Practical electrical engineering quick calculation formula collection

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Practical electrical engineering quick calculation formula collection [Copy link]

Given the capacity of the transformer,
the formula for calculating the rated current of each voltage level is:
capacity divided by voltage value, the quotient multiplied by six and divided by ten.
Note: Applicable to any voltage level.
In daily work, some electricians only calculate the 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 can be derived:
multiply the capacity coefficient.

Given the capacity of the transformer, quickly calculate the current value of the primary and secondary protection fuses (commonly known as fuses).
Formula b:
For high-voltage fuses, calculate the capacity by voltage ratio.
For low-voltage fuses, multiply the capacity by 9 and divide by 5.
Explanation:
The correct selection of fuses is extremely important for the safe operation of the transformer. When fuses are used only for the high and low voltage side protection of the transformer, the correct selection of fuses is even more important. This is a problem that electricians often encounter and need to solve.

Given the capacity of a three-phase motor, find its rated current

Formula (c): Divide capacity by kilovolts, and multiply the quotient by the coefficient 0.76.
Note:

(1) The formula is applicable to the calculation of rated current of three-phase motors of any voltage level. The formula and formula can show that the rated currents of motors with the same capacity but different voltage levels are different, that is, the voltage kilovolts are different, and when divided by the same capacity, the resulting "quotients" are obviously different. Different quotients multiplied by the same coefficient 0.76 will also result in different current values. If the above formula is called a general formula, a special calculation formula for calculating the rated current of motors of 220, 380, 660, and 3.6kV voltage levels can be derived. When using the special calculation formula to calculate the rated current of a certain three-phase motor, the relationship between capacity kilowatts and current amperes is directly multiplied, eliminating the need to divide capacity by kilovolts and multiply the quotient by the coefficient 0.76.

Three-phase 220 motor, kilowatt 3.5 amperes.
Commonly used 380 motors, one kilowatt 2 amperes.
Low-voltage 660 motors, kilowatt 1.2 amperes.
For a high-voltage 3,000-volt motor, four kilowatts per ampere.
For a high-voltage 6,000-volt motor, eight kilowatts per ampere.

(2) When using formula c, the capacity unit is kW, the voltage unit is kV, and the current unit is A. This point must be noted.

(3) The coefficient 0.76 in formula c is a comprehensive value calculated by considering the power factor and efficiency of the motor. The power factor is 0.85 and the efficiency is 0.9. These two values are more suitable for motors above tens of kilowatts, but they are larger for commonly used motors below 10kW. This is why the motor rated current calculated using formula c has an error with the value marked on the motor nameplate. This error has little effect on the rated current of motors below 10kW, such as switches, contactors, and wires.

(4) Use formula calculation skills. When using the formula to calculate the rated current of a commonly used 380V motor, first divide 0.76 by the motor power supply voltage of 0.38kV, and multiply the capacity (kW) by the quotient 2. If a 6kV motor with a larger capacity is encountered, and the kW is a multiple of 6kV, then the capacity is divided by the kilovolts, and the quotient is multiplied by the coefficient of 0.76.

(5) Error. The coefficient 0.76 in formula c is calculated by taking the motor power factor as 0.85 and the efficiency as 0.9. In this way, there is an error in calculating the rated current of motors with different power factors and efficiencies. The five special formulas derived from formula c are the multiples of capacity (kW) and current (A), which are the quotients of each voltage level (kV) minus the coefficient of 0.76. Special formulas are simple and easy to calculate mentally, but it should be noted that their errors will increase. Generally, the current calculated for a larger kilowatt is slightly larger than the nameplate; while the current calculated for a smaller kilowatt is slightly smaller than the nameplate. In this regard, when calculating the current, when the current reaches more than ten amperes or tens of amperes, it is not necessary to calculate after the decimal point. It can be rounded up and only the integer is taken, which is both simple and does not affect practicality. For smaller currents, just calculate to one decimal place.

*Measure the current to calculate the capacity

. Measure the no-load current of the motor without a nameplate and estimate its rated capacity .

Formula:
The capacity of the motor without a nameplate, measure the no-load current value,
multiply by ten and divide by eight to calculate the kilowatt of the rating.
Explanation: The formula is for the three-phase asynchronous motor without a nameplate, and its kilowatt capacity is unknown. You can estimate the kilowatt capacity of the motor by measuring the no-load current value of the motor.

