The following table shows approximate percentage wise the

Transcription

1 SHORT-CIRCUIT CALCULATION

2 INTRODUCTION Designing an electrical system is easy and simple, if only the normal operation of the network is taken into consideration. However, abnormal conditions which are likely to occur anytime must be foreseen and should be taken seriously during the design stage. A good design must not only be made simple, but most importantly, safe and reliable. An Electrical system must operate continuously during normal and healthy overload situations; its protective device must also trip expeditiously to isolate the affected parts of the system during fault conditions.

3 Why Fault Occur? Fault in power system occur because of insulation failure in plant which may be caused by a system over-voltage such as switching surge or a lightning stroke, or maybe due to broken insulators or conductors and various causes in the system. The following table shows approximate percentage wise the various causes of faults: (i) Lightning 6% (ii) Sleet, wind, Mechanical (jumping conductors) 10% (iii) Apparatus failure 10% (iv) Switching to a fault 10% (v) Miscellaneous (tree falling on lines, sabotage) 10%

4 These faults can take one of the following forms: Type Probability of failures 1. Single phase to ground faults 10% 2. Phase to phase faults 15% 3. Two-phases to ground faults 10% 4. Three-phase faults 5% Such faults cause heavy currentscalled, short-circuit currents, to flow in the system. The determination of the values of such currents enables us to make proper selection of circuit breakers, protective relays and also helps to ensure that the associated apparatus e.g. bus bars, connections, current transformers will withstand the forces which arise due to the fault currents during the period prior to the interrupting device clearing the fault.

5 Single phase to ground faults are the most common whereas the three phase short-circuit fault are the most severe faults and also the most amenable to calculations since these involve symmetrical conditions only. For unsymmetrical ground faults, line-to-ground fault on a solidly grounded system is used to size ground cable as well as to coordinate ground fault protective devices. Sources of Fault Power. All the generating and trandformers which under normal conditions take power from the system. Synchronous Machines. Under short-circuit conditions, a drop in frequency or voltage is common and in this event, synchronous machineswillfeedbackintothesystemforashortperiod. Large Induction Motor. large induction motor s flywheel effect will act as generators in the event of reduces frequency. Where machines such as these are connected, and they are of size as to have effect particularly where they are connected to a point close to that for which short-circuit values are being calculated, they should be calculated. Frequency chargers

6 FAULT CURRENTS LIMITERS: Cables Busbars Circuit breakers, Reactors Transformer P Q x G. F 1 P Q G x F 1 x F 2 G x F 3 Fig short-circuit fed from a single generator Fig short-circuit fed from a source having more then one generator

7 In Fig.1.1, letpand Qrepresents the busbars at the power station and the sub-station respectively. Also F 1 represents a feeder outgoing from the sub-station bus-bars and equipped with circuit breaker at X. IfthereisasinglefeederconnectingPandQandthegeneratingstation also consists of a single generator as in Fig 1.1, in the event of fault occurringatanypointinonfeederf 1,theshort-circuitcurrentfromthe generating station will be a certain value limited by the impedance of thegeneratorandtheimpedanceofthefeederuptothepointoffault. Now supposed that an increased load is required to be fed from the sub-station bus-bars. In that case, it may become necessary to increase the generating units at the power station and also to install a second feeder between P and Q to carry the increased load. ( It is assumed that the original generating unit and feeder between P and Q were fully loaded). With the increased load, which may be due to additional consuming apparatus, it may be similarly necessary to have more outgoing feeders F 1 andf 2 etc.asshowninfig.1.2.

8 The fault current from the generating station due to a fault occurring on,sayf 1 willbegreaterthanthatintheoriginalsystemdueto (i) large kva of the generating plant, (ii) smaller impedance of the generating plant.( the equivalent impedance of two generators in parallel is smaller than the individual impedance of each.) (iii)thesmallerimpedanceofthetwofeedersconnectingpandq since they are now in parallel. The value of fault current which can flow in any system under shortcircuit is limited only by the impedance in that system. Therefore, it is necessary in any calculation to have knowledge of these impedances.

9 Reactance. Often the resistance is so small that in most cases reactance alone is considered for calculating the fault currents. In general it is to be noted that if the reactance exceeds 3 times the resistance, the resistance may be neglected. The error in the assumption will not exceed 5%. The reactance of synchronous machines, transformers, reactors, is usually expressed as a percentage.

10 In electrical systems, power flows from the sources to loads through circuit components which has specific impedances. The power flow is principally determined by the load impedance. A FAULT can be described as a load with very low impedance. Depending upon the type of fault, Phase or ground faults can be regarded as Source Limited (phase fault) or Fault limited (ground faults). All the fault calculating methods are based upon this principle.

