Topic: what is a short circuit in an electrical circuit, what are the consequences of a short circuit.
Many have heard about an electrical short circuit, but not everyone knows the essence of this phenomenon. Let's deal with this. So, if you delve into the phrase “short circuit” itself, then you can understand that some kind of process is taking place in which something is closed along a short, namely the shortest path of flow electric current (electric charges in the conductor). Simply put, there is a path through which electricity flows, its current of charges. These are various electrical circuits, conductors of electricity. The longer this path, the more obstacles the charges need to overcome, the more electrical resistance this path. And from Ohm's law, it is known that the greater the resistance of the circuit, the less current will be in it (at a certain voltage value). Therefore, on the shortest path, there will be the maximum possible current, and this path will be short if the ends of the power source itself are shorted.
In general, we have, for example, an ordinary car battery (in a charged state). If you connect a light bulb designed for battery voltage (12 volts) to it, then as a result of the passage of a current of a certain value through this lamp, we will receive light and heat radiation. The lamp has a certain electrical resistance, which limits the current flowing through this circuit. To intentionally short circuit, we just need to take a piece of wire and connect it to the ends of the battery leads (parallel to the lamp). This wire has very little resistance compared to a lamp. Therefore, there is no special restriction that would prevent the movement of charged particles. And as soon as we close such a circuit, we get our short circuit. The wire will flow immediately large current, which can simply heat and melt this piece of wire.
As a result of such a short circuit, the conductor (its insulation) will ignite, up to a fire, if this conductor, by its ignition, transfers fire to flammable things that are nearby. In addition, such a sharp, spasmodic current flow can be harmful to the battery itself. It also starts to heat up at this time. And as you know, batteries really do not like excessive heat. At a minimum, their service life is significantly reduced after this, and as a maximum, they fail and even catch fire and explode. If such a short circuit occurs, for example, with a lithium battery in a phone (which has no electronic protection inside), strong heating occurs for several seconds, followed by a flame and an explosion.
There are some batteries that are originally designed to deliver high currents (traction batteries), but even with them a complete short circuit can lead to big trouble. Well, what happens to the voltage during a short circuit? From school physics it should be known that the greater the current, the greater the voltage drop in this section of the circuit. Therefore, when no load is connected to the power supply, you can see the maximum voltage value on it (this is source emf supply, its electromotive force). As soon as we load this power supply, a certain voltage drop immediately appears. And the greater the load, the greater the voltage drop. Since, in the event of a short circuit, the circuit resistance is practically zero, and the current strength will be the maximum possible, the voltage drop across the power source will also be maximum (near zero).
This we considered the option of a complete short circuit, which occurs directly at the terminals of the power source. Yes, here's what else to add about it. In the case of a battery, there will be a large current load on the internal parts and chemicals of the battery itself (electrolyte, plates, leads). In the event of a short circuit on such power sources as electric generators, the current load falls on the windings of these generators, which leads to its excessive heating and damage (well, those circuits that operate in the generator after this winding). A short circuit at the terminals of various power supplies leads to overheating and failure of the electrical circuits current sources and secondary winding transformer.
A short circuit may occur in the electrical circuit wiring diagrams. In this case, the consequences are also extremely negative. But at the same time, the current strength will already be, as a rule, slightly less than in the case of a short circuit at the output of the power source. For example, there is a sound amplifier circuit. Suddenly, due to poor insulation of the speakers themselves, a short circuit occurs at the sound output of this amplifier. As a result, the output transistors, the microcircuit in the last stages of sound amplification, will most likely burn out. The power supply itself in this case may not even be affected, since an excessive current load may not reach it. I think you got the point of the short circuit.
P.S. In any case, the phenomenon of an electrical short circuit leads to disastrous consequences. Normal fuses, circuit breakers, protection circuits, etc. are usually used to protect against this. Their task is to quickly break the electrical circuit with a sharp increase in current strength. That is, an ordinary fuse is, as it were, the weakest link in all electrical circuits. As soon as the current strength has increased sharply, the fusible link simply melts and breaks the circuit. This in most cases results in other circuits in the circuit remaining undamaged.
