Can someone confirm this power to thrust relation?

  • #1

Main Question or Discussion Point

This is not my homework this is just a random thought. Say you have a 1 kg helicopter hovering in midair. The question is, given perfect efficiency, how much power is it using to hover?

My approach is to think of the given information first. We know that g= 9.8. So you have a 1 kg mass that requires 9.8 N of thrust to remain hovering. Obviously there is no energy being added to the helicopter since it is standing still but there is some energy being used every second to move a certain volume of air. Let's assume that the blades are made so that they move 1 kg of air every second. So the next question is with what velocity must it blow that air to levitate? If the helicopter is in free fall then in 1 second it will attain a velocity of 9.8 m/s. It would have to blow 1 kg of air at 9.8 m/s to stop falling but it would have to blow it away at 19.6 m/s to get to the original spot. Using the equation for kinetic energy the helicopter uses 192 Joules of energy every second to hover. This of course is just 192 Watts of power. So in general 9.8 N of thrust in any situation is going to require 192 Watts. Is that correct? What is the equation that would generalize this calculation?
 

Answers and Replies

  • #2
247
1
Actually, what you have is a bit more complicated. Because the helicopter is not in motion, there is no work. Instead, the work is being done on the air by the blades to create lift, a net force, and this force equals the force of gravity.

L =1/2 * ρ * v^2 * A * CL

where

* L is lift force,
* ρ is air density
* v is true airspeed,
* A is planform area, and
* CL is the lift coefficient at the desired angle of attack, Mach number, and Reynolds number.

Another way to think of this would be to take the same helicopter and suspend it from a string. The string exerts a force equal to the mass * the force of gravity, but no work is done, and no power is gained or lost by the system.

Obviously the helicopter requires power to hover, and obviously the amount of power required is a function of its mass, but calculating the amount of power required to make it hover requires solving the above equation.

An interesting example of hovering w/o work involves a superconductor in the presence of a permanent magnet or a diamagnetic substance (water, for example) in the presence of a very high magnetic field (typically 16 Teslas or greater). Here, like the case of a string, no work is done, and yet the object "hovers".

Fish
 
  • #3
247
1
Actually, what you have is a bit more complicated. Because the helicopter is not in motion, there is no work. Instead, the work is being done on the air by the blades to create lift, a net force, and this force equals the force of gravity.

L =1/2 * ρ * v^2 * A * CL

where

* L is lift force,
* ρ is air density
* v is true airspeed,
* A is planform area, and
* CL is the lift coefficient at the desired angle of attack, Mach number, and Reynolds number.

Another way to think of this would be to take the same helicopter and suspend it from a string. The string exerts a force equal to the mass * the force of gravity, but no work is done, and no power is gained or lost by the system.

Obviously the helicopter requires power to hover, and obviously the amount of power required is a function of its mass, but calculating the amount of power required to make it hover requires solving the above equation.

An interesting example of hovering w/o work involves a superconductor in the presence of a permanent magnet or a diamagnetic substance (water, for example) in the presence of a very high magnetic field (typically 16 Teslas or greater). Here, like the case of a string, no work is done, and yet the object "hovers".

Fish
 
  • #4
757
0
in the limit of infinitely long blades and infinitely slow blade rotation, there will be no power required to hover.

in the limit of zero-length blades and infinitely high RPM, the power required is infinite.

This problem is not hard, and you had the right idea on approaching it. The helicopter needs a certain impulse to stay aloft, and the lower the energy you give the air (by moving larger volumes of air) the more "efficient" you are.
 
  • #5
563
2
Actually, what you have is a bit more complicated. Because the helicopter is not in motion, there is no work. Instead, the work is being done on the air by the blades to create lift, a net force, and this force equals the force of gravity.
Wouldn't it be correct to calculate the work being done on the air? The air is definitely in motion and is the medium that the force is acting on.
 
  • #6
803
9
As has been said, you can always make the blades longer and increase the mass of air being accelerated to gain efficiency. 100% efficiency implies infinitely long blades and/ or infinite amount of mass being accelerated infinitely slowly (0). This means NO power is being used.

You can approximate this by simply resting the helicopter on the ground.

