导线的安全电流

导线的安全电流 Safe current carrying capacity of conductors Table 1: current carrying capacity of PVC insulated copper wire or cable with different laying methods under steady-state conditions at ambient temperature (I) Laying method B1 B E Cross sectional area /mm2 current carrying capacity (I) A 0.75 - 7.6 1, 10.4, 9.6, 11.7, 11.5 1.5, 13.5, 12.2, 15.2- 16.1 2.5, 18.3, 16.52122 425232830 635293637 1044405052 1660536670 2577678488 359083104114 50 - 123123 70 - 155155 95 - 192192 120 - 221221 Method B1: wire (single core cable) is placed and protected by conduit and cable duct. Method B1: same as B1, but used for multicore cable. Method C: no conduit or conduit, cable hung on the wall side. Method E: the cable is mounted horizontally or vertically on the open cable carrier. Table 1 presents the current carrying capacity of PVC insulated conductors at ambient temperature. For other ambient temperatures, the installer shall correct the current carrying capacity values in accordance with the coefficients given in table 2. Table 2: correction factor Correction factor of ambient temperature / temperature 301.15 351.08 401 450.91 500.82 550.71 600.58 Note: current carrying capacity parameters and correction coefficients are derived from GB 5226.1-2002/IEC 60204-1:2000 Selection of cable cross section Source: Electrician's house time: 2011-03-23 read: 138 times Tab: voltage, current, cable cross section Estimation method of cable cross section Load current is estimated first 1. uses This is based on the power of electrical equipment (kW or kVA) to calculate the current (ANN) formula. The size of the current is directly related to power, voltage, and also don't rate (also known as power factor and so on). Formulas are generally available for calculation. Because the factory is usually 380/220 volt three-phase four wire system, so you can directly calculate the current according to the size of the power. 2. formula Low voltage 380/220 volt system, current per kilowatt, ann. KW, current, how to calculate? Power doubling, electric plus half. 1. Single-phase kilowatt, 4.5 ann. 2 Single phase 380, current two a.. 3 3. instructions The formula is 380/220 volts three-phase equipment in three-phase four wire systems prevail, calculated on the number of kilowatt hour. For some single or different voltage single-phase equipment, the per kilowatt security number, also explains the formula. The two formulas, the power to the motor. At 380 volts three-phase (rate 0.8), the current per kW motor is about 2. The number of kilowatts to be doubled (by 2) is current, ann. This current is also called the rated current of an electric motor. [case 1] 5.5 kW motor by "double power", the current is 11. [case 2] 40 kW water pump motor is doubled by "electric power", and the current is 80. Electric heating is a resistance furnace, such as a resistor. A three-phase 380 volt electric heating device with a current of 1.5 kilowatts per kilowatt. About half of the kilowatt count (by 1.5) is current, ann. [case 1] 3 kilowatts of electric heaters press "electric plus half", and the current is 4.5. [case 2] 15 kW resistance furnace according to "electric plus half", the current is 23. This sentence does not specifically refer to formula is also suitable for heating, lighting. Although the lighting bulb is single-phase instead of three-phase, the three-phase four wire trunk for lighting is still three-phase. This can be calculated as long as the three phases are in general equilibrium. In addition, the KVA unit for electrical appliances (such as transformer or rectifier) and kilovar units of phase-shifting capacitors (with increasing rate) are also applicable. The second half of the sentence immediately said, though the heat, but includes all the units, kilovar KVA electrical equipment, as well as in kilowatts unit heating and lighting equipment. [case 1] 12 kilowatts of three-phase (balanced) lighting mains, according to "electric plus half", the current is 18. [case 2] 30 thousand volt ampere rectifier by "electric plus half", the current is 45 ampere (refers to 380 volts three-phase AC side). [case 3] 320 thousand volt distribution transformers, according to "electric plus half", the current is 480, ANN (refers to 380/220 volts low side). [4] 100 cases kilovar phase-shifting capacitors (380 volts three-phase) by electric current is 150 plus half "is. In the 380/220 volt three-phase four wire system, two lines of single-phase equipment, one phase line and another one connected to zero line (such as lighting equipment) are single-phase 220 volt power equipment. This equipment rate mostly 1, therefore, it directly shows the single formula (per) 4.5 kilowatts". Calculation, as long as "the number of kilowatts by 4.5" is the current, ann. The same as above, it applies to all single-phase 220 volt power equipment with kva, As well as heating and lighting units in kW, and also for 220 volts dc. [case 1] 500 volt ampere (500 volt ampere) line lamp transformer (220 volt power side) press "single-phase kilowatt, 4.5 Ann", calculate current