Calculation of neutral current

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...As for calculating currents without knowing the power factor, you could assume all power factors are the same if all loads are of the same design. If not, it would be best to use approximate, typical power factors. As exemplified by this thread, accuracy of the resulting value can vary widely without knowing the power factor (or its equivalent).

Let's go to further aspects of the neutral current calculation.
What are typical power factor of typical loads such as transformers, lighting, induction motors ?
and if the manufacturers do not mark down the PF of an equipment so how we can measure the PF?

thanks
 

Smart $

Esteemed Member
Location
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Let's go to further aspects of the neutral current calculation.
What are typical power factor of typical loads such as transformers, lighting, induction motors ?
I don't have a list handy, but you may be able to glean a few by googling (internet search)...

and if the manufacturers do not mark down the PF of an equipment so how we can measure the PF?
Seldom do they mark power factor. You can use a power factor meter (or so I've read... never actually seen or used one) or a power quality analyzer.
 
Seldom do they mark power factor. You can use a power factor meter (or so I've read... never actually seen or used one) or a power quality analyzer.

I've heard about PF meter also for example Fluke's one, but never have chance to use it. Power quality analyser also I never seen and seems to be an expensive equipment.
In short, this problem can be easily handled if we have a PF meter, but in my previous reply I talked about a situation when a meter is not available but we still want to know PF. Any other calculation or approximation?

Why hardly do manufacturers mark PF on their equipment? Is is because their PF is low? Is somewhere in NEC it is stipulated that PF must be known or provided by manufacturers.
 

Smart $

Esteemed Member
Location
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I've heard about PF meter also for example Fluke's one, but never have chance to use it. Power quality analyser also I never seen and seems to be an expensive equipment.
In short, this problem can be easily handled if we have a PF meter, but in my previous reply I talked about a situation when a meter is not available but we still want to know PF. Any other calculation or approximation?
Well you could work backwards from calculating neutral current and connect two known PF loads (easiest would be PF=1 loads) to two of the lines and determine the third through measuring the neutral current then back figuring. Even not knowing the angle of neutral current there should only be two possible angles for the measured current and one should eliminate itself from contention simply by the design of the equipment under load.

Why hardly do manufacturers mark PF on their equipment? Is is because their PF is low? Is somewhere in NEC it is stipulated that PF must be known or provided by manufacturers.
I have no answer because I simply do not know why. It is not stipulated anywhere in the NEC AFAIK.
 
Well you could work backwards from calculating neutral current and connect two known PF loads (easiest would be PF=1 loads) to two of the lines and determine the third through measuring the neutral current then back figuring. Even not knowing the angle of neutral current there should only be two possible angles for the measured current and one should eliminate itself from contention simply by the design of the equipment under load.

Without knowing the angle of the third line, the i and j value of the third phase current are not known.
Even you can measure the neutral current, I do not see how we can find the PF of the third phase as we have to solve an equation of two unknowns (i (Ic) and j (Ic))
Could you please explain more to your statement "...Even not knowing the angle of neutral current there should only be two possible angles for the measured current and one should eliminate itself from contention simply by the design of the equipment under load"
What are two possible angles?

Furthermore, there are something to be considered. If you connect resistive loads to phase B and C (to get known PF of 1), the current in those two lines can be much higher than the conductor rating.
 

Smart $

Esteemed Member
Location
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Without knowing the angle of the third line, the i and j value of the third phase current are not known.
Even you can measure the neutral current, I do not see how we can find the PF of the third phase as we have to solve an equation of two unknowns (i (Ic) and j (Ic))
Could you please explain more to your statement "...Even not knowing the angle of neutral current there should only be two possible angles for the measured current and one should eliminate itself from contention simply by the design of the equipment under load"
What are two possible angles?
Let's borrow some of the example used throughout this discussion... We have Ia at 50A with PF of 0.707 lagging and Ib at 40A with PF of 0.866 lagging. Here is where it differs... we measure Ic at 40A and In at 58.86A, but we don't know the power factor or phase angle of either.

Now we draw this out graphically, starting at (0,0) we draw vectors Ia and Ib. Since we don't know the direction of Ic's vector, we place a circle of 40(A) unit radius centered at the head of Ib's vector, and likewise we place a circle of 58.86 unit radius centered at (0,0). These circles centers represent one end of their respective vectors while the periphery line represents the magnitude. The rest can be gleaned from the following depiction...

neutralcurrent5.gif





Furthermore, there are something to be considered. If you connect resistive loads to phase B and C (to get known PF of 1), the current in those two lines can be much higher than the conductor rating.
Of course, it's a matter which must be considered. We in the electrical profession often go without stating such, only because we overcome the consideration regularly, to the degree it is nearly instinctive.
 
