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
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-%CE%94_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...
I will have achieved my life's goal if I die with a smile on my face.
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.
Don't mess with B+!
(Signal Corps. Motto)
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
Any way one goes about it, this way is definitely easier than doing the math
I will have achieved my life's goal if I die with a smile on my face.
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.
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)?
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.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
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.
Typical 3Ø 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...
I will have achieved my life's goal if I die with a smile on my face.
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.
Last edited by mea03wjb; 07-19-10 at 03:05 PM. Reason: Corrected sampling time
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:
Line-to-line voltages at that instant would be:v1 (phase-neutral) = 308V
v2 (phase-neutral) = -275V
v3 (phase-neutral) = -35V
v12 = |v1 – v2| = |308V – (-275V)| = 583VCalculating 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 3Ø load (a wye connected 3Ø 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 3Ø 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.
v23 = |v2 – v3| = |(-35V) – (-275V)| = 240V
v31 = |v3 – v1| = |(-35V) – 308V| = 343V
I will have achieved my life's goal if I die with a smile on my face.
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