Line currents on a 3 phase supply with unbalanced voltage

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mull982

Senior Member
If one of the legs on a 3 phase system has a voltage higher than the other two phases how would the line current on that phase compare to the other two phases? Would it be higher on the other two phases. For instance on a 208/120 system if 2 of the phases were 120V L-G and the other phase was 150V L-G would the current be higher on the 150V leg? My guess is that it would be higher.

Would it matter weather the load was a wye or delta load?

Could it be said that the higher the L-G voltage is on a given phase the highe the curent will be fo a balanced load? Or must you also compare the relationship between the L-L voltages?
 

G._S._Ohm

Senior Member
Location
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For instance on a 208/120 system if 2 of the phases were 120V L-G and the other phase was 150V L-G would the current be higher on the 150V leg? My guess is that it would be higher.

Let's do just one case first.

For a Y load of three equal resistors with the neutral floating, once you assign voltages to two lines the third one is determined.

Let's say you have 3 ea. 1 ohm resistors, and so there is 120 volts @ 0 degrees [120/0] across one giving a phase current Ia magnitude of 120 amps, and 150 volts at 120 degrees [150/120] across the other giving a phase current Ib magnitude of 150 amps.

The vector sum of the three currents must equal zero, so then the phase current of the last phase Ic = -(Ia + Ib).

The magnitude M and phase angle pa of Ic must be solved for.

M/pa = -(120/0 + 150/120)

You need to convert each of these polar coordinates into rectangular coordinates, add them and then convert back.

Once you know Ia then the L-G voltage for this phase is just this vector multiplied by 1 ohm. I doubt it will have a magnitude of 120 v but I haven't yet worked it out.

Using Excel to do this tedious calculation and conversions will lessen the chance for error.
 

G._S._Ohm

Senior Member
Location
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I couldn't find polar to rectangular to polar on my version of Excel.

Using this link
http://www.analyzemath.com/Calculators/Polar_Rect.html
and using the j operator

-(120/0 + 150/120)
converts to
-([120 + j0] + [-75 + j130])
grouping terms,
-([120-75] + j[0 + 130])
equals
-(45 + j130)

converting back using
http://www.analyzemath.com/Calculators/Rect_Polar.html
gives
138/71
So the magnitude of the voltage on the third phase must be 138 volts.
In the links, the j value is the Y value.

If I had to work with 3 phase a lot
I'd buy or make a 3 phase generator and 3 load resistors to check my work. There are at least two ways to have a desk top setup that puts out a few watts or less, which is all you need since scaling up voltages and power level is easy on paper.

One way is to do this is to get an auto alternator from a junk yard, replace the field windings with a permanent magnet and spin the thing with an elec. drill or drill press.
 

mull982

Senior Member
Let's do just one case first.

For a Y load of three equal resistors with the neutral floating, once you assign voltages to two lines the third one is determined.

Let's say you have 3 ea. 1 ohm resistors, and so there is 120 volts @ 0 degrees [120/0] across one giving a phase current Ia magnitude of 120 amps, and 150 volts at 120 degrees [150/120] across the other giving a phase current Ib magnitude of 150 amps.

The vector sum of the three currents must equal zero, so then the phase current of the last phase Ic = -(Ia + Ib).

The magnitude M and phase angle pa of Ic must be solved for.

M/pa = -(120/0 + 150/120)

You need to convert each of these polar coordinates into rectangular coordinates, add them and then convert back.

Once you know Ia then the L-G voltage for this phase is just this vector multiplied by 1 ohm. I doubt it will have a magnitude of 120 v but I haven't yet worked it out.

Using Excel to do this tedious calculation and conversions will lessen the chance for error.

Good example that makes sense. I guess you could have determined the voltage without determining the current by simply adding the voltage vectors so that they are equal to zero in the same way that the current vectors equal to zero. So if we added 120V @0deg and 150V @120 deg we could then solve for the third voltage vectorl of 138V since the summation of all voltage vecotrs must sum to zero. Is this correct?

I'm quite sure how to go about solving for Delta load. Anybody have any ideas?
 

skeshesh

Senior Member
Location
Los Angeles, Ca
I guess you could have determined the voltage without determining the current by simply adding the voltage vectors so that they are equal to zero in the same way that the current vectors equal to zero

Not quite. That's why G S chose a simple 1ohm resistance so that he can demonstrate the approach to the solution. You would need to know the impedance of the load to get to a numerical solution.
I don't see, however, how G S's solution has answered your question. I believe a better example for your question would be to assume 120>0, 120>120, 150>-120 (mag>angle). I'll try to work it out later, pretty overloaded at work at the moment...
 

mull982

Senior Member
Not quite. That's why G S chose a simple 1ohm resistance so that he can demonstrate the approach to the solution. You would need to know the impedance of the load to get to a numerical solution.
I don't see, however, how G S's solution has answered your question. I believe a better example for your question would be to assume 120>0, 120>120, 150>-120 (mag>angle). I'll try to work it out later, pretty overloaded at work at the moment...

Dont all L-N voltages sum to zero at any given point? Maybe not?

Maybe I am thinking of L-L voltages. So do L-L voltages close in on each other and sum to zero?
 

G._S._Ohm

Senior Member
Location
DC area
Dont all L-N voltages sum to zero at any given point?
No.

Let's say you have two phases feeding two 0.5 ohm resistors tied to neutral in the middle so the current through the resistors is (V2-V1)/1 but V2 and V1 are vectors (phasors) and not scalar quantities.

