Calculation of neutral current
- Thread starter him76er
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cadpoint
Senior Member
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- Durham, NC
mcclary's electrical
Senior Member
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√((Ia?+Ib?+Ic?)-(Ia*Ib)-(Ib*Ic)-(Ia*Ic))
How about you do it
How about you do it
cadpoint
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- Durham, NC
That formula won't cut it. It is based on purely resistive loading.
Not resistive, but equal power factor, which is not the case in the OP.
The neutral current is the vectorial sum of the phase currents.
mcclary's electrical
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I have what he needs, I'm just trying to find it. I think it's on another laptop. I'll post in a bit.
No, it is based on purely resistive loading.
Equal power-factor loading also works, but is provisional on the power factors all be leading or lagging, and not a combination of leading and lagging.
mcclary's electrical
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- VA
I have the perfect app for this, but it will not let me upload for some reason, it keeps saying invalid file
Hameedulla-Ekhlas
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Smart$:
Incase we have a linear load this formula is correct otherwise for nonlinear load we have a different formula for neutral current calculation.
mcclary's electrical
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- VA
This is where it came from, although it won't let mee upload it to here for demonstration.
http://demonstrations.wolfram.com/ACThreePhaseNeutralCurrent/
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- Connecticut
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- Engineer
I'm curious where this formula came from. I know its suppose to provide an "approximation" of what neutral current will be. But if you use it in this case, you get an answer of 10Amps, when the actual answer, as Smart$ points out, is 58.865Amps.
An approximation of 10Amps vs an actual current of 58.86Amps isn't a very good approximation. Where did this formula come from?
gar
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- EE
100524-0803 EST
him76er:
Are you taking a test and this is a question on it? If so it is a vector sum problem and from that information you should be able to solve the problem.
If I did not make any mistakes the answer is approximately a lagging angle of 2 radians relative to the first specified voltage and a current approximately 15% of the average of the three current magnitudes.
This may or may not be the correct answer because I might have made some mistakes. You need to solve the problem yourself. If it is a real application, then just measure the neutral current.
Using the sum of the squares etc formula I calculate a value about 47% higher than my vector sum value.
All of my evasive answers here are in case this is a class problem. If my calculations are about correct, then they serve as a check for you after you do your own calculations.
Edit: My answer is wrong because I used both 40s as lagging.
.
him76er:
Are you taking a test and this is a question on it? If so it is a vector sum problem and from that information you should be able to solve the problem.
If I did not make any mistakes the answer is approximately a lagging angle of 2 radians relative to the first specified voltage and a current approximately 15% of the average of the three current magnitudes.
This may or may not be the correct answer because I might have made some mistakes. You need to solve the problem yourself. If it is a real application, then just measure the neutral current.
Using the sum of the squares etc formula I calculate a value about 47% higher than my vector sum value.
All of my evasive answers here are in case this is a class problem. If my calculations are about correct, then they serve as a check for you after you do your own calculations.
Edit: My answer is wrong because I used both 40s as lagging.
.
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winnie
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- Springfield, MA, USA
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- Electric motor research
The formula is exactly correct for a limited set of conditions. The formula is a useful approximation to the real world because in the real world we often end up pretty close to those conditions.
If you assume that there are no harmonics and that the three current phase angles are all 120 degrees apart, than this formula is exact. In a large system with reasonably balanced loading, or if you are calculating a small load such as a resistive heater where the elements are not evenly distributed on the phases, then the formula works.
In the example in the original post, the phase angles were wildly different than 120 degrees apart, so the formula gives a significantly incorrect result.
-Jon
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- Connecticut
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- Engineer
Jon, thanks. It appears that people are just misapplying that formula, or making the assumption that it applies for all systems.
mcclary's electrical
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- VA
No one assumed, hence my other post and post #`12. Please read everything before posting
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I did read everything.
Original Post.
Wrong equation for the problem. Gives a result of 10A, when actual answer is 58.86A.
Nice demonstration, but doesn't really help to determine the neutral current for the OP.
I've seen this neutral current equation quoted several times in recent posts, and in each case it was misapplied or based on assumptions.
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mcclary's electrical
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- VA
Understood, but I felt as though the op was trying to get us to do a homework question, so I didn't bother reading closely enough, but once that was acknowledged, your post seemed arrogantly redundant.
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Redundant maybe, arrogant, definitely not.
As I said, I've seen that formula mentioned in several recent posts in which it did not provide a correct result. I asked, since the formula seemed to be providing incorrect answer, where did it come from an what was its application. Jon was kind enough to educate me.
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