The instantaneous starting current of the 75 kW motor is 6O0A, and Q: Does it have any effect on the

Such a high-power motor is started by step-down. The instantaneous current of 600 amperes does not have much effect on a 200 kVA transformer. It mainly affects other electrical appliances under their own transformers, once the instantaneous voltage drop is too large, it will cause other electrical appliances with low-voltage protection devices to lose power, such as computer signal interruption and loss of data, and the working motor shutdown.

It has an impact on the transformer, which will cause the transformer to overcurrent or overload instantaneously, and the 75 kilowatt motor cannot be started directly, and the step-down start control circuit is required.

There is definitely an impact.

75kw motor do you use direct start? It's not suitable, man, let's make a star triangle.

Generally, the capacity of the transformer is much larger than the capacity of the motor, and 200* is just doubled, which is a bit small.

Your transformer is seriously overloaded at the moment of motor starting, it is recommended that you change it to Star Delta Star-Delta.

That is, add a few contactors.

things. If there are conditions, it is best to start soft or inverter, it is best to start inverter, the impact is relatively small, you can use it with confidence.

The rated current on the low voltage side of the 200kVA transformer is: Allowable 10% overload operation.

The transformer can last up to 120 minutes with an overload of 30%.

Overload 60% transformer can last up to 45 minutes.

75% overload: transformer can run for 20 minutes.

Overload 100% transformer can last for 10 minutes.

Overload 200% transformer can last for minutes.

According to the national standard, the 75kw motor has 600A at the moment of starting, and the transformer can carry it. But the impact on the transformer is too great.

Or consider inverter starting.

Summary. No. If the 50 transformer starts the 55 kilowatt motor, in fact, the maximum load of the transformer has been reached, and if the 75 kilowatt motor is started, the maximum load of the transformer will be exceeded, resulting in overload of the transformer, thermal runaway, and even a short circuit may occur, resulting in equipment damage.

Solution: 1. Replace the transformer with a larger capacity to meet the starting requirements of the 75 kilowatt motor; 2. Adopt the method of segmented starting, that is, start the 55 kilowatt motor first, wait for the motor speed to reach a certain value, and then start the 75 kilowatt motor; 3. Adopt frequency converter speed regulation, that is, when starting the 75 kilowatt motor, the frequency converter speed regulation is used to reduce the current when the motor is started, so as to reduce the load of the transformer. In addition, when using the transformer, you should also pay attention to the capacity, voltage, frequency and other parameters of the transformer to ensure the normal use of the transformer.

No. If you start the 75 kilowatt motor, it will exceed the maximum load of the transformer, which will lead to overload of the transformer, thermal runaway, and may even cause a short circuit, resulting in equipment damage. Workaround:

1. Replace the transformer with a larger capacity to meet the starting requirements of the 75 kilowatt motor; 2. Adopt the method of segmented starting, that is, start the 55 kilowatt motor first, wait for the motor speed to reach a certain value, and then start the 75 kilowatt motor; 3. Adopt frequency converter speed regulation, that is, when starting the 75 kilowatt motor, the frequency converter speed regulation is used to reduce the current when the motor is started, so as to reduce the load of the transformer. In addition, when using the transformer, it is also necessary to pay attention to the capacity, voltage, frequency and other parameters of the transformer mold removal to ensure the normal use of the transformer.

You've done a great job! Can you elaborate on that?

No. If the 50 transformer starts the 55 kilowatt motor, in fact, it has reached the maximum load of the transformer, and if the 75 kilowatt motor is started again, the maximum load of the transformer will be exceeded, which will lead to the overload of the transformer, and the transformer may be burned out. Therefore, when choosing a transformer, the transformer should be selected according to the actual power of the motor to ensure the normal operation of the transformer.

In addition, when installing the transformer, you should pay attention to the ambient temperature of the transformer and the wiring method of the transformer to ensure the normal operation of the transformer. In addition, regular maintenance of the transformer is also very important to effectively extend the service life of the transformer.

Summary. 1. There are many reasons for the imbalance, such as the fault of disconnection, if the wire is broken, there is no grounding or the isolation switch is not connected, it will lead to the asymmetry of the three-phase parameters, and the problem of voltage imbalance will directly occur. 2. There may be a fault in the grounding, and a phase of the line is disconnected, and the problem of one-way grounding will lead to voltage imbalance.

And after grounding, the voltage value does not change. There are two types of unidirectional grounding, one is metallic and the other is non-metallic. If it is non-metallic, there is another reason why the grounding voltage is not the phenomenon of resonance, mainly because the industrial development speed is very fast, and the power load increases rapidly, which will lead to three-phase voltage imbalance and great fluctuations.

At this time, the voltmeter will hit the head or the three-phase voltage will rise in turn, exceeding the line voltage, which is basically caused by resonance.

Hello! This is a three-phase voltage imbalance caused by load imbalance.

The normal range of 380V voltage is 7, and the normal range of 380V voltage balance is 353V 406V. Maximum deviation voltage = 380V; The minimum voltage cannot be lower than 380V-; The maximum voltage of the old barrier cannot be higher than 380V+.

You've reached the maximum dropout voltage limit at 20 volts.

1. There are many reasons for the imbalance, such as the fault of disconnection, if the wire is broken, there is no grounding or the isolation switch is not connected, it will lead to the asymmetry of the three-phase parameters, and the problem of voltage imbalance will directly occur. 2. There may be a fault in the grounding, and a phase of the line is disconnected, and the problem of one-way grounding will lead to voltage imbalance. And after grounding, the voltage value does not change.

