Mastering practical calculation formulas is an essential skill for electrical workers. However, with so many formulas available, finding the right one can be inconvenient. Below, we have compiled and collected some commonly used and practical formulas and mnemonics, along with examples and explanations.
I. Current Calculation for Lighting Circuits and Selection of Fuse Switches
Rule of thumb: To calculate the current of an incandescent lamp, divide the voltage by the power.
Calculate the current, power by voltage division, and power factor of fluorescent lamps (excluding energy-saving fluorescent lamps);
Knife switch fuses are easy to find; they should be 1.5 times the rated current.
Note: Incandescent lamps in lighting circuits are resistive loads with a power factor cosΦ=1. The rated current is equal to the power P (in watts) divided by the voltage. Fluorescent lamps are inductive loads with a power factor cosΦ of 0.4-0.6 (generally taken as 0.5), i.e., P/U/cosΦ=I.
Example 1: A lighting circuit has a rated voltage of 220V and a total power of 2200W for incandescent lamps. Calculate the total current and select the appropriate knife switch fuse.
Solution: Given U = 220V, total power = 2200W
Total current I = P/U = 2200/220 = 10A
Selecting the disconnect switch: QS = I × (1.1~1.5) = 15A
Select fuse: IR = I × (1.1~1.5) = 10 × 1.1 = 11A
(Take a coefficient of 1.1)
QS--------Disconnector
IR---------fuse
A: The circuit current is 10 amps, the knife switch should be 15 amps, and the fuse should be 11 amps.
Example 2: A lighting circuit has a rated voltage of 220V and is connected to a 440W fluorescent lamp. Calculate the total current and select the appropriate knife switch fuse. (cosΦ=0.5)
Solution: Given U = 220V, cosΦ = 0.5, total power = 440W
Total current I = P/U/cosΦ = 440/220/0.5 = 4A
Selecting the disconnect switch: QS = I × (1.1~1.5) = 4 × 1.5 = 6A
Select fuse: IR = I × (1.1~1.5) = 4 × 1.5 = 6A
A: The total current of the circuit is 4A, the knife switch should be 6A, and the fuse should be 6A.
Calculation of common 380V/220V loads
Mnemonic: Three-phase kilowatts equals twice the ampere, heat equals volt-ampere, one thousand volt-amperes equals 1.5.
A single phase is two times two and a half times four and five; if it is three and a half times eight.
Explanation: Three-phase kilowatt double ampere means a three-phase motor with a capacity of 1 kilowatt and a current of 2 amperes. 1 kV 1.5 refers to a three-phase electric heater, transformer, or capacitor with a capacity of 1 kilowatt, 1 kVA, and 1 kV capacitor current of 1.5 amperes. Single-phase double ampere means 22 times 45. If it is 38 times 2.5, it means a single-phase 220V single-phase electric welding machine with a capacity of 1 kilowatt and a current of 4.5 amperes, and a 380V single-phase electric welding machine with a capacity of 1 kVA and a current of 2.5 amperes.
Example 1: A three-phase asynchronous motor has a rated voltage of 380V, a capacity of 14 kW, a power factor of 0.85, and an efficiency of 0.95. Calculate the current.
Solution: Given U = 380VcosΦ = 0.85n = 0.95P = 14 kilowatts
Current I = P / (×U×cosΦ×n) = P / (1.73×380×0.85×0.95) = 28 (amps)
A: The current is 28 amperes.
Example 2: A three-phase 380V heater with a capacity of 10 kilowatts has a current draw.
Solution: Given U = 380VP = 10 kilowatts
Current I = P / (×U) = 10 / (1.73 × 0.38) = 15.21 (amps)
A: The current is 15 amps.
Example 3: A 380V three-phase transformer has a capacity of 20kVA. Find the current.
Solution: Given U = 380V, VS = 20KWA
Current I = S / (×U) = 20 / (1.73 × 0.38) = 30.45 (amps)
A: The current is 30 amperes.
Example 4: Given a BW0.4-12-3 capacitor, calculate the current.
