【20150817】How do people understand selecting the cable gauge based on current?

The carrying capacity of a cable is determined by the current flowing through it. Different cable diameters allow different amounts of current to pass, which means that the cable diameter must be determined based on the user’s current requirements (i.e., the load capacity).

Could you send a specific table of low-voltage and high-voltage cables along with their current conductor diameters?

Based on the power of the electrical devices, after calculating the total power, use the formula I=P/U and then multiply by a coefficient of 0.85~! If it’s too troublesome, then it’s a current of 2 amps per kilowatt~! It is the most universal, as it includes the current capacity of the discharge. 1KW=2A. There are also methods for selecting cables; calculations are based on current. The simple selection algorithm provided below takes aluminum-core wires as the basis for calculation: ten minus five, one hundred plus two; 25, 35, 43 for different ranges, and 70, 95 represent twice and a half times that value! This is a mnemonic: For BLV wires with a cross-sectional area of 10 square millimeters or less, the current they can carry is five times the wire’s diameter~! For BLV wires of 100 square millimeters or more, it is twice the wire diameter capable of carrying the current. The BLV current ratings for 25mm2 and 35mm2 are at 4 times and 3 times the breaking capacity, respectively. The current capacity of 70mm2 and 95mm2 is 2.5 times that of the wire diameter. In addition to this, for wires with copper cores, the calculation is based on the upgrade factor for aluminum wires; that is, BV-10mm2 is calculated using the current capacity of BLV-16mm2, and the same principle applies to others. When wires are enclosed in plastic or PVC pipes, the calculated current value must be multiplied by a factor of 0.8. If the wires are enclosed in steel pipes, the calculated current should be multiplied by 0.9. When wires pass through areas with high temperatures, the calculated current value must be multiplied by 0.7. If a wire is subject to all three of these conditions, first multiply by 0.9 then by 0.7, or it’s also possible to use a direct factor of 0.85. For four-core or five-core cables, the current value should be multiplied by both 0.85 and 0.7. For exposed overhead power lines, the calculation is simpler, with a factor of 0.9 applied; however, the environment also plays a role, and applying a 20% discount is a safer approach. When selecting cables, it is also necessary to consider their intended use based on the conditions at the site; for example, ordinary YJV cables are used inside cable trays. Cables with armor can be buried directly and can withstand external forces. Armored cables with tensile resistance are used in high-rise buildings and installed by direct burial. If you don’t understand these things, take a look at books on 35KV electrical engineering; they contain information on the commonly used cable types as well as the electrical equipment. The general safe calculation method for copper wires is: the safe current-carrying capacity of a 2.5 square millimeter copper power wire is 28A. The safe current-carrying capacity of a 4 mm² copper power cable is 35A. The safe current-carrying capacity of a 6 mm² copper power cable is 48 A. The safe current-carrying capacity of a 10 mm² copper power cable is 65A. The safe current-carrying capacity of a 16 mm² copper power cable is 91 A. The safe current-carrying capacity of a 25 square millimeter copper power cable is 120A. If it is an aluminum wire, its diameter should be 1.5–2 times that of a copper wire. If the current in the copper wire is less than 28A, using a value of 10A per square millimeter is definitely safe. If the current in the copper wire is greater than 120A, use 5A per square millimeter.

Give LS a thumbs up – it’s so comprehensive. Also, LS is an estimate and not very accurate, but it’s generally fine to use it this way. The cross-sectional area can be calculated based on the wire length, allowable voltage drop, and maximum current; there is a formula for accurate calculation. I haven’t memorized it in ages; it’s all the LS method’s fault :L

Please explain in detail why it is said this way.

Copper wire: S = IL / 54.4 * U`; Aluminum wire: S = IL / 34 * U’. Where: I – the maximum current flowing through the wire (A); L – the length of the wire (m); U’ – the allowable voltage drop (V); S – the cross-sectional area of the wire (mm2). This information was obtained from Baidu; these are calculated values, and the actual selection should be based on common cross-sectional areas