Unbalanced 3-phase Neutral Current Calculation

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AHarb

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I know there are plenty of threads on calculating the neutral current. I even found one that had a spreadsheet which calculated the neutral current based on given power factors and current values. I cannot figure out how to calculate the neutral current if we are give the phase angle between the currents and voltages (theta) and the current values.

I know the formula is In = sqrt(Ia^2 + Ib^2 + Ic^2 -Ia*Ib - Ia*Ic -Ib*Ic) if you're at unity (theta = 0 degrees). I've tried working it out using vector math and here is the best I can come up with for an unbalanced load.

Sqrt ( ((Ia*cos ?a) + (Ib*cos?b) + (Ic*cos?c))^2 + ((Ia*sin?a) + (Ib*sin?b) + (Ic*sin?c))^2)

I know that I'm doing something wrong because plugging in 0 degrees for theta does not work out correctly. Can anyone help me correct the formula?

Thanks in advance. Sorry if this is repetitive and somewhere else on the forums. I was unable to find the answer.

Edit: I just realized it didn't like my symbol for theta. The "?" in the formulas should be theta.
 
Last edited:
AHarb said:
I know there are plenty of threads on calculating the neutral current. I even found one that had a spreadsheet which calculated the neutral current based on given power factors and current values. I cannot figure out how to calculate the neutral current if we are give the phase angle between the currents and voltages (theta) and the current values.

I know the formula is In = sqrt(Ia^2 + Ib^2 + Ic^2 -Ia*Ib - Ia*Ic -Ib*Ic) if you're at unity (theta = 0 degrees). I've tried working it out using vector math and here is the best I can come up with for an unbalanced load.

Sqrt ( ((Ia*cos ?a) + (Ib*cos?b) + (Ic*cos?c))^2 + ((Ia*sin?a) + (Ib*sin?b) + (Ic*sin?c))^2)

I know that I'm doing something wrong because plugging in 0 degrees for theta does not work out correctly. Can anyone help me correct the formula?

Thanks in advance. Sorry if this is repetitive and somewhere else on the forums. I was unable to find the answer.

Edit: I just realized it didn't like my symbol for theta. The "?" in the formulas should be theta.
Click to expand...

Sorry I couldn't figure out how to edit my last post again. I think I may have figured out what I'm doing wrong. I think the angle in the formula above is not the angle between the voltage and the current. If it were, and we were at unity, the voltage and the current vectors would overlap each other making the angle 0 degrees. All of the voltages should always be separated by 120 degrees. I guess I'm supposed to just use the actual angle of the current vector in the formula? Sorry if I'm making this more confusing.
 
AHarb said:
... I think I may have figured out what I'm doing wrong. I think the angle in the formula above is not the angle between the voltage and the current. If it were, and we were at unity, the voltage and the current vectors would overlap each other making the angle 0 degrees. All of the voltages should always be separated by 120 degrees. I guess I'm supposed to just use the actual angle of the current vector in the formula? Sorry if I'm making this more confusing.
Click to expand...
No, that helped - cause this one loses me:

AHarb said:
...Sqrt ( ((Ia*cos ?a) + (Ib*cos?b) + (Ic*cos?c))^2 + ((Ia*sin?a) + (Ib*sin?b) + (Ic*sin?c))^2) ...
Click to expand...

Note:
Symbol "<" means "phase angle"
Symbol "#" means "not equal" (edit)

Just to clarify:
The case you are interested in is where the phase currents are unbalanced and the three phase angles are not the same

|IA| # |IB| # |IC| (edit)


and

<(VA, IA) # <(VB, IB) # <(VC, IC) (edit)


Is this true? If not, pitch the rest out:
AHarb said:
... I've tried working it out using vector math. ...
Click to expand...
This is indeed the method: (and, I'm thinking you already got this)

Note:
<(VA, IA) = <a
<(VB, IB) = <b
<(VC, IC) = <c
IX is a vector
|IX| is the magnitude, <x is the phase angle referenced to the Phase Voltage vector (VX)

Starting with:
-IN = IA + IB + IC

-IN = |IA| <(a) + |IB| <(120 +b) + |IC| <(240 + c) (polar coordinates)
I think this is what you are saying in your second post

For -IN:
Convert to rectangular coordinates. Add real and imaginary. Convert to polar coordinates.

Is this where you are headed? If so, there are a couple of threads - but not to this depth. If this is what youare after - keep going.


ice
 
Last edited:
iceworm said:
No, that helped - cause this one loses me:



Note:
Symbol "<" means "phase angle"
Symbol "#" means "not equal" (edit)

Just to clarify:
The case you are interested in is where the phase currents are unbalanced and the three phase angles are not the same

|IA| # |IB| # |IC| (edit)


and

<(VA, IA) # <(VB, IB) # <(VC, IC) (edit)


Is this true? If not, pitch the rest out:

This is indeed the method: (and, I'm thinking you already got this)

Note:
<(VA, IA) = <a
<(VB, IB) = <b
<(VC, IC) = <c
IX is a vector
|IX| is the magnitude, <x is the phase angle referenced to the Phase Voltage vector (VX)

Starting with:
-IN = IA + IB + IC

-IN = |IA| <(a) + |IB| <(120 +b) + |IC| <(240 + c) (polar coordinates)
I think this is what you are saying in your second post

For -IN:
Convert to rectangular coordinates. Add real and imaginary. Convert to polar coordinates.

Is this where you are headed? If so, there are a couple of threads - but not to this depth. If this is what youare after - keep going.


ice
Click to expand...

Thanks for the help. I think my formula was correct, but I was using the wrong angle. I think I was confusing it with the angle used when calculating power factor. Just to clarify, when calculating power factor, you use cos (theta) where theta is the angle between the voltage and the current vector, correct? For example, if Va = 208 V with at 120 degrees and the current is 10 amps at 90 degrees, is your power factor cos (120 - 90 degrees)? I was using this angle when calculating the current values. I'm pretty sure I'm supposed to be using the actual current vector's angle value.
 
AHarb said:
Thanks for the help. I think my formula was correct, but I was using the wrong angle. I think I was confusing it with the angle used when calculating power factor. Just to clarify, when calculating power factor, you use cos (theta) where theta is the angle between the voltage and the current vector, correct? For example, if Va = 208 V with at 120 degrees and the current is 10 amps at 90 degrees, is your power factor cos (120 - 90 degrees)? I was using this angle when calculating the current values. I'm pretty sure I'm supposed to be using the actual current vector's angle value.
Click to expand...
You have to use the angle relative to a reference angle, typically VAN @ 0?.
 
AHarb said:
Thanks for the help. I think my formula was correct, but I was using the wrong angle. I think I was confusing it with the angle used when calculating power factor. Just to clarify, when calculating power factor, you use cos (theta) where theta is the angle between the voltage and the current vector, correct? For example, if Va = 208 V with at 120 degrees and the current is 10 amps at 90 degrees, is your power factor cos (120 - 90 degrees)? I was using this angle when calculating the current values. I'm pretty sure I'm supposed to be using the actual current vector's angle value.
Click to expand...

Hummm .....
Yes, I'd recommend staying away from any concept deaing with power factor.

Unbalanced currents, V-I phase angles different between the three phases - loads are single phase. One could calculate a power factor, but I don't know what it would mean or what one could do with it.

ice
 
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