Unbalance neutral current in a 3 phase system
- Thread starter Gary Shumaker
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As mentioned in my response to rattus, I gather you are plotting the equidistant point from the end of the three vectors. It don't work :wink:weressl said:
Consider the following diagram. I have the base vectors in the middle, emanating from coordinate 0,0. In order to vectorially add these vectors they have to be arranged tail to head. Doing so can be done six different ways, or as in the diagram, utilizing all six. In the latter case, note how the result appears to be a slightly skewed isometric view of a cube.
Anyway, rattus is wrong about the phase angle of the resulting neutral current vector?which is 2.65A @ 199.1?.
NOTE: Because of the scale of the drawing versus the length of the neutral vector my diagram has no arrowhead on the neutral vector (the fuscia colored line).
Smart, I give you credit for figuring out what Laszlo was doing, but you are making this problem unbelievably complicated. All you need do is draw out the three phasors, head to tail, like a delta, then close the diagram with a fourth phasor which is In.
But, this is a poor way to solve the problem because in this almost balanced system the neutral phasor is so short it is hard to measure. If you use graphics, you should reduce the problem as I have done. Then, you have the phasor sum of 1.732A @ 120 and 2A @ 0. The phase angle is clearly in the first quadrant. It is not 199 degrees.
But, this is a poor way to solve the problem because in this almost balanced system the neutral phasor is so short it is hard to measure. If you use graphics, you should reduce the problem as I have done. Then, you have the phasor sum of 1.732A @ 120 and 2A @ 0. The phase angle is clearly in the first quadrant. It is not 199 degrees.
I did that, but at the same time showed it can be done six different ways!!!rattus said:
So what do you have in your reference texts that say I can't do that?
I have no idea where you are getting "the phasor sum of 1.732A @ 120 and 2A @ 0".
Earlier you said it was "3A @ 0" and "2A @ 120".
The 3A @ 0? is good, but the other should be 1A @ 120?—the result of removing the balanced portion of the circuit: 46A from all three. Therefore we have
49A – 46A = A' = 3A @ 0? = 3 + j0
47A – 46A = B' = 1A @ 120? = -0.5 + j0.866
46A – 46A = C' = 0A @ 240? = 0 + j0
The sum is 2.5 + j0.866 and represents the point at the tail end of neutral vector. The head end of neutral vector is at 0 + j0.
(0 + j0) - (2.5 + j0.866) = -2.5 - j0.866 = 2.65 @ 199.1?
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Yer right Smart:
Yer right Smart:
That is what I get for trying to do too much in my head! I should have said that In is the sum of the two phasors,
2A @ 120
3A @ 240
then,
In = -2.5 -j0.866 = 2.65A @ 199.1
You actually found the negative of In and took the liberty of negating it again to get the right phase angle.
"The sum is 2.5 + j0.866 and represents the point at the tail end of neutral vector. The head end of neutral vector is at 0 + j0."
This gets the right answer, but strictly speaking is incorrect. You should have said the sum is the negative of In and then negated it.
I believe though that we agree that the sum of ALL currents either into or out of the neutral node is zero. That is, In is that current which will satisfy Kirchoff's current law. We could have defined In as the sum of the phase currents. I am unaware of any rule or convention though. How about you?
Yer right Smart:
That is what I get for trying to do too much in my head! I should have said that In is the sum of the two phasors,
2A @ 120
3A @ 240
then,
In = -2.5 -j0.866 = 2.65A @ 199.1
You actually found the negative of In and took the liberty of negating it again to get the right phase angle.
"The sum is 2.5 + j0.866 and represents the point at the tail end of neutral vector. The head end of neutral vector is at 0 + j0."
This gets the right answer, but strictly speaking is incorrect. You should have said the sum is the negative of In and then negated it.
I believe though that we agree that the sum of ALL currents either into or out of the neutral node is zero. That is, In is that current which will satisfy Kirchoff's current law. We could have defined In as the sum of the phase currents. I am unaware of any rule or convention though. How about you?
To simply state IN as "the sum is the negative of In" does not prove such. I offered simple proof. Rejection of said proof is your prerogative, just like it is my prerogative to respectfully disagree regarding its correctness.rattus said:
Same here.
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Smart $ said:
Smart, we agree on the result, right? I am simply saying that,
2.5 +j0.866 = 2.65A @ 19.1 = -In
Negate that phasor to obtain,
In = 2.65A @ 199.1
In other words, after applying Kirchoff's current law,
In = -(Ia + Ib + Ic)
I approached the problem by finding the the sum of the currents necessary for balance. Then I didn't have to negate.
