Calculating Phase Currents for Unbalanced Loads

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jghrist

Senior Member
You can calculate the fundamental component of the currents and voltages from instantaneous values.

If two instantaneous values I1 and I2 are taken 1/4 cycle apart, then:

The magnitude of the sinusoid is I = sqrt(I2? + I1?)
The rms value will be Irms = I/sqrt(2)

The angle ? = arctan(I2/I1)
? is relative to a cosine function that has a peak at the time of the second reading I2. It doesn't really matter as long as all values measured are on the same time basis, then the relative angles of all quantities will be correct.
 

jghrist

Senior Member
The basic problem is not that you have too many unknowns for the number of equations, it is that you have too many equations for the unknowns.

Letting IA, IB, and IC be the line currents and IAB, IBC, and ICA be the delta load currents, with IA, IB, and IC known, then the node equations are:

IA + IB + IC = 0
IA = IAB - ICA
IB = IBC - IAB
IC = ICA - IBC

Three unknowns, four equations. The equations have an infinite number of solutions because you can theoretically have any amount of current "circulating" around the delta load without affecting IA, IB, and IC.

The Fermat solution is the one that minimizes the sum of the magnitudes of IAB, IBC, and ICA.
 

mea03wjb

Member
The basic problem is not that you have too many unknowns for the number of equations, it is that you have too many equations for the unknowns.

Letting IA, IB, and IC be the line currents and IAB, IBC, and ICA be the delta load currents, with IA, IB, and IC known, then the node equations are:

IA + IB + IC = 0
IA = IAB - ICA
IB = IBC - IAB
IC = ICA - IBC

Three unknowns, four equations. The equations have an infinite number of solutions because you can theoretically have any amount of current "circulating" around the delta load without affecting IA, IB, and IC.

The Fermat solution is the one that minimizes the sum of the magnitudes of IAB, IBC, and ICA.
Are you saying that this problem is impossible to solve for an exact single answer?

i.e. we can find 'a' solution problem mathematically but this may or may not correspond to the actual currents flowing in the motor windings.
 

Smart $

Esteemed Member
Location
Ohio
If the values were instantaneous, and the OP only mentioned "unbalanced load", then I must say the voltage values are unbalanced - because in a balanced three-phase voltage, the sum of the instantaneous voltages should be equal to zero!
308v + (-275) + (-35) = -2V​
... which isn't quite zero due to rounding, and perhaps a slight imbalance, but well within a nominal tolerance.

However, that reminds me that he'll likely have to use non-absolute voltage values for the line-to-line calc's:
v12 = v1 ? v2 = 308V ? (-275V) = 583V
v23 = v2 ? v3 = -275V ? (-35V) = -240
v31 = v3 ? v1 = -35V ? 308V = -343V​
and
583V + (-240V) + (-343V) = 0V​
Where'd the 2V imbalance go??? Isn't it amazing :D
 

jghrist

Senior Member
Are you saying that this problem is impossible to solve for an exact single answer?

i.e. we can find 'a' solution problem mathematically but this may or may not correspond to the actual currents flowing in the motor windings.

Yes. That's what I'm saying. It is similar to calculating the secondary load current in a delta-wye transformer when you know the primary line currents. You can calculate the line currents if you know the delta currents, but not vice-versa. In the case of a delta-wye transformer, if you add the same current (magnitude and angle) to each secondary phase current, the primary line currents do not change. The same current in each phase is a zero-sequence current and circulates in the primary delta winding.

It's a little more difficult to come up with an instance of adding zero-sequence current to a delta load, but the mathematical principle is the same.
 

rattus

Senior Member
If the values were instantaneous, and the OP only mentioned "unbalanced load", then I must say the voltage values are unbalanced - because in a balanced three-phase voltage, the sum of the instantaneous voltages should be equal to zero!

The delta phase voltages need not be unbalanced only the phase currents.

IMHO, the sum of the L-L voltages is always zero, instantaneous or RMS.
 