If you want to assume 1kg of air you will be far from 100% efficient.
 
  • #7
AlephZero
Science Advisor
Homework Helper
6,994
291
So the next question is with what velocity must it blow that air to levitate? If the helicopter is in free fall then in 1 second it will attain a velocity of 9.8 m/s. It would have to blow 1 kg of air at 9.8 m/s to stop falling
That is the right idea. Force = rate of change of momentum. (see * below for more).

but it would have to blow it away at 19.6 m/s to get to the original spot.
I don't understand that. You need a force greater than 9.8N to make the helicopter climb. But you don't need a force equal to 19.6N.

* Your energy calculations seem to be wrong. In 1 second you are blowing 1 kg of air at 9.8 m/s so the KE is (1/2).1.9.8^2 = 48.02J. Therefore the power needed to do that is 48.02W.

However you can get the same force by blowing more air per second at a lower velocity. For example if you blow 2 kg/sec at 4.9 m/s, the change in momentum of the air (and therefore the lift force) is the same, but the power is only (1/2).2.4.9^2 = 24.01W.

This is why helicopters have big rotors. It is more efficient to produce thrust and lift by moving a lot of air slowly, rather than a small amount of air fast.

@Fish4fun: you certainly need to consider the lift and drag equations to design the shape of the rotor blades to move the required amount of air, and work out the real power requirement (not assuming 100% efficiency) but you don't need to do that to answer the OP's question.
 
  • #8
rcgldr
Homework Helper
8,682
518
That is the right idea. Force = rate of change of momentum. ... velocity
One isssue here is what do you use for "velocity" of the affected air. A helicopter takes more power to hover than one moving forwards (at sufficient speed), because of the induced downwards flow of air above the rotor accelerating towards the lower pressure zone above the rotor, which increases the speed of the air flowing into the rotor wash. The rotor increases pressure without much change in velocity of the air (called pressure jump zone in the case of propellers), which then continues to accelerate downwards due to momentum and having higher than ambient pressure. One way to estimate the power involved is to note the "exit velocity" of the affected air, which it's average velocity when and where it's pressure returns to ambient. If I remember correctly, the power equals 1/2 (mass flow rate) (exit velocity)2.

Still the main point already mentioned is a larger rotor will be more efficient, involving a larger mass of air and a lower exit velocity.
 
  • #9
247
1
@Fish4fun: you certainly need to consider the lift and drag equations to design the shape of the rotor blades to move the required amount of air, and work out the real power requirement (not assuming 100% efficiency) but you don't need to do that to answer the OP's question.

AlephZero
Obviously you are right. I was caught up thinking about blade efficiency which has nothing to do with the OP's assertion of 100% efficiency.

Fish
 
  • #10
sophiecentaur
Science Advisor
Gold Member
24,425
4,403
I think there is actually NO answer to this question. All you can say is that the bigger the area of the rotor, the less work needs to be done on pushing air downwards in order to hover.
 
  • #11
If I remember correctly, the power equals 1/2 (mass flow rate) (exit velocity)2.
That is something like what I got. The units do work out as well. I guess the problem is I asked the question using a helicopter as an example. This is problematic because it's connected to its surroundings via its blades and thus can have arbitrarily high efficiency. I was originally thinking of the problem in the context of a rocket in a vacuum. I wanted to know what kind of efficiency solid state propulsion systems such as ion thrusters or nanoFETs have in order to know how efficient they must become to be able to hover the mass of the engine and fuel at Earth's surface gravity. So I guess the place to start would be given x amount of energy, what's the maximum amount of thrust one can produce by emitting mass in a vacuum?
 
  • #12
sophiecentaur
Science Advisor
Gold Member
24,425
4,403
Even with a rocket you would still have to decide on the rate of ejection of mass and the ejection speed. That would also affect the rate of energy expenditure. It would imply that the efficiency would be higher for a short hover time (i.e. ejected mass all used up quicker).
 

Related Threads on Can someone confirm this power to thrust relation?

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
17
Views
3K
Replies
9
Views
2K
  • Last Post
2
Replies
28
Views
6K
Replies
7
Views
981
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
9
Views
5K
Replies
5
Views
564
Replies
1
Views
655
Top