Overload relay setting and direction of rotation
CHIEF. LEARN · Guide 4 of 6
Why a star–delta overload relay is set to FLC ÷ √3 = 0.58 × FLC, how a fire pump ends up turning backwards, and how to reverse a star–delta motor safely.
The example throughout is the main fire pump of a platform supply vessel: a 37 kW cage motor on a 440 V, 60 Hz supply, full-load current 61 A, started star–delta from the main switchboard.
- MC
- main contactor
- DC
- delta contactor
- FLC
- full-load current
The overload relay: where it sits, why 35.2 A
MC carries the full line current, because DC takes its supply from MC’s outputs. The overload relay sits after that split, in the leads to U1, V1 and W1. Those leads always carry the winding current. In star that equals the line current; in delta it is the line current divided by √3.
At full load in delta each winding carries 61 ÷ √3 = 35.2 A. So the relay is set to 35.2 A, which is 0.58 × FLC.
Set it to the 61 A on the nameplate and the motor could run 73 % overloaded without a trip. Set it well below 35.2 A and it trips at normal full load.
Trip class 10
A class 10 relay must trip within 10 s at 7.2 times its setting: long enough for a normal start, short enough to save a stalled motor.
Phase order sets the direction
The direction is set by the phase order at U1, V1 and W1. By IEC 60034-8, L1-L2-L3 on U1-V1-W1 turns the motor clockwise, seen from the shaft end.
A correct delta always turns the same way as star: each winding gets its own phase minus a neighbouring one, which only shifts it by 30°. A wrong DC connection does not reverse the motor. It leaves a winding with both ends on one phase. After the changeover the motor growls, the current climbs towards starting current and the overload relay trips within seconds.
A fire pump turning backwards
What reverses a fire pump is the phase order swapped on both motor cables, or upstream: after a motor overhaul, a cable repair or work on the switchboard. It then runs backwards in star and in delta. A centrifugal pump turning backwards still delivers some water, roughly a third of its head, and the ammeter reads low. It is easy to miss.
How to reverse it
To reverse a star–delta motor, never swap only two leads of one cable: a winding would end up with both ends on one phase in delta. Swap the same two phases on both cables, U1↔V1 and U2↔V2, or swap two incoming phases at the starter. Isolate and lock off first, then bump the motor and watch the rotation arrow before you trust the pump.
After a motor overhaul, test the insulation resistance before the first start and check the rotation with the coupling off where you can: on some pumps a screwed-on impeller or impeller nut can work loose when it turns backwards.
Check yourself
Where does the overload relay sit in this starter, and what is it set to?
- In the supply line: 61 A
- After MC, in the winding leads: 35.2 A — correct
- After DC: 61 A
The leads from MC to U1, V1 and W1 carry winding current in both star and delta. At full load in delta that is 61 ÷ √3 = 35.2 A.
After work on the feeder cable, the fire pump gives about a third of its usual pressure and the ammeter reads low. Most likely?
- The timer is set too short
- The overload relay is set too low
- Two phases swapped: the pump turns backwards — correct
A centrifugal pump turning backwards still pumps a little, so it is easy to miss. Isolate, swap U1↔V1 and U2↔V2 together on both cables, and check the rotation arrow.
Sources
- IEC 60947-4-1, contactors and motor-starters (IEC webstore)
- IEC 60034-8, terminal markings and direction of rotation (IEC webstore)
For training only. Models are simplified; class rules and IMO texts are the authority.