Smart $

Esteemed Member
Location
Ohio
The basic problem is not that you have too many unknowns for the number of equations, it is that you have too many equations for the unknowns.

Letting IA, IB, and IC be the line currents and IAB, IBC, and ICA be the delta load currents, with IA, IB, and IC known, then the node equations are:

IA + IB + IC = 0
IA = IAB - ICA
IB = IBC - IAB
IC = ICA - IBC

Three unknowns, four equations. The equations have an infinite number of solutions because you can theoretically have any amount of current "circulating" around the delta load without affecting IA, IB, and IC.

The Fermat solution is the one that minimizes the sum of the magnitudes of IAB, IBC, and ICA.
Quite correct, whether we're discussing instant or RMS values... just that for RMS values, we have to include the phase relationships
 

rattus

Senior Member
The basic problem is not that you have too many unknowns for the number of equations, it is that you have too many equations for the unknowns.

Letting IA, IB, and IC be the line currents and IAB, IBC, and ICA be the delta load currents, with IA, IB, and IC known, then the node equations are:

IA + IB + IC = 0
IA = IAB - ICA
IB = IBC - IAB
IC = ICA - IBC

Three unknowns, four equations. The equations have an infinite number of solutions because you can theoretically have any amount of current "circulating" around the delta load without affecting IA, IB, and IC.

I don't see circulating currents in the loads. These currents are defined by the L-L voltages and the load impedances. You may see these currents in the transformer though.

The thing is we have six unknowns--three complex numbers.

Now take a look at Smart's diagram which shows the line and load current phasors. Clearly, for the general case, the intersection of the load current phasors can be anywhere within the line current delta, and this amounts to an infinite number of solutions.
 

topgone

Senior Member
Balanced rms voltage value

Balanced rms voltage value

First, it is necessary to have good data before we calculate.
Since the OP did not mention if the supply was a balanced 3-phase supply, I played around with possible values of balanced RMS voltages and simulated/plotted voltage wave swept from zero electrical degrees up to half a cyle (180 electrical degrees). I found out that the correct Line to Neutral values should have been:
V1 = 308.20 volts
V2 = -273.13 volts, and
V3 = -35.07 volts.
This values were computed using a balanced line to neutral RMS voltage of 238.62 volts (Vpeak = Vrms X 1.414 = 337.46 volts).
Above mentioned figures resolved at around 66 electrical degrees, i.e.:
V1(instantaneous) = Vpeak X sin(theta) = 337.46*sin(66) = 308.2
V2(instantaneous) = Vpeak X sin(theta-120) = 337.46*sin(66-120) = -273.0
V3(instantaneous) = Vpeak X sin(theta+120) = 337.46*sin(66+120)= -35.2
Now back to the OP.
For any phase to phase loads, it would be clear that finding the instantaneous phase current is easy (Vinst/load ohms). Bear in mind that only the magnitude of the instantaneous phase currents differ depending on the phase load of each), the angular displacement of the current wave will still be the same 120 electrical degrees. The complexity in getting the values of the instantaneous line current lies in the vectorial addition of the instantaneous currents at the delta corners of the load. Still, if you have good data, then everything can be computed as needed. I would leave the final computations to you in knowing the instantaneous line current.
Respectfully.
 

jghrist

Senior Member
I don't see circulating currents in the loads. These currents are defined by the L-L voltages and the load impedances. You may see these currents in the transformer though.

The thing is we have six unknowns--three complex numbers.

Now take a look at Smart's diagram which shows the line and load current phasors. Clearly, for the general case, the intersection of the load current phasors can be anywhere within the line current delta, and this amounts to an infinite number of solutions.

I agree that at least for passive loads, I don't see how to get circulating currents in a delta load. Perhaps with an active load like a motor, it is possible. Mathematically, it is still possible which is why you can have an infinite number of solutions to the set of equations. The same as you can graphically in Smart's diagram as you point out. Only the one shown minimizes the load currents, however, and this is probably the solution with zero circulating current (although I can't prove it).

