Let’s summarize the existing answers first: (Please add if you haven’t said everything)

1. The power of the motor is greater than 10kW;

2. The power of the motor is greater than 50kW;

3. The power of the motor is greater than 20% of the transformer capacity;

4. The voltage drop caused by the motor that is often started is greater than 10% when starting;

5. Occasionally The starting motor causes a voltage drop greater than 15% when starting.

It also seems to be an empirical formula. After substituting into the formula to calculate whether to start the step-down, I hope friends who know the formula can add it.

There seems to be no basis for the two items 1 and 2. In reality, motors with hundreds or hundreds of kilowatts are all directly started with full voltage.

For item 3, is the transformer no-load?

For items 4 and 5, how do you control them in practice? Is it an actual measurement?

In fact, the main purpose of adopting step-down startup is to avoid affecting the normal operation of other equipment. If there is no other equipment, only the transformer and the motor, then the power of the motor can be close to the capacity of the transformer. Considering factors such as power factor and efficiency, the general motor power is 80% of the transformer capacity, and can be started directly, or the transformer and the motor can be started at the same time .

In reality, when most large motors are started, the transformer has more or less other loads, so the amount of other loads is also one of the factors to be considered.

Why does the motor affect the work of other electrical equipment when it starts?

Because the starting current of the cage motor is 5-7 times its rated current, high current will cause a large voltage drop, and when the voltage drop reaches a certain value, other electrical equipment may stop work or cause a malfunction. Therefore, it is stipulated that the voltage drop caused by frequent starting of the motor should not exceed 10%, and the voltage drop caused by infrequent starting should not exceed 15%.

Why is there a voltage drop?

Because the transformer cannot provide the large current required for starting a large motor, if the transformer can provide the large current required for starting a large motor, there will be no voltage drop or generated voltage The drop is very small. This means that the size of the transformer capacity also determines the size of the voltage drop. So in practice, we must not talk about the size of the transformer capacity, but only talk about how big the motor needs to be reduced to start.

The size of the existing load of the transformer is also one of the conditions that determine the size of the voltage drop. Imagine starting the same motor separately when the transformer is no-load and full-load, the resulting voltage drop must be different. Then it is not feasible to judge whether the motor needs to be started with a reduced voltage by only talking about the motor power as a few percent of the transformer capacity without mentioning the current load of the transformer.

So far, to judge whether the motor is started with reduced voltage, at least three factors must be considered: motor power, transformer capacity, and existing load, all of which are indispensable! Then look at the various so-called judgment methods now, have these three factors been considered comprehensively?

How to comprehensively consider these three factors? For convenience, the current parameter is used. In fact, the essence of this question is: how much current? How much voltage drop will it cause? That is the problem of current and voltage drop.

Everyone knows that the starting current of the cage motor is 5-7 times the rated current.

For the transformer, how much load current will cause the voltage drop to reach 10%, 15%? This requires the parameter of impedance voltage. The impedance voltage of a general power transformer is about 5%, that is, when the output current of the transformer is the rated value, the voltage drop is 5%. This is the transformer with an output of 380V. The designed output voltage It's 400V, which happens to be 380V when fully loaded.

Assuming that as the load current increases, the voltage drop of the transformer output voltage changes linearly, then when the output current of the transformer is 2 times the rated current, the voltage drop is 5%, and when the current is 3 times the rated current is 10%, 15% at 4 times the current. Then the allowable voltage drop of the motor that starts frequently is 10%, that is, the starting current of the motor plus the existing load current, as long as it is less than 3 times the rated current of the transformer, it can be started directly.

This is the simplest and closest to the actual method of judging which method the motor uses to start.