# Calculating Phase Currents for Unbalanced Loads

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#### mea03wjb

##### Member
Hi All,

I would like to calculate the phase currents for a delta connect load - the load may or may not be balanced and the phase impedances are unknown.

I know the line currents and line to neutral voltages from the supply. Is it possible to calculate the phase currents for a delta connect load (e.g. induction motor) using this information alone? :-?

Cheers, W

#### dkarst

##### Senior Member
You should be able to write a KCL equation for each node of the delta (relating the two phase currents to the line current) and assuming you know BOTH the magnitude and phase of each of the line currents, it should be straightforward phasor math. This sure sounds like a test question of some sort?

#### dkarst

##### Senior Member
This may not be quite as straightforward as I indicated... the solution may have to be split into two problems. Could you post the real situation with values, please.

#### topgone

##### Senior Member
Hi All,

I would like to calculate the phase currents for a delta connect load - the load may or may not be balanced and the phase impedances are unknown.

I know the line currents and line to neutral voltages from the supply. Is it possible to calculate the phase currents for a delta connect load (e.g. induction motor) using this information alone? :-?

Cheers, W
Would be glad to help but can you provide some numbers for us to crank?

#### mcclary's electrical

##### Senior Member
Hi All,

I would like to calculate the phase currents for a delta connect load - the load may or may not be balanced and the phase impedances are unknown.

I know the line currents and line to neutral voltages from the supply. Is it possible to calculate the phase currents for a delta connect load (e.g. induction motor) using this information alone? :-?

Cheers, W

..........................

#### rattus

##### Senior Member
If we have the magnitudes of the line currents, it is easy enough to obtain the angles with a graphical solution. After that, I am stumped. I can write 3 equations, but there are 6 unknowns--the phase currents and their angles. And too, the determinant equals zero.

Help!

#### mivey

##### Senior Member
Thinking out loud: Connect test wye impedances and solve for impedance values. Then make a wye-delta impedance transformation.

#### mea03wjb

##### Member
I have looked at the equations and don't think you can solve them using the magnitudes only, I think some of you may be coming to the same conclusion - it looks like it should be simple but I don't think there is enough information!

It is for a system that samples the currents and voltages at regular intervals so I can give you some instantaneous values as an example (recorded from a test rig):

v1 (phase-neutral) = 308V
v2 (phase-neutral) = -275V
v3 (phase-neutral) = -35V

i1 (line) = 17.7A
i2 (line) = -2.95A
i3 (line) = -15.5A

I want it to be an online (during operation) calculation so this rules out connecting in Y, thanks anyway Mivey!

Thanks, W.

#### Smart \$

##### Esteemed Member
If we have the magnitudes of the line currents, it is easy enough to obtain the angles with a graphical solution. After that, I am stumped. I can write 3 equations, but there are 6 unknowns--the phase currents and their angles. And too, the determinant equals zero.

Help!
Well I figured out a geometric solution (assuming each line-to-line load has the same power factor)... see below. First step is arrange the line current vectors tail to head. Of course we only know the magnitudes, so they'll be arranged relative to each other but not to voltage.

Next, let each line current vector be the base of a new 30-120-30 isoceles triangle to the outside of the original triangle. The apex of each triangle is the center of a circle which passes through the its base vector endpoints. The intersection of the three circles is the point where a vector to each tail-head intersection represents the phase (line-to-line) current of the delta load.

Now the math I'll leave as an exercise for you other guys  #### Smart \$

##### Esteemed Member
....

Now the math I'll leave as an exercise for you other guys ...
FWIW, my solution appears to be an offshoot of (read: essentially the same as) a First Fernat Point.

