# Empirical formula commonly used in the hottest ele

2022-08-13
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Experience formula commonly used in electrical construction

in order to promote the discussion of electrical professionals about the capacity, current and other problems encountered in design and construction, some commonly used calculation rules and experience formulas are provided to you after sorting. I hope you will actively discuss and jointly improve:

I Current estimation of electric equipment: when you know the power of electric equipment, you can estimate its rated current:

the rated current of three-phase motor is calculated by twice the motor power, that is, multiplying each kW by 2 is the current amount of rated current. For example, if the rated power of a three-phase motor is 10 kW, the rated current is 20 amps. This estimation method is the closest to three-phase squirrel cage asynchronous motors, especially level 4 motors. For other types of motors, it can also be calculated by 8A per kW current of single-phase 220V motors

current per kW of three-phase 380V welding machine is calculated by 2.7A (DC welding machine with motor should be calculated by 2A per kW)

single-phase 220V welding machine is calculated by 4.5a per kW

single-phase incandescent lamp Iodine tungsten lamp current per kilowatt is calculated as 4.5a

note: dysprosium lamp commonly used on the construction site is 380V power supply (only two phase wires and one ground wire), and current per kilowatt is calculated as 2.7A. Calculation of rated current of three-phase motor with different voltage levels

formula: capacity divided by kilovolts, quotient factor point 76

Description:

(1) the formula is applicable to the calculation of rated current of three-phase motor at any voltage level. From the formula and formula, it can be explained that the rated current of motors with the same capacity and different voltage levels is different, that is, the voltage kilovolts are different. If the same capacity is removed, the quotient obtained is obviously different. If different quotients are multiplied by the same coefficient of 0.76, the current value obtained is also different. If the above formula is called general formula, a special formula for calculating the rated current of motors with 220, 380, 660 and 3.6kV voltage levels can be derived. When calculating the rated current of a three-phase motor with the special formula, the exhibition area shows the achievements of automobile lightweight plastic technology and innovative materials science and technology. The relationship between capacity kW and current ampere is directly multiplied, eliminating the division of capacity by kilovolts, and the quotient is multiplied by 0.76

three phase two hundred two motors, kW 3.5 amps

commonly used 380 motors, one kW two amps

low voltage 660 motor, kW 1.2 amps

high voltage three thousand volt motor, four kilowatts and one ampere

high voltage 6kV motor, eight kilowatts and one ampere

2) when using the formula, the unit of capacity is kW, the unit of voltage is kV, and the unit of current is A. This must be noted

(3) the coefficient 0.76 in the formula is the comprehensive value calculated by considering the motor power factor and efficiency. The power factor is 0.85, and the efficiency is not 0.9. These two values are applicable to motors above tens of kW, but they are larger for commonly used motors below 10kW. Therefore, there is an error between the rated current of the motor calculated by formula C and the value marked on the motor nameplate. This error has little impact on the rated current of the motor below 10kW

(4) use pithy calculation skills. When calculating the rated current of a commonly used 380V motor with a formula, first multiply the capacity (kw) by 0.76 and quotient 2 with the power supply voltage 0.38kv of the motor. In case of 6kV motor with large capacity, the kW of capacity is exactly a multiple of 6kV, then divide the capacity by kV and multiply the quotient by 0.76 coefficient

(5) error. The coefficient 0.76 in the formula is calculated by taking the motor power factor as 0.85 and the efficiency as 0.9, so there is an error in calculating the rated current of motors with different power factors and efficiencies. For the five special formulas derived from formula C, the multiple of capacity (kw) and current (a) is the quotient of the number of voltage levels (kV) minus the coefficient of 0.76. The special pithy formula is simple and easy to calculate, but it should be noted that its error will increase. Generally, if the kilowatt number is large, the calculated current is slightly larger than that on the nameplate; For those with smaller kilowatts, 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 amperes or tens of amperes, it does not need to be calculated after the decimal point. It can be rounded without rounding, and only take an integer, which is simple and does not affect practicality. For smaller currents, it is only necessary to calculate to one decimal place. 3、 Measure the current to calculate the capacity

measure the no-load current of the motor without nameplate, estimate its rated capacity

formula:

the capacity of the motor without nameplate, measure the no-load current value,

multiply ten by eight to calculate, close to the kilowatt number of the grade

Description: the pithy formula is for a three-phase asynchronous motor without nameplate. If you don't know its capacity kW, you can estimate the motor capacity kW by measuring the no-load current value of the motor

IV. given the transformer capacity, calculate the rated current on the side of each voltage grade.

formula:

divide the capacity by the voltage value, and multiply it by six divided by ten

Description: applicable to any voltage level

in daily work, some electricians are only involved in the calculation of transformer rated current of oneortwo voltage levels. Simplifying the above formula, we can deduce the formula for calculating the rated current at the side of each voltage grade:

multiply the capacity coefficient. 5、 Given the transformer capacity, quickly calculate the current value of its primary and secondary protection fuse links (commonly known as fuses)

formula:

distribution transformer high-voltage fuse links, capacity and voltage comparison

low voltage fuse link of distribution transformer, capacity multiplied by 9 divided by 5

Description:

correct selection of fuse links is of great importance to the safe operation of transformers. When the fuse is only used as the 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 need to solve