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Let's borrow some of the example used throughout this discussion... We have Ia at 50A with PF of 0.707 lagging and Ib at 40A with PF of 0.866 lagging. Here is where it differs... we measure Ic at 40A and In at 58.86A, but we don't know the power factor or phase angle of either.

Now we draw this out graphically, starting at (0,0) we draw vectors Ia and Ib. Since we don't know the direction of Ic's vector, we place a circle of 40(A) unit radius centered at the head of Ib's vector, and likewise we place a circle of 58.86 unit radius centered at (0,0). These circles centers represent one end of their respective vectors while the periphery line represents the magnitude. The rest can be gleaned from the following depiction...
Ok now I understand all. The two possible vector directions depend on the load on the phase C is inductive or capacitive and this can be obtained from the load's manufacturer?

Of course, it's a matter which must be considered. We in the electrical profession often go without stating such, only because we overcome the consideration regularly, to the degree it is nearly instinctive.
[/QUOTE]
I did not intend to critisize rather want a reminds to someone may practise this without careful consideration. Maybe my wording style is wrong or misleading. I am sorry for that.
 

Smart $

Esteemed Member
Location
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Ok now I understand all. The two possible vector directions depend on the load on the phase C is inductive or capacitive and this can be obtained from the load's manufacturer?
While the angle does depend on whether the load is overall capacitive or inductive, it is more that one angle is impossible using passive components. With passive components the angle is limited to plus or minus 90? deviation from the voltage angle. To go beyond 90? would mean the load is supplying more energy to the source than the source to the load :cool:


I did not intend to critisize rather want a reminds to someone may practise this without careful consideration. Maybe my wording style is wrong or misleading. I am sorry for that.
Not a problem, and no offense taken here. While the consideration may be nearly instinctive, there can also be a degree of compacency. Wake up calls are appreciated when not done in a condescending manner :)
 

rattus

Senior Member
What is the basis for this formula?
Can you explain the logic of formula?
I neutral =square root of the Sum of currents minus the product of the phase currents???

I believe one assumes the load currents to be separated by 120 degrees. Next add the current phasors to obtain the real and imaginary parts. Then apply the Pythagorean theorem to the real and imaginary parts to obtain In. Maybe some Smart youngster could grind that out for us?
 

Smart $

Esteemed Member
Location
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... Maybe some Smart youngster could grind that out for us?
I don't resemble that implication :roll::roll::roll:

As for the grind, less is involved if one progresses to the law of cosines rather than its precursor theorem.
 
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rattus

Senior Member
Just do a forum search with my user name and "law of cosines" as the search term... and I'm sure you'll find something useful.

I did a Google search and found my own post. I used Pythagorus, but I did not show the grind. Five years ago, it was.
 

nknfive

New member
As Jghrist said , Neutral current is vectorial sum of all the phase currents
current in red=50A at -45degree
current in yellow= 40A at -30 degree
current in blue= 40 A at 45 degree

neutral current = (50cos45-j50 sin45) + (40Cos30-j40Sin30) + (40Cos45+j40Sin45)
=(50x0.707-j50x0.707)+(40x0.866-j40x0.5)+(40x0.707+j40x0.707)
=98.28-j27.07
=101.93 at 15.399 degree lagging
 

rattus

Senior Member
Not really:

Not really:

As Jghrist said , Neutral current is vectorial sum of all the phase currents
current in red=50A at -45degree
current in yellow= 40A at -30 degree
current in blue= 40 A at 45 degree

neutral current = (50cos45-j50 sin45) + (40Cos30-j40Sin30) + (40Cos45+j40Sin45)
=(50x0.707-j50x0.707)+(40x0.866-j40x0.5)+(40x0.707+j40x0.707)
=98.28-j27.07
=101.93 at 15.399 degree lagging

Since we are given only PFs, we must reference the phase current angles to the phase voltages. You did not do this, therefore your result is wrong.
 

Lady Engineer

Senior Member
Location
New Jersey
This question was on the PE exam. The neutral current is the sum of all the currents on each leg (phase), so therefore it would be Ia<-45 + Ib<-30 +Ic<-45. You must use the phase angle with positive and neg., with cos and sine. The reason is because some phases are lagging, and some are leading. This is why your P.F. for each phase is different.

With that said, if the loads were balanced, the sum of Ia<-45 + Ib<-30 +Ic<45 would equal zero, but as you already know the neutral carries the unbalanced current when the load is unbalanced.

Sorry, this was something they drilled into us on the PE exam, and there were 5 questions on this alone. Thank God that horrible test is over....selling circus peanuts, here we come! LOL :D
 
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