The calculation is carried out to select and check the settings of relay protection and automation or to check the parameters of the equipment.
We introduce a number of assumptions that simplify the calculation and do not introduce significant errors:
- 1. Linearity of all circuit elements;
- 2. Approximate accounting of loads;
- 3. Symmetry of all elements except for places of short circuit;
- 4. Neglect of active resistances if X/R>3;
- 5. Transformer magnetization currents are not taken into account;
The calculation error under these assumptions does not exceed 2–5%.
The calculation of short circuit currents is simplified when using an equivalent circuit. The calculation of short-circuit currents is carried out in named units.
Estimated short circuit points: K1 - on LV busbars; K2 ... K5 - at the end of the overhead line.
Picture. 9.1. 10 kV equivalent circuit
Three-phase short circuit power:
where IkzVN - short-circuit current on high voltage buses.
System parameters:
Where Ucp is the average voltage, kV;
Power of a three-phase short circuit on the buses of the HV substation, MVA
EMF of the system:
Ес = Uav. (9.3)
Ec = 10.5 kV.
Parameters of power transformers:
Active resistance of the transformer, reduced to the 10.5 kV side.
Reactance of the transformer, reduced to the 10.5 kV side.
Air line parameters:
RVL = r0 l (9.6)
XVL = x0 l (9.7)
RVL = 0.72 11.21 = 8.07 Ohm
XVL \u003d 0.4 11.21 \u003d 4.48 Ohm
The parameters of the outgoing lines are given in Table 9.1.
Table 9.1. Outgoing lines parameters
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Wire brand |
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VL Nekrasovo |
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VL Borisovo |
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VL Lukino |
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VL Pozhara |
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VL Starina |
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VL Proshino |
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The calculation of short-circuit currents is performed for the voltage of the side to which the circuit resistances are reduced.
where is the total total equivalent resistance from the power source to the estimated short circuit point, Ohm.
The steady value of the current in a two-phase short circuit is determined by the value of the current of a three-phase short circuit:
Surge current:
where where is the shock coefficient.
Let's give an example of calculation for Lukino overhead line
The calculation of short circuit currents is summarized in table 9.2.
Table 9.2. Calculation of short circuit currents
|
I(3)kzmax, kA |
I(3)kzmin, kA |
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VL Nekrasovo |
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VL Borisovo |
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|
VL Lukino |
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|
VL Pozhara |
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VL Starina |
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|
VL Proshino |
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|
Busbars 10 kV |
The single-phase earth fault current is determined by the formula:
Iz(1) = 3 Uf sh? Court L (9.13)
where Uph - network phase voltage;
u - angular frequency of the mains voltage;
Court - capacity of 1 km of the network phase relative to the ground, μF / km;
L is the total length of the network, km.
But with accuracy for practical calculations, including, to address the issue of the need to compensate for the capacitive earth fault current, the calculation is made according to the formula:
Where Unom - rated voltage of the network, kV;
Lв - the total length of the overhead lines of the network, km;
Lк - total length of cable lines, km.
Let us determine the current of a single-phase earth fault for outgoing lines of 10 kV. The PUE stipulates: the value of the capacitive earth fault current for the normal mode of the network. And in this case, the normal mode of operation is the separate operation of power transformers (sectional switches are disabled).
For 10 kV outgoing lines:
According to the PUE, clause 1.2.16 Compensation of the capacitive earth fault current should be applied at values of this current in normal modes: in networks with a voltage of 3-20 kV having reinforced concrete and metal supports on overhead power lines, and in all networks with a voltage of 35 kV - more than 10 A. In our case, compensation is not required.
Hello, dear readers and visitors of the Electrician's Notes website.
I have an article on my site about . I cited cases from my practice in it.
So, in order to minimize the consequences of such accidents and incidents, it is necessary to choose the right electrical equipment. But in order to choose it correctly, you need to be able to calculate the short-circuit currents.
In today's article, I will show you how you can independently calculate the short circuit current, or short circuit current for short, using a real example.
I understand that many of you do not need to make calculations, because. Usually this is done either by designers in organizations (firms) that have a license, or by students who write the next course or diploma project. I especially understand the latter, because being a student himself (back in the year 2000), he was very sorry that there were no such sites on the network. Also, this publication will be useful for power engineers and to raise the level of self-development, or to refresh the memory of the once past material.
