Расчет электрической цепи постоянного тока онлайн: Расчет электрической цепи постоянного тока

Расчёт электрической цепи постоянного тока методом узловых и контурных уравнений.

Этот принцип основан на первом и втором законе Кирхгофа. Он не требует преобразования схемы.

Порядок расчёта:

1. Произвольно задаёмся направлением токов в ветвях. (Токи в ветвях надо направлять так, что бы хотя бы один ток выходил из узла и один входил в узел)
   

Красным выделены изменения после первого действия

Синим выделены изменения после второго пункта

2. Составляем уравнение для узлов по первому закону Кирхгофа. Их должно быть n минус 1 . (n – число узлов)
   1. Обозначаем узлы буквами.
   2. Берём один конкретный узел (Например узел А) и смотрим как направлены токи в ветвях образующих узел. Если ток направлен в узел, то записываем его со знаком  плюс, если из него то со знаком минус. 0=I1-I4-I6 (Полученное уравнение)
   3. Повторяем пункт B ещё для двух узлов.0=-I3+I4+I5(Узел В)                                         0=I3-I1-I2(Узел D)
3. Произвольно задаёмся обходом контура (по часовой или против часовой). И составляем уравнения для контуров цепи по второму закону Кирхгофа. В данном примере направление обхода контура выберем по часовой стрелке.

 3.1 Смотрим, как направлена ЭДС относительно обхода контура. Если направление обхода контура совпадает, то значение ЭДС записываем со знаком плюс (в левой части уравнения), если не совпадает, то со знаком минус (записываем также в левой части уравнения)

3.2 Смотрим, как направлено падение напряжения на сопротивлении контура.(То есть смотрим как направлены токи, только записываем в уравнение произведение тока на сопротивление через которое ток протекает в данном контуре). Если направление обхода контура совпадает, то падение напряжения записываем со знаком плюс (в правой части уравнения), если не совпадает, то со знаком минус (записываем также в правой части уравнения)

3.3 Произвести действия 3.1 и 3.2 для остальных контуров. У вас должна получится система из n уравнений, где n — количество контуров в цепи.

Контур ABDA E1=I1*(R1+R01)+I4*R4+I3*R3

Контур BCDB E2=I2*(R2+R02)+I3*R3+I5*R5

Контур ABCA 0=I6*R6-I4*R4+I5*R5

• Решаем полученную систему уравнений и находим величины токов во всех ветвях. 

Уберём лишние токи из системы используя уравнения полученные во втором пункте поскольку у нас три уравнения поэтому мы оставляем только три любых тока. Для данного примера я рекомендую оставить токи I1 I2 I4.

0=I1-I4-I6 (Узел А)

0=-I3+I4+I5(Узел В)

0=I3-I1-I2(Узел D)

Выражаем из трёх уравнений токи I3 I5 I6 через токи I1 I2 I4.

I6=I1-I4(Узел А)

I3=I1+I2(Узел D)

I5=I3-I4(Узел В)

I5=I3-I4(Узел В) В этом уравнении сразу не получилось выразить I5 через токи I1 I2 I4, поэтому вместо тока I3 подставим уравнение для узла D и получим:

I5=I1+I2-I4

Заменим токи I3 I5 I6 и получим уравнения с тремя токами :

 E1=I1*(R1+R01)+I4*R4+(I1+I2)*R3

 E2=I2*(R2+R02)+(I1+I2)*R3+(I1+I2-I4)*R5

 0=(I1-I4)*R6-I4*R4+(I1+I2-I4)*R5

Раскрываем скобки подставляем значения сопротивлений из условия и получаем например вот такие три уравнения:

40 = 71*I1 + 24*I2 + 14*I4

20 = 55*I1 + 93*I2 — 61*I4

0 = 60*I1 + 16*I2 — 81*I4

Дальше для решения системы можно воспользоваться бесплатной онлайн программой на нашем сайте.

• Если при решении системы ток получается отрицательным (со знаком —), значит его действительное направление противоположно тому направлению которое мы задали в первом действии.
• Правильность  решения можно проверить с помощью баланса мощностей.

