Ee 1352 Monograph Unitv

K.L.N. COLLEGE OF INFORMATION TECHNOLOGY Pottapalayam – 630 611, Sivagangai District.

Year : 2008 -2009,

Branch : III B.E. EEE

Semester : 6

Subject : EE 1352 Power System Analysis

Name of the Writer (Collaborator) : R. Ramesh, Professor (EEE), IT Dept. Unit : V

Date of Submission :

.1.2009

Syllabus : Vth Unit POWER SYSTEM STABILITY :

Sl.No. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

Topics Basic concepts and definitions - Rotor angle stability Voltage stability – Mid Term and Long Term stability Classification of stability – An elementary view of transient stability Equal area criterion Factors influencing transient stability Numerical integration methods Euler method Modified Euler method Runge-Kutta methods Assignment - V Unit – V Test Total

L 1 1 1 1 1 1 1 1 1 9

T 1 1 3

P -. -

EE 1352 - POWER SYSTEM ANALYSIS 1

VI semester EEE Branch (2006-2010 Batch) Unit – V – Power System Stability Syllabus : Basic concepts and definitions – Rotor angle stability – Voltage stability – Mid Term and Long Term stability – Classification of stability – An elementary view of transient stability – Equal area criterion – Factors influencing transient stability – Numerical integration methods – Euler method – Modified Euler method – Runge-Kutta methods. 5.1 Introduction The transient behavior of a power system resulting from major disturbances such as a fault followed by switching operations, sudden rejection of load or generation etc. A major disturbance upsets the balance between mechanical input and electrical output of the generator with the result that some generators may accelerate while others may decelerate. The rotor angles will undergo wide variations and in this process the “synchronism” of the system gets affected. Synchronism must be maintained by timely operation of circuit breakers. Critical clearing times of circuit breakers can be computed and protection zones of distance relays during transient swings can be adjusted. Compared to short circuit and load flow studies, transient stability studies are more complex since they involve the electromechanical dynamics of the synchronous machines and its associated controls such as excitation system and governor. 5.2 Basic concepts and Definitions : Power system stability : It is defined as that property of a power system that enables it to remain in a state of operating equilibrium under normal operating conditions and to regain an acceptable state of equilibrium after being subjected to a disturbance. A necessary condition for satisfactory power system operation is that all synchronous machine remain in synchronism or in step. This aspect of stability is influenced by the dynamics of generator rotor angles and power angle relationships. Instability may also be encountered without lose of synchronism. For example, a system consisting of synchronous generator feeding an induction motor load through a transmission line can become unstable because of the collapse of load voltage. Maintenance of synchronism is not an issue, instead, the concern in stability and control of voltage. This form of instability can also occur in loads covering an extensive area supplied by a large system. 5.3 Rotor angle stability :

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It is the ability of inter connected synchronous machines of a power system to remain in synchronism. The stability problem involves the study of the electromechanical oscillations inherent in power systems. A fundamental factor in this problem is the manner in which the power outputs of synchronous machines vary as their motor oscillate. It is usual to characterize the rotor angle stability phenomena in terms of the following two categories. 5.3.1

Small signal stability (or) Steady state stability : It is the ability of the power system to maintain synchronism under small disturbances. Such disturbances occur continually on the system because of small variations in load and generation. Instability may occur in two forms. i) Steady increase in rotor angle due to lack of sufficient synchronizing torque or ii) Rotor oscillations of increasing amplitude due to lack of sufficient damping torque. 5.3.2

Transient stability : It is the ability of the power system to maintain synchronism when subjected to a severe transient disturbance. The resulting system response involves large excursions of generator rotor angles and is influenced by the non-linear power-angle relationship. In large power systems, transient instability may not always occur as first swing instability, it could be result of superposition of several modes of oscillation causing large excursions of rotor angle beyond the first swing. In transient stability studies the study period of interest is usually limited to 3 to 5 seconds following the disturbance, although it may extent to about 10 seconds for every large systems. 5.4 Swing equation for single machine infinite bus system : The equation governing the motion of the rotor of a synchronous machine is based on the elementary principle in dynamics which states that accelerating torque is the product of the moment of inertia of the rotor times its angular acceleration. In MKS system of units this equation can be written for the synchronous generator in the form : d 2ϑ J 2 = Ta = Tm − Te N − m..................(1) dt where J = The total moment of inertia of the rotor mass in kg-m2. Θm = Angular displacement of the rotor with respect to a stationary axis in mechanical radians. t = Time, in seconds. Tm = The mechanical (or) shaft torques supplied by the prime mover in N-m. Te = Net electrical (or) electromagnetic torque in N-m. Ta = Net accelerating torque in N-m.

