Mastering practical calculation formula is the ability that electrical workers should have, but it is not convenient to find many formulas when they are used. Next, sort out some commonly used practical formulas and pithy formula, and explain them with examples.

1. Calculation of lighting circuit current and selection of fuse switch

Pithy formula: calculate the current of incandescent lamp, and use the power to divide the voltage;

Calculation of current, power voltage and power factor of fluorescent lamp (excluding energy-saving fluorescent lamp);

The safety of the switch is also required, 1.5 times of the rated current;

Note: the incandescent lamp in the lighting circuit is a resistive load, the power factor is cos Φ = 1, and its rated current is divided by the power P unit watt divided by the voltage. Fluorescent lamp is inductive load, and its power factor cos Φ is 0.4-0.6 (generally 0.5), that is, P / U / cos Φ = I.

Example 1: there is a lighting circuit with a rated voltage of 220V and a total power of 2200W for incandescent lamp. The total current is obtained by selecting the fuse of the switch.

Solution: known u = 220V, total power = 2200W

Total current I = P / u = 2200 / 220 = 10A

Selector: QS = I × (1.1 ~ 1.5) = 15A

Fuse selection: IR = I × (1.1 ~ 1.5) = 10 × 1.1 = 11A

(take coefficient 1.1)

QS ------- knife switch

IR -------- fuse

Answer: the current of the circuit is 10a, the switch is 15a and the fuse is 11a.

Example 2: there is a lighting circuit with a rated voltage of 220V, connected with a fluorescent lamp of 440W, to calculate the total current selector fuse. （cosΦ=0.5）

Solution: known u = 220V, cos Φ = 0.5, total power = 440W

Total current I = P / U / cos Φ = 440 / 220 / 0.5 = 4A

Selector: QS = I × (1.1 ~ 1.5) = 4 × 1.5 = 6A

Fuse selection: IR = I × (1.1 ~ 1.5) = 4 × 1.5 = 6A

Answer: the total current of the circuit is 4a, the knife gate is 6a, and the fuse is 6A

2. 380V / 220V common load calculation

Pithy formula: three phase kilowatt, twice a, heat, volt ampere, one thousand kilowatts is 1.5

Single phase two two times four five, if three eight times half.

Note: three phase kilowatt double a refers to three-phase motor capacity of 1kW, current of 2 A, heat, volt ampere, one thousand and one point five refers to three-phase electric heater, transformer, capacitor capacity of 1kW, 1kVA, one thousand and one point five refers to one-phase electric heater, transformer, capacitor capacity of 1KW, 1kVA, capacitance current of 1kVA, single-phase two times four five, if three-and-a-half refers to one-phase 220V capacity of 1kW, current of 4.5a, 380V single-phase welding machine 1kVA 2.5 amps.

Example 1: there is a three-phase asynchronous motor with a rated voltage of 380V, a capacity of 14kw, a power factor of 0.85 and an efficiency of 0.95?

Solution: known u = 380V cos Φ = 0.85 n = 0.95 P = 14 kw

Current I = P / (× u × cos Φ × n) = P / (1.73 × 380 × 0.85 × 0.95) = 28 (a)

A: the current is 28 amps.

Example 2: there is a three-phase 380V heater with a capacity of 10kW, to calculate the current?

Solution: known u = 380V P = 10kw

Current I = P / (× U) = 10 / (1.73 × 0.38) = 15.21 (a)

A: the current is 15 a.

Example 3: there is a 380V three-phase transformer with a capacity of 20KVA, to calculate the current?

Solution: known u = 380V s = 20kwa

Current I = s / (× U) = 20 / (1.73 × 0.38) = 30.45 (a)

A: the current is 30 a.

Example 4: there is a bw0.4-12-3 capacitor to calculate the current?

Solution: known u = 0.4kV q = 12kvar

Current I = q / (× U) = 12 / (1.73 × 0.4) = 17.3 (a)

A: the current is 17.3 amps.

Example 5: there is a single-phase 220V electric iron with a capacity of 1kW, to calculate the current?

Solution: known u = 220V P = 1000 W

Current I = P / u = 1000 / 220 = 4.5 (a)

A: the current is 4.5 a.

Example 6: there is a single-phase electric welding machine with a rated voltage of 380V and a capacity of 28kva?

Solution: known u = 380V s = 28kva

Current I = s / u = 28 / 0.38 = 73.6 (a)

Answer: the current is 73.6 a.

Note: the voltage of single-phase equipment calculated above is 220V, power factor is cos Φ = 1, current is 4.5 times of capacity, voltage of single-phase equipment is 380V, such as load of electric welding machine and portable lamp transformer, current is 2,5 times of capacity.