By the way, I already brought . If you are interested, please follow the link and read.
So, let's get down to business. A few days ago, at our enterprise, a fire broke out on the cable route near assembly shop No. 10. The cable tray burned out almost completely with all the power and control cables going there. Here is a photo from the scene.
I won’t go into “debriefing” much, but my management had a question about the operation of the introductory circuit breaker and its compliance with the protected line. In simple words, I will say that they were interested in the value of the short-circuit current at the end of the input power cable line, i.e. in the place where the fire broke out.
Naturally, shop electricians have no project documentation for calculating short-circuit currents. there was no one for this line, and I had to do the whole calculation myself, which I post in the public domain.
Data collection for calculation of short-circuit currents
Power assembly No. 10, near which a fire occurred, is powered through an automatic switch A3144 600 (A) with a copper cable SBG (3x150) from a step-down transformer No. 1 10 / 0.5 (kV) with a power of 1000 (kVA).
Do not be surprised, we still have many operating substations with isolated neutral at 500 (V) and even 220 (V) at our enterprise.
Soon I will write an article on how to network 220 (V) and 500 (V) with isolated neutral. Do not miss the release of a new article - subscribe to receive news.
The step-down transformer 10 / 0.5 (kV) is fed by the AASHv power cable (3x35) from the high-voltage distribution substation No. 20.
Some clarifications for calculating the short circuit current
I would like to say a few words about the short circuit process itself. During a short circuit, transient processes occur in the circuit, associated with the presence of inductances in it, which prevent a sharp change in current. In this regard, the short-circuit current during the transition process can be divided into 2 components:
- periodic (appears at the initial moment and does not decrease until the electrical installation is disconnected from protection)
- aperiodic (appears at the initial moment and quickly decreases to zero after the completion of the transient process)
Short circuit current I will calculate according to RD 153-34.0-20.527-98.
This regulatory document states that the calculation of the short-circuit current is allowed to be carried out approximately, but on condition that the calculation error does not exceed 10%.
I will carry out the calculation of short-circuit currents in relative units. I will approximate the values of the circuit elements to the basic conditions, taking into account the transformation ratio of the power transformer.
The target is an A3144 with a current rating of 600 (A) per switching capacity. To do this, I need to determine the three-phase and two-phase short circuit current at the end of the power cable line.
Example of calculation of short-circuit currents
We take a voltage of 10.5 (kV) as the main stage and set the basic power of the power system:
basic power of the power system Sb = 100 (MVA)
base voltage Ub1 = 10.5 (kV)
short-circuit current on the busbars of substation No. 20 (according to the project) Ikz = 9.037 (kA)
We draw up a calculation scheme for power supply.
In this diagram, we indicate all the elements of the electrical circuit and their. Also, do not forget to indicate the point at which we need to find the short circuit current. In the figure above, I forgot to indicate it, so I will explain in words. It is located immediately after the low-voltage cable SBG (3x150) before assembly No. 10.
Then we will draw up an equivalent circuit, replacing all the elements of the above circuit with active and reactive resistances.
When calculating the periodic component of the short circuit current, it is allowed to ignore the active resistance of cable and overhead lines. For a more accurate calculation, I will take into account the active resistance on cable lines.
Knowing the basic powers and voltages, we find the basic currents for each stage of transformation:
Now we need to find the reactive and active resistance of each circuit element in relative units and calculate the total equivalent resistance of the equivalent circuit from the power source (power system) to the short circuit point. (highlighted with a red arrow).
Let us determine the reactance of the equivalent source (system):
Let's determine the reactance of the cable line 10 (kV):
- Xo - specific inductive resistance for the AASHv cable (3x35) we take from the reference book on power supply and electrical equipment A.A. Fedorov, volume 2, tab. 61.11 (measured in Ohm/km)
Let's determine the active resistance of the cable line 10 (kV):
- Ro - specific active resistance for the AASHv cable (3x35) we take from the reference book on power supply and electrical equipment A.A. Fedorov, volume 2, tab. 61.11 (measured in Ohm/km)
- l is the length of the cable line (in kilometers)
Let's determine the reactance of a two-winding transformer 10 / 0.5 (kV):
- uk% - short circuit voltage of a transformer 10 / 0.5 (kV) with a power of 1000 (kVA), we take from the reference book on power supply and electrical equipment A.A. Fedorov, tab. 27.6
I neglect the active resistance of the transformer, because it is incommensurably small in relation to the reactive one.
