Is 11.60 Volts the Total or Partial Voltage Induced in a Coil?

[jearls74]

Hello everyone, i have a question about induced voltage that i need help answering, in my calculations i believe i have made a mistake about the total voltage induced in the coil. My calculations follow as: 390 RPM / 60 Seconds = 6.5 Hz x 6 pole pairs = 39 Hz x 6.28 = 244.92 radians per second x .001529 m2 = .37448268 x .5 Tesla = .18724134 x 62 turns = 11.60 volts. My question is this: is this 11.60 volts the total voltage generated in the coil ? or is it the voltage generated in one side of the coil since the coil passes through the stator in 2 places and one side is influenced by north pole and the other side is influenced by a south pole? would the 2 voltages add together to get 23.21 volts for the total voltage of the coil?

 

[m.s.j]

When you apply the area of rotating loop in formula, indeed you considered two leg of loop rotating in induced voltage calculation and you should not used any additional multiple factor.

Total induced voltage on the loop
eind = eba + ecb + edc + ead
= vBl sin θab + vBl sin θcd
= 2 vBL sinθ

If the loop is rotating at a constant angular velocity ω, then the angle θ of the loop will increase linearly with time.
θ = ωt

also, the tangential velocity v of the edges of the loop is:

v= r ω

where r is the radius from axis of rotation out to the edge of the loop and ω is the angular velocity of the loop. Hence,
eind = 2r ωBl sin ωt
since area, A = 2rl,

eind = ABω sin ωt

Finally, since maximum flux through the loop occurs when the loop is perpendicular to the magnetic flux density lines, so

Ø max=AB


Thus,

e ind= Ø max ω sin ωt


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[Bob S]

Hello jearls74-
It sounds like you are building an alternator (dynamo). Is this a permanent magnet rotor, wound-coil stator configuration? Is it a 12 pole (6 pole pair) PM rotor and a 12 pole (6 pole pair) stator? I agree with your calculation of 11.60 volts per pole.

Label the stator poles alternately "N" and "S". Connect the 6 "N" poles together in series; same for the "S" poles. Each of these circuits will produce about 70 peak volts output. Are you planning on rectifying the outputs? If so, you can 1) connect the "N" and "S" circuits in series for 140 volts output for bridge recifier, 2) connect the "N" and "S" in parallel for a 70 volt peak bridge output**, 3) connect in center-tapped (CT) (not bridge) full wave rectifier circuit with two diodes.

Look up the recommended current limit for your coils. For example, 18 Ga. wire has a current carrying capacity of about 2.3 amps.

I hope this helps.

** Postnote #1. If the paralleled coils are connected with the wrong polarity, it will be a short circuit. It is better to have separate bridge rectifiers for the "N" and "S" coils.

Bob S

 

[jearls74]

Thank You Bob, that was the answer i was looking for. I am am building a 12 pole permanet magnet dynamo with a 12 pole wound coil stator. I plan on using option #1 and winding in series to get the maximum peak voltage and and using a high amperage rectifier bridge from a 200 amp alternator. Each of the 3 phases should produce 139.20 volts, so wired in the star (wye) configuration i should get 240.81 peak volts output? is that right? Thanks again for your help and insight Bob.

 

[jearls74]

Also, i want to thank you M.S.J , i do appreciate that you took the time to answer my question even though the calculations were for a different set up of the generator. So many view my question and offered no insight. Thank you M.S.J AND Bob S.

 

[Bob S]

How are you connecting the 12 stator coils to get 3 phases?

Bob S

 

[jearls74]

Hello Bob, the stator i have constructed has 36 slots for the 3 phases to be wound in, so i have 12 slots for each phase winding, so 6 coils per phase.

 

[Bob S]

Are you building something like this axial flux alternator?

http://www.google.com/url?sa=t&sour...xZz9Aw&usg=AFQjCNF5rK4FM_dg4DFjFOBklzPG7-EWtA

See Table on page 7. How many magnets do you have in your design?

Bob S