How to calculate generator power factor

Many users do not know the power, no power, apparent power and conversion method of diesel generators. This will reduce our understanding of the generator set and affect our correct use of the generator set. Therefore, the editor of Dingbo generator manufacturer will share relevant knowledge for everyone.

 

1. Active impedance is divided into two parts: resistance and reactance. The resistance is represented by R, and the reactance is represented by X.

 

The impedance expression Z:Z=R+J(xl xc) R represents the resistance, and xl xc is the X part of the reactance. Active power is the work done by the resistance R in the circuit for 1s. Active power is represented by P and the unit is w (watts).

 

Active power formula: P=uicosφ. U is the circuit voltage, I is the effective value of the current, and Cosφ is the power factor. We often say that motor power refers to active power, such as motor power 0.75kw.

2. Reactive power

 

Reactive power is the work done by the imaginary part of the reactance X of the voltage and current impedance Z within 1s. It is represented by the symbol Q and the unit is reactive power. Reactive power formula: Q=uisin

 

 3. inspecting power

 

Apparent power is also called capacity, which represents the sum of work done by the current in the circuit in one second. In VA, it is represented by the symbol s. Apparent power formula: s=UI

 

 4. The relationship between apparent power, active power and reactive power

 

The square of apparent power is equal to the sum of squares of active power and reactive power.

 

5. Calculate the power factor of the motor

 

For example, a generator on the nameplate shows: the motor power is 0.75kw, when the power factor is 0.866, the voltage is 380V, the current is 1.7A, the apparent power s=380*1.7*√3=1118VA can be calculated, so The power factor Cosφ=P/s=750/1118=0.67.

 

6. Energy = power X time, the unit is Joule J. The unit conversion of work and watts is: 1J=1W·s.

 

 1kWh, that is, the power consumption of 1kW equipment in one hour: 1 kWh = 1000W·H.

 

 1 degree =1kW·H=1000W·3600s=3.6×10^6J