First, the line overload fire causes the electric current in the electric circuit, due to the existence of the conductor's own resistance will generate a certain amount of heat, its size is Q = I2RT. After the wire is selected, the more the load, the greater the current I power, and the more heat generated in the wire. If the current generated by the conductor is equal to the heat dissipated outside the conductor, the conductor remains relatively thermally balanced and the conductor temperature does not rise. If the current generated by the current in the conductor is greater than the heat dissipated by the conductor, the accumulation of heat will cause the temperature of the conductor to rise. Generally, the maximum temperature at which the wire works is 65 ° C. At this time, the current in the wire is a safe current, and the current is overloaded. Long-term overloading of the wire can age the insulation or cause flammable materials adjacent to the wire to ignite. The experiment proves that when the current in the wire exceeds the safe current by 2 times, the core temperature can reach 300 ° C. At this time, the odor can be smelled, the bleak of the local insulation layer of the wire is separated from the core, and even smoke occurs locally; When the safety current is 2.5 to 3 times, the core temperature can reach 700 ° C or more, and the core becomes red and the insulation layer is on fire. On the other hand, due to the long-term high temperature of the insulating layer, the organic components are gradually carbonized, and the carbonized portion may form a semiconductor, which may reduce the degree of wire insulation, which may cause short circuit between the line and the line, or leakage, and further Increase the load and produce higher temperatures. This creates a vicious circle that causes a fire. Ac Gear Motor,Ac Geared Motor,Asynchronous Ac Motor,Asynchronous Ac Gear Motor NingBo BeiLun HengFeng Electromotor Manufacture Co.,Ltd. , https://www.hengfengmotor.com
Second, the relationship between line current and load line heating and current square is proportional. The current I and the load power P have the following relationship: P=IU, U is the voltage, and is a certain value for a certain line. It can be seen that the larger the power, the more the load, the larger the current, the larger the heat generation, and the greater the fire hazard.
For several consumers in a line, the sum of their respective powers is the power of the entire line. That is: P = P1 + P2 + P3 + ... + Pn; the sum of the respective branch currents is the total current of the entire line. That is: I = i1 + i2 + i3 + ... + in, from which it seems that the more total line parallel electrical appliances, the greater the total power; the greater the total current, the more heat the line sends.
Generally, single-phase electrical appliances have rated power and rated voltage, and their rated current can be calculated by I=P/U. For three-phase electrical appliances, P=IUCOSφ can be used, but the theory is strong and the calculation is inconvenient. Here is an experience. In the formula; the three-phase electrical current is about twice the equivalent power kilowatt, and the total current of the entire line can be calculated by accumulating the current of each branch.
Third, the wire load capacity is based on Q=I2RT. When the load in the line is constant and the current is constant, the heat of the wire is proportional to the resistance of the wire itself. It is known by R=ÏL/S (Ï is the wire resistivity, L is the wire length, S is the wire cross-sectional area). The larger the cross-sectional area of ​​the wire, the larger the resistance, the less heat, and the lower the fire risk. However, the thicker the wire, the higher the cost. In addition to considering the heating conditions in the power transmission and distribution process, factors such as cost, wire strength, and voltage loss are considered. The safe current carrying capacity of different sections is specified in the professional books such as the electrician's manual. Now introduce a simple mouth can easily calculate the safe current carrying capacity of the wire.
The cross-sectional area of ​​the wire is specified (in mm2):
1, 1.5, 2.5, 4, 6, 10, 16, 25, 35, 50, 70, 95, 120, 150, 185 and other specifications.
The general summary of the summary is:
Tenth five, one hundred and two, 25, 35, four three circles;
Two and a half times 70, 95, the tube temperature is 8, 10% off,
Half of the bare wire is added, and the copper wire is upgraded.
The mouth is based on aluminum insulated conductors. The specific meanings are as follows:
Tenth Five: The safe current carrying capacity of the aluminum core wire below 10mm2 is five times the wire carrying area. For example, a 4mm2 aluminum wire has a safe current carrying capacity of 4 x 5 = 20 amps;
One hundred and two: The safe current carrying capacity of the aluminum wire above 100mm2 is twice the cross-sectional area. For example, a 150mm2 aluminum cable has a safe current carrying capacity of 150×2=300 amps;
25, 35 four three boundaries: 16mm2 and 25mm2 safe current carrying capacity is four times the cross-sectional area; 35mm2, 50mm2 is three times the cross-sectional area; such as 50mm2 aluminum cable safe current carrying capacity is 50 × 3 = 150 amps;
70, 95 two and a half: 70mm2, 95mm2 aluminum cable has a safe current carrying capacity of 2.5 times its cross-sectional area.
The pipe penetration temperature is 8 or 9 fold: if the wire is protected by the pipe, the safety current is reduced to 80% when the use level is higher; if the cable ambient temperature exceeds 25 °C, the safe current capacity is 90% of the original. If the pipe is high temperature, the safe current carrying capacity is 80% × 90% = 72%;
Add half of the bare wire: If the cable is uninsulated bare wire, its heat dissipation condition is good, and the safe current carrying capacity is increased by half. For example, 4mm2 aluminum wire insulation safe current carrying capacity is 4 × 5 (10 pm) = 20 amps, if the bare wire, its safe current capacity is increased by half to 20 + 20 × 1, 2 = 30 amps.
Copper wire upgrade calculation: Because the copper wire has good electrical conductivity, its safe current capacity is calculated at the level of the aluminum wire. For example, a 2.5mm2 insulated copper wire has a safe current carrying capacity equivalent to 4mm2 of aluminum wire of 4×5=20 amps, and so on.
According to the above port, it is convenient to calculate the safe current carrying capacity of the wire. Calculate the total load current according to the previous formula, and calculate whether the wire is overloaded. For example, the total electric appliance of a line is 94600 watts, the current is I=94600/220=430 amps, and the actual cable is 70mm2 aluminum bare cable. The actual safe current of the line is 70×2.5 (two times half of 70 and 95) +70× 2.5 × 1, 2 (naked line plus half) = 330 amps. The line load current 430A is greater than the line safety current 330A, the line temperature is higher than 65 ° C, and the line is overloaded.
4. Measures to prevent overloading of the line 1. Reasonably select the cross-sectional area of ​​the conductor and the type of the conductor when laying the circuit, and leave a certain margin when the conditions permit, in order to expand the production and increase the capacity.
2. Eliminate private connection and strictly control the load. Strengthen the management of temporary power consumption, regularly check the lines, and dismantle them on schedule. It is strictly forbidden to use the "one land, one line" system and bare wires to privately tie the trees or buildings. It is not allowed to increase the load privately without permission.
3. Regularly check the line for leakage and equipment increase and decrease. At least twice a year, measure the resistance of the line and equipment to ground, and find that the leakage is resolved in time.
4. The main line and each branch line are equipped with corresponding insurance, automatic switch or leakage protection device to cut off the power supply in time, reduce the accident range and prevent fire.