Motor current imbalance.
Motor current imbalance.
(OP)
10hp 230V 3ph motor.
The three legs are:
241.8V
242.5V
243.0V
The currents are:
31.4A
31.4A
34.5A
Rural.
Open Delta.
Sole device on the two transformers.
Rolling the leads does not change the imbalance.
The imbalance stays with the leads.
Changing the motor does not change the imbalance.
What reasons can you offer for the 10% imbalance?
The three legs are:
241.8V
242.5V
243.0V
The currents are:
31.4A
31.4A
34.5A
Rural.
Open Delta.
Sole device on the two transformers.
Rolling the leads does not change the imbalance.
The imbalance stays with the leads.
Changing the motor does not change the imbalance.
What reasons can you offer for the 10% imbalance?
Keith Cress
kcress - http://www.flaminsystems.com





RE: Motor current imbalance.
Your voltage unbalance is 0.26%.
Your current unbalance is 3.2%.
Your ratio of current unbalance / voltage unbalance is 12
What do we expect the ratio to be?
We know the positive sequence equivalent circuit.
The constant-parameter negative sequence equivalent circuit looks similar to the positive sequence except we replace R2/s with R2/(2-s).
Considering that |X1+X2| >> R2, the locked rotor impedance is approximately |X1+X2|. Since |X1+X2| >> R2/(2-s) we can also say the negative sequence impedance is approximately |X1+X2| = locked rotor impedance.
The unbalance component of the current doesn't change with load.
At full load, based on the above model, we can expect the ratio of current unbalance to voltage unbalance to be on the order of the ratio locked rotor current to full load current: 5-7. (At no-load we'd expect it to be higher by the ratio FLA/NLA.... since the negative sequence current won't change with load, but the fraction will).
Your ratio is 12 vs 6 predicted. Not perfect, but we may not be dealing with imperfect measurements, AND the model is not quite perfect. For one thing, we assumed constant parameters, but they change with frequency... as we increase frequency from locked rotor (rotor see line frequency) to negative sequence near full speed (rotor sees 2*line frequency), rotor sees a higher frequency. Deep bar effect causes R2 to go up and X2 to go down. Since the negative sequence is primarily inductive, the effect of X2 decreasing is the dominant effect and we expect a ratio somewhat higher than the locked rotor ratio of 6.
All in all, if you didn't tell us about a lead swap, a ratio of 12 doesn't seem unreasonable if the cause were simple supply system voltage unbalance.
You conducted a swap and the high current stayed on the same lead... does that mean same motor lead? If you are positive of that, then forget about supply system unbalance, it would suggest an unbalance in the motor or the lead impedance between motor and connection point. Is there a long run of cable between the connection point and the motor... there was a thread not too long ago that talked about possible unbalanced induced voltage in the long run of leads going down the hole based on positioning of the leads of a submersible
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(2B)+(2B)' ?
RE: Motor current imbalance.
thread237-267952: Earth Currents with Submersible Pumps
Now I see the subject was circulating currents induced in grounded components. It seems like these effects might also contribute to possible phase voltage or current unbalance, but I'm not sure how much.
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(2B)+(2B)' ?
RE: Motor current imbalance.
If yes, can you describe distance between location of swap and the motor... and also describe the conductor configuration.
If no, can you provide locked rotor current or KVA code.
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(2B)+(2B)' ?
RE: Motor current imbalance.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Motor current imbalance.
RE: Motor current imbalance.
However even if that assumption is not met, magnitude unbalance is still a rough indicator of voltage unbalance. If you don't believe it, try to draw an triangle that has the same length for all 3 sides but doesn't have the same angles between all sides.... Obviously it can't be done.
If you draw a closed triangle whose lengths match the three phase-to-phase voltage magnitudes reported above, and then calculate the angles between them by the law of cosines (theta = arcos{<c^2-a^2-b^2>/<-2*a_b>}), then you get vertex angles theta as follows:
59.74
60.03
60.23
If you rearrange the equilateral triangle into a star configuration, then the angle between vectors is now given by 180 – theta (where theta was vertex angle of the equilateral triangle discussed above). i.e. the angle between phase-to-phase voltage vectors would be:
120.26
119.97
119.77
I am still interested in clarification of what leads are referred to in "The imbalance stays with the leads". The T-leads or the supply "leads" ?
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(2B)+(2B)' ?