Solution: Given U = 0.4 kV and Q = 12 kVAR
Current I = Q / (×U) = 12 / (1.73 × 0.4) = 17.3 (amps)
A: The current is 17.3 amperes.
Example 5: A single-phase 220V soldering iron with a capacity of 1 kilowatt, what is the current?
Solution: Given U = 220VP = 1000 watts
Current I = P/U = 1000/220 = 4.5 (amps)
A: The current is 4.5 amperes.
Example 6: A single-phase welding machine has a rated voltage of 380V and a capacity of 28 kVA. Find the current.
Solution: Given U = 380V and VS = 28 kVA
Current I = S/U = 28/0.38 = 73.6 (amps)
A: The current is 73.6 amperes.
Note: The above calculations assume a single-phase voltage of 220V, a power factor of cosΦ=1, and a current of 4.5 times the capacity. For single-phase equipment with a voltage of 380V, such as welding machines and portable lamp transformers, the current is 2.5 times the capacity.
III. Selection of Power Distribution Transformers
Rule of thumb: Select the capacity of a power transformer.
Total equipment load factor.
Then divide by the power factor and efficiency.
Note: Total power consumption refers to the sum of the power of all equipment in the factory. Simultaneity rate refers to the ratio of the actual capacity of equipment put into operation at the same time to the total capacity of power consumption equipment, which is generally chosen to be approximately 0.7.
Power factor: Generally selected as 0.8-0.9.
Efficiency: Typically 0.85-0.9
Power transformer capacity = Total capacity of electrical equipment × Simultaneity rate / Power factor of electrical equipment × Efficiency of electrical equipment
Example: A factory has a total power equipment capacity of 700 kVA, but the actual usage is 600 kVA. Find the capacity of the transformer used.
Simultaneous rate 600kVA ÷ 700kVA = 0.86
The capacity of the power transformer is calculated as follows: (700 × 0.86) / (cosΦ = 0.85 × 0.85).
=600/0.722=830 kVA
Note: If the power factor is 0.95 and the efficiency is 0.9, its capacity is:
The capacity of the power transformer = (700 × 0.7) / (cosΦ = 0.95 × n = 0.9)
=490/0.855=576kVA
For example: power factor = 0.7, n = 0.9, simultaneity rate = 0.75, calculate the capacity.
Power transformer capacity = (700 kVA × 0.75) / (0.7 × 0.9)
=525/0.63=833kVA
Based on the above analysis, the lower the power factor, the larger the power transformer should be, and vice versa. Therefore, it is necessary to improve the power factor in order to achieve the goal of energy conservation and consumption reduction.
IV. Calculation of Power Factor
Mnemonic: How to calculate the power factor
You can check the active and reactive power meters.
Calculate the electricity consumption for the month.
The power factor can then be calculated.
Note: Some companies overlook the importance of power factor. A low power factor can increase the company's electricity costs. To reduce electricity costs, the power factor must reach 0.9. How to improve the power factor will be discussed in the next section on how to calculate and compensate for improving the power factor.
The rule of thumb states: How to calculate the power factor? You can look at the active and reactive power meters. Calculate the monthly electricity consumption to find the power factor. Some enterprises and factories do not have power factor meters, watt meters, or reactive power compensation equipment in their power distribution systems. They only have voltmeters, ammeters, active power meters, and reactive power meters, so it is more difficult to calculate the power factor. You can calculate the monthly power factor by using the monthly electricity consumption data (kWh from the active power meter) and the monthly reactive power consumption data (kvar/h).
Example: If the active power consumption is 1000 kWh and the reactive power consumption is 300 kvar/h, calculate the power factor cosΦ for the month.
Solution: cosΦ = work done / = 1000 /
=1000/1044=0.957
If the active power consumption is 1000 kWh and the reactive power consumption is 750 kvar/h, calculate the power factor cosΦ for the current month.
cosΦ = active power / = 1000 /
=1/1.22=0.81
Note: The power factor of reactive power compensation in enterprises is generally between 0.7 and 0.85, and some are below 0.65. The smaller the motor power, the lower the power factor. A large motor pulling a small cart will result in a low power factor. Generally, induction motors account for 70% of cosΦ, power transformers account for 20%, and conductors account for 10%.
