How to determine Wind turbine RPM

[alokkumar]

TL;DR

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.

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?

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).

 

[Lnewqban]

Are you going to design and to manufacture the blades of the turbine from scratch, or are you going to select an available wind turbine?

 

[russ_watters]

You posted basically the exact same question six years ago. What progress have you made since then?

 

[alokkumar]

I am almost ready with the blade design and now wants to know what rpm the generators to choose. I asked this question 6 years ago and incidentally there is no specific answer.

If there is a formula to determine the rpm of rotor blades w.r.t input velocity of wind. Then I can build electrical generator with specific poles to give desired rpm.

 

[berkeman]

2014:

I am going to design 10 kw, Permanent Magnet, Direct Drive Wind Turbine Generator.

2020:

E.g. if I choose 100rpm electrical generator which will produce 1kw power.

You posted basically the exact same question six years ago. What progress have you made since then?

Looks like he's scaled back his power by a factor of 10...

 

[alokkumar]

Are you going to design and to manufacture the blades of the turbine from scratch, or are you going to select an available wind turbine?

Design from scratch and so need the answer.

 

[jbriggs444]

Is there any relation of input wind velocity with the blade rpm?

The wind velocity and the blade tip velocity are related. A quick trip to Google says that, unsurprisingly, the ratio between the two is a factor in performance. Here is a useful link.

If you know the wind speed and the tip speed ratio, you know the tip speed.

If you know how much power you are planning to get and you know the wind speed and rough efficiency then you can calculate the required cross-sectional area of your turbine.

If you know the cross-sectional area of your turbine then you know the blade radius.

If you know the tip speed and the blade radius then you know the rpm.

 

[jrmichler]

My old copy of Marks' Standard Handbook for Mechanical Engineers Eighth Edition has a section on wind turbines which includes the following chart:

Check out a newer edition of that book, and also some of the references in the wind turbine section.

MTA: Plus what @jbriggs444 said.

 

[Lnewqban]

Design from scratch and so need the answer.

Do you have the torque curve of your generator?
Your blades will need to satisfy the demanded torque at 100 rpm.

Then, you will need to determine the axial pushing and weight of your rotor in order to decide whether or not it can be mounted directly on the generator.
If not, you will need to design support and direct coupling.

A speed limiting mechanism may be useful for storm or low load situations.

 

[alokkumar]

The wind velocity and the blade tip velocity are related. A quick trip to Google says that, unsurprisingly, the ratio between the two is a factor in performance. Here is a useful link.

If you know the wind speed and the tip speed ratio, you know the tip speed.

If you know how much power you are planning to get and you know the wind speed and rough efficiency then you can calculate the required cross-sectional area of your turbine.

If you know the cross-sectional area of your turbine then you know the blade radius.

If you know the tip speed and the blade radius then you know the rpm.

[Reply edited by a Mentor to remove a mild insult] The wind turbine is not yet built, so tip speed ratio is not known. Unless turbine assembly is ready then only tip speed can be measured. Question is simple, how to use law of conservation of energy to determine rotational body rpm.

 

[jbriggs444]

Question is simple, how to use law of conservation of energy to determine rotational body rpm.

The answer is similarly simple: "You cannot".

 

[alokkumar]

The answer is similarly simple: "You cannot".

Indeed, I am sure there must be relationship of input Kinetic Energy of parcel of air with the Rotational Energy of Wind turbine + losses and unused energy.

Simply speaking there has to be direct relationship of input energy of wind with the rotational velocity of turbine blade. If this is known then it will be clear which rpm to choose for the electric generator.

 

[alokkumar]

Do you have the torque curve of your generator?
Your blades will need to satisfy the demanded torque at 100 rpm.

Then, you will need to determine the axial pushing and weight of your rotor in order to decide whether or not it can be mounted directly on the generator.
If not, you will need to design support and direct coupling.

A speed limiting mechanism may be useful for storm or low load situations.

Torque curve of the generator will be chosen after the turbine rotor rpm is determined in relation to the input velocity of wind.

 

[alokkumar]

2014:2020:
Looks like he's scaled back his power by a factor of 10...

This isn't funny , I hope 1kw doesn't hurt.

 

[jbriggs444]

Indeed, I am sure there must be relationship of input Kinetic Energy of parcel of air with the Rotational Energy of Wind turbine + losses and unused energy.

