# How to determine Wind turbine RPM

So in this equation, the rpm keeps varying with respect to time.
Great !!

jbriggs444
Homework Helper
2019 Award
So in this equation, the rpm keeps varying with respect to time.
Great !!
Yes. If the assumptions going into the equations continue to remain correct, the blades will spin faster and faster without bound.

Of course, the assumptions will not actually remain correct. In particular, one would expect to get energy losses due to drag that scale badly with increasing rotation rate. At a high enough RPM, your efficiency will drop to zero, if the turbine does not tear itself apart first [thereby reducing efficiency to zero the hard way].

Lnewqban
Torque curve of the generator will be chosen after the turbine rotor rpm is determined in relation to the input velocity of wind.
From your old posts, I can see that you have been plannning fabricating the generator last.
I suggest working the calculations from the generator needed output power back towards the turbine, blades, bearings, base, etc.

Torque and rpm go together.
Going uphill with a bicycle, you can only reach desired rpm's of pedaling if your muscles can generate enough force or torque.
Once you reach your limit of force, if the hill gets stepper, the natural reaction is reducing rpm's of pedaling.

With enough available time, your body can produce huge amounts of work (energy), but only certain amount of power, which is delivered work in unit of time: same happens ith the wind and your turbine.

You have been considering the equation of kinetic energy, but energy is work and not power.
Since rpm's are time dependant, you need to consider power, which is work/time, like the KW's of your generator output.

http://hyperphysics.phy-astr.gsu.edu/hbase/rotwe.html#we

http://hyperphysics.phy-astr.gsu.edu/hbase/work.html#wepr

Your blades will be impulsed by wind, but slowed down by the electro-magnetic field within the generator; hence, you need to balance both effects, while adjusting things (like rotor diameter, airfoil of blades, angle of incidence) in order to achieve the desired rpm's of turbine-generator direct coupling, if at all possible.

alokkumar
cjl
My old copy of Marks' Standard Handbook for Mechanical Engineers Eighth Edition has a section on wind turbines which includes the following chart:
View attachment 267224
Check out a newer edition of that book, and also some of the references in the wind turbine section.

MTA: Plus what @jbriggs444 said.
This is a little out of date - a modern horizontal axis three bladed turbine tends to run a tip speed ratio of around 9 or 10, which is well off the right side of your chart. I'm not sure why they list a "modern multiblade turbine" at a TSR of only 2 or so - that's never been true of any decent horizontal axis design, even going back a few decades to small designs in the realm of a few hundred KW.

That having been said, TSR is absolutely a great place to start for this. You can't just ask what RPM your turbine will have - that depends on blade angle, generator characteristics, aerodynamic properties, etc. Instead, you should assume a tip speed ratio (5-7 might be a good place to start) and set up your blade angles appropriately. You can then figure out what kind of torque you need your generator to have based on power and RPM, and work from there.

berkeman
cjl
Yes. If the assumptions going into the equations continue to remain correct, the blades will spin faster and faster without bound.

Of course, the assumptions will not actually remain correct. In particular, one would expect to get energy losses due to drag that scale badly with increasing rotation rate. At a high enough RPM, your efficiency will drop to zero, if the turbine does not tear itself apart first [thereby reducing efficiency to zero the hard way].
The main thing that keeps it from spinning faster and faster isn't the drag, it's the fact that as it spins faster, the angle of attack of the blades decreases. For a set blade angle, as the RPM increases (at fixed wind speed), the blades will create less and less power and the lift vector will rotate more and more in plane, so you really don't have to worry about this boundless RPM case.

EDIT: Also, with regards to the original post, give me a bit to compose my response. I'll try to get something together soon though.

russ_watters, Lnewqban and jbriggs444
So far we have ##P=346 Watts## arriving in the air stream. But we are only going to capture some 35 to 38 percent of that. Call this capture efficiency ##e_c=0.38##. We also lose 30% to friction. Let us call that mechanical efficiency ##e_m=0.70##.

You've stated that 100% of the remaining input energy is used to generate electrical energy. But that is a problem. Because if 100% of the remaining energy goes into generating power then there is no energy left to increase the rotation rate of the turbine. That leads to a dead end: RPM = 0.

