# Small wind turbine

we need to make a wind turbine using this motor http://www.kidwind.org/xcart/product.php?productid=15&cat=79&page=1" [Broken]

the rules can be found here: http://www.mrmaloney.com/nssl/events/2009-2010/0910_4-jan-wind.pdf" [Broken]

We were thinking, since they will be using P=(I^2)R instead of P=IV to calculate power that we could use a step down transformer to raise our current and lower our voltage. This seems like it should work but at the same time I feel like it would be magically be making more power.

We would also like to use some kind of CVT if possible, so that our fan can start up fast and keep going while generating more power. Anyone know of a way to make/buy one this small?

Any input that could help us out is greatly appreciated.

oh and i hope this is in the right section, i wasnt sure

Last edited by a moderator:

## Answers and Replies

This seems like it should work but at the same time I feel like it would be magically be making more power.

There are two problems with using an electrical transformer between the DC generator and 100 ohm load resistor. A transformer needs an AC signal to operate propertly. Also it cannot magically make more power as you properly question.

Apply Ohm's Law to the Load Resistor:

V = IR, but V is produced by the generator and R is fixed, so write:

I = V/R

Now you can calculate the current that would flow in the resistor for different voltage output of the motor. More current means more average power in the resistor, so work backward in your design from there.

could we use a bridge rectifier to convert the voltage to ac before the transformer. Would this not give us a greater power output? I think im missing something crucial because it doesnt quite make sense.

I know we want our motor spinning as fast as possible for the most power but we're just trying to find a way to really bump it up. thanks for the help

These are good questions. But what happens is that the components you insert in the system generate some heat internally, the temperature rises, and this is a loss of power. The load resistor itself will be dissipating all the input power as heat. It will get warm. The heat flows to the surroundings and the power is gone.

So every time you try to boost the power with another non-ideal transformer, you actually reduce the power to the load. You could pulse the DC into the transformer but it won't boost power, and it will add complication and expense.

The DC generator is already a power transformer. You put in mechanical power and take out electrical power. Focus on the power in blades, the design of the blades, and the gearbox design. Can you find a formula for wind power at the blades using google search? It's out there ...

so basically we're gonna lose more power than we would appear to increase it. makes sense i think.

Just picking some very random numbers for conversation sake, if we had say 1 volt and 1 amp, and we stepped down the voltage to .5 volts, we would have 2 amps. in theory we would start off with 100 watts ((1^2)100) and end up with 400watts ((2^2)100).
This part is what is confusing me. It doesnt seem like it should work like this.. I understand that we will lose energy to heat but this math doesnt seem right

On another note, is there any way to buy or make a cvt(continuously variable transmission) small enough to work with this? Or should we just mess around with gearing until we find the best gear ratio? Or even make/buy some kind of gearbox that allows us to change gears on the fly? Therefore we can use a lower gear to get it spinning and then switch it higher once it is already going to get the motor spinning faster.

The load is fixed at 100 ohms. Put 1 volt DC across it and compute the amps I = V/R. Put 0.5 volt DC across it and the amps go down (not up). You can't get more power out of a transformer than you put into it. Do you have the DC generator in hand? A volt and amp meter? What about a way to measure rpm? These are the basics you'll need to start your design.

How much time is allowed to get the machine started and up to speed? How much to measure power output? These times will determine whether there is any benefit to looking into a variable transmission or cvt. Chances are it is too much effort for no real gain ...

oh ok it makes more sense now. I know you cant get more power out than you started with which is why it was confusing me. i have the generator and a multimeter, but nothing for rpm. honestly im not even sure what youd use for that.

It is 30 seconds to align and test it, and then 30 seconds to measure output. It says the circuit will be grounded in between so i think they will be 2 seperate runs, meaning we will have 30 seconds to measure power, starting from a dead stop. Therefore we want this thing to get spinning fast as fast as possible and then we can change the gear ratio to get more power output, which made me think a cvt would be ideal. I just dont know if a cvt would be practical at this small scale. Our other option would be to just have a single gear ratio, but if its too much our turbine will take the whole time just to get up to speed.