Let's determine the reactance of the cable line 0.5 (kV):
- Xo - resistivity for the SBG cable (3x150) we take from the reference book on power supply and electrical equipment A.A. Fedorov, tab. 61.11 (measured in Ohm/km)
- l is the length of the cable line (in kilometers)
Let's determine the active resistance of the cable line 0.5 (kV):
- Ro - resistivity for the SBG cable (3x150) we take from the reference book on power supply and electrical equipment A.A. Fedorov, tab. 61.11 (measured in Ohm/km)
- l is the length of the cable line (in kilometers)
Let's determine the total equivalent resistance from the power source (power system) to the short circuit point:
Find the periodic component of the three-phase short circuit current:
Find the periodic component of the two-phase short circuit current:
Short circuit current calculation results
So, we calculated the current of a two-phase short circuit at the end of a power cable line with a voltage of 500 (V). It is 10.766 (kA).
The introductory circuit breaker A3144 has a rated current of 600 (A). The setting of the electromagnetic release is set to 6000 (A) or 6 (kA). Therefore, we can conclude that in the event of a short circuit at the end of the input cable line (in my example, due to a fire), the damaged section of the circuit was disconnected.
The obtained values of three-phase and two-phase currents can also be used to select settings for relay protection and automation.
In this article, I did not perform the calculation for the shock current at short circuit.
P.S. The above calculation was sent to my management. For an approximate calculation, it will fit perfectly. Of course, the low side could be calculated in more detail, taking into account the resistance of the contacts of the circuit breaker, the contact connections of the cable lugs to the busbars, the arc resistance at the fault point, etc. I will write about this some other time.
If you need a more accurate calculation, you can use special programs on your PC. There are many of them on the Internet.
The calculation of short-circuit currents is carried out to select and check the electrodynamic and thermal resistance of electrical devices and conductors, design and configure relay protection.
The power sources of the place of short circuit are generators of power plants, power systems and electric motors with voltage over 1000 V, if they are connected to the place of short circuit directly, cable lines, conductors or through linear reactors. The feeding action of electric motors is taken into account only at the initial moment of a short circuit.
To calculate short-circuit currents, a design diagram is drawn up corresponding to the normal mode, which is compiled on the basis of an analysis of the SES circuit and is a single-line electrical circuit.
The design diagram indicates all power sources and network elements, outlines the necessary places in which the calculation of short-circuit currents will be performed. The parameters of power sources and SES elements are given in the initial data. For synchronous generators and electric motors with voltages over 1000 V, the EMF is taken equal to the supertransient EMF E".
As an example, in fig. 3.4 shows the design scheme for the power supply circuit shown in fig. 3.2. According to the design scheme, an equivalent circuit is made up. In this case, all electromagnetic connections between circuit elements are replaced by electrical ones by means of equivalent transformations. Next to each element of the circuit in the numerator is indicated its serial number. n, and in the denominator - the value of the resistance (Ohm) or relative basic units, reduced to the base stage. The transformation stage is usually taken as the base stage, on which the short-circuit current is calculated. Base stage voltage U b is taken equal to the average (rated) voltage U n steps of transformation in accordance with the scale: 230; 154; 115; 37; 10.5; 6.3 kV.
The calculation of short-circuit currents can be made in physical units or relative basic units. When calculating the short-circuit current in relative units, it is convenient to take the base power as a multiple of 10 (for example, 100 or 1000 MB × A), or the power of the power system that supplies the enterprise with electricity, or the rated power of any element of the solar power plant. If the calculation of the short-circuit current is carried out approximately using the calculation curves, then the base power must be taken equal to the power of the supply system.
