There are formulas, but in actual use, general electricians use the "empirical formula" that is easy to remember: that is, each square millimeter of copper core wire can pass a current of about 4.5-5A; then if it is a single-phase circuit, the load current per 1KW is about 4.5A, if it is a three-phase balanced load, the load current per 1KW is about 2A; in this way, it can be calculated that the copper core wire per square millimeter cross-sectional area can carry a single-phase 1KW Load or 2.5KW three-phase balanced load, and so on, you can know how much load the cable core can carry.
Generally, electricians can choose the cross-section of the cable core using the experience value of the following jingle description: [Jingle]: ten under five, one hundred and two, fifty-three and four, and 25% off the buried casing. 【Meaning】: A wire with a total current of less than 10A can pass a current of 5A per square millimeter; a wire with a total current of more than 100A can pass a current of 2A per square millimeter; a wire with a total current of 10-50A can pass a current of 2A per square millimeter A current of 4A can be passed; a wire with a total current of 50-100A can pass a current of 3A per square millimeter; after this calculation, if it is buried or the casing is attached, the current value that can be passed should be multiplied by 0.75.
According to the calculation, the current that a four-square copper wire can withstand is 16-20A; a single-phase load of four kilowatts can choose an insulated copper wire of 4mm^2 (a certain synchronization factor can be considered at this time).
One, four square copper wire. How many loads can it withstand?
Answer: According to the current-carrying capacity of the commonly used wires in the "Modern Electrician Handbook" of Guangdong Science and Technology Press:
The copper core plastic insulated wire of 500V and below is laid in the air, the working temperature is 30℃, and the current-carrying capacity under the long-term continuous 100% load is as follows:
4 square millimeters - 39A
Because it is a household, it is single-phase, and its maximum power (Pm) is:
Pm=voltage U×current I=220V×39A=8580W
Taking the safety factor as 1.3, then it works for a long time, and the allowable power (P) is:
P=Pm÷1.3=8580÷1.3=6600 watts
"Four squares" of copper wire. Can withstand a load of 6600 watts.
2. The electricity consumption of my house is four kilowatts. How much square copper wire do I need?
answer:
1. First calculate the single-phase current according to the size of the power,
Current (A) = equipment power (W) ÷ rated voltage (V) = 4000W ÷ 220V = 18.2A
2. According to the calculated current, multiply the coefficient 1.3 (mainly considering the overcurrent) to obtain the current value of the wire to be selected, and find out the cross-sectional value of the wire to be used from the "Electrical Manual".
18.2A×1.3＝23.7A
