# Is the movement of a turbine itself a loss of efficiency?

1. Mar 14, 2016

### sean39

Is the movement of the turbine itself a loss of efficiency since that is energy that is not being converted into other (e.g., electrical) energy?

For example, if a wind turbine is not hooked up to anything, it will spin freely, but none of the air's kinetic energy is converted into electrical or other forms of energy. If the turbine is hooked up to a weak magnet, some of the kinetic energy is converted, and the resistance of the magnet causes the turbine to slow a little. If the turbine is hooked up to a strong magnet, even more of the energy is converted and the turbine slows a lot. So is a stronger magnet that hardly allows the turbine to move at all the most efficient design? Or is there a sweet spot where the resistance of the magnet is "just right"? Since I see lots of windmills spin fairly fast, I assume I am thinking about this all wrong.

2. Mar 14, 2016

### Staff: Mentor

There is a sweet spot, but it is more about what the turbine is doing to the wind than the other way around. A turbine spinning at constant rpm isn't consuming energy to remain spinning, except for the near zero friction in the bearings. The issue is that to extract energy from the wind, you have to slow the air down, which makes it pile up and go around the turbine instead of through it. Google Betz's law.

Last edited: Mar 14, 2016
3. Mar 14, 2016

### CWatters

+1

It's a bit like an electric motor. In the right units the power output of a motor is..

Power = torque * rpm

If you load up a motor so it stops/stalls the torque is typically a maximum but the rpm is zero so output work is zero. Likewise with no load the rpm is very high but the torque is zero so the power is also zero. Somewhere in the middle is a sweet spot.

4. Mar 14, 2016

### sean39

Ok. I kind of understand, but not quite. I'm going to switch between another analogy. Hope that doesn't confuse the question too much. Here's the analogy:

Say you use a certain amount of force to pushing a weight up a slight incline. Compare this with using the same amount of force to push the same weight up a steep incline. Ignoring friction, the weight will reach the same height under either scenario. It will have traveled further on a slight incline versus the distance traveled on the steep incline, but in the end the two weights will be at the same height and have the same potential energy. So under the frictionless scenario, it seems like there is no sweet spot since the potential energy generated is the same whether there is a lot of resistance or almost none. (Don't know if I'm technically using the word 'resistance' correctly here).

If you then consider friction, I would think that you get more friction pushing the weight up the slight incline because you encounter friction over a longer distance. With the wind turbine, seems like the same thing would apply. More revolutions mean more friction, so seems like it would be more efficient to use the strongest magnet possible to the point where the turbine very nearly stops since this minimizes friction. So I don't see why there is a sweet spot.

5. Mar 14, 2016

### CWatters

6. Mar 14, 2016

### CWatters

Power = torque * angular velocity (or rpm if you refer). If the turbine almost stops then the angular velocity is small.

7. Mar 14, 2016

### sean39

Thanks, it's been a very long time since I took physics, so I apologize that I am having trouble with these concepts. From your answers, I can't tell whether I have made my question clear, or if I am just not recognizing the answer in your response.

I think I get the equation above. So applying the same amount of power to twist a tight rubber band generates more torque but fewer RPMs, while a more elastic rubber band generates less torque but greater RPMs. I don't see how this implies a sweet spot. The equation seems indifferent to the "resistance" of the rubber band if that is the right word.

8. Mar 14, 2016

### Staff: Mentor

But that's just it: the friction here is negligible. Just a few percent.

9. Mar 14, 2016

### sean39

Ok, but it's still something. And less motion = less friction, so shouldn't a machine that exerts force over a shorter distance against higher resistance be more efficient than a machine that exerts the same amount of force over a larger distance?

The force required to push a bowling ball up a 45 degree incline to a height of 1 Meter on earth would be equivalent to the force required to push the same bowling ball up a 45 degree incline to a height of about 6 meters on the moon. Each bowling ball would have the same potential energy at the top of their respective ramps because of the difference in gravity coming back down. But wouldn't the machine on Earth be slightly more efficient because the bowling ball encounters sqrt(2) meters of friction each way up and down the ramp, whereas the bowling ball on the moon would encounter 6*sqrt(2) meters friction each way.

So doesn't this mean machines like windmills, water wheels, etc. should use as much resistance in the generator so that the moving part of the machine moves as little as possible? Please let me know if I am using terminology incorrectly. Again, I am obviously not a student of physics, and I don't know if I am getting my question across clearly.

10. Mar 14, 2016

### Staff: Mentor

Yes. But you are only describing the drivetrain of the generator -- that has nothing at all to do with the wind.

11. Mar 15, 2016

### sean39

Thank you, but can you please elaborate to explain what I am misunderstanding?

