Handbook for calculating short circuit currents. Calculation of short circuit currents. Network short circuit protection
Hello, dear readers and visitors of the Electrician's Notes website.
I have an article on my website about. I cited cases from my practice.
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 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... This is usually done either by designers in licensed organizations (firms), or by students who are writing their next coursework or diploma project. I especially understand the latter, because... Being a student myself (back in the year 2000), I really regretted that there were no such sites on the Internet. This publication will also be useful for energy workers to raise the level of self-development, or to refresh their memory of previously passed material.
By the way, I already brought it. If anyone is interested, follow the link and read.
So let's get down to business. A few days ago at our enterprise there was a fire on a cable route near assembly workshop No. 10. The cable tray with all the power and control cables running there was almost completely burnt out. Here is a photo from the scene.
I won’t go into much detail about the debriefing, but my management had a question about the operation of the input circuit breaker and its suitability for 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 happened.
Naturally, shop electricians have no design documentation for calculating short-circuit currents. there was no money for this line, and I had to make the entire calculation myself, which I am posting in the public domain.
Collection of data for calculating short-circuit currents
Power assembly No. 10, near which the fire occurred, is fed through a circuit breaker 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).
Don’t be surprised, we still have many operating substations at our enterprise with an isolated neutral at 500 (V) and even 220 (V).
Soon I will write an article about how to connect to a 220 (V) and 500 (V) network with an isolated neutral. Don't miss the release of a new article - subscribe to receive news.
The 10/0.5 (kV) step-down transformer is powered by the AAShv power cable (3x35) from the high-voltage distribution substation No. 20.
Some clarifications for calculating 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 due to the presence of inductances in it that 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 the 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 short-circuit current can be carried out approximately, but provided that the calculation error does not exceed 10%.
I will calculate short circuit currents in relative units. I will approximately bring 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 rated current 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 calculating short circuit currents
We take a voltage of 10.5 (kV) as the main stage and set the base power of the power system:
base 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) Is = 9.037 (kA)
We draw up a design diagram 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. I forgot to indicate it in the picture above, so I’ll explain it 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 permissible not to take into account the active resistance of cable and overhead lines. For a more accurate calculation, I will take into account the active resistance on the cable lines.
Knowing the base powers and voltages, we will find the base currents for each transformation stage:
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 by 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 reactance for cable AAShv (3x35) is taken from the reference book on power supply and electrical equipment by A.A. Fedorov, volume 2, table. 61.11 (measured in Ohm/km)
Let's determine the active resistance of the cable line 10 (kV):
- Rо - specific active resistance for cable AAShv (3x35) is taken from the reference book on power supply and electrical equipment by A.A. Fedorov, volume 2, table. 61.11 (measured in Ohm/km)
- l — cable line length (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), taken from the reference book on power supply and electrical equipment by A.A. Fedorov, table. 27.6
I neglect the active resistance of the transformer, because it is disproportionately 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) is taken from the reference book on power supply and electrical equipment by A.A. Fedorov, table. 61.11 (measured in Ohm/km)
- l — cable line length (in kilometers)
Let's determine the active resistance of the cable line 0.5 (kV):
- Ro - resistivity for the SBG cable (3x150) is taken from the reference book on power supply and electrical equipment by A.A. Fedorov, table. 61.11 (measured in Ohm/km)
- l — cable line length (in kilometers)
Let's determine the total equivalent resistance from the power source (power system) to the short-circuit point:
Let's find the periodic component of the three-phase short circuit current:
Let's find the periodic component of the two-phase short circuit current:
Results of calculation of short circuit currents
So, we have calculated the two-phase short circuit current at the end of a power cable line with a voltage of 500 (V). It is 10.766 (kA).
The input 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 be used to select settings for relay protection and automation.
In this article, I did not calculate the shock current during a short circuit.
P.S. The above calculation was sent to my management. For an approximate calculation it is quite suitable. Of course, the low side could be calculated in more detail, taking into account the resistance of the circuit breaker contacts, the contact connections of the cable lugs to the busbars, the arc resistance at the fault point, etc. I'll 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.
Electrical energy carries a fairly high danger, from which neither workers at individual substations nor household appliances are protected. Short circuit current is one of the most dangerous types of electricity, but there are methods on how 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. As a result of a short circuit, wires may melt or certain electrical appliances may burn out.
Photo - timing diagramIt consists of two main components - the aperiodic current component and the forced periodic component.
Formula – periodic
Formula – aperiodic According to the principle, the most difficult thing to measure is 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 peculiarity is the non-typical occurrence of this current in networks. The diagram of its formation will help show the principle of operation of this flow.
