68 electrician’s tips outline commonly used electrical technology theory, data, construction operating procedures, instrumentation usage methods, etc.
This tip comes from the power industry peer compiled some information, and several in the power industry have a profound knowledge of the predecessors, thank you for bringing us a lot of conveniences, and then expressed loyal thanks.
Then there are these 68 tips just throw bricks to introduce jade, more is to hope that the vast number of peers can publish some insights and insights in practice, collect similar tips used in practical work, to make it more rich and practical.
1, simple estimation of wire load flow of
105, 100 2, 25354 boundaries, 7095 times the half, temperature 80%, copper upgrade calculation
Explanation: The load flow of aluminum conductors below 10mm2 is calculated at 5A/square mm; The aluminum wire load is calculated at 4A/square mm; the aluminum wire load of35mm2 is calculated at 3A/square mm;70mm2,95mm2 The aluminum wire load is calculated at 2.5A/square mm, and the “copper upgrade”: for example, to calculate the copper wire load flow of 120mm2, 150mm2 aluminum wire can be selected, for aluminum wire the load rate of, affected by temperature, and finally multiplied by 0.8 or 0.9(by geographic location).
2, known transformer capacity, the voltage level side rated current
Description: suitable for any voltage level.
Tip: The capacity is divided by the voltage value, and its business is multiplied by six by ten.
Example: Visual current I – Visual power S/1.732s10KV s 1000KVA/1.732s10KV s 57.736A
Estimated I s 1000KVA/10KVs6/10 s 60A
3, roughly verify the accuracy of low-voltage single-phase power meter method One by
one, close the switch and time again.
Time simultaneous number of revolutions to record the six-point revolution value.
There are several meter dials, kilowatt-hours of turning.
This value is reduced by a hundred times, roughly equal to the number of records.
4, known three-phase motor capacity, and then rate the current port tip: capacity divided by kV, a business factor of 76.
Known three-phase 22 motor, kilowatt 3.5 amperes.
1KW ÷ 0.22KV x 0.76≈1A
known high-voltage 3 kV motor, four kW one ampere.
4KW ÷ 3KV x 0.76≈1A
Note: The tip is suitable for three-phase motor-rated current calculations of any voltage level. When used, the capacity unit is kW, the voltage unit is kV, and the current unit is A.
5, to measure the secondary side current of the power transformer, to calculate the load capacity it contains is known to be the secondary pressure of the secondary pressure, measured the current for kilowatts.
The voltage level is 400 volts, one amp 0.6 kW.
The voltage level is 3 kV, 4.5 kW.
The voltage level is 6 kV, an integer of 9 kW.
The voltage level is ten kV, one amp fifteen kilowatts.
The voltage level is 35,000, one amp 55 kW.
6, known small 380V three-phase cage-type motor capacity, the minimum capacity of its power supply equipment, load switch, protection melt current value


directly start the motor, the capacity is not more than ten kilowatts;


Six-fold kW select switch, five times kW melt.


Power supply equipment kVA needs to be three times the kilowatt.


Description: The motor described in the mouth tip is a small 380V rat cage-type three-phase motor, the starting current of the motor is very large, generally 4-7 times the rated current. The maximum electrical motivation capacity started directly with a load switch should not exceed 10kW, generally at 4.5kW or less, and the open load switch (glue cover porcelain bottom isolation switch) is generally used Small-capacity motors of 5.5kW and below are not frequently started directly, and closed load switches (iron shell switches) are generally not started directly infrequently with motors below 10kW. The load switch is composed of a simple isolation switch gate and a fuse or melt, and it is advisable to select a 6x switch rated for power, and to avoid a high current when the motor is started, a fuse of 5x the rated power should be selected as appropriate, i.e. rated current(A) The melt rated current (A) for short-circuit protection. Finally, select the appropriate power supply, which should have an output power of no less than 3 times the rated power.


7, to detect the nameless 380V single-phase welding transformer’s empty load current, to calculate its rated capacity port


tip: 388 welding machine capacity, empty load current multiplied by five.


A Single-phase AC welding transformer is a special-purpose buck transformer, compared with ordinary transformers, the basic principle of its working is about the same. To meet the requirements of the welding process, welding transformers work in a short-circuit state, requiring a certain arc voltage when welding. As the welding current increases, the output voltage drops sharply. Depending on the P-UI(power is certain, voltage is inversely proportional to current). When the voltage drops to zero (i.e. the secondary side short circuit), the secondary side current is not too large and so on, that is, the welding transformer has the external characteristics of a steep drop, the welding transformer’s steep drop outside characteristics are obtained by the pressure drop generated by the electrical resistance coil. When empty, because there is no welding current passing through, the electro-resisting coil does not produce a voltage drop, at this time the empty load voltage is equal to the secondary voltage, that is to say when the welding transformer is empty and the normal transformer is empty. The unloaded current of the transformer is generally about 6% to 8%of the rated current(the state stipulates that the empty current should not be greater than 10%of the rated current).


8, judge AC and DC Electric pen to judge AC


DC, AC bright DC dark,


AC tube bright, DC tube bright end.


Description: When identifying intersections and DIRECT, it is best to compare between “two electricities”, so it is obvious. When measuring AC, both ends of the tube are lit at the same time, and when measuring DC, only one end of the tube is extremely bright.