90000 Short Circuit Current Calculation at Various Point of Electrical Circuits (Isc) 90001 90002 Available fault current calculation 90003 90004 Infinite Bus Method 90005 90006 Short Circuit Current Calculation at Various Point of Electrical Circuits (Isc) 90007 1. Fault Current At Transformer 90008 90007 Inputs: 90008 90011 90012 Transformer Kva 90013 90012 Secondary Voltage of Transformer (Line-Line) 90013 90012 Secondary Voltage of Transformer (Line-Neutral) 90013 90012 Transformer Impedance (Zt) 90013 90012 Transformer Secondary Full Load Current I (Line-Line) 90013 90012 Transformer Secondary Full Load Current I (Line-Neutral) 90013 90024 90011 90012 KVA Short Circuit At Primary Side 90013 90012 Source Impedance (Zu) 90013 90012 Total Impedance (Z = Zu + Zt) 90013 90012 Short Circuit Current RMS Symmetrical 90013 90012 Fault Current At Transformer Secondary (Isc (L-L) = I (L-L) / Total Impedance) 90013 90012 Fault Current At Transformer Secondary (Isc (L-N) = I (L-N) / Total Impedance) 90013 90024 90007 2.Fault Current At Main Panel 90008 90007 Inputs: 90008 90011 90012 Cable Length from Transformer to Main Panel 90013 90012 Type of Cable 90013 90012 No of Conductor / Phase 90013 90012 Size of Phase Conductor 90013 90012 Size of Neutral Conductor 90013 90012 Type of Race Way 90013 90012 Phase Conductor Constant (C1) 90013 90024 90011 90012 Neutral Conductor Constant (C2) 90013 90012 F1 (L-L) [F (L-L) = (1.732 X Length X Isc) / (C1 X No of Conductor X Volt)] 90013 90012 F2 (L-N) [F (L-L) = (1.732 X Length X Isc) / (C2 X No of Conductor X Volt)] 90013 90012 M1 (L-L) [M = (1 / (1 + F1)] 90013 90012 M2 (L-N) [M = (1 / (1 + F2)] 90013 90012 Fault Current at Main Panel Disconnection (L-L) [Isc X M1] 90013 90012 Fault Current at Main Panel Disconnection (L-N) [Isc X M2] 90013 90024 90007 3. Fault Current At Sub Panel 90008 90007 Inputs: 90008 90011 90012 Cable Length from Main Panel to Sub Panel 90013 90012 Type of Cable 90013 90012 No of Conductor / Phase 90013 90012 Size of Phase Conductor 90013 90012 Size of Neutral Conductor 90013 90012 Type of Race Way 90013 90012 Phase Conductor Constant (C1) 90013 90024 90011 90012 Neutral Conductor Constant (C2) 90013 90012 F1 (L-L) [F (L-L) = (1.732 X Length X Isc) / (C1 X No of Conductor X Volt)] 90013 90012 F2 (L-N) [F (L-L) = (1.732 X Length X Isc) / (C2 X No of Conductor X Volt)] 90013 90012 M1 (L-L) [M = (1 / (1 + F1)] 90013 90012 M2 (L-N) [M = (1 / (1 + F2)] 90013 90012 Fault Current at Sub Panel Disconnection (L-L) [Isc X M1] 90013 90012 Fault Current at Sub Panel Disconnection (L-N) [Isc X M2] 90013 90024 90007 4. Fault Current At Distribution Board 90008 90007 Inputs: 90008 90011 90012 Cable Length from Sub Panel to Distribution Board 90013 90012 Type of Cable 90013 90012 No of Conductor / Phase 90013 90012 Size of Phase Conductor 90013 90012 Size of Neutral Conductor 90013 90012 Type of Race Way 90013 90012 Phase Conductor Constant (C1) 90013 90024 90011 90012 Neutral Conductor Constant (C2) 90013 90012 F1 (L-L) [F (L-L) = (1.732 X Length X Isc) / (C1 X No of Conductor X Volt)] 90013 90012 F2 (L-N) [F (L-L) = (1.732 X Length X Isc) / (C2 X No of Conductor X Volt)] 90013 90012 M1 (L-L) [M = (1 / (1 + F1)] 90013 90012 M2 (L-N) [M = (1 / (1 + F2)] 90013 90012 Fault Current at Distribution Board Disconnection (L-L) [Isc X M1] 90013 90012 Fault Current at Distribution Board Disconnection (L-N) [Isc X M2] 90013 90024 90006 Short Circuit Current Calculation at Various Point of Electrical Circuits (Isc) 90148 90004 Related EEP’s content with sponsored links 90005 90151 .90000 Direct On Line (DOL) Motor Starter 90001 90002 90002 Direct Online Motor Starter — Square D 90004 Different starting methods are employed for starting induction motors because Induction Motor draws more starting current during starting. To prevent damage to the windings due to the high starting current flow, we employ different types of starters. 90005 90004 The simplest form of motor starter for the induction motor is the 90007 Direct On Line starter 90008. The Direct On Line Motor Starter (DOL) consist a MCCB or Circuit Breaker, Contactor and an overload relay for protection.Electromagnetic contactor which can be opened by the thermal overload relay under fault conditions. 90005 90004 Typically, the contactor will be controlled by separate start and stop buttons, and an auxiliary contact on the contactor is used, across the start button, as a hold in contact. I.e. the contactor is electrically latched closed while the motor is operating. 90005 90012 90013 Principle of Direct On Line Starter (DOL) 90014 90004 To start, the contactor is closed, applying full line voltage to the motor windings.The motor will draw a very high inrush current for a very short time, the magnetic field in the iron, and then the current will be limited to the Locked Rotor Current of the motor. The motor will develop Locked Rotor Torque and begin to accelerate towards full speed. 90005 90004 As the motor accelerates, the current will begin to drop, but will not drop significantly until the motor is at a high speed, typically about 85% of synchronous speed. The actual starting current curve is a function of the motor design, and the terminal voltage, and is totally independent of the motor load.90005 90004 The motor load will affect the time taken for the motor to accelerate to full speed and therefore the duration of the high starting current, but not the magnitude of the starting current. 