If there is a power factor meter on the distribution panel, the value can be directly displayed, or the instantaneous power factor can be calculated from the readings of the voltmeter, ammeter, and power meter on the distribution panel.
That is: cosΦ = P/(×U×I)
In the formula, P is the power meter reading (kW), U is the voltage reading (kV 0.38KV), and I is the ammeter reading (A).
V. Selection of Thermal Elements for Electric Motor Contactors
Rule of thumb: When selecting the current rating for a motor, double the rated current.
To calculate the current in an electric motor, one kilowatt equals one current.
When selecting a thermal element for the electric motor, calculate based on 1.2 times the rated current.
Set to one times the rated current, ensuring overload protection.
Explanation: An AC contactor is a control electrical device used to connect and disconnect the load current of a motor. Generally, the rated current of an AC contactor is selected based on 1.3-2 times the rated current of the motor. The rule of thumb, "Select the motor's rated current, calculate twice the rated current," means that the rated current of the AC contactor should be twice the motor's rated current. The rule of thumb, "Select the motor's thermal element, calculate 1.2 times the rated current, set it to 1 times the rated current, overload protection is guaranteed," means that the rated current of the motor's thermal element should be selected based on 1.2 times the motor's rated current, and setting it to 1 times the motor's rated current ensures overload protection of the circuit.
For example: A three-phase asynchronous motor has a rated voltage of 380 volts and a capacity of 10 kilowatts.
With a power factor of 0.85 and an efficiency of 0.95, calculate the motor current and select the AC contact heating element and its setting value.
Solution: (1) Empirical formula: 10 kW × 2 = 20 (A)
(2) Given U = 380VP = 10 kW, cosΦ = 0.85, n = 0.95
Current I = P / (×U×cosΦ×n)
=10/(1.73×0.38×0.85×0.95)=20 (amps)
Select an AC contactor: KM = Ke × (1.3 - 2) = 20 × 2 = 40 (amps)
Select CJ10-40
Selecting a heating element: FR = Ic × (1.1~1.25) = 20 × 1.25 = 25 (amps)
Select JR16-20/30, JR set to 20 amps.
A: The motor current is 20 amps, so select a 40 amp contactor. The rated current of the thermal element is 25 amps, so set it to 20 amps.
VI. Calculation of safe current for insulated conductors
Rule of thumb (1): Ten down five, one hundred up two, two five three five four three boundary, seven zero, nine five two and a half times, bare wire plus half, copper wire upgrade calculation, conduit temperature eight or nine times.
Explanation: "Ten below five" refers to a conductor cross-section of less than 10 square millimeters, with a safe current of 5 amps per square millimeter; "Hundred above two" refers to a conductor cross-section of more than 100 square millimeters, with a safe current of 2 amps per square millimeter; "Two-five, three-five, four-three boundary" refers to conductor cross-sections of 16 and 25 square millimeters, with a safe current of 4 amps per square millimeter, and conductor cross-sections of 35 and 50 square millimeters, with a safe current of 3 amps per square millimeter; "Seven zero, nine five two and a half" refers to a safe current of 2.5 amps per square millimeter; "Bare wire plus half, copper wire upgraded calculation" means that for bare conductors, the safe current can be calculated by multiplying the insulated conductor by 1.5 times, and for copper conductors of the same cross-section, the safe current is calculated as one line higher than the aluminum conductor; "Conduit temperature eight or nine times" means that the conductor is multiplied by a coefficient of 0.8 when in conduit, and multiplied by a coefficient of 0.9 when used in high-temperature environments.
Rule 2: Multiply 2.5 by 9, upgrade by 1 in sequence, 35 lines multiplied by 3.5, pairs reduced by half, high temperature 90%, copper wire upgrade, bare wire added by half, conductors in conduit 20%, 30%, 40%, 80%, 70%, 60% fold, don't forget.
