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Question about star conection?
By Jerry, Section Controls Posted on Mon Feb 12th, 2007 at 04:19:32 AM MST
Its the 20 volts equals 17 volts thing.

I'm brain storming about this quandry??????
Its accepted that if the voltage accross one phase of star 3 phase is 10 volts then the voltage accross any 2 phases will be 17.
As near as I can tell with my math skills this is 85% of 20 volts. We know if the coils were in proper phase 10v + 10v would equal 20 volts but since these 2 phase are out of sync by 120 degrees the voltage comes out 17 volt or about 85% of 20v.
I'm thinking that the amperage may be less also?
If each phase produces 5 amp each and the phases are wired in sires then one would think then the voltage would be 17v but is the amperage, 5 amps as it would be if the phases were in phase or dose the amperage have the same 85% of 5 amps isue just like the voltage does?
If so then 85% of 5 amps equals 4.25 amps and 85% of 20 volt equals 17 volts.
If the losses did not exsists then 20v X5 amps equals 100 watts. On the other hand 4.25 amps X 17 volts equals 72.25 watts.
I know there is another phase here but I was just thinking and wondering if there was an amperage diferance to?
Thanks for the answer that may clear my head.
JK TAS Jerry

Question about star conection? | 8 comments (8 topical, 0 editorial)

Re: Question about star conection? (3.00 / 0) (#1)
by Flux on Mon Feb 12th, 2007 at 01:10:31 AM MST
(User Info)

Your voltage ratios are correct for a sine wave, if the waveform is distorted then the factor will be other than 1.7

This only applies to voltage. Current is determined by the emf( open circuit voltage) divided by total circuit resistance.

The circuit resistance is also affected by the type of connection, a star connected winding will effectively have 3 times the equivalent resistance of a delta.

These normal ratios apply when loaded with resistors. When you start loading into rectifiers you not only have to think about the resistance but the conduction angle during when the diodes conduct, it is very different from resistance loading and a machine efficiency is lower with rectifier loads.

It seems from the tests I did that the equivalent resistance may be lower with your Jerry connection and that may be why you get more out. It does come at a price to low load efficiency, but again your method comes out better than delta.

Flux

Re: Question about star conection? (3.00 / 0) (#2)
by electrondady1 on Mon Feb 12th, 2007 at 07:26:15 AM MST
(User Info)

your preaching to the choir here jerry , i think i was the one who came up with the term "jerry rigged".
the only place i can see a down side is if you will need to trans port low voltage dc any distance, there would be line loss .

[ Parent ]

Re: Question about star conection? (3.00 / 0) (#4)
by Jerry on Mon Feb 12th, 2007 at 09:09:38 AM MST
(User Info) http://www.dplusv.com/Photo-03.html

Funny you should mention that. My dad was a preacher and he was very pastionet about his belifes also. I think he would preach to anybody that would listen.

Thanks for the info guys. Now off to the lab.

JK TAS Jerry

Airheads Page

[ Parent ]

Re: Question about star conection? (3.00 / 0) (#3)
by finnsawyer on Mon Feb 12th, 2007 at 08:39:25 AM MST
(User Info)

You are making this overly complicated. The resistances of the two phases add directly, while the voltages add as vectors due to the phase difference (that's why the 1.73 factor). So, the whole shebang acts like a voltage source of 1.73 times a single phase voltage with resistance twice that of one phase. Just put that 'model' for the alternator output for the time it's valid in series with the rest of your circuit. Technically, the coils also would have inductance, which would complicate matters considerably, but at the frequencies usually encountered can be ignored.

GeoM

Re: Question about star conection? (3.00 / 0) (#5)
by racer on Mon Feb 12th, 2007 at 12:15:13 PM MST
(User Info)

The way I understand the star wiring setup with all coils 120 degrees out of phase from each other and the second and the third legs are actually causes an increase in resistance to equal the total of all three phases and causes the voltage output to drop to 1.732 net increase in voltage.

Now with that being said what happens in Ed Lenz 12/36 setup? This is what I'm trying to find out because it seems to me that all of the coils in the star wiring are pushing electrons in the same direction. The coils in his setup look to be only 60 degrees out of phase and he states a 2.3? increase in voltage . Would this mean that the net effective resistance in his star wiring design equals closer to only two coil sets in the star design? If so this should allow for a higher Amp caring capacity in Star setup because of lower resistance.

Re: Question about star conection? (3.00 / 0) (#6)
by Flux on Tue Feb 13th, 2007 at 01:24:52 AM MST
(User Info)

racer
Your first paragraph is confusing to put it mildly.

You are correct that in Eds conventional 3 phase winding it is wound with a displacement of 60 deg. This is sorted out in the connections by reversing the start and finish of the middle phase. This comes out as a winding with 120 deg displacement.

An alternative and possibly less confusing way would have been to take the starts from alternate coils ( not adjacent ones).This would then give a conventional 120 deg displacement without needing to reverse one phase. The idea of winding with 60 deg displacement and reversing the middle phase is the one usually adopted but you will meet both methods.

Flux

[ Parent ]

Re: Question about star conection? (3.00 / 0) (#7)
by GerryS on Tue Feb 13th, 2007 at 06:33:45 PM MST
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I'm confused how the third phase voltage adds in. If the first phase gives 10 volts, then including the second phase yields 17 volts. What about the third phase? Is the total voltage:

10+0.7(10)+0.7(10)=24V

or is the total voltage always 85% of the total? In this case:

(10+10+10)x0.85 = 25.5?

Re: Question about star conection? (3.00 / 0) (#8)
by TomW on Tue Feb 13th, 2007 at 06:48:53 PM MST
(User Info)

Gerry;

Do to the phasing angle [spacing] only adjacent phases add. one phase is dropping and the other is rising and they "add" in the area occupied by both pulses. This only happens on 2 at a time due to the phase angle. I used to have a nice graphic that showed how this works but I can't seem to find it. You also are only tapping voltage across pairs of coils so the third one is not even in the active circuit.

Good luck getting a handle on it and hope this does not just muddy the waters.

Cheers.

TomW

"Education consists mainly of what we have unlearned."--Mark Twain
[ Parent ]

Question about star conection? | 8 comments (8 topical, 0 editorial)

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