You slipped the negation in without fully explaining it. Your phasor diagram would look funny with the arrowhead at the origin.
I'm OK with the above, as long as it is [mentioned and...] understood that your last formula is regarding the phasor relationship and not a simple algebraic equation. Without noting and comprehending such, someone could come along and calculate the problem as In = -(Ia + Ib + Ic) = -49A - 47A - 46A = -44A, which all us knowledgeable types know to be an incorrect result.rattus said:
Seems to me we're both coming up a little short on complete explanations :grin:
In my mind it wouldn't :wink:
It meets the head to tail requirement of vector addition AND is accurately descriptive in the application of Kirchoff's current law.
A little trick:
A little trick:
The algebraic formula,
In = sqrt(Ia^2 + Ib^2 + Ic^2 -Ia*Ib - Ia*Ic -Ib*Ic),
is a little tricky to evaluate even with a calculator. We can however simplify the problem greatly.
Subtract the smallest phase current from Ia, Ib, and Ic. This leaves two currents which we will call I1 and I2. Then,
In = sqrt(I1^2 + I2^2 - I1*I2)
From Smart's example, let I1 = 1 and I2 = 3, then
In = sqrt(2^2 + 3^2 - 3*2) = sqrt(4 + 9 - 6) = sqrt(7) = 2.65A
A little trick:
The algebraic formula,
In = sqrt(Ia^2 + Ib^2 + Ic^2 -Ia*Ib - Ia*Ic -Ib*Ic),
is a little tricky to evaluate even with a calculator. We can however simplify the problem greatly.
Subtract the smallest phase current from Ia, Ib, and Ic. This leaves two currents which we will call I1 and I2. Then,
In = sqrt(I1^2 + I2^2 - I1*I2)
From Smart's example, let I1 = 1 and I2 = 3, then
In = sqrt(2^2 + 3^2 - 3*2) = sqrt(4 + 9 - 6) = sqrt(7) = 2.65A
I'm not sure exactly what it is you want me to diagram.rattus said:
As to negating IN, it is most obvious when stated initially as a vectorial equation conforming to Kirchoff's current law.
IA + IB + IC + IN = 0 vectorially
Therefore,IA + IB + IC = ?IN vectorially
In a head to tail graphical representation, no matter where the first vector starts it will always be the same point where the last vector ends, i.e. the four vectors will form a quadrilateral [for those who don't know, a quadrilateral is a four-sided, closed polygon].
Thought I was wrong!
Thought I was wrong!
Smart, you tricked me. I was responding to Laszo's question where,
Ia = 46A
Ib = 47A
Ic = 49A
2.65A @ 40.9 degrees is correct.
You defined the currents as,
Ia = 49A
Ib = 47A
Ic = 46A
Magnitude doesn't change; angle does.
Thought I was wrong!
Smart $ said:
Smart, you tricked me. I was responding to Laszo's question where,
Ia = 46A
Ib = 47A
Ic = 49A
2.65A @ 40.9 degrees is correct.
You defined the currents as,
Ia = 49A
Ib = 47A
Ic = 46A
Magnitude doesn't change; angle does.
I revised my Excel calculator to also plot the vectors...
Click here to download the Excel file (Excel 2007 converted to Excel 97-2003 format)
EDIT I discovered an error in my calculator. I had forgot that I was working with positive phase angles and subtracted the lag angle due to power factor. When working with positive angles the lag angle gets added... so I corrected the formula. If you downloaded prior to this edit, please download again.
I also re-formatted the diagram for a better appearance. I should note however, I have yet to find a method of scaling both axes the same, so the vector angles may be skewed somewhat. I considered making fixed scale axes, but then the vectorial representation may be too small or too big. To have the scale and angle relationships appear more accurate will require manual resizing of the "chart".
Click here to download the Excel file (Excel 2007 converted to Excel 97-2003 format)
EDIT I discovered an error in my calculator. I had forgot that I was working with positive phase angles and subtracted the lag angle due to power factor. When working with positive angles the lag angle gets added... so I corrected the formula. If you downloaded prior to this edit, please download again.
I also re-formatted the diagram for a better appearance. I should note however, I have yet to find a method of scaling both axes the same, so the vector angles may be skewed somewhat. I considered making fixed scale axes, but then the vectorial representation may be too small or too big. To have the scale and angle relationships appear more accurate will require manual resizing of the "chart".
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cschmid
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with all the add ons for Excel what one are you using for the math you are doing?
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