First Time Wind TurbineDesign

[dsmith1427]

After a lot of reading and research, this is my first attempt at a wind generator and I'm looking for some feedback.  While I want to know if my numbers are in the ballpark, I would really like to know if my methodology is correct.  Anyway, here it is...

According to Historical Wind Data from an NOAA website the average wind speed for my area is 4.6 meters per second (10.4 mph).  I used this wind data to start my calculations.  Wanting to keep the project on the small and inexpensive side, I utilized materials around my shop to make 558 mm blades (1.116 meter diameter).  I mounted the blades to a hub and took it outside on a breezy day.  I was satisfied with the results.  See the following YouTube Video.  

www.youtube.com/user/dsmith1427

Now,  I want to size a PMA for this turbine.  System definitions for the PMA include the following:

Average Wind Velocity = 4.6 Meters per second

Cut in speed = 315 RPM

Dual Rotor

Eight Poles (16 N52 magnets)

Six Coils (16 gage magnet wire)

Magnet Dimensions = 25.4 mm x 25.4 mm x 12.7 mm thick

Blades = 3

Tip Speed Ratio =  4

Battery = 12 volts

Turns of 16 gage wire per coil = 125

Sweep Diameter = 1.116 meters

Charge Voltage = 13.2 v

The following method was used to calculate the numbers.

Power in the Wind/Power to Harvest

If the average wind speed is 4.6 meters per second (mps) and the sweep diameter is 1.116 meters, there is about 65 watts of power in the wind.  The following formula is used for the calculation:

P   = 0.5 * R * AS * V^3

where,

P   = Power (Watts)

R   = Air Density (1.23 kg per cubic meter)

AS  = Swept Area

V   = Wind Velocity (Meters per Second)

However, there are system inefficiencies and I am setting my efficiency factor at 0.3; therefore, I hope to harvest about 19 watts of power.  From this power, the plan is to generate 1.5 amps of current at 13.2 volts to charge a 12 volt battery.  In order to generate the current, the next step calculates the cut in speed needed to begin charging the battery.

Cut In Speed

The following equation was used to calculate the cut-in speed:

C = (VW * TSR * 60) / (Pi * D)

where,

C   = Cut in Speed (RPM)

VW  = Wind velocity in meters per second (4.6)

TSR = Tip Speed Ratio (For a three bladed system, I chose TSR = 4)

Pi  = 3.14159

D   = Diameter of the wind turbine in meters (1.116)

From the above formula, cut-in speed is 315 revolutions per minute (RPM).

Number of Turns per Coil

Since cut-in speed and the power are now known, the next step is to find the number of turns of magnet wire required to generate 13.2 volts.  To get the total number of turns, the following equation was used:

N =  E / ((B * AM) / P)

Where,

N = Total number of turns (for all coils)

E = Voltage generated at cut in speed (13.2)

B = Magnetic flux in teslas ( 1 tesla = 12,000 guass)

AM = Area of the Magnetic Pole in square meters

P = Period (the time it takes the turbine to complete one rotation)

Using the above formula, 5,907 turns of wire are needed.  In the calculations, 0.66 tesla was used for the magnetic field.  0.660Tesla was used because the magnetic field is measured at the poles.  The magnets I selected are rated 13,200 gauss - BrMax.  I know the coil will not see 13,200 gauss so I swagged it.  Since I have eight poles and six coils, I determined 125 turns are needed per coil.  The turns per coil was found by dividing the total number of turns by the product of the coils and poles.  The formula for this calculation is as follows:

TC = N/(P * C)

Where,

TC = Turns per coil (123, round up to 125)

N = Total number of turns (5,907)

P = total number of poles (

C = Total number of coils (6)

The next step is to package 125 turns of 16 gauge magnet wire into a coil for magnets 25.4 mm x 25.4 mm x 12.7 mm.  On this board, I read the air gap between the rotors should be about the same as the magnet thickness.  Also, the plan view width of the coil "leg" should be equal to the size of the magnet (25.4).  Since the diameter of 16 gauge magnet wire is 1.67 mm, I am hoping to fit 16 turns per plan view layer.  

big pic 1

16 * 1.67 mm = 26.7 mm (just over 25.4 mm)

Next, I determined the side view thickness of my coils in a two step process.  First, I found out how many layers of side view turns are required by dividing the turns required per coil (125) by the number of turns per plan view layer (15).  

8 = 125 turns / 16 turns (7.8 rounded up to

Secondly, since I have eight side view layers of wire at 1.67 mm each, the thickness of my coil should be approximately 13.4 mm.  I will allow one millimeter of casting material on each side of my coil making the stator thickness about 15.4 mm.  Adding two millimeters of clearance to each side of the stator will result in an air gap of 16.4 mm.  As stated above, the goal was to keep the air gap between the rotors the same as the thickness of the magnets.  While it is a little larger then 12.7 mm I am hoping performance will still be acceptable.

Thanks for your input

i fixed up your post best i could made your youtube url a link and found the photo you wanted to display in your files and made it a link it was to large pixels wise to make it display in the story we have strict rules about that here. less than 150kb file size and less than 640 x 480 pixels.
Kurt

[kurt]

holding a wind turbine in your hands and letting it spin up in the wind is a very dangerous thing to do good way to get badly hurt....