Basic impedance modulus Z b to the place of short circuit, current I b and power S b are determined by the formulas:
For three-phase two-winding transformers, the active R t and inductive X t resistances, reduced to the higher voltage winding and used in the calculation of the reduced resistances, are given in table. P1.5. For three-phase three-winding transformers, the values of active R t.v., R t.s, R so-called and inductive X t.v., X t.s, X the so-called winding resistances of higher, medium and lower voltages, necessary to calculate the reduced resistances, are indicated in the work. Inductive reactances of reactors X p are given in the work and in table. P1.9.
When calculating the short-circuit current, the EMF of all sources is assumed to be in phase. Therefore, the calculation is carried out using the superposition method: the current from each power source at the fault location is calculated separately, and then the resulting current is found by arithmetic summing the components from the individual sources.
The effective value of the periodic component of the current of a three-phase short circuit in physical units:
When powered by a synchronous generator or electric motor with a voltage of 1000 V:
where ; ; ; p- the number of series-connected active resistances from the power source to the place of the short circuit; m- the number of series-connected inductive reactances from the power source to the place of the short circuit.
The effective value of the periodic component of the current of a three-phase short circuit in relative basic units:
When powered from the power system:
; (3.31)
when powered by a synchronous generator or electric motor with a voltage of over 1000 V:
where is the power of a three-phase symmetrical short circuit,
; ; 
. (3.33)
The transition from short-circuit current and power in relative units to current and power in physical units is made according to the formulas:
If< 0.3 или < 0.3, то активное сопротивление R s is not taken into account when calculating the periodic component of the current of a three-phase short circuit.
The periodic component of the two-phase short circuit current:
Surge current of a three-phase symmetrical short circuit when powered from the power system:
where is the shock coefficient.
Decay time constant of the aperiodic component of the current of a three-phase symmetrical circuit when powered from the power system:
where f- power supply frequency, Hz.
Quadratic current impulse. The overheating temperature of the conductor by the current in the steady state with respect to the ambient temperature is determined from the heat balance equation, i.e. equality of the amounts of released and dissipated heat. Due to the short duration of the short circuit process, heat removal is not taken into account, since the process is considered adiabatic. The total impulse of the quadratic short-circuit current (heating current):
where IN c.p - quadratic current pulse from the periodic component of the short-circuit current; IN c.a – quadratic current pulse from the aperiodic component of the short-circuit current.
In general:
, (3.39)
where m- the number of segments of the discrete time interval when the integral is replaced by a finite sum, ; ε is the symbol of the integer part of the quotient; D t– discrete time interval of splitting the dependency ; - the average value of the short-circuit current on n-th discrete time interval.
Once, a fitter told a lady who was not very knowledgeable in electrical engineering the reason for the loss of light in her apartment. It turned out to be a short circuit, and the woman demanded that it be lengthened immediately. You can laugh at this story, but it’s better to consider this trouble in more detail. Electrical specialists, even without this article, know what this phenomenon is, what it threatens and how to calculate the short circuit current. The information below is addressed to people who do not have a technical education, but, like everyone else, are not immune from the troubles associated with the operation of equipment, machines, production equipment and the most common household appliances. It is important for every person to know what a short circuit is, what are its causes, possible consequences and methods to prevent it. You can not do in this description without getting acquainted with the basics of electrical science. A reader who does not know them may get bored and not finish reading the article to the end.
Popular exposition of Ohm's law
No matter what the nature of the current in an electrical circuit is, it only occurs if there is a potential difference (or voltage, it's the same thing). The nature of this phenomenon can be explained by the example of a waterfall: if there is a level difference, the water flows in some direction, and when not, it stands still. Even schoolchildren know Ohm's law, according to which the current is greater, the higher the voltage, and the less, the higher the resistance included in the load:
I is the magnitude of the current, which is sometimes called "current strength", although this is not a completely competent translation from German. It is measured in Amperes (A).
In fact, the current itself does not have power (that is, the cause of acceleration), which is exactly what manifests itself during a short circuit. This term has already become familiar and is often used, although teachers of some universities, having heard the words “current strength” from the lips of a student, immediately put “bad”. “But what about the fire and smoke coming from the wiring during a short circuit? - the stubborn opponent will ask, - Is this not strength? There is an answer to this comment. The fact is that ideal conductors do not exist, and their heating is due precisely to this fact. If we assume that R=0, then no heat would be released, as is clear from the Joule-Lenz law below.