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The mechanical torque Tm and electrical torque Te are considered positive for synchronous generator. These means that Tm is the resultant shaft torque which tends to accelerate the rotor in the positive θm direction of rotation as shown in fig. 9.1 (a). Under steady state operation the generator Tm and Te are equal and the accelerating torque Ta is zero. In this case there is no acceleration or deceleration of the rotor masses which include the rotor of the generator and prime mover are said to be in synchronism with the other machines operating at synchronous speed in the power system. The electrical torque Te corresponds to the net air gap power in the machine and thus accounts for the total output power of the generator plus I2R losses in the armature winding. In the synchronous motor the direction of power flow is opposite to that in the generator. Accordingly for a motor both Tm and Te are reversed in sign as shown in fig. 9.1 (b). Te is then corresponds to the air gap power supplied by the electrical system to drive the rotor while Tm → the counter torque of the load and the rotational losses tending to retard the rotor. In our discussion Tm is considered to be constant since the prime mover is controlled by governors. Since θm is measured with respect to a stationary references axis on the stator, it is an absolute measure of rotor angle. θm increases continuously with time even when synchronous speed is constant. Since the rotor speed relative to synchronous speed is more convenient to measure the rotor angular position with respect to reference axis which rotates at synchronous speed. We define θm = ωsm + δm . . . . . (2) where ωsm is the synchronous speed of the machine in mech. Radians per second. δm is the angular displacement of the rotor in mech. radians from the synchronously rotating axis. dθ m dδ = ω sm + m dt dt . . . . . (3) & (4) 2 2 d θm d δ = dt dt 2

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dθ m is constant and dt dδ m dδ m equals the synchronous speed only when is zero. Therefore represents the dt dt deviation of the rotor speed from synchronism and the units of measure are mechanical radians per second. From equation (3) we can say the rotor angular velocity

Equation (4) represents rotor acceleration in mechanical radian per second squared. Substitute equation (4) in equation (1) d 2δ m J = Ta = Tm − Te N − m..........(5) dt 2 It is convenient for notational purposes to introduce dϑ ω m = m for angular velocity of the rotor. dt From elementary dynamics power = Torque x angular velocity. Multiply equation (5) by ωm , d 2δ m Jω m = Pa = Pm − PeW . . . . (6) dt 2 where Pm is the shaft power input to the machine Pe is the electrical power crossing its air gap Pa is the accelerating power which account for any unbalance between those two quantities usually rotational loss and I2R losses are neglected and think Pm as power supplied by the prime mover and Pe → electrical power output. Coefficient Jωm is the angular momentum of the rotor at synchronous speed ωsm and is denoted by M called inertia constant of the machine. Jωm = M Equation (6) becomes d 2δ m M = Pa = Pm − Pe . . . . . (7) dt 2 In practice M cannot be a constant as it is the coefficient. M is expressed in Joule-second / mech. radian. We write as d 2δ m M = Pa = Pm − Pe . . . . . (8) dt 2 M is not constant because ωm=ωsm under all condition. One more constant which is very much encountered in the stability studies is H – constant defined as Stored kinetic energy in megajoules at synchronous speed H = ————————————————————————— Machine rating in MVA

