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Compiled by an expert in electrical engineering – Basic knowledge of wires and cables (reposted). What is the maximum current that can flow through a 1mm2 power cable, and what is the corresponding power? For example, for 2.5 mm2 wires, how is it determined what size of wire is needed in construction projects? ① For wires with cross-sectional areas of 1.5, 2.5, 4, 6, and 10 mm2, multiply their cross-sectional area by 5. ②For 16 and 25 mm2 wires, multiply their cross-sectional area by 4. ③For 35 and 50 mm2 wires, multiply their cross-sectional area by 3. ④For 70 and 95 mm2 wires, their cross-sectional area can be multiplied by 2.5. ⑤For wires of 120, 150, and 185 mm2, their cross-sectional area can be multiplied by 2. At an operating temperature of 30°C, the current-carrying capacity under a continuous 90% load for a long period is as follows: 1.5 square millimeters – 18A; 2.5 square millimeters – 26A; 4 square millimeters – 26A; 6 square millimeters – 47A; 10 square millimeters – 66A; 16 square millimeters – 92A; 25 square millimeters – 120A; 35 square millimeters – 150A. The power is calculated as P = voltage U × current I = 220 volts × 18 amps = 3960 watts. According to the national standard GB4706.1-1992/1998, the allowable current values for wires are as follows (for some cases): For copper-core wires, the allowable continuous current depends on the cross-sectional area of the wire: 2.5 square millimeters – 16A to 25A; 4 square millimeters – 25A to 32A; 6 square millimeters – 32A to 40A. For aluminum-core wires, the allowable continuous current is: 2.5 square millimeters – 13A to 20A; 4 square millimeters – 20A to 25A; 6 square millimeters – 25A to 32A. [Example] 1. Each computer consumes approximately 200–300 watts of power (about 1–1.5 amps). Therefore, 10 computers would require a copper-core wire with a cross-sectional area of 2.5 square millimeters to supply power; otherwise, a fire might occur. 2. Three large air conditioners consume approximately 3000W of power (about 14A), so each air conditioner requires its own copper wire with a cross-sectional area of 2.5 square millimeters for power supply. 3. The wiring used for power supply to homes these days is generally 4 square millimeter copper wire; therefore, the total power consumption of household appliances connected simultaneously should not exceed 25A (i.e., 5500 watts). It is pointless to replace the wires in a house with 6 square millimeter copper wire, as the wire that reaches the electricity meter is already 4 square millimeters in diameter. 4. In early housing (15 years ago), the incoming wires were usually 2.5 square millimeter aluminum wires; therefore, the number of household appliances that could be turned on at the same time could not exceed 13A (i.e., 2800 watts). 5. Home appliances with high power consumption include: air conditioners at 5A (1.2 horsepower), electric water heaters at 10A, microwave ovens at 4A, rice cookers at 4A, dishwashers at 8A, washing machines with drying function at 10A, and electric kettles at 4A. In fires caused by electrical issues, 90% are resulting from overheating of connectors; therefore, all connectors must be welded. Contact components that cannot be welded need to be replaced every 5–10 years (such as sockets and circuit breakers). According to national standards, the allowable long-term current is 25–32 A for 4-square millimeter wires, and 32–40 A for 6-square millimeter wires. These are actually theoretical safety values; the actual maximum values are higher than these. For copper wires with a cross-sectional area of 2.5 square millimeters, the maximum power that can be handled is 5,500 watts; for 4 square millimeter wires, it’s 8,000 watts, and for 6 square millimeter wires, 9,000 watts is no problem. A digital meter with a capacity of 40A can handle 9,000 watts without any issues, and mechanical meters can handle up to 12,000 watts without being damaged. There’s a rule for determining the current-carrying capacity of copper wire cables: the estimation formula is – multiply 2.5 by 9, then decrease the result by 1 for each additional unit in cross-sectional area. Thirty-five times 3.5, pair them up and subtract 0.5. If conditions change, a discount is applied; at high temperatures, copper upgrades at a 10% discount. Two, three, four roots in the tube; 80%, 70%, 60% capacity for current conduction. Note: This section’s mnemonic does not specify the current-carrying capacity (safe current) for various insulated wires (rubber- and plastic-insulated wires) directly; instead, it is expressed as \"the cross-sectional area multiplied by a certain factor\", which can be calculated mentally. “\"2.5 down multiplied by 9, then subtract one and move forward in sequence\" refers to aluminum-core insulated wires with cross-sections of 2.5 mm or less; their current-carrying capacity is approximately 9 times the value of