motor
I have a rather small motor and a matching energy source.
When the motor starts, it needs a (short) amount of time to 'rev up', to accelerate at its max speed.
Is there any cause for this other than mechanical inertia and friction? Would, say, a pullcord attached wound in the rotation's direction, pulled as the motor start, serve to bypass this delay?
Please explain it like I'm five (but good at googling terminology).
The average current of the motor may remain below 0.5 Amperes as noted, but transient currents could rise as high as 8.6 Amperes - the starting current of the motor would approximate the stall current rating.
Secondly, a DC motor such as the one listed, would generate significant back-EMF (voltage spikes of reverse polarity) from the coil commutation.
Each of those factors may trigger protection circuitry of the power supply. From the description of the intermittent operation of the motor, it appears that the initial starting current drawn by the motor is causing the power supply to either enter short-circuit protection, or over-temperature protection. The supply then recycles, and restarts the motor, and the cycle repeats.
While current limiting may be an option, operation will still be marginal. Possible mitigation options:
Even if you could make a motor work within the 1 Ampere limit with no load, any increase in load would inherently increase the current drawn, so measuring the average current with no load on the motor has no value for the intended operation.
There is no way to calculate the Starting Current or Locked Rotor Current (LRA) without more information!
Single-phase or three-phase? NEMA Motor Design B, C or D?
What does academic education's sake mean? A voltage of 15V with a power of 132kW is meaningless for an induction motor. You just can't make up numbers. You are also using \$P = V\ I\$, which is DC power.
You'd be better off looking up a motor nameplate and going from there.
Take a 150hp, 1789rpm, 460V, Design B, Code G, 3-phase induction motor. So rated current is 163A, with a power factor of 0.897 lagging and an efficiency of 96.2%.
Code G gives you locked rotor kVA on a per hp basis. Locked rotor kVA will allow you to calculate LRA. Code G = 5.6 up to but not including 6.3. Worst case = 6.3.
$$150hp \times 6.3 = 945 kVA$$
$$ S = \sqrt {3}\ V_{Line}\ I_{Line} $$ $$ I_{Line} = \frac {S} {\sqrt {3}\ V_{Line}} = \frac {945 kVA} {\sqrt {3} \times 460V} = 1,186A $$
LRA will be between 1,102A and < 1,186A vs 163A or 676% to 728% of full-load current.
Best Answer
This delay is due to inertia, not to friction.
A spinning motor has rotational kinetic energy, which is equal to \$0.5\omega^2 I\$, where \$I\$ is the moment of inertia of the motor, and \$\omega\$ is its angular speed.
To make a motor spin, you have to supply this energy.
The rate at which you can supply energy is limited. All motors and power sources have some limit to the power, that is the rate of change of energy, that they can produce. This means that for any given motor and power source, there's a lower limit to the time it takes to spin up to a given speed.
If you can increase the rate at which you supply energy to the motor, by using a pull-cord wrapped round the shaft for instance, then you can reduce the time it takes for the motor to get up to speed.
There may be other more convenient ways to increase the power. If the limitation is the power supply, then using an energy store like a capacitor to temporarily deliver more power can help. If the limitation is the continuous thermal rating of the motor, then it's usually OK to overload the motor briefly at startup and rely on its thermal mass to limit the temperature rise. If the limitation is the maximum rated current (or the stall current at the rated voltage) of the motor, then it's a bad idea to supply more current, as it could risk demagnetising the field magnets.