The main reason for the disruption of the normal operation of the power supply system (PSS) is the occurrence of short circuits (SC) in the network or electrical equipment due to damage to the insulation or improper actions of the maintenance personnel. To reduce the damage caused by the failure of electrical equipment during the flow of short-circuit currents, as well as to quickly restore the normal operation of the solar power plant, it is necessary to correctly determine the short-circuit currents and select electrical equipment, protective equipment and means of limiting short-circuit currents.
short circuit is called a direct connection between any points of different phases, phase and neutral wire or phase to earth, not provided for by the normal operating conditions of the installation.
The main types of short circuits in electrical systems:
3. Single-phase short circuit, at which one of the phases is shorted to a neutral wire or ground. Symbol single-phase short circuit points 
Currents, voltages, powers and other quantities related to a single-phase short circuit are indicated 
,
,
etc.
There are other types of short circuits associated with wire breaks and simultaneous short circuits of wires of various phases.
A three-phase short circuit is symmetrical, since with it all three phases are in the same conditions. All other types of short circuits are asymmetric, since with them the phases do not remain in the same conditions, as a result of which the systems of currents and voltages turn out to be distorted.
When a short circuit occurs, the total electrical resistance of the power supply system circuit decreases, as a result of which the currents in the branches of the system increase sharply, and the voltages in certain sections of the system decrease.
Elements of electrical systems have active and reactive (inductive or capacitive) resistances, therefore, in the event of a sudden disruption of normal operation (in the event of a short circuit), the electrical system is an oscillatory circuit. Currents in the branches of the system and voltages in its individual parts will change for some time after the occurrence of a short circuit in accordance with the parameters of this circuit. Those. during the short circuit in the circuit of the damaged section, a transient process occurs.
During a short circuit in each of the phases, along with the periodic component of the current (component of the variable sign current), there is an aperiodic component of the current (component of the constant sign), which can also change sign, but at longer intervals compared to the periodic one.
Instant value full current Short circuit for an arbitrary point in time:
where 
- aperiodic component of the short-circuit current at the moment of time 
;
- angular frequency alternating current; 
- phase angle of the source voltage at the moment of time ;
- current shift angle in the short circuit relative to the source voltage; - time constant of the short circuit; 
- inductance, inductive and active resistance of the short circuit circuit.
Periodic component 
short circuit current (Fig. 1) is the same for all three phases and is determined for any moment of time by the value of the ordinate of the envelope divided by 
. Aperiodic component 
short circuit current is different for all three phases (see Fig. 2) and varies depending on the moment of occurrence of short circuit.
Rice. 3. Change in time of the periodic component of the short circuit current:
a) when powered by generators without ATS; b) when powered by generators with ATS; c) when powered by the power system.
The amplitude of the periodic component changes in the transient process in accordance with the change in the EMF of the short circuit source (Fig. 3). With a source power commensurate with the power of the element where the short circuit is considered, as well as the absence of AVR generators, the EMF of the source decreases from the initial value 
to steady state 
, as a result of which the amplitude of the periodic component changes from 
(supertransient short-circuit current) up to 
(sustained short circuit) (Fig. 3a).
In the presence of AVR generators, the periodic component of the short circuit current changes, as shown in fig. 3, b. The decrease in the periodic component in the initial period of the short circuit is explained by the inertia of the action of the ARV device, which starts working through 0.08-0.3 s after the onset of the short circuit. With an increase in the excitation current of the generator, its EMF increases and, accordingly, the periodic component of the short-circuit current up to a steady value.
If the power of the source is significantly greater than the power of the element where the short circuit is considered, which corresponds to a source of unlimited power, for which the internal resistance is zero, then the EMF of the source is constant. Therefore, the periodic component of the short-circuit current is unchanged during the transient process (Fig. 3, c), i.e.
