# B Energy and power

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1. Mar 27, 2016

### Thom_Silva

Hi,

Imagine this situation:

I'm pushing against a wall, and i obviously making a force. The wall doesn't move, so no work is being done, nevertheless i wasted energy. Is there way to calculate how much energy was spent assuming that we know the force being applied to the wall.

Another problem related to this:

I'm building a small plane that has to lift a certain weight and stay hovering in the same place. I have to choose an engine to do so, and all i have are the power outputs of each engine. When the plane is hovering is there any work being done ? The engine will clearly be using energy, but apparently no work is being done, because the plane stays in the same place. How am i supposed to choose an engine that could do that activity?

Thank you very much, and sorry for the stupid question xD

2. Mar 27, 2016

### Simon Bridge

Not really - the energy being expended is in the muscles rather than in the wall and you basically have to measure it. Technically you can model the muscle system and use the characteristics iof the model to work out energy expended due to exertion. I think you need ergonomics - a branch of biology.
The trick with this sort of calculation is always which work you are talking about.

The trick, once again, is to identify which work you are talking about
- the engine clearly is doing work, but it is not doing work on the weight (i.e. against gravity)
If the engine is powerful enough to lift the weight into position, then it will use less energy to hold it in that position.
To figure out where the work is going, look at how the plane generates lift.
A helecopter will do work pushing the blades against air resistance, accelerating the air downwards to generate lift, and also work against friction in the moving parts, make a noise, and just heating everything up. All this without otherwise accelerating or changing height.

The kind of work involved in moving stuff about is mechanical work ... there are other ways to use energy.

3. Mar 27, 2016

### Thom_Silva

What you are explaining me leads me to the concept of thrust. Is there any relation between thrust and power?

4. Mar 27, 2016

### Staff: Mentor

When using air to produce thrust, you can relate the force to the work done on the air (force times distance over time is also pressure times volumetric flow rate).

5. Mar 27, 2016

### Thom_Silva

Yes, i got that. My problem is that i have a electric engine but i only have the power outputs. I will attach it to propellers, how can i know the amount of thrust the engine will provide ? I guess i'll do this by trial an error xD

6. Mar 27, 2016

### Staff: Mentor

I'm not sure you can equate power to thrust without knowing the characteristics of the propeller.

7. Mar 27, 2016

### Staff: Mentor

I just described in words how to do the calculation. There are some intricacies to it, but a first approximation should be reasonable.

How big are the propellers? Do you know what their characteristics are (pitch? Efficiency?) What is the power output of the motor? At what RPM? What can you put together from that information based on what I have told you?

8. Mar 27, 2016

### Staff: Mentor

We'll see what information the OP has, but with power and diameter (and, say, 50% efficiency), you can calculate thrust. The pitfalls would come into play if the propeller doesn't well match the motor: rpm and torque = power for the motor, rpm and pitch = power for the propeller. If they don't have the same power at the same rpm, they'll have to find a different operating point.

9. Mar 27, 2016

### rootone

1. Yes work is being done but almost all of it is your body converting chemical energy into heat, the wall will experience some small amount of pressure which might contribute to structural change so it's eventual disintegration at some point in the future might happen sooner.
2. The engine is providing sufficient power to exactly balance gravity, so the 'plane' is stationary', that means it is doing work.

(However, sitting in your chair has the same effect)

10. Mar 28, 2016

### Simon Bridge

I see you got some help ...

Usually an electric engine has power inputs: in a DC motor that is two leads that are attached to a power source.
The "output" is usually an axle that you attach the propeller to.

How much thrust you can get from a given engine depends mainly on the propeller and how it is set up.
Without that information there is just no way to tell.

The maximum thrust would be if the entire max-rated power input to the motor was converted to thrust: this would require a magic lossless conversion.
... I don't recall offhand typical efficiencies for converting power to thrust using a prop though. But see resources like this:
http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node86.html

To help properly, we'd need the details of your project. There should be enough here now to get you going.

11. Mar 28, 2016

### Limmin

Just curious, I wonder if you could get an approximation of the mechanical work done by the propeller (when under load) by measuring the voltage and amps supplied by the electrical controller. Clearly some would be lost to internal friction / conversion efficiencies, etc.

