# Turbine Speed Control

#### lenoplix

Joined Jun 11, 2020
16
Hi guys,

So i am currently working on a vertical axis wind turbine but faving a minor issue. The turbine rotates due to the wind force applied on its blades. So my question is how can you limit the speed of an already rotating shaft using a DC motor? My thoughts are to have the motor mounted to the shaft directly, then set a voltage to be supplied to the DC motor which in theory makes the motor supply a certain torque, and that would reduce the speed of the turbine. Is this analogy correct?

#### Irving

Joined Jan 30, 2016
814
The DC motor acts as a generator so you can control speed by loading the motor. The more current you pull the more load you are applying (subject to not overloading it). Applying a voltage to turn the motor in the opposite direction to the wind force will increase the loading but uses power and further stresses the motor. I'm not sure that's a sensible approach.

If the wind is too strong you have to take other measures such as turning the head out of the wind or feathering the blades or having a mechanical brake.

• lenoplix

#### lenoplix

Joined Jun 11, 2020
16
Thanks for replying so fast. The point of using the DC motor is to calculate the required torque to stop the turbine. That way I can ibtain the torque output of the turbine at any speed of rotation and wind speed. Do you have any other suggestion for doing so? Getting a torque meter is way too expensive given that the hughest torque generated according to ANSYS simulations is 0.05 Nm

#### Irving

Joined Jan 30, 2016
814
Ah, I see. Well if you put a DC motor on the turbine shaft and load it with a dump resistor R the the power in the resistor is V^2/R and power is revs x torque so if you know the motor characteristics (torque = I . Kt) you'll be able to get a rough approximation for the torque.

R = 0 will give the motor stall torque which may be enough to lock the rotor, esp if it's only 0.05Nm which is puny!

• lenoplix

#### lenoplix

Joined Jun 11, 2020
16
So what you are saying is that I load the motor with resistances in parallel for dump and make R > 0, that way since the turbine is rotating the motor when the motor is not loaded electrically and then an electric load is applied there would be no stalling since the motor is rotating opposite to the turbine shaft. Is that right?

#### Irving

Joined Jan 30, 2016
814
The motor always rotates in the same direction as the shaft, but applies a counter (braking) torque when loaded by virtue of the current in the winding opposing the motion generating it.

When the motor is unloaded (open circuit) and being rotated by the turbine it will generate a voltage V = S/Kv where S is revs and Kv is motor constant. If you put a load resistor R onto the motor it will apply a braking torque to the shaft. The torque is given by T = I . Kt = V/R . Kt In an ideal motor Kt and Kv are the same but in a real motor you have to look at the motor performance curves to get the values. If you know the motor 'no load' and 'stall' torque and associated currents then you can estimate Kt. You need to choose a motor you can get performance curves for and that has a stall torque > than your estimated turbine Max torque. Then a load resistor approaching 0 will be able to slow the turbine rotation close to zero. Of course you can't actually stop it as motor volts then = 0 and no power is generated in the motor, but by bringing R close to 0 you can extrapolate the turbine Max torque.

#### lenoplix

Joined Jun 11, 2020
16
So there is no battery to feed the motor? Just resistors on the motor?

#### Irving

Joined Jan 30, 2016
814
So there is no battery to feed the motor? Just resistors on the motor?
Yes. But note, motor's normal operating speed should be in line with turbine typical speed. You can't attach a motor with no load speed of 5000rpm to a turbine that normally runs at say 300rpm. If you can't get a gearbox, look at HTD pulleys and belt drive as reduction of 10:1 to 20:1 are easily do-able.

In lab condition, stall motor by dynamometer
and measure torque for different currents through motor.
Ideally yes, but OP doesn't have access to such. So has to interpret by motor curves or direct measurement. One way to directly measure is as follows. Fit a pulley, radius r, to motor and arrange physically to lift a weight off ground to a height of 1 or more metres (this generally doesn't work well if directly on high-speed motor shaft, it needs gearing down with gearbox or belt drive as above). Attach weight m to cord and time how long, t secs, it takes to raise weight by h metres. Measure voltage v, current i while doing so over a range of weights (experiment to find an 'easy' weight, a 'hard' weight and an in-between one, plus no weight and the smallest weight that the motor just can't lift)
Use equations below (use g = 9.8):

torque = r . m . g in Nm

Plot torque vs i to give Kt (as gradient of best-fit straight line)

Other useful info:
power out = m.g.h/t in Watts
power in = v . i in Watts
Efficiency = Power out/Power in
speed = 30 . h / (t . pi . r) in rpm
= h / ( t . r ) in rads/sec
power out = speed (rads/sec) . torque

• lenoplix

#### drc_567

Joined Dec 29, 2008
855
There may be another approach to a braking mechanism for your apparatus.
It would be necessary to build an electromagnet that could be powered from the DC motor operated as a generator. The advantage of this particular method would be convenient speed control using adjustable resistors to achieve current regulation in the electromagnet. ... It would require an aluminum disk mounted to the primary motor shaft.
Aluminum Disk Magnetic Brake Last edited:
• lenoplix

#### lenoplix

Joined Jun 11, 2020
16
Yes. But note, motor's normal operating speed should be in line with turbine typical speed. You can't attach a motor with no load speed of 5000rpm to a turbine that normally runs at say 300rpm. If you can't get a gearbox, look at HTD pulleys and belt drive as reduction of 10:1 to 20:1 are easily do-able.

Ideally yes, but OP doesn't have access to such. So has to interpret by motor curves or direct measurement. One way to directly measure is as follows. Fit a pulley, radius r, to motor and arrange physically to lift a weight off ground to a height of 1 or more metres (this generally doesn't work well if directly on high-speed motor shaft, it needs gearing down with gearbox or belt drive as above). Attach weight m to cord and time how long, t secs, it takes to raise weight by h metres. Measure voltage v, current i while doing so over a range of weights (experiment to find an 'easy' weight, a 'hard' weight and an in-between one, plus no weight and the smallest weight that the motor just can't lift)
Use equations below (use g = 9.8):

torque = r . m . g in Nm

Plot torque vs i to give Kt (as gradient of best-fit straight line)

Other useful info:
power out = m.g.h/t in Watts
power in = v . i in Watts
Efficiency = Power out/Power in
speed = 30 . h / (t . pi . r) in rpm
= h / ( t . r ) in rads/sec
power out = speed (rads/sec) . torque
So i should find resistors of high rated wattage/ohmage so that they load the motor thus reducing the speed of the turbine. After that as the turbine spins the resistors will slow it down to a certain speed, at that speed I would obtain V and I over the resistors of the DC motor and plug it in the equation T=(V/R).Kt and this should work at every speed no? Does a potentiometer work as if to mimic a variable resistance? rather than add more resistors?
Sorry for my lame questions I am kind of new to the motor field. And can you link me to some page with such a setup?