# Force needed to turn a windmill generator

1. Oct 16, 2011

### einergy

Many sites describe their windmills in terms of wind speed, minimum or max etc.
Can anyone share with me how much force in terms of how much torque or kg force is needed to get it rotating? Is there any conversion formula to it? I'm sure different size of generator would require different force and at different force will give a different rpm.

Anywhere to find this information? Assume a very small unit like 600W generator or maybe a better one at 2KW. What would be the force needed and what rpm would it go at?

2. Oct 16, 2011

### Bob S

You can approach this problem from two directions.

(A) The energy in the wind impinging on a HAWT (horizontal axis wind turbine) is equal to the kinetic energy of the air per unit time (=power) on the swept frontal area of the turbine blades. The incident wind power is proportional to the wind velocity cubed. The Betz limit constrains the maximum extracted power to 59% of this. Actual experience shows that about 40% of the wind energy can be extracted, when the blade tip speed is roughly 6 or 8 times the wind speed (depending on blade design). So you can calculate power, RPM, and torque for any wind speed from this.

(B) Secondly, the power (watts) extracted from a wind turbine alternator is equal to the angular velocity (radians per second, = 2 pi RPM/60) times the torque (Newton-meters). For the turbine speed to be constant, (A) must equal (B). You can adjust the alternator output by varying the rotor dc excitation (for a wound rotor, not permanent magnet), which will change the voltage output at any RPM. For a permanent magnet alternator without friction connected to a resistive load, both the voltage and current output will be proportional to RPM, so the power is propottional to RPM squared. Changing the load resistance will change the load line (output current at any RPM) and change the windmill RPM.

3. Oct 16, 2011

### einergy

Thanks Bob for your reply. Although the answer was not so direct, but now at least I got some idea how to compute it to my applications. Appreciate your sharing.
Cheers,

4. Oct 17, 2011

### W R-P

A more pragmatic approach is to isolate the generator and :
1) perform a Prony brake test
or
2) couple it to a motor(with a motor performance chart provided) to find the minimum torque(provided by the motor) required to start the generator.
Hope this helps

5. Oct 17, 2011

### AlephZero

In practice the "starting torque" will depend on a lot of messy details rather than some "simple" theory, so as #4 says the best way is measure it.

Since the generator has a finite number of poles, the torque required to turn it at a constant slow RPM will not be constant. Also, if the rotor is turing slowly, it doesn't store much energy acting as a flywheel.

So the bottom line is that if you apply a constant torque, there is a minimum speed that the rotor can turn continuously without stopping when it hits the "speed bump" of a peak in the magntic forces.

The mimimum starting torque is the torque required to accelerate the rotor up to that speed from rest, before it reaches the first "speed bump" and is stopped again. This means the starting torque is usually higher than the torque to maintain continuous slow rotation.

None of the above has anything to do with friction in the bearings etc, which should be negligible.

6. Oct 17, 2011

### einergy

Thanks for the above guidance.
I have prototyped a small buoyancy device which can produce a constant rotational force of about 4+ kg based on current size ( expandable depending how many tube I add) . The force was measured using a spring guage i borrowed.

This device will keep turning a shaft in one direction no matter which direction the core is moving (up or down).

The drive wheel is able to drive a generator. For my proto, I used a little generator from a car and sure it can turn.

The device had been in place for almost 2 years and yesterday after all the soaking in the medium, the device was still functioning fine. So before I spend further funds to try to patent it, I want to be sure it can be sized up to turn a commercial generator. Thats why I would like to know what turning force ( in kg) is needed to turn a decent generator and how the rpm thing works .Finally to calculate the power that can be generated. The input cost is SGD$0.007 per shot. Output from each shot - I can turn a driver wheel (dia. 130mm) driving a small wheel (dia.22mm) on the generator. I didn't measure how many rpm that was as was too fast and I haven't got a counter tool. Maybe someone can calculate that. In the typical structure in my country that I can fix my device to, I can get 240 or more revolutions. Anyone can help to calculate how much electricity can be generated ? 240 revolution at 4kg force. Would the electricity output with this performance be more than my input of SGD$0.007 ? To buy a generator to find out the answer would be .... money badly spent.

7. Oct 20, 2011

### Staff: Mentor

You are not using meaningful units. Post #2, item (B) describes the calculation and what you need. I'll repeat and show where your errors are:

1. You need revolutions per minute (RPM). You have just given us "revolutions".
2. You need torque in N-M (force in Newtons times lever arm length in meters). You have just given a force in kg (which should be converted to Newtons).