Due to the high voltage during a short circuit, the resistance of the sources is short-circuited over a short distance or “short circuit” - that’s why this phenomenon got its name. There is a short circuit current of three-phase, two-phase and single-phase - here the classification occurs according to the number of closed phases. In some cases, the short circuit may be shorted between phases and to ground. Then, to determine it, you will need to separately take into account grounding.
Photo – result of short circuit You can also distribute short circuits according to the type of electrical equipment connection:
- With grounding;
- Without him.
To fully explain this phenomenon, we suggest considering 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 circuit, the total voltage in the network is equal to the difference in EMF at the power source and the voltage reduction in local electrical networks. Based on this, Ohm's formula can be used to determine the short-circuit current:
R = 0; Ikz = Ɛ/r
Here r is the short-circuit resistance.
If you substitute certain values, you can determine the fault current at any point along the entire power line. There is no need to check the short circuit multiplicity here.
Calculation methods
Let's assume that a short circuit has already occurred in a three-phase network, for example, at a substation or on the windings of a transformer, how then the short circuit currents are calculated:
Formula - three-phase fault current Here U20 is the voltage of the transformer windings, and Z T is the resistance of a certain phase (which was damaged in the short circuit). If the voltage in the networks is a known parameter, resistance must be calculated.
Each electrical source, be it a transformer, a battery terminal, or electrical wires, has its own nominal resistance level. In other words, everyone has their own Z. But they are characterized by a combination of active and inductive resistances. There are also capacitive ones, but they are not important when calculating high currents. Therefore, many electricians use a simplified method for calculating this data: an arithmetic calculation of the direct current 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 = ε/r = 12 / 10 -2
Based on this, the short circuit strength of this battery is 1200 Amperes.
In this way, it is also possible to calculate the short-circuit current for a motor, generator and other installations. But in production it is not always possible to calculate acceptable parameters for each individual electrical device. In addition, it should be taken into account that with asymmetrical short 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 cross-section of the cable, a short circuit can pass unnoticed by the wiring. The optimal short circuit duration is up to 5 seconds. Taken from Nebrat’s book “Calculation of short circuits in networks”:
| Section, mm 2 | Short circuit duration permissible for a specific type of wire | |||
| PVC insulation | Polyethylene | |||
| Copper veins | 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 conditional duration of a short circuit in normal operation, the amperage on the busbars and various types of wires.
If there is no time to calculate data using formulas, then special equipment is used. For example, the Shch41160 indicator is very popular among 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 household and industrial networks. Such a meter can be purchased at special electrical stores. This technique is good if you need to quickly and accurately determine the current level of a loop or section of circuit.
The “Emergency Emergency” program is also used, which can quickly determine the thermal effect of a short circuit, loss rate and current strength. The check is carried out automatically, known parameters are entered and it calculates all the data itself. This is a paid project, the license costs about a thousand rubles.
Video: protecting the electrical network from short circuits
Protection and equipment selection guidelines
Despite the danger of this phenomenon, there is still a way to limit or minimize the likelihood of emergency situations. It is very convenient to use an electrical apparatus to limit short circuits; this can be a current-limiting reactor, which significantly reduces the thermal effect of high electrical impulses. But this option is not suitable for domestic use.
Photo - diagram of the short-circuit protection unit At home, you can often find the use of automatic circuit breakers and relay protection. These releases have certain restrictions (maximum and minimum network current), if exceeded, the power is turned off. The machine allows you to determine the permissible ampere level, which helps increase safety. The choice is made among equipment with a higher protection class than necessary. For example, on a 21 amp network, it is recommended to use a 25 A circuit breaker.
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 short circuit are generators of power plants, power systems and electric motors with voltages over 1000 V, if they are connected to the short circuit directly, by cable lines, conductors or through line reactors. The feeding effect of the electric motors is taken into account only at the initial moment of the short circuit.
To calculate short-circuit currents, a design diagram corresponding to the normal mode is drawn up, which is compiled based on an analysis of the SES diagram and is a single-line electrical diagram.
The calculation diagram indicates all power sources and network elements, and outlines the necessary places where the short-circuit currents will be calculated. The parameters of power supplies and SES elements are given in the source 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 diagram for the power supply circuit shown in Fig. 3.2. Based on the calculation diagram, an equivalent circuit is drawn up. In this case, all electromagnetic connections between the circuit elements are replaced by electrical ones through equivalent transformations. Next to each element of the diagram, its serial number is indicated in the numerator n, and the denominator is the resistance value (Ohm) or relative basic units reduced to the base stage. The transformation stage at which the short-circuit current is calculated is usually taken as the base stage. Base stage voltage U b is taken equal to the average (nominal) voltage U n stages of transformation in accordance with the scale: 230; 154; 115; 37; 10.5; 6.3 kV.
