Electrician oral formulas: Based on the theoretical formulas, simplified formulas and empirical formulas in electrical engineering, combined with the essence of mathematical operation rules and oral arithmetic skills, one hundred and one electrician quick calculation formulas were created. In order to meet the demand for on-site quick calculation skills required by electricians, it can quickly improve the quality of personal electrician work.
Instructions for calculating the current carrying capacity of conductors
The next five are 10, and the second is 1 0 0.
2 5, 3 5, four and three realms.
7 0, 95, two and a half times.
The temperature of the pipe is 19% off.
Add half of the bare wire.
Copper wire upgrade is included.
Explanation: Tips for calculating the current carrying capacity of conductors
1. Purpose: The current carrying capacity (safety current) of various conductors can usually be found in the manual. But by using formulas and some simple mental arithmetic, you can calculate directly without looking up tables. The carrying capacity of a conductor is related to the carrying surface of the conductor, as well as the conductor's material (aluminum or copper), model (insulated wire or bare wire, etc.), laying method (open laying or through pipe, etc.) and ambient temperature (about 25 degrees or (larger), etc., there are many influencing factors, and the calculation is more complex.
10 is five, 1 0 0 is two.
2 5, 3 5, four and three realms.
7 0, 95, two and a half times.
The temperature of the pipe is 19% off.
Add half of the bare wire.
Copper wire upgrade is included.
4. Note: The formula is based on the condition that the aluminum core insulated wire is exposed and the ambient temperature is 25 degrees. If the conditions are different,
the formula will be explained separately. Insulated wires include various types of rubber insulated wires or plastic insulated wires. The formula does not directly indicate the carrying capacity (current, amperes) of various cross-sections, but "multiplies the cross-section by a certain multiple" to express it. To do this, you should first be familiar with the arrangement of wire cross-sections (square millimeters):
1 1.5 2.5 4 6 10 16 25 35 50 7O 95 l20 150 185...
The cross-sectional area of ??the aluminum core insulated wire manufactured by the manufacturer usually starts from 2.5, the copper core insulated wire starts from 1; the bare aluminum wire starts from 16; the bare copper wire starts from 10.
① This formula points out: The carrying capacity of aluminum core insulated wire, in amperes, can be calculated according to the multiple of the cross-section number. The Arabic numerals in the formula represent the wire cross-section (square millimeters), and the Chinese characters represent multiples. The relationship between the cross sections and multiples of the formula is arranged as follows:
..10 16-25 35-50 70-95 120....
Five times four times three times two One and a half times two times
Now it will be clearer when compared with the formula. It turns out that "10 under five" means that the cross-section is from 10 to below, and the current carrying capacity is five times the cross-section number. "100
上二" (pronounced 百上二) means that the cross-section is more than 100, and the carrying capacity is twice the number of cross-sections. Sections 25 and 35 are the boundaries between four times and three times. This is the "Four and Three Boundaries of Formula 25 and 35". The cross-sections 70 and 95
are 2.5 times. From the above arrangement, it can be seen that except for those below 10 and above 100, the wire cross-section in the middle is the same multiple of each two specifications.
The following is an example of an exposed aluminum core insulated wire with an ambient temperature of 25 degrees:
Example 1 6 square millimeters, press 10 times five, and the current carrying capacity is 30 A .
Example 2150 square millimeters, based on 100, the current carrying capacity is 300 A.
For example, if it is 370 square millimeters, based on 70 and 95 2 and a half times, the current carrying capacity is calculated to be 175 A.
It can also be seen from the above arrangement that the multiple decreases with the increase of the cross-section. At the junction of multiple changes, the error is slightly larger. For example, sections 25 and 35
are the boundary between four times and three times. 25 belongs to the range of four times, but is close to the side that changes to three times. According to the formula, it is four times, that is, 100 A. But the actual value is less than four times (97 A according to the manual). 35
On the contrary, according to the formula, it is three times, that is, 105 amps, but it is actually 117 amps. However, this has little impact on usage. Of course, if you can have a good idea, when choosing the wire cross-section, do not let it reach 100 amps if it is 25
, and slightly exceed 105
amps if it is 35
It will be more accurate. Similarly, the wire position of 2.5 square millimeters at the initial (left) end of five times is actually more than five times (up to more than 20 amps). However, in order to reduce the power loss in the wire, it is usually not used to be so large. Manual Generally, it is only marked with 12
An.