Explanation: The phrase "2.5 down, multiply by nine, upgrade and subtract one in sequence" means that for conductors with a cross-section of 2.5 square millimeters, the safe current per square millimeter is 9 amps. For conductors with a cross-section of 2.5 square millimeters or larger (starting from 4 square millimeters), the safe current decreases by 1 amp for each increase in wire size, up to 2.5 square millimeters. "35 wires multiplied by 3.5, pairs grouped together, half the current" means that for conductors with a cross-section of 35 square millimeters, the safe current carrying capacity per square millimeter is 3.5 amps. For conductors larger than 35 square millimeters, two wire sizes are grouped together, and the safe current decreases by 0.5 amps, and so on upwards. "High temperature, 90% discount, copper wire upgrade, bare wire, add half, conductors in conduit, 20%, 30%, 40%, 80%, 70%, 60% discount" means that for two wires in a conduit, multiply by a coefficient of 0.8; for three wires, multiply by a coefficient of 0.7; for four wires, multiply by a coefficient of 0.6.
Note: The above rules (1) and (2) are based on aluminum insulated wires at a temperature of 25 degrees Celsius.
VII. Calculation of Conductor Cross-Section Selection for 380V Three-Phase Motor
Mnemonic: When selecting wires for a motor, calculate by adding or subtracting the cross-sectional coefficient. 2.5, 2 and 4 are 3, 6 and above are all 5, and so on. 100 is 200, then return 100. Upgrade by subtracting, and choose the level one with larger capacity and smaller capacity.
Note: Selecting the conductor cross-section for a 380V three-phase asynchronous motor is a common problem encountered in electrical work. This rule of thumb can be used to select the conductor cross-section. The rule of thumb is based on aluminum insulated conductors. When using copper insulated conductors, the current carrying capacity of the copper conductor can be determined by subtracting one wire number from the aluminum conductor of the same cross-section, taking into account the conductor's use in conduit and high-temperature environments.
"2.5" means that the 2 in 2.5 square millimeters of wire plus a coefficient of 2 equals the motor capacity, i.e., 2.5 + 2 = 4.5 (kW). A 2.5 square millimeter insulated aluminum wire can be used for motors of 4.5 kW and below. If copper insulated wire is used, a 1.5 square millimeter copper insulated wire can be selected. "423" means that a 4 square millimeter wire plus a coefficient of 3 equals the motor capacity, i.e., 4 + 3 = 7 (kW), which can be used for a 7 kW motor. "6 and above all 5" means that for wires with a cross-section of 6 square millimeters or more, the coefficient is 5.
For example: 6 square millimeters plus a coefficient of 5 = 6 + 5 = 11 (kilowatts), 10 square millimeters + 5 = 15 (kilowatts), 16 square millimeters + 5 = 21 (kilowatts), 25 square millimeters + 5 = 30 (kilowatts), 35 square millimeters + 5 = 40 (kilowatts), 50 square millimeters + 5 = 55 (kilowatts), 70 square millimeters + 5 = 75 (kilowatts), 95 square millimeters + 5 = 100 (kilowatts).
"100% return, upgrade reduction, large wire capacity small level selection" means that a conductor cross-section of 120 square millimeters can be used for a 100 kW three-phase 380V motor. If the conductor cross-section is above 120 square millimeters, the motor capacity is calculated according to the smaller wire cross-section.
For example, a 120 square millimeter insulated aluminum wire can power a 100 kW motor; a 150 square millimeter insulated aluminum wire can power a 120 kW motor; a 185 square millimeter insulated aluminum wire can power a 150 kW motor; and a 240 square millimeter insulated aluminum wire can power a 185 kW motor. Due to the skin effect of motors, the larger the wire cross-section, the smaller its current coefficient.
