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ELECTRICAL PROTECTION IV Dr. Oriedi Akumu Building 20-G01M akumuao@tut.ac.za Consultation Hours: Monday: 7:00-10:00 Wednesday:7:00-12:00 Thursday:7:00-10:00 ELECTRICAL PROTECTION IV LEARNING OUTCOMES Student should be able to: explain the concepts and methods for calculating unsymmetrical short circuit currents ASSESSMENT CRITERIA Unsymmetrical faults are defined Use of sequence networks in fault current calculation is explained Calculation of line-to-ground faults, line-to-line faults, and line-to-line-to-ground faults is explained Effect of delta-star transformer on fault currents is determined ELECTRICAL PROTECTION IV Text Book: Power System Analysis & Design Authors: J. D. Glover, M. S. Sarma and T. J. Overbye Assessment Type: Semester Tests Class Tests Assignments Design Project using Power World ELECTRICAL PROTECTION IV Symmetrical Components A mathematical model for representing an unbalanced system of ‘n’ phasors as ‘n’ systems of balanced phasors. N/B: This is only a model for analysis since symmetrical components do not exist in real life Unsymmetrical fault calculation is important for: Determination of protective relay setting Transient stability studies for interconnected systems ELECTRICAL PROTECTION IV Symmetrical Components The key idea of symmetrical component analysis is to decompose the system into three sequence networks. The networks are then coupled only at the point of the unbalance (i.e., the fault) The three sequence networks are known as the positive sequence (this is the one we’ve been using) negative sequence zero sequence ELECTRICAL PROTECTION IV Symmetrical Components Three phase voltages (or currents) can be resolved into three sets of sequence components: Phase “a” Zero sequence components Phase “b” Positive sequence components Same phase sequence as original phasor Phase “c” Negative sequence components Opposite phase sequence to original phasor ELECTRICAL PROTECTION IV Symmetrical Components Defined by: or Where Also or ELECTRICAL PROTECTION IV Symmetrical Components Similarly for currents: and In 3-phase Y connected systems: In a balanced Y-connected system, and in any three-phase system with no neutral path, such as a -connected system or a three-wire Y-connected system with an ungrounded neutral, line currents have no zero-sequence component. ELECTRICAL PROTECTION IV Symmetrical Components Example Solution ELECTRICAL PROTECTION IV Symmetrical Components Example Solution ELECTRICAL PROTECTION IV Sequence Impedances Each piece of equipment has 3 values of impedance: positive, negative and zero sequence impedances (Z1, Z2 and Z0, respectively) Z1 and Z2 of linear, symmetrical and static circuits (transformers, static loads, transmission lines …) are equal, and are the same as those used in analysis of balanced conditions The positive and negative sequence impedances of rotating machines are normally different The zero sequence impedance depends on the path taken by the zero sequence current ELECTRICAL PROTECTION IV Synchronous generators: Z2 < Z1 Z0 = variable value = Z1 if its value is not given Any impedance Ze in the earth connection introduces an impedance of 3Ze per phase Transformers: Z1 = Z2 = impedance of transformer Z0 = Z1, if there is circuit for earth current = infinity, if there is no through circuit for earth current Transmission lines: Z1 = Z2 = impedance of the line Z0 = variable value = 3Z1, if its value is not given ELECTRICAL PROTECTION IV Symmetrical components of generator Zero sequence network Positive sequence network 𝑉𝑛 = 𝑍𝑛 𝐼𝑛 = 3𝑍𝑛 𝐼0 Negative sequence network Voltage drop in neutral due only to zero sequence current ELECTRICAL PROTECTION IV Zero sequence circuits for three phase transformers ELECTRICAL PROTECTION IV Unsymmetrical Fault Analysis The following assumptions are made to compute subtransient fault current for short circuits in power systems: Transformers are represented by their leakage reactances. Winding resistances, shunt admittances, and delta/star phase shifts are neglected. Transmission lines are represented by their equivalent series reactances. Series resistance and shunt admittances are neglected. Synchronous machines are represented by constant voltage sources behind subtransient reactances. Armature resistance, saliency and saturation are neglected Non-rotating impedance loads are neglected Induction motors are either neglected (especially for small motors rated less than 40kW), or represented in the same way as synchronous machines. ELECTRICAL PROTECTION IV Unsymmetrical Fault Analysis In addition to previous assumptions, the following assumptions should be made with regards to unsymmetrical fault analysis: The power system operates under balanced steady-state conditions before the fault occurs. Thus the zero-, positive-, and negative sequence networks are uncoupled before the fault occurs. During unsymmetrical faults they are interconnected only at the fault location depending on the type of fault. Prefault load current is neglected. Because of this, the positive sequence internal voltages of all machines are equal to the prefault voltage VF. Therefore, the prefault voltage at each bus in the positive-sequence network equals VF. ELECTRICAL PROTECTION IV Unsymmetrical Fault Analysis Development of three sequence networks: ELECTRICAL PROTECTION IV Positive Sequence Network ELECTRICAL PROTECTION IV Negative Sequence Network ELECTRICAL PROTECTION IV Zero Sequence Network How? ELECTRICAL PROTECTION IV To do further analysis we first need to calculate the Thevenin equivalents as seen from the fault location. In this example the fault is at the terminal of the right machine (bus 2) so the Thevenin equivalents are: Zth j 0.2 in parallel with j0.455 Zth j 0.21 in parallel with j0.475 ELECTRICAL PROTECTION IV Single-Line-Ground Faults 𝑓 𝐼𝑎 𝑓 𝐼𝑏 𝑓 𝐼𝑏 𝐼𝑓𝑎 = 0 0 Constraint 1 𝐼𝑓0 𝐼𝑓+ 𝐼𝑓− 1 1 1 = 1 𝛼 3 1 𝛼2 1 𝛼2 𝛼 𝐼𝑓𝑎 𝐼 𝑓 0 + − 0 ⟹ 𝐼𝑓 = 𝐼𝑓 = 𝐼𝑓 = 3 𝐼𝑎 0 ELECTRICAL PROTECTION IV Single-Line-Ground Faults f Va Vaf f Vb f Vc f Z f Ia 1 1 2 1 1 = 3𝑍𝑓 𝐼𝑓+ 0 V 1 f V f 2 V f Constraint 2 This means Vaf V f0 V f V f The only way these two constraints can be satisified is by coupling the sequence networks in series ELECTRICAL PROTECTION IV Single-Line-Ground Faults With the sequence networks in series we can solve for the fault currents (assume Zf=0 for bolted fault) 𝐼𝑓0 = 𝐼𝑓+ = 𝐼𝑓− = 𝑉𝑓 𝑗 𝑍 0 + 𝑍 + + 𝑍 − + 3𝑍𝑓 1.050 I I I j1.96427 p.u. j (0.25 0.13893 0.14562 3Z f ) 0 f f f I af 3 I 0f j 5.8928 p.u. ELECTRICAL PROTECTION IV Line-Line Faults I af 0, I bf I cf , I 0f 0 Vbg Vcg if Z F 0 𝑉𝑏𝑔 − 𝑉𝑐𝑔 = 𝑍𝐹 𝐼𝑏 ELECTRICAL PROTECTION IV Line-Line Faults Using the current relationships we get I 0f 1 0 1 1 1 2 f I f 1 I b 3 1 2 I f I f b 0 If 0 If 1 f 2 Ib 3 Hence I f I f If 1 f 2 Ib 3 Constraint 1 ELECTRICAL PROTECTION IV Line-Line Faults Using the voltage relationships we get f V f0 V 1 ag 1 1 1 V f 1 2 Vbgf 3 2 Vcgf V f 1 Hence 1 f Vag 2 Vbgf 3 1 f V f Vag 2 Vbgf 3 V f Constraint 2 V f V f ELECTRICAL PROTECTION IV Line-Line Faults To satisfy I f I f & V f V f the positive and negative sequence networks must be connected in parallel Doesn’t play role as I0=0 Why? ELECTRICAL PROTECTION IV Line-Line Faults 𝐼𝑓+ = −𝐼𝑓− = Solving the network for the currents we get 1.050 If 3.691 90 j 0.1389 j 0.1456 I af 1 1 1 0 0 f 2 6.39 I 1 3.691 90 b f 1 2 3.69190 6.39 I c ELECTRICAL PROTECTION IV 𝑉𝑓 𝑗 𝑍+ + 𝑍− Line-Line Faults Solving the network for the voltages we get Vf Why “-”? V f 1.050 j 0.1389 3.691 90 0.5370 j 0.1452 3.69190 0.5370 Vaf f Vb f Vc 1 0 1.074 1 1 2 0.537 0.537 1 1 2 0.537 0.537 ELECTRICAL PROTECTION IV Double Line-Ground Faults I af 0 Vbgf Vcgf Z f ( I bf I cf ) ELECTRICAL PROTECTION IV Double Line-Ground Faults From the current relationships we get I af f Ib f I c 0 I 1 f 1 1 2 1 I f 1 2 I f Since I af 0 Constraint 1 I 0f I f I f 0 Note, because of the path to ground the zero sequence current is no longer zero. ELECTRICAL PROTECTION IV Double Line-Ground Faults From the voltage relationships we get f V f0 V 1 ag 1 1 1 V f 1 2 Vbgf 3 1 2 f V f Vbg Since f Vbg f Vcg Vf Vf Constraint 2 Then Vbgf V f0 ( 2 )V f But since 1 2 0 2 1 Vbgf V f0 V f ELECTRICAL PROTECTION IV Double Line-Ground Faults Vbgf V f0 V f Zf I I f b f c Also since I bf I 0f 2 I f I f I cf I 0f I f 2 I f Adding these together (with 1) 2 Vbgf Z f 2 I 0f I f I f with I 0f I f I f V f0 V f 3Z f I 0f Constraint 3 ELECTRICAL PROTECTION IV Double Line-Ground Faults Taking all constraints into consideration: 𝐼𝑓+ 𝑉𝐹 = + 𝑍 + 𝑍 − 𝑍 0 + 3𝑍𝐹 ELECTRICAL PROTECTION IV Double Line-Ground Faults V 1 . 05 0 F I f j 4.547 p.u. 0 Z Z // Z j 0.13893 j 0.092 V f 1.05 4.547 90 j 0.1389 0.4184 I f 0.4184 / j 0.1456 j 2.874 Why “-”? I 0f I f I f j 4.547 j 2.874 j1.673 Converting to phase: I bf 1.04 j 6.82 I cf 1.04 j 6.82 ELECTRICAL PROTECTION IV Unsymmetrical Fault Summary SLG: Sequence networks are connected in series, and also in parallel to three times the fault impedance VF I I I j ( Z Z Z 0 3Z f ) 0 f f I 3 I f a f 0 f ELECTRICAL PROTECTION IV LL: Positive and negative sequence networks are connected in parallel; zero sequence network is not included since there is no path to ground 3VF I I j(Z Z ) f b f c ELECTRICAL PROTECTION IV DLG: Positive, negative and zero sequence networks are connected in parallel, with the zero sequence network including three times the fault impedance VF I 0 Z Z // Z 3Z F f ELECTRICAL PROTECTION IV