Aperiodic component of short-circuit current 
is different in all phases and may vary depending on the moment of occurrence of the short circuit and the previous mode (within the period). The decay rate of the aperiodic current component depends on the ratio between the active and inductive resistance of the short circuit circuit, i.e. from constant 
: the greater the active resistance of the circuit, the more intense the attenuation. The aperiodic component of the short circuit current noticeably manifests itself only in the first 0.1-0.2 s after the onset of the short circuit. Usually is determined by the largest possible instantaneous value, which (in circuits with predominant inductive resistance 
) takes place at the moment when the source voltage passes through the zero value ( 
) and no load current. Wherein 
.In this case, the total short-circuit current has the greatest value. These conditions are calculated when determining short-circuit currents.
The maximum instantaneous short circuit current occurs after about half a period, i.e. 0.01 s after the occurrence of a short circuit. The highest possible instantaneous short circuit current is called shock current 
(Fig. 3). It is determined for the moment 
With:
where 
- shock coefficient, depending on the time constant of the short circuit.
The effective value of the total short-circuit current for an arbitrary moment of time is determined from the expression:
(3.4)
where 
- effective value of the periodic component of the short circuit current; 
- effective value of the aperiodic component equal to
(3.5)
The largest effective value of the shock current for the first period from the beginning of the short circuit process:
(3.6)
Short-circuit power for an arbitrary moment of time:
(3.7)
Short circuit power supplies. When calculating short circuit currents, it is assumed that the power sources of the short circuit are turbo and hydro generators, synchronous compensators and motors, and asynchronous motors. The influence of asynchronous motors is taken into account only at the initial moment of time and in those cases when they are connected directly to the place of short circuit.
Determined quantities. When calculating short-circuit currents, the following quantities are determined:
- the initial value of the periodic component of the short circuit current (the initial value of the supertransient short circuit current);
- shock short-circuit current, necessary for checking electrical devices, tires and insulators for electrodynamic stability;
- the largest effective value of the shock short-circuit current, necessary for checking electrical apparatus for stability during the first period of the short-circuit process;
- meaning for 
, necessary to check the circuit breakers for the current they turn off;
- the effective value of the steady-state short-circuit current, which is used to check electrical apparatus, busbars, bushings and cables for thermal stability;
- short circuit power for time ;determined to test circuit breakers according to the maximum permissible disconnected power. For high-speed circuit breakers, this time can be reduced to 0.08 s.
Assumptions and design conditions. To facilitate the calculation of short-circuit currents, a number of assumptions are made:
1) EMF of all sources are considered to be in phase;
2) EMF sources, significantly remote from the place of short circuit ( 
), are considered unchanged;
3) do not take into account transverse capacitive short circuit circuits (except for overhead lines 330 kV above and cable lines 110 kV above) and magnetizing currents of transformers;
4) the active resistance of the short circuit circuit is taken into account only with the ratio 
,
where 
and 
- equivalent active and reactive resistances of a short-circuited circuit;
5) in some cases, the influence of loads is not taken into account (or taken into account approximately), in particular, the influence of small asynchronous and synchronous motors.
In accordance with the purpose of determining short circuit currents, design conditions are established, which include drawing up a design scheme, determining the short circuit mode, short circuit type, location of short circuit points and estimated short circuit time.
When determining the short circuit mode, depending on the purpose of the calculation, the possible maximum and minimum levels of short circuit currents are determined. So, for example, checking electrical equipment for the electrodynamic and thermal effects of short-circuit currents is carried out according to the most severe mode - the maximum, when the highest short-circuit current flows through the element under test. On the contrary, according to the minimum mode corresponding to the smallest short-circuit current , calculate and check the performance of devices relay protection and automation.