②
From here on, the formula is to deal with changes in conditions. This sentence: The temperature of the pipe is 10% off, which means that if the pipe is laid (including the laying of grooved plates, that is, the wires are covered with protective sheaths and are not exposed), after calculating according to ①, the temperature will be 20% off (multiplied by 0.8) if the environment If the temperature exceeds 25
degrees, the price should be calculated according to ①, and then a 10% discount will be applied. (Multiply by 0.9).
As for ambient temperature, it refers to the average maximum temperature in the hottest month of summer. In fact, the temperature changes. Generally, it does not affect the current carrying capacity of the conductor very much. Therefore, discounts will only be considered when the temperature in some high-temperature workshops or hotter areas exceeds 25 degrees.
There is also a situation where both conditions change (the temperature of the pipe is higher). Then calculate according to ① and get 20% off, then 10% off. Or simply calculate it with a 30% discount (i.e. 0.8 ×
0.9=0.72, about 0.7). This can also be said to be the temperature of the pipe, which means 80% off.
For example: (aluminum core insulated wire) 10 square millimeters, the tube (20% off) 40 A (10 × 5× 0.8 = 40)
The tube is high temperature (7 (fold) 35 A (1O × 5 × 0.7=35)
95 square millimeters, pipe (20 fold) 190 A (95×2.5×0.8=190)
High temperature (10% off), 214 A (95 × 2.5 × 0.9=213.8)
The pipe is worn and the temperature is high (30% off). 166 A (95 × 2.5 × 0.7 = 166.3)
② For the current carrying capacity of the bare aluminum wire, the formula points out that half is added to the bare wire, that is, add half after the calculation in ① (multiply by l.5) . This means that the current carrying capacity of aluminum core insulated wires with the same cross-section can be increased by half compared to bare aluminum wires.
Example 1 16 square millimeters of bare aluminum wire, 96 A (16 × 4 × 1.5 = 96). High temperature, 86 A (16 × 4 × 1.5 × 0.9=86.4)
Example 2 35 mm2 bare aluminum wire, 150 A (35 × 3 × 1.5=157.5)
Example 3120 square millimeters of bare aluminum wire, 360 A (120 × 2 × 1.5 = 360)
③As for the current carrying capacity of copper wires, the formula points out that copper wire upgrades are counted. That is, the cross section of the copper conductor is upgraded to one level in the order of cross section arrangement, and then calculated according to the corresponding aluminum wire conditions.
Example 1: 35 square meters of bare copper wire at 25 degrees, upgraded to 50 square millimeters, and then 50 square millimeters of bare aluminum wire at 25 degrees is calculated as 225 A (50 × 3 × 1.5)
Example 2: 16 square millimeters of copper insulated wire at 25 degrees, based on the same conditions as 25 square millimeters of aluminum insulation, calculated as 100 A (25 × 4)
Example 3: 95 square millimeters of copper insulated wire at 25 degrees , through the pipe, according to the same conditions as the 120 square millimeter aluminum insulated wire, the calculation is 192 A (120 × 2 × 0.8)
Given the transformer capacity, find the rated current of each voltage level side
Tip a:
The capacity divided by the voltage value, its quotient multiplied by six divided by ten.
Description: Suitable for any voltage level.
In daily work, some electricians are only involved in the calculation of rated current of transformers of one or two voltage levels. Simplifying the above formula, the formula for calculating the rated current of each voltage level side can be derived:
The capacity coefficient is multiplied to calculate.
Knowing the capacity of the transformer, quickly calculate the current value of the primary and secondary protection fuses (commonly known as fuses)
Tip b:
High-voltage fuses of distribution transformers , the capacity voltage is compared to find.
For distribution transformer low-voltage fuse links, multiply the capacity by 9 divided by 5.
Note:
The correct selection of fuses is closely related to the safe operation of the transformer. When only using fuses to protect the high and low voltage sides of the transformer, the correct selection of the melt is even more important. This is a problem that electricians often encounter and have to solve.
Given the capacity of the three-phase motor, find its rated current
Formula (c): Divide the capacity by the number of kilovolts, and the business multiplication coefficient is seventy-six.
