evidenced by the noise.
The true answer is probably somewhere in between.

 

[jack action]

In solution # 1, you assume velocity is constant and the mass isn't. In solution #2, you assume a mass $\Delta m$ accelerates from $0$ to $v$ during an interval $\Delta t$.

Solution #1:
$$dE = Fdx = \left(\frac{dm}{dt}v + m\cancelto{0}{\frac{dv}{dt}}\right)dx = \frac{\rho Avdt}{dt}vdx = \rho Av^3dt$$
$$\Delta E = \rho Av^3 \Delta t$$
$$P = \frac{dE}{dt} = \rho Av^3$$
Solution #2:
$$dE = Fdx = \left(\cancelto{0}{\frac{dm}{dt}}v + \Delta m \frac{dv}{dt}\right)dx =\Delta m \frac{dx}{dt}dv = (\Delta m) vdv$$
$$\Delta E = \frac{1}{2}(\Delta m) (v^2 - (0)^2)$$
$$P = \frac{\Delta E}{\Delta t} = \frac{1}{2}\frac{\Delta m}{\Delta t} v^2 = \frac{1}{2}\rho Av^3$$
But the correct way is to imagine a control volume around your blade and you will find out that actually $P = 2\rho Av_i^3 = \frac{1}{4}\rho A w^3$, depending which velocity you imply by "causing the air below to flow uniformly at speed v".

 

[phantomvommand]

In solution # 1, you assume velocity is constant and the mass isn't. In solution #2, you assume a mass $\Delta m$ accelerates from $0$ to $v$ during an interval $\Delta t$.

Solution #1:
$$dE = Fdx = \left(\frac{dm}{dt}v + m\cancelto{0}{\frac{dv}{dt}}\right)dx = \frac{\rho Avdt}{dt}vdx = \rho Av^3dt$$
$$\Delta E = \rho Av^3 \Delta t$$
$$P = \frac{dE}{dt} = \rho Av^3$$
Solution #2:
$$dE = Fdx = \left(\cancelto{0}{\frac{dm}{dt}}v + \Delta m \frac{dv}{dt}\right)dx =\Delta m \frac{dx}{dt}dv = (\Delta m) vdv$$
$$\Delta E = \frac{1}{2}(\Delta m) (v^2 - (0)^2)$$
$$P = \frac{\Delta E}{\Delta t} = \frac{1}{2}\frac{\Delta m}{\Delta t} v^2 = \frac{1}{2}\rho Av^3$$
But the correct way is to imagine a control volume around your blade and you will find out that actually $P = 2\rho Av_i^3 = \frac{1}{4}\rho A w^3$, depending which velocity you imply by "causing the air below to flow uniformly at speed v".

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Thanks for the very informative response. May I know why $T = w \dot m$? Why not $T = v_i \dot m$? In the control volume method, will the power change based on whether mass/volume is kept constant, like how solution 1 and 2 yield different results?

 

[jack action]

May I know why $T = w \dot m$? Why not $T = v_i \dot m$?

Because you consider the velocity where the airflow crosses the control volume.

will the power change based on whether mass/volume is kept constant, like how solution 1 and 2 yield different results?

What mass or volume are referring to?

 

[phantomvommand]

Because you consider the velocity where the airflow crosses the control volume.

What mass or volume are referring to?

Sorry, I realized the control flow method gives a more careful treatment than assuming constant velocity (in soln 1) or constant acceleration (in soln 2).
In this control flow method, is more air constantly being added to the flow tube? I am referring to the following link:
https://www.spinningwing.com/the-helicopter/momentum-theory/
It suggests that there is an increase in momentum of system as the flowtube lengthens. What is really happening here? I know momentum has to increase because there is a net force, but how?

 

[jack action]

What is done in your link, is that they imagine there is a constant area tube at the end of our control volume filled with moving air. The length of that imaginary tube is $vdt$, i.e the length filled by the moving air during time $dt$. So the mass of air in that tube is $m = \rho V = \rho A vdt$, leading to $dp = d(mv) = \rho Av^2dt$, thus $\frac{dp}{dt} = \rho Av^2$.

It is another way to approach the problem and help visualize the phenomenon described more mathematically in the link I presented.