So let us assume instead that while we are busy spinning up the turbine the generator will not be drawing off any energy. That means that the energy flow rate ##P_\text{rot}## going into increasing the rotation rate of the turbine will be given by:$$P_\text{rot} = P e_c e_m$$

If we substitute in the percentages, we down to 92 Watts going into increasing the rotation rate.

You've already done a good job with the moment of inertia calculation. Let's call the moment of inertial ##I##. Three blades times ##\frac{1}{3}mr^2## for each blade and ##m=4 \text{ kg}## and ##r=1.2 \text{ m}## so we have ##I=5.76 \text{ kg m}^2##

After ##t## seconds at an energy flow rate of ##P_\text{rot}## we will have accumulated rotational kinetic energy:$$KE=\frac{1}{2}I\omega^2=tP_{rot}$$If we solve for ##\omega## that gives:$$\omega=\sqrt{\frac{2tP_\text{rot}}{I}}$$Let us fill in the givens:
##P_\text{rot}##=92 Watts
##t## =10 seconds
##I## = 5.76 ##\text{ kg m}^2##
If I have not screwed up, that comes to
$$\omega = 17 \text{ rad/sec}$$after ten seconds. If I have not screwed up, that is 162 RPM.

Of course, we have also determined that your 1 kw turbine is only gathering 92 watts. And that is before the 85% generator efficiency is factored in. That may be a problem.
@jbriggs444

A quick question. In a practical wind turbine system. The electrical generator is drawing continous energy out of the turbine system.
So in that case wouldn't the rpm of the overall turbine assemble (including generator housing) will reduce ??

I/2*I*omega*omega = t * P(rot) - Energy drawn by generator ??

So I guess omega (i.e. rotational rpm of overall assembly) will reduce.

While chosing appropriate generator for turbine. Do I need to consider rpm 162RPM( in your example). Or generator chosen should have much lower rpm ? Because actual assembly will rotate with much lower than 162rpm (because electrical generator is drawing the energy out continuously)

Your help on this would be much appreciated!

jbriggs444
Homework Helper
2019 Award
A quick question. In a practical wind turbine system. The electrical generator is drawing continous energy out of the turbine system.
So in that case wouldn't the rpm of the overall turbine assemble (including generator housing) will reduce ??

I/2*I*omega*omega = t * P(rot) - Energy drawn by generator ??

So I guess omega (i.e. rotational rpm of overall assembly) will reduce.

While chosing appropriate generator for turbine. Do I need to consider rpm 162RPM( in your example). Or generator chosen should have much lower rpm ? Because actual assembly will rotate with much lower than 162rpm (because electrical generator is drawing the energy out continuously)
Yes, certainly. This entire calculation was based on a length of time pulled straight out of thin air and an assumption that the generator would not be drawing power during the spin-up time. The 162 RPM figure is, accordingly, a complete fiction.

As long as you are drawing power out at the same rate that the wind is putting power in, the turbine RPM can be anything that you please. You cannot derive it from this kind of simplistic energy balance.

The better way to determine RPM and the one that was first suggested to you is to assume a particular tip speed ratio and see where that leads.

cjl
OK, I'll try to answer this as best I can (with the caveat that, although I am a wind turbine engineer, there are a lot of differences between the giant ones I work on and small ones for individual generation)

Summary:: I am designing micro Wind turbine however I am unclear what rpm electrical generator should be chosen for a given power rating of wind turbine.

E.g. if I choose 100rpm electrical generator which will produce 1kw power. Then I have to be sure that my wind turbine system at least rotates around 100rpm. Otherwise i have to put gear boxes. Which I want to avoid.
Generally, RPM is mostly determined by rotor diameter. For large wind turbines, you'd tend to run a maximum tip speed of around 75-90m/s, so your RPM will be whatever it needs to be to achieve that tip speed. I'd imagine that maybe small ones run a bit slower than this, but honestly I'm not sure. It doesn't really relate to power (other than that obviously larger turbines tend to both spin slower and make more power). If you're running similar tip speeds to large turbines, I'd expect one spinning 100RPM to be around 15-18m diameter. That having been said, that'd be awfully large for a 1kW turbine, since usually that would be more like a 2m diameter or so to get that kind of power. If you have a 100RPM 1kW generator, you'll either be running a fairly low tip speed, a gearbox, or a very low rotor loading.

