P (electric power) = √ 3× 380 (three-phase voltage )× I (current )× cosφ, where COS φ is the reactive loss of inductive load, and a conservative constant of 0.85 is enough:
√ 3 = 1.732 This is the basic mathematical knowledge, and the current (i) is deduced:
I = p (power) /√3×380×0.85(cosφ). Use current to find out the diameter of the power line (note that this is the line diameter, not the cross-sectional area, and the specifications of the power line are all expressed by the cross-sectional area). Line diameter formula:
D = 0.8 ×√ I, and then calculate the cross-sectional area of the conductor with the diameter of the conductor. This is a mathematical formula: radius × radius × 3.14; All right, that's it. The specifications of the power cord used have been worked out.
Very simple, you need to remember two formulas, one:
P (electric power) = √ 3× 380 (three-phase voltage )× I (current )× cos φ.
Another formula: d = 0.8 ×√ i.