# Torque needed to rotate low-speed shaft

1. Jun 18, 2014

### boyle007

Hi,

I'm interested in wind power/energy and I would like to know how much torque is required to rotate a low-speed shaft of either a 2mw or 3mw wind turbine at an rpm of 20. Also, I'd like to know if the weight of components (e.g. the shafts, gears, or any other components) affect how much torque is required for the same rpm (20).

Thank you

2. Jun 18, 2014

### Simon Bridge

Welcome to PF;
It takes zero net torque to rotate a shaft at a constant rpm. A non-zero torque would accelerate the shaft.

The applied torque - provided by the wind or whatever, accelerates the shaft until it is equal to the friction/losses (which depend on the rotation speed).

It is impossible to calculate that without a LOT more detail - it basically depends on the specific machine.

notes:
Part of the mission here is to help people learn the language of science and engineering - so let me just tweak your language a bit here:

- the "rpm" goes after the number; it is the units, not the thing.
You wouldn't talk about a person who has a "kg of 75" right?
The thing, in this case, is "rotation rate" or "angular speed", so you write: "a rotation rate of 20rpm" or similar.

- there is no such thing as "2mw or 3mw turbines" ... you can have 2mW or 2MW and the difference is quite important. Though I realize that some documentation can be lazy about writing SI units, in general, you shouldn't be.

I have a feeling you mean the industrial 2-3MW windmills - that correct?

3. Jun 18, 2014

### Chronos

Torque is a second order derivative. There is no torque without acceleration, or deceleration.

4. Jun 18, 2014

### boyle007

Thank you for the corrections. I meant 2-3 MW industrial turbines and I also meant 20 rpm.

I was always confused about the idea of torque at a constant rpm and I thank you for the above answers. I know I might be asking in the wrong way, but what would be the force required to keep a shaft rotating at a more or less constant speed? If I can detect the rpm, can I extrapolate from knowing only the rpm, the force required to maintain this rpm?

5. Jun 19, 2014

### Simon Bridge

I know what you mean - the force you use to turn the shaft at a constant rpm would be the applied torque.
This is opposed by drag and friction, which provides a counter-torque that depends on speed.
If you apply a constant torque to the shaft, it will accelerate until the counter-torque matches, and then continue at a constant speed ...

Short answer - no. You cannot I extrapolate from knowing only the rpm, the force required to maintain this rpm. The torque is the sort of thing you measure.

6. Jun 19, 2014

### CWatters

If the output from the generator is 2MW then at least that much power must be delivered from the blades to the generator. In practice it must be more due to losses in the generator etc. So you can calculate a lower bound for the torque between the two.

Power = Torque * Angular Velocity
Torque = Power/Angular Velocity

7. Jun 19, 2014

### CWatters

If the output from the generator is 2MW then at least that much power must be delivered from the blades to the generator. In practice it must be more due to losses in the generator etc. So you can calculate a lower bound for the torque between the two.

Power = Torque * Angular Velocity
Torque = Power/Angular Velocity

8. Jun 19, 2014

### Simon Bridge

9. Jun 19, 2014

### boyle007

First, I would like to thank all who answered this query.

Please tell me if I am getting this right: If one knows the power being generated and the rpm’s (of the high speed shaft), then one can calculate the approximate applied torque.

So I can use the equation of

M = P/n * 30/π where M = applied torque, P = power generated, and n = rpm of high speed shaft

to calculate the applied torque. I know this would be approximate, but I’m just looking to be able to calculate a rough estimate of the applied torque, if I know rpm’s and power output. I hope I understand it correctly now.

Also, I had another question about weight of components. My initial hypothesis was that if one decreases the weight of the components, then one would need less force (torque) to obtain the same power output. But I have a feeling that my hypothesis is incorrect.

10. Jun 19, 2014

### boyle007

Excuse me:

In the above equation, I meant Pi - not 30/n - it should be 30/Pi

thank you

11. Jun 20, 2014

### CWatters

Correct.

Torque (in Newton Meters) = Power (in Watts) * angular velocity (in Radians/second)

= 30p/nπ

which is what you got.

Changing the weight of the rotating parts won't effect the torque if it's operating at constant rpm. However it will effect the torque during acceleration and deceleration. I suspect the worse case for a wind turbine might be if the turbine has to shut down due to high winds? The generator is already acting as a "brake" on the rotor and torque will be high.. but then another "brake" has to be applied to slow the turbine to a stop. Heavier blades would tend to increase the inertia and torque under those conditions. (I'm assuming that the blade pitch can't be changed to slow the rotor).

12. Jun 20, 2014

### boyle007

thanks very much for that answer. now i can understand the wind turbine problem.

As far as weight goes, if lighter components for a wind turbine means less torque needed for acceleration, then why can't we try to find the lightest components, which would mean lighter winds for power? e.g., fan blades today are made of light plastic instead of the heavier metal you find with older fans, so why can't we use lighter turbine blades, or a lighter low speed shaft?

13. Jun 20, 2014

### CWatters

Strength is also important. Modern turbine blades are indeed made of composite materials for their strength weight ratio.

14. Jun 20, 2014

### mrspeedybob

The energy used to accelerate the rotating assembly is negligible compared to the output of the wind turbine. Besides that, it's not lost. That is the kinetic energy of the wind turbine. When the wind stops you get that energy back as the blades slow down.

That being said, Lighter components will put less stress on bearings resulting in lower friction losses, more durable (or less expensive) bearings, and less stress on the support structure.

BTW. Fans are made of plastic instead of metal because it's cheaper, not because it's better. If you look at industrial fans, they are made of metal because industrial customers are looking for the best combination of price, efficiency, and durability instead of just the lowest price like a typical Wal-mart customer

15. Nov 27, 2014

16. Nov 27, 2014

17. Nov 27, 2014

Hello
I have one question ....if wind power is constant and generates constant torque which applies a shaft(with no opposite torque ,no friction);this torque accelerates shaft and w(rotation velocity) increase ....so our mechanical power (p=Tw) can be more than our input power .... i really confused....plz solve my problem

18. Nov 27, 2014

### jack action

Without any load on the turbine, it would accelerate until it would break apart. But in real life, as the speed increases, some inefficiencies comes into play and the torque produced by the turbine drops to a point that it is in equilibrium with the losses again (which are usually much larger at that stage). At any time, this would be true:
$$P_{out} = P_{turbine} - P_{losses} - I\alpha\omega$$
Where:
• $P_{out}$ is the output power (load);
• $P_{turbine}$ is the power produced by the turbine;
• $P_{losses}$ is power of the losses (ex.: friction);
• $I$ is the inertia of the turbine;
• $\alpha$ is the angular acceleration of the turbine;
• $\omega$ is the angular velocity of the turbine.

19. Nov 28, 2014

### Simon Bridge

If the wind can generate a constant torque, then it must be able to supply infinite power.
Since this is clearly not the case, it follows that the wind cannot generate a constant torque.
The torque, in an ideal windmill, depends on the angular velocity of the blades.
I think I missed that out earlier.