their cross-sectional area. For a 2.5mm’ wire, the current-carrying capacity is 2.5×9=22.5(A). The multiple relationship between the current-carrying capacity and the number of cross-sections for wires of 4 mm² and above is arranged in ascending order of wire size, with the multiple decreasing by 1 each time; that is, 4×8, 6×7, 10×6, 16×5, 25×4. “\"Fifteen times three point five, in pairs subtracted by five,\" means that the current-carrying capacity of a 35mm” wire is 3.5 times the area of its cross-section; that is, 35×3.5=122.5(A). For wires of 50 mm and above, the multiplicative relationship between current-carrying capacity and cross-sectional area changes such that wire sizes are grouped in pairs, with the multiplier decreasing by 0.5 for each subsequent pair. That is, the current-carrying capacity of 50 and 70 mm² wires is 3 times their cross-sectional area; the current-carrying capacity of 95 and 120 mm² wires is 2.5 times their cross-sectional area, and so on. “If conditions change, a discount is applied; at high temperatures, a 10% discount applies and the copper grade is upgraded. The above formula is established for aluminum-core insulated wires installed exposed in an environment with a temperature of 25°C. If aluminum-core insulated wires are installed exposed in areas where the ambient temperature remains above 25°C for extended periods, the current-carrying capacity of such wires can be calculated using the method outlined above, with a 10% reduction applied to the resulting value. When copper-core insulated wires are used instead of aluminum wires, their current-carrying capacity is slightly higher than that of aluminum wires of the same specification; in such cases, the current-carrying capacity can be determined by using the aforementioned method and increasing the wire size by one grade. For example, the current-carrying capacity of a 16 mm² copper wire can be estimated based on that of a 25 mm² aluminum wire. Calculating the cable current-carrying capacity to select a cable (selecting a cable based on current): The current-carrying capacity of a conductor is related to its cross-sectional area, as well as factors such as the material and type of the conductor, the method of installation, and the ambient temperature. There are many influencing factors, making the calculations complex. The current-carrying capacity of various wires can usually be found in manuals. But by using a mnemonic along with some simple mental arithmetic, the calculation can be done directly without needing to consult a table. 1. Mnemonic for the multiple relationship between the current-carrying capacity of aluminum-core insulated wires and their cross-sectional area: 5 times for values below 10, 2 times for values above 100; for 25 and 35, it’s 4 or 3 times; for 70 and 95, it’s 2.5 times. Tube insertion, temperature: 20-30% discount. Add half of the bare wire. Copper wire upgrade counts. It should be noted that the mnemonic does not directly indicate the current-carrying capacity (in amps) for various cross-sections; instead, it is expressed by multiplying the cross-sectional area by a certain factor. To this end, the nominal cross-sectional areas (in square millimeters) of the commonly used wires in our country are listed as follows: 1, 1.5, 2.5, 4, 6, 10, 16, 25, 35, 50, 70, 95, 120, 150, 185……) (1) The first rule states that the current-carrying capacity of aluminum-core insulated wires (in amperes) can be calculated by multiplying it by the cross-sectional area. The Arabic numerals in the mnemonic represent the wire cross-section (in square millimeters), while the Chinese numerals indicate the multiple. Arranging the cross-sectional areas and the corresponding multiple values according to the mnemonic gives us: 1–10, 16, 25, 35, 50, 70, 95, 120. These represent five times, four times, three times, two and a half times, and two times the cross-sectional area respectively. By comparing this with the mnemonic, it becomes clearer; the phrase “five times for cross-sectional areas below 10” means that for cross-sectional areas less than 10, the current-carrying capacity is five times the value of the cross-sectional area. “\"100 upper two\" (read as hundred upper two) refers to the current-carrying capacity at a cross-section of over 100 being twice the value corresponding to that cross-section on www.gczjy.com. Slices of 25 and 35 are the boundaries between four times and three times. This is the mnemonic: “25, 35, the four and three realms.” While sections 70 and 95 are two and a half times. As can be seen from the above arrangement: except for values below 10 and above 100, the cross-sectional area of the wires in the middle follows the same multiple pattern for every two specifications. For example, in the case of aluminum-core insulated wires, the calculation of the current-carrying capacity at an ambient temperature of no more than 25°C is as follows: when the cross-sectional