POWER SYSTEM ANALYSIS EET 3083 1 EET 308

POWER SYSTEM ANALYSIS EET 308/3 1 EET 308 POWER SYSTEM ANALYSIS

CHAPTER 4 SYMMETRICA L FAULT 2 EET 308 POWER SYSTEM ANALYSIS

On completion of this lesson, a student should be able to: Ability to analyze fault current in Symmetrical Fault 3 EET 308 POWER SYSTEM ANALYSIS

TOPIC OUTLINE 4. 1 Introduction 4. 2 Balance Three-phase Fault 4. 3 Short-circuit Capacity 4. 4 Systematic Fault Analysis Using Bus Impedance Matrix 4. 5 Selection of Circuit Breaker 4 EET 308 POWER SYSTEM ANALYSIS

Introduction 5 EET 102 ELECTRIC CIRCUIT II

Cause of fault 6 EET 102 ELECTRIC CIRCUIT II

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Effect of faults 8 EET 102 ELECTRIC CIRCUIT II

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Faults in power systems are divided into three-phase balanced fault and unbalanced faults. The information from fault studies are used for proper relay setting and coordination. Three-phase balanced fault – the information is used to select and set phase relay. Fault analysis also used to determine the rating of protective switchgear. The magnitude of the fault current depends on the internal impedance of the generator and the network. 10 EET 308 POWER SYSTEM ANALYSIS

4. 2 BALANCED THREE-PHASE FAULT Xd” – subtransient reactance Xd’ – transient reactance Xd – synchronous reactance Generally, subtransient reactance is used for determining the interrupting capacity of the circuit breakers. While transient reactance is used for relay setting and coordination. If the fault impedance is zero, the fault is referred as the bolted fault or the solid fault. Since the impedance of a new path is usually low, an excessive current may flow. 11 EET 308 POWER SYSTEM ANALYSIS

4. 2 BALANCED THREE-PHASE FAULT Symmetrical AC component of the fault current: 1) Subtransient: first cycle or so after the fault – AC current is very large and falls rapidly; 2) Transient: current falls at a slower rate; 3) Steady-state: current gets back to normal. 12 EET 308 POWER SYSTEM ANALYSIS

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ASSUMPTION BEEN MADE Load current is neglected (no load condition) All prefault bus voltages are assumed to be equal to 1. 0 per unit. Shunt capacitances are neglected Resistances are neglected All generators are running at their rated voltage and rated frequency with their emf in phase. 14 EET 308 POWER SYSTEM ANALYSIS

QUIZ 15 EET 308 POWER SYSTEM ANALYSIS

The one-line diagram of a system as shown in Figure 2(a) is initially at no load with generators operating at their rated voltage with their emfs in phase. In the system, all resistances are neglected. If a three-phase balanced fault occurs at bus 4, determine the short-circuit current and short-circuit MVA at bus 4. Choose 100 MVA as your base MVA. [Gambarajah satu-talian bagi satu sistem seperti yang ditunjukkan dalam Rajah 2(a) pada awalnya adalah pada keadaan tanpa beban dengan penjana-penjana beroperasi pada voltan terkadar dengan daya-daya elektromagnetik adalah sefasa. Dalam sistem tersebut, semua rintangan diabaikan. Jika satu kerosakan seimbang tiga-fasa berlaku pada bas 4, tentukan arus litar-pintas dan MVA litar-pintas pada bas 4. Pilih 100 MVA sebagai MVA asas anda. ] 90 MVA, 30 k. V, Xd’= 25% 80 MVA, 30 k. V, Xd’= 25% 16 XL= 150Ω 100 MVA, 30/400 k. V, XT 1= 15% EET 308 POWER SYSTEM ANALYSIS XL= 150Ω 50 MVA, 400/30 k. V, XT 2= 10%

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4. 3 SHORT CIRCUIT CAPACITY Measure the electrical strength of the bus. Also called as short-circuit MVA Stated in MVA Determines the dimension of bus bar and the interrupting capacity of a circuit breaker Short-circuit capacity (SCC) or short-circuit MVA at bus k is given by; VLk = Line-to-line voltage in k. V at bus k Ik(F) = Fault current at bus k in Ampere. Equation 1 19 EET 308 POWER SYSTEM ANALYSIS

4. 3 SHORT CIRCUIT CAPACITY The symmetrical fault current in p. u. Vk (0)= per unit prefault bus voltage. Xkk = per unit reactance to the point of fault. System resistance and capacitance are neglected and only inductive reactance is allowed. The base current is given by; SB in MVA VB in k. V 20 EET 308 POWER SYSTEM ANALYSIS

4. 3 SHORT CIRCUIT CAPACITY Fault current in Ampere; Equation 2 Substituting Equation 2 into Equation 1; 21 EET 102 ELECTRIC CIRCUIT II

4. 3 SHORT CIRCUIT CAPACITY If the rated voltage equal to base voltage, VL = VB ; The prefault voltage is usually assumed to be 1. 0 pu. So; 22 EET 102 ELECTRIC CIRCUIT II

EXAMPLE 4. 2 Determine the short circuit capacity on bus 3 in Example 4. 1. j 0. 2 j 0. 8 j 0. 4 1 2 j 0. 4 3 I 3(F) 23 EET 308 POWER SYSTEM ANALYSIS Zf = j 0. 16

SOLUTION The equivalent impedance from source to the fault point From previous example 4. 1 Base MVA = 100 MVA The short circuit capacity at bus 3; 24 EET 308 POWER SYSTEM ANALYSIS

