How to calculate the current carrying capacity of high-voltage power cables?

High-voltage power cable current carrying capacity calculation method? In theory, the calculation formula for the current carrying capacity of high-voltage cables:
Maximum current capacity = cable breakage area (mm2) × allowable maximum current density of this cable material (A/mm2)
Calculation of wire cross-sectional area and current carrying capacity:
Generally, the safe current carrying capacity of the copper wire current-carrying wire is determined according to the maximum core temperature, cooling conditions and laying conditions allowed; the safe current carrying capacity of the general copper wire is 5~8A/mm2, and the safe current carrying capacity of the aluminum wire is 3~5A/mm2.
Such as: 2.5 mm2 BVV copper wire safe current carrying capacity recommended value 2.5 × 8A / mm2 = 20A 4 mm2 BVV copper wire safe current carrying capacity recommended value 4 × 8A / mm2 = 32A
Calculate the cross-sectional area of ​​the copper wire. Use the recommended value of the safe current carrying capacity of the copper wire to 5~8A/mm2, and calculate the upper and lower ranges of the cross-sectional area S of the selected copper wire: S==0.125 I ~0.2 I(mm2) S----- Copper wire cross-sectional area (mm2) I-----load current (A)
Power calculation General load (can also be used as electrical appliances, such as lighting, refrigerators, etc.) is divided into two types, one type of resistive load and one type of inductive load. For the calculation of the resistive load: P = UI For the calculation of the fluorescent lamp load: P = UIcosф, where the power factor of the fluorescent lamp load is cosф = 0.5. Different inductive loads have different power factors. When calculating household appliances, the power factor cosφ can be taken as 0.8.
That is to say, if the total power of a household appliance is 6000 watts, the maximum current is I=P/Ucosф=6000/220*0.8=34(A) However, in general, it is impossible for home appliances to be used at the same time. , so add a common coefficient, the common coefficient is generally 0.5. Therefore, the above calculation should be rewritten as I = P * common coefficient / Ucos ф = 6000 * 0.5 / 220 * 0.8 = 17 (A) That is, the total current value of this family is 17A. The total brake air switch cannot use 16A, it should be greater than 17A.
Many people have raised such doubts about the current carrying capacity of high-voltage cables:
1. Is the same section of high-voltage cable and ordinary cable current carrying capacity the same?
This problem must take into account some of the special features of high-voltage cable materials:
Because of the high insulation level, the external film is thick, and the heat dissipation is not as good as the low-voltage cable, the high-voltage cable has a slightly lower current carrying capacity than the ordinary cable.
2. Why is the cable of the high-voltage side-in transformer as long as YJV-8.7/10 3×95 and some of the low-voltage side outlets after the transformer transformation are YJV22—4×180+1×95, and the cross-sectional area of ​​the cable on the low-voltage side is higher than that on the high-voltage side. Big?
Because theoretically the input power of the transformer is equal to the output power. The power P=UI, the voltage on the high voltage side is high, so the passing current is small, and the voltage on the low voltage side is low, so the passing current is large. If the design is reasonable, you said that the high-voltage side uses 95 cables and the low-voltage side uses 4X180 cables. The current ratio on both sides is about 1:9, the voltage ratio should be 9:1, the low voltage is 400 volts, and the high voltage is 3600 volts. The capacity is about 1 mega volt ampere. Http://news.chinawj.com.cn Editor: (Hardware Business Network Information Center) http://news.chinawj.com.cn

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