Analysis of the empirical formulas commonly used in electrical construction

1. Current estimation of electrical equipment: When the power of the electrical equipment is known, its rated current can be estimated. The rated current of the three-phase motor is calculated according to the motor power twice, that is, the current per kilowatt is 2 is the current of the rated current, for example. A three-phase motor is rated at 10 kW and has a rated current of 20 amps. This estimation method is the closest to the three-phase squirrel-cage asynchronous motor, especially the four-stage, and can be used for other types of motors. Single-phase 220V motor is calculated according to 8A per kilowatt current; three-phase 380V electric welding machine is calculated according to 2.7A per kilowatt current (with motor type DC welding machine should be calculated according to 2A per kilowatt); single-phase 220V welding machine is calculated according to 4.5A per kilowatt; Single-phase incandescent lamp and iodine-tungsten lamp are calculated according to 4.5A per kilowatt; Note: The common xenon lamp on the construction site is 380V power supply (only two phase lines, one ground wire), and the current is calculated according to 2.7A per kilowatt.

2. The rated current calculation of the three-phase motor of different voltage levels: the capacity is divided by the kilovolt number, and the quotient coefficient is seven or six.

Note (1) The port is suitable for the calculation of the rated current of three-phase motors of any voltage class. The formula and the port can indicate that the rated currents of motors with different voltage levels are different, that is, the voltage is not the same, and the same capacity is removed. The resulting "quotient" is obviously different, and the quotient is different. The number is multiplied by the same coefficient of 0.76, and the resulting current values ​​are also different. If the above port is called a general port, it can be deduced to calculate the rated current for the 220, 380, 660, 3.6kV voltage class motor. When calculating the rated current of a three-phase motor with a dedicated calculation port, the capacity is kilowatt and current amperes. The relationship is directly multiplied, eliminating the capacity divided by kilovolts and the quotient multiplied by a factor of 0.76. Three-phase two hundred and two motors, 5,000 amps. Commonly used three hundred and eight motors, one kilowatt two amperes. Low-voltage six hundred and six motors, kWh two amps. High-voltage three-kV motor, four kW-ampere. High voltage six kilovolt motor, eight kilowatts an ampere.

(2) When using the mouth, the unit of capacity is kW, the unit of voltage is kV, and the unit of current is A. This point must be noted.

(3) The coefficient of 0.76 in the mouth is a comprehensive value considering the calculation of the motor power factor and efficiency. The power factor is 0.85 and the efficiency is not 0.9. These two values ​​are more suitable for motors with tens of kilowatts or more, and are larger for commonly used motors of 10 kW or less. Therefore, the rated current of the motor calculated by the port c is inaccurate with the value marked on the nameplate of the motor. This error has little effect on the rated current of the motor below 10 kW, the contactor, the wire, and the like.

(4) Using oral calculation skills. When calculating the rated current of a commonly used 380V motor with a port, first use the motor to connect the power supply voltage to 0.38kV to remove 0.76, quotient 2 to multiply the capacity (kW). In the case of a 6kV motor with a large capacity, the capacity kW is exactly a multiple of 6kV, then the capacity is divided by the kilovolts and the quotient is multiplied by the 0.76 factor.

(5) Error. The coefficient of 0.76 in the mouth is calculated by taking the motor power factor of 0.85 and the efficiency of 0.9. Therefore, there is an error in calculating the rated current of the motor with different power factor and efficiency. The five dedicated ports derived from the port c, the multiple of the capacity (kW) and the current (A), are the quotient of the number of voltage levels (kV) minus 0.76. The special port is easy to calculate, but it should be noted that the error will increase. Generally, the current is larger than the one on the nameplate; the smaller the kilowatt is, the calculated current is slightly smaller than that on the nameplate. In this regard, when calculating the current, when the current reaches more than ten amps or tens of amps, it is not necessary to count after the decimal point. It can be rounded up and rounded up, taking only integers, which is simple and does not affect the practicality. For a smaller current, you only need to count one decimal.

3. Measure the current to determine the no-load current of the nameplate motor and estimate its rated capacity. The capacity of the unlicensed motor is measured, and the value of the no-load current is measured. It is calculated by multiplying ten by eight, and is close to the level of kilowatts.

