Calculation of various motor currents and rated currents

Low-voltage 380/220V three-phase four-wire system

It is a power supply system widely used in all parts of our country

The nameplates of various low-voltage electrical appliances generally indicate the capacity

How to quickly calculate the rated load current according to the capacity

To be fitted with appropriate fuses, switches, wires, etc.

It is the most common calculation problem that electricians encounter

Today, let us know

How to Calculate the Current of an AC Motor!

three-phase motor

The current 380V three-phase motor, the power factor is generally about 0.8, its rated current is about twice the rated capacity, when the power of the motor is below 2KW, it can be considered as 2.5 times.

For a 220V three-phase motor, its rated current is about 3.5 times the rated capacity. When the power of the motor is below 2KW, it can be considered as 4 times.

The specific calculation formula is I=KP

In the formula, I--- the rated current of the three-phase asynchronous motor (A)

P---motor power (KW) K---coefficient

Related Q&A:

Three-phase motor current calculation formula I=P/1.732/U/cosΦ, what is cosΦ, and why is there a cosΦ?

Is the formula I=P/1.732/U right or wrong? Under what circumstances will cosΦ be used.

The formula I=P/1.732/U is wrong, the correct one should be I=P/(1.732*U*cosΦ).

cosΦ is the cosine of the phase difference (Φ) between the voltage and current in an AC circuit, which is called the power factor, and is represented by the symbol cosΦ. In numerical terms, the power factor is the ratio of active power to apparent power, that is, cosΦ=P/ S

The size of the power factor is related to the load nature of the circuit. For example, the power factor of resistive loads such as incandescent light bulbs and resistance furnaces is 1. Generally, the power factor of circuits with inductive or capacitive loads is less than 1. Power factor is an important technical data of power system. Power factor is a factor that measures the efficiency of electrical equipment. The low power factor indicates that the reactive power used by the circuit for the conversion of the alternating magnetic field is large, thereby reducing the utilization rate of the equipment and increasing the power supply loss of the line. Therefore, the power supply department has certain standard requirements for the power factor of the power consumption unit.

What does P in cosΦ=P/S mean?

P represents active power, and active power is also called average power. The instantaneous power of alternating current is not a constant value. The average value of power in one cycle is called active power. It refers to the power consumed by the resistance part in the circuit, and for the motor, it refers to its output.

To calculate the three-phase current, the power factor of the power consumption unit should be considered, so the power factor is represented by cosΦ.

If cosΦ is 1, it can be ignored in the formula.

cosΦ (that is, the power factor) analysis and translation:

The power factor of the motor is not a constant, it is related to the quality of manufacture and the size of the load rate. In order to save electric energy, the state requires the motor products to increase the power factor, from 0.7 to 0.8 to the current 0.85 to 0.95, but the load rate is controlled by the user, so it is not uniform. In the past, the power factor was often taken as 0.75 in the calculation of motor current, and now it is often taken as 0.85.

The current is 144.34A, what size copper core cable should I choose? What is the calculation formula?

Requires 50 square meters of line.

The formula for calculating is: 10 down five, 100 up 2; 2535 four and three realms; 7095 twice and a half;

Single-phase motor current

The rated current of a 220V single-phase motor is about 8 times the rated capacity. Such as the electrical load of single-phase 220V motors in fans, hand drills, blowers, washing machines, and refrigerators

The calculation formula is I=8P

where I---single-phase asynchronous motor

The rated current (A) P---the power of the motor (KW)

Both star connection and corner connection are P=1.732*U*I*power factor

When the star is connected, the line voltage = 1.732 phase voltage, the line current = phase current

When connecting, line voltage = phase voltage, line current = 1.732 phase current

Notice

These formulas must be familiar to you, the key is to understand clearly whether the measured (or given) current or voltage is a line or phase problem.

In the actual three-phase load circuit, due to the convenience of measuring the external power supply of the load, if it is a star connection, the voltage generally refers to the line voltage, and if it is a delta connection, the current generally refers to the line current, so the first calculation is used. Formula (square root 3 times).

If there are special cases, both phase voltage and phase current, then the second calculation formula must be used (3 times relationship)

Calculation of rated current

For AC three-phase four-wire power supply, the line voltage is 380, the phase voltage is 220, and the line voltage is the root 3-phase voltage

For a motor, the voltage of a winding is the phase voltage, the voltage of the wire is the line voltage (referring to the voltage between the A-phase B-phase C-phase, the current of a winding is the phase current, and the current of the wire is the line current

When the motor is star connected:

line current = phase current;

Line voltage = root 3-phase voltage.