Short circuit currents can be calculated in physical units or relative basic units. When calculating the short-circuit current in relative units, it is convenient to take as the base power a power that is 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 design curves, then the base power should be taken equal to the power of the supply power system.
Basic Impedance Module Z b to the point of short circuit, current I b and power S b are determined by the formulas:
For three-phase two-winding transformers, the active value R t and inductive X t of resistance, reduced to the higher voltage winding and used in calculating the reduced resistance, are given in table. P1.5. For three-phase three-winding transformers, the active values R t.v, R t.s. R so-called and inductive X t.v, X t.s. X The so-called resistances of the windings of high, medium and low voltages, necessary for calculating the given resistances, are indicated in the work. Reactor inductances 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 performed using the superposition method: the current from each power source at the short circuit is calculated separately, and then the resulting current is found by arithmetic summing the components from the individual sources.
Effective value of the periodic component of the three-phase short circuit current 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 short circuit; m– the number of series-connected inductive reactances from the power source to the short circuit.
Effective value of the periodic component of the three-phase short circuit current in relative basic units:
· when powered from the power system:
; (3.31)
· when powered by a synchronous generator or electric motor with voltage 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 carried out 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 three-phase short circuit current.
Periodic component of the two-phase short circuit current:
Surge current of three-phase symmetrical short circuit when powered from the power system:
where is the impact 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 – supply frequency, Hz.
Square current pulse. The temperature of conductor overheating by current in steady state in relation to the ambient temperature is determined from the heat balance equation, i.e. equality of the amounts of generated 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. Total pulse of square-law short-circuit current (heating current):
Where IN kp – quadratic current pulse from the periodic component of the short-circuit current; IN k.a – quadratic current pulse from the aperiodic component of the short-circuit current.
In general:
, (3.39)
Where m– the number of segments of a discrete time interval when replacing the integral with a finite sum, ; ε – symbol of the integer part of the quotient; D t– discrete time interval of dependence splitting; – average value of short-circuit current on n-th discrete time interval.
In this article, I will consider example of calculating the current of a single-phase short circuit (SC) using in the first version the lookup tables presented in [L1], and in the second option the lookup tables from [L2].
Methods for determining the magnitude of the single-phase short-circuit current and the given reference tables for all elements of a short-circuited circuit can be found in the article:
Initial data:
- oil transformer with voltage 6/0.4 kV, power 1000 kVA with winding connection diagram - Y/Yo.
- From the transformer to the ASU, a AAShvU 3x95 cable with a length of 120 m is used.
- From the ASU to the engine, a AASHvU 3x95+1x35 cable with a length of 150 m is used.
Fig. 1 - Design diagram of the electrical network. engine
Option I
1. Calculation of the single-phase short circuit current will be performed using the formula of the approximate method at high power of the supply power system (Xc< 0,1Хт) [Л1, с 4 и Л2, с 39]:
- Uph – phase voltage of the network, V;
- Zt – total resistance of the transformer to the current of a single-phase short circuit to the body, Ohm;
- Zpt – total resistance of the phase-zero loop from the transformer to the short-circuit point, Ohm.
2. Using Table 2 [L1, p. 6], we determine the resistance of the transformer at a secondary voltage of 400/230 V, Zt/3 = 0.027 Ohm.
3. We determine the total resistance of the phase-zero circuit for the section from the pipeline to the short-circuit point using formula 2-27 [L2, p. 40]:
- Zpt.ud.1 = 0.729 Ohm/km – total resistivity of the phase-zero loop for cable brand AAShvU 3x95, determined according to Table 12 [L1, p. 16];
- l1 = 0.120 km – length of section No. 1.
- Zpt.ud.2 = 0.661 Ohm/km – total resistivity of the phase-zero loop for cable brand ААШвУ 3х95+1х35, determined according to Table 13 [L1, p. 16];
- l2 = 0.150 km – length of section No. 2.
4. Determine the single-phase short circuit current:
I draw your attention to the fact that when determining the magnitude of the single-phase short-circuit current using the approximate method, the contact resistances of busbars, devices, and current transformers are not taken into account in this method, since the arithmetic sum of Zt/3 and Zpt creates a certain reserve [L2, p. 40].
Option II
Let's determine the single-phase short circuit current using the reference tables from [L2].
1. Using Table 2.4 [L2, p. 29], we determine the transformer resistance Zt/3 = 33.6 mOhm.
2. We determine the total resistance of the phase-zero circuit for the section from the pipeline to the short-circuit point using formula 2-27 [L2, p. 40]:
- Zpt.ud.1 = 0.83 mOhm/m – total resistivity of the phase-zero loop for cable brand AAShvU 3x95, determined according to Table 2.11 [L2, p. 41];
- l1 = 120 m – length of section No. 1.