RE: Motor current imbalance.
Label the three phase to phase voltages as VA, VB, VC. (sorry about the notation, these are phase to phase voltages even though they carry only one letter).
Define the angle of V as angle 0. If we draw a vector diagram starting with the equilateral triangle with VA on the horizontal pointing to the right, and VB, VC pointing CCW around the triangle, then we can rearrange into a star configuration and we see the phase to phase voltages as follows:
"VA" = 241.8 angle 0
"VB" = 242.5 angle 119.7678358
"VC" = 243 angle 239.7411481
Let q = exp(i*120 degrees)
Positive sequence voltage is
V1 = (Va + q^2*Vb + q*Vc)/3 = 242.43 - 0.69i
Negative sequence voltage is
V2 = (Va + q*Vb + q^2*Vc)/3 = -0.2827 + 0.4909i
|V1|/|V2| = 0.00233671
0.23% - just a hair lower than we computed examining the magnitudes alone (0.26%).
I am always amazed at how closely the short-cut maximum-deviation from average works in computing ratio of negative to positive sequence. In this case it works very well. If you include the effects of voltage phase deviation, the computed true ratio negative to positive is a little bit lower than estimation based on using the magnitudes alone.
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(2B)+(2B)' ?
RE: Motor current imbalance.
How large and where in the circuit is this open delta transformer?
Rafiq Bulsara
http://www.srengineersct.com
RE: Motor current imbalance.
If readings are taken under load, then all of the effects of this unbalanced delta (be it impedance mismatch or ratio mismatch, or upstream neutral voltage drop, or whatever) are reflected in the voltage unbalance measurement. For that matter, assuming the source of unbalance originates in the supply rather than the motor, the motor doesn't know the difference whether an open delta transformer causes 0.23% unbalance, or an unbalanced single-phase load on 3 phase system cause the 0.23% mismatch, or something else. All it cares about is the voltage that it sees at its terminals under load. At least that's the way I think about it.
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(2B)+(2B)' ?
RE: Motor current imbalance.
Rafiq Bulsara
http://www.srengineersct.com
RE: Motor current imbalance.
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(2B)+(2B)' ?
RE: Motor current imbalance.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Motor current imbalance.
"VC" = 243 angle 239.7411481
should have been:
"VC" = 243 angle 240.03
This results in |V1|/|V2| = 0.00287
Still pretty close to the estimate based on magnitudes.
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(2B)+(2B)' ?
RE: Motor current imbalance.
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(2B)+(2B)' ?
RE: Motor current imbalance.
Theta_a = arccos{(Qa^2-Qb^2-Qc^2)/(-2*Qa*Qb)}
should've been
Theta_a = arccos{(Qa^2-Qb^2-Qc^2)/(-2*Qb*Qc)}
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(2B)+(2B)' ?
RE: Motor current imbalance.
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(2B)+(2B)' ?
RE: Motor current imbalance.
Sorry for the delay I was out quoting a crazy job for the third time. Then they wanted to "see me" which hacks 3 hours out of my day for a 15 minute face to face...
The answers you've asked for and further info are:
1) All measurements taken with the motor fully loaded.
2) The voltages are measured while running.
3) The leads have been rolled until the lowest value on the high current leg is achieved.
4) The high current leg is always the same supply lead.
5) The two pole mounted POCO transformers solely running this pump are radically different sizes.
6) The POCO pole is about 10 feet away from the starter.
7) The motor is about 80 feet down some triplex 6AWG from the starter where the rolling took place.
Keith Cress
kcress - http://www.flaminsystems.com
RE: Motor current imbalance.
The voltage unbalance is not tremendously high (0.26%). But the motor is expected to react with a current unbalance that is higher by a factor of around 6 (negative sequence impedance ~locked rotor impedance is around 1/6). You actually saw current unbalance a factor of 12 higher than voltage unbalance. Possible reasons for that factor of 12 instead of 6:
1 - We do know that the negative sequence impedance is in fact expected to be a little bit lower than locked rotor impedance due to deep bar effect which decreases X2 further at negative sequence rotor frequencies (~2*LF) than at locked rotor frequencies (1*LF). So expect X- < 1/6...ratio > 6.
2 - Perhaps some small measurement errors. We are dealing with small differences between large numbers... small errors have big effect.
The ratio of 12 doesn't seem unreasonable to me - a simple expected result of supply imbalance. There might be something else going on to explain the difference, but I sort of doubt it.
