Actually it is an NEC issue. "110.10 Circuit Impedance and Other Characteristics. The overcurrent protective devices, the total impedance, the component short-circuit current ratings, and other characteristics of the circuit to be protected shall be selected and coordinated to permit the circuit-protective devices used to clear a fault to do so without extensive damage to the electrical components of the circuit." If you look at page 50 in Bussmann's SPD you will find a Copper, 75? Thermoplastic Insulated Cable Damage Table. You will find that #14 wire is good for only 3400 amperes Maximum Short-Circuit Withstand Current in 1/4 cycle.Originally posted by Roger:
This is not an NEC issue however it is a pretty common engineers specification.
Yes. Reference: Ohm?s Law. But would the higher value of current (in the locked rotor condition) exceed the rated FLA? Only if the applied voltage exceeded the value upon which the rated FLA was based. If you connected a motor rated at 460 volts to a 480 volt panel, and if you oversized the conductors, you could get more than the rated FLA.Originally posted by bennie: Would a motor fed with #12, with a FLA of 6 amps, have an increase in locked rotor current than it would with #14?
Yes. A fault internal to the motor would see higher current, if the applied voltage were higher, which would be the case if the branch circuit conductors had a lower resistance, which would be the case if you oversized them."can a larger than necessary conductor raise the level of damage by a fault?
Not exactly. The motor winding still has some impedance, even without the rotor moving.A locked rotor is a short circuit, or a line to line fault.
Not quite true. The windings of a motor have a significant amount of impedance, as compared, say, to a wrench dropped across the motor terminals. The impedance is primarily inductive, and is the result of wrapping the coils of wire. If you physically prevent a motor from turning, and then hit the ?on? button, it will draw between 6 to 14 times the normal running current. This is far below what would be considered a line-to-line fault. A wrench in contact with the terminals could draw more than 1000 times the normal running current. The instantaneous trip feature of the circuit breaker will terminate this event. It?s purpose is personnel safety, not the protection of cables or motor windings.Originally posted by bennie:A locked rotor is a short circuit, or a line to line fault.
The devices you are describing here are the thermal overloads. Motor protection in a jammed-rotor condition comes not from the breaker, but from the thermal overloads.Short circuit devices are calculated to open the circuit before heat can damage the winding. The time/trip curve is selected based on the withstand rating of the winding insulation.
Replace the #12 branch circuit conductors with superconductive bus bars, and you still won?t defeat the protective system. When the thermal overloads see current values high enough and long enough to risk damaging the motor, they trip. When the circuit breaker sees fault-levels of current, it will trip. They will do their jobs no matter what had caused the high current levels.Any radical departure from the design values will defeat the protective system.
Wayne I am curious why you used the term "strategies"?Originally posted by awwt:
There are many strategies for protecting a motor load with #14 or other supply conductors:
Starting heaters;
Fuses & heaters;
Circuit breakers;
Thermal sensors.