The tails of the three windings are connected and the potential is zero, so the voltage across the windings is 220 volts

When the motor is angularly connected:

Line current = root 3-phase current;

Line voltage = phase voltage. The winding is directly connected to 380, and the current of the wire is the vector sum of the currents of the two windings

The power calculation formula p = square root of three UI times the power factor is correct

Use a clamp-type ammeter card to measure the line current on any line of ABC

Three-phase calculation formula

P=1.732×U×I×cosφ

(Power factor: resistive load = 1, inductive load ≈ 0.7 ~ 0.85, P = power: W)

Single-phase calculation formula: P=U×I×cosφ

The selection of the air switch should be based on the load current, and the capacity of the air switch is about 20-30% larger than the load current.

The formula is generic:

P=1.732×IU×power factor×efficiency (three-phase)

Single-phase does not multiply by 1.732 (root sign 3)

The choice of the air switch is generally 1.2-1.5 times the overall rated current.

?The empirical formula is:

380V voltage, 2A per kW, 660V voltage, 1.2A per kW, 3000V voltage, 4 kW 1A,

6000V voltage, 8kW 1A.

Above 3KW, current=2*power;

3KW and below current = 2.5*power

?Power factor (divide the active power by the reactive power, find the arc tangent and then find the sine value)

Power factor cosΦ=cosarctg (reactive power/active power)

Apparent power S

Active power P

Reactive power Q

Power factor cosΦ

Apparent power S = (square of active power P + square of reactive power Q) and then square root

And power factor cosΦ = active power P / apparent power S

?For the calculation formulas of active power, reactive power and power factor, please explain in detail.

?(The transformer is a single-phase transformer)

In addition, the reduction of reactive power will also reduce the active power?

Conversely, will the increase of reactive power also increase the active power?

?Active power = I*U*cosφ, that is, the rated voltage multiplied by the rated current and then multiplied by the power factor

in watts or kilowatts

Reactive power=I*U*sinφ, the unit is var or thousand var.

I*U is the capacity, and the unit is volt-ampere or kVA.

When the reactive power decreases or increases, the active power remains unchanged. However, when the reactive power decreases, the current decreases and the line loss decreases, and vice versa, the line loss increases.

?What is leading or lagging reactive power?

Why does the asynchronous motor absorb the lagging reactive power from the grid when it is running?

The inductive current is the leading current, and the capacitive current is the lagging current.

Reactive power is more abstract, it is the electric power used for the exchange of electric and magnetic fields in the circuit, and used to establish and maintain the magnetic field in electrical equipment. It does not do external work, but transforms into other forms of energy.

All electrical equipment with electromagnetic coils will consume reactive power to establish a magnetic field. For example, a 40-watt fluorescent lamp requires more than 40 watts of active power (the ballast also needs to consume a part of the active power) to emit light, and also requires about 80 watts of reactive power for the coil of the ballast to establish an alternating magnetic field.

Because it does not do external work, it is called "reactive work". The symbol of reactive power is represented by Q, and the unit is var (Var) or kilovar (kVar).

The three-phase motor 11KW motor is marked with a current of 22.6A?

There is also a single-phase 0.75KW marked with 4.5A, how to calculate the speed of 2860rmin?

?Three-phase motor power calculation P = UI * power factor (0.75)

Rated current

I=P/(U*0.75)/1.732=11/(0.38*0.75)/1.732=22.28A

0.75KW single phase (220V)

Calculated as: 0.75/(0.22*0.75)=4.54A

The above 0.22 and the first 0.38 are kilovolts of voltage, 0.22KV=220V0.38KV=380V

2860r/min refers to the speed of the motor per minute for a two-pole motor.

?For example, 1450 rpm is a four-pole motor. It is mainly for the selection of special motors with different requirements for rotation speed.

For example, the two-pole motor is used in axial flow fans and some equipment that requires higher speed, and the four-pole ordinary motor is often equipped with a reducer for a flexible assembly line and some glass edging and angle grinding machines, etc.

?How to calculate the wattage of a motor.

?A motor, it has the following information:

1430 RPM, 380V, 7.5A for 5 minutes, how many kilowatts is it?

?According to the three-phase motor power formula P=1.732UIcosφ can be calculated

P——Power WU——Voltage VI——Current AcosφThe power factor is generally 0.8

The result P=1.732*380*7.5*0.8=3948.96W is about 4KW

There is also an empirical formula, that is, the current of a 380V three-phase motor is twice the power, that is, 4A for 2KW, 6A for 3KW, 8A for 4KW... 100A for 50KW

?How many capacitors does a three-phase motor use to improve the power factor?

?Generally, the natural power factor of a three-phase motor is about 0.7. If you want to increase the power factor to 0.9, the compensation capacitance is: Q≈0.5P, that is, the compensation amount of the capacitor is about half of the rated power of the motor.