# Buildings

In buildings 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 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.

## 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 (Ia/In) are in the range of app. 1,5 - 3  times the motor rated current (In). 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 [kgm2] (which can be estimated in the first step) should be present for an offer.

 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 x In motor rated speed (synchronous) nn 1.000 rpm moments of inertia motor JM 150 kgm2 fan JL 2500 kgm2 start voltage Us 0,61 x Un start time ts 80 s

## Torque

The starter starts the motor with reduced voltage, 61% of the rated voltage, in this example (Us/Un = 0,61). The torque speed curve of the motor will be reduced by the factor Ts/TDOL = (Us/Un × F)2.
Ts/TDOL = (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 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 = 150 + 2.500 kgm2 = 2.650 kgm2.

## Current

The motor current is IMot = Us/Un × IDOL × F = 0,61 × 5 × 0,91 = 2,8 × 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= 0,61 × 2,8 = 1,7 × 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 = 2,8 × In

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 (Ia/In) 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 (IDOL/In).

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/nn = 1) at a torque requirement of 30 % (this corresponds to the power consumption of the pump).

## Start 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 x In motor rated speed (synchronous) nn 1.500 rpm moments of inertia motor JM 80 kgm2 driven machine JL 50 kgm2 start voltage Us 0,61 x Un start time ts 9 s

## Torque

The starter starts the motor with reduced voltage, 61 % of the rated voltage, in this example (Us/Un = 0,61). The torque speed curve of the motor will be reduced by the factor Ts/TDOL = (Us/Un × F)2.
Ts/TDOL = (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 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 + 50 kgm2 = 120 kgm2.

## Current

The motor current is IMot = Us/Un × IDOL × F = 0,61 × 5 × 0,91 = 2,8 × 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= 0,61 × 2,8 = 1,7 × 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 = 2,8 × In

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 (Ia/In) are in the range of app. 1 - 1,5 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 600 rpm moments of inertia motor JM 450 kgm2 rotating converter JL 500 kgm2 start voltage Us 0,47 × Un start time ts 15 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 = 450 + 500 kgm2 = 950 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.