How to calculate the current of an AC motor
The 380/220V three-phase four-wire system is a widely used power supply system throughout my country. The nameplates of various low-voltage electrical appliances generally indicate their capacity. How to quickly calculate the rated load current based on the capacity in order to select appropriate fuses, switches, wires, etc., is a common calculation problem encountered by electricians.
1 Three-phase motor
A 380V three-phase motor typically has a power factor of around 0.8. Its rated current is approximately twice its rated capacity. When the motor's power is below 2KW, it can be considered as 2.5 times its rated capacity.
For a 220V three-phase motor, its rated current is approximately 3.5 times its 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 represents the rated current (A) of the three-phase asynchronous motor.
P --- Power of the motor (kW) K --- Coefficient
Related questions and answers:
The formula for calculating the current of a three-phase motor is I=P/1.732/U/cosΦ. What is cosΦ, and why is there a cosΦ?
Is the formula I=P/1.732/U correct or incorrect? In what situations would cosΦ be used?
The formula I=P/1.732/U is incorrect. The correct formula should be I=P/(1.732*U*cosΦ).
In an AC circuit, the cosine of the phase difference (Φ) between voltage and current is called the power factor, denoted by the symbol cosΦ. Numerically, the power factor is the ratio of active power to apparent power, i.e., cosΦ = P/S
The power factor is related to the nature of the circuit load. For example, the power factor of resistive loads such as incandescent light bulbs and electric furnaces is 1, while the power factor of circuits with inductive or capacitive loads is generally less than 1. The power factor is an important technical data point for power systems. It is a coefficient that measures the efficiency of electrical equipment. A low power factor indicates that the circuit uses a large amount of reactive power for alternating magnetic field conversion, thus reducing equipment utilization and increasing power supply losses. Therefore, power supply departments have certain standard requirements for the power factor of electricity users.
What does S represent in cosΦ=P/S?
S represents active power, also known as average power. The instantaneous power of alternating current is not a constant value; the average power over one cycle is called active power. It refers to the power consumed by the resistive part in the circuit, and for a motor, it refers to its output force.
When calculating three-phase current, the power factor of the electricity user must be taken into account, so the power factor is represented by cosΦ.
If cosΦ is 1, it can be ignored in the formula.
Analysis of cosΦ (i.e., power factor):
The power factor of an electric motor is not a fixed value; it depends on the manufacturing quality and the load factor. To save energy, the government mandates that electric motors have a higher power factor, increasing it from 0.7-0.8 to 0.85-0.95. However, the load factor is determined by the user and is not uniformly applied. In the past, a power factor of 0.75 was often used in motor current calculations, and now 0.85 is still commonly used.
If the current is 144.34A, what size copper core cable should be selected? What is the calculation formula?
50 square meters of cable is needed.
The calculation formula is: 10 down five, 100 up two; 2535 four three boundary; 7095 two and a half times; high temperature in conduit eight or nine times; bare wire add half; copper wire upgrade.
2. Single-phase motor current
The rated current of a 220V single-phase motor is approximately eight times its rated capacity. Examples of single-phase 220V motors used in appliances such as electric fans, electric drills, blowers, washing machines, and refrigerators have a relatively high electrical load.
The formula is I=8P
In the formula, I represents a single-phase asynchronous motor.
Rated current (A) P --- Motor power (KW)
Both star and delta connections use P = 1.732 * U * I * power factor.
In a star connection, the line voltage equals 1.732 phase voltages, and the line current equals the phase current.
When connected at a delta connection, the line voltage equals the phase voltage, and the line current equals 1.732 times the phase current.
Notice
You are probably familiar with these formulas. The key is to understand whether the measured (or given) current and voltage are for a line or a phase.
In actual three-phase load circuits, 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. Therefore, the first calculation formula (√3) is used for both.
If there are special cases where both are phase voltage and phase current, then the second calculation formula (3 times the ratio) must be used.
Calculation of rated current
For a three-phase four-wire AC power supply, the line voltage is 380V, the phase voltage is 220V, and the line voltage is the square root of the three-phase voltage.
For an electric motor, the voltage across a winding is the phase voltage, and the voltage across a conductor is the line voltage (referring to the voltage between phases A, B, and C). The current in a winding is the phase current, and the current in a conductor is the line current.
When the motor is star-connected:
Line current = phase current;
Line voltage = √3 phase voltage.
The tail wires of the three windings are connected together, and the potential is zero, so the voltage of the windings is 220 volts.