Note:
(1) The formula is applicable to the calculation of rated current of three-phase motors of any voltage level. It can be explained from the formulas and formulas that the rated currents of motors with the same capacity and different voltage levels are different, that is, the voltages in kilovolts are different. If divided by the same capacity, the "quotients" obtained are obviously different. Different quotients If the number is multiplied by the same coefficient of 0.76, the current value obtained is also different. If the above formula is called a general formula, we can derive a special calculation formula for calculating the rated current of a motor with voltage levels of 220, 380, 660, and 3.6kV. When using the special calculation formula to calculate the rated current of a certain three-phase motor, the capacity in kilowatts and the current in amperes The relationship is directly multiplied, eliminating the need to divide the capacity by the number of kilovolts and multiply the quotient by the coefficient 0.76.
Three-phase 220 kilowatt motor, 3.5 amps.
A commonly used 380 motor is one kilowatt and two amps.
Low voltage 660 motor, kilowatt 1.2 amps.
High voltage three thousand volt motor, four kilowatts and one ampere.
High-voltage six-kilovolt motor, eight kilowatts per ampere.
(2) Tip c When using, the unit of capacity is kW, the unit of voltage is kV, and the unit of current is A. You must pay attention to this point.
(3) The coefficient 0.76 in formula c is a comprehensive value calculated taking into account the motor power factor and efficiency. The power factor is 0.85 and the efficiency is not 0.9. These two values ??are more suitable for motors above tens of kilowatts, but are larger for commonly used motors below 10kW. This means that there is an error between the rated current of the motor calculated using formula c and the value marked on the motor nameplate. This error has little impact on the rated current of motors below 10kW, such as switches, contactors, wires, etc.
(4) Use formula calculation skills. When using the formula to calculate the rated current of a commonly used 380V motor, first divide the 0.38kV power supply voltage number of the motor by 0.76 and multiply the capacity (kW) number by the quotient 2. If you encounter a 6kV motor with a larger capacity, and the kW capacity is exactly a multiple of the 6kV number, then the capacity is divided by the kilovolts, and the quotient is multiplied by a 0.76 coefficient.
(5) Error. The coefficient 0.76 in formula c is calculated by assuming that the motor power factor is 0.85 and the efficiency is 0.9. In this way, there will be errors in calculating the rated current of motors with different power factors and efficiencies. According to the five special formulas derived from formula c, the multiple of capacity (kW) and current (A) is the quotient of each voltage level (kV) minus the 0.76 coefficient. The special formula is simple and easy to calculate mentally, but it should be noted that the error will increase. Generally, for larger kilowatts, the calculated current is slightly larger than what is on the nameplate; for smaller kilowatts, the calculated current is slightly smaller than what is on the nameplate. In this regard, when calculating the current, when the current reaches more than ten amperes or dozens of amperes, it is not necessary to calculate after the decimal point. You can round off and only take integers, which is simple and does not affect practicality. For smaller currents, it only needs to be calculated to one decimal place.
Measure the current and find the capacity
Measure the no-load current of the motor without a nameplate and estimate its rated capacity
Tip:
None To calculate the capacity of a brand motor, measure the no-load current value.
Multiply ten and divide by eight to calculate, which is close to the kilowatt rating.
Explanation: The secret is that for a three-phase asynchronous motor without a nameplate, if you don’t know its capacity in kilowatts, you can estimate the motor’s capacity in kilowatts by measuring the no-load current value of the motor.
Measure the current on the secondary side of the power transformer and calculate its load capacity
Function:
If the secondary voltage of the distribution transformer is known, measure the current Asking for kilowatts.
The voltage level is 400 volts, 0.6 kilowatts per ampere.
The voltage level is three thousand volts, one amp is 4.5 kilowatts.
The voltage level is six thousand volts, and one ampere is an integral number of nine thousand watts.
The voltage level is ten thousand volts, one ampere is fifteen thousand watts.
The voltage level is thirty-five thousand, and one amp is fifty-five kilowatts.
Note:
(1) In daily work, electricians often encounter questions from superior departments and managers about the operation of the power transformer. What is the load? The electrician himself often needs to know what the load on the transformer is. The load current is easy to know. You can directly look at the ammeter set on the power distribution device, or use a corresponding clamp-type ammeter to measure it. The load power cannot be directly seen or measured. You need to rely on this formula to calculate, otherwise it will be complicated and time-consuming to use conventional formulas to calculate.
