1. Junior Member
Join Date
Feb 2005
Location
Posts
23

## neutral current calculation....again

Hi,
I read through all 150 or so postings on the unbalanced neutral thread and I examined the xls calculators but I have a lingering question that I hope you all can answer.

My scenario is this....120/208V, 3 phase 4 wire system. Three individual single phase motors, all at different pf's.

I plug the variables into the neutral calculator to get an answer.....no problem.

My question is this....why does the spreadsheet utilize the phase angles as leading the phase voltage? For instance if one of my motors has a pf of .707 on phase A, shouldn't the calculator use an angle of -45 degrees?

BB

2. Originally Posted by basicbill

My scenario is this....120/208V, 3 phase 4 wire system. Three individual single phase motors, all at different pf's.
Sorry but I have to be clear, these are 120V motors?

3. Junior Member
Join Date
Feb 2005
Location
Posts
23
yup....3 - single phase, 120 volt motors.

4. Originally Posted by basicbill
...why does the spreadsheet utilize the phase angles as leading the phase voltage?

5. Junior Member
Join Date
Feb 2005
Location
Posts
23
The neutral calculator spreadsheet is at ...

http://forums.mikeholt.com/showthrea...culator&page=8

If you unlock it with 'abracadabra' you can see the formula used to calculate the phase angle.

Once again, it appears that the phase angles are considered to be leading and not lagging as would be appropriate.

Thanks again.

6. Originally Posted by basicbill
The neutral calculator spreadsheet is at ...

http://forums.mikeholt.com/showthrea...culator&page=8

If you unlock it with 'abracadabra' you can see the formula used to calculate the phase angle.

Once again, it appears that the phase angles are considered to be leading and not lagging as would be appropriate.

Thanks again.
Ahhaa! That's the calculator I made. Good thing I edited the post, or else I wouldn't have remembered why

Explanation: When working in the positive phase displacement angle domain, all phasor angles are reversed (i.e. multiplied by -1). The reason for this is because phase angle displacement is actually in the negative domain. We always assign 0° to Phase A... but Phase B lags Phase A by 120°, which is technically a phasor angle of -120°, and Phase C is -240°.

A lagging power factor has a negative phase angle displacement, which gets added to the voltage phase angle. For example, a lagging PF of .707 is -45°. If this load is connected to Phase A, the phase angle of the current would be 0° + (-45°) = -45°. When working in the positive domain for phase angle displacement it would be 0° + (-1 × -45°) = 45° for the lagging PF.

I actually prefer working the actual (negative) domain for phase angles to avoid the confusion. Unfortunately, I seem to be in the minority... so I made the calculator work in a manner that others seemed to prefer :cool:

7. Junior Member
Join Date
Feb 2005
Location
Posts
23
thanks Smart.
If I understand you correctly, you are saying that our phasor representations, where A phase is drawn at 0, B at 120 and C at 240 ..... with a ccw rotation actually should represent lagging currents as lagging ccw?

<light bulb slowly illuminating>

8. Originally Posted by basicbill
thanks Smart.
If I understand you correctly, you are saying that our phasor representations, where A phase is drawn at 0, B at 120 and C at 240 ..... with a ccw rotation actually should represent lagging currents as lagging ccw?

<light bulb slowly illuminating>
Yes.

In the positive phase angle domain, voltage phasors initially A@0°, B@120°, and C@240° would rotate CW. That is, each phasor would rotate about its tail point in the CW direction as time elapses. At t=1/3 cycle, voltage phasor A would be at -120°, voltage phasor B would be at 0°, and voltage phasor C at +120°. For a lagging current, the phase angle would be to the CCW side of the voltage angle.

If you prefer everything to be in the negative domain, the calculator shouldn't be to hard to change. If you need my help, just "holler"... :grin:

9. Senior Member
Join Date
Nov 2004
Posts
3,172

## Wait a minute:

Originally Posted by basicbill
thanks Smart.
If I understand you correctly, you are saying that our phasor representations, where A phase is drawn at 0, B at 120 and C at 240 ..... with a ccw rotation actually should represent lagging currents as lagging ccw?

<light bulb slowly illuminating>
Smart must be left handed. I cannot see that phase sequence has anything to do with power factor.

An inductive impedance carries a positive phase angle. Dividing this impedance into the applied voltage yields a current which lags the voltage. For example, each of these cases yields a current lagging by 45 degrees.

V @ 0/Z @ 45 = I @ -45
V @ 120/Z @ 45 = I @ 75
V @ -120/Z @ 45 = I @ -165

The current angle is the difference in the voltage angle and impedance angle.

Also, PF is the same for + or - angles--always positive.

KISS = Keep It Simple Smart.

10. Originally Posted by rattus
Smart must be left handed. I cannot see that phase sequence has anything to do with power factor.

An inductive impedance carries a positive phase angle. Dividing this impedance into the applied voltage yields a current which lags the voltage. For example, each of these cases yields a current lagging by 45 degrees.

V @ 0/Z @ 45 = I @ -45
V @ 120/Z @ 45 = I @ 75
V @ -120/Z @ 45 = I @ -165

The current angle is the difference in the voltage angle and impedance angle.

Also, PF is the same for + or - angles--always positive.

KISS = Keep It Simple Smart.
Your taking the easy way out...

Is 120° actually 120° or is it -240°...

...and is -120° actually 240° or is it -120°

I can look at your equations and by the result, I know your phasor rotation is CCW.

If you can teach everyone to use 0°, -120, and -240° for phases A, B, and C voltages, respectively, then I can keep it as simple as it gets... :grin:

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•