Alternator Calculation Conundrum

Hi folks; I've been teaching myself alternator design, trying to come up with a reasonably large alternator to run from a wind turbine to power my ocean-going raft. I've been composing my work as I go, mainly to keep track of it, to back up from my many brain-numbing confusions, but with a view to eventually posting it as a guide for future explorers. I'm an electronics tech, learned my trade in the late sixties, when we had real electricity, but I never got into electrical machine design, and it's proving to be a much deeper mudpool than I thought it would be. I asked some questions a short while ago, and received some welcome help, but soon wandered inadverdently from the trail again. So, I thought I'd post what I've got, to ask for any critique or corrections any kind soul would care to make. I think it's all good up to the current determination, where I'm getting numbers I don't believe. But, it could all be wrong! Any comments at all will be humbly and gratefully heard.

Thanks in anticipation, here it all is...

How to calculate the voltage and current output of an axial-flux alternator (All in SI units)

Example:

Design considerations: We need 120 Volts 3-phase AC at 40 Amps, at 180 rpm. The alternator needs to produce 170 peak Volts AC at 57 peak Amps, because the formulas all give instantaneous, peak values. The results will be reduced by a factor of 0.707 to give standard RMS value (this is valid because the current and voltage are pure sinewave). Having AC output allows transforming to another voltage if required. 3-phase produces a smoother output, and also allows the use of transmission cable whose conductors are 75% of the size of those used in single phase. The frequency of the AC is not important, so the rotation speed can be low, which puts much less stress on everything involved, and keeps the noise levels down. 24 poles (electromagnets) are used because more poles give more effective generation at low speed. Elecromagnets enable control of the exciter field and thus control of the output voltage; useful for automatic voltage regulation. If output voltage control is not required, permanent magnets can be used instead; change the figures for B in the formula to suit the field strength of the magnet used. Neo blocks are about 0.6 Tesla.

The rotor disk holds 24 electromagnets, spaced 1cm apart, whose cores are mild steel 4mm thick. The field strength B is 1 Tesla, which is the approximate optimum for mild steel, taken from the B/H curve graph. From the same B/H curve graph, a magnetic field intensity H of 800 A-t/m is required to maintain a flux density B of 1T. H is A-t/m; MMF per unit length (ampere-turn/meter); H = NI/l; 100 turns x 32 milliamps / 0.004m = 800A.t/m

The electromagnet core is 2cm x 6cm x 0.4cm; perimeter of 16cm. Using wire gauge tables, 32 milliamps can be carried safely without overheating by AWG 34 enamelled winding wire, which is 0.16mm OD, and has a resistance of 0.856 ohm/m. 25 turns will fill the first layer, and 4 layers of 25 turns each will make 100 turns. Each turn is 16cm, 100 turns = 16m. At 0.856 ohms/m, 16m is 13.7 ohms. E = iR; so 0.032amp x 13.7 = 0.438 volt drop over each coil. The 24 coils are in series; 0.438 volts each gives 10.5V as the maximum exciter voltage. 10.5V @ 32mA = 337 milliwatts exciter power draw.

The stator disk has 18 air-cored coils, all spaced 4.3cm apart, each 1cm thick, 5cm wide x 9cm long. Each coil has a median area of 22.5cm² and is oval-shaped, arranged with the long axis in the radius direction. All coil centres are at 26.6cm radius. The 18 stator coils are arranged in 3 groups; 6 in series in each of 3 phase legs. Required phase voltage is 170V, so each of the 6 coils needs 28.33V across it.

The magnets and coils are in close enough proximity to ignore the separation distance in the calculations. In practice, as long as they're within a couple of millimetres, these figures will be fairly accurate.

Output VOLTAGE is calculated as follows:

Volts per turn: V = -N (∆Φ/Δt) This relationship has been found experimentally and is referred to as Faraday's law.

(The minus sign before the N can for these calculations be ignored)

N is one turn: (assumed for simplicity)

Calculate ∆Φ: (how much the flux changes, minimum to maximum, in Webers. Assume minimum is zero)

Magnetic flux Φ = B x Area.

B is 1T; median Area of coil is 22.5cm² = 0.00225m²; (ie area of loop drawn through the centre of the winding in plan view)

Φ = B x A; = 1 x 0.00225; Flux rises from 0 to 0.00225 Webers.

Calculate Δt: (how fast the flux changes from minimum to maximum, in seconds)

Rotor speed is 180rpm. Coils are mounted at 26.2cm radius on rotor; circumference at radius is 1.67m; x 180/60 = 5m/sec. Each coil is 0.05m wide and, as the rotor disk rotates, is half-covered by a magnet in 0.025/5 = 0.005 seconds (from not covered to half-covered is the maximum change, ie when the magnet is directly over one side or leg of the coil)

V=−N (∆Φ/Δt); = -N (0.00225/0.005); = -N x 0.45.

A magnet of 1 Tesla flux density and 12cm² area, moving past a 4.3cm coil at 5m/sec, generates 0.45V (peak) per turn.

So, for the required 28.33 V across each individual coil, 63 turns per coil are needed (in round figures).

Output CURRENT is calculated as follows:

E = iR = - A ∆B/∆t: The induced current depends on both the change in magnetic field and the area of the coil. Each turn of the coil contributes to area A, so the induced voltage will be proportional to the number of turns in the coil. But the induced current does not increase with increasing number of turns because that increases the length of wire and so increases the resistance. Consequently, the resistance used in the formula is just the resistance of one turn, not the whole coil.

Amps per turn: I = V / R (current is the same regardless of the number of turns in the coil)

From above calculations, for one turn;

V is 0.45

Calculate R: R is the DC resistance of the wire in the coil, taken from standard wire tables. AWG 12 enamelled winding wire can carry a current of 41A without overheating; it is 2.05mm OD, and has a resistance of 0.0052 ohm/m.

The stator coil is 1.6cm thick, so 8 turns of 2mm diameter wire will fill the first layer, and 8 layers of 8 turns each will suffice for 63 turns. The inner perimeter of the 2cm x 6cm coil is 16cm. 8 layers is a 1.6cm thick winding, so the outside dimension of the wound coil is 5.2cm x 9.2cm and the perimeter of the outside layer is 28.8cm. The average length of a turn is thus between 16cm and 28.8cm, let's allow 25cm. At 0.0052 ohms/m, 25cm is 0.0013 ohms.

So, the instaneous peak current induced in one turn is;

I = V / R; = 0.45 / 0.0013; = 346 amp (peak). ***SURELY THIS CAN'T BE CORRECT?***

Converting peak values to RMS values:

The 16 coils are arranged in 3 groups of 6 in series and in phase. Each group is swept by a magnetic field twice every revolution (induction occurs when the magnet arrives and also when it leaves). So there are 6 pulses of induction (plus and minus once for each group) every rev; 3 revs per second is 18 pulses of 250 amps (peak) per second.

RMS current is 0.707 x 346 = 244.6 amps, and RMS voltage is 0.707 x 28.33 = 20.03 volts.

Thanks again for reading all this,

regards, Ormusgold