is 2.3. [case 2] 1000 watt light, according to "single-phase kilowatt, 4.5 Ann", the current is 4.5. For the single phase voltage is lower, no mention in the formula. You can take the 220 volt as the standard to see how much the voltage drops and the current increases in turn. For example, 36 volts voltage, to 220 volts as the standard, it is reduced to 1/6, the current should be increased to 6 times, that is, the current per kilowatt of 6*4.5=27. For example, 36 volts, 60 watts of row lights, each current is 0.06*27=1.6, ANN, 5, a total of 8 security. In the 380/220 volt three-phase four wire system, the two lines of the single-phase equipment are connected to the phase line, which is customarily called the single-phase 380 volt power equipment (actually connected to the two phases). When the equipment in kilowatts units, mostly 1 rate, but also directly explained: "formula 380 single-phase current two, half". It also includes 380 volt single-phase equipment in KVA units. As long as the kilowatt or KVA is multiplied by 2.5, the current is the ampere. [case 1] 32 kW molybdenum wire resistance furnace, single-phase 380 volts, press "current two and a half", the current is 80. [example 2] 2 thousand volt line transformer, primary phase 380 volts, press "current two and a half", the current is 5. [case 3] 21 thousand volt alternating current welding transformer, primary connection single-phase 380 volts, press "current two and a half", calculate current is 53. To estimate the load current in the current section according to the selected corresponding wire, there are several aspects to consider is the mechanical strength of the conducting wire two is the current density of wire selection wire section (safety cut-off flow), three is the allowable voltage drop Estimation of voltage drop 1. uses According to the load torque on the line, the voltage loss on the power supply line is estimated, and the power supply quality is checked. 2. formula The benchmark data an estimation of voltage loss, through some simple calculations, which can estimate the voltage loss on power supply line. Press loss according to "kilowatt meter", 2.5 aluminium wire 20 - 1. The section increases, the load moment is large, and the voltage is lower than the square. 1. Three phase four wire, 6 times meter, copper wire multiply 1.7. 2 The inductive load high pressure loss, under section 10 influence is small, in terms of rate 0.8, 10 increase from 0.2 to 1. 3 3. instructions The calculation of voltage loss is related to many factors, and the calculation is complicated. When estimating, the line has selected the lead and the section according to the load situation, namely the related condition already basically has. The voltage loss is measured by "a few percent of the rated voltage". The main list of formulas for voltage loss estimation the most basic data, the number of "load moment" voltage loss will be 1%. When the load moment is large, the voltage loss increases correspondingly. For some reason, the load moment of the line should be worked out first. The so-called load moment is the load (kw) multiplied by the length of the line (the length of the line is the length of the laying line of the guide line "m", that is, the path of the traverse, irrespective of the number of wires leading to the line). ) the unit is kW m.". For a radial line, the calculation of the load moment is simple. The load moment at the installation position of the equipment shall be counted as follows: starting from the line power point, according to the line branch, divide it into three segments. In each segment of the line, three loads (10, 8, 5 kW) are passed, so the load torque is: Paragraph 1: 10* (10+8+5) =230 kW M. Second paragraph: 5* (8+5) =65 kW M. Third paragraph: 10*5=50 kW M. The total load moment at the 5 kW facility is 230+65+50=345 kW M. The following formulas are described: First of all, the basic basis for calculating voltage loss is load torque: kW M. Next, a baseline data is presented: 2.5 mm square wire, 220 volt single-phase load, resistive (rate of 1), every 20 kW. M "load moment voltage loss is 1%. This is the 2.5 wire 20 formula - 1. When the voltage loss is 1%, the load moment of the cross section is larger, and it varies according to the direct ratio. For example, 10 square millimeter aluminum wire, the cross section is 2.5 square millimeter 4 times, then 20*4=80 kilowatt meter, namely this kind of conductor load moment is 80 kilowatt meters, the voltage loss is only 1%. The rest of the sections follow analogies. When the voltage is not 220 volts, but other values, such as 36 volts, it is gray 36 volts, equivalent to 220 volts 1/6. At this point, the load torque of this line voltage loss of 1% is not 20 kW meters, but should be reduced by 1/6 square, 1/36, which is 20* (1/36) =0.55 kW m.. That is to say, at 36 volts, the voltage loss is reduced by 1% per 0.55 kW M. (i.e., 550 watts per watt). "Voltage reduction square low" is not only applicable to lower rated voltage, but also can be applied to higher rated voltage. At