When the motor is angle-connected:
Line current = √3 phase current;
Line voltage equals phase voltage. The windings are directly connected to a 380° connection; the current in the conductor is the vector sum of the currents in both windings.
The power calculation formula p = √3 UI multiplied by the power factor is correct.
Using a clamp meter, the current measured on any of lines A, B, or C will be the line current.
1. Three-phase calculation formula:
P = 1.732 × U × I × cosφ
(Power factor: resistive load = 1, inductive load ≈ 0.7 to 0.85, P = power: W)
The formula for calculating single-phase P is: P = U × I × cosφ
The selection of circuit breakers should be based on the load current, and the circuit breaker capacity should be about 20 to 30% larger than the load current.
The formula is universal:
P = 1.732 × IU × power factor × efficiency (three-phase)
For single-phase, do not multiply by 1.732 (√3).
Generally, the circuit breaker should be selected to be 1.2-1.5 times the total rated current.
The empirical formula is:
380V voltage, 2A per kilowatt; 660V voltage, 1.2A per kilowatt; 3000V voltage, 1A per 4 kilowatts.
6000V voltage, 8kW 1A.
For power above 3KW, current = 2 * power;
Current for 3KW and below = 2.5 * power
2. Power factor (divide active power by reactive power, find the arctangent value, 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 cos@ (I can't type the symbol, so I'll use @ instead)
Apparent power S = (square root of active power P + square root of reactive power Q)
The power factor cosφ = active power P / apparent power S
3. Please provide the formulas for calculating active power, reactive power, and power factor in detail. (The transformer is a single-phase transformer.) Also, will a decrease in reactive power also decrease active power? Conversely, will an increase in reactive power also increase active power?
Active power = I * U * cosφ, which is the product of rated voltage, rated current, and power factor.
The unit is watt or kilowatt.
Reactive power = I * U * sinφ, the unit is VAR or kilovar.
I*U represents capacity, measured in volt-amperes (VA) or kilovolt-amperes (kVA).
When reactive power decreases or increases, active power remains unchanged. However, when reactive power decreases, current decreases and line losses decrease; conversely, when reactive power increases, line losses increase.
4. What is leading or lagging reactive power? Why does an asynchronous motor absorb lagging reactive power from the power grid during operation?
Inductive current is leading current, and capacitive current is lagging current.
Reactive power is a rather abstract concept. It refers to the electrical power used for the exchange of electric and magnetic fields within a circuit, and for establishing and maintaining magnetic fields in electrical equipment. It does not perform work externally, but is instead converted into other forms of energy.
Any electrical device with an electromagnetic coil needs to 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 consumes some active power) to emit light, and about 80 VAR of reactive power to power the ballast coil to establish an alternating magnetic field.
Because it does not perform work externally, it is called "reactive power". The symbol for reactive power is Q, and the unit is var (Var) or kilovar (kVar).
5. A three-phase 11KW motor is labeled with a current of 22.6A. How do you calculate the current of a single-phase 0.75KW motor labeled with a current of 4.5A and a speed of 2860r/min?
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)
The calculation is: 0.75 / (0.22 * 0.75) = 4.54A
The 0.22 above and the 0.38 in the first point refer to voltage in kilovolts (kV). 0.22 kV = 220 V, 0.38 kV = 380 V
2860 r/min refers to the rotational speed per minute of a two-pole motor.
A motor with a speed of 1450 rpm is a four-pole motor. It is mainly selected for motors with different speed requirements.
Two-pole motors are used in axial flow fans and some equipment requiring high speeds, while four-pole ordinary motors are often paired with speed reducers for use on assembly lines and in glass edging and corner grinding machines, etc.
6. How is the wattage of an electric motor calculated? An electric motor has the following information: 1430 RPM, 380V, 7.5A for 5 minutes. How many kilowatts is it?
The power of a three-phase motor can be calculated using the formula P=1.732UIcosφ.
P – Power; WU – Voltage; VI – Current; Acosφ – Power factor is generally taken as 0.8.
The result is P = 1.732 * 380 * 7.5 * 0.8 = 3948.96 W, approximately 4 kW.
Another rule of thumb is that the current of a 380V three-phase motor is twice its power, which means 4A for 2KW, 6A for 3KW, 8A for 4KW, and so on, up to 100A for 50KW.
7. How many capacitors are needed to improve the power factor of a three-phase motor?
The natural power factor of a typical three-phase motor is around 0.7. If you want to increase the power factor to 0.9, the compensation capacitance is Q≈0.5P, which means that the compensation amount of the capacitor is about half of the rated power of the motor.
