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Topic Title: Unbalanced voltage in an induction motor Topic Summary: Symmetrical components results Created On: 06 December 2012 08:43 PM Status: Post and Reply |
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I have a problem on a customers site with an imbalance in the LV phases from the supply. One phase is about 5% down on the other 2 so its causing excessive currents in the 110kW motor. I understand its to do with the negative phase component and proceeded to calculate the positive and negative components and then the negative current but i'm not sure if i'm getting the right answer, which i understyand to be somewhere in the region of 6-7 time FLC. My customer is pushing for a report he can then pass to the supply authority to prove the tripping problem is related to their supply. Any technical geeks who can point me in the right direction.
Cheers |
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Please provide line voltages and currents when motor is running. What protection relay has tripped and the relay settings?
Regards Chris Chew |
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the line voltages are:
r-y 439 y-b 417 b-r 439 the currents when checked during loading were: r-220a y-180a b-150a The currents when checked off load were: r-85a y-70a b-66a |
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and the trip device is a combined overload and thermal circuit breaker molded case part number schneider electric GV7RE220
cheers |
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I did calculate the negative sequence component to be 200a which is near enough FLC but i don't know if its right
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Please give the relay setting.
Regards Chris Chew |
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Hmm. My gut feel was that you have >2% unbalanced voltages, but actually this may not be so.
The following MATLAB code, assuming that the voltages are still 120 degrees apart, shows that the actual unbalance (based on negative sequence) is only 1.69%, so technically this is still within EN50160 limits of 2%, and the utility isn't bound to do anything about it. But, clearly, this is not desirable! a = exp(j*pi*2/3) a2 = exp(-j*pi*2/3) abc_to_PosNegZero = 1/3 * [ 1 a a2 ; 1 a2 a ; 1 1 1 ] PosNegZero_to_abc = inv(abc_to_PosNegZero) p1 = 1*exp(j*0) p2 = conj(a) p3 = conj(a2) % Example customer premesis measurements Vabc = [439*p1 417*p2 439*p3].' PosNegZero = abc_to_PosNegZero*Vabc Unbalance_pct = abs(PosNegZero(2))/abs(PosNegZero(1))*100 Unbalance_pct = 1.6988 ------------------------- Dr. Andrew Roscoe http://personal.strath.ac.uk/andrew.j.roscoe |
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Overload is set at 110% FLC so 210a
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What is the load profile of the motor? Is the load constant or changing according to its burden? It is quite obvious that the relay setting is too low for the motor operation. Is there any means of monitoring the motor temperature?
Regards Chris Chew |
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Andrew,
Those voltages were taken during off load periods, once the mixer is loaded then the voltages drop but the difference remains the same. Loading voltages: r-y 419 y-b 399 b-r 419 Also if it is proven to be more than 2% imbalanced then do the supply authority just need to improve to better than 2% ie:1.99%, which won't improve the situation. If this is the case then we'll need a generator set. Cheers Doc |
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the mixer load profile is as almost S6 type loading, with continuous running and intermittent loading to full capacity then steady off loading before starting the cycle again. i have another plant doing exactly the same work but with a balanced supply and the current loading there doesn't trip the mixer at all.
Cheers |
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The report based on the voltage and currents recorded will provide a good basis for the utility to rectify the voltage unbalance. The motor trippings need to be emphasized so that they can regard this case as urgent because of loss of production.
Regards Chris Chew |
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thanks Chris, I'll push the customer to get this rectified based on what I've done so far. As it happens, my customers energy supplier is the also their customer but i think they are dragging their heels hoping the problem will just go away but if anything its getting worse. I'm going to site on Tuesday so hopefully is can meet with who ever and iron this one out.
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Unbalanced voltage in an induction motor
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