U is the same potential difference, also called voltage. It is measured in Volts (we have V, abroad V). It is also called electromotive force (EMF).
R - electrical resistance, that is, the ability of the material to prevent the passage of current. For dielectrics (insulators) it is large, although not infinite, for conductors it is small. It is measured in ohms, but is estimated as a specific value. It goes without saying that the thicker the wire, the better it conducts current, and the longer it is, the worse it is. Therefore, resistivity is measured in ohms multiplied by a square millimeter and divided by a meter. In addition, its value is affected by temperature, the higher it is, the greater the resistance. For example, a gold conductor 1 meter long and 1 sq. mm at 20 degrees Celsius has a total resistance of 0.024 Ohm.
There is also the formula of Ohm's law for a complete circuit, the internal (intrinsic) resistance of the voltage source (EMF) is introduced into it.
Two simple but important formulas
It is impossible to understand the reason why a short circuit current occurs without mastering another simple formula. The power consumed by the load is equal (excluding reactive components, but about them later) to the product of current and voltage.
P - power, Watt or Volt-Amp;
U - voltage, Volt;
I - current, Ampere.
Power is never infinite, it is always limited by something, therefore, with its fixed value, as the current increases, the voltage decreases. The dependence of these two parameters of the working circuit, expressed graphically, is called the current-voltage characteristic.
And one more formula necessary in order to calculate the short circuit currents is the Joule-Lenz law. It gives an idea of how much heat is generated when resisting a load, and is very simple. The conductor will heat up with an intensity proportional to the magnitude of the voltage and the square of the current. And, of course, the formula is not complete without time, the longer the resistance is heated, the more heat it will release.
What happens in the circuit during a short circuit
So, the reader can consider that he has mastered all the main physical laws in order to figure out what the magnitude (okay, let it be) of the short circuit current can be. But first you need to decide on the question of what, in fact, it is. A short circuit (short circuit) is a situation in which the load resistance is close to zero. We look at the formula of Ohm's law. If we consider its version for a section of the circuit, it is easy to understand that the current will tend to infinity. In the full version, it will be limited by the resistance of the EMF source. In any case, the short-circuit current is very large, and according to the Joule-Lenz law, the larger it is, the more the conductor along which it goes is heated. Moreover, the dependence is not direct, but quadratic, that is, if I increases a hundredfold, then ten thousand times more heat will be released. This is the danger of a phenomenon that sometimes leads to fires.
The wires glow red hot (or white hot), they transfer this energy to walls, ceilings and other objects that they touch and set them on fire. If the phase in some device touches the neutral conductor, a short-circuit current of the source occurs, closed to itself. The combustible base of electrical wiring is a nightmare for fire inspectors and the cause of many fines imposed on irresponsible owners of buildings and premises. And the fault, of course, is not the laws of Joule-Lenz and Ohm, but the insulation dried up from old age, inaccurate or illiterate installation, mechanical damage or overloading of the wiring.
However, the short-circuit current, no matter how large, is also not infinite. The size of the troubles that he can do is affected by the duration of heating and the parameters of the power supply scheme.
AC circuits
The situations discussed above were of a general nature or concerned DC circuits. In most cases, both residential and industrial facilities are supplied with electricity from an alternating voltage network of 220 or 380 volts. Trouble with DC wiring most often occurs in cars.
There is a difference between these two main types of power supply, and a significant one. The fact is that the passage of alternating current is prevented by additional components of resistance, called reactive and due to the wave nature of the phenomena arising in them. Inductances and capacitances react to alternating current. The short circuit current of the transformer is limited not only by active (or ohmic, that is, one that can be measured with a pocket tester) resistance, but also by its inductive component. The second type of load is capacitive. Relative to the active current vector, the vectors of the reactive components are rejected. The inductive current lags behind, and the capacitive current leads it by 90 degrees.
An example of the difference in the behavior of a load that has a reactive component is a conventional speaker. Some lovers of loud music overload it until the magnetic field knocks the diffuser forward. The coil flies off the core and immediately burns out, because the inductive component of its voltage decreases.