Then the calculation is just power = torque * RPM

8. Oct 20, 2011

### einergy

Thanks Russ for pointing out my errors.

I will find ways to measure these factors (rpm, torque) to calculate the power that can be generated.

Thanks to all for your kind and valuable inputs. It helped me a lot.

9. Nov 7, 2011

### einergy

Hi I'm back. Based on the formula, I got a number 7103. (240rpm x 3kg x 9.866 - to convert to Newtons ,I suppose)

I hope someone can enlighten me.

What does the 7103 means? Power ?
If I keep this power generating for 1 hour, does it mean I get 7.103 kwh of electricity?

I hope its true. Means my input of 42 cents/hr can generate ,at current rate to sell back to the grid , 85 cents of electricity ,by a gadget I'm making, translating to 100% profit.
Then its worth furthering my R & D to make this work and scale up in size to make good economic returns.

10. Nov 7, 2011

### Staff: Mentor

The calculation is incorrect, so the answer is meaningless. Please read the wiki on "torque" and gain an understanding of what it is: torque is force times the length of the lever arm...

....so where, exactly, are you applying that 29.4N force?

11. Nov 7, 2011

### einergy

Hi Russ,
Will get in touch with you via mail. Will put the description into drawing to show you what I trying to do and hope to get the right way to calculate.
Will do that after knock off from work.
thanks

12. Feb 23, 2012

### Whitewolf4869

hears some simple facts for you and this is how i went about figuring out what i needed to build a wind gen with a say alternator An alternator needs to turn at a certain rpm before it will produce full power around 800 rpm and are usually around 60 amps.Thats fast so you need to use a little math and figure out how fast your prop will turn in a say 10 mile per hour breeze use gear reduction belts and pulleys like the ones used on a car work fine for your own use or for a prototype dont forget you need a field current in the alternator so you need to connect it to a battery before it can produce power. Your wind gen cant start under load so it needs to start turning first. This is the point where you need to make a mechanism that censes a wind speeds that produce enough hp to say charge batteries at say 10 amps 12 volts and completes the circuit (volts x amps = watts) 750 watts =1 hp

13. Mar 15, 2012

### Bob S

The basic horizontal axis wind turbine (HAWT) is only about 40% efficient. The theoretical (Betz) limit is about 59%, but other factors limit it to about 40%. The HAWT is typically most efficient when the turbine blade tip speed is about 6 times the wind speed, so small diameter blades have a much higher RPM than large ones.

The incident kinetic power P in wind (in mks units) incident on a HAWT with frontal area A = pi R2 is
$$P=\frac{1}{2}\rho A v^3 \space \space \space watts$$
where ρ is the air density (about 1.2 kilograms per cubic meter) and v is the wind velocity (meters per second).

Dividing the above equation by the velocity v will give the frontal wind force in Newtons.

To design a 2kW HAWT, start by calculating the frontal area needed (A = pi R2) for the expected wind velocity, then calculate the blade tip velocity and the RPM. This will give you the design parameters for the alternator.

Bob S

14. Mar 30, 2012

### syed nasir

Can any one help me to find how much wind or air is required to rotate turbine of 7.5 feet diameter to produce 1 rpm

15. Mar 30, 2012

### Bob S

The blade tip speed has to be at least equal to the wind speed. 1 RPM is equal to 0.1 radians per second. Because the blade radius is about 1 meter, the wind speed should be at least 0.1 radians/sec x 1 meter/radian = 0.1 meters/sec..

You should have a wind speed of at least 5 meters/sec. Using a blade area of A = pi R2 = 3.14 meters2, the output power would be (using efficiency η = 0.25)$$P=\frac{1}{2} \eta \rho A v^3 = \text { 0.5 x .25 x 1.2 x 3.14 x 5^3 = 60 watts}$$
This is barely enough to overcome friction.

16. Apr 3, 2012

### syed nasir

it is a kind of vertical turbine with chimney of 500 ft .so we think this chimney connected to it will cause some suction which will also increase its wind speed.

i have sent one attach file plz see it and reply me .i will be waiting for your kind reply

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17. Apr 3, 2012

### Bob S

18. Apr 4, 2012

### syed nasir

the turbine is not in chimney it is in casing from which the chimney of 5 feet diameter is going up word it is basically near mountain and the chimney is you can say going upward on mountain i have tried to describe little in this file

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19. Apr 6, 2012

### syed nasir

20. Apr 7, 2012

### syed nasir

sir i am sending some pic of the turbine plzzzzzzzzz see and answer for question

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