 # Recycling

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

## Shredders

Shredders usually feature controllable hydraulic couplings between the electric motor and the mill. Therefore, the characteristics of the shredder does not influence the start process. We only need to consider the moment of inertia of the motor and the very small moment of inertia of the coupling half at the motor side for the start calculation. This results in a very short start time of app. 5-15 seconds. The start will be carried out almost without load.The coupling generates only a very small counter torque. Therefore the start current can be greatly reduced. Typical start currents (Ia/In) for the start of a shredder range at app. 1 - 1,5 times of the motor start current at direct on line (IDOL/In). Below you will find a characteristic example of 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 1.500 rpm moments of inertia motor JM 80 kgm2 coupling JL 5 kgm2 start voltage Us 0,47 × Un start time ts 8 s ## Torque The starter starts the motor with reduced voltage, 47 % of the rated voltage, in this example (Us/Un = 0,47). The torque speed curve of the motor will be reduced by the factor Ts/TDOL = (Us/Un × F)2.
Ts/TDOL = (0,47 × 0,9)2 = 0,18 (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 + 5 kgm2 = 85 kgm2. ## Current The motor current is IMot = Us/Un × IDOL × F = 0,47 × 5 × 0,9 = 2,1 × 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= 0,47 × 2,1 = 1 × 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 = 2,1 × In

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

## Mills

Mills exist in a variety of different designs and types. We have prepared detailled information for the start procedure of a shredder. Needless to say, our starter can also be used for further mill types, if the start torque requirement of the mill is not too high. High moments of inertia and therefore long start times do not pose a problem for our starter. If a controlable (f. ex. hydraulic) coupling is mounted between mill and motor, the mill can be started with a very low start current (Ia/In) in the range of app. 1-2 times of the motor start current at direct on line (IDOL/In). We will gladly provide you with an offer and technical support.

## Controlable couplings (f.ex. hydraulic)

If a controllable coupling (f.ex. hydraulic) is located between the electrical motor and the driven load, the characteristic of the load does not influence the start process at all. We only need to consider the moment of inertia of the motor and the very small moment of inertia of the coupling half at the motor side for the start calculation. This results in a short start time of app. 5-15 seconds. The start will be carried out almost without load. The coupling generates only a very small counter torque. Therefore the start current can be greatly reduced. Typical start currents (Ia/In) for the start of controllable coupled drives are in the range of app. 1-1,5 times of the motor rated current (In). Below you will find a characteristic example of 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 1.500 rpm moments of inertia motor JM 80 kgm2 coupling JL 5 kgm2 start voltage Us 0,47 × Un start time ts 8 s ## Torque The starter starts the motor with reduced voltage, 47 % of the rated voltage, in this example (Us/Un = 0,47). The torque speed curve of the motor will be reduced by the factor Ts/TDOL = (Us/Un × F)2.
Ts/TDOL = (0,47 × 0,9)2 = 0,18 (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 + 5 kgm2 = 85 kgm2. ## Current The motor current is IMot = Us/Un × IDOL × F = 0,47 × 5 × 0,9 = 2,1 × 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= 0,47 × 2,1 = 1 × 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 = 2,1 × 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. 