The one (input kinetic energy) amounts to a rate at which energy is delivered.
The other (rotational energy of the turbine) amounts to the amount of energy currently on hand.

The one relates to the rate of change of the other.

Simply speaking there has to be direct relationship of input energy of wind with the rotational velocity of turbine blade. If this is known then it will be clear which rpm to choose for the electric generator.

That is not correct.

 

[alokkumar]

The one (input kinetic energy) amounts to a rate at which energy is delivered.
The other (rotational energy of the turbine) amounts to the amount of energy currently on hand.

The one relates to the rate of change of the other.

That is not correct.

What is the source of rotational energy on turbine ??

I guess it is the kinetic energy of parcel of air.

=1/2 *m*v*v

As per law of conservation of energy. It is the KE of wind which converts into rotational KE of wind turbine + losses.

So isn't there a relationship?
I mean a formula to equate all these ?

Any other principle which I am missing

 

[jbriggs444]

What is the source of rotational energy on turbine ??

I guess it is the kinetic energy of parcel of air.

=1/2 *m*v*v

It is the kinetic energy of the stream of air from the time the turbine was erected to the present time. Multiplied by the efficiency with which that energy was harvested. Less the losses and power output from the time the turbine was erected to the present time. It has no immediate relationship to the energy in a particular parcel of air.

Conservation of energy: It takes zero net energy to maintain the rotation of a turbine that is already rotating.$$\text{KE}_{\text{final}} = \text{KE}_{\text{initial}} + E_{\text{in}} - E_{\text{out}}$$$$\frac{1}{2}I\omega_{\text{final}}^2 = \frac{1}{2}I\omega_{\text{initial}}^2 + E_{\text{in}} - E_{\text{out}}$$Knowing how much energy is input only helps tell you how the current rotational energy is changing. Not how much current rotational energy there is.

 

[alokkumar]

It is the kinetic energy of the stream of air from the time the turbine was erected to the present time. Multiplied by the efficiency with which that energy was harvested. Less the losses and power output from the time the turbine was erected to the present time. It has no immediate relationship to the energy in a particular parcel of air.

Conservation of energy: It takes zero net energy to maintain the rotation of a turbine that is already rotating.$$\text{KE}_{\text{final}} = \text{KE}_{\text{initial}} + E_{\text{in}} - E_{\text{out}}$$$$\frac{1}{2}I\omega_{\text{final}}^2 = \frac{1}{2}I\omega_{\text{initial}}^2 + E_{\text{in}} - E_{\text{out}}$$Knowing how much energy is input only helps tell you how the current rotational energy is changing. Not how much current rotational energy there is.

Assume a closed door experiment.
Turbine initial rpm = 0 rad/s
Wind velocity = 0 m/s

After some time t sec if we blow wind with 5 m/s then turbine will start to rotate. What will be the rotational rpm of turbine ?

 

[jbriggs444]

Assume a closed door experiment.
Turbine initial rpm = 0 rad/s
Wind velocity = 0 m/s

After some time t sec if we blow wind with 5 m/s then turbine will start to rotate. What will be the rotational rpm of turbine ?

As I understand the setup, we have a turbine in a wind tunnel. It is motionless. We start a wind blowing at 5 m/s. We maintain this for t seconds. We quickly measure the rotation rate and ask: how fast is the turbine now rotating?

We assume that the wind tunnel is much wider than the turbine (free flow, not ducted flow).

Questions for you:

How efficient is the turbine at harvesting the energy from the wind?
What is the turbine's cross-sectional area?
How much energy is in t seconds worth of 5 m/s wind? [Assume, for the sake of argument that we are capturing only wind energy in the area directly swept out by the turbine].
How much energy does that mean is captured by the turbine?
What is the moment of inertia of the turbine?
How much energy are you drawing out of the turbine while this is going on?
How much energy is lost to friction in the bearings?
How much energy is carried away in the downstream wind? What impact does this have on achievable efficiency?
What rotation rate does this mean the turbine has after t seconds?

If the wind continues like this for another two minutes, two hours, two days, two weeks, two months or two years, do any of the assumptions that went into this calculation change? What rotation rates result if they do not?

 

[etotheipi]

I don't know how useful this is, but the maximum theoretical power extractable from air passing through a turbine of cross-sectional area $S$, with initial speed $v$, is $P_{max} = \frac{16}{27} \frac{1}{2}\rho S v^3$, where the coefficient $\frac{16}{27}$ is the Betz limit. In a steady state (constant rotational energy of turbine), I suppose you could set that equal to the sum of the useful power and the dissipative losses for a vague upper estimate.

