Air separation

In air separation the following machines are usually operated, for which we can configure our starters:

Turbo compressors

The moment of inertia of a turbo compressor normally ranges between 2 to 6 times the moment of inertia of the driving motor. Most of the compressor drives start with a starting time between app. 20 to 50 seconds. The torque speed curves (counter torque) of nearly all turbo compressors have a quadratic shape. They can be started with reduced load by regulating vanes and/or blades. By this a lower counter torque is reached which can further reduce the start current. Normal start currents for turbo compressors (Ia/In) are in the range of app. 1,5 - 2,5 times the motor rated current (In). Below you will find a characteristic example for a calculation. When starting a turbo compressor one has to observe that the critical speed will be run through fast enough to avoid mechanical vibration problems.

 motor rated power Pn 2.000 kW rated voltage Un 6.000 V motor rated current In 250 A motor start current DOL Ia 5 × In motor rated speed (synchronous) nn 1.500 rpm moments of inertia motor JM 80 kgm2 compressor JL 300 kgm2 start voltage Us 0,57 × Un start time ts 30 s

Torque

The starter starts the motor with reduced voltage, 57 % of the rated voltage, in this example (Us/Un = 0,57). The torque speed curve of the motor will be reduced by the factor Ts/TDOL = (Us/Un × F)2.
Ts/TDOL = (0,57 × 0,91)2 = 0,27 (F is a factor depending 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.

Start time

The start time ts is calculated from the acceleration torque Ta and the total moment of inertia J = JM + JL = 80 + 300 kgm2 = 380 kgm2.

Current

The motor current is IMot = Us/Un × IDOL × F = 0,57 × 5 × 0,91 = 2,6 × 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= 0,57 × 2,6 = 1,5 × 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 = 2,6 × In

For that reason these starting methods create a higher mains voltage drop.

Screw compressors

Amongst others screw compressors are applied as refrigeration compressors for air condition (chiller). The moment of inertia of a screw compressor is significantly lower that the moment of inertia of the driving motor. Most of the screw compressors will start with start times between app. 10 to 20 seconds. They can be started with reduced load. This results in a lower counter torque which can further reduce the start current. Normal start currents for screw compressors (Ia/In) range from app. 1 -3 times the motor rated current (In). Below you will find a characteristic example for a calculation.

 motor rated power Pn 2.000 kW rated voltage Un 6.000 V motor rated current In 250 A motor start current DOL Ia 5 × In motor rated speed (synchronous) nn 3.000 rpm moments of inertia motor JM 45 kgm2 compressor JL 8 kgm2 start voltage Us 0,63 × Un start time ts 17 s

Torque

The starter starts the motor with reduced voltage, 63 % of the rated voltage, in this example (Us/Un = 0,63). The torque speed curve of the motor will be reduced by the factor Ts/TDOL = (Us/Un × F)2.
Ts/TDOL = (0,63 × 0,91)2 = 0,33 (F is a factor depending 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.

Start time

The start time ts is calculated from the acceleration torque Ta and the total moment of inertia J = JM + JL = 45 + 8 kgm2 = 53 kgm2.

Current

The motor current is IMot = Us/Un × IDOL × F = 0,63 × 5 × 0,91 = 2,9 × 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= 0,63 × 2,9 = 1,8 × 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 = 2,9 × In

For that reason these starting methods create a higher mains voltage drop.

Other machines

Machines driven by electric motors exist in a variety of different designs and types. For the most common ones we have prepared detailled information for the start procedure. Needless to say, our starter can also be used for further types of machines. We will gladly provide you with an offer and technical support.