3. Selection of distribution power transformer

Formula: select capacity of power transformer,

Total electrical equipment capacity at the same time.

Besides power factor and efficiency.

Note: the total power consumption equipment refers to the sum of the power of all the equipment in the plant, and the simultaneous rate refers to the ratio of the actual capacity of the equipment put into operation at the same time to the capacity of the power consumption equipment, generally about 0.7

Power factor: 0.8-0.9 in general

Efficiency: generally 0.85-0.9

Power transformer capacity = total capacity of electric equipment × simultaneity rate / power factor of electric equipment × efficiency of electric equipment

For example, there is a factory with a total capacity of 700 KVA of electric equipment, and the actual consumption is 600 KVA, so the capacity of the electric transformer is calculated.

Simultaneous rate: 600KVA ÷ 700kva = 0.86

Power transformer capacity = (700 × 0.86) / (COS Φ = 0.85 × 0.85)

=600 / 0.722 = 830kva

Note: if the power factor is 0.95 and the efficiency is 0.9, its capacity is:

Power transformer capacity = (700 × 0.7) / (COS Φ = 0.95 × n = 0.9)

=490 / 0.855 = 576kva

For example, if the power factor = 0.7, n = 0.9 and the rate is 0.75, calculate the capacity.

Power transformer capacity = (700kva × 0.75) / (0.7 × 0.9)

=525 ／ 0.63 = 833kva

According to the above analysis, the lower the power factor, the larger the selection of power transformer, and vice versa, it is concluded that the purpose of energy saving and consumption reduction can only be achieved by improving the power factor.

4. Calculation of power factor

Pithy formula: how to find the power factor,

It can be seen from the active and reactive meters,

Calculate the electricity consumption of the current month,

The power factor can be calculated.

Note: some enterprises ignore the power factor, which can lead to the increase of power cost. In order to reduce the power cost, the power factor must reach 0.9. How to improve the power factor will be discussed in the next section.

In the pithy formula: how to calculate the power factor can be seen from the active and reactive meters. The power factor can be calculated by calculating the electricity consumption of the current month. Some enterprises have no power factor meter, power meter or reactive compensation equipment in the distribution system, but only equipped with voltmeter, ammeter, active watt hour meter and reactive watt hour meter, so it is difficult to calculate the power factor, which can be obtained through the active watt hour The power factor of the current month shall be calculated according to the kilowatt / hour consumption of the current month and kvar / hour reactive power meter.

For example: 1000 kW / h is used for the active power and 300 kvar / h is used for the reactive power meter in the current month, and the power factor cos Φ in the current month is calculated.

Solution: cos Φ = active power / = 1000/

=1000/1044=0.957

If 1000 kW / h is used for the active power and 750 kvar / h is used for the reactive power meter, the power factor cos Φ of the current month shall be calculated.

Cos Φ = active power / = 1000/

=1/1.22=0.81

Note: the power factor of enterprise reactive power compensation is generally 0.7-0.85, and some of them are below 0.65. The smaller the motor power is, the lower the power factor is. The power factor of big horse pulling car is low. Generally, induction motor accounts for 70% of cos Φ, power transformer 20% and conductor 10%.

If there is a power factor meter on the distribution panel, it can be seen directly, or the instantaneous power factor can be calculated by the indication number of voltmeter, ammeter and power meter on the distribution panel

Namely: cos Φ = P / (× u × I)

Where p is the power meter (kw), u is the voltage indicator (kV 0.38kv), and I is the ammeter indicator (a).

5. Selection of thermal elements of motor contactor

Pithy formula: select the current of the motor, double the rated current,

The motor seeks the current, one thousand watts is equal to one current.

Heat element of motor, one point two times of rated current calculation,

Double current setting, overload protection is guaranteed.

Note: AC contactor is a kind of control appliance for switching on and off the load current of motor. Generally, the rated current of AC contactor is 1.3-2 times of the rated current of motor. In the formula, the motor is selected for current connection and the double rated current is calculated, which means that the rated current of AC contactor selected for motor is 2 times of the rated current of motor. Select the motor in the formula, select the thermal element, calculate by 1.2 times of the rated current, and set by one time of the rated current. Overload protection is guaranteed, which means that the rated current of the thermal element of the motor is selected by 1.2 times of the rated current of the motor, and setting by one time of the rated current of the motor is the overload protection of the circuit.

For example, there is a three-phase asynchronous motor with a rated voltage of 380V and a capacity of 10kW,

The power factor is 0.85 and the efficiency is 0.95. Calculate the motor current and select the AC contact thermal element and setting value.