Selecting the type of short circuit determined by the purpose of calculating short-circuit currents. To determine the electrodynamic resistance of devices and hard tires, a three-phase short circuit is taken as the calculated one; to determine the thermal resistance of devices, conductors - a three-phase or two-phase short circuit, depending on the current. Checking the breaking and making abilities of the devices is carried out according to a three-phase or according to single-phase current Short circuit to earth (in networks with high earth fault currents) depending on its value.
The choice of the type of short circuit in the calculations of relay protection is determined by its functional purpose and can be three-, two-, single-phase and two-phase short circuit to ground.
Locations of short circuit points are chosen in such a way that during a short circuit the electrical equipment under test, the conductors are in the most unfavorable conditions. For example, to select switching equipment, it is necessary to choose the location of the short circuit directly on their output terminals, the choice of section cable line produce a short circuit current at the beginning of the line. The locations of short circuit points in the calculations of relay protection are determined according to its purpose - at the beginning or end of the protected area.
Estimated short circuit time. The actual time during which a short circuit occurs is determined by the duration of the protection and disconnecting equipment,
. (3.8)
In the calculations, the reduced (fictitious) time is used - the period of time during which the steady-state short-circuit current releases the same amount of heat that the actually passing short-circuit current should release during the actual short-circuit time.
The reduced time corresponding to the total short-circuit current,
. (3.9)
where 
- the reduced time for the periodic component of the short circuit current;
- the reduced time for the aperiodic component of the short circuit current.
With real time 
with the reduced time for the periodic component of the short circuit current is determined by nomograms.
With real time 
With 
, where 
- the value of the given time for 
With.
Determination of the reduced time for the aperiodic component , and is produced at 
according to the formula:
, (3.10)
where 
- the ratio of the initial supertransient current to the current established at the fault location ( 
).
At 
- according to the formula:
. (3.11)
With real time more than 1 sec. or 
the reduced time of the aperiodic component of the short-circuit current ( ) can be neglected.
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, we had a fire at our enterprise. cable route near workshop assembly No. 10. Burnt out almost completely cable tray with all the power and control cables. Here is a photo from the scene.
I won’t go into the “debriefing” much, but my management had a question about the operation of the introductory circuit breaker and its correspondence to the protected line. In simple words I will say that they were interested in the magnitude 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, no project documentation for shop electricians, according to the calculations of 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.
In that normative document it is said 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's define reactance 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.
Required to complete calculation three-phase current short circuit (TKZ) on the buses of the designed ZRU-6 kV 110/6 kV substation "GPP-3". This substation is powered by two 110 kV overhead lines from the 110 kV GPP-2 substation. ZRU-6 kV "P4SR" receives power from two power transformers TDN-16000/110-U1, which work separately. When one of the inputs is disconnected, it is possible to supply power to the de-energized busbar section by means of a sectional switch in automatic mode (ATS).
Figure 1 shows the design scheme of the network
Since the chain from I s.sh. "GPP-2" to I s.sh. "GPP-3" is identical to chain II s.sh. from "GPP-2" to II s.sh. "GPP-3" calculation is carried out only for the first chain.
The equivalent circuit for calculating short-circuit currents is shown in Figure 2.
The calculation will be made in named units.
2. Initial data for calculation
- 1. System data: Ikz=22 kA;
- 2. Data for overhead lines - 2xAS-240/32 (Data are given for one AC-240/32 circuit, RD 153-34.0-20.527-98, Appendix 9):
- 2.1 Direct sequence inductive reactance - Х1ud=0.405 (Ohm/km);
- 2.2 Capacitive conductivity - wsp=2.81x10-6 (S/km);
- 2.3 Active resistance at +20 С per 100 km of the line - R=R20C=0.12 (Ohm/km).
- 3. Transformer data (taken from GOST 12965-85):
- 3.1 TDN-16000/110-U1, Uin=115 kV, Un=6.3 kV, on-load tap-changer ±9*1.78, Uk.in-nn=10.5%;
- 4. Data of the flexible conductor: 3хАС-240/32, l=20 m. (To simplify the calculation, the resistance of the flexible conductor is not taken into account.)
