Is this calc right?

[mbrooke]

Ok, I am a tad confused here. Shouldn't a lower power factor on one leg actually cause the neutral current to drop- rather than rise? Either my vector math is off or this spread sheet is off- but I can not figure out which one.

 

[Ingenieur]

The i and j add incorrectly, should be negative
and the resultant angle is -105

 

[mbrooke]

The i and j add incorrectly, should be negative
and the resultant angle is -105

Ok, thats what I suspected. Guess the spread sheet is off then.


Anyone know how to fix a spread sheet? I have the ability to edit this one.

 

[david luchini]

Ok, I am a tad confused here. Shouldn't a lower power factor on one leg actually cause the neutral current to drop- rather than rise?

The angle is incorrect, but the magnitude is right...123.4A. The neutral current is larger than the balanced 100A, in this case.

 

[Ingenieur]

upload the excel spreadsheet
xls is an allowed file format

 

[jumper]

Your version was not the author's latest. I just compared. You could contact him for info.

 

[mbrooke]

Your version was not the author's latest. I just compared. You could contact him for info.

I'll send you a PM to get the right person

 

[Smart $]

While I have more enhanced "calculators", the following is the latest version of the Neutral Calculator.

https://drive.google.com/file/d/0By0rzU3UuNh7NWgzSEFRQzNjVVU/view?usp=sharing

The magnitudes and angles are correct, recheck your math guys.

 

[Smart $]

The i and j add incorrectly, should be negative
and the resultant angle is -105

Which of the following are equivalent? Which are opposites?

1. -123.49A@-105°
2. 123.49A@-285°
3. 123.49A@75
4. -123.49A@255°

 

[Ingenieur]

Which of the following are equivalent? Which are opposites?

1. -123.49A@-105°
2. 123.49A@-285°
3. 123.49A@75
4. -123.49A@255°

???
resolve each into rectang complex coordinates
are they all equal?

 

[david luchini]

The magnitudes and angles are correct, recheck your math guys.

The magnitude is correct, the angle is wrong.

 

[mivey]

The i and j add incorrectly, should be negative
and the resultant angle is -105

Ia+Ib+Ic+In=0

In=-(Ia+Ib+Ic)

 

[mivey]

The magnitude is correct, the angle is wrong.

Signs are to math as blackheads are to teenagers.

The angle is correct.

 

[Ingenieur]

Ia+Ib+Ic+In=0

In=-(Ia+Ib+Ic)

nope
Ia + Ib + Ic = 0

 

[mivey]

nope
Ia + Ib + Ic = 0

Kirchhoff is rolling in his grave.

Re-think what unbalanced current means and what it means when we say the neutral carries the unbalanced current.

 

[david luchini]

Signs are to math as blackheads are to teenagers.

The angle is correct.

OK, In=Ia+Ib+Ic. The angle is now incorrect...nothing to to with signs.

 

[Ingenieur]

Kirchhoff is rolling in his grave.

Re-think what unbalanced current means and what it means when we say the neutral carries the unbalanced current.

nope
Ia, Ib, Ic are entering the node
In leaving
opposite signs (by convention and power flow)

Ia + Ib + Ic -In =0
Ia + Ib + Ic = In
not In = -(I a + Ib + Ic)

 

[Smart $]

OK, In=Ia+Ib+Ic. The angle is now incorrect...nothing to to with signs.

That's not a proper application of Kirchoff's Current Law (and just outright wrong ).

 

[Smart $]

nope
Ia, Ib, Ic are entering the node
In leaving
opposite signs (by convention and power flow)

Ia + Ib + Ic -In =0
Ia + Ib + Ic = In
not In = -(I a + Ib + Ic)

The empirical value(s) get the sign change where appropriate. Not the vectorial representatives. K's Law says all current entering and leaving a node sum to zero. Making one or more variables a negative is making the result something other than the sum. Entering and leaving is merely stated as a required event for the occurrence of the anomaly.

 

[Smart $]

???
resolve each into rectang complex coordinates
are they all equal?

I already know the answer. I was asking you...