#### topgone

##### Senior Member
Okay, here it is:
Vab = 533.47 /30 deg
Vbc = 476.31 /-90 deg
Vca = 60.62 /150 deg

Iab = 2.95 / 22.64 deg amperes
Ibc = 0
Ica = 15.5 / 157.36 deg amperes

#### Smart \$

##### Esteemed Member
Okay, here it is:
Vab = 533.47 /30 deg
Vbc = 476.31 /-90 deg
Vca = 60.62 /150 deg

Iab = 2.95 / 22.64 deg amperes
Ibc = 0
Ica = 15.5 / 157.36 deg amperes
Don't forget he said instantaneous values (actually, just instant values, being of a single point in time). Instant voltages and currents have no phase angle. The line-to-line voltages are simply the arithmetic difference in potential, and currents evaluate as if DC.

So now we're back sort of to what Mivey suggested, but doing it theoretically and mathematically. Google "wye-delta transform" or "...transformation". The following link takes you to the wikipedia page on the topic:

http://en.wikipedia.org/wiki/Y-Δ_transform

Using such we can mathematically determine an equivalent wye-configured resistor circuit, then transform it into a delta-configured resistor circuit and obtain our "phase" instant currents...

#### rattus

##### Senior Member
Not very elegant, but:

Not very elegant, but:

FWIW, my solution appears to be an offshoot of (read: essentially the same as) a First Fernat Point.
After finding the angles geometrically, one can simply draw a 120 degree wye on a sheet of tracing paper, then lay this over the line current delta sketch and wiggle it around until each leg of the wye intersects a vertex of the delta. Now scale the load current phasors and measure the angles if you wish. Probably good to 5% or so.

#### Smart \$

##### Esteemed Member
After finding the angles geometrically, one can simply draw a 120 degree wye on a sheet of tracing paper, then lay this over the line current delta sketch and wiggle it around until each leg of the wye intersects a vertex of the delta. Now scale the load current phasors and measure the angles if you wish. Probably good to 5% or so.
A caveman could do it Seriously though, with scale, straight edge, a drafting compass, and a sheet of paper, it can be done rather easily. Six lines, six arcs... pictured below, but I included some extra lines just for the heck of it, I guess :roll:

Any way one goes about it, this way is definitely easier than doing the math  #### mea03wjb

##### Member
Ok thanks Smart - I will have a look at your link and see if I can calculate a solution.

Is assuming the PF is equal in each phase a valid assumption for the induction motor load (I am not challenging your original statement, I am a novice and I just want to understand more!) ? Thanks.

#### jghrist

##### Senior Member
I would like to calculate the phase currents for a delta connect load - the load may or may not be balanced and the phase impedances are unknown.

I know the line currents and line to neutral voltages from the supply. Is it possible to calculate the phase currents for a delta connect load (e.g. induction motor) using this information alone?
Please define what you mean by "line current" and "phase current". By line current, do you mean the current in the line between the source and the load (each phase)? By phase current, do you mean the current in the delta load (between phases)?
It is for a system that samples the currents and voltages at regular intervals so I can give you some instantaneous values as an example (recorded from a test rig):

v1 (phase-neutral) = 308V
v2 (phase-neutral) = -275V
v3 (phase-neutral) = -35V

i1 (line) = 17.7A
i2 (line) = -2.95A
i3 (line) = -15.5A
You can't really do much with a single instantaneous value. Consider that any sinusoidal value will have a value of zero twice per cycle, so if you measure zero amps, the magnitude of the sinusoid could be zero or 10000000.

If the values are sinusoidal (constant magnitude and phase with no distortion), you can define the magnitude and phase with two values 1/4 cycle apart.

#### Smart \$

##### Esteemed Member
Ok thanks Smart - I will have a look at your link and see if I can calculate a solution.

Is assuming the PF is equal in each phase a valid assumption for the induction motor load (I am not challenging your original statement, I am a novice and I just want to understand more!) ? Thanks.
Typical 3? loads, such as a motor, can be assumed to have the same power factor per line or phase. That said, when evaluating instant values, there is no such thing as power factor. Each instant is processed as if DC. Power factor is only a concern when you average the measured values over time. Starting with raw instant values, that data can be used to determine the power factor. There should be no reason to make power factor assumptions at that stage (unless your processing speed isn't up to snuff for the task).