Measuring the secondary current of the power transformer and calculating its load capacity
:
Given the secondary voltage of the distribution transformer, calculate the kilowatts of the measured current.
Voltage level 400V, 1A 0.6 kilowatts. Voltage level 300V
, 1A 4.5 kilowatts.
Voltage level 600V, 1A integer 9 kilowatts.
Voltage level 1000V, 1A 15 kilowatts.
Voltage level 35000, 1A 55 kilowatts.
Note:
(1) In daily work, electricians often encounter superiors, managers, etc. asking about the operation of power transformers and what the load is. Electricians themselves often need to know the load of the transformer. The load current is easy to know, just look at the ammeter installed on the distribution device, or use the corresponding clamp ammeter to measure it, but the load power cannot be directly seen and measured. This requires calculation based on this formula. Otherwise, it is complicated and time-consuming to calculate using conventional formulas.

(2) "The voltage level is 400 volts and the power is 0.6 kilowatts per generator." When the load current on the secondary side of the power transformer (voltage level 400 V) is measured, the ampere value is multiplied by the coefficient 0.6 to obtain the load power in kilowatts.

Measure the current of the incandescent lighting circuit and calculate its load capacity . The lighting
voltage is 220V and the voltage is 220W.
Note: Industrial and mining enterprises mostly use 220V incandescent lamps for lighting. The lighting power supply circuit refers to the circuit from the distribution board to each lighting distribution box. The lighting power supply trunk line is generally three-phase four-wire. Single-phase can be used when the load is below 4kW. The lighting distribution circuit refers to the circuit from the lighting distribution box to lighting facilities such as lighting fixtures or sockets. Regardless of the power supply or distribution circuit, as long as the current value of a phase line is measured with a clamp-on ammeter and then multiplied by a coefficient of 220, the product is the load capacity of the phase line. Measuring the current to calculate the capacity can help electricians quickly adjust the problem of unbalanced load capacity of the three-phase lighting trunk line, and can help electricians analyze the reasons why the protective fuse in the distribution box often blows, the reasons why the distribution wires heat up, and so on.

Measure the no-load current of the 380V single-phase welding transformer without a nameplate, and calculate the basic rated capacity
with the formula:
380V welding machine capacity, no-load current multiplied by five.
The 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 under a short-circuit state, and a certain arc-starting voltage is required 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 a steep drop external characteristic. The steep drop external characteristic of the welding transformer is obtained by the voltage drop generated by the reactor coil. When no-load, since there is no welding current passing through, the reactor coil does not produce a voltage drop. At this time, the no-load voltage is equal to the secondary voltage, that is, the no-load current of the welding transformer is the same as that of the ordinary transformer when no-load. The no-load current of the transformer is generally about 6%~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 of the formula and formula.
***

Given the capacity of a 380V three-phase motor, find the rated current and setting current of its overload protection thermal relay element
. Formula:
Motor overload protection, thermal relay thermal element;
rated current capacity two and a half times, two kilowatts setting.
Explanation:
(1) Motors that are prone to overload, may fail to start due to severe starting or self-starting conditions, or need to limit the starting time, should be equipped with overload protection. Motors that run for a long time without supervision or motors of 3kW and above should also be equipped with overload protection. Overload protection devices generally use thermal relays or circuit breaker time-delay overcurrent releases. At present, thermal relays produced in China are suitable for light load starting, long-term operation or intermittent long-term operation of motor overload protection.

(2) The structure and principle of thermal relay overload protection devices are very simple, but the selection of thermal elements is very delicate. If the level is selected too high, it must be adjusted to the lower limit, which often causes the motor to stop secretly, affecting production and increasing maintenance work. If the level is selected too low, it can only be adjusted to the upper limit, often the motor will not work when overloaded, or even burn the motor. (3) To correctly select the overload protection thermal relay for 380V three-phase motors, it is necessary to understand that the same series of thermal relays can be equipped with thermal elements with different rated currents. The setting current of the thermal element is set at "twice the kilowatts"; the rated current of the thermal element is selected at "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-controlled AC contactor. The
formula is:
remote-controlled motor contactor, twice the capacity depends on the level;
frequent starting forward and reverse rotation, depending on the level basis to upgrade one level.
Explanation:
(1) Currently, the commonly used AC contactors are CJ10, CJ12, CJ20 and other series, which are more suitable for the control of the starting of general three-phase motors.