Note: The mouth is a three-phase asynchronous motor without a nameplate. I don't know what the kilowatt-number of the capacity is. It can be used to estimate the kilowatt-hour of the motor capacity by measuring the no-load current of the motor. 4. Know the transformer capacity, and find the rated current port capacity of each voltage level divided by the voltage value, and divide the quotient by six by ten. Description: Applicable to any voltage level. In daily work, some electricians only involve the calculation of the rated current of a transformer of one or two voltage levels. By simplifying the above port, it is possible to derive the port for calculating the rated current of each voltage level side: the capacity factor is multiplied. 5. Knowing the capacity of the transformer, the current value of the first and second protection fuses (commonly known as fuses) is matched with the high-voltage fuse, and the capacity voltage is compared. Distribution transformer low-voltage fuse-link, the capacity multiplied by 9 divided by 5. Explain that the correct selection of the fuse-link has a great relationship to the safe operation of the transformer. When only fuses are used for high and low voltage side protection of the transformer, the correct selection of the melt is more important. This is a problem that electricians often encounter and have to solve. 6. Measure the secondary current of the power transformer, calculate the load capacity of the port, and calculate the secondary voltage of the transformer. The voltage level is four hundred volts, one ampere and six kilowatts. The voltage level is three thousand volts, one ampere of 4.5 kilowatts. The voltage level is six kilovolts, an integer of nine kilowatts. The voltage level is ten kilovolts, one to fifteen kilowatts. The voltage level is 35,000 and 1 amp is 55 kW. Explain that in the daily work, the electrician often encounters the higher-level department, the management personnel asks about the operation of the power transformer, what is the load? The electrician himself often needs to know what the load of the transformer is. The load current is easy to know. Look directly at the ammeter set on the power distribution unit, or use the corresponding clamp-type ammeter to measure the load power, which cannot be directly seen and detected. This depends on the calculation of the mouth, otherwise it is complicated and time-consuming to calculate with the conventional formula.

7. Measure the current of the incandescent lighting line and calculate its load capacity.

The lighting voltage is two hundred and twenty, one and two hundred and twenty watts.

Explain the lighting of industrial and mining enterprises, and use 220V incandescent lamps. The lighting power supply line refers to the line from the distribution board to each lighting distribution box. The lighting supply main line is generally three-phase four-wire, and single-phase can be used when the load is below 4kW. Lighting distribution lines are lines that are connected from lighting distribution boxes to lighting fixtures such as luminaires or outlets. Regardless of the power supply or distribution line, as long as the current value of a certain phase line is measured with a clamp-type ammeter, and then multiplied by 220 coefficient, the product number is the load capacity of the phase line. Measuring the current capacity, can help the electrician to quickly adjust the three-phase load capacity imbalance of the lighting main line, can help the electrician to analyze the reason why the protective melt is often blown in the distribution box, the cause of the heating of the distribution wire, and so on. 8, known 380V three-phase motor capacity, seeking its overload protection thermal relay components rated current and setting current port 诀

Motor overload protection, thermal relay thermal components; two-and-a-half of the stream capacity, twice the kilowatts setting.

Note (1) Motors that are easily overloaded may fail to start due to severe starting or self-starting conditions, or if the starting time needs to be limited, overload protection shall be installed. It is also advisable to install overload protection for long-running unsupervised motors or motors of 3 kW and above. The overload protection device generally uses a time delay overcurrent release of a thermal relay or a circuit breaker. At present, the thermal relays produced in China are suitable for light load starting, motor overload protection for long-term work or intermittent long-term work. (2) Thermal relay overload protection device, the structural principle is very simple, but the optional heat regulating component is very delicate. If the grade is selected, it has to be adjusted to the lower limit, which often causes the motor to stop, affecting production and increasing maintenance work. If the level is chosen to be small, it can only be adjusted to a high limit. When the motor is overloaded, it will not operate or even burn the motor. (3) Correctly calculate the overload protection thermal relay of 380V three-phase motor. It is necessary to clarify that the same series of thermal relays can be equipped with thermal components with different rated currents. The setting current of the hot element is set according to “double kWh”; the rated current of the hot element is selected according to “two and a half times of the current capacity”; the model specification of the thermal relay, that is, its rated current value should be greater than or equal to the rated current value of the thermal element.

9. Measure the no-load current of the 380V single-phase welding transformer without nameplate, and calculate the rated capacity of the base 诀 380 welder capacity, and the no-load current is multiplied by five. The single-phase AC welding transformer is actually a special-purpose step-down transformer, and its basic working principle is roughly the same as that of the ordinary transformer. In order to meet the requirements of the welding process, the welding transformer works in a short-circuit state, and it is required to have a certain arc-ignition voltage during welding. When the welding current increases, the output voltage drops sharply. When the voltage drops to zero (ie, the secondary side is short-circuited), the secondary side current is not too large, etc., that is, the welding transformer has a steep drop external characteristic, and the welding transformer is The steep drop characteristic is obtained by the voltage drop generated by the reactance coil. At no load, since no welding current passes, the reactance coil does not generate a voltage drop. At this time, the no-load voltage is equal to the secondary voltage, that is to say, the welding transformer is unloaded at the same time as the ordinary transformer is no-load. The no-load current of the transformer is generally about 6%~8% of the rated current (the nationally specified no-load current should not exceed 10% of the rated current). This is the theoretical basis for the oral and formula.


