# Water / Waste water

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

## 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:**

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}= 0,61 × 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.

## Piston pumps

Normally piston pumps do not have an especially high moment of inertia with reference to the motor speed. It is usually located in the range of app. 10 - 40% of the moment of inertia of the driving motor. Therefore, piston pump drives can be started with a relatively shot start time of app. 5-15 seconds. Piston pumps can be started nearly without load by pressure release. This is achieved by opening valves and results in a very low counter torque which greatly reduces the start current of the piston pump. Start currents (I_{a}/I_{n}) during the start of piston pumps are ususally in the range of app. 1 - 2 times of 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} |
||||

piston pump | J_{L} |
35 | kgm^{2} |
||||

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

start time | t_{s} |
9 | 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} = 150 + 35 kgm^{2} = 185 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.

## 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.

## Fans / Blowers

Fans (sometimes also called blowers) usually have a very high moment of inertia; it is often many times higher than the moment of inertia of the driving motor. Therefore most of the fan drives start with a very long start time of up to app. 100 seconds. The torque speed curves (counter torque) of nearly all fans have a quadratic shape. Most fans can be started with reduced load by adjusting the blades. This adjustment provides a lower counter torque which results in a lower start current. Normal start currents for fans (I_{a}/I_{n}) are in the range of app. 1,5 - 3 times the motor rated current (I_{n}). Below you will find a characteristic example for a calculation. Due to the possibly high moment of inertia of the fan and the resulting very long start time the moment of inertia J [kgm^{2}] (which can be estimated in the first step) should be present for an offer.

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} |
||||

fan | J_{L} |
2500 | kgm^{2} |
||||

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

start time | t_{s} |
80 | 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 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 + 2.500 kgm^{2} = 2.650 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:**

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}= 0,61 × 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.

## 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.