How to calculate max current potential in a windmill generator....

In summary, to calculate the current a windmill generator can produce, you need to know the voltage, the number of turns, the resistance of the coil, and the inductance of the coil.
  • #1
Steven Robinson
7
3
Hi, new member with a question that I might be asking wrong. A week ago I read an article or something on building a windmill generator, which I come back to every few years. So, this week, I was reading about how to calculate the voltage based on loops, turns, Teslas, area, RPMs and such, and I keep falling down when it comes to current and torque.

I think it must be a common question, how much CURRENT can I produce from my generator that produces 12 - 15 volts? So far, I have read things about the size and cross section of the wire used in the coil, the number of turns, and the speed at which the field changes, etc. But invariably, when it comes down to it, it seems you have to know the load on the circuit, and the torque (of the turning windmill?), or the resistance of the coil, which you calculate if you already know the current...

I am trying to find out if there is a formula or calculator that will tell me, for a give set of magnets, a given set of coils with a given set of turns, using a given gauge of wire, turning a given RPM, my generator can produce a maximum sustainable current before things start melting, or slowing down.

I am not asking for a specific answer, just some guidance on what I should be looking for, because I keep finding the same articles and I am going around in circles. It seems to me if you want to size a generator, at some point the size of the wiring as it relates to the number of turns and the speed of the rotation have to come together to produce the current and therefore Watts. But I can't find the connection or I am not recognizing it when I see it.

Thank you.

PS, been out of high school for 30 years, not bad at trig, terrible at calculus, remember the very basics on electricity.
 
Engineering news on Phys.org
  • #3
Thanks for moving the post.

My best guess is that I have to calculate the impedance of the coil in the generator, and then based on the voltage, calculate the amps? Or do I still need a load of some kind?
 
  • #4
This is why realworld engineering problems are complicated (open ended). I would treat your system as a linear series of power conversions, but I'll leave it to you to list them and pick which is the start and which is the end. What you seem to be doing is starting at the middle and thus being unsure of which way to go.
 
  • Like
Likes Steven Robinson
  • #5
Thanks, Russ.

Here are the calculations I have run:

Rotors: 32 magnets, N42 grade, 1/2 in thick, wedge shaped, 2 inches long, 1.57 inches at the top, .78 inches at the bottom, arranged N/S in two discs facing each other, 1/2 in apart. According to the K&J Magnets website calculator, this should produce an average field strength of .4936 T. Theses magnets make a disc 8" in diameter on the outside, 4 inch in diameter on the inside, six inches in diameter average. The Area of the magnet is 2.3562 square in or .00152 sq.meter. The width at the six in radius is 1.175 in or .0298 m, giving a total circumference at that radius of 9.424 in, with 8 pairs of magnets, and therefore 8 sections of coil. The rotors will spin at 200 RPMs, 3.33 rps, so the 6" radius will be moving at 28.27 in/s or .718 m/s. The magnet will take .0415 seconds to pass a coil.

T=.4936
Delta t = .0415 s ( in know I mixed a lot of units, I am sparing you the conversions )
Voltage target is 6 V (over a single magnet, because I am guessing it will be -6V of the next, opposite magnet, give a total potential of 12V, right?)

Calculate V=-N(BA/T)
6=-N(.4936*.00152/.0415)
6/.0181=-N

N=331 turns, divided into 8 parts, gives 41 turns per each coil section, to produce 6 volts. 41 turns is technically possible to fit between 1/2 in gap, but would not allow three phases of coils. For this exercise let's assume a single phase.

How am I doing so far?

Next, I need to calculate the Inductance, resistance, and them impedance, right?

Wire gauge is 22, which is .64516 mm diam. Total length will be 31.809 meters, plus another .319 m for the legs between coil bundles. Online calculators give the wire resistance as 1.684 ohms. Inductance for the coil through air based on size and turns on an online calculator is .0218H. 8 coils at 200 RPMs is 26.66 Hz.

r=1.684
26.66Hz
.0218H
2*pi*26.66*.0218H = 3.65 ohms or inductive reactance

Impedance is pythangorian (3.65^2+1.684^2)^.5 = 4.02 ohms. Let's say 4.