90005 90004 Provided the torque developed by the motor exceeds the load torque at all speeds during the start cycle, the motor will reach full speed. If the torque delivered by the motor is less than the torque of the load at any speed during the start cycle, the motor will stops accelerating. If the starting torque with a DOL starter is insufficient for the load, the motor must be replaced with a motor which can develop a higher starting torque.90005 90004 The acceleration torque is the torque developed by the motor minus the load torque, and will change as the motor accelerates due to the motor speed torque curve and the load speed torque curve. The start time is dependent on the acceleration torque and the load inertia. 90005 90004 90007 DOL starting have a maximum start current and maximum start torque. 90008 90005 90004 This may cause an electrical problem with the supply, or it may cause a mechanical problem with the driven load.So this will be inconvenient for the users of the supply line, always experience a voltage drop when starting a motor. But if this motor is not a high power one it does not affect much. 90005 90012 90013 Parts of DOL Starters 90014 90034 Contactors & Coil 90035 90036 90036 DOL part -Contactor 90004 Magnetic contactors are electromagnetically operated switches that provide a safe and convenient means for connecting and interrupting branch circuits. 90005 90004 Magnetic motor controllers use electromagnetic energy for closing switches.The electromagnet consists of a coil of wire placed on an iron core. When a current flow through the coil, the iron of the magnet becomes magnetized, attracting an iron bar called the armature. An interruption of the current flow through the coil of wire causes the armature to drop out due to the presence of an air gap in the magnetic circuit. 90005 90004 Line-voltage magnetic motor starters are electromechanical devices that provide a safe, convenient, and economical means of starting and stopping motors, and have the advantage of being controlled remotely.The great bulk of motor controllers sold are of this type. 90005 90004 Contactors are mainly used to control machinery which uses electric motors. It consists of a coil which connects to a voltage source. Very often for Single phase Motors, 230V coils are used and for three phase motors, 415V coils are used. The contactor has three main NO contacts and lesser power rated contacts named as Auxiliary Contacts [NO and NC] used for the control circuit. A contact is conducting metal parts which completes or interrupt an electrical circuit.90005 90046 90047 NO-normally open 90048 90047 NC-normally closed 90048 90051 90012 90034 Over Load Relay (Overload protection) 90035 90004 Overload protection for an electric motor is necessary to prevent burnout and to ensure maximum operating life. 90005 90004 Under any condition of overload, a motor draws excessive current that causes overheating. Since motor winding insulation deteriorates due to overheating, there are established limits on motor operating temperatures to protect a motor from overheating.Overload relays are employed on a motor control to limit the amount of current drawn. 90005 90004 90007 90061 The overload relay does not provide short circuit protection. This is the function of over current protective equipment like fuses and circuit breakers, generally located in the disconnecting switch enclosure. 90062 90008 90005 90004 The ideal and easiest way for overload protection for a motor is an element with current-sensing properties very similar to the heating curve of the motor which would act to open the motor circuit when full-load current is exceeded.The operation of the protective device should be such that the motor is allowed to carry harmless over-loads but is quickly removed from the line when an overload has persisted too long. 90066 90005 90068 90068 DOL part — Termal Overload Relay 90004 Normally fuses are not designed to provide overload protection. Fuse is protecting against short circuits (over current protection). Motors draw a high inrush current when starting and conventional fuses have no way of distinguishing between this temporary and harmless inrush current and a damaging overload.Selection of Fuse is depend on motor full-load current, would «blow» every time the motor is started. On the other hand, if a fuse were chosen large enough to pass the starting or inrush current, it would not protect the motor against small, harmful overloads that might occur later. 90005 90004 The overload relay is the heart of motor protection. It has inverse-trip-time characteristics, permitting it to hold in during the accelerating period (when inrush current is drawn), yet providing protection on small overloads above the full-load current when the motor is running.Overload relays are renewable and can withstand repeated trip and reset cycles without need of replacement. Overload relays can not, however, take the place of over current protection equipment. 90005 90004 The overload relay consists of a current-sensing unit connected in the line to the motor, plus a mechanism, actuated by the sensing unit, which serves, directly or indirectly, to break the circuit. 