Common sense of wire and cable calculation

Common sense of wire and cable calculation

What is the maximum current and power that a square wire can pass? How much wire is used for 100KW of electricity for project construction? Large wires cause waste, and small wires have more serious consequences. Here I share with you my estimation experience.

Aluminum core wire and cable current carrying capacity standard cable carrying capacity port decision:

　　 Two and five times down and multiply by nine, and go up and down by one.

　　 Thirty-five times three and five, both groups minus five points.

　　 The conditions are changed and the conversion is added, and the high temperature 10% copper upgrade.

　　 The number of piercing pipes is two, three, four, and the full load current is 770%.

Explanation:

　　 The formula in this section does not directly indicate the current carrying capacity (safe current) of various insulated wires (rubber and plastic insulated wires), but is expressed by "cross section multiplied by a certain multiple", which is obtained through mental calculation.

　　 "Multiply two and five times by nine, and go up and down by one" refers to various cross-section aluminum core insulated wires of 2.5mm and below, and their current carrying capacity is about 9 times the number of cross-sections.

　　 Such as 2.5mm’ wire, the current carrying capacity is 2.5×9=22.5(A).

　　The multiple relationship between the current-carrying capacity of 4mm’ and above conductors and the number of cross-sections is to go up the line number, and the multiples are successively reduced by 1, namely 4×8, 6×7, 10×6, 16×5, 25×4.

　　 "Thirty-five times 3.5, subtracting a point for both pairs in groups" means that the carrying capacity of a 35mm wire is 3.5 times the number of sections, that is, 35×3.5=122.5(A).

For conductors of 　　50mm’ and above, the multiple relationship between the current-carrying capacity and the number of cross-sections becomes a set of two wire numbers, and the multiples are successively reduced by 0.5. That is, the current-carrying capacity of 50 and 70mm' wires is 3 times the number of cross-sections; the current-carrying capacity of 95, 120mm" wires is 2.5 times the number of cross-sectional areas, and so on.

　　 "Conditions are changed and discounted, high temperature 10% copper upgrade". The above formula is determined by the aluminum core insulated wire and the open coating at an ambient temperature of 25°C.

If the aluminum core insulated wire is exposed to the area where the ambient temperature is higher than 25℃ for a long time, the current carrying capacity of the wire can be calculated according to the above formula calculation method, and then a 10% discount is enough; when the copper core insulated wire is not used, Its current-carrying capacity is slightly larger than that of the aluminum wire of the same specification. According to the above formula method, the current-carrying capacity of one wire larger than the aluminum wire can be calculated. For example, the current-carrying capacity of 16mm’ copper wire can be calculated as 25mm2 aluminum wire.

　　In actual calculations, we have to distinguish three-phase electricity and single-phase electricity. The conversion between the two types of power and current is different.

　　 1. Calculation formula of three-phase resistance type electric power = 1.732*line voltage U*line current I (star connection)

　　 Three-phase resistance type electric power calculation formula = 3*phase voltage U*phase current I (angle connection)

　　 Three-phase motor power calculation formula = 1.732*line voltage U*line current I*power factor COSΦ (star current=I, voltage=U, resistance=R, power=P)

　　2. The relationship between single-phase electric power P and line voltage U, line current I and power factor cosφ is:

P=U*I*cosφ So, I=P/(U*cosφ). Among them, incandescent lamps, electric furnaces, electric soldering irons, etc. can be regarded as resistive loads, and if cos φ=1, P=UI