area is 6 square millimeters, the calculated current-carrying capacity is 30 amps; when it is 150 square millimeters, the capacity is 300 amps; and when it is 70 square millimeters, the capacity is 175 amps. It can also be seen from these figures that the multiplier decreases as the cross-sectional area increases, and there is a slightly larger error at the points where the multiplier changes. For example, cross-sections 25 and 35 represent the boundary between four times and three times the value. Section 25 falls within the four-times range; according to the formula, it should be 100 amps, but according to the manual it is 97 amps. Conversely, for section 35, the formula gives 105 amps, while the table indicates 117 amps. However, this has little impact on usage. Of course, if one has a clear idea of the requirements, it would be more accurate to ensure that for a 25 mm² wire the current rating does not reach 100 amps, while for a 35 mm² wire the current rating can slightly exceed 105 amps. Similarly, for a 2.5 square millimeter wire, the position is at five times the starting point; in practice the current is more than five times that value (it can reach over 20 amps). However, to reduce power losses within the wire, such high currents are generally not used, and the manuals usually specify only 12 amps. (2) The next three lines of the mnemonic are for dealing with changes in conditions. “For installation in tubes or at elevated temperatures, a discount of 20% or 10% applies: if the wires are installed in tubes (including installations using troughs or similar methods, meaning the wires have a protective covering and are not exposed), a 20% discount is applied after the calculation; if the ambient temperature exceeds 25°C, a 10% discount is applied after the calculation. If the wires are installed in tubes and the temperature is also above 25°C, then a 20% discount followed by a 10% discount is applied, or alternatively, a 30% discount can be applied directly. Regarding the ambient temperature, it is defined as the average maximum temperature of the hottest month in summer. In fact, the temperature varies, and under normal circumstances, its impact on the current-carrying capacity of the wire is not significant. Therefore, a discount is only considered in certain warm workshops or hotter areas where the temperature exceeds 25°C significantly. For example, the calculation of the current-carrying capacity of aluminum-core insulated wires under different conditions: when the cross-sectional area is 10 square millimeters and the wire is installed in a conduit, the current-carrying capacity is 10×5×0.8 = 40 amps; in high-temperature conditions, it is 10×5×0.9 = 45 amps; and when the wire is in a conduit and exposed to high temperatures, the current-carrying capacity is 10×5×0.7 = 35 amps. (3) Regarding the current-carrying capacity of bare aluminum wires, the rule of thumb states to \"add half to the value for bare wires\", that is, add half to the calculated value. This means that, compared to aluminum wires with an exposed core, insulated aluminum wires of the same cross-section can have a current-carrying capacity that is half as high. For example, in calculating the current-carrying capacity of bare aluminum wires: when the cross-sectional area is 16 square millimeters, the current-carrying capacity is 16×4×1.5 = 96 amps; at high temperatures, it is 16×4×1.5×0.9 = 86.4 amps. (4) Regarding the current-carrying capacity of copper wires, the rule of thumb states that \"upgrade the copper wire grade\", that is, increase the cross-sectional category of the copper wire by one level and then calculate based on the corresponding conditions for aluminum wires. For example, for a bare copper wire with a cross-sectional area of 35 square millimeters and an ambient temperature of 25°C, the current-carrying capacity is calculated as follows: if it is upgraded to a bare aluminum wire with a cross-sectional area of 50 square millimeters, then the value becomes 50×3×1.5=225 amps. For cables, it is not covered in the mnemonic. For high-voltage cables that are generally buried directly in the ground, it is generally possible to use the multipliers mentioned in the first mnemonic phrase for the calculations. For example, the current-carrying capacity of a 35 square millimeter high-voltage armored aluminum-core cable laid underground is 35×3=105 amps. 95 square millimeters is approximately 95×2.5≈238 amps. In a three-phase four-wire system, the cross-sectional area of the neutral wire is usually chosen to be around 1/2 of the cross-sectional area of the phase wires. Of course, it must also be no smaller than the minimum cross-section permitted by the mechanical strength requirements. In a single-phase circuit, since the load current flowing through the neutral wire and the phase wires is the same, the cross-sectional area of the neutral wire should be the same as that of the phase wires.