Solution: (1) empirical formula: 10kW × 2 = 20 (a)

(2) Known u = 380V P = 10kW cos Φ = 0.85 n = 0.95

Current I = P / (× u × cos Φ × n)

=10 / (1.73 × 0.38 × 0.85 × 0.95) = 20 (a)

Select AC contactor: km = Ke × (1.3-2) = 20 × 2 = 40 (a)

CJ10 -- 40

Heat selecting element: FR = IC × (1.1 ~ 1.25) = 20 × 1.25 = 25 (a)

Select jr16-20 / 30, Jr is set according to 20 A

Answer: motor current is 20 A, choose 40 a contactor, rated current of thermal element is 25 A, set to 20 a.

6. Safe current calculation of insulated conductor

Pithy formula (1): ten down five, one hundred up two, two five three five four three boundaries, seven zero, nine five and a half times, the bare wire plus half, copper wire upgrade calculation, pipe temperature eight, nine fold.

Note: the safety current per square millimeter is 5 a when the cross-section of the conductor is less than 10 square millimeters; the safety current per square millimeter is 2 a when the cross-section of the conductor is more than 100 square millimeters; the safety current per square millimeter is 4 a when the cross-section of the conductor is 16 square millimeters and 25 square millimeters; the safety current per 1 square millimeter is 4 a when the cross-section of the conductor is 35 square millimeters And 50 square mm, the safety current per 1 square mm is 3 amperes; 70, 95 times and half refer to the safety current per 1 square mm is 2.5 amperes; bare wire plus half, copper wire upgrade refers to the bare wire of cross-section, the safety current can be calculated by multiplying the insulated wire by 1.5 times, and the copper wire of the same cross-section is calculated by the large wire size of aluminum wire; the threading temperature is 80, 90 percent The guide line crossing factor is 0.8, and the practical factor in high temperature places is 0.9. www.diangon.com

Pithy formula 2: two point five down nine times, upgrade minus one sequence right, 35 lines multiplied by three point five, double double double group halved, high temperature 90 fold, copper wire upgrade, bare wire plus half, wire threading two, three, four, eight, seven, six fold don't forget.

Note: in the pithy formula, two point five is nine times the whole. The sequence of upgrading minus one is that the cross section of the conductor is 2.5mm2, and the safety current of every 1mm2 is 9A. The cross section of the conductor is more than 2.5mm2, that is to say, the safety current of the conductor is reduced by 1 A for every grade increase of 4mm2, until 2.5mm2; the 35 lines are multiplied by three point five, and the two lines are halved in groups It refers to the conductor with a cross section of 35mm2 and a safe cut-off flow of 3.5A for every 1mm2 and above. The two grades of wire numbers are a group, the safe current is reduced by 0.5A, and it is calculated upward in turn; the high temperature is reduced by 10%, the copper wire is upgraded, the bare wire is increased by half, and the two wires with two, three, four, eight, seven and six turns of the conductor are not forgotten to be the guide wire, and the multiplication coefficient is 0.8 Three lines of conduit shall be multiplied by a factor of 0.7, and four lines of conduit shall be multiplied by a factor of 0.6.

Note: the above formula (I) and (II) are based on aluminum insulated wire, and the temperature is 25 ℃. Formula (I) safety current coefficient table of conductor section:

Formula (II) safety current coefficient table of conductor section:

7. Selection and calculation of conductor section of 380V three-phase motor

Pithy formula: select the lead wire for the motor, add and subtract the section coefficient, 2.5, 2, 4 for 3, 6 for 5, up, 12 for 100, upgrade and reduce, and select the first level of large capacity and small line.

Note: it is a common problem for electricians to select the conductor section of 380V three-phase asynchronous motor. According to this formula, the conductor section can be selected. The formula is based on the aluminum insulated conductor. When using copper insulated conductor, the carrying capacity of copper conductor can be calculated according to the section of aluminum conductor with the same section and small wire size, and the use of conductor in pipe and high-temperature place shall be considered.

2.5.2 it means that the capacity of the motor is 2.5 mm2 conductor plus coefficient 2, that is, 2.5 + 2 = 4.5 (kw), and the 2.5 mm2 insulated aluminum conductor can be used for the motor of 4.5kw and below. If copper insulated conductor is used, 1.5mm2 copper insulated conductor can be selected; 4.2-3 refers to 4mm2 conductor plus factor 3 refers to motor capacity, i.e. 4 + 3 = 7 (kw), which can be used by 7KW motor; 6-up-all-5 refers to 5 for wires with more than 6mm2 cross section.