90005 90061 Overload relays can be classified as being thermal, magnetic, or electronic: 90062 90078 90047 90007 Thermal Relay 90008: As the name implies, thermal overload relays rely on the rising temperatures caused by the overload current to trip the overload mechanism.Thermal overload relays can be further subdivided into two types: melting alloy and bimetallic. 90048 90047 90007 Magnetic Relay 90008: Magnetic overload relays react only to current excesses and are not affected by temperature. 90048 90047 90007 Electronic Relay: 90008 Electronic or solid-state overload relays, provide the combination of high-speed trip, adjustability, and ease of installation. They can be ideal in many precise applications. 90048 90091 90012 90013 Wiring of DOL Starter 90014 90034 1.Main Contact 90035 90046 90047 Contactor is connecting among Supply Voltage, Relay Coil and Thermal Overload Relay. 90048 90047 L1 of Contactor Connect (NO) to R Phase through MCCB 90048 90047 L2 of Contactor Connect (NO) to Y Phase through MCCB 90048 90047 L3 of Contactor Connect (NO) to B Phase through MCCB. 90048 90051 90107 NO Contact (- || -): 90108 90046 90047 (13-14 or 53-54) is a normally Open NO contact (closes when the relay energizes) 90048 90047 Contactor Point 53 is connecting to Start Button Point (94) and 54 Point of Contactor is connected to Common wire of Start / Stop Button.90048 90051 90107 NC Contact (- | / | -): 90108 90046 90047 (95-96) is a normally closed NC contact (opens when the thermal overloads trip if associated with the overload block) 90048 90051 90034 2. Relay Coil Connection 90035 90046 90047 A1 of Relay Coil is connecting to any one Supply Phase and A2 is connecting to Thermal over Load Relay’s NC Connection (95). 90048 90051 90034 3. Thermal Overload Relay Connection: 90035 90046 90047 T1, T2, T3 are connect to Thermal Overload Relay 90048 90047 Overload Relay is Connecting between Main Contactor and Motor 90048 90047 NC Connection (95-96) of Thermal Overload Relay is connecting to Stop Button and Common Connection of Start / Stop Button.90048 90051 90012 90013 Wiring Diagram of DOL Starter 90014 90140 90140 Direct On Line Starter — Wiring Diagram 90013 Working principle of DOL Starter 90014 90004 The main heart of DOL starter is Relay Coil. Normally it gets one phase constant from incoming supply Voltage (A1) .when Coil gets second Phase relay coil energizes and Magnet of Contactor produce electromagnetic field and due to this Plunger of Contactor will move and Main Contactor of starter will closed and Auxiliary will change its position NO become NC and NC become (shown Red Line in Diagram).90012 90066 90005 90012 90034 Pushing Start Button 90035 90004 When We Push the start Button Relay Coil will get second phase from Supply Phase-Main contactor (5) -Auxiliary Contact (53) -Start button-Stop button-96-95-To Relay Coil (A2) .Now Coil energizes and Magnetic field produce by Magnet and Plunger of Contactor move. Main Contactor closes and Motor gets supply at the same time Auxiliary contact become (53-54) from NO to NC. 90005 90012 90034 Release Start Button 90035 90004 Relay coil gets supply even though we release Start button.When We release Start Push Button Relay Coil gets Supply phase from Main contactor (5) -Auxiliary contactor (53) — Auxiliary contactor (54) -Stop Button-96-95-Relay coil (shown Red / Blue Lines in Diagram). 90005 90004 In Overload Condition of Motor will be stopped by intermission of Control circuit at Point 96-95. 90005 90012 90034 Pushing Stop Button 90035 90004 When we push Stop Button Control circuit of Starter will be break at stop button and Supply of Relay coil is broken, Plunger moves and close contact of Main Contactor becomes Open, Supply of Motor is disconnected.90005 90165 90165 DOL — Wiring scheme 90012 90013 Motor Starting Characteristics on DOL Starter 90014 90046 90047 Available starting current: 100%. 90048 90047 Peak starting current: 6 to 8 Full Load Current. 90048 90047 Peak starting torque: 100% 90048 90051 90012 90013 Advantages of DOL Starter 90014 90078 90047 Most Economical and Cheapest Starter 90048 90047 Simple to establish, operate and maintain 90048 90047 Simple Control Circuitry 90048 90047 Easy to understand and trouble-shoot.90048 90047 It provides 100% torque at the time of starting. 90048 90047 Only one set of cable is required from starter to motor. 90048 90047 Motor is connected in delta at motor terminals. 90048 90091 90012 90013 Disadvantages of DOL Starter 90014 90078 90047 It does not reduce the starting current of the motor. 90048 90047 90007 High Starting Current: 90008 Very High Starting Current (Typically 6 to 8 times the FLC of the motor). 90048 90047 90007 Mechanically Harsh: 90008 Thermal Stress on the motor, thereby reducing its life.90048 90047 90007 Voltage Dip: 90008 There is a big voltage dip in the electrical installation because of high in-rush current affecting other customers connected to the same lines and therefore not suitable for higher size squirrel cage motors 90048 90047 90007 High starting Torque: 90008 Unnecessary high starting torque, even when not required by the load, thereby increased mechanical stress on the mechanical systems such as rotor shaft, bearings, gearbox, coupling, chain drive, connected equipments, etc.leading to premature failure and plant downtimes 90007. 