10. Calculating the current carrying capacity of the wire The current carrying capacity of the wire is related to the wire cross section. It is also related to the material, type, laying method and ambient temperature of the wire. The factors affecting it are more and the calculation is more complicated. The current carrying capacity of various conductors can usually be found in the manual. But with the use of mouth and then with some simple mental arithmetic, you can directly calculate, do not have to look up the table. 1. The relationship between the current carrying capacity of the aluminum core insulated wire and the cross-section of the cross-section is 10, five, 100, two, 25, 35, four, three, 70, 95, two and a half. Pipe, temperature, eight or nine fold. Add half of the bare wire. Copper wire upgrade count. It is indicated that the current carrying capacity (A) of various cross-sections is not directly indicated, but is multiplied by a certain multiple of the cross-section. To this end, the nominal cross-section (square mm) of commonly used wires in China is arranged as follows: 1, 1.5, 2.5, 4, 6, 10, 16, 25, 35, 50, 70, 95, 120, 150, 185... (1) The first sentence indicates that the aluminum core insulated wire current carrying capacity (A) can be calculated as a multiple of the cross section. The Arabic numerals in the mouth indicate the wire cross section (square millimeters), and the Chinese character numbers indicate multiples. The relationship between the cross section of the mouth and the multiple is arranged as follows: 1 to 1016, 2535, 5070, 95120 or more, 5 times, 4 times, 3 times, 2 times and 2 times, and now it is more clear with the mouth and mouth, and the mouth is "10 to 5" means The cross-section is below 10, and the current carrying capacity is five times the cross-sectional value. "100 upper two" (reading two hundred and two) means that the current carrying capacity of 100 or more in cross section is twice the cross-sectional value. Sections 25 and 35 are four and three times the boundary. This is the mouth of "25, 35, four or three circles." The cross-sections 70 and 95 are two and a half times. It can be seen from the above arrangement that, except for 10 or less and 100 or more, the cross section of the middle wire is the same multiple for each of the two specifications. For example, aluminum core insulated wire, the calculation of the current carrying capacity when the ambient temperature is not more than 25 ° C: when the cross-section is 6 mm 2 , the calculated carrying capacity is 30 amp; when the cross-section is 150 mm 2 , the calculated carrying capacity is 300 amp; When the cross section is 70 mm 2 , the current carrying capacity is 175 amps. It can also be seen from the above arrangement that the multiple decreases as the cross section increases, and the error is slightly larger at the junction of the fold transition. For example, sections 25 and 35 are four times and three times the boundary, 25 is four times the range, which is calculated as 100 amps by mouth, but 97 amps according to the manual; and 35 is opposite, 105 amps by mouth, but Check the table for 117 amps. However, this has little effect on the use. Of course, if you can "have a number in the chest", when you select the cross section of the wire, 25 does not let it fill 100 amps, and 35 can be slightly more than 105 amps more accurate. Similarly, the 2.5 mm square wire is at the beginning of five times, and the actual is more than five times (up to 20 amps or more). However, in order to reduce the power loss in the wire, the current is usually not used so much. 12 amps.

(2) The last three sentences are the treatment of conditional changes. "Pipe, temperature, eight or nine fold" means: if the pipe is laid (including the laying of the trough plate, that is, the wire is covered with a protective cover layer, it is not clear), after the calculation, another 20% off; if the ambient temperature exceeds 25 ° C, after the calculation, then make a 10 fold, if the pipe is laid, the temperature is more than 25 ° C, then hit a 20% discount and then 10% off, or simply press once to make a 30% discount calculation. Regarding the ambient temperature, the average maximum temperature of the hottest month in summer is specified. In fact, the temperature is variable. In general, it affects the current carrying of the wire and is not very large. Therefore, discounts are only considered when there are more than 25 °C in some warm workshops or hotter areas. For example, the calculation of the download flow rate of the aluminum insulated wire under different conditions: when the cross section is 10 square millimeters, the current carrying capacity is 10×5×0.8═40 amp; if it is high temperature, the current carrying capacity is 10×5×0.9. ═ 45 amp; if the tube is hot and high, the current carrying capacity is 10 × 5 × 0.7 ═ 35 amps. (3) For the current carrying capacity of bare aluminum wire, the mouth indicates that “naked wire plus half” is calculated and then added half. This means that the same cross-section bare aluminum wire is compared with the aluminum core insulated wire, and the current carrying capacity can be increased by half. For example, the calculation of the current carrying capacity of bare aluminum wire: when the cross-section is 16 square millimeters, the current carrying capacity is 16×4×1.5═96 amps. If the temperature is high, the current carrying capacity is 16×4×1.5×0.9=86.4 amps. . (4) For the current carrying capacity of the copper wire, the port indicates “the copper wire upgrade calculation”, that is, the cross-sectional arrangement order of the copper wire is increased by one step, and then calculated according to the corresponding aluminum wire condition. For example, the bare copper wire with a cross-section of 35 square millimeters has an ambient temperature of 25 ° C, and the current carrying capacity is calculated as: 50 × 3 × 1.5 = 225 amps by upgrading to a bare aluminum wire of 50 square millimeters. For the cable, it is not described in the mouth. Generally, the high-voltage cable directly buried in the ground can be directly calculated by using the relevant multiple in the first sentence. For example, a 35 mm square high-voltage armored aluminum core cable is buried at a current carrying capacity of 35 × 3 = 105 amps. 95 square millimeters is about 95 x 2.5 ≈ 238 amps. The zero-line cross section in the three-phase four-wire system is usually selected as about 1/2 of the phase line cross section. Of course, it must not be less than the minimum cross-section allowed by mechanical strength requirements. In a single-phase line, since the load currents passing through the neutral line and the phase line are the same, the zero line cross section should be the same as the phase line cross section.

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