11 FAULTS CALCULATING METHODS: Ohms Method The basic formula for all current flow is Ohms Law, Volts = Ampere x Ohms. This is easily applied to single phase circuits operating at only one voltage level. However, when several different voltages occur, the situation becomes more complicated. Each location in the circuit where faults must be calculated has an specific operating voltage. All circuit impedances must be referred to the fault voltage before calculating the short-circuit current.

12 Formula 1 - is based for referring impedance to a new voltage at which the fault is being calculated. When all the impedances have been referred to the fault voltage, they can be added. New Voltage Ohms at new V = Ohms at old V x ( Formula 1) Old Voltage Example: Calculate the two faults indicated on the following diagram. 2

13 480 volts Fault 1 - I SC1 = = 96 Amps 5 ohms Fault 2 - refer 5 ohms from 480V to 240V 240 V 240V = 5 x = 5 x V= V Assume perfect transformer (Z = 0%) Total Impedance = = 4.25 ohms 240 volts Fault 2 - I SC2 = = 56.5 Amps 4.25 ohms 2

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15 This conversion is necessary when a transformer must be included or motor contribution is being considered. The formula is derived from the fact that the impedance of the transformer or motor is the only limitation to the circulation of current with a winding shorted or rotor locked. The rating of voltage required to circulate full rated current to the rated voltage is the Per Unit Impedanceof the particular transformer of motor. Consider the following system consisting of a utility in-feed through the cable to a transformer. The step by step solution is shown below. Example:

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24 (Formula 5) (Formula 6) (Formula 9)

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26 The circuit used to illustrate the Ohmic calculations is used below to illustrate Per Unit calculations.

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29 So far only three phase faults have been considered. In all cases, the fault impedance was assumed to be zero. However, three phase faults do not often occur. Ground faults are by far the most common and they generally have arc and contact impedance at the point of the fault. Impedance of the ground return path may also be an important factor. The fact that the ground faults are rarely bolted (near zero fault impedance), does not mean the case may be ignored. Bolted phase to ground faults are generally the theoretical worst case. For this reason it is necessary to calculate them and understand the theory behind the calculations. The study of ground faults necessarily leads us to symmetrical component analysis. The theory presented here is brief and no mathematical proof is included.

30 The bolted 3-phase faults which have been considered thus far were calculated with the assumption that all voltages and impedances were balanced, i.e., a symmetrical fault. When it becomes necessary to analyze an unsymmetrical fault (phase to phase, phase to ground, etc.) the calculations become quite complicated. To simplify the analytical process, the unbalanced system of vectors is reduced into three balanced vector systems which are known as POSITIVE, NEGATIVE, and ZERO phase sequence components. The positive sequence components consist of three vectors equal in magnitude, 120 degrees displaced and rotating ABC. The negative sequence components are three equal vectors, 120 degrees displaced and rotating ACB. The zero sequence components consists of three vectors equal in magnitude and in phase. Each of these sets of voltage vectors is shown in the following diagram.

31 Now that three sets of sequence voltages have been assumed, It follows that they are the result of sequence currents flowing through sequence impedances. Positive sequence currents only flow through positive sequence impedance, etc. It is then possible to draw an impedance diagram for each sequence in order to calculate faults. The Theory and generation of these diagrams is best described in applied Protective relaying references. It will simply be stated here that the impedance diagram is drawn and connected as shown in the following sketches.

32 IMPEDANCE DIAGRAM

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34 Since delta-delta and delta-wye grounded transformers are most common, the equivalent diagram of each is shown below:

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36 For most circuits, the positive and negative sequence impedances can be assumed equal. The zero sequence impedance will depend upon the location of delta windings in the circuit. The term Zn refers to impedance in the neutral or return path from the fault to the transformer neutral connection. Z f refers to the impedance of the fault arc. The example assumes Zn and Z f are zero, i.e., a bolted phase to ground fault.

37 Step 1. Draw the impedance diagram. Note that Z 0 only includes the transformer. Z 0 is assumed equal to Z 1. Step 2. The impedance of each system element has already been calculated in the 3-phase example. All of the following impedances have been referred to the 480 volts level. Utility = ohms (slide 16) Cable = ohms (slide 17) Transformer = ohms (slide 18)

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40 Example: The basic form of the Ohmic impedance diagram is suitable for this example. Step 1. Draw the impedance diagram. Step 2. The impedance calculated for the 3-phase fault are valid for this condition. Utility Cable = 0.1 0/1 ohm = /1 ohm Transformer = 0.5 0/1 ohm Step 3. Calculate the total impedance. Z t1 = /1 ohm = Z t2 Z t2 = /1 ohm Z t0 = 0.5 0/1 ohm Z T = /1 ohm

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