There are two types of unidirectional grounding, one is metallic and the other is non-metallic. If it is non-metallic, there is another reason why the grounding voltage is not the phenomenon of resonance, mainly because the industrial development speed is very fast, and the power load increases rapidly, which will lead to three-phase voltage imbalance and great fluctuations. At this time, the voltmeter will hit the head blindly or the three-phase source-resistant empty voltage will rise in turn, exceeding the line voltage, which is basically caused by resonance.

1. Once there is an imbalance, it will bring harm to equipment, transformers, etc., whether in industrial or domestic electricity, it will lead to asymmetric operation, increase the loss of transformer, and cause the problem of no-load loss or load loss. At this time, the current of the transformer is too large, and some parts will heat up instantaneously, which will directly lead to the problem of burning. 2. It will also have a great impact on the equipment that uses electricity, because at this time, the temperature will rise and the efficiency will decrease, which will not only increase the energy consumption and cause vibration, but also lead to overload and short circuit.

3. In addition, it will also have a great impact on wires and cables, at this time, the load increases, the current is unbalanced, and the bearing will also increase.

To calculate the setting current of a 55 kW motor, it is necessary to know its rated power, rated voltage, and power factor. Assuming that the rated voltage of this 55 kW motor is 660V and the power factor is, then its setting current can be calculated using the following formula:

Setting current = rated power ( 3 rated voltage power factor) According to the above formula, you can get Xiang Tremor to:

Setting current = 55000 3 660 Therefore, the setting current of a 55 kW motor with a voltage of 660V is about. It should be noted that this is only a theoretical calculation value, and in the actual loose banquet wide operation, it is also necessary to consider various factors such as the load situation of the motor and the operating environment.

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The maximum allowable starting current of 660V +37 kW motor, 660V voltage motor current is 1kw,. Electric current refers to the flow of electric charge in a conductor, usually represented by the symbol i, the unit is ampere a, charge is the basic unit of electricity, and current is the amount of charge passing through the conductor per unit time. <>

What is the maximum allowable starting current of 660 volts + 37 kilowatt motor.

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660 volt + 37 kilowatt motor Huisun starting current maximum allowable, 660V voltage motor current 1kw,. Current refers to the flow of charge in a conductor, usually represented by the symbol i, the unit is ampere a, the charge is the basic unit of electricity, and the current is the amount of charge passing through the conductor per unit time. <>

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The rated operating current of 37KW660V is, and the star connection of this motor is 660V, and the corner connection is 380V conversion formula660 In addition to the angle connection, the starting current of the motor refers to the line current that is transmitted to Bizhi when the rated voltage of the motor is turned on from zero speed. <>

Hello, I would like to know the maximum starting current.

660 volt 37 kilowatt motor starting current 480 amps phase current 45 amps will affect the motor.

Dear, yes. Pro, it will affect, excessive starting current will increase the power loss on the motor and the line<> isn't this rated current?

Dear, the maximum starting current is about 180A.

What is the maximum allowable starting current of 660 volts + 37 kilowatt motor.

Hello dear, to calculate the starting current of a 660 volt and 37 kilowatt motor, you need to know the rated current of the motor, the starting mode and the negative speed no-load situation. Generally speaking, the starting current of the motor will be several times higher than its rated current, so it is necessary to consider the compass and the maximum allowable current when the motor starts to determine the rated capacity and protection measures of the power supply and electrical equipment. In the absence of specific motor and load information, we can use an empirical formula to estimate the starting current of the motor.

According to the NEMA standard, the starting current of a three-phase motor is usually 5-7 times its rated current. Thus, for a 660 volt, 37 kW motor with a current rating of about 56 amps (37 kW 660 volts), its starting current may reach around 280 amps to 392 amps (5 to 7 times the rated current). It should be noted that the above is only an estimate, and the actual starting current of the motor will be affected by a variety of factors, such as the load situation, power supply voltage, starting mode, etc., so the specific starting current needs to be determined by the actual test or the technical parameters provided by the motor manufacturer.

In addition, when the motor is started, the corresponding protection measures should be selected according to the actual situation, such as current protection, fuse, overload protector, etc., to protect the safe operation of the motor and electrical equipment. I'm glad to answer for you, and I hope I can help you.

Instantaneously from about 1000A according to the load size in about 3-8 seconds to below 150A starting current refers to the impulse current of electrical equipment (inductive load) at the beginning of starting, is the motor or inductive load energized moment to the operation of the short period of time in a short period of time change, this current is generally 4-7 times the rated current, the state regulations, for the safety of the operation of the line and the normal operation of other electrical equipment, high-power engines must be equipped with starting equipment to reduce the starting current. Inrush current is the maximum instantaneous current that passes through the input voltage before it reaches a steady state when the input voltage is turned on or off at specified intervals. The common starting methods of AC motor include direct start, string resistance start, self-coupling transformer start, star-delta decompression start and inverter start to reduce the impact on the power grid

Motor starting current calculation method:

Due to the different powers, types, and power supply voltages of the motors, as well as the differences in starting methods, there is no unified calculation formula for the size of the starting current of the motor. For example, the following 380V three-phase motor often uses direct start, and the motor greater than this power requires the use of star-delta conversion step-down start, auto-decompression start, soft starter starter, frequency conversion start and other methods. There is a difference between the instantaneous maximum and the average value during start-up when the starting current is magnitude.

In short, it is possible to estimate the average starting current value of 3 to 5 times the rated current of the motor. For common 220V single-phase motors, the rated current is estimated at kW 5 amps, and for 380V three-phase motors, the rated current is estimated at kW 2 amps, multiplied by 3 to 5 as the starting current.