90008 90048 90091 90012 90013 Features of DOL starting 90014 90046 90047 For low- and medium-power three-phase motors 90048 90047 Three connection lines (circuit layout: star or delta) 90048 90047 High starting torque 90048 90047 Very high mechanical load 90048 90047 High current peaks 90048 90047 Voltage dips 90048 90047 Simple switching devices 90048 90051 90012 90034 Direct On Line Motor Starter (DOL) is suitable for: 90035 90046 90047 A direct on line starter can be used if the high inrush current of the motor does not cause excessive voltage drop in the supply circuit.The maximum size of a motor allowed on a direct on line starter may be limited by the supply utility for this reason. For example, a utility may require rural customers to use reduced-voltage starters for motors larger than 10 kW. 90048 90047 DOL starting is sometimes used to start small water pumps, compressors, fans and conveyor belts. 90048 90051 90012 90034 Direct On Line Motor Starter (DOL) is NOT suitable for: 90035 90046 90047 The peak starting current would result in a serious voltage drop on the supply system 90048 90047 The equipment being driven can not tolerate the effects of very high peak torque loadings 90048 90047 The safety or comfort of those using the equipment may be compromised by sudden starting as, for example, with escalators and lifts.90048 90051 .90000 Transformer Short-Circuit Current Calculation and Reinforcement Solutions 90001 90002 About Thesis 90003 90004 This thesis has three main goals. The first goal is to introduce types of short-circuits. The second one is to introduce the transformer short-circuit current calculations. And the last one is to find suitable reinforcement methods for the transformers which are running now. 90005 90006 Transformer Short-Circuit Current Calculation and Reinforcement Solutions (on photo: High Current Short Circuit Test Facility; credit: CPRI) 90004 Using a comparative approach to analytic research, the advantages and disadvantages of different reinforcement methods can be analyzed.90005 90004 The result shows that 90010 the neutral reactor is the best choice to reinforce the S / C withstand capability 90011: low cost, easy maintenance, high technique maturity and so on. 90005 90013 90014 Short circuit phenomenon 90015 90004 A short circuit (S / C) in an electrical circuit is a part of the circuit that for some reasons has become 90010 «shorter» than it should be 90011. 90005 90004 The current in an electrical circuit flows the easiest way and if two points in a circuit with different potentials are connected with low electrical impedance the current is taking a shortcut between the two points.90005 90004 The consequences of an S / C can be everything from just a minor malfunction to a disaster. The consequences are dependent of the system’s capacity for driving current in an S / C situation and how long time the S / C current is allowed to flow. 90005 90024 Fault types 90004 In almost every electric circuit there has to be some kind of protection against S / C currents. When circuits are analyzed mathematically, an 90010 S / C is usually described by zero impedance between two nodes in the circuit 90011.90005 90004 In reality it is impossible that the impedance should be zero and therefore the calculations will not give the «real» value but in most cases the highest possible value. 90005 90004 To get right results of a calculation it is also important to know all parameters of the circuit. Especially in S / C situations the behavior of the circuits are «strange» and there is no linearity between the voltage of the system and the current flowing. 90005 90004 A short circuit may lead to an electric arc if the current driving capacity of the system is «enough» and protecting devices do not disconnect the circuit.The arc is a channel of hot ionized plasma that is highly conductive. Even short arcs can remove significant amount of materials from the contacting points. 90005 90035 90004 The temperature of the electrical arc is very high 90010 causing the metal on the contact surfaces to melt 90011. 90005 90040 90041 Transformer Short Circuit Current Calculation and Solutions — Ling Song .90000 Short circuit phenomenon you should understand 90001 90002 Why short circuit is so important? 90003 90004 A short circuit in an electrical circuit is a part of the circuit that for some reasons has become «shorter» than it should be. The current in an electrical circuit flows the easiest way and if two points in a circuit with different potentials are connected with low electrical impedance the current is taking a shortcut between the two points. 90005 90006 Short circuit phenomenon you should understand 90004 The consequences of an short circuit can be everything from just a minor malfunction to a disaster.The consequences are dependent of the system’s capacity for driving current in an short circuit situation and how long time the short circuit current is allowed to flow. In almost every electric circuit there has to be some kind of protection against short circuit currents. 90005 90004 When circuits are analyzed mathematically, 90010 the short circuit is usually described by zero impedance between two nodes in the circuit 90011. 