(2) "The voltage level is 400 volts, and one generator is 0.6 kilowatts." When the load current on the secondary side of the power transformer (voltage level 400V) is measured, the ampere value is multiplied by the coefficient 0.6 to obtain Load power in kilowatts.
Measure the current of the incandescent lighting circuit and calculate its load capacity
Function:
The lighting voltage is 220, and one amp is 220 watts.
Note: Industrial and mining enterprises mostly use 220V incandescent lamps for lighting. The lighting power supply line refers to the line from the distribution panel to each lighting distribution box. The lighting power supply main line is generally three-phase and four-wire. Single-phase can be used when the load is less than 4kW. Lighting distribution lines refer to the lines connected from the lighting distribution box to lighting facilities such as luminaires or sockets. Regardless of power supply or distribution lines, just use a clamp ammeter to measure the current value of a phase line, and then multiply it by a factor of 220. The product is the load capacity of the phase line. Measuring the current to find the capacity number can help electricians quickly adjust the unbalanced three-phase load capacity of the lighting main line. It can also help electricians analyze the reasons why the protective melt in the distribution box often fuses, the reasons why the distribution wires heat up, etc.
Measure the no-load current of the 380V single-phase welding transformer without a nameplate, and calculate the base rated capacity
Function:
Three hundred and eight welding machine capacity, empty Multiply the current carrying current by five.
Single-phase AC welding transformer is actually a special-purpose step-down transformer. Compared with ordinary transformers, its basic working principle is roughly the same. In order to meet the requirements of the welding process, the welding transformer works in a short-circuit state and requires a certain arc 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 a steep drop. The steep drop external characteristic is obtained by the voltage drop generated by the reactance coil. When no-load, since there is no welding current passing through, the reactance coil does not produce a voltage drop. At this time, the no-load voltage is equal to the secondary voltage, which means that the welding transformer is the same as the ordinary transformer when it is no-load.
The no-load current of the transformer is generally about 6% to 8% of the rated current (the state stipulates that the no-load current should not be greater than 10% of the rated current). This is the theoretical basis for formulas and formulas.
Given the capacity of a 380V three-phase motor, find the rated current and setting current of its overload protection thermal relay element
Function:
Motor overload protection, thermal relay Thermal element;
The current capacity is two and a half times, and the setting is twice the kilowatt number.
Note:
(1) Overload protection should be installed for motors that are prone to overload and may fail to start due to serious starting or self-starting conditions, or if the starting time needs to be limited. Motors that run unsupervised for a long time or motors with 3kW and above should also be equipped with overload protection. Overload protection devices generally use thermal relays or time-delay overcurrent releases of circuit breakers. At present, the thermal relays produced in my country are suitable for light load starting, long-term operation or intermittent long-term operation of motor overload protection.
(2) The thermal relay overload protection device has a very simple structural principle, but the optional heating element is very subtle. If the level is selected high, it must be adjusted to the low limit, which often causes the motor to stop unexpectedly and affects production. , increased maintenance work. If the level is selected small, it can only be adjusted to the high limit. Often the motor will not move when it is overloaded, or even burn the motor. (3) To correctly calculate and select the overload protection thermal relay of a 380V three-phase motor, it is necessary to understand that the same series of thermal relays can be equipped with thermal elements with different rated currents. The rated current of the thermal element is calculated according to "twice the kilowatt number"; the rated current of the thermal element is calculated according to "two and a half times the current capacity"; the model specification of the thermal relay, that is, its rated current value should be greater than or equal to the rated current value of the thermal element.
Given the capacity of a 380V three-phase motor, find the rated current level of its remote control AC contactor
Rule:
Remote control motor contactor, twice the capacity Depends on the level;
Step by step to start forward and reverse, and rely on the level to level up.
Explanation:
(1) Currently commonly used AC contactors include CJ10, CJ12, CJ20 and other series, which are more suitable for the starting control of general three-phase motors.
Given the capacity of a small 380V three-phase cage motor, find the minimum capacity of its power supply equipment, load switch, and protection melt current value
Rule:
Direct Starting motor, the capacity does not exceed ten kilowatts;
Six times the kilowatt is selected for the switch, and five times the kilowatt is equipped with the melt.
The power supply equipment needs to be three times larger in kilowatts than kilovolt amperes.