So, V2 = 1/0 [1 volt at zero degrees, your reference phase] and V1 equals 1/120 [1 volt at 120 degrees].
In rectangular coordinates, using
http://www.analyzemath.com/Calculators/Polar_Rect.html
these values are
V2 = 1+ j0 and V1 = -0.5 + j0.866

Using the rules for complexor arithmetic,
V2 - V1 = (1+ j0) - (-0.5 + j0.866)
= 1 + 0.5 + j0.866
= 1.5 + j0.866

Converting back to polar using
http://www.analyzemath.com/Calculators/Rect_Polar.html
the current is
1.73/30 and the L-L voltage is the same. The L-N voltages are 1 volt.

The trick is to remember what are vector quantities and what are scalar quantities. Scalars are easy to deal with.

You've heard of pilots flying by instruments?
Well, unless you have a lab setup to confirm these results you really are depending totally on the correctness of your calculations, unless you also confirm these answers graphically, with a ruler and protractor.
The best way is to build a model of these 3 phase setups, with a toy 3 phase generator, load resistors and load inductors. You can scale up from there.
 
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mull982

Senior Member
No.

Let's say you have two phases feeding two 0.5 ohm resistors tied to neutral in the middle so the current through the resistors is (V2-V1)/1 but V2 and V1 are vectors (phasors) and not scalar quantities.

So, V2 = 1/0 [1 volt at zero degrees, your reference phase] and V1 equals 1/120 [1 volt at 120 degrees].
In rectangular coordinates, using
http://www.analyzemath.com/Calculators/Polar_Rect.html
these values are
V2 = 1+ j0 and V1 = -0.5 + j0.866

Using the rules for complexor arithmetic,
V2 - V1 = (1+ j0) - (-0.5 + j0.866)
= 1 + 0.5 + j0.866
= 1.5 + j0.866

Converting back to polar using
http://www.analyzemath.com/Calculators/Rect_Polar.html
the current is
1.73/30 and the L-L voltage is the same. The L-N voltages are 1 volt.

The trick is to remember what are vector quantities and what are scalar quantities. Scalars are easy to deal with.

You've heard of pilots flying by instruments?
Well, unless you have a lab setup to confirm these results you really are depending totally on the correctness of your calculations, unless you also confirm these answers graphically, with a ruler and protractor.
The best way is to build a model of these 3 phase setups, with a toy 3 phase generator, load resistors and load inductors. You can scale up from there.

O.k. but when drawing out the phasors of L-L voltages they must form a closed triangle therefore I would assume that L-L voltages always must sum to zero?
 

G._S._Ohm

Senior Member
Location
DC area
O.k. but when drawing out the phasors of L-L voltages they must form a closed triangle therefore I would assume that L-L voltages always must sum to zero?
My text shows the L-L voltages as being shifted 120 degrees and radiating out from the center. I guess then they do sum to zero.

For three phase work I recommend getting to the answer at least two different ways. This 3 phase stuff is as confusing as it gets, mostly for "bookkeeping" reasons, making sign mistakes, etc.

My text shows L-L voltages going into a Y connected load showing impedances. No source is shown.
For delta loads the line currents are shown and the load admittances [1/impedances] are used.
I've never confirmed this but I think all cases can be derived from these two general cases and the several pages of equations that govern the behavior of these basic circuits.

As more problems are posted I guess I will find out, but I'd feel better about this if someone has a lab setup. You don't need kw to confirm the results and if your load is resistive you don't even care about the freq.

Something else occurs to me.
The Schaum's Outline Series publishes worked-out problems and I'm sure they have one with worked-out power problems. Schaum's now belongs to McGraw Hill so I'd ask them specifically to recommend a text with 3 phase problems with resistive/reactive loads.
 
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rcwilson

Senior Member
Location
Redmond, WA
Here's my simple analysis:

A 120/208 V system has four wires, 3 phase + 1 neutral. If one phase-neutral voltage is higher than the others, the current drawn by the resistive loads on that phase is more. There is no effect on the other two phases. The additional current flows out the neutral connection.

Do all phase-phase voltages add up to zero? Yes.

Draw any three vectors with a common point to represent the phase to neutral voltages. Now draw the three phase-phase vectors. Start at the end of the B phase vector and draw a line to A phase to represent Vab. No draw one from A to C to represent Vca. Draw the third vector from C to B to represent Vbc. You are back where you started. Since the vectors are added together by drawing each tail on the point of the previous vector you have just proved that the three phase-phase voltages of any three phase system will always add to zero. It doesn't matter if one is 2 volts and the others are 208 V. The vectors may not be equal or 120 degrees apart, but they will add up to zero.
 

skeshesh

Senior Member
Location
Los Angeles, Ca
A 120/208 V system has four wires, 3 phase + 1 neutral. If one phase-neutral voltage is higher than the others, the current drawn by the resistive loads on that phase is more. There is no effect on the other two phases. The additional current flows out the neutral connection.
I think this comment is a bit misleading. It is really the load impedance alone that will determined the current, unless you're assuming a completely balanced system. I don't think in practice it will be difficult to determine which phase has a higher voltage by measuring the currents

Do all phase-phase voltages add up to zero? Yes.

Draw any three vectors with a common point to represent the phase to neutral voltages. Now draw the three phase-phase vectors. Start at the end of the B phase vector and draw a line to A phase to represent Vab. No draw one from A to C to represent Vca. Draw the third vector from C to B to represent Vbc. You are back where you started. Since the vectors are added together by drawing each tail on the point of the previous vector you have just proved that the three phase-phase voltages of any three phase system will always add to zero. It doesn't matter if one is 2 volts and the others are 208 V. The vectors may not be equal or 120 degrees apart, but they will add up to zero.

All true but I don't think KVL will help solve the original questions.

Mull, I think G.S. is right in that it's necessary to whip out the power systems book and just go over the math a couple of times. Also if you want to avoid the pain with the signage and imaginery number, try out scilab or other free software that can easily produce accurate results when you use matricies.
 
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