Start›Ing will prepare a start calculation for each starter. Even in the offer stage we gladly provide this calculation which is an integral part of our service and naturally rendered free of charge. Below you can find a typical example for a start calculation.
|motor rated power||Pn||2.000||kW|
|motor rated current||In||250||A|
|motor start current DOL||Ia||5||× In|
|motor rated speed (synchronous)||nn||1.500||rpm|
|moments of inertia|
|start voltage||Us||0,61||× Un|
The starter starts the motor with reduced voltage, 61 % of the rated voltage, in this example (Us/Un = 0,61). The torque speed curve of the motor will be reduced by the factor Ts/TDOL = (Us/Un × F)2.
Ts/TDOL = (0,61 × 0,91)2 = 0,31 (F is a factor dependent on the motor). The torque of the motor must always be higher than the counter torque. The start voltage Us/Un has been determined accordingly. The difference between the reduced motor torque Ts and the counter torque is the acceleration torque Ta.
The start time ts is calculated from the acceleration torque Ta and the total moment of inertia J = JM + JL = 80 + 50 kgm2 = 120 kgm2.
The motor current is IMot = Us/Un × IDOL × F = 0,61 × 5 × 0,91 = 2,8 × In
Mains start current:
A transformer has the same power at primary and secondary, which leads to the fact that the products of current and voltage are the same.
Un × Is = Us/Un × Imot => Is = Us/Un × IMot / Un
Is= 0,61 × 2,8 = 1,7 × In
The mains start current is therefore much lower than the motor start current.
For other start methods (electronic soft starter, starting reactor etc.) the following is valid: The mains current equals the motor current:
Is = IMot = 2,8 × In
For that reason these starting methods create a higher mains voltage drop.
If we know the configuration and the technical data of your feeding MV grid, we will prepare a network analysis which shows as a result the voltage drop during the start.