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(2B)+(2B)' ?
RE: Motor current imbalance.
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(2B)+(2B)' ?
RE: Motor current imbalance.
Now I am still wondering about the difference between 6 and 12. Some additional thoughts:
1: The number 6 came from my assumption about ratio of LRC/FLA. But for some motors that ratio could easily be 8, which may explain part of the difference.
2: We have 0.26% voltage unbalance at ground level... so what is the voltage balance by the time the current flows thru 80' of 6AWG down to the motor. Does the voltage unbalance tend to get bigger or smaller as the unequal currents travel along the cables (creating unequal voltage drops)? My gut feel is that the voltage unbalance would tend to grow smaller (the leg with the highest voltage with respect to some ficticious neutral tends to draw the most current... has a higher voltage drop... and get closer to the other phases), but I'm not positive about that. I'd be interested to hear if anyone else wants to weigh in on that question.
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(2B)+(2B)' ?
RE: Motor current imbalance.
Tells me there is nothing wrong with the motor.
are the probable culprits.
I will leave formulas to Pete.
Rafiq Bulsara
http://www.srengineersct.com
RE: Motor current imbalance.
If you look at the vector diagrams of an open delta connection you will see that the open delta forms a virtual transformer across the open side with the same characteristics as the two equal transformers.
But in the real world, you may have differing voltages due to the transformer regulation and different per unit loading on the transformers. This may cause phase angle errors. Neutral voltage drops will also cause phase angle errors.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Motor current imbalance.
Yes, imho there is no "problem" just the field conditions and reality.
If overload is the concern, there are other means to address that.
Rafiq Bulsara
http://www.srengineersct.com
RE: Motor current imbalance.
Rafiq Bulsara
http://www.srengineersct.com
RE: Motor current imbalance.
Rafiq Bulsara
http://www.srengineersct.com
RE: Motor current imbalance.
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(2B)+(2B)' ?
RE: Motor current imbalance.
http
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(2B)+(2B)' ?
RE: Motor current imbalance.
It's kind of a lame result promulgated by the pump makers. The only real alternative is to stick on the next larger motor and run it lightly loaded. But as a well guy you find yourself quoting against the next guy who doesn't use the larger motor that costs another half k dollars.
I'm glad I don't have to work in that arena.
BTW good luck with your "problem".
e-pete; I would expect down that little distance of grossly over-sized wire you'd see the same exact voltage. That would be my guess.
Keith Cress
kcress - http://www.flaminsystems.com
RE: Motor current imbalance.
My "problem" is of a different kind, convincing another engineer that perfect coordination does not exist in real world, especially when using MCCBs!
Rafiq Bulsara
http://www.srengineersct.com
RE: Motor current imbalance.
80 feet -> 0.04 ohms
The current difference among phases is 3A.
The associated difference in voltage drop is 3A*0.04 Ohms = 0.12 volts.
That sounds small, but the range of measured phase-to-phase voltages is only 0.6 (242.5 - 243.1). The 0.12 volts difference in voltage drop is 20% of the 0.6 volt variation between highest and lowest phases. At first glance I thought it might explain portion of the difference between 6 (or 8 or 10) and 12.
But voltage divider analysis shows that we do in fact expect the voltage unbalance to be lower at the motor terminals than at the supply. We can show this by looking at voltage divider which applies to both positive and negative sequence circuits:
Vmotorterminal/Vsource = Zmotor / (Zmotor + Zcable).
Zcable is the same for positive and negative sequence assuming symmetrical layout of the three conductors. Zmotor is much lower for negative sequence than for positive sequence (our factor of 6 lower). So the ratio Zmotor / (Zmotor + Zcable) = Vmotorterminal/Vsource is lower for negative sequence than positive sequence. Therefore the ratio of negative sequence to positive sequence voltage (i.e. the unbalance) is lower at the motor terminals than at the supply. So, it matches the intuition, the voltage unbalance would be lower at the motor terminals than at the supply... but it doesn't help to explain the difference between 6 and 12.
It was just a curiosity anyway - nothing to change the conclusion in this case.
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(2B)+(2B)' ?
RE: Motor current imbalance.
"but the range of measured phase-to-phase voltages is only 0.6
should have been
"but the range of measured phase-to-phase voltages is only 1.2
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(2B)+(2B)' ?