Based on some years of experience at a Shipyard with an ungrounded electrical system, I generally found that separate phases of separate equipment in separate areas of the plant would simultaneously blow across ground when their insulation became weakened and ionize sufficiently. Just like lightening exploding under similar circumstances. To add to the problem, the systems utilized circuit breakers and when the equipment did fail as mentioned above, it normally took out a main air breaker in addition to the molded case circuit breakers in each area of the plant. In your case, I wonder what other equipment not directly associated with the MOV fault but indirectly tied to your distribution system might have blown at the same time as I suspect this happened. Bob Bitner

 

Responding to Bill Mostia's Thu, Mar 15,1:34pm requesting additional detail. We're getting well beyond "Elecricity for the Uninitiated", but I'll try. There are many types of electrical failures in a 3-phase system: SLG (single-line to ground); LLG (double-line to ground); LLS (line-to-line short); and LLL, (3-phase short), are some of the well known types. Of course most members of the list are familiar with the last one. The current in a 3-phase short is essentialy determined by the source phase to ground voltage divided by the sum of the circuit element impedances between the source and the point of failure. This is called a 'symmetrical' fault. All of the others are considered 'asymmetrical' faults, and their determination is not as simple as the LLL. Thus electrical engineers resort to a concept called symmetrical components. It assigns to each cable, generator, transformer, etc, three impedances.They are: Z1 (the positive sequence); Z2 (the negative sequence); and Zo (the zero sequence). Then by judiciously combining these three impedances, any of the faults described above can be solved. Let the fault current equal If. Then, as an example for LLL, the current If(3)=Ey/Z1, where Ey is the phase to neutral voltage. For the LLS, the current If(2)=SQRT(3)xEy/(Z1+Z2); the LLG is more involved, but can be found from If(1)=3xEy/(Z1+Z2+Zo+3Zn), where Zn represents the equivalent system impedence between the 'effective' neutral of the system and ground. For any system for which there is a deliberate connection between neutral and ground, i.e., solidly, via a resistor, or via an inductance, then Zo+3Zn will be a positive impedence. Note that the distributed ground capacitance, Zc, is also present, but it is in parallel with Zo+3Zn. Zo+3Zn is normally much smaller than Zc so it essentially negates any deleterious capacitive effects. So, Ik(1) is finite. However, when the system is ungrounded, Zo+3Zn becomes infinite and Zc, the distributed ground capacitance becomes the prominent impedance. Of course, you can now see that if the quantity Zc approaches the sum of Z1+Z2, then Ik(1) can reach infinity because the sum (Z1+Z2-Zc) approachs zero. In fact if phase A becomes grounded, then phase A capacitance is effectivly shorted to ground, but both the intact phases B and C will exhibit very high voltage with respect to ground. There is a formula for B and C voltages, which I can supply upon request. Needless to say they contain the denominator (Z1+Z2-Zc). Phew! I bet some of you are muttering 'Fantastic' at this point. The above said, then how did the MOV's explode? Essentially, the voltage triangle representing the phase to phase voltages, A-B-C, remains intact. Digressing a moment... although not stated I'd bet my career on the fact that the phase-to-phase MOV's remained intact. Returning to the discussion at hand. Phase A is now at ground potential. Phases B and C are at some magntude greater than the expected 480/SQRT(3) or 277V. From a practical viewpoint, the 'overvoltage' could be from 3 to 5 times greater than the expected 277V. Remember, in my war story, I measured 800-1,200 Volts or 2.89 to 4.33 times the expected value. If the MOV's were typcally specified for 15% above nominal, then one, more likely both, would be triggered into conduction. Thereby producing first the SLG and then cascading into the LLG fault. If this discussion, together with my previous ones is still unclear, please let me know. Regards, Phil Corso, PE (Boca Raton, FL)

 

Correction to my Fri, Mar 16, 3:44pm response to Bill Mostia: Third paragraph, last (long) sentence: Change "the LLG is more... to "the SLG is more..." Regards, Phil Corso, PE (Boca Raton, FL)