#68022  07/27/06 02:22 PM
Calculating neutral loads.

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Registered: 08/15/05
Posts: 248
Loc: Waterford, MI, USA


Whats the formula to find the amps a neutral is carrying? And if 220 doesn't need a nuetral to work,.. is there current on the nuetral? So I need some schooling an nuetrals. Any and all help is appreciated.
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#68023  07/27/06 06:15 PM
Re: Calculating neutral loads.

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Registered: 05/18/02
Posts: 1779
Loc: N.C.


Trick, Whats the formula to find the amps a neutral is carrying? are you asking about three phase or single phase? And if 220 doesn't need a neutral to work,.. is there current on the nuetral? Since a neutral is not needed in this voltage (not true of all world voltage systems)source, it doesn't exist to have current flowing on it. Roger

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#68024  07/27/06 06:55 PM
Re: Calculating neutral loads.

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Registered: 08/15/05
Posts: 248
Loc: Waterford, MI, USA


Single phase.
Three phase wouldn't hurt either.
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#68025  07/28/06 04:14 AM
Re: Calculating neutral loads.

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Registered: 12/12/04
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For a wye system the nuetral current is calculated by:
N= Sq root of (A^+B^+C^)(AB+AC+CB)
^=squared
For split phase(single phase) the nuetral current is calculated by:
AB or BA.
Does this make sense?

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#68026  07/28/06 11:37 AM
Re: Calculating neutral loads.

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Registered: 08/15/05
Posts: 248
Loc: Waterford, MI, USA


nope it don't make sense.
The Wye formula I see and understand.. except for, What is A, and B?
The single phase. Same situation, what is the A and B refering to. Volts, Watts, amps?
And with this slash () are you tring to say A, minus B?
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#68027  07/28/06 12:55 PM
Re: Calculating neutral loads.

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Registered: 12/12/04
Posts: 1091
Loc: nowhere usa


Trick, the A,B,C are the currents in phase A, Phase B, and Phase C.
So for a split(single)phase service, the nuetral current is the unbalnced current flowing between phase A and phase B.
So if 30 amps were flowing on phase A, and 15 amps were flowing on Phase B, the nuetral current would be: Yes it is a minus...
AB=N 3015=15
On a wye service, if we had 150amps on A, 110amps on B, and 75amps on C.
N= Sq root of (A^+B^+C^)(AB+AC+CB) N=Sq root of (150^+110^+75^)((150*110)+(150*75)+(75*110) N=Sq root of (22500+12100+5625)((16500)+(11250)+(8250)) N=Sq root of (40225)(36000) N=Sq root of (4225) N=65 amps
How does that look?
[This message has been edited by Dnkldorf (edited 07282006).]

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#68028  07/28/06 02:13 PM
Re: Calculating neutral loads.

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Registered: 05/18/02
Posts: 1779
Loc: N.C.


Trick, to go along with Dnk's posts here are some graphics for help in visualizing the single phase neutral BTW, I like to refer to the ends of single phase windings as legs, not phases. So with this said, A and B in Dnk's post would relate to legs 1 and 2 Roger [This message has been edited by Roger (edited 07282006).]

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#68029  07/30/06 01:55 PM
Re: Calculating neutral loads.

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Registered: 08/15/05
Posts: 248
Loc: Waterford, MI, USA


Thats great guys. I see now. Thank you.
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#68030  07/30/06 02:44 PM
Re: Calculating neutral loads.

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#68031  07/30/06 05:12 PM
Re: Calculating neutral loads.

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Registered: 11/29/02
Posts: 773
Loc: Oradell NJ USA


All the above assumes that all loads are pure resistors, like incandescent light bulbs. If we talk about a computer server farm fed by 3 phase 120V, things get more complicated. Computer power supplies draw large spikes of current at the very peak of voltage of the phase they are connected to. When the voltage on phase A peaks, computers on phase A draw their spikes of current and return them to the neutral. Later, 120 degrees of the 60Hz cycle, phase B peaks, and computers on phase B draw their current spikes, and return them to the neutral. Likewise, 120 degrees later for phase C. No overlap of these current spikes (in terms of time). Spikes last about 5% of the cycle time. Note that the neutral will see 3 times as much spike current than any one phase (assuming equal phase loading). Thus you'd need a thicker wire for the neutral.
[This message has been edited by wa2ise (edited 07302006).]

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