Motor starting and generator voltage dip

What the generator sees

A generator cannot follow a sudden load instantly. The voltage drops first, and the automatic voltage regulator pulls it back within about a second. A simple estimate treats the generator as a voltage behind its transient reactance x and the starting motor as an impedance. With k the starting current divided by the generator’s rated current, the dip is x·k ÷ (1 + x·k).

For our 625 kVA main generator x = 0.18 and k = 360 A ÷ 820 A = 0.44, so x·k = 0.08 and a direct-on-line start dips the MSB by about 7 %. In star it is under 3 %. Once the AVR has caught up, the running pump leaves the MSB at 440 V.

The limits

Two different limits apply. Consumers on board must ride through transient dips of up to 20 % that recover within 1.5 s (IEC 60092-101, IACS UR E5). The generator has its own test: running at no load and switched suddenly onto 60 % of its rated current at a power factor of 0.4 or less, its voltage must not fall below 85 % and must be back within ±3 % in 1.5 s (IACS UR E13). This motor’s direct-on-line start, about 45 % of rated current, is inside that test.

Contactors are only required to close reliably between 85 % and 110 % of their coil voltage (IEC 60947-4-1). A deep dip makes contactors elsewhere chatter or drop out, lights flicker and running motors slow down and draw more current, which deepens the dip.

A smaller generator

The same start on a smaller generator is a different story. A 200 kVA set has a rated current of 262 A, so k = 1.4. With the same x, a direct-on-line start dips the voltage by about 20 %: right at the ride-through limit and below the 85 % contactors need. In star it stays under 10 %.

Check yourself