# 3ph Star Delta transition time

• posted

I am looking to use a 5.5kw 3 ph 380v motor to power a bow thruster with 12 inch diameter kaplan prop. If I use a star delta starter to reduce the initial start up current, how long should the transition between the star and delta connections be?? I have no idea whether it is parts of a sceond or many seconds. TVMIA

• posted

The transition is done at a particular RPM. The time for a motor to accelerate to this point depends on the load characteristics such as torque vs speed and inertia.

Props tend not to be terribly high inertia devices, so they will come up to the speed relatively quickly. I'm not familiar with the above equipment specs, but I would think that the switch over would occur within a second or two at the most.

• posted

On Sun, 19 Nov 2006 13:47:46 -0800 Paul Hovnanian P.E. wrote: | snipped-for-privacy@hotmail.com wrote: |> |> I am looking to use a 5.5kw 3 ph 380v motor to power a bow thruster |> with 12 inch diameter kaplan prop. If I use a star delta starter to |> reduce the initial start up current, how long should the transition |> between the star and delta connections be?? I have no idea whether it |> is parts of a sceond or many seconds. |> TVMIA | | The transition is done at a particular RPM. The time for a motor to | accelerate to this point depends on the load characteristics such as | torque vs speed and inertia. | | Props tend not to be terribly high inertia devices, so they will come up | to the speed relatively quickly. I'm not familiar with the above | equipment specs, but I would think that the switch over would occur | within a second or two at the most.

And even have a lower transition to running RPM ratio?

• posted

The answer to your question is in the question itself. Set the starter to the fastest time that satisfies the requirement for reduced starting current! Whether the motor is turning or not is irrelevant. Bear in mind that the available starting torque on the wye connection will be about 1/3 of maximum, so expecting the motor to achieve anything near full speed is pointless.

Is there a specific specification on permissible starting current, or on starting current steps, or is this just sort of a "feel-good" design?

It is important to understand that all these reduced-voltage starting schemes are for the benefit of the power supply and possibly the driven load. The motor heating is unaffected. The only real benefit to the motor might be reduced mechanical stress, but the vast majority of induction motors are designed for full-voltage starting anyway. The motor will be happiest when it sees full voltage as quickly as possible.

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