Power to keep helicopter hovering

[walking]

Homework Statement

Find the power required to keep a helicopter hovering

Relevant Equations

P=Fv

My thinking was that the power produced by a falling object is P=Fv=(-mg)(-gt)=mg^2t. So it depends on t, ie the "counter power" must also depend on t if it is to balance it, right? But author's solution is a constant. I know I am making a mistake somewhere.

Note: Apologies for the lack of working out. I took these questions down a while ago and am posting them one at a time to avoid spamming. I will include working out for future questions once I get through this "batch".

 

[kuruman]

Still, according to our rules, you need to post your best effort. What is the origin of the force that holds the helicopter up against the force of gravity? Start from there.

 

[haruspex]

the power produced by a falling object

The air is not falling under gravity from rest. It is being accelerated down by the blades from rest to a nonzero velocity in a negligible time.

you need to post your best effort

As I read the post, @walking is only asking what is wrong with the approach described.

 

[kuruman]

As I read the post, @walking is only asking what is wrong with the approach described.

Perhaps I misunderstood what OP meant by "lack of working out". If that is the case, I apologize for my remark. Nevertheless, I attempted to redirect OP's thinking away from a free fall approach.

 

[anuttarasammyak]

A body needs no power but force or momentum gain to keep it airborne. You may know it by airborne magnet by superconductivity or a model plane hanged by threads from the ceiling. No energy is consumed there. Helicopter takes the way to get momentum by pushing air. Thus pushed air have kinetic energy, so by energy conservation law helicopter has to consume energy or power.

 

[Lnewqban]

Homework Statement:: Find the power required to keep a helicopter hovering
Relevant Equations:: P=Fv

View attachment 279993
My thinking was that the power produced by a falling object is P=Fv=(-mg)(-gt)=mg^2t. So it depends on t, ie the "counter power" must also depend on t if it is to balance it, right? But author's solution is a constant. I know I am making a mistake somewhere.

The problem that I see with your thinking is the following:
You are calculating the power of impact of the helicopter against the ground, after free-falling for certain time (no air resistance) or from certain height, which is the same.
Of course, that velocity of impact would depend on the falling time or height.

Nevertheless, your helicopter is not falling; it is instead trying to climb higher and higher in a medium of air, which keeps moving downwards because it is not a very solid substance to cling to.

In reference to your earlier thread, the situation of your helicopter is somehow similar to the situation of a man trying to climb a moving stair in the wrong direction at the same speed the steps are moving down: he is not moving vertically from its position.

The only way the hovering helicopter can counteract the relentless effect of gravity is by inducing a force of equal magnitude and opposite direction to its own weight.
The blades do that by changing the momentum of an endless cylinder of air.
Note that that effect does not depend on the height of the helicopter or the time it would take for the helicopter or the mass of air to reach the ground.

The blades move a certain amount of mass of air by unit of time; therefore, they induce a mass flow rate, which depend on the density of the air, its change in velocity moving down from the blades, and the area covered by the blades disk.

The required power to make all the above happens can be estimated by calculating how much kinetic energy that mass of air gains per unit of time.

Please, see:
https://www.grc.nasa.gov/www/k-12/airplane/thrsteq.html

http://hyperphysics.phy-astr.gsu.edu/hbase/ke.html#c3

Air mass flow rate:
$\dot m=\frac {dm}{dt}=\rho VA_{disk}$

$E_k=\frac 1 2 mV^2$

$Power=\frac 1 2 \dot mV^2=\frac 1 2 \rho A_{disk}V^3=\frac 1 2 \rho \pi r_{blades}^2V^3$

 

[walking]

Still, according to our rules, you need to post your best effort. What is the origin of the force that holds the helicopter up against the force of gravity? Start from there.

Yeah, sorry about that. Since I'm posting one at a time now (to avoid spamming) and taking time to read responses etc I figured it won't take that much extra to rework the problems, so that is what I'll be doing from now on.

The origin of the force is the blades pushing down the air. Of course! I don't know why I was thinking about free fall which is a completely different scenario.

The air is not falling under gravity from rest. It is being accelerated down by the blades from rest to a nonzero velocity in a negligible time.

I was imagining the plane falling not the air. My thinking was that the plane is constantly falling and this is the only motion that needs to be opposed to keep it hovering. So we just need to provide a "counter power" that is equal to the power generated by the falling plane.

I think I know why this approach is wrong now: I was confusing force and power. The force is mg regardless of how the power is generated, yes, but here power is generated by pushing air down not by free fall. So we need to calculate the power from that scenario rather than imagining the plane falling down. At least I think that was my mistake (as in my reply to kuruman)?

The problem that I see with your thinking is the following:
You are calculating the power of impact of the helicopter against the ground, after free-falling for certain time (no air resistance) or from certain height, which is the same.
Of course, that velocity of impact would depend on the falling time or height.

Nevertheless, your helicopter is not falling; it is instead trying to climb higher and higher in a medium of air, which keeps moving downwards because it is not a very solid substance to cling to.

Yep, makes sense now.

In reference to your earlier thread, the situation of your helicopter is somehow similar to the situation of a man trying to climb a moving stair in the wrong direction at the same speed the steps are moving down: he is not moving vertically from its position.

I think I understand this but isn't the man walking up stairs that are already moving downwards, whereas the plane is pushing air that is static?

Thank you so much for the detailed explanation and link by the way, I finally managed to solve it:

I didn't directly use the downward speed to get mass flow rate etc but same thing. Basically I assumed that one "circular sweep" of air is moved down L in time t. So the mass flow rate is ALp/t where A is the area of the circle and the change in momentum is (ALp/t)(L/t). I guess I avoided speed just to make to make it clearer to understand due to being new to mass flow rate. I then reintroduced speed here (which is just L/t) to get Apv^2. This is the force and we need $Apv^2=mg$ or $v=\sqrt{\frac{mg}{Ap}}=13.922$. Then $P=Fv=mgv=2.4831\cdot 10^5=2.5\cdot 10^5 W$.

 

[Lnewqban]

Yes, the example of the man climbing the moving stairs was not the best; I just tried to show that something had to be moving downwards for the helicopter to keep a spatial position.
I am glad that you could understand the whole problem.

I just want to note that the calculated power is not all the power the engine of our helicopter must generate in real life: the process of generating lift at the blades consumes more energy that does not directly translate into change of momentum of the air.
Also, the ideal cylinder of air does not take that shape in real life, as some amount of air goes back up around the tips of the blades, driven by the area of low pressure that is located above the blade's disk.