this point the square rises. For example, single-phase 380 volts, due to the voltage of 380 volts to 220 volts 1.7 times, so the voltage loss of 1% load torque should be 20*1.7 square =58 kilowatts. As can be seen from the above formula: "section increases load torque and reduce low voltage square". The reference data is "2.5 aluminium wire 20 - 1". [case 1] a 220 volt light outlet with 2.5 mm2 aluminium wire with a load moment of 76 kW m.. Since 76 is 3.8 times 20 (76/20=3.8), the voltage loss is 3.8%. [case 2] a 4 metre long aluminium wire laying 40 metre long line, supplying 220 volts, 1 kW single-phase arc furnace 2, estimating voltage loss is: Load torque 2*40=80 kW m.. Calculate again 4 square millimeter aluminium wire voltage to lose 1% load moment, according to "section increases, load moment big" principle, compare 4 and 2.5, cross section increases to 1.6 times (4/2.5=1.6), accordingly, load moment increases 20*1.6=32 kW meter (this is 1% of the voltage loss). Finally, the 80/32=2.5 is calculated, that is, the line voltage loss is 2.5%. When the line is not single-phase, but three-phase four lines, the three-phase four wire generally requires three-phase load is balanced. Its voltage is corresponding to the single-phase. If the phase is 220 volts, the corresponding three-phase is 380 volts, or 380/220 volts. The same is 2.5 square millimeter of aluminium wire, and the load moment of the voltage loss 1% is 6 times of the datum data, that is 20*6=120 kw M. As for the cross section or voltage change, the value of the load moment will change accordingly. When the wire is not aluminum, but copper wire, it should be the aluminum load torque data multiplied by 1.7, such as "2.5 aluminum 20 - 1" to the same section of copper wire, the load torque was changed to 20*1.7=34 kW M. the voltage loss was only 1%. [example 3] an example of an illumination branch in front of which, if copper wire, is 76/34=2.2, i.e., the voltage loss is 2.2%. For the circuit supplied by the electric furnace, if the copper wire is the 80/ (32*1.7) =1.5, the voltage loss is 1.5%. [case 4] a 50 volt three-phase wire laying 380 volt three-phase line, 30 meters long, supply a three-phase electric furnace of 60 kw. Voltage loss estimation is: Load torque: 60*30=1800 kW m.. Then calculate load moment 50 mm aluminum at 380 volts three-phase voltage loss case 1%: according to the "load moment section increases, since 50 is 20 times 2.5, so by 20, according to the" three-phase four line 6 times, and by 6, so the load moment increased to 20*20*6=2400 kilowatt meter. Finally, 1800/2400=0.75, that is, voltage loss of 0.75%. The above is based on the resistive load. For the inductive load (such as motor), calculation method is more complex than the above. But the formula first pointed out: the same load moment - kW. M, voltage loss of load impedance is higher than resistive. It is related to the size of the section and the distance between the laying of the wires. For 10 mm2 and below wire, the influence is smaller, can not increase. For a section of 10 mm2 or more, the line can be calculated as follows: first, calculate the voltage loss by one or two, and then increase "0.2 to 1", which means 0.2 to 1 times increase, then multiply 1.2 to 2. This can be determined according to the size of the section, the larger section of the larger. For example, 70 mm 1.6150 mm by a mere 2. The above refers to the line overhead or stent case. For cables or wear lines, because the line distance is small is not affected, still according to the provisions, the estimate will not increase or only for large section conductors increase slightly (less than 0.2). [case 5] if the 20 kW is 380 volt three-phase motor, the line is 3*16, the aluminum wire bracket is exposed, then the voltage loss is estimated to be: the known load moment is 600 kW m.. The calculation section 16 square mm aluminum 380 volt three-phase load moment, voltage loss of 1%: since 16 is 6.4 times 2.5, three-phase load moment is 6 times of single phase, thus increasing load moment: 20*6.4*6=768 meters 600/768=0.8 kilowatts. That is to say, the estimated voltage loss is 0.8%. But now it is the motor load, and the conductor cross section is more than 10, so it should be increased. According to the cross section, considering 1.2, and estimated to be 0.8*1.2=0.96, the voltage loss can be considered to be about 1%. The above is the estimation method of voltage loss. Finally, let's talk about some of these issues: First, the voltage loss on the line is large and the quality is not good. Generally, the principle of 7~8% is. (strictly speaking, the voltage loss is based on the rated voltage of the electrical equipment (such as 380/220 volts), allowing 5% (2.5% of the illumination) below this rated voltage. But the specified voltage busbar distribution transformer and the rated voltage is 5% higher than (400/230 V), so start from the transformer to the electrical equipment of the whole line, a total loss of 5%+5%=10% in theory, but usually only allow 7~8%. This is because the minus voltage