So current in the coil is 6 volts/4 ohms = 1.5 A. Since I have two opposing magnets, the total difference is 12 v so 3 amps?
 
Last edited:
  • #6
Steven Robinson said:
I think it must be a common question, how much CURRENT can I produce from my generator that produces 12 - 15 volts? So far, I have
russ_watters said:
This is why realworld engineering problems are complicated (open ended). I would treat your system as a linear series of power conversions, but I'll leave it to you to list them and pick which is the start and which is the end. What you seem to be doing is starting at the middle and thus being unsure of which way to go.
Did you understand the comment by @russ_watters ? You can't really do a detailed design of the generator portion of the windmill until you define a bunch of other parameters of the installation. Can you list what some of those other parameters might be, and take a first cut at the values? Those will feed directly into the generator's physical and electrical design...
 
  • Like
Likes Steven Robinson and russ_watters
  • #7
What are those parameters?, I don't understand what you are looking for. I am not actually building a generator, I am trying to understand more about how the current is calculated in the coils. What are the specific "parameters" you consider important that I need to start with?

If you mean requirements, site, wind speed, blade diameter, air density, and that stuff, then let's establish an idealized environment so that we can do the math.

The first requirement is that I want to use the generator to charge a 12 volt deep cycle battery. I assume this means I need to generate more than 12 volts, similar to a car alternator, like 14 volts, right? I don't know how many amps the battery is, or how many I need to generate. I assume since "trickle" chargers exist, that the amperage can vary, and with that variance, the time of charging varies. So let's assume that the system needs to "keep" the batter charged, not fully charge it every day. Maybe allow three days to charge it.

Requirement: 14 volt DC output, to charge a 12 volt deep cycle RV battery, fully charged over three days.

Parameter: Windmill blades are three feet long, for a six foot diameter circle, and the wind speed and air density is steady and can turn the blades at a regular 200 RPMs, 24x7, with as much torque as required to do the job.

Parameter: Generator will be 3 ph, Y design, axial flux, permanent magnets, with rectifiers to convert the AC into DC. Also, the coils with be core-less or air-core, no iron laminates or ferrite. The stator and rotors should be 18" in diameter or less, to avoid creating problems with the blades.

Unknown at this time - size and strength of magnets, number of magnets, number of turns on the three coils, gauge of the wire.

What other parameters do we need to get started?
 
  • Like
Likes berkeman
  • #8
Steven Robinson said:
What are those parameters?

The first requirement is that I want to use the generator to charge a 12 volt deep cycle battery.
This is the starting point: the requirement of the project/generator. But it is not complete:
I assume this means I need to generate more than 12 volts, similar to a car alternator, like 14 volts, right? I don't know how many amps the battery is, or how many I need to generate. I assume since "trickle" chargers exist, that the amperage can vary, and with that variance, the time of charging varies. So let's assume that the system needs to "keep" the batter charged, not fully charge it every day. Maybe allow three days to charge it.
You'll need to get more specific. We could guess for you, but you're better off picking one yourself, even if it is just a guess. Try googling "deep cycle battery capacity", picking one as an example, and we'll go from there.
Parameter: Windmill blades are three feet long, for a six foot diameter circle, and the wind speed and air density is steady and can turn the blades at a regular 200 RPMs, 24x7, with as much torque as required to do the job.

Parameter: Generator will be 3 ph, Y design, axial flux, permanent magnets, with rectifiers to convert the AC into DC. Also, the coils with be core-less or air-core, no iron laminates or ferrite. The stator and rotors should be 18" in diameter or less, to avoid creating problems with the blades.
Where did all of that come from? This is a real project, isn't it? When I described above the "power conversions" and "direction" of the problem, I was referring to everything from the input power to the output power and everything in between. What the battery gets is the output. What the turbine bladed grab from the air is the input. You've made some assumptions about the input that may or may not relate well to your output. That isn't a good idea.
 
