Calculating voltage of pm generator

[watts up?]

I'm attempting to build my own pm generator and am trying to calculate voltage output. I see the formula for voltage output is N = number of turns of a coil, A = area of the coil and M = rate of magnetic field change in Tesla's/sec.
So V=NxAxM Ok I get most of that,but what is eluding me is how to calculate the area of the coil ? Does it mean the wire diameter x the length of wire used, or length x width of the coil (in the case of a square or rectangular coil) ?
Thanks very much for any and all help.
Watts up.

 

[Hesch]

I see the formula for voltage output is N = number of turns of a coil, A = area of the coil and M = rate of magnetic field change in Tesla's/sec.

Yes. It could be rewritten:

V = dΨ_v/dt ,

ψ is the flux = B * A , Ψ_v = Ψ*N.

or length x width of the coil (in the case of a square or rectangular coil) ?

Yes.

Your problem is to calculate B. B = μ * H, and some magnets are specified by their B-field. But when you build in the magnets in connection with some iron circuit, the specified B-field will change ( probably raise ).

To make the generator effective, ( optimized ), you must regard the width of the airgap, between the magnets and the coil, to be not too small and not too large. You must use magnetizing characteristics to calculate the optimal width. Optimizing the airgap means to create as much magnetic energy in the airgap as possible, because it's the change in this energy that makes the electrical power.

 

[watts up?]

Ok thank you much for your reply !

 

[Hesch]

The magnetic energy in the airgap is calculated as:

E = ½*B*H * volume_airgap

If you enlarge the the volume of the airgap ( making the width larger ) the B and H-field will be decreased, and vica versa. If the width is zero, the energy is zero. Therefore: Not too small and not too large.

 

[cordin]

Hi, I have the same question actually. Could I possibly add to this thread with a concept of my own, which may shed some light on the subject? I would like to add a drawing here, because the answer would really depend on the configuration?

 

[cordin]

Hi, I have the same question actually, but am still in the dark as to Faraday's proper application. Could I possibly add to this thread with a concept of my own, which may shed some light on the subject? I would like to add a drawing here, because the answer would really depend on the configuration?

 

[cordin]

Here is the concept attached. The target is 220 V; I have 3 rings of solid copper coil of 220, 300 and 400 windings respectively joined in series. The cross section area is about 3 mm^2 with a very thin insulation, which I hope can handle over 25 Amps. These are embedded in the (blue) casing, 2 to the top and the outer one to the side. Generator magnets revolve around the coils and consist of 3 rings (each 4 block magnets in height), 24, 36 and 44, totaling 416. I have arranged them to alternate poles every half revolution. From what I understand, for 50 Hz I require 3000 rpm on a 2-pole system for this configuration. There is a 1 mm air gap.

There are a number of things I still am not sure of though:
1. Firstly, would the magnetic flux be effective at all in this configuration?
2. Is the air gap correct? Here I am totally in the dark concerning B and H fields!
3. How does one properly calculate the magnetic flux area here?

I am not a physics pro, so I only understand things in simple terms.
Any assistance would be greatly appreciated!

 

[cordin]

I forgot to mention the magnet specs: 9x9x10mm long Ceramic grade 8 blocks, From a website app, I get a field strength 1mm from the surface of about 0.1158 Tesla.

 

[Hesch]

Could I possibly add to this thread with a concept of my own,

Well, I think you must post a new thread, but I'm not a "member of the staff".
I think you are disrupting this thread.

 

[cordin]

OK, my apologies. I have tried a new thread, but it's almost impossible to get a useful response from anyone. I would just like to discuss this idea briefly to check if I am on the right track!

 

[Hesch]

I have tried a new thread, but it's almost impossible to get a useful response from anyone.

Try again.