Given the capacity of a small 380V three-phase squirrel cage motor, calculate the minimum capacity of its power supply equipment, load switch, and protective fuse current value. The
formula is:
for direct starting of motors, the capacity should not exceed 10 kW;
select a switch for six times the kW, and a fuse for five times
the kW. The kilovolt-ampere of the power supply equipment should be three times the kW.
Note:
(1) The motor directly started in the formula is a small 380V squirrel cage three-phase motor. The motor starting current is very large, generally 4 to 7 times the rated current. The maximum capacity of the motor directly started with a load switch should not exceed 10kW, and is generally less than 4.5kW. In addition, open-type load switches (rubber-covered porcelain-bottomed isolating switches) are generally used for infrequent direct starting of small-capacity motors of 5.5kW and below; closed-type load switches (iron-shell switches) are generally used for infrequent direct starting of motors below 10kW. Both require fuses for short-circuit protection, and the motor power should not exceed 30% of the capacity of the power supply transformer. In short, remember that there are conditions for direct starting of motors with load switches!

(2) Load switches are composed of simple isolating switch blades and fuses or fuses. In order to avoid large currents when the motor starts, the capacity of the load switch, i.e. the rated current (A); the rated current (A) of the fuse for short-circuit protection, are selected according to "six times the kilowatts for switches and five times the kilowatts for fuses". Since iron shell switches and rubber-covered porcelain-bottom isolating switches are manufactured according to certain specifications, the current value calculated by the formula must be close to the switch specifications. Similarly, when selecting fuses, they should be selected according to product specifications.

Given the capacity of a cage motor,
the formula for calculating the action time and setting current of the star-delta starter (QX3, QX4 series) is as follows:
When the motor is started in star-delta, the starting time is easy to set;
multiply the capacity by two, and add four seconds to the product.
When the motor is started in star-delta, the thermal element is overload protected;
the setting current is the phase current, the capacity multiplied by eight and divided by seven.
Note:
(1) The QX3 and QX4 series are automatic star-delta starters, which consist of three AC contactors, a three-phase thermal relay and a time relay, with a start button and a stop button. Before using the starter, the time relay and thermal relay should be properly 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 using the formula.

(2) When adjusting the time relay, do not connect the motor to operate temporarily, and test whether the action time of the time relay can be consistent with the starting time of the controlled motor. If not, the action time of the time relay should be fine-tuned again, and then the test should be carried out. However, the interval between the two tests should be at least 90s to ensure that the bimetallic time relay automatically resets.

(3) Adjustment of the thermal relay. Since the thermal element in the thermal device of the QX series starter is connected in series in the phase current circuit of the motor, and the motor is connected in a triangle when running, the phase current of the motor when running is 1/√3 times the line current (i.e. the rated current). Therefore, the set current value of the thermal element of the thermal relay should be calculated by the formula "capacity multiplied by eight divided by seven". According to the calculated value, adjust the set current knob of the thermal relay to the corresponding scale-center line scale. If the calculated value is not within the rated current adjustment range of the thermal element of the thermal relay, that is, greater than or less than the upper or lower limit value marked on the scale of the adjustment mechanism, it is necessary to replace the appropriate thermal relay or select an appropriate thermal element.

Given the capacity of a squirrel-cage motor, calculate the set current of the circuit breaker release that controls it.
The formula is:
the set current of the circuit breaker release is times the capacity;
the instantaneous is generally twenty, and twenty-four for smaller motors;
the delayed release is three and a half times, and the thermal release is two times.
Note: (1) Automatic circuit breakers are often used as circuit breakers that are not frequently operated on the lines that supply power to squirrel-cage motors. If the operation is frequent, a contactor can be added in series to operate it. The circuit breaker uses the electromagnetic release (instantaneous) as short-circuit protection and the thermal release (or delayed release) as overload protection. Calculating the set current value of the circuit breaker release is a common problem encountered by electricians. The formula gives the multiple relationship between the set current value and the kilowatt capacity of the squirrel-cage motor controlled.