[electronbaby]

TSR of 4 seems a little on the slow side.

[Flux]

On first glances you seem to be on the right lines but there is an awful lot of material to go through and I just haven't done it.

You seem to be correct in the power in the wind and at that wind speed 30% efficiency may be possible.

You are also correct to use about 1/2 the magnet Br figure as you should be working half way down the curve. 660mT is probably reasonable.

What is more difficult to deal with is how the alternator will match the prop at your chosen wind speed. I will see if I can find time to run some of your figures and see what it looks like. You have been incredibly thorough and i am inclined to think you are close.

I do agree with Roy that tsr4 is slow but if you reach tsr4 at your optimum wind speed with a prop of nominal tsr6, cutting in at something like tsr7 or a bit higher than you are probably very close. The tsr drops very rapidly above cut in unless you have some clever electronic loading scheme. With nominal tsr6 you will hit stall somewhere below tsr4 so you may be expecting to stall somewhat above 15mph if you have g0t it right.

10 plus mph average is actually a good wind area and with the typical Rayleigh distribution you would probably be better off designing your optimum wind speed for something like 15mph.

If you are cut in and doing something by 8mph and you are on peak at 15mph then you may find that your average power works out considerably better than your prediction based on 10. something mph. The mean of the cubes is a lot higher than the cube of the mean.

Flux

[Flux]

I have had a quick look at this. Your cut in speed looks ok but I don't get 125 turns for that cut in speed, it would give you nearer 24v by my estimates. I make it nearer 70 turns for 12v.

I think you could use thicker wire with perhaps#14.

With your 125 turns and 16 wire it will be too powerful for a 1.1m prop and will stall.

If you use a more realistic size of wire that you could get in then perhaps the resistance would come up enough to be a better match and reducing the turns to get a realistic cut in speed may get you near.

From instinct I suspect that your 1" x 1" a 1/2" magnets would be a reasonable choice for the 1.1m prop.

Flux

[Flux]

Having had another look I have realised how small your prop is. That alternator is way too big for a tiny prop.

Your actual tsr is likely to be 4 not a figure running near stalled as I first thought.

The alternator would better suit a 1.5 m prop or even bigger.

Don't expect anything useful from a 1.1m diameter at your nominal 10mph, it will be cut in but not doing anything significant. I think as it stands with your alternator it will hit hard stall and limit you to a few tens of watts whatever the wind speed.

To get useful power from tiny diameter machines you need to match for higher wind speed unless you are happy with a few watts.

If you have built the alternator as you describe then it would suit 24v far better with that prop.

Flux

[dsmith1427]

Thanks.  Your input is greatly appreciated.  On another note, I am a little unclear on the application of TSR.  I understand what it is and how it applies to the cut in speed equation:

Cut-InSpeed = (60 * Windspeed * TSR) / (Pi * Sweep_Diameter)

However, how does one determine the TSR?  I read that three bladed systems recommended a TSR equal to four or five.  In the same document, it stated the angle of the blade at 0.75R and a TSR of four should be approximately 6 degrees.  Is this angle based on the airfoil and the angle at which the airfoil optimizes the lift to drag ratio?  Any insight is appreciated.  Once again, thanks and have a great day!

Don

[Flux]

In general low tsr props have more blades. For tsr over 10 you may choose 1 blade. For tsr over 7 you may choose 2 blades. For the range below tsr8 most choose 3 blades as they have no balance and vibration problems. If you want to go below tsr4 then structurally it may be simpler to use more small blades then 3 wide ones.

Typically most machines work with tsr between 5 and 7. Tsr4 is slow and puts more restraints on the choice of alternator.

For a given tsr there is an optimum angle for the blades but this angle depends on the blade profile, some high lift blades with curved mean line will need to be set at a less angle than common NACA or Clark Y profiles.

If in doubt then choose tsr 6 or 7 for axial flux machines. If you have iron cored alternators you may choose something lower to help start up.

I wouldn't choose tsr from your formula. You need to be running near the correct tsr in the efficient working range of your alternator but at cut in it is better to be above the ideal tsr as it will fall quickly with load. If you choose tsr6 for your working region then you will be probably at tsr8 at cut in. At 10mph you will be down to something like your design figure and it will fall in higher winds as the load comes on unless you have some clever tracking scheme.

If you choose your tsr to be correct at cut in you will drop below it over the whole of the working range and you will be stalled far too early. It makes no sense to match the prop to the alternator losses below cut in.

By definition tsr is the ratio of tip speed to wind speed but you want this to occur at the peak power point on your prop's curve, this the value of tsr normally quoted and is what you should aim for in the range of useful lower winds.

Flux

[ghurd]

What Flux said,

plus windstuffnow Ed has a $5 blade program.  It helped me understand everything that was happening with my windmills.  Many of the values (such as number of blades) can be changed showing the effect on the rest of the variables.

Worth $5 for sure.  Saved me that much in paper and calculator batteries.  

G-