8. Calculation of Cross-Section Selection for Low-Voltage 380V/220V Three-Phase Four-Wire Overhead Conductors
Rule of thumb: For overhead lines, select the cross-section by multiplying the load factor. Multiply the load factor by four for three-phase lines and by twenty-four for single-phase lines. Divide the result by 1.7 to get the cross-sectional area of the copper wire.
Note: Selecting the conductor cross-section for low-voltage overhead line installation is a common practical issue encountered in electrical work. Choosing a conductor cross-section that is too large will result in waste and high investment, while choosing a conductor cross-section that is too small will not meet the requirements for power supply safety and voltage quality. Selecting the conductor cross-section according to the rule of thumb can meet the power supply safety requirement of 5% voltage drop.
In the formula, for overhead lines, the cross-section is determined by multiplying the load factor to get the conductor cross-section. The load factor for the power supply is calculated, and then multiplied by the factor to get the conductor cross-section to be selected. For three-phase load factor multiplication by four, and for single-phase load factor 24, this refers to a three-phase four-wire power supply system. For three-phase 380V power supply, the load factor is calculated and then multiplied by the factor of 4 to get the conductor cross-section to be selected. For single-phase 220V power supply, the load factor is calculated and then multiplied by the factor of 24 to get the conductor cross-section to be selected. When selecting copper wire, the cross-section can be obtained by dividing the calculated conductor cross-section by 1.7.
Example 1: A three-phase four-wire power supply system uses a 380V overhead line, 200 meters long, with a transmission capacity of 30 kilowatts and an allowable voltage drop of 5%. Calculate the cross-sectional area of the conductor.
Solution: Three-phase four-wire power supply
S = P × coefficient × M
=30×4×0.2
=24 square millimeters
Copper wire S = Aluminum wire S/1.7 = 24/1.7 = 16 square millimeters
S—Cross-section of the conductor
M – Distance between load cells (kW/km)
A: For the conductor cross-section, choose 25 square millimeters for aluminum conductors and 16 square millimeters for copper conductors.
Example 2: A three-phase four-wire power supply system uses a 380V overhead line, 350 meters long, to transmit 30 kilowatts of power. What is the cross-sectional area of the conductor?
Solution: S=4
S = 4 × 40 × 0.35
S = 56 square millimeters, adjusted to 70 square millimeters.
Copper wire = 70/1.7 = 41.1 square millimeters, set to 50 square millimeters.
A: Choose 70 square millimeter aluminum wire or 50 square millimeter copper wire.
Example 3: A single-phase 220V lighting circuit is 100 meters long, transmits 20 kilowatts of power, and has a permissible voltage drop of 5%. What conductor cross-section should be selected?
Solution: S = coefficient × P × M
S = 24 × 20 × 0.1
S = 48 square millimeters, adjusted to 50 square millimeters.
Copper wire = 50/1.7 = 29.4 square millimeters, set to 35 square millimeters.
Note: Based on the above rules of thumb, the selection of conductors according to voltage drop coefficient and economic density generally meets the requirements of power supply technology. For a complete and ideal selection, please consult relevant materials.
Different output capacities and transmission distances are recommended based on different rated voltages:
Aluminum core paper insulated, PVC insulated armored cables and cross-linked polyethylene insulated cables
Persistent flow rate in air (25℃)
Note: 1. The current carrying capacity of copper core cables is the value in the table multiplied by a coefficient of 1.3;
2. The current carrying capacity in this table is the capacity of a single cable;
3. Single-core plastic cables are arranged in a triangular pattern, with the center distance equal to the cable's outer diameter.
Long-term allowable current carrying capacity and its correction factor: For aluminum core insulated, PVC insulated, armored, and cross-linked PVC insulated cables, the long-term allowable current carrying capacity of the soil when directly buried underground (25℃) is 80℃cm/W.
Note: 1. The current carrying capacity of copper core cables is the value in the table multiplied by a coefficient of 1.3;
2. The current carrying capacity in this table is the capacity of a single cable;
3. Single-core plastic cables are arranged in a triangular pattern, with the center distance equal to the cable's outer diameter.
