Alternator RPMs

[TomG]

If we gear up the input to an alternator (by *2 for example), how does this change the design requirements of the alternator? (Assuming we want to charge the same battery bank)

We've doubled the RPM and halved the torque, so the power available to the alt is the same. (Assuming perfect gears)

The RPM has doubled, so the speed of the magnets past the coils has doubled. This will double the voltage generated, so we can halve the number of turns/coil to get back to the same voltage.

But if we halve the number of turns/coil, the coils will be half as thick. So we can either close the airgap (not sure of the effect of this?) or double the thickness of the wire we're using (or just wind two-in-hand).

That doubles the cross-sectional area of the wire. Combining that with the halving of the number of turns, the resistance of each coil will now be 1/4 of what it was.

There are 3 loading cases I'm interested in here.

1) Load resistance = infinity

No current flows, so we're just measuring the open-circuit voltage. Easy, it should be double what the original alt made.

2) Load resistance = 0

Short circuit, so there's no voltage difference. We measure the short-ciruit current. No power will be produced in the external load: all the power will dump as heat inside the coils. Given that the resistance is 1/4 what it was, and Power=R*I^2, I'd expect the current to be double what it was in the original alt, meaning the same power (ie all of it) is being dumped in the coils. Is that right?

3) Load resistance = same as coils

Perfectly matching the load to the alt, to get the maximum power from the load. This means the load should be 1/4 of the perfect load for the original alt. Since the voltage of the battery bank is the same, and its resistance is 1/4 what it was, the current through it should be 4 times what it was. Which can't be right, because that would mean the load is getting 4 times more power than the original alt! And Since the power being supplied to the new alt is the same, that can't happen!

Clearly I've misunderstood something. How can good ol' V=I*R hold across the load, when V is unchanged, and R is 1/4 what it was?

Can someone with better understanding than mine share please?

Thanks!

[Flux]

If you double the speed you use half the turns of twice csa wire ( 2 in hand if you wish). Absolutely no point in changing the air gap, that should be set to run magnets at near half Br.

The voltage will now be the same for a given prop speed and cut in will be the same.

Resistance will be 1/4  of what it was previously. Short circuit current should be 4 times what it was before but why you would want to know it I can't imagine.

Maximum power point is a myth, but for wind you do tend to end up working at about 50% efficiency to match the prop. In any other application you work with high efficiency and internal impedance would be a fraction of the load. For any other application it would be crazy to do otherwise, you would burn the thing out and regulation would be dreadful.

So effectively if you double the speed you could get about 4 times the power but the prop would now be stalled. You need a larger prop that runs slower so you need an even bigger speed increase to maintain the intended alternator speed.

Not sure where this gets you as you have only asked part of the question, I can only give you information based on the alternator as would happen on a bench test.

Flux

[Flux]

You got me confused with all this maximum power point nonsense.

In real life it will be thermally limited. With the same volts and 1/4 the resistance you can produce twice the current for the same heating. So power is double at twice speed.

Flux

[TomG]

You got me confused with all this maximum power point nonsense.

?? Not sure why it's nonsense. As I understand it, the maximum external power you can draw (under fixed conditions) is when the external resistance matches the internal resistance. If that's not true, could you explain why, 'cos I must have another misunderstanding!

In real life it will be thermally limited.

Isn't that a question of implementation? This is sort of a bench-test thought experiment. As long as I know how many watts are being dumped in the coils, preventing it burning is not something I'm too worried about.

With the same volts and 1/4 the resistance you can produce twice the current for the same heating. So power is double at twice speed.

So in cases 2 and 3, the current will be double what it was in the original alt?

That's what I expected in case 2.

In case 3, I think I might have found my mistake. The EMF in the coils has to drive current through not only the internal resistance but also the external resistance. Hence the total resistance of the circuit is 1/2 of the resistance of the coils in the original alt. So the current is double that of the original alt, not quadruple.

Which means we're demanding twice the power of the original alt, so we'll stall the input. The input would need double the torque to keep up.

Is that all correct?

So, if we don't have twice the torque available, we could either use smaller coils (1-in-hand rather than 2-in-hand), or use magnets with half the power (either thinner or lower grade)? Either of which should bring the power demands of the alt back to those of the original one?

[Flux]

This maximum power transfer thing is ok as far as it goes, it's fine with electronic circuits at low power but it is never applied to power machinery.

If you had a power station alternator loaded in this way you would only get half volts when fully loaded so your lights would be dim and people in general would be very unhappy with a 2:1 change in volts from no load to full load. Even more serious is the problem of heat from losses. A typical large alternator is likely over 95% efficient and still needs very elaborate cooling to keep it from frying. there is no way it could dissipate 50% of rated power as losses. The manufacturers are contracted to supply a machine of guaranteed efficiency and a small fraction of one percent under claimed performance would be subject to a heavy penalty. The generating company doesn't want to use turbine power to heat the power station more than necessary.

Even small diesel driven alternators are in the 80% region. This crappy scheme of loading wind turbine alternators down to about 50% to make some sort of use of the blade power is unique and only works because wind power is very low duty cycle and even then there is constant worry about burn out. Put it on a bench test and it will burn out for sure in a short time.

Your final point, yes you require twice torque at the same speed which the prop can't supply ( at the same wind speed and tsr).

You can save copper and magnet for the same prop size and power out by using a speed increasing transmission if you tolerate all the associated problems. That was what had to be done before neo magnets.

Flux

[TomG]

Yay, thanks for sharing knowledge Another little query cleared up. I love this place!

Ahem.

Slightly off topic now, but I'm pretty sure you know more power electrics than me, so:

why don't we run alts at 90+% efficiency, instead of 50%? Battery banks have pretty low resistance (unless we add an in-line power resistor), so we'd need really low resistance coils. Few turns, many-in-hand. Which will give us lower voltages, so we'd need higher RPM to reach cutin voltage.

Is this to do with matching the (mostly linear) power curve of the alt to the (cubic) power curve of the blades? A more efficient alt will require the same torque but at higher RPMs, hence demanding more power at any given RPM, which stalls the blades in anything less than a hurricane? But increasing the diameter of the blades would give us even lower RPM, so we can't collect more power that way without adding gearing?

[Flux]

Yes indeed it is all to do with this matching of a linear alternator characteristic to the prop's cube law curve.

If you get things right in low wind at cut in, you hit stall not much beyond cut in.

In the old days before decent permanent magnets, the field and iron losses prevented you having significant output below 10/12 mph. From there upwards the match is much better on the steep part of the prop curve.

The problem with wound field machines was that they curved over as the reactance dominated and it was difficult to hold the prop loaded, stall was never an issue.

With the coming of decent neo magnets and the elimination of iron in the coils it becomes possible to start producing from about 6mph, but the stall issue suddenly present problems never noticed before.

Most starting out with PMAs may never have known otherwise but to me the sacrifice you make in high winds is very noticeable with performances way below what I was obtaining.

That is probably why I have concentrated on this load matching issue.

Flux