So I need formula of rotational wind turbine system. Does it have any relation to input Kinetic energy of wind? Or some other formula.

How to determine rpm of rotational body such as wind turbine blades. Is there any relation of input wind velocity with the blade rpm?
As I mentioned yesterday, often turbines spin such that the tip speed is about 8x the wind speed or so. Generally, the higher the L/D ratio of your blades, the faster you want to spin them, so for a small turbine, you'd probably want to go a bit slower because you probably can't achieve quite as high a L/D. For a basic home design, I might initially design for a tip speed ratio of 4-5 or so and go from there.

There's also a concern with overspeeding - large turbines can pitch the blades out above a certain wind speed to prevent the turbine from spinning too fast. However, I'm assuming you're setting this up to be largely passive. If you're running a lower TSR, that helps a lot since that just means you'll be spinning a lot slower in general, but you still need to pay attention to how fast you can spin your generator and what the structural constraints are.

I also know that some early turbines were "stall-regulated" designs, where the blades were intentionally designed to stall above a particular wind speed to prevent them from going any faster. I don't know the details of exactly how this was achieved though, since as I said, that hasn't been a thing for a long time (much longer than it has been since I've been in the industry).

If the rpm formula in relation to input wind velocity is known then I can choose exact matching rpm for electrical generator for a given power output.

As per law of conservation of energy

Input Kinetic Energy (wind) = Rotational Kinetic Energy of wind turbine + Energy unused by turbine (approx. max < 41 % betz law)

But input KE of wind is unknown because (1/2 mv*v) mass of parcel of air is unknown.

Rotational energy of rotational body(1/2 * I*omega*omega).
The KE of the incoming air (along with the Betz limit) does let you know something about what your generator's power curve should look like. Also, you can use it to size your turbine if you also know the typical mean wind speed in your area - if your area averages 7m/s, you probably want to design your turbine to hit full power at no more than 10m/s or so, because you'll spend so little time up at those high wind speeds above 10m/s that it's not worth designing the turbine to run faster/more powerfully up there. Then, if you know you want 1kW, and you know you want to hit that at 10m/s, you can figure out what kind of diameter you need (based on the incoming KE of the wind) and what the associated RPM will be (based on your design tip speed ratio). Also, although the betz limit theoretically allows for 59% efficiency, I'd be very hesitant to assume more than 30% or so for a home design.

Hopefully this is at least a little bit helpful?

Last edited:
BvU, Tom.G, russ_watters and 1 other person
cjl
@jbriggs444

A quick question. In a practical wind turbine system. The electrical generator is drawing continous energy out of the turbine system.
So in that case wouldn't the rpm of the overall turbine assemble (including generator housing) will reduce ??
Ideally, the generator should be drawing energy off at the same rate that the wind is adding it, so at a fixed wind speed, you'll end up holding a steady generator speed as well. If the wind speed increases, now you're adding energy faster than you're drawing it off, so it'll accelerate. If the wind speed decreases, the reverse will happen.

I think it's probably not useful to consider the rotational inertia too much here - especially with a very small turbine, the amount of kinetic energy in the rotor will be quite small relative to the rate of energy addition by the air and energy draw from the generator, so if conditions change, it should relatively quickly adjust to the new equilibrium point.

It is worth mentioning that in the ideal case, power available from the wind will scale as windspeed to the third power, while a simple generator will output voltage proportional to RPM. Connected to a simple resistive load, this generator would generate power proportional to RPM squared. Since you want your RPM proportional to wind speed, this means that you'll be trying to pull too much power at low RPM (which is bad because you'll stall the blades) and too little at high RPM (losing out on a lot of potential power). To really get the best, you'll want some kind of electronics to make sure you're pulling power from it proportional to RPM^3 to match the wind energy available.

Tom.G and Lnewqban