loss of transformer and transformer due to the low rate. ) but this 7~8% is the entire line from the low-voltage side of the distribution transformer to the electrical equipment that is being calculated. It usually consists of outdoor overhead lines, indoor trunk lines and feeder lines. Should be the results of each paragraph added, all about 7~8%. Two, estimating voltage loss is the design work, mainly to prevent the quality of the voltage when the future use of poor phenomenon. Because there are many factors affecting the calculation (such as the main trunk load calculation accuracy, stability of the transformer power supply side voltage etc.), therefore, the calculation is very accurate in general have little significance, as long as you can. For example, the relationship between the cross section can be simplified to 4 to 2.5, 1.5 times, 6 to 2.5 times, 16 to 2.5 times to 6 times, 2.5 times. It would be more convenient to calculate. Three, in the estimation of motor line voltage loss, there is another situation is to estimate the motor voltage loss when starting. If the loss is too large, the motor will not start directly. Because the current when starting, low rate, general provisions when starting voltage loss of up to 15%. The starting voltage loss calculation is more complicated, but the calculation formula the results presented above can be used to determine the aluminum, 25 mm less than the general section if they meet the requirements of 5%, can also meet the direct starting requirements: 35, 50 mm square wire if the voltage loss in 3.5% to 70, can also be satisfied; and 95 mm square wire if the voltage loss is less than 2.5%, and can also be satisfied; 120 mm square wire if the voltage loss is less than 1.5. To be content with. These 3.5%, 2.5%, and 1.5% are just 5%, five, seventy percent off, and seven, so they can be simply as follows: "more than 35, seven, five, seventy percent off."". Four, if in use actually found the voltage loss is too large, affecting the quality of electricity, can reduce the load (will transfer part of load to other lighter lines, or an additional loop), or cross section segment increased (increased in front of the best route) to solve. For motor circuits, cables can also be used to reduce voltage losses. When the motor can not start directly, in addition to the above solutions, you can also use step-down starting equipment (such as star delta starter or self - coupling starter, etc.) to solve. Select the section according to the current 1. uses The flow of various wires (safe electricity) is usually found in the manual. But the use of formulas with some simple arithmetic, can be calculated directly, without look-up table. Wire section and wire cutting flow, and wire material (aluminum or copper) and type (insulated wire or wire laying method, etc.) (surface or through pipe etc.) and ambient temperature (about 25 DEG C or greater) and so on, there are many factors affecting the complicated calculation. 2. formula Relationship between cross sectional volume of aluminum core insulated wire and cross section: square of S (sectional) =0.785*D (diameter) 10 5100 two, 25, 35, four, 95 realms, 70, Two and a half times. 1. Tube and temperature 11% off. 2 Add half bare. 3 Copper upgrade. 4. 3. instructions Formula is conditions with aluminum core insulation wire, deposited in the environmental temperature of 25 DEG C for quasi. If the conditions are different, as otherwise stated. Insulated wires include various types of rubber insulated wires or plastic insulated wires. Formulas for various sections of the cut-off flow (current, an) is not directly pointed out, but with the "section multiplied by a certain factor" to express. To do this, one should be familiar with the arrangement of the wire cross section (square millimeters): 11.5, 2.5461016253550, 7095120150185...... The cross section of the aluminum core insulation line is usually started at 2.5, and the copper core insulated wire starts at 1. The bare aluminum wire starts at 16 and the bare copper wire starts at 10. This formula pointed out: aluminum core insulation wire cut flow, ANN, you can press the "many times" section to calculate the number of. Arabia digital said formulas in the wire section (mm2), Chinese characters digital times. The formula "section and multiple relationship" arranged following: ...... 