12. Mar 28, 2016

### rootone

Different propeller characteristics would be more efficient in some cases than others, a ducted fan would be the most efficient, flat blades instead of aerodynamic ones would work,but would be very inefficient..
I doubt that there could be some kind of 'works for all cases' type of calculation that would be reasonably accurate..

13. Mar 28, 2016

### David Lewis

Simon Bridge wrote: "If the engine is powerful enough to lift the weight into position, then it will use less energy to hold it in that position."

David Lewis wrote: You're absolutely correct but, ignoring ground effect, the power needed to climb slowly usually turns out to be roughly the same as the power needed to hover. How much energy you need to hover depends on how long you want to do it.

The easiest way to determine the thrust of a propulsion system is refer to the manufacturer or vendor performance tables.

14. Mar 29, 2016

### Simon Bridge

Note:
The power needed to lift can, in principle, be made arbitrarily small... just by lifting very very slowly.
You have very little control over the power needed to keep the blades turning.
Thus you can, in principle, lift so slowly that the power to lift is negligible compared with that to turn the blades.
Depending on the prop, and the load, that can be quite easy to do.

You could measure the increased power drain under load to get an idea of what power is used turning the prop, lifting, etc but not sure how that helps OP.
You can also measure the efficiency of the motor under different loads by getting it to lift weights through a known height.
You can also fix the motor+prop to a test-bed using a newton-meter and just measure the thrust for different input power.

15. Mar 29, 2016

### David Lewis

Simon Bridge wrote: "The power needed to lift can, in principle, be made arbitrarily small... just by lifting very very slowly."

David Lewis wrote: You're correct. Say you have a crane with a motorized winch. When the vertical speed of the weight is zero, power consumed goes to zero. For a helicopter, it changes slightly. Even if your vertical speed is zero, you still need power to lift.

In a free-body diagram of a hovering helicopter, lift equals weight. When the helicopter ascends, lift still equals weight. Nothing much changes in terms of power required. (We're ignoring initial acceleration, accumulation of potential energy, aerodynamic drag of the helicopter and advance ratio of the rotor in this simplified calculation.)

You can prove this by noting the difference in throttle setting when a hovering helicopter begins a slow ascent. You'll probably find that the difference is small.

Last edited: Mar 29, 2016
16. Mar 29, 2016

### Simon Bridge

When the helecopter ascends at a constant speed, then lift equals weight.
From stationary, then you need to increase power to accelerate to the desired lift speed... the slower the desired lift speed the smaller the power requirement to accelerate to it.

17. Mar 29, 2016

### David Lewis

That's an important point. My free-body analysis is only valid when the helicopter is in equilibrium (at rest or moving at constant speed).

18. Mar 29, 2016

### Simon Bridge

Don't forget, even though the force is the same, the motion is different.
If the copter rises at constant speed v, it uses additional energy at rate P=mgv ... when hovering, v=0.

19. Mar 29, 2016

### David Lewis

Thank you, Simon. I had been wondering about that. Another key fact I overlooked.

20. Mar 29, 2016

### Staff: Mentor

I think David's previous point was that the helicopter requires a certain input power to move a certain amount of air to generate the lift required to keep itself aloft. The additional power to make it rise can be very small compared to the power needed to make it hover.

21. Mar 29, 2016

### Simon Bridge

That was my take also. Please see post #14 ...

22. Mar 30, 2016

### David Lewis

Right, but one paradox left. I assume in order for a rotor to produce more thrust, it must be fed more power. However, it looks like whether the helicopter is hovering or climbing, thrust is the same. So only thing I can figure (for a given amount of thrust) is the rotor unavoidably consumes more power when it's moving than when it's static.

23. Mar 30, 2016

### jbriggs444

If I understand the concern correctly, relevant notions are "parasitic drag" and "induced drag". While hovering in place [and assuming negligible downwash velocity], the drag on the rotor is primarily parasitic. In order to climb, the relative downwash speed must increase. This increases induced drag.

24. Mar 31, 2016

### Staff: Mentor

The way I would say it is that when the helicopter is hovering, it accelerates the air from stationary to whatever velocity (momentum) is needed to provide the lift. But if the helicopter is rising, the air is already moving past it and therefore requires more energy to add the same amount of velocity.

This is the same as why even without friction, accelerating an object at constant acceleration requires linearly increasing power.