VI. measure the secondary side current of the power transformer and calculate its load capacity.

formula:

know the secondary voltage of the distribution transformer, and measure the current to calculate kW

voltage level 400 V, 1 a 0.6 kW

the voltage level is 3000 volts, one ampere, 4.5000 watts

voltage level: 6kV, one ampere integer: 9kw

the voltage level is 10kV, one ampere and fifteen thousand watts

voltage level is 35000, one ampere, fifty-five thousand watts

Description:

electricians often encounter higher-level departments and managers in their daily work, asking about the operation of power transformers and the load? Electricians often need to know the load of the transformer. The load current is easy to know. Directly look at the ammeter set on the power distribution device, or measure it with the corresponding clamp ammeter. The load power can not be seen and measured directly. This needs to be calculated by this pithy formula, otherwise it will be complex and time-consuming to calculate with conventional formula

VII. Measure the current of incandescent lamp lighting line, and calculate its load capacity

lighting voltage is 220, one ampere, 220 watts

note: 220V incandescent lamps are mostly used for lighting in industrial and mining enterprises. The lighting power supply line refers to the line from the distribution board to each lighting distribution box. The lighting power supply main line is generally three-phase four wire, and single-phase can be used when the load is below 4kw. Lighting distribution line refers to the line from lighting distribution box to lighting facilities such as illuminators or sockets. No matter the power supply or distribution line, as long as the current value of a phase line is measured with a clamp ammeter, and then multiplied by the 220 coefficient, the product is the load capacity of the phase line. Measuring the current and calculating the capacity can help the electrician quickly adjust the imbalance of three-phase load capacity of the lighting trunk line, and help the electrician analyze the reasons for the frequent fusing of the protective melt in the distribution box and the heating of the distribution wire, etc

VIII. Given the capacity of 380V three-phase motor, calculate the rated current and setting current of its overload protection thermal relay element

pithy formula:

motor overload protection, thermal relay thermal element

two and a half times of the current capacity and twice the kilowatt number setting

Description:

(1) motors that are prone to overload, which may fail to start due to serious starting or self starting conditions, or need to limit the starting time, should be equipped with overload protection. Overload protection should also be installed for motors that operate unsupervised for a long time or motors of 3KW and above. The overload protection device generally adopts the time-delay overcurrent release of thermal relay or circuit breaker. At present, the thermal relay produced in China is suitable for motor overload protection of light load starting, long-term operation or intermittent long-term operation

(2) the structure and principle of thermal relay overload protection device are very simple, but the optional heat regulating elements are very subtle. If the grade is selected to be large, it must be adjusted to the low limit, which often causes the motor to stop secretly, affects production, and increases maintenance work. If the grade is selected to be small, it can only be adjusted to the high limit, and the motor often does not act when overloaded, or even burns the motor. (3) To correctly calculate and select the overload protection thermal relay of 380V three-phase motor, it is still necessary to understand that the thermal relay of the same series of models can be equipped with thermal elements with different rated currents. The setting current of the thermal element is set by twice the kW; The rated current of the thermal element is calculated and selected according to two and a half times of the signal current capacity; The model and specification of thermal relay, that is, its rated current value should be greater than or equal to the rated current value of thermal element

IX. measure the no-load current of 380V single-phase welding transformer without nameplate, calculate the base rated capacity

pithy formula:

three hundred and eight welding machine capacity, and multiply the no-load current by five

single phase AC welding transformer is actually a step-down transformer for special purposes, and its basic working principle is roughly the same as that of ordinary transformer. In order to meet the requirements of the welding process, the welding transformer works in the short-circuit state, and it is required to have a certain arc striking voltage during welding. When the welding current increases, the output voltage drops sharply. When the voltage drops to zero (that is, the secondary side is short circuited), the secondary side current will not be too large, etc., that is, the welding transformer has the external characteristics of steep drop, which is obtained by the voltage drop generated by the reactance coil. When no-load, because no welding current passes through, the reactance coil does not produce voltage drop. At this time, the no-load voltage is equal to the secondary voltage, that is, when the welding transformer is no-load, it is the same as when the ordinary transformer is no-load. The no-load current of the transformer is generally about 6% to 8% of the rated current (the no-load current specified by the State shall not be greater than 10% of the rated current). This is the theoretical basis of formulas and formulas

in order to promote the electrical professionals to discuss the capacity, current and other problems encountered in the design and construction, some commonly used calculation rules and experience formulas are provided to you after finishing. I hope you will actively discuss and jointly improve:

I Current estimation of electric equipment: when you know the power of electric equipment, you can estimate its rated current:

the rated current of three-phase motor is calculated by twice the motor power, that is, multiplying each kW by 2 is the current amount of rated current. For example, if the rated power of a three-phase motor is 10 kW, the rated current is 20 amps. This estimation method is the closest to three-phase squirrel cage asynchronous motors, especially level 4 motors. For other types of motors, it can also be calculated by 8A per kW current of single-phase 220V motors

current per kW of three-phase 380V welding machine is calculated by 2.7A (DC welding machine with motor should be calculated by 2A per kW)

single-phase 220V welding machine is calculated by 4.5a per kW

single-phase incandescent lamp Iodine tungsten lamp current per kW is calculated as 4.5a

note: dysprosium lamp commonly used on the construction site is 380V power supply (only two phase wires and one ground wire

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