Example for 4. 2 topic 25 EET 102 ELECTRIC CIRCUIT II

EXAMPLE 4. 1 The one line diagram of a simple three-bus power system is shown in the Figure 4. 1. Each generator is represented by an emf behind the transient reactance. All impedances are expressed in per unit in a common 100 MVA base, and for simplicity, resistances are neglected. The following assumption are made: shunt capacitances are neglected and the system is considered on no load. all generators are running at their rated voltage and rated frequency with their emf in phase Determine the fault current, the bus voltages and the line currents during the fault when a balanced three-phase fault with a fault impedance Zf = j 0. 16 p. u. fault occurs on a) bus 3 b) bus 2 c) bus 1 26 EET 308 POWER SYSTEM ANALYSIS

EXAMPLE 4. 1 j 0. 1. j 0. 2 j 0. 1 j 0. 2 j 0. 8 1 2 j 0. 4 Figure 4. 1 One-line diagram of a simple power system 3 27 EET 308 POWER SYSTEM ANALYSIS

SOLUTION (a) Fault on bus 3 j 0. 2 j 0. 8 1 j 0. 4 Impedance network with fault at bus 3 j 0. 4 2 j 0. 4 3 Zf = j 0. 16 28 EET 308 POWER SYSTEM ANALYSIS

SOLUTION Thevenin’s theorem states that the changes in the network voltages caused by the added branch (the fault impedance) is equivalent to those caused by the added voltage V 3(0) with all other sources short-circuited. j 0. 2 j 0. 8 1 2 j 0. 4 All prefault voltages are assumed to be equal to 1. 0 per unit: V 1(0) = V 2(0) = V 3(0) = 1. 0 pu 29 EET 308 POWER SYSTEM ANALYSIS j 0. 4 3 Vth = V 3(0) I 3(F) Zf = j 0. 16 Thevenin’s equivalent network

SOLUTION using ∆ to Y transformation j 0. 2 j 0. 4 1 2 j 0. 1 3 Vth I 3(F) 30 EET 308 POWER SYSTEM ANALYSIS j 0. 16

SOLUTION combining the parallel branches, Thevenin impedance is j 0. 24 j 0. 1 3 Vth Z 33 is the Thevenin impedance viewed From the faulted bus. 31 EET 308 POWER SYSTEM ANALYSIS I 3(F) j 0. 16

SOLUTION From figure, the fault current is: Z 33 = j 0. 34 3 Vth I 3(F) 32 EET 308 POWER SYSTEM ANALYSIS j 0. 16

SOLUTION The current division between the two generators are The bus voltage changes 33 EET 308 POWER SYSTEM ANALYSIS

SOLUTION The bus voltages during the fault The short circuit line current are 34 EET 308 POWER SYSTEM ANALYSIS

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SOLUTION (b) Fault on bus 2 j 0. 4 j 0. 8 1 2 j 0. 4 36 j 0. 4 3 EET 308 POWER SYSTEM ANALYSIS Zf = j 0. 16 Impedance network with fault at bus 2

SOLUTION j 0. 2 j 0. 4 j 0. 8 1 2 Vth j 0. 4 I 2(F) Thevenin’s equivalent network 37 EET 308 POWER SYSTEM ANALYSIS 3 Zf = j 0. 16

SOLUTION The equivalent impedance between bus 1 and 2 j 0. 4 2 1 Vth Zf = j 0. 16 38 EET 308 POWER SYSTEM ANALYSIS I 2(F)

SOLUTION Combining parallel branches from ground to bus 2 results The fault current is Z 22 = j 0. 24 2 Vth Zf = j 0. 16 39 EET 308 POWER SYSTEM ANALYSIS I 2(F)

SOLUTION The current division between the two generators are The bus voltage changes 40 EET 308 POWER SYSTEM ANALYSIS

SOLUTION The bus voltages during the fault The short circuit-current in the line are 41 EET 308 POWER SYSTEM ANALYSIS

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SOLUTION (c) Fault on bus 1 j 0. 2 j 0. 4 j 0. 8 2 1 Zf = j 0. 16 j 0. 4 3 43 EET 308 POWER SYSTEM ANALYSIS Impedance network with fault at bus 1

SOLUTION j 0. 2 j 0. 4 j 0. 8 1 2 Vth j 0. 4 I 1(F) j 0. 4 Zf = j 0. 16 3 Thevenin’s equivalent network 44 EET 308 POWER SYSTEM ANALYSIS

SOLUTION Combining parallel branches between bus 1 and 2 j 0. 4 1 2 Vth I 1(F) 45 EET 308 POWER SYSTEM ANALYSIS j 0. 4 Zf = j 0. 16

SOLUTION Combining parallel branches from ground to bus 1 results Z 11 = j 0. 16 The fault current is 1 Vth I 1(F) 46 EET 308 POWER SYSTEM ANALYSIS Zf = j 0. 16

SOLUTION The current division between the two generators are The bus voltage changes 47 EET 308 POWER SYSTEM ANALYSIS

SOLUTION The bus voltages during the fault The short circuit-current in the line are 48 EET 308 POWER SYSTEM ANALYSIS

For More Accurate Calculation Pre fault bus voltages can be obtained from the power flow solution. Include the load currents into fault analysis where loads are expressed by a constant impedance evaluated at the pre fault bus voltages. 49 EET 102 ELECTRIC CIRCUIT II