superheat6k
Well-Known Member
An induction motor will produce a variable amount of power and hence amps, subject to the load to which it it subjected, and they can be overloaded, although most will sustain a level of overload before overheating.
Depending upon the coils layout a motor will spin at a certain no load speed. With no load applied the current draw and hence power will be at its minimum, typically ~1/3 it's design maximum run amps, also known as Rated Load Amps or RLA. At minimum load the motor is also inefficient because it does not require 1/3 of its power rating just to keep the shaft spinning, and this inefficiency manifests as heat, although the motor can easily dissipate this heat.
As a load is applied it tends to slow the shaft, however induction motors are self governing due to an effect called induced current. The internal induced current opposes the forward driving current being fed in by the supply. As the shaft slows the induced current reduces, allowing the supply current to increase, this increases the shaft speed, in turn causing the induced current to increase, until the two balance, but at an increased supply level, and slightly reduced shaft speed. As load continues to increase this effect continues until the motor reaches its rated current. In selecting a motor the designer should use a motor that will satisfy the maximum applied load just below the point at which the motor reaches the FLA point. The reduction in shaft speed between minimum and maximum load is known as slip.
So what happens in an overload situation ?
The pure electrical resistance of a motor is far lower than that determined by considering that dictated by ohm's law, but as the motor runs the opposing current creates inductive resistance which adds to the pure resistance, so at any point in its operation, regardless of load, Ohm's Law is always satisfied. If the motor is overloaded, e.g. In the case of a windlass the cable becomes gradually snagged, the motor will be forced to slow down, in turn the inductive resistance will fall away to zero as the shaft stops, leaving the pure electrical resistance only opposing the current flow. At this point the motor will be heavily overloaded, and will draw Locked Rotor Amps (LRA). Indeed at initial start up the motor will also draw close to LRA as the power is first applied. LRA is typically 5 - 7 times the FLA, and whereas the motor windings can tolerate such an overload for a few seconds, the massive current is not producing any motive power, just torque, where it is trying to turn the load, and with it lots of heat, and sufficient to heat the field winding wires to the temperature at which the insulations will start to melt, and if this happens burnout will ensue. Once the insulations have broken down, the coils wires can short out, and the whole thing massively overheats and self destructs.
Between the FLA point, where the motor is producing its maximum rated power and the LRA point, when the shaft has stopped, the overload will progressively increase, and the motor will tolerate a certain level of overload without harm, e.g. ~10%.
So to answer your original question, subject to the applied load somewhere between 1/3 x FLA and 1 x FLA, which for your 1,000 watt motor running on a nominal 12v supply will be ~25 - 30 amps minimum, and ~85 - 90 amps maximum, allowing for the fact that at minimum load cables losses will be less, and the supply voltage will be closer to 13.5v, and at FLA point (1,000 watts) the cables losses will mean the local voltage will likely have dropped at least 1.5 - 2 volts to around 11.5 - 12v.
Hot spots on your supply cabling and distribution are adding to the overall circuit resistance, so are actually reducing the voltage available at the motor, in turn this will cause the motor amps to increase to maintain a particular load. So a 1,000 watt 12v motor will become an 830 watt motor if the supply it is seeing drops to 10v, whilst still not exceeding the FLA rating of the motor.
I hope this helps.
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