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1 1 2 Jω sm Mω sm 2 2 H= = MJ / MVA S mach S mach where Smach three phase rating of the machine in MVA 2H M = S mach MJs/mach.rad ω sm Substitute this in equation (7) we get P P − Pe 2 H d 2δ m = a = m . . . . (8) 2 ω sm dt S mach S mach δm → expressed in mech. rad. ωsm → expressed in mech. rad./sec. We can write the above equation as 2 H d 2δ = Pa = Pm − Pe p.u. . . . . . (9) ω s dt 2 δ and ωs are in mech (or) electrical degrees or radians. Thus for a system with a frequency of f herts, 2 H d 2δ = Pa = Pm − Pe 2πf dt 2 H d 2δ = Pa = Pm − Pe perunit.......(10) πf dt 2 when δ is in electrical degrees. Equation (10) is called the swing equation o the machine is the fundamental equation which governs the rotational dynamics of the synchronous machine in stability studies. When swing equation is solved we obtain the expression for δ as a function of time. A graph of the solution is called the swing curve of the machine and inspection of the swing curves of all the machines of the system will show whether the machines remain in synchronism after a disturbance. Multimachine system In a multimachine system a common system base must be chosen. Let Smach = machine rating (base) Ssystem = system base Equation (10) can be written as S mach  H mach d 2δ  S = ( Pm − Pe ) mach (or)  2  S system  πf dt  S system H system d 2δ = Pm − Pe p.u. in system base πf dt 2

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 S mach  where H system − H mach   = Machine inertia constant in system base.  S system  --------------------------------------------------------------------------------------------------------Solved problems : Example : A 50 Hz, four pole turbogenerator rated 100 MVA, 11 kV has an inertia constant of 8.0 KJ/MVA. a) Find the stored energy in the rotor at synchronous speed. b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find the rotor acceleration elec.deg./sec2 neglecting mechanical and electrical losses. Solution : Stored energy in MJ H = ———————————— Machine rating in MVA a) Stored energy in MJ = H (inertia constant) x MVA = 8 x 100 = 800 MJ. b) Pa = Pm – Pe = 80 – 50 = 30 MW Md 2δ Pa = dt 2 H 8 M = S mach = x100 = 0.089MJ − s / elec. deg ree 180 f 180 x50 d 2δ d 2δ Pa 30 Pa = M 2 ⇒ 2 = = = 337.08elec. / deg ree M 0.089 dt dt d 2δ Acceleration = 337.08elec. deg ree / sec 2 dt 2 ------------------------------------------------------------------------------------------------------Typical values of H Type of machine Inertia constant H in MJ/MVA Water wheel generator Slow speed < 200 rpm 2.3 High speed > 200 rpm 2.3 Synchronous capacitor Large 1.25 Small 1.00 Synchronous motor with load varying from 2.00 1.0 to 5.0 and higher for flywheels Turbine generator Condensing 1800 rpm 9.6 3000 rpm 7.4 Non condensing 3000 rpm 4.1

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 I a  1 1 1   I a 0   I  = 1 a 2 a   I   b    a1  2  I c  1 a a   I a 2  ∴ I a = I a 0 + I a1 + I a 2 I b = I a 0 + a 2 I a1 + aI a 2

I a2

I c = I a 0 + aI a1 + a 2 I a1 Given that, I a 0 = 350∠90° = 0 + j 350 I a1 = 600∠ − 90° = 0 − j 600 = 250∠90° = 0 + j 250

∴ aI a1 = 1∠120° x600∠ − 90° = 600∠30° = 519.62 + j 300 a 2 I a1 = 1∠240° x600∠ − 90° = 600∠150° = −519.62 + j 300 aI a 2 = 1∠120° x 250∠90° = 250∠210° = −216.51 − j125 a 2 I a 2 = 1∠240° x 250∠90° = 250∠330° = 216.51 − j125 I a = I a 0 + I a1 + I a 2 = j 350 − j 600 + j 250 = 0 I b = I a 0 + a 2 I a1 + aI a 2 = j 350 − 519.62 + j 300 − 216.51 − j125 = −736.13 + j 525 = 904.16∠145° A I c = I a 0 + aI a1 + a 2 I a 2 = j 350 + 519.62 + j 300 + 216.51 − j125 = 736.13 + j 525 = 904.16∠35° A ----------------------------------------------------------------------------------------------------------Power in terms of Symmetrical Components If the symmetrical components of currents and voltages are known, the power in a 3 phase circuit can be computed directly from the components. The total complex power flowing into a three phase circuit through three lines a,b * * * and c is S = P + jQ = Va I a + Vb I b + Vc I c Where Va, Vb and Vc are voltages to neutral at the terminals and Ia, Ib and Ic are the currents flowing into the circuit in the three lines. A neutral connection may or may not present. In matrix notation, *

S = [Va

Vb

I a  Va  Vc ]  I b  = Vb   I c  Vc 

T

I a  I   b  I c 

*

Where the conjugate of a matrix is composed of elements that are the conjugates of the corresponding elements of the original matrix. Now use the symmetrical components of the voltages and currents in the above equation.