Explanation:
(1) The direct starting motor described in the formula is a small 380V squirrel cage three-phase motor. The starting current of the motor is very large, usually 4 times the rated current. ~7 times. The maximum capacity of the motor started directly by the load switch should not exceed 10kW, generally below 4.5kW, and the open load switch (plastic cover porcelain bottom isolating switch) is generally used for small capacity motors of 5.5kW and below for infrequent operation. Direct starting; enclosed load switch (iron case switch) is generally used for infrequent direct starting of motors below 10kW. Both need to have melt for short circuit protection, and the motor power should not be greater than 30% of the capacity of the power supply transformer. In short, remember that there are conditions for direct starting of the motor with a load switch!
(2) Load switches are composed of simple isolating switch blades and fuses or melts. In order to avoid large current when starting the motor, the capacity of the load switch, that is, the rated current (A); the rated current of the melt for short-circuit protection (A), are calculated according to "six times the kilowatt switch, five times the kilowatt with the fusion piece". Since the iron shell switch and the rubber-covered porcelain-bottom isolating switch are manufactured according to certain specifications, the current value calculated using the formula needs to be close to the switch specifications. The melt selection is also calculated and should be selected according to product specifications.
Given the cage motor capacity, calculate the operating time and thermal element setting current of the star-delta starter (QX3, QX4 series)
Function:
Star-delta motor starts, and the starting time is easy to adjust;
Multiply the square root of the capacity by two, and add four units of seconds to the product.
Motor starting star-delta, overload protection thermal element;
Setting current phase current, capacity multiplied by eight divided by seven.
Description:
(1) QX3 and QX4 series are automatic star-delta starters, consisting of three AC contactors, a three-phase thermal relay and a time relay It consists of a start button and a stop button. Before using the starter, the time relay and thermal relay should be appropriately adjusted. Both of these tasks are performed at the starter installation site. Most electricians only know the capacity of the motor, but not the normal starting time of the motor and the rated current of the motor. The action time of the time relay is the starting time of the motor (the time from starting to the speed reaching the rated value). This time value can be calculated by formula.
(2) When adjusting the time relay, do not connect the motor for operation temporarily, and test whether the action time of the time relay can be consistent with the starting time of the controlled motor. If it is inconsistent, the action time of the time relay should be fine-tuned and then tested. However, the interval between two tests must be at least 90s to ensure that the bimetal time relay automatically resets.
(3) Adjustment of the thermal relay. Since the thermal elements in the thermal relay of the QX series starter are connected in series in the motor phase current circuit, and the motor is connected in a triangle when running, the motor will The phase current is 1/√3 times the line current (that is, the rated current). Therefore, the setting current value of the thermal element of the thermal relay should be calculated using the formula "capacity multiplied by eight divided by seven". According to the calculated value, adjust the setting current knob of the thermal relay to the corresponding scale - around the center line scale. If the calculated value is not within the rated current adjustment range of the thermal relay's thermal element, that is, greater than or less than the high or low limit value marked on the adjustment mechanism's scale, the appropriate thermal relay needs to be replaced, or an appropriate thermal element needs to be selected.
Given the capacity of the cage motor, calculate the setting current of the circuit breaker release that controls it
Function:
The setting current of the circuit breaker release The current capacity is times;
The instantaneous capacity is generally twenty, and the smaller motor is twenty-four;
The delay trip is three and a half times, and the thermal tripper is twice as much.
Explanation: (1) Automatic circuit breakers are commonly used as circuit breakers that operate infrequently on lines that supply power to squirrel-cage motors. If the operation is frequent, a contactor can be added in series for operation. The circuit breaker uses the electromagnetic release (instantaneous) for short-circuit protection, and uses the thermal release (or time-delay release) for overload protection. The calculation of the set current value of the circuit breaker tripper is a problem often encountered by electricians. The formula gives the multiple relationship between the set current value and the kilowatt capacity of the cage motor being controlled.
(2) "Three and a half times the delay tripping, and exactly twice the thermal tripping" refers to the automatic circuit breaker as an overload protection. The current setting value of the delay tripping can be Select 1.7 times the rated current of the controlled motor, which is 3.5 times the kilowatt number. The current setting value of the thermal tripper should be equal to or slightly larger than the rated current of the motor, that is, it should be selected as twice the kilowatt capacity of the motor.
Given the capacity of the asynchronous motor, calculate its no-load current
Function:
Calculate the no-load current of the motor at about 20% off the capacity;
The new large pole is 40% off, and the old small pole has more kilowatts.