10*5 16, 25*4 35, 45*3 70, and 95*2.5 120*2...... Now more and more clearly as the control, the original "10 under five" refers to the section from 10 below the cut-off flow is five times the number of cross section. "100 on two" refers to the cross section of more than 100, the flow is two times the number of sections. Sections 25 and 35 are four and three times the dividing line. This is the "25, 35 and four three formulas". And the cross section 70 and 95 is two point five times. From the above arrangement it can be seen that in addition to the following 10 and more than 100, the middle section of the wire is often two kinds of specifications are of the same kind of multiples. The Ming aluminum core insulation wire, the ambient temperature is 25 DEG C, for example: [case 1] 6 square millimeter, press "10, next five" calculate, the interception flow is 30. [case 2] 150 square millimeters, press "100 on two", count the flow of 300. [case 3] 70 square millimeters, according to "70, 95, two and a half" count, the cut-off flow is 175. It can also be seen from the above arrangement that the multiples decrease with the increase of the cross section. At the junction of the multiple transition, the error is slightly greater. For example, section 25 and 35 is the boundary between the four and three times the range of 25 is four times, but close to the side of change to three times, which according to the formula is four times that of 100, but less than four times (according to the manual for the 97), and 35 on the contrary, according to the formula is three times 105 is 117, the actual security, but this did not affect the use of. Of course, if a few chest, in the lead section, 25 do not let it full to the 100, 35 can be slightly more than 105. It is more accurate. Also, the location of the line 2.5 square mm in five times of the beginning of the end (left), the actual is more than five times (up to 20 above), but in order to reduce the power loss in the wire, usually without so much, the manual also only 12. The following is from this, formulas for processing conditions change. Its name is "wear tube, temperature, eight or nine fold" refers to: if the pipe laying (including groove plate, etc., laying, that is, wire plus protective sleeve layer, unknown exposed), according to the calculation, and then call twenty percent off (by 0.8). If the ambient temperature is over 25 degrees centigrade, you should call "ten percent off" (0.9). About the environmental temperature, according to the Regulations refers to the highest average on the hottest summer temperatures. In fact, the temperature is changing, and in general, it affects the conductor closure is not very large. Therefore, only for certain high temperature workshops or hotter areas more than 25 degrees before considering discounts. Another situation is that the two conditions are changed (tube and temperature higher), then calculated by twenty percent off, and then call ten percent off. Or simply count thirty percent off at a time (0.8*0.9=0.72, about 0.7). This can also be said to "wear tube, temperature, eight or nine fold" mean. For example: (aluminum core insulated wire) 10 mm2, worn tube (twenty percent off), 40 ANN (10*5*0.8=40) High temperature (ten percent off) 45 ANN (10*5*0.9=45) Tube piercing and high temperature (thirty percent off) 35 ANN (10*5*0.7=35 ANN) 95 mm2, worn tube (twenty percent off) 190 ANN (95*2.5*0.8=190) High temperature (ten percent off) 214 ANN (95*2.5*0.9=213.8) Tube piercing and high temperature (thirty percent off) 166 ANN (95*2.5*0.7=166.3) The interception of bare wire, pointed out that "bare formula plus half", according to the calculation after half (by 1.5). This refers to the same cross section of the aluminum core insulation core, compared with the bare aluminum, the cut flow can be increased by half. [case 1] 16 mm2, bare aluminium wire, 96 an (16*4*1.5=96) High temperature, 86 An (16*4*1.5*0.9=86.4) [case 2] 35 mm2, bare aluminium wire, 158 an (35*3*1.5=157.5) [case 3] 120 mm2, bare aluminium wire, 360 an (120*2*1.5=360) For the interception of copper wire, copper is pointed out that the formula upgrade "section is copper wire section according to order level, and then calculate the corresponding condition of aluminum. [case 1] 35 mm2, bare copper wire, 25 deg.. Upgrade to 50 mm2, then press 50 mm2 of bare aluminum, 25 degrees to 225 50*3*1.5. [case 2] 16 mm2 copper insulated wire 25 degrees centigrade. According to the same condition of 25 mm2 aluminum insulated wire, it is calculated to be 100 A. (25*4). [case 3] 95 square millimeter copper insulated wire, 25 degrees, through tube. According to the same condition of 120 mm2 aluminum insulated wire, it is calculated to be 192 A. (120*2*0.8). Incidentally: for cable, not introduced in the formula. Generally, direct buried high-voltage cables can be calculated directly in terms of the relevant multiples. For example, 35 square meters of high voltage armored aluminum core cables are installed at about 105 35*3. 95 mm2, about 238 95*2.5. The estimation formulas and above have different approaches but equally satisfactory results: Multiply by two point five by nine, and go down a clockwise number. Thirty-five by three point five, both in group minus five points. Conditions have changed and converted, high temperature ten percent off copper upgrade. Intubation root number two three four, 87 forty percent off full flow. 2.5 square *9, 4 square *8, 6 square *7, 10 square *6, 16 square *5, 25 square *4, 35 square *3.5 50 and 70 square *3 95 and 120 square *2.5...... Finally, it is explained that the current estimation cross section is suitable for the near power supply (not far from the power supply) and the voltage drop is suitable for long distance.