3 phase motor, high resistance & high amps.

[Dave0098]

at work the other day, a 3 phase motor was running very high amps, when everything checked out we found a loose connection on 1 of the phases back at the DB, cable was melted and burnt. now for me i always thought high resistance would mean less amps would flow, the other electrician put the high amps down to the high resistance. all of the fuses where fine so i dont think the motor was trying to run on 2 phases which i could understand giving high amps.

anyone feel like they could explain this?

 

[Notsosmart]

The higher resistance means that you have a greater volt drop over the length of cable.

The motor draws power from the mains so if you transpose P=IV then I=P/V Therefore if you have a higher volt drop then a greater current will flow, usually disipated as heat at the high resistance point; so causing insulation degredation & burning with eventual cable failure.

You would also get slight phasing of the motor due to the imbalanced voltage, causing poor flux density across the affected poles. This will also raise the power required to work the motor compounding the original problem.

 

[Dave0098]

Hey thanks Notsosmart, thats really cleared that up for me, its all good experience! sometimes you just need to take a step back and have a little think about it or in my case as the question!!!

but yep, thanks again.

Dave

 

[NoSparks]

The higher resistance means that you have a greater volt drop over the length of cable.

The motor draws power from the mains so if you transpose P=IV then I=P/V Therefore if you have a higher volt drop then a greater current will flow, usually disipated as heat at the high resistance point; so causing insulation degredation & burning with eventual cable failure.

You would also get slight phasing of the motor due to the imbalanced voltage, causing poor flux density across the affected poles. This will also raise the power required to work the motor compounding the original problem.

I'm sorry but I can't agree your first analysis. Your equation P= IV and its transposition are valid but you forget the equation R=V/I In this case R has increased (contact resistance at the burned terminal) and since V (the mains voltage, not the cable drop part of it) remains at 240v (one phase) so to maintain the equation I falls. Now return to your equation and since V is a constant P drops. ie the power consumed will fall.

But this is not the whole story. Your second analysis is I believe closer to the point. The inballance will cause different current flows in the coils than normal, the motor will slow, and the back emf generated by the motor will fall thereby counteracting the loss of voltage at the motor, allowing more current to flow especially throgh the other two phases. This is known as negative feedback, and it would take me another page or two to fully analyse the reaction of the motor. The overall effect will be an increase in the current in the two good phases and a reduced current in the failing phase.

The fact that the fuses didn't blow simple indiocates that the current was within the rupture current of the fuses.

Now maybe somebody familiar with the design of motors and the effectes of falling back emf will tell me where the flaws are in my arguments.!






The burned out cable it typical of any situation where the contacts have high resistance (bad contact) amd therefore start generating heat.