How much current can a 4 square bv line withstand at 220 V voltage emergency How to calculate? There should be a formula

How much current can a 4 square bv line withstand at 220 V voltage emergency How to calculate? There should be a formula

The current carrying capacity is related to the ambient temperature. The pipe is divided into steel pipe and plastic pipe. The current carrying capacity is 15mm. Unit a: two single cores 1 square 25 degrees 1430 degrees 1335 degrees 1240 degrees 111.5 square 25 degrees 1930 degrees 1835 degrees 1740 degrees 152.5 square 25 degrees 2630 degrees 2435 degrees 2240 degrees 204 square 25 degrees 3530 degrees 32

What is the maximum power consumption of 6 square and 10 square bv lines? 3Q

According to the pure national standard, 6 square meters is 42a, single-phase with 9kw, three-phase with 24kw, 10 square meters of 60A, single-phase with 13kw, three-phase with 35kw. This is in the distance of not more than 80 meters. You can also calculate by yourself. One square of copper wire bears the safety current, which is generally calculated according to 4A current, For example, 1.5 square meters is calculated according to 2.5. This is the formula of factory calculation. It works very well. Answer: the formula of estimation: 2.5 times 9, up minus 1. Go straight. 35 times 3.5, double pairs in groups minus 1.5, Note: (1) the formula in this section does not directly point out the current carrying capacity (safety current) of various insulated wires (rubber and plastic insulated wires), but "cross section multiplied by a certain multiple", which is obtained by mental calculation. It can be seen from table 53 that the multiple decreases with the increase of cross section, For example, the current carrying capacity of 2.5mm 'conductor is 2.5 × 9 = 22.5 (a). The multiple relationship between the current carrying capacity and the number of sections of 4mm' and above conductor is arranged upward along the wire number, and the multiple is reduced by 1, that is, 4 × 8, 6 × 7, 10 × 6, 16 × 5, 25 × 4, It means that the current carrying capacity of 35mm "conductor is 3.5 times of the number of cross-sections, that is 35 × 3.5 = 122.5 (a). From the multiple relationship between the current carrying capacity and the number of cross-sections of 50 mm 'and above conductor, two wire numbers become a group, The current carrying capacity of 50 and 70mm 'conductor is 3 times of the number of cross-sections, that of 95 and 120mm' conductor is 2.5 times of the number of cross-sections, and so on. The above formula is based on the condition that the aluminum core insulated wire is exposed at 25 ℃ for a long time, The current carrying capacity of the conductor can be calculated according to the above formula, and then 10% discount can be made. When the copper core insulated wire is used instead of aluminum wire, its current carrying capacity is slightly larger than that of aluminum wire of the same specification, the current carrying capacity of one wire number larger than that of aluminum wire can be calculated according to the above formula. For example, the current carrying capacity of 16mm 'copper wire can be calculated as 25mm2 aluminum wire