Dimensioning a generator

Hi!

I'm working on a project where we are building a windturbine from the bottom up. Myself and a colleague are designing a double stator axial flux generator for our turbine. Our requirements are that we deliver about 10 W at nominal rotation speed. It's a direct axis driven generator propelled by a VAWT. The issue that has been bothering us is how we determine the size of the generator. We found some rather advanced articles on the back-torque of the machine but are still a bit lost. So to put our concerns into a question. What is a common design parameter that makes a generator too heavy/too light to drive?

Attachments

• Axial generator drawing 1.pdf
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Nidum
Gold Member
Designing a generator for a specific purpose usually requires considerable technical knowledge .

I see though that you are building a very small system intended to deliver just 10 W . At that power level there are many ready made dynamos and motors that could be made use of at very little cost .

I'm aware that we are essensially building a bike dynamo. But the project is to build all the components ground up (including coils etc). We already have a simulink model giving us the emf from the rotation speed which tells us how many poles and so forth we will need. We just feel that we are lacking some knowledge on how to model the back-torque to make sure our turbine can drive it once we assemble the whole machine
.

anorlunda
Staff Emeritus
Must you calculate the torque from first principles, or just estimate it?

If efficiency is 100%, then torque*speed = voltage*current (after suitable units conversion.) Most generators are pretty efficient, so a guess of 75% efficiency for your home made generator seems to be pretty conservative. So just divide the required 10 watts by 0.75 to get estimated torque*speed.

To improve efficiency, my guess is that the bearings are more important than the weight.

Is your apparatus required to be self starting? You may need significantly more torque to spin it up starting from zero RPM than to keep it running at rated RPM, and the acceleration does depend on weight. If self start is not required, then a push by the hand to start it spinning would be allowed. Alternatively, he electric load could be removed for startup and switched in after achieving rated speed. Both alternatives mean manual intervention.

Onjii
Nidum
Gold Member
You have the rotation speed . Make a conservative guess at the dynamo efficiency .

Shaft torque * (revs/sec) * 2 * Pi = dynamo output power / dynamo efficiency

Torque in N-m and power in W .

Example : shaft speed = 50 revs/sec . (3000 rpm)
output power = 10 W
efficiency = 0.4

Shaft torque * 50 * 2 * 3.142 = 10 / 0.4

Shaft torque = 0.08 N-m

Onjii
Appreciate the help guys!
I think we will settle with approximating the torque like you just described.
Thanks!