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POWER SYSTEM ANALYSIS EET308/3 EET308 POWER SYSTEM ANALYSIS
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CHAPTER 4 SYMMETRICAL FAULT EET308 POWER SYSTEM ANALYSIS
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On completion of this lesson, a student should be able to:
Ability to analyze fault current in Symmetrical Fault EET308 POWER SYSTEM ANALYSIS
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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 EET308 POWER SYSTEM ANALYSIS
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Introduction EET 102 ELECTRIC CIRCUIT II
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Cause of fault EET 102 ELECTRIC CIRCUIT II
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EET 102 ELECTRIC CIRCUIT II
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Effect of faults EET 102 ELECTRIC CIRCUIT II
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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. EET308 POWER SYSTEM ANALYSIS
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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. EET308 POWER SYSTEM ANALYSIS
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4.2 BALANCED THREE-PHASE FAULT
Symmetrical AC component of the fault current: Subtransient: first cycle or so after the fault – AC current is very large and falls rapidly; Transient: current falls at a slower rate; Steady-state: current gets back to normal. EET308 POWER SYSTEM ANALYSIS
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EET 102 ELECTRIC CIRCUIT II
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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. EET308 POWER SYSTEM ANALYSIS
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QUIZ EET308 POWER SYSTEM ANALYSIS
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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, 30kV, Xd’= 25% XL= 150Ω 80 MVA, 30kV, Xd’= 25% 100 MVA, 30/400 kV, XT1= 15% XL= 150Ω 50 MVA, 400/30 kV, XT2= 10% EET308 POWER SYSTEM ANALYSIS
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EET308 POWER SYSTEM ANALYSIS
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EET308 POWER SYSTEM ANALYSIS
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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 kV at bus k Ik(F) = Fault current at bus k in Ampere. Equation 1 EET308 POWER SYSTEM ANALYSIS
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4.3 SHORT CIRCUIT CAPACITY
The symmetrical fault current in p.u. System resistance and capacitance are neglected and only inductive reactance is allowed. The base current is given by; Vk (0)= per unit prefault bus voltage. Xkk = per unit reactance to the point of fault. SB in MVA VB in kV EET308 POWER SYSTEM ANALYSIS
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4.3 SHORT CIRCUIT CAPACITY
Fault current in Ampere; Substituting Equation 2 into Equation 1; Equation 2 EET 102 ELECTRIC CIRCUIT II
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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; EET 102 ELECTRIC CIRCUIT II
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EXAMPLE 4.2 Determine the short circuit capacity on bus 3 in Example 4.1. j0.2 j0.4 j0.8 1 2 j0.4 j0.4 3 I3(F) Zf = j0.16 EET308 POWER SYSTEM ANALYSIS
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SOLUTION The equivalent impedance from source to the fault point
Base MVA = 100 MVA The short circuit capacity at bus 3; From previous example 4.1 EET308 POWER SYSTEM ANALYSIS
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Example for 4.2 topic EET 102 ELECTRIC CIRCUIT II
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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 = j0.16 p.u. fault occurs on bus 3 bus 2 bus 1 EET308 POWER SYSTEM ANALYSIS
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One-line diagram of a simple power system
EXAMPLE 4.1 j0.1. j0.2 j0.1 j0.2 j0.8 1 2 j0.4 j0.4 Figure 4.1 One-line diagram of a simple power system 3 EET308 POWER SYSTEM ANALYSIS
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Impedance network with fault at bus 3
SOLUTION (a) Fault on bus 3 j0.2 j0.4 j0.8 1 2 j0.4 j0.4 Impedance network with fault at bus 3 3 Zf = j0.16 EET308 POWER SYSTEM ANALYSIS
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Thevenin’s equivalent network
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 V3(0) with all other sources short-circuited. j0.2 j0.4 j0.8 1 2 j0.4 j0.4 All prefault voltages are assumed to be equal to 1.0 per unit: 3 Vth = V3(0) I3(F) Zf = j0.16 V1(0) = V2(0) = V3(0) = 1.0 pu Thevenin’s equivalent network EET308 POWER SYSTEM ANALYSIS
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SOLUTION j0.2 j0.4 using ∆ to Y transformation 1 2 j0.2 j0.2 j0.1 3
Vth I3(F) j0.16 EET308 POWER SYSTEM ANALYSIS
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SOLUTION combining the parallel branches, Thevenin impedance is j0.24
3 Vth Z33 is the Thevenin impedance viewed From the faulted bus. I3(F) j0.16 EET308 POWER SYSTEM ANALYSIS
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SOLUTION From figure, the fault current is: Z33 = j0.34 3 Vth I3(F)
EET308 POWER SYSTEM ANALYSIS
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SOLUTION The current division between the two generators are
The bus voltage changes EET308 POWER SYSTEM ANALYSIS
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SOLUTION The bus voltages during the fault
The short circuit line current are EET308 POWER SYSTEM ANALYSIS
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EET 102 ELECTRIC CIRCUIT II
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Impedance network with fault at bus 2
SOLUTION (b) Fault on bus 2 j0.2 j0.4 j0.8 1 2 j0.4 Zf = j0.16 j0.4 Impedance network with fault at bus 2 3 EET308 POWER SYSTEM ANALYSIS
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Thevenin’s equivalent network
SOLUTION j0.2 j0.4 j0.8 1 2 Vth j0.4 j0.4 I2(F) Zf = j0.16 3 Thevenin’s equivalent network EET308 POWER SYSTEM ANALYSIS
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SOLUTION The equivalent impedance between bus 1 and 2 j0.2 j0.4 j0.4 2
Vth Zf = j0.16 I2(F) EET308 POWER SYSTEM ANALYSIS
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SOLUTION Combining parallel branches from ground to bus 2 results
The fault current is Z22 = j0.24 2 Vth Zf = j0.16 I2(F) EET308 POWER SYSTEM ANALYSIS
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SOLUTION The current division between the two generators are
The bus voltage changes EET308 POWER SYSTEM ANALYSIS
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SOLUTION The bus voltages during the fault
The short circuit-current in the line are EET308 POWER SYSTEM ANALYSIS
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EET 102 ELECTRIC CIRCUIT II
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Impedance network with fault at bus 1
SOLUTION (c) Fault on bus 1 j0.2 j0.4 j0.8 1 2 j0.4 Zf = j0.16 j0.4 3 Impedance network with fault at bus 1 EET308 POWER SYSTEM ANALYSIS
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Thevenin’s equivalent network
SOLUTION j0.2 j0.4 j0.8 1 2 Vth j0.4 j0.4 I1(F) Zf = j0.16 3 Thevenin’s equivalent network EET308 POWER SYSTEM ANALYSIS
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SOLUTION Combining parallel branches between bus 1 and 2 j0.2 j0.4
Vth I1(F) Zf = j0.16 EET308 POWER SYSTEM ANALYSIS
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SOLUTION Combining parallel branches from ground to bus 1 results
The fault current is Z11 = j0.16 1 Vth I1(F) Zf = j0.16 EET308 POWER SYSTEM ANALYSIS
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SOLUTION The current division between the two generators are
The bus voltage changes EET308 POWER SYSTEM ANALYSIS
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SOLUTION The bus voltages during the fault
The short circuit-current in the line are EET308 POWER SYSTEM ANALYSIS
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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. EET 102 ELECTRIC CIRCUIT II
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