(2) "Time-delay release three and a half times, thermal release two times" means that the current setting value of the time-delay release of the automatic circuit breaker used as overload protection can be selected as 1.7 times the rated current of the controlled motor, that is, 3.5 times the kilowatts. The current setting value of the thermal release should be equal to or slightly greater than the rated current of the motor, that is, selected as twice the kilowatts of the motor capacity.

Given the capacity of an asynchronous motor, the formula for calculating its no-load current
is:
the no-load current of the motor is calculated at about 80% of the capacity;
the new large pole number is 60% less, and the old small pole number is more kilowatts .
Explanation:
(1) When the asynchronous motor is running at no load, the current passing through the three-phase winding is called the no-load current. Most of the no-load current is used to generate a rotating magnetic field, 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 at no load. This part is the active component of the no-load current, which can be ignored because it accounts for a very small proportion. Therefore, the no-load current can be considered as all 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 beneficial to the power supply of the power grid. If the no-load current is large, because the conductor area of the stator winding is fixed and the current allowed to pass is fixed, the active current allowed to flow through the conductor can only be reduced, and the load that the motor can drive will be reduced, the motor output will be reduced, and when carrying too large a load, the winding will easily heat up. However, the no-load current cannot be too small, otherwise it will affect other performances of the motor. Generally, the no-load current of a small motor is about 30%~70% of the rated current, and the no-load current of a large and medium-sized motor is about 20%~40% of the rated current. The specific no-load current of a motor is generally not marked on the motor nameplate or product manual. However, electricians often need to know this value to judge the quality of the motor repair and whether it can be used.

(2) The formula is a formula for quickly calculating the specific value of the motor no-load current on site. It is obtained from a large number of test data. It conforms to the principle that "the no-load current of the motor is generally 1/3 of its rated current". At the same time, it conforms to the practical experience that "the no-load current of the motor can be used as long as it does not exceed the capacity kilowatts" (referring to old and small-capacity motors after repair). The formula "calculate at about 80% of the capacity" means that the no-load current value of a general motor is about 0.8 times the rated capacity kilowatts of the motor. The no-load current of a medium-sized, 4- or 6-pole motor is 0.8 times the kilowatts of the motor capacity; for a new series, large capacity, small pole number 2-stage motor, its no-load current is calculated as "60% less for a new large pole number"; for an old, old series, smaller capacity, large pole number 8-pole or above motor, its no-load current is calculated as "small pole more kilowatts", that is, the no-load current value is approximately equal to the kilowatts of the capacity, but generally less than the kilowatts. Using the formula to calculate the no-load current of the motor, the calculated value has a certain error from the actual value marked in the motor manual, but the formula can fully meet the needs of electricians' 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) output line.
Formula:
For the secondary side power supply of the transformer, it is best to use a circuit breaker;
the instantaneous release setting value is three times the capacity kVA.
Note:
(1) When the circuit breaker is used as the switch of the secondary side power supply line of the power transformer, the instantaneous action setting value of the circuit breaker release is generally set according to
*****

