Calculation of current by power - Solutions - Huaqiang Electronic Network

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1. Use:
This is the calculation of the current (A) according to the power (kW or kVA) of the electrical equipment.
The magnitude of the current is directly related to power, and is also related to voltage, phase, and force rate (also called power factor). Generally, there are formulas for calculation. Since the factory is usually a 380/220 volt three-phase four-wire system, the current can be directly calculated according to the power.
2. Mouth: The current per KW of the low voltage 380/220 volt system is safe.
Kilowatt, current, how to calculate?
The power is doubled and the electric heat is added half.
Single phase kilowatts, 4.5 amps.
Single phase 380, current two and a half.
3. Description: The port is based on the three-phase equipment in the 380/220V three-phase four-wire system, and the number of amps per kW is calculated. For some single-phase devices with different single-phase or voltage, the number of amps per kW is additionally described.
1 In these two sentences, electric power refers to the motor. When the 380V three-phase (force rate is about 0.8), the current per kW of the motor is about 2 amps. That is, the doubling of the kilowatts (by 2) is the current, . This current is also called the rated current of the motor.
[Example 1] The 5.5 kW motor has a current of 11 amps according to "power doubling".
[Example 2] The 4 kW pump motor calculates the current as 80 amps according to "power doubled".
Electrothermal means an electric resistance furnace heated by electric resistance or the like. Three-phase 380 volt electric heating equipment, the current per kilowatt is 1.5 amps. That is, "half watts plus half" (by 1.5), is the current, safety.
[Example 1] The 3 kW electric heater has a current of 4.5 amps according to "electric heating plus half".
[Example 2] The electric current of 1 5 kW is calculated to be 2 3 amps according to "electric heating plus half".
This sputum does not specifically refer to electric heating, but also applies to lighting. Although the illuminated bulb is single-phase rather than three-phase, the three-phase four-wire trunk that supplies the illumination is still three-phase.
As long as the three phases are substantially balanced, this can also be calculated. In addition, electrical appliances (such as transformers or rectifiers) in kilovolt-amperes and phase-shifting capacitors in kilowatts (for increasing force) are also suitable. That is to say, although the latter part of the sentence is about electric heating, it includes all electrical equipment in kilovolt-amperes, thousands of units, and electric heating and lighting equipment in kilowatts.
[Example 1] 1 2 kW three-phase (balanced) The lighting main line is calculated as "electric heating plus half" and the current is 1 8 amps.
[Example 2] The rectifier of 30 kVA is calculated to have a current of 45 amps by "electric heating plus half". (refers to the 380 volt three-phase AC side)
[Example 3] The distribution transformer of 3 2 0 kVA is calculated as “electric heating plus half” and the current is 480 amps (referring to the low voltage side of 380/220 volts).
[Example 4] A 100-kilowatt phase-shift capacitor (380 volts three-phase) has a current of 150 amps according to "electric heating plus half".
2. In the 380/220 volt three-phase four-wire system, two lines of single-phase equipment, one connected to the phase line and the other connected to the zero line (such as lighting equipment) are single-phase 220 volt power equipment. The force rate of this type of equipment is mostly 1, so the mouth is directly stated as "single phase (each) kW 4.5 amps". When calculating, just “multiply the kilowatt number by 4.5” is the current, safety. As above, it is suitable for all single-phase 220 volt electrical equipment in kilovolt-amperes, as well as electrothermal and lighting equipment in kilowatts, and also for 220 volts DC.
[Example 1] A 500 volt (0.5 kVA) line light transformer (220 volt power supply side) has a current of 2.3 amps as "single phase (each) kW 4.5 amps".
[Example 2] The 1000-watt floodlight has a current of 4.5 amps according to "single-phase kilowatts, 4.5 amps". For single-phase with lower voltage, it is not mentioned in the mouth. You can take 220 volts as a standard to see how much the voltage is reduced and how much the current is increased in turn. For example, the voltage of 36 volts, which is reduced to 1/6 by 220 volts, should be increased to 6 times, that is, the current per kilowatt is 6 × 4.5 = 27 amps. For example, a 36 volt, 60 watt line of light has a current of 0.06 × 27 = 1.6 amps, and 5 amps have a total of 8 amps.
3 In the 380/220 volt three-phase four-wire system, the two lines of the single-phase device are connected to the phase line, which is customarily called a single-phase 380-volt power device (actually connected to two phase lines). When the device is in kilowatts, the force rate is mostly 1, and the mouth is also directly stated: "single phase 380, current is two and a half". It also includes 380 volt single-phase equipment in kilovolt-amperes. When calculating, just multiply the kilowatt or kilovolt ampere by 2.5 is the current, safety.
[Example l] The 32 kW molybdenum wire resistance furnace is connected to a single phase of 380 volts, and the current is 80 amps according to the current of two amps.
[Example 2] A 2 kVA lamp transformer, the primary is connected to a single phase of 380 volts, and the current is 5 amps according to the current.
[Example 3] 21 kVA AC welding transformer, the primary phase is 380 volts, and the current is 53 amps according to the current.
Note 1: The current is calculated according to “power doubled”, and there is some error with the current on the motor nameplate.
Generally, the kilowatt is larger, the calculated current is slightly larger than that on the nameplate, and the kilowatt is smaller, the calculated current is slightly smaller than that on the nameplate. In addition, there are some factors that affect the current, however, As an estimate, the impact is not large.
Note 2: When calculating the current, when the current reaches more than ten amps or tens of angstroms, it is not necessary to count to the decimal point and can be rounded to an integer. This is simple and does not affect the practicality. For a smaller current, it is only necessary to calculate a decimal number.