You can treat three complex numbers as six unknowns, but then you can double the number of equations by separating the real and imaginary components. Each equation would apply to both the real and the imaginary components. For instance, if for complex currents,

IA = IAB - ICA,

Then

IAreal = IABreal - ICAreal

and

IAimag = IABimag - ICAimag
 

jghrist

Senior Member
For any phase to phase loads, it would be clear that finding the instantaneous phase current is easy (Vinst/load ohms).
But what if we don't know the load ohms?
Bear in mind that only the magnitude of the instantaneous phase currents differ depending on the phase load of each), the angular displacement of the current wave will still be the same 120 electrical degrees.
Who says the unbalanced delta loads have the same power factor? If the solution is defined by the Fermat Point as posited by Smart, then the angle between the load currents is 120?. See http://mathworld.wolfram.com/FermatPoints.html This would seem to rule out the Fermat point being the correct solution if the loads are not the same power factor.
 

rattus

Senior Member
I agree that at least for passive loads, I don't see how to get circulating currents in a delta load. Perhaps with an active load like a motor, it is possible. Mathematically, it is still possible which is why you can have an infinite number of solutions to the set of equations. The same as you can graphically in Smart's diagram as you point out. Only the one shown minimizes the load currents, however, and this is probably the solution with zero circulating current (although I can't prove it).

You can treat three complex numbers as six unknowns, but then you can double the number of equations by separating the real and imaginary components. Each equation would apply to both the real and the imaginary components. For instance, if for complex currents,

IA = IAB - ICA,

Then

IAreal = IABreal - ICAreal

and

IAimag = IABimag - ICAimag

I would argue that a motor under steady state conditions behaves like a passive RL load, and ideally the load currents would be balanced. Still don't see any possibility of a circulating current in the load.

Simply put, the number of solutions is infinite because in the general case, the number of possible values for each load impedance is infinite for any set of line currents.
 

mea03wjb

Member
The end?

The end?

Simply put, the number of solutions is infinite because in the general case, the number of possible values for each load impedance is infinite for any set of line currents.

Thanks everyone for their input - everyone seems to agree that there can be no single solution with the information available (line currents and line to neutral voltages).

In my scenario the load impedances are unknowns and there is a good chance they will not be balanced since my application is one of fault detection and diagnosis.

The main thing is when I go to my supervisors I can pull together enough info from these discussions, equations and diagrams here to explain why it can't be done!

Cheers, W.
 

Smart $

Esteemed Member
Location
Ohio
Thanks everyone for their input - everyone seems to agree that there can be no single solution with the information available (line currents and line to neutral voltages).

In my scenario the load impedances are unknowns and there is a good chance they will not be balanced since my application is one of fault detection and diagnosis.

The main thing is when I go to my supervisors I can pull together enough info from these discussions, equations and diagrams here to explain why it can't be done!

Cheers, W.
Perhaps no ultimately accurate solution, but there is defintely a reasonably accurate solution. It has and is being done all the time with power distribution monitoring equipment.

Do you think GF detection devices are setup to distinguish whether a load is delta- or wye-connected on a wye system??? What about combinations loads thereof on the same monitored lines???

IMO it is entirely possible to do within a less than 100% certainty level (exactly what percentage of certainty is entirely another matter in itself :D). The only way to certify with near 100% certainty that your phase current values are accurate for a delta-connected 3? load is to monitor the actual phase currents.

However, as I see it, the question is merely whether it is feasible to do with a soft-coded versus a hard-coded solution. In other words, should you replace your current volt/ampere data device with a full-featured monitoring device, or rely on your math and data processing skills and utilizing a soft-coded solution? Being that you had to ask about how to determine load phase currents to begin with is not a good sign of your knowledge and skills being adequate for the task (sorry, no offense intended). That said, even full-featured monitoring equipment make assumptions of likelihoods and those assumptions are hard-coded into the device (the manufacturers just don't tell you this).
 
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mea03wjb

Member
In response to your comments:

I am involved in research into this area - therefore the intention is not to use an off-the-shelf product to provide an answer but to work through the problem myself and develop a solution.

