Current Equation from Magnetic Field

[Venomshocker]

I am interested in finding out if there is an equation that you can use to calculate the amount of 'current or voltage' produced in a single copper wire by knowing the 'strength' and 'speed' of a moving permanent magnetic field that is perpendicular to said wire. Application is that of a permanent magnet generator.

Now I am assuming that the 'size(as in square area)' of the magnetic field, and also the 'length' and 'diameter/gauge' of the copper wire will also have some relevance.

This simply just seems to be the right hand rule, except I am having difficulty finding the appropriate equation. Seems everything out there is in reference to forces on charged particles. :/

 

[sreid]

What you are looking for is

V = Blv

V = Voltage
B = Field strength in Tesla (1 Tesla = 10,000 Gauss)
l = Wire length in meters
v = Velocity of the wire in meters/second

The gage of the wire only effects the source impedance (bigger wire, less resistance).

 

[electricpete]

Or more generally, Faraday's law:

V = N * dPhi/dt
where Phi is the flux passing through the area bounded by the conductor loop. Phi = Integral B dot dA

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[electricpete]

N is number of turns
V = voltage induced in the loop
B is magnetic flux density

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[Venomshocker]

Thanks guys, that's exactly what I was looking for.

My next question is how do you determine how much current a given generator can pump out?

I am aware that the load/impedance determines the current but ultimately there has to be some kind of limit.(I=V/R)

Given the following values pulled of various real world generators:

500MVA @ 15kV ~>33.4kA 15kV/33.4kA= 0.45
722MVA @ 19kV ~>38kA 19kV/38kA = 0.5
1530MVA @ 27kV ~>56.7kA 27kV/56.7kA = 0.48

Now I didn’t take PF into consideration, but Ball Park is it safe to assume for roughly every 1Volt you create with a generator, it can supply roughly 0.48A?

Again current is totally dependent on load….but how are VA ratings for generators determined? Is there some sort of consensual load?

Can various identical output voltage sources supply different amounts of current if the load is identical?

 

[sreid]

Ideally for a generator that is used to create power, the internal resistance should be zero. In reallity, the internal resistance is a small fraction of the load resistance, say 3%. The reason is that you want the power being generated to be used in the load, not being dissipated in the generator.

On the other hand, some generators may have a high resistance. A Tachometer (really a generator) is used to generate a voltage with little to no power requirement.

 

[electricpete]

For a synchrounous generator, typically the VA rating would based on the steady state thermal limits of the stator insulation.

You can get little more detailed and talk about a generator capability curve which describes the limits in the P-Q plane. For the overexcited region, the field windings are usually limiting. For the underexcited region, the core end-iron heating from axial flux is usually limiting. In between these two regions, close to unit power factor, the VA limit applies based on thermal limit of the stator coil insulation.


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[Venomshocker]

What I am interested in is, ultimatley how to calculate how much current a given 'SET' voltage can put out from a permanent magnet generator, assuming ideal situation with superconducting wires, and unity PF.

Im looking for ball park numbers/ratios/formulas.

 

[electricpete]

So, throw all heating constraints out the window?

Ok, the superconducting device still has self inductance. The next constraint that I know of will have to do with the fact that there is a limit of real power which can be pushed through an inductance.

P = |V1| |V2| * sin(delta) / |XL|
where P is power, V1 is internal voltage, V2 is external infinite bus system voltage |XL| exp(delta) is the inductive reactance from the generator to the bus. Once delta reaches 90 degrees, you have a pole slip with big damage. Stable limit for genrators are much lower.

I have the sense you are searching for something else... a V*A limit inherent in the magnetic system. I dont think it exists except to the exxtent that you consider realistic effects such as heating and self inductance.

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[Venomshocker]

Yes, throw all heating constraints out the window.

From what I took in school, there is something known as "Regulation Up" characteristic. At a SET PF, with increasing load and current, eventually the 'terminal voltage' begins to decrease. This happens even at unity PF.

I am curious as to WHY this happens? I am assuming 'terminal voltage'= 'output voltage' of the generator. Is this also due to self inductance?

 

[IRstuff]

I don't know if I have the right concept, but consider motion in a gravitational field. For small objects and small motions in Earth's field, you ignore the Earth and treat it as a constant. However, when you consider an object like the Moon in the same field, there are all sorts of interactions, tidal bulges and locking, etc.

Likewise, if the current-induced magnetic field from the wires is small, there's no big impact to the overall system calculation. But, if the currents get really large, with their larger magnetic fields, you'd expect either some sort of shielding against the PM field to occur, or, that there might be sufficient interaction with the PM to cause the PM to heat up and lose strength.

TTFN

FAQ731-376

 

[Venomshocker]

You cannot heat up a PM with an rapidly changing opposing magnetic field can you?

 

[electricpete]

"At a SET PF, with increasing load and current, eventually the 'terminal voltage' begins to decrease. This happens even at unity PF.....I am curious as to WHY this happens? I am assuming 'terminal voltage'= 'output voltage' of the generator. Is this also due to self inductance?"

If there is a generator hooked to a simple constant impedance load, terminal voltage will droop as load increases.

If the generator is part of the network, there may be other factors controlling or affecting terminal voltage.

Either way, there is a real power limit created by self-inductance associated with:
Pmax = |V1| |V2| * sin(delta) / |XL|
where sin(delta)<1

In the isolated generator scenario as terminal voltage V2 droops, Pmax decreases.

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[electricpete]

I should have started my last post by saying: "Yes, you're right about the terminal voltage droop with increasing load"

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