Motor introduction:

       It is a device that converts electrical energy into mechanical energy . It uses energized coils (that is, stator windings) to generate a rotating magnetic field and acts on the rotor (such as a squirrel-cage closed aluminum frame) to form a magneto-electric power rotation torque . Motors are divided into DC motors and AC motors according to the power used . Most of the motors in the power system are AC motors , which can be synchronous motors or asynchronous motors (the motor stator magnetic field speed and rotor rotation speed do not maintain synchronous speed). The motor is mainly composed of a stator and a rotor. The direction of the current-carrying wire in the magnetic field is related to the direction of the current and the direction of the magnetic field line (magnetic field direction). The working principle of the motor is that the magnetic field acts on the force of the current to make the motor rotate.

The motor will learn many calculation formulas during the learning process:

The relationship between the number of pole pairs and torque

n=60f/ pn : motor speed  60 :  60 seconds  f: China's current uses 50Hz p: number of motor pole pairs  1 pair of pole pairs Motor speed: 3000 rpm ; 2 pairs of pole logarithms Motor speed: 60×50/2 = 1500 rpm _

In the case of constant output power, the more the number of pole pairs of the motor, the lower the speed of the motor, but the greater its torque. Therefore, when selecting a motor, consider how much starting torque the load requires.

The speed n=(60f/p)×(1-s) of the asynchronous motor is mainly related to the frequency and the number of poles.

The speed of a DC motor has nothing to do with the number of poles. Its speed is mainly related to the voltage of the armature, the magnetic flux, and the structure of the motor. n=(motor voltage-armature current*armature resistance)/(motor structure constant*magnetic flux).

torque formula

T=9550*P output power/N speed

The formula for calculating wire resistance is :

The resistivity of copper wire ρ=0.0172, R=ρ×L/S (L=length of wire, unit: meter, S=cross section of wire, unit: m㎡) calculation formula of magnetic flux :

B is the magnetic induction intensity, S is the area. The law of the known magnetic field is: Φ=BS

Calculation formula of magnetic field strength: H = N × I / Le

In the formula: H is the magnetic field strength, the unit is A/m; N is the number of turns of the excitation coil; I is the excitation current (measured value), the unit is A; Le is the effective magnetic circuit length of the test sample, the unit is m.

Magnetic induction calculation formula: B = Φ / (N × Ae) B=F/IL u magnetic permeability pi=3.14  B=uI/2R

In the formula: B is the magnetic induction intensity, the unit is Wb/m^2; Φ is the induced magnetic flux (measured value), the unit is Wb; N is the number of turns of the induction coil; Ae is the effective cross-sectional area of the test sample, the unit is m ^2.

induced electromotive force

1) E=nΔΦ/Δt (universal formula) {Faraday's law of electromagnetic induction, E: induced electromotive force (V), n: number of turns of the induction coil, ΔΦ/Δt: rate of change of magnetic flux}

Rate of change = change in magnetic flux / time change in magnetic flux = magnetic flux after change - magnetic flux before change

2) E = BLV vertical (cutting the magnetic induction line movement) {L: effective length (m)}

3) Em=nBSω (the maximum induced electromotive force of the alternator) {Em: peak value of the induced electromotive force}

4) E=BL2ω/2 (one end of the conductor is fixed and cut with ω rotation) {ω: angular velocity (rad/s), V: velocity (m/s)}

Three-phase calculation formula:

P＝1.732×U×I×cosφ

COSφ is the rated power factor of the motor, and the rated power factor refers to the phase difference between the stator phase voltage and the phase current when the motor is running under the rated working condition.

(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 breaker should be based on the load current, and the capacity of the air breaker 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 (square root 3)

The selection of the circuit breaker is generally 1.2-1.5 times the overall rated current.

The empirical formula is:

380V voltage, 2A per kilowatt, 660V voltage, 1.2A per kilowatt,

3000V voltage, 4 kW 1A, 6000V voltage, 8 kW 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 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@ (the symbol cannot be typed out, use @ instead)

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

And the power factor cos@=active power P/apparent power S

Please explain in detail the calculation formulas for active power, reactive power and power factor. (The transformer is a single-phase transformer)

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

Answer: Active power = I*U*cosφ, that is, rated voltage multiplied by rated current and multiplied by power factor

Units are watts or kilowatts

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

I*U is the capacity , the unit is VA or KVA .

When the reactive power decreases or increases, the active power remains unchanged. But when the reactive power decreases, the current decreases and the line loss decreases, otherwise, the line loss increases.