CAUGHT FLAT FOOTED

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sundowner

sundowner

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West Wisconsin
Ok Guys I'LL admit I should know this one... I can't seem to find it though. Here's my question: In a three phase lighting panel three hots are run to three loads... one nuetral is used for the three loads, how do I calc the nuetral current. Lets say the three loads are all the same. (Some of this is hypothetical, some of it picking on the guy who has'nt been all the way around the block yet).

Thanks for an explanation

Steve
 
Re: CAUGHT FLAT FOOTED

if, all three phases are present, and equal, and they are all resistive then the current on the neutral is zero.

if they are not balanced then use this formula.
Neutral Current = Square root of (A squared + B squared + C squared) - (AB)+(BC)+(CA)

if you have harmonics........ then i am lost.
 
Re: CAUGHT FLAT FOOTED

Thanks jbwhite, I knew I'd get a quick answer.

What would you call the addition though, could I call it adding the currents tangentially, or would I be more correct in saying that they are adding logorithmically.

Thanks again

Steve
 
Re: CAUGHT FLAT FOOTED

lets wait for an EE to chime in.

all I know about it is... it is not addition. whether logrim or tangent.... hmmmm i just use the formula that someone else wrote.
 
Re: CAUGHT FLAT FOOTED

yea! what don said....

and let me add that is why i write alot of RFIs that begin.. "request engineering support for..."
 
Re: CAUGHT FLAT FOOTED

First, if the load currents are equal in magnitude and phase relative to the phase voltages, they add to zero. They don't have to be resistive.

However, 3rd harmonic currents do not cancel as do the fundamental currents.

The math is called vector algebra although these are not true vectors; they are phasors. You need to be up on trig for this.

Anyone up for an example?

[ December 10, 2005, 08:18 PM: Message edited by: rattus ]
 
Re: CAUGHT FLAT FOOTED

ok rattus,
just when i thought i had learned enough about this save the phone number of a good EE.
Lets see an example.
 
Re: CAUGHT FLAT FOOTED

jb, I was hoping someone else would do it so I could pick on them, but I will do it, and let them pick on me.

Let's say the fundamental currents are:

Ia = Ip @ 0 deg.
Ib = Ip @ -120
Ic = Ip @ -240

That is, Ip is the RMS value of each phase current in the wye, and the separation is 120 degrees. This means the load is perfectly balanced. We don't know the PF and don't care.

Now sketch out the phasor diagram. It looks like the letter "Y" lying on its side--one leg pointing due east. Then using complex notation,

Ia = Ip + j0
Ib = Ip[-cos(60) -jsin(60)] = Ip[-0.5 -j0.866]
Ic = Ip{-cos(60) +jsin(60)] = Ip{-0.5 +j0.866}

Ia + Ib + Ic = 0

The operator "j" indicates the positive vertical component of a phasor, and "-j" indicates the negative vertical component of a phasor.

You can do this graphically and see for yourself. Just draw the phasor diagram as precisely as you can and break the 3 legs into horizontal and vertical components. You will see that the horizontal components cancel as do the vertical components.

Be aware that this exercise completely ignores any harmonic currents.
 
Re: CAUGHT FLAT FOOTED

Originally posted by hardworkingstiff:
Originally posted by jbwhite:


if they are not balanced then use this formula.
Neutral Current = Square root of (A squared + B squared + C squared) - (AB)+(BC)+(CA)
Click to expand...
Missing some parenthesis?
Click to expand...
I agree. It should be The sqaure root of:(A^+B^+C^)-(AB+BC+CA)
 
Re: CAUGHT FLAT FOOTED

3.jpg
 
Re: CAUGHT FLAT FOOTED

Originally posted by georgestolz:
3.jpg
Click to expand...
I believe this formula works well on currents which are spaced exactly 120 degrees apart. This is the case if the PFs of the three currents are equal. If the PFs are nominally equal, the formula should provide a good approximation.
 
Re: CAUGHT FLAT FOOTED

The more general formula, where ^ is squared, capital letters the magnitude as above, lowercase letters are the phase angles:

sqrt(A^ + B^ + C^ + 2ABcos(a-b) + 2BCcos(b-c) + 2ACcos(a-c) )

When all the phase angles are 120 degrees apart, it becomes the other formula (since all the cos are -0.5).
 
Re: CAUGHT FLAT FOOTED

This post is for SUNDOWNER

Is this a delta 120/240
or a wye 120/208?

the 120/208 even if you have ballanced the load on the hots the grounded conductor can be carrying 200% of the hot load if harmonics/induced loads IE gas lighting,computers ect
 
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