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Va  1 1 V  = 1 a 2  b  Vc  1 a

1  Va 0   I a  1 1    a  Va1  and  I b  = 1 a 2  I c  1 a a 2  Va 2 

1   I a0  a   I a1  a 2   I a 2 

Va  I a  V  = AV , and  I  = AI  b  b Vc   I c 

∴ S = [ AV ] x[ AI ] T

*

The reverse rule of matrix algebra states the transpose of the product of two matrices is equal to the product of the transposes of the matrices in reverse order. According to this rule, [ AV ] T = V T AT ∴ S = V T AT A * I * Noting that AT = A and a and a2 are conjugates we get,

S = [Va 0

1 1 Va 2 ] 1 a 2 1 a

Va1

1  1 1 a  1 a a 2  1 a 2

1 I a0  a 2   I a1  a   I a 2 

*

1 0 0 A A = 30 1 0 0 0 1  T

*

S = 3[Va 0 Va1

I a0  Va 2 ]  I a1   I a 2 

*

So complex power is

Va I a* + Vb I b* + Vc I c* = 3V0 I 0* + 3V1 I 1* + 3V2 I 2* If the symmetrical components of currents and voltages are known, the power consumed by a three phase circuit can be computed directly from the sequence components. Thus the above relation is very useful for computing power in unbalanced power system. ----- --Sequence Impedance In any part of circuit, the voltage drop caused by current of a certain sequence depends on the impedance of the part of the circuit to current of that sequence. The impedance of any section of a balanced network to current of one sequence may be different from impedance to current of another sequence. The sequence impedance of an equipment or component of a power system are the positive, negative and zero sequence impedances. They are defined as follows :

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The impedance of a circuit element when positive sequence currents alone are flowing is called positive sequence impedance. Similarly, when only negative sequence currents are present, the impedance is called negative sequence impedance. When only zero sequence currents are present the impedance is called zero sequence impedance. The impedance of any element of a balanced circuit to current of one sequence may be different from impedance to current of another sequence. The single phase equivalent circuit of power system (impedance or reactance diagram) formed using the impedances of any one sequence only is called the sequence network for the particular sequence. Therefore the impedance or reactance diagram formed using positive sequence impedance is called positive sequence network. Similarly the impedance or reactance diagram formed using negative sequence impedance is called negative sequence network. The impedance or reactance diagram formed using zero sequence impedance is called zero sequence network. The sequence impedances and networks are useful in the analysis of unsymmetrical faults in the power system. In unsymmetrical fault analysis of a power system, the positive, negative and zero sequence networks of the system are determined and then they are interconnected to represent the various unbalanced fault conditions. Each sequence network includes the generated emfs and impedances of like sequences. Also, the sequence network carries only the currents of like sequence. Sequence impedance and networks of a generator Consider the three phase equivalent circuit of a generator shown in fig. 1. The neutral of the generator is grounded through a reactance Zn. When the generator is delivering a balanced load or under symmetrical fault, the neutral current is zero. But when the generator is delivering an unbalanced load or during unsymmetrical faults the neutral current flows through Zn. The generator is designed to supply balanced three phase voltages. Therefore the generated emfs are positive sequence only. Fig. 1 Three phase equivalent circuit of a generator grounded through a reactance.

Let Ea, Eb, Ec = Generated emf per phase in phase a, b and c respectively. Z1 = Positive sequence impedance per phase of generator. Z2 = Negative sequence impedance per phase of generator. Zg0 = Zero sequence impedance per phase of generator. Zn = Neutral reactance. Z0 = Total zero sequence impedance per phase of zero sequence network of generator. The positive sequence network consists of an emf in series with positive sequence impedance of the generator. The negative and zero sequence network will not have any sources but include their respective sequence impedance. The positive, negative and zero sequence current paths are shown in fig. 2. The positive, negative and zero sequence networks of the generator are shown in fig. 3.