Explanation:
(1) When the asynchronous motor runs without load, the current passing through the three-phase winding is determined, which is called no-load current. Most of the no-load current is used to generate a rotating magnetic field, which is called the no-load excitation current, which is the reactive component of the no-load current. There is also a small part of the no-load current used to generate various power losses (such as friction, ventilation and core loss, etc.) when the motor is running without load. This part is the active component of the no-load current, because it accounts for a small proportion. Negligible. Therefore, no-load current can be considered as reactive current. From this point of view, the smaller it is, the better, so that the power factor of the motor is improved, which is good for power supply to the grid. If the no-load current is large, since the conductor carrying area of ??the stator winding is certain and the current allowed to pass through is certain, the active current allowed to flow through the conductors can only be reduced, and the load that the motor can drive will be reduced. When the motor output is reduced and the load is too large, the windings tend to heat up. However, the no-load current cannot be too small, otherwise it will affect other properties of the motor. Generally, the no-load current of small motors is about 30% to 70% of the rated current, and the no-load current of large and medium-sized motors is about 20% to 40% of the rated current. The specific no-load current of a certain motor is generally not marked on the motor's nameplate or product manual. But electricians often need to know what this value is, and use this value to judge the quality of the motor repair and whether it can be used.
(2) The formula is to quickly calculate the specific value of the motor’s no-load current on site. It is obtained from numerous test data. It complies with "the no-load current of a motor is generally 1/3 of its rated current". At the same time, it is in line with practical experience: "The no-load current of the motor can be used if it does not exceed the capacity of kilowatts" (referring to old, small-capacity motors after overhaul). The formula "ask for about 20% off the capacity" means that the no-load current value of a general motor is about 0.8 times the rated capacity of the motor in kilowatts. The no-load current of medium-sized, 4- or 6-pole motors is 0.8 times the kilowatt capacity of the motor; for new series, large-capacity, small-pole 2-level motors, the no-load current calculation is based on the "New large pole number is 40% less" "; For old, old-fashioned series, smaller capacity, more than 8-pole motors with a large number of poles, the no-load current is calculated according to "small pole multi-kilowatts", that is, the no-load current value is approximately equal to the capacity kilowatts, But it is generally less than kilowatts. Use the formula to calculate the no-load current of the motor. There is a certain error between the calculated value and the measured value marked in the motor manual, but the formula value can fully meet the needs of electricians in daily work.
Given the capacity of the power transformer, calculate the setting current value of the instantaneous release of the automatic circuit breaker on its secondary side (0.4kV).
Function:
For power supply on the secondary side of the distribution transformer, it is best to use a circuit breaker;
Instantaneous trip setting value, three times the capacity of kVA.
Explanation:
(1) When the circuit breaker is used as a power supply line switch on the secondary side of the power transformer, the instantaneous action setting value of the circuit breaker tripper is generally as follows
< p>*****Electricians need to be familiar with application tips
Use low-voltage test pen skillfully
Low-voltage test pen is an auxiliary safety tool commonly used by electricians .
Used to check whether conductors below 500V or the shells of various electrical equipment are charged. An ordinary low-voltage test pen can be carried with you. As long as you master the principles of the test pen and combine it with familiar electrical engineering principles, you can use it flexibly.
(1) Tips for judging alternating current and direct current
The electric pen can judge AC and DC, AC is bright and DC is dark,
AC neon tube is bright all over the body, and DC neon tube is bright. One end.
Note:
First of all, we would like to inform readers that before using the low-voltage test pen, it must be tested on a confirmed charged object; the test pen must not be used until it is confirmed that it is normal. When distinguishing between AC and DC, it is best to compare between the "two currents" so that it is obvious. When measuring alternating current, both ends of the neon tube light up at the same time; when measuring direct current, only one end of the neon tube lights up at all.
(2) Tips for judging the positive and negative poles of DC:
Use an electric pen to judge the positive and negative poles. Be careful when observing the neon tube.
The bright front end is the negative pole, and the back end is the negative pole. Bright is the positive pole.