Types of short circuit
Short circuit current can occur in different circuits connected to different sources of direct or alternating current. The easiest way is with the usual plus, which suddenly connected with the minus, bypassing the payload.
But with alternating current, there are more options. A single-phase short-circuit current occurs when a phase is connected to a neutral or grounded. In a three-phase network, unwanted contact between two phases may occur. A voltage of 380 or more (when transmitting power over long distances through power lines) volts can also cause unpleasant consequences, including an arc flash at the time of switching. It can close all three (or four, together with the neutral) wires at the same time, and the three-phase short circuit current will flow through them until the protective automatics work.
But that's not all. In the rotors and stators of electrical machines (motors and generators) and transformers, sometimes such an unpleasant phenomenon as an interturn short circuit occurs, in which adjacent wire loops form a kind of ring. This closed loop has extremely low AC resistance. The strength of the short circuit current in the turns increases, this causes the entire machine to heat up. Actually, if such a disaster occurs, you should not wait until all the insulation melts and the electric motor smokes. The windings of the machine need to be rewound, this requires special equipment. The same applies to those cases when, due to the “interturn”, a short-circuit current of the transformer has arisen. The less the insulation burns, the easier and cheaper rewinding will be.
Short circuit current calculation
No matter how catastrophic this or that phenomenon, its quantitative assessment is important for engineering and applied science. The short circuit current formula is very similar to Ohm's law, it just needs some explanation. So:
I short circuit = Uph / (Zn + Zt),
I k.z. - value of short circuit current, A;
Uph - phase voltage, V;
Zn is the total (including the reactive component) resistance of the short-circuited loop;
Zt is the total (including the reactive component) resistance of the power transformer (power), Ohm.
Impedances are defined as the hypotenuse of a right triangle, the legs of which are the values of active and reactive (inductive) resistance. It's very simple, you need to use the Pythagorean theorem.
Somewhat more often than the short-circuit current formula, experimentally derived curves are used in practice. They represent dependences of the value of I k.z. on the length of the conductor, the cross section of the wire and the power of the power transformer. Charts are a collection of descending exponential lines, from which it remains only to choose the appropriate one. The method gives approximate results, but its accuracy is well suited to the practical needs of power supply engineers.
How is the process
Everything seems to happen instantly. Something hummed, the light dimmed and immediately went out. In fact, like any physical phenomenon, the process can be mentally stretched, slowed down, analyzed and divided into phases. Before the onset of the emergency moment, the circuit is characterized by a steady current value that is within the nominal mode. Suddenly, the impedance drops sharply to a value close to zero. The inductive components (electric motors, chokes and transformers) of the load at the same time, as it were, slow down the process of current growth. Thus, in the first microseconds (up to 0.01 sec), the short-circuit current of the voltage source remains practically unchanged and even slightly decreases due to the onset of the transient. At the same time, its EMF gradually reaches zero, then passes through it and is set at some stabilized value, which ensures the flow of a large I short circuit. The current itself at the time of the transient process is the sum of the periodic and aperiodic components. The shape of the process graph is analyzed, as a result of which it is possible to determine a constant value of time, depending on the angle of inclination of the tangent to the acceleration curve at the point of its inflection (first derivative) and the delay time, determined by the value of the reactive (inductive) component of the total resistance.
Surge short-circuit current
In the technical literature, the term "short-circuit shock current" is often found. You should not be afraid of this concept, it is not at all so terrible and has nothing to do with electric shock. This concept means the maximum value of I k.z. in the alternating current circuit, which usually reaches its value in half a period after the emergency situation has arisen. At a frequency of 50 Hz, the period is 0.2 seconds, and its half is 0.1 seconds, respectively. At this moment, the interaction of conductors located close to each other reaches its maximum intensity. The shock short-circuit current is determined by the formula, which in this article, intended not for specialists and not even for students, does not make sense. It is available in specialized literature and textbooks. In itself, this mathematical expression is not particularly difficult, but it requires rather voluminous comments that deepen the reader into the theory of electrical circuits.