9, clever use of electric pen for low-voltage nuclear phase judgment two lines are the


same, two hands each holding a pen,


two feet and ground insulation, two pens each touch a key line, with the eye to


see a pen, not bright with the same bright for the same difference.


Description: When testing this, remember that the feet must be insulated from the ground. Because most of our country is 380/220V power supply, and transformers generally use neutral point direct ground, so when doing testing, the human body and the earth must be insulated, to avoid forming a circuit, so as not to misjudge;


10, clever use of electric pen to judge DC positive and negative electrode


electro-pen to judge positive and negative poles, observe the tube to be careful,


the front end bright is negative, the back end bright is positive.


Description: The front end of the tube refers to one end of the tip of the electric pen, the back end of the tube refers to one end of the handgrip, the front end is bright for the negative pole, and vice versa. Be aware during testing: the supply voltage is 110V and above; If a person is insulated from the earth, one hand touches either pole of the power supply, the other holds a measuring pen, and the metal head of the pen touches the other pole of the power supply under test, and the front end of the tube is bright, The measured power supply is negative, and if the rear end of the tube is bright, the measured power supply is positive, which is based on the principle of DC one-way flow and electron flow from negative to positive.


11, clever use of electric pen to judge whether dc power supply is grounded, positive and negative ground difference The DC coefficient of the


substation, the pen does not light up;


If the light is near the tip of the pen, there is a ground fault at the positive pole; If the light is near the finger


end, the ground fault is at the negative pole.


Description: The DC coefficient of power plant and substation is insulated to the ground, the person stands on the ground, uses the electric pen to touch the positive or negative pole, the tube should not be lit, if lit, then the DC system is grounded phenomenon;


12, clever use of a pen to judge the 380/220V three-phase three-wire power supply line phase fault


star-shaped joint three-phase line, the pen touched two brights, the


remaining one weak brightness, the phase wire Grounded;


If it is barely lit, the metal is grounded.


Description: The secondary side of the power transformer is generally connected to a Y-shaped, in the neutral point is not grounded in the three-phase three-wire system, with the electric pen touch three-phase lines, there are two usually slightly brighter, while the brightness on the other is weaker, indicating that the weak brightness of the phase line has ground phenomenon, but not too serious; The only remaining one, which is barely visible, is a metal ground fault on this phase line.


13, the motor wiring


tips:2.5 plus three,4 plus four;6 after plus six,25 Five;120 wires with hundreds This recipe is for


three-phase 380-volt motor wiring. The wire is made of aluminum core insulation (or plastic wire) through the tube.


First, learn about the general motor capacity ( kW ) arrangement:


0.8 1.1 1.5 2.2 3 4 5.5 7.5 1O 1317 22 30 40 55 75 100


“2.5” Add three, representing 2.5 square millimeters of aluminum core insulation line through the pipe laying, can be equipped with “2.5 plus three” kilowatts of electric motor, that is, up to a maximum of 5.5 kilowatts of the electric machine.


“4 plus four” is 4 square millimeters of aluminum core insulation wire, pipe laying, can be equipped with “4 plus four” kilowatts of electric motors. That is, motors up to 8 kW (products are only 7.5 kW) are available.


“6 plus six” means that from 6 square millimeters, and later can be equipped with “plus six” kilowatts of electric motor. That is, 6 square millimeters can be matched with 12 kW,10 sq mm with 16 kW,and16 mm2 can be matched with 22 kW.


“25 five” means that starting at 25 square millimeters, the addition is changed from six to five. That is, 25 square millimeters can be matched with 30 kW,35 sq mm with 40 kW,and50 sq mm with 50 sq mm 55 kW,70 sq mm with 75 kW.


“120 wires with hundreds” (read “12 wires with hundreds”) is to say that the motor is as large as 100 kilowatts. Instead of “increasing” the conductor cross-section to attach the motor, the 120-square-millimeter conductor can only be equipped with a 100-kilowatt motor.


14, according to the power calculation of the current port


tip: power doubled, electric heat plus a half. Single-phase kW,4.5 amps. Single-phase 380, current two and a half amps.


Explanation: Power refers specifically to the motor at 380V three-phase (power of about 0.8), the electric motor per kilowatt of current is about 2 Ann To double the “kW” (multiplied by 2) is the current (amp). This current is also known as the rated current of the motor; A three-phase 380-volt electric device with a current of 1.5 amps per kilowatt The “kW plus half” (multiplied by 1.5)is the current (amp); In a three-phase four-wire system, two lines of a single-phase device, one of the junctions, and the other of the zero-wires (e.g. lighting) are single-phase 220 volts of electricity The device. Most of the power of this device is 1KW, so the tip directly states “single-phase ( 4.5 amps per kW). ”。 When calculating, as long as “multiply the kilowatt by 4.5” is the current, Ann. As above, it applies to all single-phase 220 volts of electrical equipment in kilowatts, as well as to electric heating and lighting in kilowatts, and also to DC in 220 volts; In a 380/220 volt three-phase four-wire system, two lines of a single-phase device are connected to a phase line, which is traditionally referred to as a single-phase 380-volt electrical device (actually connected to a two-phase line). )。 When this equipment is in kilowatts, the power is mostly 1KW, and the port tip directly states “single phase 380, the current is two and a half amps” ”。 It also includes 380-volt single-phase devices in kilowatts. The calculation is calculated as long as “multiplying the kilowatt by 2.5 is the current (amp).