# Wood / Paper

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

## Hackers

Hackers are used to shred wood in the wood/paper industry. They are applied for example in the chipboard, pelletizing, pulp and paper industry. A hackers produces chips from wood (f.ex. logs) using a rotor assembled with knifes. Hackers have a high moment of inertia due to their massive execution. It is multiple times higher than the moment of inertia of the driving motor. Therefore hacker drives start with a very long starting time between app. 50 and 120 seconds. Hackers are started in empty condition (no logs in the machine). This results in a very low counter torque and therefore in an extensive reduction of the start current. Normal start currents for hackers (I_{a}/I_{n}) are in the range app. 1 - 2 times the motor rated current (I_{n}). Below you will find a characteristic example for a calculation.

motor rated power | P_{n} |
1.500 | kW | ||||

rated voltage | U_{n} |
6.000 | V | ||||

motor rated current | I_{n} |
190 | A | ||||

motor start current DOL | I_{a} |
5 | x I_{n} |
||||

motor rated speed (synchronous) | n_{n} |
1.000 | rpm | ||||

moments of inertia | |||||||

motor | J_{M} |
110 | kgm^{2} |
||||

hacker | J_{L} |
1.500 | kgm^{2} |
||||

start voltage | U_{s} |
0,47 | x U_{n} |
||||

start time | t_{s} |
100 | s |

Torque

The starter starts the motor with reduced voltage, 47% of the rated voltage, in this example (U_{s}/U_{n} = 0,47). The torque speed curve of the motor will be reduced by the factor T_{s}/T_{DOL} = (U_{s}/U_{n} × F)^{2}.

T_{s}/T_{DOL} = (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 U_{s}/U_{n} has been determined accordingly. The difference between the reduced motor torque T_{s} and the counter torque is the acceleration torque T_{a}.

## Start time

The start time t_{s} is calculated from the acceleration torque T_{a} and the total moment of inertia J = J_{M} + J_{L} = 110 + 1.500 kgm^{2} = 1.610 kgm^{2}.

## Current

The motor current is I_{Mot} = U_{s}/U_{n} × I_{DOL} × F = 0,47 × 5 × 0,9 = 2,1 × I_{n}

**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.

U_{n} × I_{s} = U_{s}/U_{n} × I_{mot} => I_{s} = U_{s}/U_{n} × I_{Mot} / U_{n}

I_{s}= 0,47 × 2,1 = 1 × I_{n}

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:

I_{s} = I_{Mot} = 2,1 × I_{n}

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

## Refiner

Refiners are used to shred (grind) fibers (f.ex. pulp or waste paper) in the wood and paper industry. Refiners normally have a relatively high moment of inertia, which is usually higher than the moment of inertia of the driving motor. Most of the refiners start with a starting time of app. 30 to 50 seconds. They are started in empty condition (no fibers in the refiner). This results in a very low counter torque of the refiner, therefore an extensive reduction of the start current is possible. Normal start currents for refiners (I_{a}/I_{n}) are in the range of app. 1 - 2,5 times the motor rated current (I_{n}). Below you will find a characteristic example for a calculation.

motor rated power | P_{n} |
4.000 | kW | ||||

rated voltage | U_{n} |
6.000 | V | ||||

motor rated current | I_{n} |
500 | A | ||||

motor start current DOL | I_{a} |
5 | × I_{n} |
||||

motor rated speed (synchronous) | n_{n} |
1.500 | rpm | ||||

moments of inertia | |||||||

motor | J_{M} |
200 | kgm^{2} |
||||

refiner | J_{L} |
760 | kgm^{2} |
||||

start voltage | U_{s} |
0,5 | × U_{n} |
||||

start time | t_{s} |
40 | s |

Torque

The starter starts the motor with reduced voltage, 50 % of the rated voltage, in this example (U_{s}/U_{n} = 0,5). The torque speed curve of the motor will be reduced by the factor T_{s}/T_{DOL} = (U_{s}/U_{n} × F)^{2}.

T_{s}/T_{DOL} = (0,5 × 0,9)^{2} = 0,2 (F is a factor depending on the motor). The torque of the motor must always be higher than the counter torque. The start voltage U_{s}/U_{n} has been determined accordingly. The difference between the reduced motor torque T_{s} and the counter torque is the acceleration torque T_{a}.

## Start time

The start time t_{s} is calculated from the acceleration torque T_{a} and the total moment of inertia J = J_{M} + J_{L} = 200 + 760 kgm^{2} = 960 kgm^{2}.

## Current

The motor current is I_{Mot} = U_{s}/U_{n} × I_{DOL} × F = 0,5 × 5 × 0,9 = 2,25 × I_{n}

**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.

U_{n} × I_{s} = U_{s}/U_{n} × I_{mot} => I_{s} = U_{s}/U_{n} × I_{Mot} / U_{n}

I_{s}= 0,5 × 2,25 = 1,1 × I_{n}

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:

I_{s} = I_{Mot} = 2,25 × I_{n}

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

## 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 (I_{a}/I_{n}) for the start of controllable coupled drives are in the range of app. 1-1,5 times of the motor rated current (I_{n}). Below you will find a characteristic example of a calculation:

motor rated power | P_{n} |
2.000 | kW | ||||

rated voltage | U_{n} |
6.000 | V | ||||

motor rated current | I_{n} |
250 | A | ||||

motor start current DOL | I_{a} |
5 | × I_{n} |
||||

motor rated speed (synchronous) | n_{n} |
1.500 | rpm | ||||

moments of inertia | |||||||

motor | J_{M} |
80 | kgm^{2} |
||||

coupling | J_{L} |
5 | kgm^{2} |
||||

start voltage | U_{s} |
0,47 | × U_{n} |
||||

start time | t_{s} |
8 | s |

Torque

The starter starts the motor with reduced voltage, 47 % of the rated voltage, in this example (U_{s}/U_{n} = 0,47). The torque speed curve of the motor will be reduced by the factor T_{s}/T_{DOL} = (U_{s}/U_{n} × F)^{2}.

T_{s}/T_{DOL} = (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 U_{s}/U_{n} has been determined accordingly. The difference between the reduced motor torque T_{s} and the counter torque is the acceleration torque T_{a}.

## Start time

The start time t_{s} is calculated from the acceleration torque T_{a} and the total moment of inertia J = J_{M} + J_{L} = 80 + 5 kgm^{2} = 85 kgm^{2}.

## Current

The motor current is I_{Mot} = U_{s}/U_{n} × I_{DOL} × F = 0,47 × 5 × 0,9 = 2,1 × I_{n}

**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.

U_{n} × I_{s} = U_{s}/U_{n} × I_{mot} => I_{s} = U_{s}/U_{n} × I_{Mot} / U_{n}

I_{s}= 0,47 × 2,1 = 1 × I_{n}

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:

I_{s} = I_{Mot} = 2,1 × I_{n}

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 (I_{a}/I_{n}) in the range of app. 1-2 times of the motor start current at direct on line (I_{DOL}/I_{n}). We will gladly provide you with an offer and technical support.

## 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.

## Any questions?

Do you need support to start your drives? We are here to help. Please contact us.