And that takes us to the question of why you are processing instantaneous values? Seems to me that would be better handled by a dedicated power monitor...

#### mea03wjb

##### Member
I think we're getting there......

I think we're getting there......

Thanks everyone so far for their input!

Firstly, people have asked for clarification on some of the terms I've used:

Line current = current measured on each of the 3 lines (phases) between source and load.

Phase current = current flowing through the 3 windings (phases) of the induction motor load (i.e. current flowing in the different sides of the delta load).

Secondly, I was asked to provide example figures for people to have a look at - that is why I gave some instantaneous values. I am actually able to sample the data at up to 20kHz (5kHz realistically) through a dedicated sensor box.

Therefore, each of the currents and voltages will be available as a discrete waveform (0.0002 secs between each data point). Does this make the calculation any easier?

The graphical solution seems fine for one-offs but I wanted to implement a system to calculate the phase currents automatically in software based on the measured line quantities - that is why I'm trying to do it mathematically.

--

In terms of the Fermat solution, I think the idea is to calculate the centre point of the triangle made by the 'line' current vectors and use this with the original vector values to calculate the new 'phase' current values. This is what topgone has done in a previous post.

If this was done at each time instant (every 0.0002 seconds) with the instantaneous 'line' values, then the instantaneous 'phase' values could be calculated at each time instant. Then if this process was repeated which each set of instantatneous (sampled) values over an extended period of time (say 1 second) the waveform for each of the 3 'phase' currents would be produced.

Does anyone see a flaw in this approach?

Cheers,
W.

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#### Smart \$

##### Esteemed Member
.... I am actually able to sample the data at up to 20kHz (5kHz realistically) through a dedicated sensor box.

Therefore, each of the currents and voltages will be available as a discrete waveform (0.0002 secs between each data point). Does this make the calculation any easier?

The graphical solution seems fine for one-offs but I wanted to implement a system to calculate the phase currents automatically in software based on the measured line quantities - that is why I'm trying to do it mathematically.

--

In terms of the Fermat solution, I think the idea is to calculate the centre point of the triangle made by the 'line' current vectors and use this with the original vector values to calculate the new 'phase' current values. This is what topgone has done in a previous post.

If this was done at each time instant (every 0.0002 seconds) with the instantaneous 'line' values, then the instantaneous 'phase' values could be calculated at each time instant. Then if this process was repeated which each set of instantatneous (sampled) values over an extended period of time (say 1 second) the waveform for each of the 3 'phase' currents would be produced.

Does anyone see a flaw in this approach?

Cheers,
W.
First let's make a distinction here. The graphical solution (including any math solution associated thereto) is for RMS data, not instantaneous data.

Values of an instant are processed with basic math. "Phase" voltage is the absolute difference of line-to-neutral values. For example, your earlier post gave the following values for a single instant:

v1 (phase-neutral) = 308V
v2 (phase-neutral) = -275V
v3 (phase-neutral) = -35V
Line-to-line voltages at that instant would be:
v12 = |v1 ? v2| = |308V ? (-275V)| = 583V
v23 = |v2 ? v3| = |(-35V) ? (-275V)| = 240V
v31 = |v3 ? v1| = |(-35V) ? 308V| = 343V​
Calculating instant phase current also uses basic math, but gets slightly more complex because it requires the use all six discrete values to compute each phase current. With a delta connected 3? load (a wye connected 3? load would be a different calculation), you simply do a wye-delta transform substituting a correlated "vi" for each "R" in the formula. A wye-connected 3? load would be calculated differently because each line current goes through three resistances (except when one line current is zero; if two are zero, so would be the third, or there would be a ground fault), rather than two as in a delta configuration, and the neutral point is a floating voltage.

#### topgone

##### Senior Member
Don't forget he said instantaneous values (actually, just instant values, being of a single point in time). Instant voltages and currents have no phase angle. The line-to-line voltages are simply the arithmetic difference in potential, and currents evaluate as if DC.
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!

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