- 5. Data of the current limiting reactor - RBSDG-10-2x2500-0.2 (taken from GOST 14794-79):
- 5.1 Rated current reactor - Inom. = 2500 A;
- 5.2 Nominal power losses per reactor phase - ∆P= 32.1 kW;
- 5.3 Inductive resistance - Х4=0.2 Ohm.
3. Calculation of element resistances
3.1 System resistance (for voltage 115 kV):
3.2 Overhead line resistance (for voltage 115 kV):
Where:
n - Number of wires in one overhead line VL-110 kV;
3.3 Total resistance to the transformer (for voltage 115 kV):
X1.2 \u003d X1 + X2 \u003d 3.018 + 0.02025 \u003d 3.038 (Ohm)
R1.2=R2=0.006 (Ohm)
3.4 Transformer resistance:
3.4.1 Active resistance of the transformer (OLTC is in the middle position):
3.4.2 Active resistance of the transformer (OLTC is in the extreme "minus" position):
3.4.3 Active resistance of the transformer (OLTC is in the extreme "positive" position):
Minimum inductive resistance of the transformer (OLTC is in the extreme "minus" position)
The maximum inductive resistance of the transformer (OLTC is in the extreme "positive" position)
The value included in the formula above is the voltage corresponding to the extreme positive position of the on-load tap-changer, and it is equal to Umax.VN = 115 * (1 + 0.1602) = 133.423 kV, which exceeds the highest operating voltage of electrical equipment equal to 126 kV (GOST 721-77 " Power supply systems, networks, sources, converters and receivers electrical energy. Rated voltages over 1000 V"). The voltage UmaxVN corresponds to Uk%max=10.81 (GOST 12965-85).
If Umax.VN, it turns out more than the maximum allowable for this network (Table 5.1), then Umax.VN should be taken according to this table. The value Uk% corresponding to this new maximum value Umax.VN is determined either empirically or found from the applications of GOST 12965-85.
3.4.5 Resistance of the current-limiting reactor (at a voltage of 6.3 kV):
4. Calculation of three-phase short-circuit currents at point K1
4.1 Total inductive reactance:
X∑=X1.2=X1+X2=3.018+0.02025=3.038 (Ohm)
4.2 Total active resistance:
R∑=R1.2=0.006 (Ohm)
4.3 Total impedance:
4.4 Three-phase short circuit current:
4.5 Surge short-circuit current:
5. Calculation of three-phase short-circuit currents at point K2
6.1.1 The value of the total resistance at point K2, we lead to a network voltage of 6.3 kV:
6.1.2 The current at the place of a short circuit, reduced to an effective voltage of 6.3 kV, is equal to:
6.1.3 Surge short-circuit current:
6.2 Resistance on the busbars of the ZRU 6 kV with the on-load tap-changer of the T3 transformer set to the minus position
6.2.1 The value of the total resistance at point K2 is reduced to a network voltage of 6.3 kV:
6.2.2 The current in the place of a short circuit, reduced to an effective voltage of 6.3 kV, is equal to:
6.2.3 Surge short-circuit current:
6.3 Resistance on the busbars of the ZRU 6 kV with the on-load tap-changer of the T3 transformer installed in the positive position
6.3.1 The value of the total resistance at point K2, we lead to a network voltage of 6.3 kV:
6.3.2 The current at the place of the short circuit, reduced to an effective voltage of 6.3 kV, is equal to:
6.3.3 Surge short-circuit current:
The results of the calculations are entered in table РР1.3
Table РР1.3 - Calculation data for three-phase short-circuit currents
| Transformer tap position | Short circuit currents | Short circuit point | ||
|---|---|---|---|---|
| K1 | K2 | K3 | ||
| Tap changer in middle position | Short circuit current, kA | 21,855 | 13,471 | 7,739 |
| Surge current short circuit, kA | 35,549 | 35,549 | 20,849 | |
| Short circuit current, kA | - | 13,95 | 7,924 | |
| Surge current short circuit, kA | - | 36,6 | 21,325 | |
| Tap changer in positive position | Short circuit current, kA | - | 13,12 | 7,625 |
| Surge current short circuit, kA | - | 34,59 | 20,553 | |
7. Calculation of short circuit current made in Excel
If you perform this calculation using a piece of paper and a calculator, it takes a lot of time, besides, you can make a mistake and the whole calculation will go down the drain, and if the initial data is constantly changing, this all leads to an increase in design time and unnecessary waste of nerves.