Explanation:
The front end of the neon tube refers to the tip of the test pen, and the back end of the neon tube refers to the end held by the hand. The bright front end is the negative electrode, and the opposite is the positive electrode. When testing, please pay attention to: the power supply voltage is 110V and above; if the person is insulated from the earth, touch any pole of the power supply with one hand, and hold the test pen with the other hand. The metal head of the pen touches the other pole of the power supply under test, and the front end of the neon tube If it lights up, the power supply being measured is the negative pole; if the rear terminal of the neon tube lights up, the power supply being measured is the positive pole. This is based on the principle of one-way flow of DC and the flow of electrons from the negative pole to the positive pole.
(3) Tips to determine whether the DC power supply is grounded or not, and the difference between positive and negative poles
The DC coefficient of the substation will not light up when touched with an electric pen;
If the light is close to the tip of the pen, there is a ground fault in the positive pole;
If the light is close to the finger end, the ground fault is in the negative pole.
Explanation:
The DC coefficient of power plants and substations is insulated from the ground. When a person stands on the ground and touches the positive or negative electrode with an electric test pen, the neon tube is not It should be bright. If it is bright, it means that the DC system is grounded; if it is bright at the end near the pen tip, it means the positive electrode is grounded; if it is bright at the end close to the finger, it is the negative electrode.
(4) Tips for judging the same phase and different phases
To judge whether the two wires are the same or different, hold a pen in each hand,
The two feet are insulated from the ground. Each of the two strokes touches an important line.
Looking at a pen with your eyes, if it does not light up in the same phase, it means it is different.
Note:
During this test, remember that both feet must be insulated from the ground. Because most of the power supply in our country is 380/220V, and the neutral point of the transformer is generally directly grounded, so when doing the test, the human body and the earth must be insulated to avoid forming a loop and misjudgment; when testing, two pens are on and off. The lights show the same, so just look at one.
(5) Tips for judging phase line grounding faults of 380/220V three-phase three-wire power supply lines
Star connection three-phase line, two lights will light up when the electric pen touches two wires,
If the brightness of the remaining one phase wire is weak, the phase conductor is grounded;
If there is almost no brightness, the metal is grounded.
Explanation:
The secondary side of the power transformer is generally connected in a Y shape. In a three-phase three-wire system where the neutral point is not grounded, touch the three phase wires with a test pen When there are two wires that are slightly brighter than usual, and the brightness on the other wire is weaker, it means that the weak phase wire has a grounding phenomenon, but it is not too serious; if two wires are very bright, and the remaining one is almost If no light is visible, there is a metal ground fault in this phase wire.
Artificial respiration is the only method for on-site first aid after electric shock
After the person who receives an electric shock is disconnected from the power source, the physiological state should be determined immediately. Only through correct judgment can the rescue method be determined.
(1) Determine whether there is consciousness or not. The rescuer gently taps or shakes the injured person's shoulders (be careful not to use excessive force or shake the head to avoid aggravating possible trauma) and shout loudly in the ear. If there is no response, immediately pinch the Renzhong point with your fingers. When there is no response to calls or stimulation, it can be determined that consciousness has been lost. This judgment process should be completed within 5S.
When the person who gets an electric shock has lost consciousness, call for help immediately. Lie the person who received the electric shock on his back on a solid surface, with his head flat, his neck not higher than his chest, and his arms flat on both sides of his body. Untie his tight shirt, loosen his belt, take out his dentures, and remove foreign matter from his mouth. . If the person is face down due to electric shock, the head, head, and trunk should be turned over at the same time as a whole without twisting to avoid aggravating possible injuries to the neck. The method of flipping is: the rescuer kneels next to the electrocuted person's shoulders, first raises the electrocuted person's hands above his head, straightens the legs, and puts one leg on top of the other. Then hold the neck of the person who was electrocuted with one hand, hold the shoulder of the person who was electrocuted with one hand, and turn the whole body over at the same time.
(2) Determine whether there is breathing. While keeping the airway open, ways to determine whether there is breathing include: using your eyes to observe the rise and fall of the electrocuted person's chest and abdomen; putting your ears close to the electrocuted person's mouth and nose to listen for the sound of breathing; using your face or hands Place a piece of tissue paper close to the person's mouth and nose to see if gas is discharged; place a piece of tissue paper on the person's mouth and nose to see if the paper moves. If there is no rise and fall in the chest and abdomen, no exhalation, no gas discharge, and the paper does not move, it can be determined that the person who was electrocuted has stopped breathing. This judgment is completed within 3~5S.
Hope this helps!