Get hold of a 100 ohm resistor and measure its actual value. It will be within a few percent error of the resistor they use for the competition.

Then hook up the generator leads to the load resistor and spin the generator with a variable speed drill. Measure the voltage across the resistor and try not to exceed 1.5{V} to 1.6{V} initially. The power in the resistor will be the voltage squared over the resistance, or compute the current from V and R, then power will be voltage times current (the same in both cases).

I'm trying to think of a simple way for you to measure rpm at the same time as the voltage output from the DC generator. Do you have access to a frequency counter on your meter?

what exactly am i trying to find by using the drill. I get what youre saying, but whats the point? I dont believe the multimeter has a frequency counter, its a pretty cheap one.
THe website that sells the motor/generator has a graph and table for voltage output and such at certain rpms. i assume this would help?

Aren't we basically just looking for the best way to spin the motor as fast as possible for as long as possible? So make the blades the size of the fan put a big gear on the turbine and a small one on the generator. I believe this the most basic way to do it, but i think there is a better way, which is why i am curious about cvt, shiftable transmission(like a 5 speed car), etc. We want to win. We tend to think outside the box and they dont like that so we just want to go and absolutely destroy everyone. i really appreciate all the help

what exactly am i trying to find by using the drill. THe website that sells the motor/generator has a graph and table for voltage output and such at certain rpms. i assume this would help?

Yes, this would help. You want to understand how the DC machine operates as a power transformer. This will help answer your transformer question above while giving more insight into the system at the same time.

The DC machine has two air gap equations which are best written in standard SI units, so power is in watts {W} on both sides of the air gap.

Torque-Current Relation:

$T = kI$

Voltage-Speed Relation:

$V = k \omega$

where T is input shaft torque in newton-meter, $\omega$ is shaft speed in radian per second, V is terminal output voltage in volts and I is generator output current.

If you wish, find the factor k by converting rpm to radian per second in the graph of voltage versus rpm. Then find the slope of the line as the change in volts (rise) over the change in radian speed (run). This gives the air gap constant k in volt-second per radian, which will help you determine the torque and speed of the generator when coupled to the 100 ohm load through the air gap.

Also if you can measure resistance try to measure the winding resistance of the DC generator. Turn the shaft in discrete small steps around a circle, and take measurements at different shaft positions. Average the readings and you may get an OK approximation of the armature resistance.

The power in the wind itself can be estimated. Here is a link with some formulas:

http://www.windstuffnow.com/main/wind.htm

If you put a big prop on the little generator it has large moment of inertia, taking more time to spin up to speed. It might be better to funnel the air into a smaller prop similar to the design shown in the top photo here:

http://www.airflowsciences.com/Services/PhysicalModeling/Power/

If you wish, start looking for a spefication of the air flow output from a 24" box fan. I can't find data that fits the formula on windstuffnow ...

i believe funneling the air is against the rules.

but couldnt we just put a big gear on the fan and a really small one on the motor? therefore the fan doesnt have to spin very fast to get the generator going good.

But im still set on making a cvt, or at least trying. its probably not possible at this scale but i believe its the best design to use

so did a little more research and found that cvt on the small scale are too heavy and ineffecient. so now our options become:

- some create/ buy an automatic transmission
-make/buy a manual transmission
-choose a single gear ratio that will get us the best average

and ideas on most efficient ways to gear it. just straight gears, pulleys, etc?

Kidwind uses 6:1 gearbox. Look at some of the Construction Instruction pdf's for ideas:

http://www.kidwind.org/PDFs.html [Broken]

Last edited by a moderator:
do you think using stator veins to straighten the airflow would be considered a structure intended to focus the moving air mass?

Check out the following link for a small CVT designed for a bicycle.

http://www.fallbrooktech.com/03_bicycle.asp [Broken]

Last edited by a moderator: