# Marine (onshore / offshore)

Offshore power supplies are isolated networks. If an electric motor is started within an isolated network, the feeding diesel generators have to provide the high start power of the electric motor for a short time. The diesel generators have to be designed for this temporary high load. Our starter reduces the start current (and therefore the start power) of the motor which in turn reduces the load of the isolated network . The voltage drop of the power supply on board will be minimized, when the motor is started.

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

## Bow and stern thrusters

Bow and stern thrusters ususally have low moments of inertia; they are normally slightly smaller than the moment of inertia of the driving motors. Therefore most thrusters start with a relatively short start time. Their average start times range from app. 5 to 10 seconds. The torque speed curves of nearly all thrusters have a quadratic shape. Many thrusters can be started with reduced load by adjusting the pitch control to zero pitch. This results in a very low counter torque of the thruster therewith extensively reducing the start current. The normal start currents of thrusters (I_{a}/I_{n}) are in the range of 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} |
2.000 | kW | ||||

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

motor rated current | I_{n} |
250 | 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} |
150 | kgm^{2} |
||||

thruster | J_{L} |
70 | kgm^{2} |
||||

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

start time | t_{s} |
9 | 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} = 150 + 70 kgm^{2} = 220 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.

## Main propulsion of vessels

The main propulsion of a vessel normally has a low moment of inertia; in general, it is lower than the moment of inertia of the driving motor. Therefore most main propulsions start with a short start time. Usual start times range between app. 5 to 10 seconds. The torque speed curves of nearly all vessel propulsions have a quadratic shape (counter torque). Only main propultion systems, which can minimize the counter torque by pitch control, gain an effective reduction of the start current when using a starter. In case that pitch control is possible, normal start currents for main propulsions (I_{a}/I_{n}) are in the range of 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} |
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} |
1.000 | rpm | ||||

moments of inertia | |||||||

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

propeller | J_{L} |
140 | kgm^{2} |
||||

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

start time | t_{s} |
9 | 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} = 300 + 140 kgm^{2} = 440 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.

## Rotating converters

Rotating converters are a combination of an electrical motor and a generator with their shafts tightly coupled. Both machines run with a speed of 600 rpm; the motor is a 10 pole machine, the generator a 12 pole machine (or vice versa). The converter is aiming to change the net frequency from 50 to 60 Hz (or vice versa). The converter can for instance be used to provide power supply to foreign vessels in a harbour. For the start calculation the moment of inertia of both machines must be taken into account. The start time of rotating converter normally ranges between app. 10 - 20 seconds. They can be started nearly without load, because the generator will only be loaded after the start; only a very low counter torque due to friction has to be considered which greatly reduces the start current. Normal start current for converters (I_{a}/I_{n}) are in the range of app. 1 - 1,5 times the motor rated current (I_{n}). Below you will find a characteristic example for 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} |
600 | rpm | ||||

moments of inertia | |||||||

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

rotating converter | J_{L} |
500 | kgm^{2} |
||||

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

start time | t_{s} |
15 | 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} = 450 + 500 kgm^{2} = 950 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.

## Centrifugal pumps

Normally centrifugal pumps do not have an especially high moment of inertia. It is usually located in the range of app. 20-50% of the moment of inertia of the driving electric motor. Therefore, pump drives can be started with a relatively short start time of 5-15 seconds. The torque speed curves of nearly all centrifugal pumps have a quadratic shape (counter torque). The pump should be started with closed discharge valve, which reduces the power consumption of the pump during the start. The counter torque of the pump will be minmized which results in a further reduction of the start current. Start currents (I_{a}/I_{n}) during the start of centrifugal pumps are ususally in the range of app. 1,5-3 times of the motor start current at direct on line (I_{DOL}/I_{n}).

In the following chart a typical pump characteristics is shown:

The corresponding power consumption looks like this:

Therefore the power consumption of the pump is 30 % of its maximum power consumption, if the pump is started with a closed discharge valve. (flow rate Q=0). In the following torque chart the torque speed curve of the pump has its usual quadratic shape. It ends (speed n/n_{n} = 1) at a torque requirement of 30 % (this corresponds to the power consumption of the pump).

## Start 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 | x I_{n} |
||||

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

moments of inertia | |||||||

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

driven machine | J_{L} |
50 | kgm^{2} |
||||

start voltage | U_{s} |
0,61 | x U_{n} |
||||

start time | t_{s} |
9 | s |

Torque

The starter starts the motor with reduced voltage, 61 % of the rated voltage, in this example (U_{s}/U_{n} = 0,61). 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} = (0,61 × 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 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 + 50 kgm^{2} = 120 kgm^{2}.

## Current

The motor current is I_{Mot} = U_{s}/U_{n} × I_{DOL} × F = 0,61 × 5 × 0,91 = 2,8 × I_{n}

**Mains start current:**

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

I_{s}= 0,61 × 2,8 = 1,7 × 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} = 2,8 × I_{n}

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

## Other pumps

Pumps exist in a variety of different designs and types. For the most common ones (centrifugal and piston pump) we have prepared detailled information for the start procedure. Needless to say, our starter can also be used for further pump types. We will gladly provide you with an offer and technical support.

## 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 (I_{a}/I_{n}) are in the range of app. 1,5 - 2,5 times the motor rated current (I_{n}). 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 | 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} |
||||

compressor | J_{L} |
300 | kgm^{2} |
||||

start voltage | U_{s} |
0,57 | × U_{n} |
||||

start time | t_{s} |
30 | s |

Torque

The starter starts the motor with reduced voltage, 57 % of the rated voltage, in this example (U_{s}/U_{n} = 0,57). 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} = (0,57 × 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 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 + 300 kgm^{2} = 380 kgm^{2}.

## Current

The motor current is I_{Mot} = U_{s}/U_{n} × I_{DOL} × F = 0,57 × 5 × 0,91 = 2,6 × I_{n}

**Mains start current:**

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

I_{s}= 0,57 × 2,6 = 1,5 × 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} = 2,6 × I_{n}

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

## Other compressors

Compressors exist in a variety of different designs and types. For the most common ones (turbo, screw and piston) we have prepared detailled information for the start procedure. Needless to say, our starter can also be used for further compressor types. 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 (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:**

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.

## 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.