It seems to me that we are trying to perform two steps: (1) convert kinetic energy of the air into the kinetic energy of the drivetrain, and (2) convert the kinetic energy of the drivetrain into some other form of energy (e.g. electricity). I understand that the process of converting kinetic energy of the air into kinetic energy of the drivetrain involves slowing the air, with some loss of efficiency to heat. Likewise, converting the kinetic energy of the drivetrain into electrical energy, I assume, involves slowing the drivetrain with some loss to heat.

In step 1, the windmill speeds up as the air is slowed (or vice versa if you prefer). So the more the wind is slowed, the more kinetic energy is transferred to the windmill. I think the equations and article you linked have something to do with the point at which optimal efficiency occurs. Is there a simple conceptual explanation of why there is a sweet spot (i.e, why it isn't just optimal to use a windmill that slows the wind in the extreme) rather than a linear relationship where the loss of kinetic energy of air = gain in kinetic energy of windmill + heat loss?

Likewise, does this also apply to step 2? As the energy of the drivetrain is converted into some other form of energy, isn't greater efficiency achieved as the drivetrain slows?

12. Mar 15, 2016

### OrangeDog

The sweet spot is due to a few things and isn't as straightforward as youd think.
Going in order of operations:

1) The wind passes along the blades of the turbine. Depending on the orientation of the wind relative to the chord line (think of this as the centerline) of the blade cross section (this is called an airfoil) lift and drag is generated. Some energy is lost due to heat and some energy is lost due to drag.

2) The blade rotates, some energy is lost in the shaft. I am not sure how this is measured or if it is significant at all.

3) The rotation of the shaft generates a magnetic field which is converted into an electric field at the rotor/stator. There are losses here as well, though I cannot comment on how they are computed.

4) The transmission of the electrical energy to a distribution/storage center incurs losses too. This due mostly to electrical resistance

5) Wider distribution results in lossess too.

Now, within the context of the turbine, the "sweet spot occurs for 2 main reasons:

1) The efficiency of an electric motor is proportional to its revolutions per minute.
2) The efficiency of a wing is proportion to its Reynolds number and relative orientation (also known as the angle of attack) of the wing cross section (aka, airfoil). The wing shape is a big deal too, but that isnt as easy to quantify.The Reynolds number is proportional to the fluid velocity and the length of the airfoil, assuming the properties of the fluid don't change significantly.

The result of these two facts is that you have two different efficiency curves that you have to optimize. The net efficiency can generally be taken as the product of the aerodynamic and electrical efficiency curves.

13. Mar 15, 2016

### sean39

Ok. Now I think we're getting somewhere.

If f(x) is strictly increasing and g(x) is strictly increasing, then f(x) * g(x) will be strictly increasing. So for f(x) * g(x) to have a maximum, these functions are either going in different directions (i.e., one is increasing and the other is decreasing with speed), or at least one of the functions itself has a maximum. Which is it?

14. Mar 15, 2016

### OrangeDog

They can be thought of as skewed parabolas with different maximums at different RPMs. So you have to maximize their product. There are other parameters too, such has how long a certain windspeed is maintained throughout the year. As a result you'll have to optimize over time as well. So you might have a maximum efficiency at one RPM, but that RPM is only in operation for 1% of the year, you would be much more inclined to maximize efficiency at an RPM that might be observed in 20% of the data. In this sense you are solving the inverse problem to the question you posed.

15. Mar 15, 2016

### sean39

Ok. Thanks. This is really, really helpful. I think I have a much better understanding now.

So from a theoretical standpoint, isn't a motor just something that takes some form of energy and converts it into mechanical energy (i.e., motion)? Would all motors have a parabola-shaped function? Can a bowling ball perched on a ramp be considered a motor? If the initial force to push down the ramp is the input, does this ramp have a parabola-shaped function of efficiency? Can this be explained in non-technical terms (e.g., at very slow speeds, there is a substantial loss of power because the friction between the ball and ramp is high relative to the energy of the ball in motion. While at very high speeds, there is a substantial loss of power because of what? Something to do with turbulence maybe?

16. Mar 15, 2016

### Nidum

17. Mar 15, 2016

### Staff: Mentor

No, the drivetrain is spinning at constant speed, so there is no conversion to kinetic energy in the drivetrain. The kinetic energy of the air does work on the drivetrain, which does work on the generator: a force over a distance (rotation).
The air gets slowed (loss of kinetic energy), the drivetrain does not.

But in neither case is conversion to heat a major factor. Betz's law isn't about how much of the kinetic energy in the wind is lost as heat, it is about how much of the kinetic energy in the wind has to stay kinetic energy in the wind and can't be harnessed.
Yes: neither necessarily involve any heat loss. Betz's law is about conversion efficiency, but what is left isn't converted to heat, it just stays kinetic energy (as said above).