amperage on a neutural

[gmtt]

SPARKYNSJ , what makes you think it is too much current thru the neutral? The neutral current of the panel is the vector sum of the two-phase currents and I won't discount any value you measured in the neutral line.
Let's say, of the VA at an angle 0 deg, VB at angle 120 deg and VC at angle -120 deg from a Y transformer, phases VA and VC are connected to the panel in question. Let's say phase A load is pure resistive such that phase A current is IA at angle 0 and phase C load is pure inductive such that the phase C current is IC at angle ?30 deg. Adding IA at angle 0 and IC at angle ?30 will result a substantial amount current flowing thru the panel neutral wire.

 

[chris kennedy]

FWIW, I received a PM from the OP stating that this is indeed 2 ungrounded and the grounded conductors of a 208y/120 system.

 

[roger]

FWIW, I received a PM from the OP stating that this is indeed 2 ungrounded and the grounded conductors of a 208y/120 system.

That makes more sense.

Roger

 

[Dennis Alwon]

FWIW, I received a PM from the OP stating that this is indeed 2 ungrounded and the grounded conductors of a 208y/120 system.

I guess I sensed it.

 

[chris kennedy]

I guess I sensed it.

Your my favorite superhero.

 

[SPARKYNSJ]

i measured the amperage again. The panel is being fed from a 3 phase 4 wire 120/208 main. The subpanel is being fed by a 70 amp breaker w/ 2 2/0's hots and 1 2/0 neutural. one hot has 22 amps the other 41, the neutral is carrying 30-35 amps. how is this possible? Always thought that the neutural would carry the difference between both hots..

 

[LarryFine]

I guess I sensed it.

A human non-contact voltage tester.

 

[roger]

i measured the amperage again. The panel is being fed from a 3 phase 4 wire 120/208 main. The subpanel is being fed by a 70 amp breaker w/ 2 2/0's hots and 1 2/0 neutural. one hot has 22 amps the other 41, the neutral is carrying 30-35 amps. how is this possible? Always thought that the neutural would carry the difference between both hots..

You would be correct if you were dealing with a single phase 120/240 volt system but, since you are dealing with a wye you would have to figure it as follows.

SQRT I?A + I?B - (IA x IB)

SQRT 22 x 22 + 41 x 41 - (22 x 41)

SQRT 484 + 1681 - 902

SQRT 2165 - 902 = 1263

SQRT 1263 = 35.54 amps

Roger

 

[gar]

091208-2116 EST

SPARKYNSJ:

If you draw a vector diagram and assume both phases only have resistive loads from the hot lines to neutral and the phases are 120 deg apart, then the result is:

Draw a vertical line P1 to P2 with a length of 41. Classify P1 as the south end and P2 as the north end. Next draw a line from P2 to the south-east with an included angle of 60 deg between this new vector, called P2 to P3, and the vector P1-P2. Make P2-P3 with a length of 22. Now measure the length from point P1 to point P3. Its value is 35.54871, and this corresponds to your reading of 35 A. The phase angle of the current in the neutral is 90-57.6 = 33.4 deg from the voltage P2 to P1.

.

 

[Smart $]

... Always thought that the neutural would carry the difference between both hots..

Electrically it carries the unbalanced current. When the current through the "hots" is in phase through a neutral-connected junction, the amount of current directed through the neutral conductor is exactly the arithmetic difference. The more the "hots" currents are out of phase through the neutral-connected junction, the less mutual current there is going through the "hot" conductors, so the neutral conductor has to make up for that.

In your case, the voltages are 120? out of phase with respect to the neutral. The respective "hot" currents will likely be out-of-phase accordingly, handling the same type of load (i.e. having nearly the same power factor). This is why you get the sum for neutral current if both "hots" are connected to the same phase. Being 120? out-of-phase yields a neutral current somewhere in between the arithmetic sum and difference.

Roger has given an example of the basic math solution... but such method does not account for differing power factors.

 

[jghrist]

FWIW, I received a PM from the OP stating that this is indeed 2 ungrounded and the grounded conductors of a 208y/120 system.

So we have about 60 posts based on the OP saying "3 wire 100 amp panel, 120/240 volt" and now he PMs you to say it is completely different?

I say we let him figure it out himself.

 

[SPARKYNSJ]

Neutural

Neutural

Thanks for the answers guys.. I have to study them to fully understand the math. Thanks again.

 

[mull982]

091208-2116 EST

SPARKYNSJ:

If you draw a vector diagram and assume both phases only have resistive loads from the hot lines to neutral and the phases are 120 deg apart, then the result is:

Draw a vertical line P1 to P2 with a length of 41. Classify P1 as the south end and P2 as the north end. Next draw a line from P2 to the south-east with an included angle of 60 deg between this new vector, called P2 to P3, and the vector P1-P2. Make P2-P3 with a length of 22. Now measure the length from point P1 to point P3. Its value is 35.54871, and this corresponds to your reading of 35 A. The phase angle of the current in the neutral is 90-57.6 = 33.4 deg from the voltage P2 to P1.

.

Shouldn't the included angle between these two vectors be 120deg instead of 60deg due to the fact that they are L-N loads?

 

[Smart $]

Shouldn't the included angle between these two vectors be 120deg instead of 60deg due to the fact that they are L-N loads?

120? when tail to tail... but when doing vector addition graphically, they are arranged tail to head and thus a 60? included angle.

 

[mull982]

120? when tail to tail... but when doing vector addition graphically, they are arranged tail to head and thus a 60? included angle.

Yup you are right that is my mistake.

 

[gar]

091209-2110 EST

mull982:

Try to visualize it this way:

Two sine waves are being added together.

Call the angle of the vector P1-P2 0 deg and indicate phasing by putting an arrow at the P2 end.

Next create a second vector P2'-P3 with its arrow at P3. This vector is at an angle of -120 deg relative to the first vector. To add these two move P2'-P3 so that P2' is coincident with P2.

The resultant current vector is P1-P3.

You can turn these three vectors into two right triangles. From this you can solve for the magnitude and angle of P1-P3 relative to P1-P2. Or you can use a CAD program for a graphical solution.

Consider the case where the 120 deg is changed to 180 and you will see that the second vector is parallel to the first and in the opposite direction. Thus, the magnitude of the second subtracts from the first.


Another way:
Go to http://en.wikipedia.org/wiki/List_of_trigonometric_identities
A little more than half way down the long page, and just above "Other sums of trigonometric functions" find the sentence "More generally, for an arbitrary phase shift, we have", and you will see the equation for the sum of two sine waves of different amplitude and phase shift. Note: the vector diagram is easier.

.

 

[gar]

091210-2250 EST

A question for you to consider.

Why is it possible to use vector math to solve problems about the relationship of sine waves of the same frequency but possibly of different amplitudes and phases?

To answer this consider what is the result of of adding sine waves of the same frequency. My previous post may give a hint.

.

 

[paulcbrowne]

Check for harmonics

Check for harmonics

You didn't say what your 3rd leg measured. It sound liek your legs are not balanced.

I have measured this sort of imbalance with computer equipment, particularly with older switching power supplies. Usually, it's a phenomenon measurable only in the 3rd order harmonics using something like a Fluke Power Quality meter.