Low Fault Level Effect on LRA and FLA

If the fault is large enough, (a) circuit protective device(s) will disconnect it, and the <...motors...> will not start.

<...FLA... LRA...> are characteristic design features of the <...motors...> at various voltages, and are unrelated to a <...1kA...fault...> in the supply cables.

https://www.acronymfinder.com/LRA.html

https://www.acronymfinder.com/FLA.html

RLA = Rotor Locked amps

FLA = Full load Amps

Quite.

It remains impossible to relate these features to fault currents.

Assuming an induction 3 phase motor, with normal starting current...

It is likely that an extremely low available fault current such as 1kA symmetrical (assumed) is due to very small generator system or to extremely high impedance between generation and consumer.

In the case of a small generator, i.e. 450kW to 900kW, the generator voltage regulator may not have the range to develop the voltage (motor accel torque), usually a field forcing circuit and voltage restrained overcurrent protection on the stator leads is required to supply the current to bring to motor up to speed from a stop.

In the case of a very weak supply system, the system impedance between the source generators and starting motor is very high. The voltage at the motor terminals could be so low at start the the motor would not be able to accelerate the load in the time required to avoid extreme rotor and stator heating and/or motor current protection activation.

The fault current is listed in amperes, and there is no voltage reference. Assuming a 600hp ~ 600kVA motor, and assuming the fault current is 1000A at 415V basis, it amounts to 718kVA fault source, less than twice the motor rating, so normally not enough to get the motor started.

Reduced starting methods would generally not work, unless the rotor was wound and had stepped resistors, or was specially designed to safely take the heat generated during an extended start. The actual driven load speed characteristic is needed to analyze the starting capability.

A variable frequency drive for the motor (i.e PWM inverter) will work in this case, providing the motor driven load breakaway torque is low. You do not have much of a margin available, limited by your source and the hardware current rating of the inverter, often a limit you encounter when driving constant torque loads

Hi All,

Not the ideal power supply.

Just bit more info the utility supply is 22kV 15km from source sharing power with 14 other customers we transform the 1600kVA TX 4% Z to 11kV to 400metres to a underground coal mine transform again 800kVA 4% Z down to 1000V. The machine is a standard design continuous miner DOL.

You used your utility fault contribution to calculate the 1000 ampere 3 phase bolted fault contribution?

i would think you would get closer to 10,000 amperes symmetrical, even with the 800 kVA transformer.

i would assume given this configuration (not knowing the 1000 sc avail) you would have no trouble starting your motor if unloaded. Your locked rotor available torque is maybe slightly less than you might get with a larger transformer, but assuming the transformer is a unit transformer, no other large rotating machinery, it is a generous rating for a unit transformer, applications quite common in industry. They (unit transformer) perform quite well as a replacement for a reduced voltage starter, often a cost effective option when considering various options to reduce flicker on the upstream power system.

One wonders what all those <...1kA...> are heating? That's one hell of an arc!

Thank goodness someone else is paying the electricity bill...

In my opinion, the short-circuit current at 800 kVA 11/1 kV transformer terminal could be 10 kA.

The rated current of 450 kW/1000V induction motor it is about 320 A and at 3 times starting current [developing 25% of rated starting torque]it produces 15% voltage drop.

By reducing the load [including inertia of the motor and of the load] to a less than 25% one start could be done.

Hi All,

I have done a spreadsheet which I need to attach ( how does one do it on this site) to show that fault level at source plays a major part in starting large (450kW) motors on DOL. The kA rating at the motor supply transformer (10kA) will not help when the Intake supply is on 1.4kA it pulls the voltage down.

The only part missing is the short circuit available at the utility connection to make the calculation. You have provided all the other details, we assume your mining machine is a constant torque load with somewhat normal 150% or less breakaway torque and negligible inertia reflected back through the gearbox to the motor.

It would be helpful to know if this is a problem with new equipment, and whether you have similar installations that work ok under similar conditions. Or if this is a suspected cause of otherwise unexplained incidents.

It needs to be noted that <...reducing the load [including inertia of the motor and of the load]...> has no effect on the <...LRA...> or the <...FLA...> as these remain features of the particular motor employed.

Just the time spent at those current levels...

Sorry! I forgot to divide by √3 [1000/√3].So the short-circuit current it is only 6500 A. However, if the motor start current will be 6*Irated =1890 A then Zstartmotor=0.3 ohm and at start [s=1] the current will be 1460 A that means the torque could be 60% of the start rated torque. The voltage drop at the motor terminals it could be 226 V[=22.6 %].

The OP states that the SC current is 1000A but gives no voltage reference or location? It is implied to be at the motor, which is possible, but appears to be improbable...