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Note : The positive and negative sequence currents are balanced currents and so they will not pass through neutral reactance. The reactances in positive sequence network is subtransient, transient or synchronous depending on whether subtransient, transient or steady state conditions are being studied. Under no load condition the emf Ea is the induced emf per phase. Under load or fault condition Ea is replaced by Eg’ for transient state and Ea is replaced by Eg” for subtransient state.

Fig. 2 Sequence current paths in a generator

Fig. 3 Sequence networks of a generator

On examining the zero sequence current paths (refer fig. 2 ©) the current through neutral reactance is 3 Ia0. The zero sequence voltage drop from point-a to ground is − 3I a 0 Z n − I a 0 Z g 0 .

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The zero sequence network is a single phase network and assumed to carry only the zero sequence current of one phase. Hence the zero sequence current of one phase, must have an impedance 3Z0 + Zg0.

∴Total zero sequence impedance per phase of a generator grounded through reactance Z0 = 3 Zn + Zg0.

With reference to fig. 3, the equations for the phase-a component voltages are, Va1 = E a − I a1 Z 1 Va 2 = − I a 2 Z 2 Va 0 = − I a 0 Z 0 The zero sequence network of generator when the neutral is solidly grounded (i.e. directly grounded) and when the neutral is ungrounded are shown in Fig. 4 and Fig. 5 respectively. In these cases there is no change in positive and negative sequence network.

Fig. 4 : Zero sequence network of a generator when the neutral is solidly grounded.

Fig. 5 : Zero sequence network of a generator when the neutral is ungrounded.

Note : The sequence networks of synchronous motor is same as that of generator when the directions of currents in the sequence network of generator are reversed. Sequence impedances and networks of transmission lines The impedance per phase of transmission line for balanced currents is independent of phase sequence. This is due to the symmetry of transmission lines. Therefore the impedances offered by the transposed transmission lines for positive and negative sequence currents are identical. The zero sequence current is identical (both in magnitude and phase) in each phase conductor and returns through the ground, through overhead wires or through both. The ground wires being grounded at several towers, the return currents in the ground wire may not be uniform along the entire length of transmission line. But for positive, negative sequence currents there is no return currents and they have a phase difference of 120°. Therefore the magnetic field due to zero sequence current is different from the magnetic field caused by either positive or negative sequence current. Due to the difference in the magnetic field, the zero sequence inductive reactance is 2 to 3.5 times the positive sequence reactance. Let, Z1 = Positive sequence impedance of transmission line. Z2 = Negative sequence impedance of transmission line. Z0 = Zero sequence impedance of transmission line. 12

The positive, negative and zero sequence impedances of transmission lines are represented as a series impedance in their respective sequence networks as shown in fig. 6.

Fig. 6 Positive, Negative and zero sequence networks of a transmission line. Sequence impedances and networks of a transformer When the applied voltage is balanced, the positive and negative sequences of linear, symmetrical, static devices are identical. Therefore in a transformer the positive and negative sequence impedances are identical. Even though the zero sequence impedance may slightly differ from positive and negative sequence impedance, it is normal practice to assume the zero sequence impedance as equal to positive or negative sequence impedance. [For all types of transformers the series impedances of all sequences are assumed equal]. Note : When the neutral of star connection is grounded through reactance Zn then 3Zn should be added to zero sequence impedance of transformer to get the total zero sequence impedance. Let, Z1 = Positive sequence impedance of transformer. Z2 = Negative sequence impedance of transformer. Z0 = Zero sequence impedance of transformer. The positive, negative and zero sequence impedances of transformer are represented as a series impedance in their respective sequence networks as shown in fig. 7

Fig. 7 Positive and Negative sequence networks of transformer The zero sequence network of the transformer depends on the type of connections (Y or Δ) of the primary and secondary windings and also on the grounding of neutral in Y connection. The following general observations can be made for zero sequence currents in transformers. 1. When magnetizing current is neglected, the primary winding will carry current only if there is a current flow in the secondary winding. Therefore the zero sequence current can flow in the primary winding of

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2. 3. 4.

a transformer only if there is a path for zero sequence current in secondary winding or vice-versa. If the neutral point in the Y connected winding is not grounded then there is no path for zero sequence current in star connected winding. The zero sequence current flows in the star connected winding and in the lines connected to the winding only when the neutral point is grounded. The zero sequence current can circulate in the delta connected winding but the zero sequence current cannot flow through the lines connected to the delta connected winding.