Useful short circuit
It would seem that the obvious fact is that a short circuit is an extremely bad, unpleasant and undesirable phenomenon. It can lead, at best, to a de-energization of the facility, shutdown of emergency protective equipment, and, at worst, to burnout of the wiring and even a fire. Therefore, all efforts must be focused on avoiding this scourge. However, the calculation of short-circuit currents has a very real and practical meaning. A lot of technical means have been invented that operate in the mode of high current values. An example is a conventional welding machine, especially an arc welding machine, which almost short-circuits the electrode with grounding at the time of operation. Another issue is that these regimes are of a short-term nature, and the power of the transformer makes it possible to withstand these overloads. When welding at the point of contact of the end of the electrode, huge currents pass (they are measured in tens of amperes), as a result of which enough heat is released to locally melt the metal and create a strong seam.
Protection methods
In the very first years of the rapid development of electrical engineering, when mankind was still boldly experimenting, introducing galvanic devices, inventing various types of generators, motors and lighting, the problem arose of protecting these devices from overloads and short circuit currents. Its simplest solution was to install fusible elements in series with the load, which were destroyed by resistive heat if the current exceeded the set value. Such fuses serve people today, their main advantages are simplicity, reliability and low cost. But they also have disadvantages. The very simplicity of the “cork” (as the holders of fusible rates were called for their specific shape) provokes users, after it burns out, not to philosophize slyly, but to replace the failed elements with the first wires, paper clips, and even nails that come to hand. Is it worth mentioning that such protection against short-circuit currents does not fulfill its noble function?
At industrial enterprises, circuit breakers began to be used to de-energize overloaded circuits earlier than in apartment shields, but in recent decades, "plugs" have been largely replaced by them. "Automatic machines" are much more convenient, they can not be changed, but turned on by eliminating the cause of the short circuit and waiting for the thermal elements to cool down. Their contacts sometimes burn out, in which case it is better to replace them and not try to clean or repair them. More complex differential automata at a high cost do not last longer than usual ones, but functionally their load is wider, they turn off the voltage in case of minimal current leakage “to the side”, for example, when a person is struck by an electric current.
In everyday life, experimenting with a short circuit is not recommended.
Electric energy carries a rather high danger, from which neither employees of individual substations nor household appliances are protected. Short circuit current is one of the most dangerous types of electricity, but there are methods to control, calculate and measure it.
What it is
Short circuit current (SCC) is a sharply increasing shock electrical impulse. Its main danger is that, according to the Joule-Lenz law, such energy has a very high rate of heat release. A short circuit can melt wires or burn out certain electrical appliances.
Photo - timing diagramIt consists of two main components - the aperiodic component of the current and the forced periodic component.
Formula - Periodic
Formula - aperiodic According to the principle, it is most difficult to measure the energy of aperiodic occurrence, which is capacitive, pre-emergency. After all, it is at the moment of the accident that the difference between the phases has the greatest amplitude. Also, its feature is the non-typical occurrence of this current in networks. The scheme of its formation will help to show the principle of operation of this flow.
The resistance of the sources due to the high voltage during a short circuit closes at a short distance or "short" - therefore this phenomenon has received such a name. There is a three-phase short circuit current, two-phase and single-phase - here the classification occurs according to the number of closed phases. In some cases, the short circuit can be shorted between phases and to ground. Then, in order to determine it, it will be necessary to take into account grounding separately.
Photo - the result of short circuit You can also distribute the short circuit according to the type of connection of electrical equipment:
- With grounding;
- Without him.
To fully explain this phenomenon, we propose to consider an example. Let's say there is a specific current consumer that is connected to a local power line using a tap. With the correct scheme, the total voltage in the network is equal to the difference in EMF at the power source and the decrease in voltage in local electrical networks. Based on this, Ohm's formula can be used to determine the strength of the short circuit current:
R = 0; Ikz = Ɛ/r
Here r is the short circuit resistance.
If we substitute certain values, then it will be possible to determine the fault current at any point along the entire power line. There is no need to check the multiplicity of the short circuit.
Calculation methods
Assume that a short circuit has already occurred in a three-phase network, for example, in a substation or on the windings of a transformer, as then the calculation of short circuit currents is made:
Formula - three-phase short circuit current Here U20 is the voltage of the transformer windings, and Z T is the resistance of a certain phase (which was damaged in a short circuit). If the voltage in the networks is a known parameter, the resistance must be calculated.