  • #9
Not a real project. As I said, I have visited this subject more than once over my life from the old “windstuffnow” page that is defunct to a journal article I just read from Taxila, Pakistan about axial flux, air core designs. I just picked a few diameters that would be close to what I would do in real life. For instance, I doubt I could build a sturdy windmill blade longer than three feet. I don’t know how to actually design a blade yet, or calculate the availe power in the wind, so I need to set those parameters in a default way. Several websites I have crossed suggest 200 rpm is an acceptable range for a home wind mill to get power from. So, use that as an idealized input, strictly for the math.

The original WindStuffNow design was a single magnetic rotor and single stator. However, after discovering the magnetic field fall off as the distance from the magnet grows, an knowing that a coil of wire has a thickness, I figure using two magnetic rotors with the stator coil in the middle would give more consistent voltage, as the field would be nearly uniform for much of the coil area.

So, we have an idealize rotation, a max possible diameter of the rotor and stator, and a voltage requirement. Like I said, voltage on a coil is easy, but it’s the amps that make electricity useful, by and large, IMHO. So, a 12 volt 120 Ah battery can accept up to 30 amps during charging, according to a convenient web site. Does that mean it would literally take four hours to charge? Probably a little more since the charging would lose some power to heat. You tell me, I haven’t found the answer.

But let’s say that’s how it works, so to charge a 120Ah battery at 12 volts in 24 hours we’d need 5 amps. 2.5 amps for 48. So let’s add 2.5 amp, 12 volts to the requirement.

Now what? In real life, I would collect a year of wind data at my site before doing this, BTW. So don’t worry. The weather station is on the Xmas list.
 
  • Like
Likes Tom.G

Related to How to calculate max current potential in a windmill generator....

1. How do I determine the maximum current potential in a windmill generator?

To calculate the maximum current potential in a windmill generator, you will need to know the wind speed, the diameter of the windmill blades, and the efficiency of the generator. You can use the formula I = (0.5 x p x A x v^3 x e), where I is the current potential, p is the air density, A is the swept area of the blades, v is the wind speed, and e is the efficiency of the generator. By plugging in these values, you can determine the maximum current potential of the windmill generator.

2. How does the wind speed affect the maximum current potential in a windmill generator?

The wind speed is a crucial factor in determining the maximum current potential of a windmill generator. As the wind speed increases, the amount of kinetic energy that can be harnessed by the blades also increases, resulting in a higher current potential. However, there is a limit to how much energy a windmill can extract from the wind, so at very high wind speeds, the current potential may plateau or even decrease due to the strain on the blades.

3. What is the role of air density in calculating the maximum current potential in a windmill generator?

Air density is a measure of how much air is present in a given volume. This is important because the density of air affects the amount of kinetic energy that can be extracted by the windmill blades. Higher air density means more air molecules passing through the blades, resulting in a higher current potential. Air density can vary depending on factors such as altitude and temperature.

4. Does the diameter of the blades affect the maximum current potential in a windmill generator?

Yes, the diameter of the blades plays a significant role in determining the maximum current potential of a windmill generator. The larger the diameter of the blades, the more surface area there is for wind to push against, resulting in more kinetic energy being harnessed and a higher current potential. However, larger blades also mean more weight and drag, so there is a trade-off to consider.

5. Can the efficiency of the generator impact the maximum current potential in a windmill generator?

Absolutely, the efficiency of the generator directly affects the amount of electricity that can be produced from the kinetic energy of the wind. A higher efficiency means more energy can be converted into electricity, resulting in a higher current potential. It is essential to choose a generator with a high efficiency rating to maximize the current potential of a windmill generator.

Similar threads

  • Electrical Engineering
Replies
4
Views
241
  • Electrical Engineering
Replies
26
Views
2K
Replies
13
Views
1K
Replies
4
Views
827
  • Electrical Engineering
Replies
9
Views
889
  • Electrical Engineering
Replies
5
Views
254
  • Electrical Engineering
Replies
5
Views
2K
  • Electrical Engineering
Replies
2
Views
653
  • Electrical Engineering
Replies
28
Views
13K
  • Electrical Engineering
Replies
14
Views
1K
Back
Top