90005 90004 In reality it is impossible that the impedance should be zero and therefore the calculations will not give the «real» value but in most cases the highest possible value.To get right results of a calculation it is also important to know all parameters of the circuit. 90005 90004 Especially in short circuit situations the behavior of the circuits are «strange» and there is no linearity between the voltage of the system and the current flowing. 90005 90004 90010 Contents: 90011 90005 90021 90022 The need for transformer short-circuit current calculation 90023 90022 Symmetrical components 90023 90022 Two kinds of short circuit 90021 90022 DC circuits 90023 90022 AC circuits 90021 90022 Single-phase circuits 90023 90022 Three-phase circuits 90023 90036 90023 90022 Development of short-circuit current 90023 90036 90023 90036 90043 90044 1.The need for transformer short-circuit current calculation 90045 90004 Today more than ever before, the electricity grid is developing so quickly — the power plant capacity, the substation capacity and electricity loads, as well as load density, sustainably grow. 90005 90004 Take China as an example. The number of 500 kV substations in the North China power grid is almost 2 times higher than in the past decade. The number has grown from 48 to 97; the substation capacity has increased from 90010 52,069,000 kVA to 157,960,000 kVA 90011.90005 90004 As a result, 90010 the short-circuit currents in the power grid increase year by year 90011. Based on the statistical analysis of the State Grid Corporation of China (SGCC), the short current current accidents of power transformers (Size ≥ 110 kV) happened 125 times. The total power capacity influenced by the short circuit accidents is 7,996 MVA in 1995 ~ 1999. The number represents 37.5% of all power accidents and 44% of the transformers accidents. 90005 90004 The short circuit current is an important specification and standard for equipment and conductors in the power industry, and 90010 short circuit current withstand capability 90011 of the main devices decides whether the grid could run more safely or not.So it’s significant to calculate the short circuit current and offer some possible solutions. 90005 90004 90010 The correct calculation can help us to: 90011 90005 90021 90022 Specify fault ratings for electrical equipment (e.g. short circuit withstand ratings) 90023 90022 Help identify potential problems and weaknesses in the system and assist in system planning 90023 90022 Form the basis for protection coordination studies 90023 90036 90004 Go back to contents ↑ 90005 90043 90044 2.Symmetrical components 90045 90004 In the practical work, engineers often use «90010 symmetrical components 90011» to analyze the three-phase power system. It was invented by a Canadian electrical engineer Charles L.Fortescue in 1913. Mr Fortescue’s original purpose was to analyze the operation of the electrical motors. 90005 90004 The theory was not used for the power system until 1937. The analytical technique was adopted and advanced by engineers at General Electric and Westinghouse and after World War II it was an 90010 accepted method for asymmetric fault analysis 90011.90005 90004 Now it’s a common tool used to analyze the faults of three-phase power system. 90005 90004 The basic setting for the theory is that any unbalanced system quantities (current or voltage) could be decomposed into 3 symmetrical sets of balanced vectors: 90005 90021 90022 Positive sequence components, 90023 90022 Negative sequence components and 90023 90022 Zero sequence components. 90023 90036 90097 90097 Figure 1 — Sequence components to represent the three-phase electrical system 90004 The positive sequence component of the current shown in Figure 1 above is balanced in magnitude with a 120 degree phase separation and counter-clockwise rotation, just like the original balanced system.90005 90004 The negative sequence component of the current is balanced in magnitude with a 120 degree phase separation, but has the opposite rotation, in this case, clockwise. 90005 90103 90004 The zero sequence components have 90010 equal magnitudes, but zero phase separation 90011. 90005 90108 90004 Here, we denote the positive sequence with the subscript «1». Likewise, the negative sequence is denoted with the subscript «2» and the zero sequence is denoted with the subscript «0». 90005 90004 Under a no fault condition, the power system is considered to be essentially a symmetrical system 90010 and therefore only positive sequence currents and voltages exist 90011.At the time of a fault, positive, negative and possibly zero sequence currents and voltages exist. 90005 90004 Using real world phase voltages and currents along with Fortescue’s formulas, all positive, negative and zero sequence currents can be calculated. Protective relays use these sequence components along with phase current and / or voltage data as the input to protective elements. 90005 90043 90118 Principles of Symmetrical Components (VIDEO) 90119 90004 90121 90122 90005 90004 90125 90122 90005 90004 Go back to contents ↑ 90005 90043 90002 3.Two kinds of short circuit 90003 90044 3.1 DC circuits 90045 90004 What circuit information is needed to do an short circuit calculation for a DC circuit? In an electrical circuit the current is dependent of the 90010 electromotive force (emf), the electromagnetic field, and the total impedance of the circuit 90011. 