导线截面积计算方法（Traverse area calculation method）

导线截面积计算方法（Traverse area calculation method） 导线截面积计算方法(Traverse area calculation method) Safe load flow of 2.5 square mm copper power cord - 28A. Safe load flow of 4 square mm copper power lines - 35A. Safe load flow of 6 square mm copper power line - 48A. Safe load flow of 10 square mm copper power wire - 65A. Safe load flow of 16 square mm copper power cord - 91A. Safe load flow of 25 square mm copper power lines - 120A. If it is aluminum, the diameter of the line is 1.5 to 2 times that of copper wire. If the copper current is less than 28A, it is safe to take 10A per square millimeter. If the copper wire current is greater than 120A, take 5A per square millimeter. The electric current that can be normally passed by the cross-sectional area of the conductor can be selected according to the total current amount that it needs to guide. Generally, it can be determined as follows: Ten, five, two, five, three, five, four, half, half, and copper wire upgrade. Explain to you, it is 10 square of aluminium wire, square millimeter number multiplied by five is ok, if the copper wire, and make a file, such as copper, 2.5 square is calculated at four square. More than one hundred are cross-sectional area multiplied by 2, below 25 square multiplied by 4, 35 square more than 3 times, suhuang and 95 square multiplied by 2.5, so a few words of formula should be very easy to remember, Description: only estimates, not exactly. In addition if according to indoor remember wire and copper wire under 6 square millimeter, every square current is not more than 10 a safe, from this perspective, you can choose to 1.5 square copper wire or aluminium wire 2.5 square. Within 10 meters, the current density of the wire is 6A/m2, which is suitable, 10-50 meters, 3A/square mm, 50-200 meters, 2A/square mm, and more than 500 meters is less than 1A/square millimeter. From this Angle, if it's not very far, you can choose a 4 square copper wire or 6 square aluminum wire. If it's really 150 meters of power supply (not tall buildings), you must use a four square copper wire. The impedance of a conductor is proportional to its length, inversely proportional to its diameter. When using the power supply, pay special attention to the input and output conductor's wire and line diameter problems. The accident was caused by an overheating of the ambassador wire. The wire diameter is generally calculated as follows: The copper wire: S = IL / 54.4 * U ` Aluminium wire: S = 34 * U ` IL / The maximum current (A) passing through the wire L -- length of wire (M) U ` - allowing power down (V) S -- the sectional area of the wire (MM2) Description: 1, U ` voltage drop can be used in the system as a whole range of devices (such as detector) for the system power supply with the power supply voltage rating taken together to choose. 2. The calculated sectional area depends on it. Estimation of flow rate of insulated conductor The relationship between the flow rate and the cross section of the aluminum core insulated conductor Wire section (mm 2) 1 1.5 2.5 4 6 10 16 25 35 50 70 95 120 The current load is a multiple of 9, 8, 7, 6, 5, 3.5, 3 and 2.5 Load flow (A) 9 14 23 32 48 60 90 100 123 238 300 Estimate the formula: take a number of nine to reduce the number. Thirty-five times three five, Both groups are reduced by five. The conditions have changed to add, the high temperature 90% copper upgrade. The tube root is two, three, four, eight, seven, six fold full load. Show: (1) this section formula of all kinds of insulated wire (rubber and plastic insulated wire) carrying capacity (safety current) is not directly points out, but "section multiplied by a certain ratio", by mental arithmetic. It can be seen from table 5 3 that the multiple decreases with the increase of the cross section. "Two point five take nine, go up to reduce one shun number to walk" say is 2.5mm "and below all kinds of section aluminium core insulated wire, its carrying capacity is about 9 times of cross section number. For example, the flow rate is 2.5 x 9 = 22.5 (A). The relationship between the load flow and the number of sections of the wires from 4mm to the above is the line number, which is the line number, which is the number of times, which is 4 times 8, 6 times 7, 10 times 6, 16 times 5, 25 times 4. "35 by three point five, both group minus five", said is 35 mm "wire carrying capacity for the number of section 3.5 times, namely, 35 * 3.5 = 122.5 (A). From more than 50 mm 'and conductor, its carrying capacity and the section number to multiples of relationship between two into A set of line number, multiples, in turn, minus 0.5. Namely, 50, 70 mm' conductor is three times the number of the carrying capacity of cross section, 95, 120 mm wire carrying capacity" is the cross section of the accumulation of 2.5 times, so on. "Conditions have changed to add, the high temperature 90% copper upgrade". The above formula is the aluminum core insulation wire, Ming fu in the conditions of the environment temperature 25 ?. If aluminum core insulation wire Ming fu in the ambient temperature above 25 ? region for a long time, wire carrying capacity according to the above formula calculation method to calculate, then ninety percent discount; When it is not aluminum wire, but copper insulated wire, it has a slightly larger flow rate than that of the same specification, but can calculate the load flow of a line number larger than the aluminum line. If the capacity of the copper wire is 16mm, it can be calculated according to the 25mm2 aluminum line. The copper wire is 6 amperes per square millimeter. Aluminium wire is 5 amps per square millimeter (for quick estimation) 4 square copper wire: 4 * 6 = 24A 6 square copper wire: 6 * 6 = 36A 10 square copper wire: 10 * 6 = 60A 16 square copper wire: 16 * 6 = 96A 4 square aluminum: 4 * 5 = 20A 6 square aluminum: 6 * 5 = 30A 10 square aluminum: 10 * 5 = 00A 16 square aluminum: 16 * 5 = 90A The capacity of transformer is known, and the rated current of each voltage grade Formula a: The capacity divided by the voltage value is multiplied by six divided by ten. Description: applicable to any voltage grade. In daily work, some electricians only involve a voltage rating of one or two types of transformer rated current. By simplifying the above rules, we can derive the formula for calculating the rated current of each voltage level. Multiply the volume coefficient. The capacity of the transformer is known as the current value of the first and second protection of the fused body (commonly known as fuse). Formula b: The voltage of the transformer is compared with that of the transformer. It's going to be a low pressure fuse, and it's going to be 9 divided by 5. Description: It is very important to choose the safe operation of the transformer. The correct selection of melt is more important when the transformer is only used for high and low voltage side protection. This is a problem that electricians often encounter and solve. The capacity of three-phase motor is known to be rated current Formula (c) : capacity divided by kilovolts, quotient point seven and six. Description: (1) the formula is applicable to the rated current of three phase motors of any voltage grade. By the formula and the formula can explain different capacity of the same voltage grade of the rated current of the motor is not the same, namely the kv voltage number, remove in the same capacity, the "quotient" is obviously not the same, is not the same quotient to take the same coefficient of 0.76, the current value is not the same. If the above formula is called the general formula, it can be deduced to calculate 220, 380, 660, 3.6 kV voltage and the rated current of the motor is special calculation formula, with a special calculation formula to calculate the rated current of a three-phase motor, kw direct relations with current ampere capacity multiples, eliminating the capacity divided by the number of kV, quotient by coefficient of 0.76. Three phase two hundred two electric motor, kilowatt three point five amperes. Usually three hundred and eight motors, one kilowatt and two ampere. Low pressure six hundred six motor, kilowatt hour two ampere. High voltage 3 kv motor, 4 kilowatt 1 ampere. High pressure 6 kv motor, 8 kilowatt 1 ampere. (2) when the formula c is used, the capacity unit is kW, the voltage unit is kV, and the current unit is A, which must be noted. (3) the coefficient 0.76 in the formula c is the comprehensive value of the calculation of motor power factor and efficiency. The power factor is 0.85, the efficiency is not 0.9, the two numerical comparisons apply to the electric motor which is more than a few hundred kilowatts, and it is larger for the commonly used 10kW motor. It would have to use the formula to calculate the rated current of a motor and c motor nameplate on the numerical error, the error of under 10 kw motor according to the rated current to switch, contactor, wire, etc. (4) using the technique of verbal formula. When calculating the rated current of the commonly used 380V motor, the power supply voltage of 0.38 kV was removed, and the number of quotient (kW) was removed. If the capacity of 6kV motor, the capacity kW number is a multiple of 6kV, then the capacity is divided by kilovolts, the quotient times 0.76 coefficient. (5) error. It is calculated that the power factor of the motor is 0.85 and the efficiency is 0.9, so the rated current of the motor with different power factors and efficiency is in error. The five special keys derived from the formula c, the volume (kW) and the multiple of current (A), are the quotient of the voltage grade (kV) to remove the 0.76 coefficient. The special formula is simple and easy to calculate, but it should be noted that the error will increase. The typical kilowatt is a little larger than the nameplate; The smaller number of kilowatts, the current is slightly smaller than the nameplate. In this case, when the current is calculated, when the current reaches more than ten or dozens of amps, it is not necessary to calculate the decimal point. It can be four and five, and it's only integers, so it's simple and it doesn't matter. For the smaller current, you just have to get a decimal