How a star–delta starter works

One motor, two ways to connect it

Inside the motor are three identical windings: U, V and W. Each has a start, marked 1, and an end, marked 2. All six ends come out to the terminal box.

Every winding is built for one voltage. On this motor it is 440 V. The links in the terminal box decide how much voltage each winding actually gets.

The nameplate says Δ 440 V / Y 760 V: delta on a 440 V supply, star on a 760 V supply. Same motor, same windings, only the links change.

Why a motor draws so much current to start

A stopped induction motor behaves like a transformer with its secondary short-circuited. The rotor bars are that shorted secondary. The moment the supply comes on, the rotating field sweeps past the bars at full speed and drives a huge current through them, and the stator has to supply it.

Little limits that current except the leakage reactance of the windings. That is why a cage motor started direct-on-line draws six to seven times its full-load current: about 360 A for this motor instead of 61 A.

As the rotor speeds up, the field slips past the bars more slowly, the rotor current falls, and the stator current falls with it. Near full speed only the working current is left. A pump is light at low speed, so on this one that takes under a second.

Most of the starting current is reactive: at standstill the power factor is only about 0.3. A generator can deliver that only by letting its voltage sag for a moment. That sag is the real problem on a ship.

Star: 254 V per winding

In star the ends U2, V2 and W2 are joined into one star point. Each winding sits between one phase and that point.

The voltage across it is the line voltage divided by √3: 440 ÷ 1.73 = 254 V. The current in each supply line is the same as the current in its winding.

Delta: 440 V per winding

In delta each winding sits between two phases and gets the full 440 V. This is how this motor is meant to run on the ship’s supply.

Two windings share every supply line, so the line current is √3 times the winding current. At full load that is 61 A in the cable and 35.2 A in each winding.

Star at the start: a third of the current

Started direct-on-line in delta, this motor draws about 360 A from the MSB.

In star each winding gets 1/√3 of the voltage, so its current falls by √3. And in star the line current equals the winding current, while in delta it is √3 times bigger. √3 × √3 = 3: a star start draws a third of the current.

The generator sees a third of the surge, and the voltage dip on the MSB falls with it.

The price: a third of the torque

Torque grows with the square of the voltage. 1/√3 of the voltage gives (1/√3)² = 1/3 of the torque, at every speed.

A centrifugal pump needs almost no torque at standstill: its load torque rises with the square of the speed. So a third of the motor torque is still enough to run it up. Star torque and pump load meet at about 94 % speed, and that is as fast as star will take it.

A loaded reciprocating compressor, a winch with a load on the hook or a positive-displacement pump such as a screw pump starting against full discharge pressure needs high torque from standstill. In star it stalls or hangs at low speed, draws heavy current, and trips on overload.

Check yourself