Volt Drop Calculation with Auto transformer

[magneticted]

Calculating volt drop at say 6Kv bus, with 126MVA, and a motor rating of 7.7Mw 6Kv FLC 832A and starting current of 4401 a ( Is/In 5,29) and volt drop on 6kV net not to exceed 15% then on DOL we get 27% drop.

So if we introduce a auto transformer with a rating of 9MVA, impedance of 6% and say 70% tap, do we calculate the volt drop with the motor Is / In of 5.29, or do we calculate with Is / In 2.6 ( 0.7 x O.7 x 5.29 )

 

[LionelHutz]

use 3.7 (0.7 x 5.29)

The voltage is reduced to 70% so the current is also reduced to 70%.

The torque the motor produces drops by 0.7 x 0.7

In reality, the starting voltage drop will lower the current even further but it's a safe first approximation to ignore this effect.

Do make sure the motor still produces enough starting torque with the lower voltage though. If you use the 70% tap and drop the source voltage 15% then you will be applying 55% of rated voltage to the motor. This means the motor produces 0.55 x 0.55 = 0.3 = 30% of it's full-voltage starting torque. Lowering the starting torque that much means the motor may stall before it reaches full speed.

 

[GGOSS]

The current drawn form the supply will be 0.7 x 0.7 x LRC plus a little extra to cover the losses. The current delivered to the motor will be 0.7 x LRC.

The current drawn from the supply is what you need to woprk with.

 

[LionelHutz]

GGOSS is right about the line current. I went and gave the motor current.