Starting kVA and frequency for water pump

[Electricalsmall]

Hello,
I am not an expert in motor control but I have a simple question to be clarified. I need to determine the starting kVA and frequency for water pump which is rated at 110kW with auto-transformer starter. However, I have no more information about this like the setting of voltage tap (say, 50%, 65% or 80%), torque or starting time. How can I determine the starting kVA and frequency? I just want to know a rough figure for multiplying the full load kVA to find out the starting kVA. If I estimate this as follows:
Full load kVA x 4 = starting kVA by assuming at 80% voltage tap.
Is it a safe estimation?
Please give me some idea.

On the other hand, I also want to more about the auto-transformer starter and soft starter as I am not very familiar about them. Please give me some idea on them and their application

 

[electricpete]

In the US, the nameplate code letter tells you starting kva per horsepower at normal voltage. Reduce this starting kva by the square of the factor that voltage is reduced below normal (starting motor acts like constant-impedance.... 80% voltage gives 64% kva.)

 

[JBD]

From Square D publication #8600PD9201:
"With autotransformer starting, the line current is always less than the motor current during starting by an amount equal to the transformation ratio. For example, when a motor is started on the 65% taps, motor current is 65% of line voltage starting values, while the line current is only 65% of 65% (42%) of line voltage starting values. The difference between line and motor current is due to the transformer in the circuit."

Also the starting torque is directly related to the line current so a 65% tap will yield: 65% motor voltage, 42% line current, 65% motor current, and 42% motor starting torque.

NEMA rating for starting frequency are:
max. 200HP Above 200HP
On 15 sec 30 sec
Off 3 min. 45 sec 30 sec
Repeat 14 times (total of 15) 2 times (total of 3)
Rest 2 hours 1 hour
Tap setting 65% 65%

 

[TheDOG]

The figures that JBD uses do not take into account the losses for the transformer. These are usually anywhere from 5% to 20% and must be added to the line current.
The starting kVA is very difficult to determine because of motor power factor during start - as a rule of thumb, 4 x running kVA is a good estimate.
The difference between soft starters and Autotransformers are many and varied. A soft starter uses back to back SCR's, and is a "straight through device". This means that motor current will be equal to line current, unlike the Autotransformer (Va in = Va out). A centrifigul pump will typically require around 3.5 x FLC for 15 seconds (approx 65% tap) to get to full speed. This is dependant on motor LRC, LRT and motor load.
The advantage the soft starter has over the Autotransformer is the flexibility to change the motor current during start (motor torque), by simple programming. An Autotransformer starter is limited by the tappings available (usually 50%, 65% and 80%).
On the other hand, the Autotransformer will allow a lower line current for torque produced at the motor, provided the motor can get to speed with the reduced voltage.

 

[cbarn24050]

Hi, an electronic soft starter wil enable you to control the starting current, most pumps need between 2.5 to 4 times flc.A soft starter will also allow a soft stop option which can be usefull in some pumping situations.

 

[Marke]

All motors are different and have different starting characteristics depending on their design. The Locked rotor current of the motor can vary between 550% and 900% of the rated current of the motor and the Locked rotor torque can vary from 60% to 300% of the rated full load motor torque.
To start a machine, the motor must deliver enough torque for the motor to accellerate the load to full speed. i.e. minimum torque requirement is a function of the load.
The motor acts as a transducer converting electrical energy into mechanical energy, i.e. amps into newton meters.
Under full voltage starting conditions, a given motor will always produce the same torque and draw the same current, independant of driven load, until it reaches full speed.
Reducing the voltage applied to the motor during start, reduces the starting current, and also reduces the start torque. The start current reduces by the voltage reduction while the torque reduces as the voltage reduction squared.
The minimum start current is that required by the motor used to generate enough torque to start the load. Because of the wide variation in motor characteristics, the start current or start KVA for a given load can vary dramatically, i.e. by a greater that 2 : 1 ratio!
You could look at my web site at

http://www.lmphotonics.com

for more info on this. Also download the software Electrical calculations to try different combinations of motors, loads and starters to see the effects.
Mark Empson
Mark Empson

http://www.lmphotonics.com