90005 90004 In a battery the emf-value is dependent of the charge of the battery. The internal impedance of the battery is also a changing parameter and dependent of the charge, the temperature, and the age of the battery and so on.90005 90004 In a DC circuit 90010 the resistance is the current limiting factor together with the emf 90011 in steady-state which means «after a while». 90005 90004 In the beginning of a transient, like an short circuit situation, also the inductance of the circuit is limiting. Any inductance in the circuit will smooth up the rise of the current. The current is increasing exponentially due to the relation between the inductance and the resistance of the circuit. 90005 90147 90147 Figure 2 — The current in an inductor 90004 Direct current causes different problems from AC when trying to interrupt high value currents since the arc extinction is more difficult.AC passes through zero every half period thus helping the breaking of current. 90005 90103 90004 90010 A circuit breaker for a certain AC current is usually not able to break the same magnitude of DC current. 90011 The difficulties of breaking a DC circuit increases with the ratio of inductance versus resistance in the circuit. Inductances are always in opposition to changes of current. 90005 90108 90004 Go back to contents ↑ 90005 90043 90044 3.2 AC circuits 90045 90004 Alternating current circuits (AC) are more complex to solve than direct current circuits (DC).There are more parameters affecting the results and in fast changing situations the first values ​​of current are strongly dependent of the phase of the active voltage source. 90005 90043 90118 3.2.1 Single-phase circuits 90119 90004 Most large power networks are three-phase but especially in low voltage systems most of the connected circuits are single-phase. When calculating short circuit currents the situation is dependent on 90010 how near the generator or transformer the fault occurs 90011.90005 90004 Not only due to the increasing impedance in the end of the network but also to the fact that generators and transformers are acting «strange» when they are not loaded symmetrically in all phases. 90005 90004 In some cases the circuit may be fed from a single-phase transformer with a current carrying capacity that is not enough to make the three-phase system behave «strange». 90005 90004 The fact that the short circuit current is easier to calculate far from a transformer or a generator is because 90010 the line impedances are playing an important role in the process and the impedances are often easier to know than the voltage in the beginning of the circuit 90011.90005 90004 With longer lines the currents decreases and the voltage from the source will not change very much. 90005 90004 In single-phase low voltage circuits that are commonly used in households the short circuit currents must be disconnected for different reasons. One reason is because of the touch voltage that may occur during a contact between phase and protective earth. 90005 90004 The protective earth in a circuit is used to prevent exposed conductive parts from getting a dangerous potential referred to earth.When a direct contact between phase and exposed conductive parts is established by a fault situation the potential will rise to a dangerous level for persons to touch and therefore the circuit must be disconnected by protection devices 90010 like fuses and circuit breakers 90011. 90005 90187 90187 Figure 3 — Phase-to-earth short circuit current (single-phase) 90004 In household situations the maximum time for disconnection is 90010 normally 0.4 seconds 90011. To access the clearance time under fault conditions the prospective fault current must be determined by measurement or calculation.It is the prospective current that will flow when the end of the cable being protected is connected to the protective earth conductor that is of concern. 90005 90004 With long cable runs this prospective current can be found to be comparatively low. 90005 90103 90004 It should be remembered however that 90010 the first problem with long cable runs is the possibility of excessive voltage drop, and cables should be selected first for current rating, and then checked for voltage drop before determining the fault prospective 90011.90005 90108 90004 Go back to contents ↑ 90005 90043 90118 3.2.2 Three-phase circuits 90119 90004 Three-phase electric power is a common method of 90010 AC electric power generation, transmission, and distribution 90011. It is a type of polyphase system and is the most common method used by electrical grids worldwide to transfer power. 90005 90004 It is also used to power large motors and heavy loads. A three-phase system is usually more economical than an equivalent single-phase or two-phase system at the same voltage because it uses less conductor material to transmit electrical power.90005 The three-phase system was independently invented by Galileo Ferraris, Mikhail Dolivo-Dobrovolsky and Nikola Tesla in the late 1880s. 90004 Most single-phase circuits are just a part of a three-phase network. In a three-phase system various types of short circuit can occur. 