I would like to ask what the rated current is for a 150kw, 380-volt three-phase motor? Are the rated currents different for triangular and star connections?

I hadn’t known it after working for 10 years; it was only when my apprentice asked me that I remembered to seek help from everyone. Thank you for your advice

The rated current is equal to the motor power divided by (sqrt(3) * rated voltage * power factor); typically, the rated current refers to the line current. Relationship between line current and phase current: In delta connection, the line current is equal to sqrt(3) * phase current ； Star connection: The line current is equal to the phase current.

:lol Not bad, not bad; I’ll learn it too*

Rough calculation: for voltages above 7.5 KV, rated power * 2; for wire selection, rated power * 2.5

If two motors differ only in the winding connection while being otherwise identical, their rated currents are essentially equal.

In other words, the delta connection has a value that is \sqrt{3} times larger than that of the star connection. I used your formula to do the calculations; with a power factor of 0.76, the current is approximately 300 A

For Y-series AC motors, the power factor is approximately 0.84–0.89 for 2-pole motors and approximately 0.76–0.89 for 4-pole motors. I’m not sure whether the motor in use is 4-pole.

The rated current of a 150kw motor generally does not exceed 290A; it depends mainly on the motor model, its specifications, and the number of poles. The star and delta connection methods also result in different values due to the differences in rated voltage used.

Generally, for one kilowatt, there are two currents, and the total value is slightly lower

For Y-series ordinary asynchronous motors, generally the higher the capacity of the motor, the higher its power factor. If the motor does not have a nameplate, an approximate value can be used; for a typical 150KW motor, the power factor can be taken as 0.85. The calculation formula is: Power = 1. 732*Voltage*Current*Power Factor ； Using this calculation, the rated current of this motor is approximately 270 amps. The rated current represents the current supplied to the motor when it is operating at full load under normal conditions. There is no difference in the current supplied in delta or star wiring configurations; it remains the same in both cases

For the same motor, when the stator windings are connected in a Y configuration, the voltage applied to the stator windings is the phase voltage; whereas when the stator windings are connected in a delta configuration, the voltage applied to them is the line voltage. In both star and delta connections, the winding voltage ratio is as shown in http://jpkc.w*t.net.cn/Dgdz/wlkc/N05/N05_03_clip_image006.gif. This results in a corresponding reduction in phase current, line current, and torque when operating in star connection. It is typically used as a no-load voltage-reduction starting method for motors with a power of 3 kW or more, which normally operate in delta connection. Let: the line voltage of the power supply be http://jpkc.w*t.net.cn/Dgdz/wlkc/N05/N05_03_clip_image008.gif, and the impedance of each phase of the motor be ׀ Z ׀. When the stator windings are connected in a star configuration, that is, during reduced-voltage starting: http://jpkc.w*t.net.cn/Dgdz/wlkc/N05/N05_03_clip_image012.gif. When the stator windings are connected in a delta configuration, that is, during operation at full voltage: http://jpkc.w*t.net.cn/Dgdz/wlkc/N05/N05_03_clip_image014.gif. By comparing these two situations, it can be seen that http://jpkc.w*t.net.cn/Dgdz/wlkc/N05/N05_03_clip_image016.gif the current during reduced-voltage starting is 1/3 of the current during direct starting. Since torque is proportional to the square of voltage, the starting torque also decreases to http://jpkc.w*t.net.cn/Dgdz/wlkc/N05/N05_03_clip_image018.gif compared to that during direct starting. 　　Therefore, this method is only suitable for starting under no-load or light-load conditions. Even for motors of the same model, differences such as those in the air gap can result in varying power factors among individual motors. The motor’s operating current can be roughly estimated using the formula power * 2; however, for accuracy, measurements are necessary. This post was last edited by zhx7540 on 2008-4-4 09:20]

The current calculated is: I=150000/1.732*380*0.8=284.89(A)； In the star connection of the motor, the current in each phase winding is: 1.732*284.89=493.43(A) ； When the motor is connected in star configuration, the current in each phase winding of the motor is 284.89(A), which is the rated current.

In the star connection of the motor, the voltage across each phase winding is: 380(V)； When the motor is connected in star configuration, the voltage across each phase winding of the motor is 1/1.732 times the rated voltage; that is, 380/1.732 = 219.4(V) ； Motors of this size are equipped with voltage reduction devices; a reactor is connected in series with the three power leads of the motor, and an autotransformer-based voltage reduction starter is used, with a time relay controlling the switching. I repaired a 250KW starting cabinet – after the switching process, the contactor in the power supply had its contacts for one phase damaged. The motor started normally, but once load was applied, the current rose sharply. The fault was identified using a clamp meter, and after replacing the faulty parts, everything returned to normal.

Whether in delta or star connection, power = square root of 3 * voltage * current * power factor. If the power factor is 0.85, the current is 268A based on this calculation

For a standard three-phase asynchronous motor, the current can be estimated using the rule of \"2 amperes per kilowatt\", meaning 2 amps for each kilowatt of capacity.

In a star connection, the line voltage is equal to the square root of 3 times the phase voltage, and the line current is equal to the phase current. In a delta connection, the phase voltage is equal to the line voltage, while the line current is three times the phase current. The power calculation formula is P=3*U_phase*I*power factor; the power factor is approximately 0.84–0.89 for 2-pole motors and 0.76–0.89 for 4-pole motors. A rough estimation suggests that 1 KW corresponds to 2 amperes of current, which is roughly 300 A, a value slightly higher than the actual rated current.

According to the formula, it should be around 270A. For the same motor at the same voltage, the currents in delta and star connections are different. When connected in a delta configuration, the phase current flowing through the load is 1.732 times that in a star configuration.