12Views4Replies

# Please help with ac generator calculations

Hi I want to make a ac generator and I’ve been trying to use faraday’s law to solve for the coils I need. I don’t know of this is really possible but if anyone can help me or show me how to properly use the equations for faraday’s law I would greatly appreciate it. The specs are listed below I just really want to figure out what gauge wire and at how many windings per coil I would need

__I’m wanting the following specs: __

Less than 10 inch diameter stator

12 volts

Produce around 50 amps

Have 6 or 9 coils to be used in 3 phase

Thank you for your help!

## Discussions

15 days ago

I forgot to mention the Wikipedia article for, "American wire gauge",

https://en.wikipedia.org/wiki/American_wire_gauge#...

It has a table with a column in it, telling us the resistance per unit length, for different AWG sizes of copper wire.

Note that all those numbers should be in good agreement with the formula for bulk resistance,

R = rho*(L/A)

and R/L = rho/A

I mentioned previously.

However it is kind of handy to have these numbers in a table, sorted by AWG sizes, since those sizes are how copper wire is actually specified, and bought and sold, in our world today, for some reason.

AWG sizes always kind of seemed like voodoo to me, so that is actually another reason the AWG table is handy, so I can translate AWG numbers into real diameters, in millimeters, or fractions of an inch, that I could actually see, from putting some calipers on a piece of wire, or imagine.

15 days ago

Well, each coil has some amount of magnetic flux, Phi, linked through it, and the magnitude of that just depends on the angle of the stator, theta. Essentially Phi is a function of theta. Moreover theta is probably changing at a constant rate; i.e. the stator is turning at some constant angular speed, omega.

Phi = Phi(theta)

omega = (d theta/ dt) = constant

I want to know the time derivative of magnetic flux through the coil, because (dPhi/dt) is the main factor in my Faraday's Law equation

Vcoil = -N*(d Phi/dt)

with the other factor, N, being the number of turns.

From the chain rule for derivatives,

(d Phi/dt) = (d Phi/d theta)*(d theta/dt) = omega*(d Phi/d theta)

I can see the magnitude of (d Phi/dt) is directly proportional to angular speed, and that's handy to know. It suggests I can get more voltage, simply by turning the turning the rotor faster. For example, turn the rotor twice as fast, get twice as much voltage.

I still have to puzzle out the magnitude of (d Phi/ d theta). It might be a good guess to say the maximum for this is going to be the flux maximum density seen in the center of the coil, multiplied by the coil's area. That is what it would be for a uniform field, but maybe the field is strong in the center, and weaker at the edges, so the total maximum flux through the coil is only perhaps only half what it would be for uniform field throughout.

How does this flux change with angle? I don't know, but maybe a sinusoid would be a good guess for this. I mean I know this function is periodic in theta, with peaks and troughs. So it is probably something like that.

Another way to find out how much flux I can get through a coil, is to just puzzle it out experimentally. I build my magnetic circuit first, including a rotor a stator, and magnets, and at least one test coil. Then measure the open circuit voltage across this coil, with an oscilloscope or similar tool, while driving the rotor at some known constant speed.

Regarding these questions of what gauge wire to use, for that I suggest writing some expressions for how much power you want to waste in heating the coils, through the necessary action of driving current through them.

For example, suppose I want the output voltage of this generator to be 12 volts, supplying a current of 50 amperes.

How much voltage is it reasonable to expect is dropped across the coil windings? Maybe 2 volts, 4 volts?

Supposing it was 4 volts, which corresponds to electrical efficiency of 75%; i.e 4 V dropped across the copper windings, 12 V across the load, 16 V total.

What is the total resistance in those windings? Answer: (4V/50 A) = 8/100 ohm = 0.08 ohm = 80 milliohm

Then I consider my coil geometry, like if that whole volume is almost filled with copper magnet wire. How few turns, of wire how fat, would give a resistance that small?

I dunno. Do you want me to explain the math for how to do that?

You know, the volume of a wire is its length mulitplied by its cross section, L*A, and that volume does not change too much, when you wind it around in loops into a coil. The electrical resistance of a length of wire also depends on length and cross section area, but differently, as

R = rho*(L/A)

where rho is resistivity, a property of of the conductor itself, with units of ohm*meter, in SI. The value of rho for copper electrical wire is well known, 1.724e-8 ohm*m, I think. I copied that number from a page at engineeringtoolbox, here

https://www.engineeringtoolbox.com/resistivity-conductivity-d_418.html

At this point it kind of hard for me to guess what else to suggest.

You know, I can't just like tell you, which gauge copper wire to use, not without doing a whole bunch of math myself, and as I have pointed out there are also other unknowns you have not mentioned, e.g. how fast the rotor is turning, and how much flux you're pushing through those coils.

However, I am hopeful that you can find some hints in what I have written here, and get started on solving this problem yourself.

15 days ago

50A all up or 50A per phase?

Either way I struggle to get it working with your size limits.

At these powers and 12V you need massive cable sizes and I doubt you get anywhere near enough turns within the space you provide.

Reply 15 days ago

50 A all together and I can do up to a 12 inch stator but I don’t wanna go any bigger than that. I know it’s a big task to as of but I just seeing if it’s possible I want to make a portable ac generator just a design in my head