Electronic – how to compute the average power from AC generator

When one reads the data sheet, it answers some questions, and prompts others.

The output is rated at 500W in a 14m/s wind speed. In higher winds, up to its 60m/s survival speed, you may get more, depending on whether it has any controls that limit it. At lower winds speeds, down to the cut-in speed of a few m/s, you will get less, a lot less.

The output voltage is rated as 24v charging, but no wind speed is mentioned. You might expect that is also at 14m/s, at which you will be getting 500W, or around 20 amps.

What happens at higher or lower wind speeds? Is the voltage controlled to be constant, or does it vary with the wind speed? Does it have a built in charging control, that can limit the current, and to what chemistry of batteries?

Does the turbine charge at 24v? Or is it 'suitable for charging at 24v' ('when used with our optional ...')

You need to grub around on their website for more details, and if you can't find them, contact the manufacturer. If it was stated to be a simple permanent magnet generator, then it would be possible to calculate what would happen versus wind speed, at least to first order. As the '24v charging' statement implies some sort of controller in there, you need to find out what it has been designed to do, because frankly, anything the designer thought was useful could be programmed in, as long as the electrical power output doesn't exceed the wind power input ("ye canna break the laws of physics cap'n")

Bear in mind that these guys want to sell their stuff, so it's in their interest to help you use it. If they're smart, then there is already an 'applications' page on their website, showing you how it's intended to be used.

I find it convenient to think of problems like this in terms of energy conservation. So let's put some numbers to it.

Mechanical power is force \$F\$ times velocity \$v\$:

$$ P = Fv \tag 1$$

or for a rotating system, torque \$\tau\$ times angular velocity \$\omega\$:

$$ P = \tau \omega \tag 2$$

For electrical power, it's current \$I\$ times voltage \$E\$:

$$ P = IE \tag 3$$

For the moment let's assume no losses in the system. The electrical power put out by the generator must equal the mechanical power put into the turbine.

$$ P_\text{turbine} = P_\text{generator} $$

Thus by equations 2 and 3:

$$ \tau \omega = IE \tag 4 $$

In low wind, you want to keep torque \$\tau\$ low. Otherwise the turbine may not even overcome static friction, and not turn at all.

While the product of torque and angular velocity is fixed by the amount of power available from the wind, their proportionality is not. Theoretically, you could add a mechanical gearbox so the turbine spins with less torque but at a faster angular velocity, ideal for low wind.

In high wind you wouldn't want the turbine to spin too fast, so the gearbox could work the other way: spin with more torque but a lower angular velocity.

The ideal gearbox would have no losses, and be continuously variable so the turbine can spin at the optimal speed in all wind conditions. Unfortunately such a gearbox would be very expensive and complex, and even so probably not all that efficient.

Fortunately, you can implement the "gearbox" on the electrical side of the system.

In a generator, the output voltage is proportional to the speed at which it is turning, and the current is proportional to the torque. So,

$$ E \propto \omega $$ $$ I \propto \tau \tag 5$$

If you want to reduce torque, you must reduce current. But you don't get to choose the power: that's decided by the wind. So to maintain power, according to equation 3, you must simultaneously increase voltage.

Generators produce voltage as described by Faraday's law of induction:

The induced electromotive force in any closed circuit is equal to the negative of the time rate of change of the magnetic flux enclosed by the circuit.

In the generator, this magnetic flux is the rotor moving through the stator field. Your permanent magnet stator provides the magnetic flux, the rotor windings are the thing enclosing the flux, and the rotation of the rotor makes the flux change. So reducing the voltage could be accomplished by:

• spinning the rotor slower
• decreasing the strength of the stator field
• decreasing the number of windings on the rotor

Spinning the rotor slower is the opposite of what we hoped to accomplish, but the other two options are valid. Using weaker magnets would decrease the strength of the stator field, and perhaps allow cheaper magnets. Using fewer turns in the rotor windings accomplishes the same thing, and allows thicker wire to be used, which reduces resistive losses in the wire.

So there's the answer to your first question: weaker magnets would indeed reduce torque, and thus the turbine would spin faster. You've effectively modified the "electrical gearbox" of the generator.

The trouble is now, what happens if the wind is strong? It would be more convenient if the "gearbox" was variable, but we can't exactly mechanically modify the generator to suit every wind condition.

There's another way to adjust the proportionality of voltage and current in an electrical system while keeping power constant: a switch-mode power supply (or perhaps it should be called a converter).

As luck would have it, the generator itself is already a big part of such a converter. You can also think of the generator as a motor which is always trying to spin too fast, so you must apply the battery as a brake. All you need to do is alternate between shorting the generator terminals together, and connecting the generator to the battery to charge it. And this is at the core of what a charge controller does.