Based on the above observations the zero sequence network of 3 phase transformer can be obtained for any configuration. The zero sequence network for nine possible configuration are presented in table 1 shown below. The arrows on the windings, indicate path for zero sequence current and the absence of arrows indicate that there is no path for zero sequence currents.

Table 1 Zero sequence network of three phase transformer

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Sequence impedances and networks of loads 15

In balanced Y or Δ connected loads the positive, negative and zero sequence impedances are equal. When the neutral point of star connected load is grounded through a reactance Zn then 3Zn is added to the zero sequence impedance of load to get the total zero sequence impedance of load. Let, ZL1 = Positive sequence impedance of load. ZL2 = Negative sequence impedance of load. ZL0 = Zero sequence impedance of load. The positive, negative and zero sequence impedances of transmission lines are represented as a shunt impedance in their respective sequence networks as shown in fig. 8.

Fig. 8 Positive and Negative sequence network of load The zero sequence network of the 3 phase load depends on the type of connection, i.e. Y or Δ connection. The zero sequence current will flow in network only if a return path exists for it. The zero sequence network for various types of loads are shown in table below : Table 2 : Zero sequence networks of loads

Single line to ground fault on an unloaded generator 16

Review Questions (2 marks) 1. How do short circuits occur on a power system ? 2. Distinguish between symmetrical and unsymmetrical short circuits. 3. What are the applications of short-circuit analysis? 4. Define short circuit capacity of a power system. 5. What are the different types of fault which occur in a power system? 6. Define symmetrical fault. 7. Name any two methods of reducing short circuit current. Review Questions (16 marks) 1. (a) Explain the need for short circuit studies. (b) Draw the oscillogram when an unloaded alternator is subjected to symmetrical short circuit and hence explain how to obtain xd, xd’ and xd” from it. 2. With the help of a detailed flow chart, explain how a symmetrical fault can be analyzed using ZBUS. 3. What are the various types of faults ? Discuss the frequency of occurrence and severity. 4. Explain the difference between direct axis transient reactance and direct axis sub-transient reactance of alternators. 5. Explain the method of calculation of fault currents in the transmission system using Thevenin’s theorem. 6. A Generator-transformer unit is connected to a line through a circuit breaker. The unit ratings are : Generator : 10 MVA, 6.6 kV, Xd” = 0.1 p.u., Xd’= 0.20 p.u. and Xd = 0.80 p.u. Transformer : 10 MVA, 6.9/33 kV, reactance = 0.08 p.u. The system is operating at no load at a line voltage at 30 kV, when a three phse fault occurs on the line just beyond the circuit breaker. Find (a) the initial symmetrical rms current in the breaker. (b) the maximum possible DC off-set current in the breaker (c) the momentary current rating of the breaker (d) the current to be interrupted by the breaker and the interrupting kV and (e) the sustained short circuit current in the breaker. 8. The system shown in figure is delivering 50 MVA at 11 kV, 0.8 lagging power factor into a bus which may be regarded as infinite. Particulars of various system components are : Generator : 60 MVA, 12 kV, Xd’ = 0.35 p.u. Transformers (each) : 10 MVA, 12/66 kV, reactance 0.08 p.u. Line : Reactance : 12 ohms, resistance negligible. Calculate the symmetrical current that the circuit breakers A and B will be called upon to interrupt in the event of a three phase fault occurring at F near the circuit breaker B.

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9. In the system configuration of figure, the system impedance data is given below : Transient reactance of each generator = 0.15 p.u. Leakage reactance of each transformer = 0.05 p.u., Z12 = j0.1, Z13 = j0.12, Z23 = j0.08 p.u. For a solid 3 phase fault on bus 3, find all bus voltages and S C currents in each component. Assume prefault voltages to be 1 p.u. and prefault currents to be zero.

10. For a fault (solid) location shown in figure, find the short circuit currents in line 1-2 and 1-3. Prefault system is on no-load with 1 p.u. voltage and prefault currents are zero. Use ZBUS method and compute its elements by the current injection technique.

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