 

[alokkumar]

As I understand the setup, we have a turbine in a wind tunnel. It is motionless. We start a wind blowing at 5 m/s. We maintain this for t seconds. Then we stop the wind. We are asked: how fast is the turbine now rotating.

We assume that the wind tunnel is much wider than the turbine (free flow, not ducted flow).

Questions for you:

How efficient is the turbine at harvesting the energy from the wind?
What is the turbine's cross-sectional area?
How much energy is in t seconds worth of 5 m/s wind?
How much energy does that mean is captured by the turbine?
What is the moment of inertia of the turbine?
How much energy are you drawing out of the turbine while this is going on?
How much energy is lost to friction in the bearings?
How much energy is carried away in the downstream wind? What impact does this have on achievable efficiency?

Great let me answer !

How efficient is the turbine at harvesting the energy from the wind?
= 35 to 38%
What is the turbine's cross-sectional area?
= radius of blade = 1.2 m [Cross section = (22/7)*1.2*1.2 ]

How much energy is in t seconds worth of 5 m/s wind?
= Very good question. This is exactly what i am asking too !
Since the mass of air in t secs is not know so how can we determine KE of wind ??
KE = 1/2 * mass of air * velocity of air square

How much energy does that mean is captured by the turbine?
= as said efficiency approx 35 to 38%What is the moment of inertia of the turbine?
I = 1/3 * no. of blade * mass of blade * radius of blade square
no. of blade = 3
mass of blade = 4 kg
radius of blade = 1.2 m

How much energy are you drawing out of the turbine while this is going on?
= Entire energy output of the Electrical Generator with efficiency 85% will be fed into batteries

How much energy is lost to friction in the bearings?
= Loss is around 30%

How much energy is carried away in the downstream wind? What impact does this have on achievable efficiency
= already said
Energy in Wind (100%) = Energy in Rotational Turbine (38%) + Losses

The Energy of Rotational Turbine will further have losses of friction + bearing+ electrical around 30%

Hope this helps !

 

[jbriggs444]

How much energy is in t seconds worth of 5 m/s wind?
= Very good question. This is exactly what i am asking too !
Since the mass of air in t secs is not know so how can we determine KE of wind ??
KE = 1/2 * mass of air * velocity of air square

Great! Here is a well defined question. We can work it and find the answer.

We are considering a parcel of air whose cross-sectional area is given by the cross section of the turbine. We know the radius of the blades. We can call that "$r$". And we can call the cross sectional area $A$.$$A=\pi r^2$$It is good practice to leave quantities in symbolic form as long as possible. We can substitute in the measured values (for instance 1.2 meters for r) as a final step.

Let us consider a parcel that is long as air can move in 1 unit of time (1 second). We know that the wind velocity is 5 m/s. But let us use $v$ to denote wind velocity and $V$ to denote the volumetric flow rate (volume per second).

The volume of air that passes through the turbine in 1 second is cross sectional area times wind velocity:$$V=Av=\pi r^2v$$The mass of air that passes through the turbine in 1 second is the volumetric flow rate times the density of air. We could look up the density of air and get a number. But let us leave it as $\rho$ instead. The mass flow rate $M$ is then given by$$M=\rho V=\rho\pi r^2v$$The energy delivered per second is the mass flow rate times the energy per unit mass. Let us call the energy flow rate $P$ for power. Then we have:$$P=\frac{1}{2}Mv^2=\frac{1}{2}\rho \pi r^2v^3$$Now we are in a position to substitute in our known values.
$\rho$ = 1.225 kg/m^3
$\pi$ = 3.14
$r$ = 1.2 m
$v$ = 5 m/s
If I have not screwed up the calculations, that gives:$$P = 346\ \text{Watts}$$Note that this is before we've applied the correction for Betz' law. It is also before we have multiplied by $t$ seconds.

Are you with me this far? Shall we keep going?

 

[alokkumar]

This is absolutely clear...
Please continue..

 

[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.

 

[alokkumar]

Thats brilliance !

Great job and many thanks...
You are genius... !
I was missing the last calculation of equating KE with Power into time.
Thanks again !

 

[alokkumar]

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

 

[jbriggs444]

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.

Please, see:
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.

 

[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.

 

[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.