- Zpt.ud.2 = 1.45 mOhm/m – total resistivity of the phase-zero loop for cable brand AAShvU 3x95+1x35, determined according to Table 2.10 [L2, p. 41].
Please note that in this table the value of Zpt.ud. given for cables regardless of the cable sheath material.
If you look at [L1, p. 16], then in table 13 for 4-core cables with an aluminum sheath 3x95+1x35, Zpt.ud. = 0.661 mOhm/m. I accept Zpt.ud.2 = 1.45 mOhm/m, in order to clearly see how much the value of the single-phase short-circuit current will differ from the calculation according to “Option I”. In practice, it is better to combine lookup tables from [L1 and L2].
3. Determine the single-phase short circuit current:
As can be seen from the calculation results (option I: Ik = 1028 A; option II: Ik = 627 A), the obtained values of the single-phase short-circuit current are almost 2 times different. Decide for yourself which reference tables to use to calculate the single-phase short-circuit current; in any case, this is an approximate method, therefore, if you need exact values of the single-phase short-circuit current, you should calculate using the formula presented in GOST 28249-93.
Today I would like to present to your attention a method for calculating short-circuit currents. The most important thing is without any water and each of you will be able to use it with a minimum of effort, and some of you will receive my next program, with which it will be even easier to count.
This is the second article devoted to short circuit currents. I drew your attention to the protection of extended electrical networks and the fact that in such networks, sometimes, it is not so easy to select protection against short circuit currents. That's what a designer is for, to solve such issues.
The theory for calculating short-circuit currents can be found in the following documents:
1 GOST 28249-93 (Short circuits in electrical installations. Calculation methods in AC electrical installations with voltage up to 1 kV).
2 RD 153-34.0-20.527-98 (Guidelines for the calculation of short circuit currents and the selection of electrical equipment).
3 A.V. Belyaev (Selection of equipment, protection and cables in 0.4 kV networks).
I couldn’t find anything on the Internet where everything was clearly laid out from “A” to “Z.”
I think you will agree with me that short-circuit currents are not so easy to calculate, since the designer does not always thoroughly have all the necessary information. This calculation method is simplified, because it does not take into account the contact resistance of circuit breakers, fuses, busbars, and current transformers.
Perhaps later I will take all these resistances into account, but, in my opinion, these values have little effect on the final result.
Sequence of calculation of short circuit currents.
1 Collection of initial data on the transformer:
Ukz— transformer short circuit voltage, %;
RK— transformer short circuit losses, kW;
Uvn– rated voltage of the HV windings of the step-down transformer; kV;
Unn (El)– rated voltage of the LV windings of the step-down transformer; IN;
Eph– phase voltage of the LV windings of the step-down transformer; IN;
Snt– rated power of the transformer, kVA;
Zt– total resistance of the step-down transformer with single-phase short-circuit current, mOhm;
Active and inductive resistance of transformers 6 (10)/0.4 kV, mOhm
2 Collection of initial data on the supply line:
Type, cable cross-section, number of cables;
L– line length, m;
Ho– inductive reactance of the line, mOhm/m;
Zpt– total resistance of the phase-zero loop from the transformer to the short-circuit point, measured during testing or found from calculation, mOhm/m;
3 Other data.
Where– impact coefficient.
After collecting the initial data, you can proceed directly to the calculations.
Active resistance of the step-down transformer, mOhm:
Transformer resistance
Inductive reactance of step-down transformer, mOhm:
Active resistance of the supply line, mOhm:
RTo= Rud.k*l/ NTo
Inductive reactance of the supply line, mOhm:
XTo=Hood.k*l/ Nk
Total active resistance, mOhm:
RΣ = RT+RTo
Total inductive reactance, mOhm:
XΣ =XT+XTo
Total resistance, mOhm:
Three-phase short circuit current, kA:
Impact current of three-phase short-circuit, kA:
Single-phase short circuit current, kA:
Zpt=Zpt.ud.*L
Having calculated the short circuit currents, you can begin to select protective devices.
Using this principle, I made my new program for calculating short-circuit currents. Using the program, all calculations can be performed much faster and with minimal risk of errors that may occur during manual calculations. For now, this is still a beta version, but nevertheless, I think it’s a completely working version of the program.
Appearance of the program:
Below in the program are all the necessary tables for selecting the necessary parameters of the transformer and supply line.
Also, along with the program, I am attaching a sample of my calculation so that the calculation can be quickly completed and provided to all interested authorities.
It is worth noting that I have another small program - interpolation. It is convenient, for example, to find the specific load of apartments for given values.
I look forward to your feedback, wishes, suggestions, clarifications.
To be continued... there will be more
Is it necessary to take into account the resistance of switching devices when calculating the short circuit?