Electricians need to be familiar with the application formulas
and skillfully use low-voltage test pens
. Low-voltage test pens are an auxiliary safety tool commonly used by electricians. They are used to check whether conductors below 500V or the casings of various electrical equipment are charged. An ordinary low-voltage test pen can be carried with you. As long as you master the principle of the test pen and combine it with the familiar electrical principles, you can flexibly use it.
(1) Formula for judging AC and DC: The test
pen judges AC and DC. AC is bright and DC is dark
. AC neon tubes are bright all over, and DC neon tubes are bright at one end.
Note:
First of all, let the readers know that before using a low-voltage test pen, you must test it on a confirmed charged body; you must not use it before confirming that the test pen is normal. When judging AC and DC, it is best to compare the "two electricity" so that it is very obvious. When measuring AC, both ends of the neon tube are lit at the same time, and when measuring DC, only one end of the neon tube is lit.
(2) Formula for judging the positive and negative poles of DC:
The test pen judges the positive and negative poles. Be careful when observing the neon tube.
The front end is bright if it is negative, and the back end is bright if it is positive.
Note:
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. If the front end is bright, it is the negative pole, otherwise it is the positive pole. When testing, please note: the power supply voltage is 110V or above; if the person is insulated from the ground, one hand touches one pole of the power supply, and the other hand holds the test pen, the metal head of the test pen touches the other pole of the power supply being tested, the front end of the neon tube is bright, and the power supply being tested is the negative pole; if the back end of the neon tube is bright, the power supply being tested is the positive pole, which is based on the principle of unidirectional flow of direct current and the flow of electrons from the negative pole to the positive pole.
(3) To determine whether the DC power supply is grounded, the difference between positive and negative grounding is the
DC coefficient of the substation, the test pen does not light up when touched;
if it is bright near the tip of the pen, there is a grounding fault at the positive pole;
if it is bright near the finger end, the grounding fault is at the negative pole.
Note:
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 pole with a test pen, the neon tube should not light up. If it does, it means that the DC system is grounded. If it lights up near the tip of the pen, the positive pole is grounded; if it lights up near the finger, the negative pole is grounded.
(4) To judge whether
two lines are the same or different, hold a pen in each hand,
insulate the two feet from the ground, touch one of the two pens to a line, and
watch one pen with your eyes. If the same phase does not light up, it is different.
Note:
When doing this test, remember that the two feet must be insulated from the ground. Because most of China is powered by 380/220V, and transformers generally use direct neutral point grounding, so when doing the test, the human body must be insulated from the ground to avoid forming a loop and avoid misjudgment. When testing, the display of two pens is the same whether they are lit or not, so you only need to watch one.
(5) Tips for judging the grounding fault of the phase line of the 380/220V three-phase three-wire power supply line : If
the three-phase line is connected in star shape, two of them are brighter than usual when touched by the test pen, and
the brightness of the remaining one is weak, the phase conductor is grounded;
if it is almost invisible, it is a metal grounding fault.
Explanation:
The secondary side of the power transformer is generally connected in a Y shape. In a three-phase three-wire system with an ungrounded neutral point, when the three phase lines are touched by the test pen, two of them are slightly brighter than usual, and the brightness of the other one is weaker, which means that the phase line with weak brightness is grounded, but it is not too serious; if two of them are very bright, and the remaining one is almost invisible, then this phase line has a metal grounding fault.

First aid for electric shock: artificial respiration
should be performed immediately after the person is disconnected from the power source. Only after a correct judgment can the rescue method be determined.
(1) Determine whether the person is conscious. The rescuer gently pats or shakes the shoulder of the person who has been electrocuted (be careful not to use too much force or shake the head to avoid aggravating possible external injuries) and calls loudly in the ear. If there is no response, immediately pinch the Ren Zhong point with your fingers. If there is no response to the call and no response to the stimulation, it can be determined that the person has lost consciousness. This judgment process should be completed within 5 seconds.
If the person who has been electrocuted has lost consciousness, call for help immediately. Lay the person on his back on a solid surface, with his head flat, the neck not higher than the chest, and his arms flat on both sides of the torso. Unbutton the tight top, loosen the belt, remove the dentures, and remove foreign objects in the mouth. If the person who has been electrocuted is facing down, the head, shoulder, and torso should be turned over as a whole at the same time, and should not be twisted to avoid aggravating possible injuries to the neck. The method of turning over is as follows: the rescuer kneels beside the shoulder of the person who has been electrocuted, first raises the person's hands above his head, straightens his legs, and puts one leg on top of the other. Then, one hand holds the person's neck, the other hand holds the person's shoulder, and the whole body turns over at the same time.
(2) Determine whether the person is breathing. While keeping the airway open, the methods for determining whether the person is breathing are as follows: use your eyes to observe whether the chest and abdomen of the person who has been electrocuted are rising and falling; put your ears close to the mouth and nose of the person who has been electrocuted to listen for breathing sounds; put your face or hands close to the mouth and nose of the person who has been electrocuted to test whether there is any gas discharge; put a thin piece of paper on the mouth and nose of the person who has been electrocuted to observe whether the paper moves. If there is no rise and fall of the chest and abdomen, no exhalation, no gas discharge, and the paper does not move, it can be determined that the person who has been electrocuted has stopped breathing. This determination is completed within 3 to 5 seconds.

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