Each electrical source, be it a transformer, a battery contact, electrical wires, has its own nominal resistance level. In other words, Z is different for everyone. But they are characterized by a combination of active and inductive resistances. There are also capacitive ones, but they do not matter when calculating high currents. Therefore, many electricians use a simplified way to calculate this data: the arithmetic calculation of the DC resistance in series-connected sections. When these characteristics are known, it will not be difficult to calculate the impedance for a section or an entire network using the formula below:
Full Grounding Formula Where ε is the EMF and r is the resistance value.
Considering that during overloads the resistance is zero, the solution takes the following form:
I \u003d ε / r \u003d 12 / 10 -2
Based on this, the short-circuit strength of this battery is 1200 amperes.
This way you can also calculate the short circuit current for the motor, generator and other installations. But in production it is not always possible to calculate the permissible parameters for each individual electrical device. In addition, it should be taken into account that in case of asymmetric circuits, the loads have a different sequence, which requires knowing cos φ and resistance to take into account. For the calculation, a special table GOST 27514-87 is used, where these parameters are indicated:
There is also the concept of a one-second short circuit, here the formula for the current strength during a short circuit is determined using a special coefficient:
Formula - short circuit coefficient It is believed that, depending on the cable cross-section, a short circuit can pass unnoticed by the wiring. The optimal duration is up to 5 seconds. Taken from Nebrat's book "Calculation of short circuit in networks":
| Section, mm 2 | Short-circuit duration allowed for a particular type of wire | |||
| PVC insulation | Polyethylene | |||
| strands of copper | Aluminum | Copper | Aluminum | |
| 1,5 | 0,17 | No | 0,21 | No |
| 2,5 | 0,3 | 0,18 | 0,34 | 0,2 |
| 4 | 0,4 | 0,3 | 0,54 | 0,36 |
| 6 | 0,7 | 0,4 | 0,8 | 0,5 |
| 10 | 1,1 | 0,7 | 1,37 | 0,9 |
| 16 | 1,8 | 1,1 | 2,16 | 1,4 |
| 25 | 2,8 | 1,8 | 3,46 | 2,2 |
| 35 | 3,9 | 2,5 | 4,8 | 3,09 |
| 50 | 5,2 | 3 | 6,5 | 4,18 |
| 70 | 7,5 | 5 | 9,4 | 6,12 |
| 95 | 10,5 | 6,9 | 13,03 | 8,48 |
| 120 | 13,2 | 8,7 | 16,4 | 10,7 |
| 150 | 16,3 | 10,6 | 20,3 | 13,2 |
| 185 | 20,4 | 13,4 | 25,4 | 16,5 |
| 240 | 26,8 | 17,5 | 33,3 | 21,7 |
This table will help you find out the expected short circuit duration in normal operation, the amperage on tires and various types of wires.
If there is no time to calculate data using formulas, then special equipment is used. For example, the Sh41160 pointer is very popular with professional electricians - this is a 380/220V phase-to-zero short-circuit current meter. The digital device allows you to determine and calculate the short circuit strength in domestic and industrial networks. Such a meter can be bought in special electrical stores. This technique is good if you need to quickly and accurately determine the current level of a loop or circuit segment.
The Avral program is also used, which can quickly determine the thermal effect of a short circuit, the loss rate and the current strength. The check is performed automatically, known parameters are entered and it itself calculates all the data. This project is paid, the license costs about a thousand rubles.
Video: protecting the electrical network from a short circuit
Protection and equipment selection guide
Despite the danger of this phenomenon, there is still a way to limit or minimize the likelihood of accidents. It is very convenient to use an electrical apparatus for limiting short circuits, this can be a current-limiting reactor, which significantly reduces the thermal effect of high electrical impulses. But for domestic use, this option is not suitable.
Photo - diagram of the protection unit against short circuit At home, you can often find the use of a machine and relay protection. These releases have certain limits (maximum and minimum mains current), above which they turn off the power. The machine allows you to determine the allowable level of amperes, which helps to increase safety. The choice is made among equipment with a higher protection class than necessary. For example, in a network of 21 amperes, it is recommended to use a machine to turn off 25 A.
