Three phase symmetrical load, each phase resistance R = 6 Ω, inductive reactance XL = 8 Ω, access to 380V three-phase three wire power supply, calculate the power consumption of star and triangle connection After carefully studying the two brothers' opinions, there is still a little doubt about why they did not mention the power factor Power of star connection: P = 3uicos, φ = 3 * 220 * 22 * 0.6 = 8712

Three phase symmetrical load, each phase resistance R = 6 Ω, inductive reactance XL = 8 Ω, access to 380V three-phase three wire power supply, calculate the power consumption of star and triangle connection After carefully studying the two brothers' opinions, there is still a little doubt about why they did not mention the power factor Power of star connection: P = 3uicos, φ = 3 * 220 * 22 * 0.6 = 8712

The impedance of the load is
Z = √ (R ^ + XL ^) = 10 Ω
The power factor of the load is
cosφ = 1/√1+(Z/R)^ = 0.6
The power of three-phase symmetrical load is
P = √ 3uicos φ (U is line voltage, I is line current)
Star time
i = 220/10 =22A
And line current = phase current
p =√3UIcosφ =√3*380*22*0.6=8687.712W
Triangle time
i = 380 /10 =38A
And the line current = 3 times of the phase current
I = 1.732*38 = 65.816A
p =√3UIcosφ =√3*380*65.816*0.6=25990.48W
In fact, "spx8655" is also correct, because its power is obtained through the square of current * resistance, and the resistance does not contain reactive power component, so there is no need to use the power factor. As for the answer, there is a small error, because the root sign 3 in the calculation only uses three decimal places of 1.732, and there is an error. The root sign 3 is really 1.732050808
There are some mistakes in the concept of "Feihu 09". P means active power. If it is the total power, it should be apparent power. S means volt ampere, not watt
I'll give you an answer in another way, just to provide you with another way of thinking,

2. There is a three-phase motor, each phase equivalent resistance R = 29 Ω, equivalent inductance XL = 21.8 Ω. The winding is star connected and connected to the three-phase power supply with line voltage UL = 380V. Try to find the phase current, line current and power input from the power supply of the motor

For star connection, line voltage UL = 380V, so phase voltage U = 380 / 1.732 = 220V. Phase impedance r = root (29 ^ 2 + 21.8 ^ 2) = 36.28 Ω, phase current I = u / r = 6a, line current I = phase current I = 6a, power P = 3 * u * I = 4kw

Known power 500kW, length 1000m how much cable

What are the voltage and power factor respectively? The selection of cable is mainly related to the current rather than the power. We assume that it is a 10KV asynchronous motor with power factor of 0.85, and the current is 500 / 10 / 1.732/0.85 = 33.96a. Then we can use a 25mm2 high-voltage cable, such as yjv22-8.7/15 3 * 25mm2, assuming that it is a transformer