The idea is to find an elegant and effective solution to the problem - without making the assumptions that other approaches may take. Therefore, I am not interested in approximate solutions. You may say that this is pointless if the approximate solution is OK - but is it accurate enough to be used for determining a developing turn-to-turn stator fault in its early stages (e.g. 0 - 5% shorted turns)? I have no interest in GF detection as this is provided by basic protection devices and should already be in place on machine in industrial environments.

Is the partial solution accurate enough to be compared with mathematical model outputs and drive optimisation routines?

Is it right to base a system on this partial solution? I don't think so.

You have hit the nail on the head - I have only been looking into this problem and been working in this field for just over a year - I am by no means an expert. However the initial question was valid.....

Being that I had to ask about how to determine load phase currents to begin with is not a good sign of my knowledge and skills being adequate for the task?

No offense taken - I am yet to be provided with the answer. :)
 

mivey

Senior Member
No offense taken - I am yet to be provided with the answer. :)
You did not catch it. If you want to use the impedances:

Use the line currents and line-neutral voltages (be sure to get the phase angles) to come up with a test set of wye impedances (not real ones, just ones that you will use as a test case).

Transform these test impedances to delta impedances. Given line-neutral voltages, you can find the line-line voltages. Use the line-line voltages and the delta impedances to get the phase currents.
 

Smart $

Esteemed Member
Location
Ohio
In response to your comments:

I am involved in research into this area - therefore the intention is not to use an off-the-shelf product to provide an answer but to work through the problem myself and develop a solution.

The idea is to find an elegant and effective solution to the problem - without making the assumptions that other approaches may take. Therefore, I am not interested in approximate solutions. You may say that this is pointless if the approximate solution is OK - but is it accurate enough to be used for determining a developing turn-to-turn stator fault in its early stages (e.g. 0 - 5% shorted turns)? I have no interest in GF detection as this is provided by basic protection devices and should already be in place on machine in industrial environments.

Is the partial solution accurate enough to be compared with mathematical model outputs and drive optimisation routines?

Is it right to base a system on this partial solution? I don't think so.

You have hit the nail on the head - I have only been looking into this problem and been working in this field for just over a year - I am by no means an expert. However the initial question was valid.....

Being that I had to ask about how to determine load phase currents to begin with is not a good sign of my knowledge and skills being adequate for the task?

No offense taken - I am yet to be provided with the answer. :)
With your intention better understood, my first question: Is monitoring phase current (rather than line current) a possibility? Such would require the CT's (or whatever sensing element you are using) be installed at/near the load. or five to six wires run between wherever measurement takes place and the load.
 

mea03wjb

Member
You did not catch it. If you want to use the impedances:
Use the line currents and line-neutral voltages (be sure to get the phase angles) to come up with a test set of wye impedances (not real ones, just ones that you will use as a test case).

Mivey, please can you expand on this point?

I used wye-delta transforms (replacing R's with (v/i)'s) which resulted a completely random phase currents (in line with what was said about the problem having infinite solutions).

I only used the instantaneous values - How are the phase angles used? How do you come up with the test set of wye impedances?

---

Smart, I can't really use the phase currents since the idea is to use commonly available signals (e.g. line quantities) and not to use additional sensors/hardware. If I had the phase quantities everything would be so much easier though!
 

mivey

Senior Member
Mivey, please can you expand on this point?

I used wye-delta transforms (replacing R's with (v/i)'s) which resulted a completely random phase currents (in line with what was said about the problem having infinite solutions).

I only used the instantaneous values - How are the phase angles used? How do you come up with the test set of wye impedances?
Instantaneous won't work unless you want to assume some impedance angles. You would probably be safe enough to assume the voltages have a uniform 120 degree displacement, but that might be a stretch for the currents.

You have to have the phase angles. When you have the current phasor (line current) and voltage phasor (line-neutral) for each line, each impedance in the wye will be the voltage phasor divided by the current phasor. I'm too lazy to write out the impedance conversion so here is a wiki link:
http://en.wikipedia.org/wiki/Y-Δ_transform
 
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