For example: 6 mm2 plus factor 5 = 6 + 5 = 11 (kw), 10 mm2 + 5 = 15 (kw), 16 mm2 + 5 = 21 (kw), 25 mm2 + 5 = 30 (kw), 35 mm2 + 5 = 40 (kw), 50 mm2 + 5 = 55 (kw), 70 mm2 + 5 = 75 (kw), 95 mm2 + 5 = 100 (kw).

The first level selection of "one hundred two returns to one hundred", upgrading and reducing, line large capacity and small size refers to that the conductor section of 120 mm2 can be used for 100 kW three-phase 380 V power supply motor, and the conductor section is more than 120 mm2, and the motor capacity is calculated according to the first level of line number section.

For example: 120 mm2 insulated aluminum wire can provide 100 kW motor capacity; 150 mm2 insulated aluminum wire can provide 120 kW motor capacity; 185 mm2 insulated aluminum wire can provide 150 kW motor capacity; 240 mm2 insulated aluminum wire can provide 185 kW motor capacity; because of the skin effect of the motor, the larger the wire section, the smaller the current coefficient.

8. Section selection calculation of LV 380V / 220V three-phase four wire overhead conductor

Formula: select the cross section of the overhead line, multiply the load distance coefficient, multiply the three-phase load distance by four, single-phase load distance by twenty-four, divide the number by 1.7 to get the copper wire cross-sectional area.

Note: the selection of conductor section for the installation of LV overhead lines is a practical problem often encountered in electrical work. The selection of conductor section causes waste, high investment and small selection of conductor section, which can not meet the requirements of power supply safety and power supply voltage quality. The selection of conductor section according to the formula can meet the power supply safety requirements of 5% of voltage loss.

In the formula, the cross-section of the overhead line is selected, and the multiplication of the load distance coefficient refers to the cross-section of the conductor, and the load distance of the power transmission personnel is calculated, and then the multiplication of the load distance coefficient is the conductor to be selected; the three-phase load distance is multiplied by four, and the single-phase load distance 24 refers to the three-phase four wire system power supply, three-phase 380V, and the cross-section of the conductor to be selected by the multiplication of the load distance coefficient 4, single-phase 220V power supply, and then the multiplication of the load distance by the coefficient 24 For the section of wire to be selected, when selecting copper wire, the section of copper wire can be divided by 1.7 according to the calculated section of wire.

Example 1: there is a three-phase four wire system for power supply, 380V overhead lines, 200m in length, 30kW in transmission power, 5% in allowable voltage loss?

Solution: three phase four wire power supply

S = P × coefficient × M

=30×4×0.2

=24 mm2

Copper wire s = aluminum wire s / 1.7 = 24 / 1.7 = 16 mm2

S - conductor section

M -- load distance (kW / km)

Answer: the cross section of the conductor is 25 mm2 aluminum conductor and 16 mm2 copper conductor.

Example 2: three phase four wire system power supply, 380 V overhead line, length 350 meters, transmission power 30 kW, to find the conductor cross-section?

Solution: S = 4

S=4×40×0.35

S = 56 mm2, set to 70 mm2

Copper wire = 70 / 1.7 = 41.1 mm2, set to 50 mm2

Answer: choose 70 mm2 aluminum wire or 50 mm2 copper wire

Example 3: there is a single-phase 220 V lighting circuit with a length of 100 meters, a transmission power of 20 kilowatts, and a allowable voltage loss of 5%. Select the conductor section?

Solution: S = coefficient × P × M

S=24×20×0.1

S = 48 mm2, set to 50 mm2

Copper wire = 50 / 1.7 = 29.4mm2, set to 35mm2

Note: according to the above experience formula, it basically meets the requirements of power supply technology according to the voltage loss coefficient according to the principle of wire selection and the economic density selection coefficient. To achieve the complete ideal selection, please refer to the relevant data.

Different output capacity transmission distance is recommended according to different rated voltage:

Long term allowable carrying capacity of aluminum core paper insulated, PVC insulated armored cable and XLPE insulated cable in air (25 ℃)

Note:

1. The current carrying capacity of copper core cable is 1.3 times of the value in the table multiplied by the coefficient;

2. The current carrying capacity in this table is the capacity of a single cable;

3. The single core plastic cable is arranged in triangle, and the center distance is equal to the outer diameter of the cable.

The long-term allowable carrying capacity and its correction coefficient of cable the long-term allowable carrying capacity of aluminum core insulated, PVC insulated, armored cable and cross-linked PVC insulated cable is 80 ℃ cm / W when buried directly in the ground (25 ℃)

Note:

1. The current carrying capacity of copper core cable is 1.3 times of the value in the table multiplied by the coefficient;

2. The current carrying capacity in this table is the capacity of a single cable;

3. The single core plastic cable is arranged in triangle, and the center distance is equal to the outer diameter of the cable.