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EMF expression in electrical generator

  • Thread starter Galfer
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  • #1
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Hi all!

I'm studying electrical generators, and while I was trying to come up to the expression of EMF induced in a conductor I got two different results following two different ways, and I can't find the error.

With sinusoidal B, with [tex]e=bvl[/tex] I get an RMS value [tex]E=2K_{f}f\phi[/tex], while with [tex]e=\frac{d\Lambda}{dt}[/tex] I get [tex]E=\pi K_{f}f\phi[/tex]

Following the calculations; can you please tell me where the error is?

Case 1:

p = polar pairs
r = radius
ƒ = frequency
Kf = form factor

[tex]\tau=\frac{\pi r}{p}[/tex]
[tex]v=2f\tau[/tex]
[tex]E=K_{f}B_{m}lv=K_{f}\frac{\phi}{\tau l}l2f \tau=2K_{f}f\phi[/tex]

Case 2:

[tex]e=\frac{d\Lambda}{dt}=\frac{1}{2}\frac{d\phi}{dt}[/tex]
[tex]E_{M}=\sqrt{2}E=\frac{1}{2}\phi_{M}\omega=\frac{1}{2}\sqrt{2}K_{f}\phi2\pi f=\pi f\sqrt{2}K_{f}\phi[/tex]
[tex]E=\pi K_{f}f\phi[/tex]

I thought it could be a [tex]\frac{2}{\pi}[/tex] due to a mean value, but the flux is the same in both the expressions...
 
Last edited:

Answers and Replies

  • #2
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Just adding some notation and clarifications about my calculations:

[tex]v = conductor\; velocity[/tex]
[tex]\tau = polar\; pitch[/tex]
[tex]l=rotor\; and\; conductor\; length[/tex]
[tex]\Lambda=magnetic\; flux\; of\; B\; linked\; with\; the\; coils[/tex]
[tex]\phi = polar\; magnetic\; flux[/tex]
[tex]B_{m} = mean\; value\; of\; B[/tex]

[tex]B_{m}=\frac{\phi}{\tau l}[/tex]

And the passage from Λ to 0.5Φ:

[tex]e=\frac{d\Lambda}{dt}=\frac{1}{2}\frac{d\phi}{dt}[/tex]

is because with one conductor I've half a coil.

I can't solve it, if you've any ideas it'd be great.
 

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