Therefore, I decided to perform this calculation using an Excel spreadsheet, so that I no longer waste my time on recalculating the TKZ and protect myself from unnecessary errors, with its help you can quickly recalculate the short circuit currents, changing only the initial data.
I hope that this program will help you and you will spend less time designing your facility.
8. References
- 1. Guidelines for the calculation of short-circuit currents and the choice of electrical equipment.
RD 153-34.0-20.527-98. 1998 - 2. How to calculate the short circuit current. E. N. Belyaev. 1983
- 3. Calculation of short circuit currents in power networks 0.4-35 kV, Golubev M.L. 1980
- 4. Calculation of short circuit currents for relay protection. I.L. Nebrat. 1998
- 5. Rules for the installation of electrical installations (PUE). Seventh edition. 2008
Hello dear friends! In this article, you will learn what short circuit current is, its causes and how to calculate it. A short circuit occurs when current-carrying parts of different potentials or phases are connected to each other. A short circuit can also form on the equipment case that is connected to the ground. This phenomenon also characteristic of electrical networks and electrical receivers.
Causes and effects of short circuit current
The causes of a short circuit can be very different. This is facilitated by a humid or aggressive environment in which the insulation resistance deteriorates significantly. A short circuit can be the result of mechanical influences or human errors during repair and maintenance. The essence of the phenomenon lies in its name and is a shortening of the path along which the current passes. As a result, current flows past the resistive load. At the same time, it increases to unacceptable limits if the protective shutdown does not work.
Short-circuit currents have an electrodynamic and thermal effect on equipment and electrical installations, which ultimately leads to their significant deformation and overheating. In this regard, it is necessary to pre-calculate the short-circuit currents.
How to calculate short circuit current at home
Knowledge of the magnitude of the short circuit current is essential to ensure fire safety. It is obvious that if the measured short circuit current is less than the current setting of the maximum protection of the machine or 4 times the value of the fuse current, then the response time (burnout of the fuse link) will be longer, and this, in turn, can lead to excessive heating of the wires and their fire.
How can this current be determined? There are special techniques and special devices for this. Here we consider the question of how to do this, having only or even a voltmeter. It is obvious that this method has not very high accuracy, but still sufficient to detect the discrepancy between the overcurrent protection and the value of this current.
How to do it at home? It is necessary to take a sufficiently powerful receiver, for example, an electric kettle or iron. It would be nice to have a trio. We connect our consumer and a voltmeter or multimeter to the tee in voltage measurement mode. We write down the steady value of the voltage (U1). We turn off the consumer, and record the voltage without load (U2). Next, we make a calculation. You need to divide the power of your consumer (P) by the difference in the measured voltages.
Ic.c.(1) = Р/(U2 – U1)
Let's count on an example. Kettle 2 kW. The first measurement is 215 V, the second measurement is 230 V. According to the calculation, it turns out 133.3 A. If, for example, there is a VA 47-29 machine with characteristic C, then its setting will be from 80 to 160 Amperes. Therefore, it is possible that this machine will work with a delay. According to the characteristics of the machine, it can be determined that the response time can be up to 5 seconds. Which is basically dangerous.
What to do? It is necessary to increase the magnitude of the short circuit current. You can increase this current by replacing the wires of the supply line with a larger cross section.
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, disconnection 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 has been invented technical means operating 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.