90005 90004 For example, short circuit current can be 90010 phase-to-earth 90011 (80% of faults), 90010 phase-to-phase 90011 (15% of faults — this type of fault often degenerates into a three-phase fault) and t 90010 hree-phase 90011 (only 5% of initial faults).These different short-circuit currents are shown in Figure 4. 90005 90222 90222 Figure 4 — The fault types 90004 In China, there is another rough classification which is based on the number of the fault phase: three-phase fault, double-phase fault and single phase fault due to phase-to-earth fault which may happen for two phases. 90005 90004 90010 The primary characteristics of short-circuit currents are: 90011 90005 90021 90022 90010 Duration 90011 — The current can be self-extinguishing, transient or steady-state 90023 90022 90010 Origin 90011 — it may be caused by mechanical reasons (break in a conductor, accidental electrical contact between two conductors via a foreign conducting body such as a tool or an animal), internal or atmospheric overvoltage, and insulation breakdown due to heat, humidity or a corrosive environment 90023 90022 90010 Location 90011 (inside or outside a machine or an electrical switchboard) 90023 90036 90004 The consequences of short circuit are depending on the type and duration of the fault and the short-circuit power available.Locally at the fault point there may occur electrical arcs causing damage to insulation, welding of conductors and fire. 90005 90004 Voltage drops occur in other networks during the time of short circuit and shutdown of a part of a network may include also «healthy» parts of the network depending on the design of the whole network. 90005 90004 Go back to contents ↑ 90005 90043 90118 3.3 Development of short-circuit current 90119 90004 A simplified AC network can be represented by a source of AC power, some kind of switching device, a 90010 total impedance Z 90255 N 90256 90011 that represents all the impedances upstream of the switching point and a load, represented by its impedance (see Figure 5).90005 90004 In a real network the 90010 total impedance Z 90255 N 90256 is made up of the impedances of all components upstream 90011. The components are for example generators, transformers, wires, circuit-breakers and metering systems. 90005 90004 When a fault with negligible impedance occurs between 90010 A 90011 and 90010 B 90011 a short-circuit current limited only by Z 90255 N 90256 flows in the circuit. The 90010 short-circuit current I 90255 sc 90256 90011 develops under transient conditions depending on the relation between inductances and resistances in the whole circuit.90005 90277 90277 Figure 5 — The simple short circuit 90004 If the circuit is mostly resistive the waveform of the current is following the waveform of the voltage but if there are inductances in the circuit the waveform of the current will differ from the waveform of the voltage during a transient time of the process. 90005 90004 In an inductive circuit the current can not begin with any value but zero. The influence of inductances is described by 90010 reactance X 90011 in AC circuits with a fixed frequency of the voltage.90005 90004 In low voltage systems where cables and conductors represent most of the impedance 90010 it can be regarded as mostly resistive 90011. In power distribution networks the reactance is normally much greater than the resistances. 90005 90004 Generally the total impedance Z in steady-state in an AC circuit is made up of the total resistance R and the total reactance X as the following relation shows. 90005 90004 90292 90292 90005 90004 In the simplified circuit above the voltage is constant and so is the total impedance.In faults far from generators and transformers where most of the impedance consists of impedances from wires the calculations can be done with a good result and the transient current is almost the same as if the current would flow for a longer time. 90005 90004 The meaning of far from is not necessarily physical but means that 90010 generator or transformer impedances are less than the impedance of the elements from wires 90011. 90005 90103 90004 90010 The impedance elements from wires are constant at a constant temperature 90011 but the impedances of generators vary during a short-circuit and the impedances of transformers change if the transformers are asymmetrically loaded with high currents.90005 90108 90307 90307 Figure 6 — The currents continue symmetrically 90004 Figure 6 shows the 90010 current in beginning of a short-circuit far from the generator 90011. The short-circuit starts at a moment when the current normally is zero and continues symmetrically. 90005 90313 90313 Figure 7 — The currents continue asymmetrically 90004 Figure 7 shows the current when the short-circuit starts at a moment when the voltage is zero and the current is also starting from zero but asymmetrically during a transient time.90005 90043 90044 IEC 61439 — Short-circuit withstand tests (VIDEO) 90045 90004 Short-circuit withstand testing, examples of 90010 rated conditional short-circuit (Icc) tests 90011 on functional units of a low-voltage assembly (Protective device trip units are enabled) 90005 90004 90325 90122 90005 90004 Go back to contents ↑ 90005 90004 90331 90010 Reference // 90011 Transformer Short Circuit Current Calculation and Solutions by Ling Song 90334 90005 .