Ee 1352 Monograph Unitiv

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 : IV

Date of Submission :

.1.2009

Syllabus : IVth Unit FAULT ANALYSIS - SYMMETRICAL COMPONENTS AND UNBALANCED FAULT :

Fundamentals of symmetrical components - sequene impedances & sequence networks — single line to ground fault — line fault - Double line to ground – Unbalanced fault analysis using bus impedance matrix. EE 1352 - POWER SYSTEM ANALYSIS VI semester EEE Branch (2006-2010 Batch) Unit – IV – Fault Analysis – Symmetrical Components and Unbalanced Fault 1

Syllabus : Fundamentals of symmetrical components - sequene impedances & sequence networks — single line to ground fault — line fault - Double line to ground – Unbalanced fault analysis using bus impedance matrix. Fundamentals of symmetrical components : Introduction : The unsymmetrical faults are the faults in which the fault currents in the three phases are unequal. The different types of unsymmetrical faults are 1) Single line to ground fault. (LG) 2) Line to line fault. (LL) 3) Double line to ground fault. (LLG) 4) One or two open conductors fault. Since any unsymmetrical fault causes unbalanced currents to flow in the system, the unsymmetrical faults are analyzed using symmetrical components. Fortescue’s Theorem : The method of symmetrical components consists of resolving an unbalanced set of ‘n’ related phasors into ‘n’ sets of balanced phasors called symmetrical components of the original unbalanced phasors. Phase sequence : Phase sequence of the phasors is the order in which they pass through a positive maximum. Thus a phase sequence ‘abc’ implies that the maxima of the three phasors occur in the order ‘a’ followed by ‘b’ then by ‘c’. If ‘abc’ is taken as the positive phase sequence then ‘acb’ represents the negative phase sequence. The direction of rotation of the phasors in anticlockwise direction is considered as positive. ‘a’ – operator : The phasor ‘a’ is an operator which has a magnitude of unity and phase angle of 120°. When it is operated upon a voltage/current phasor, it rotates the original phasor by 120° electrical degrees without changing the magnitude of the phasor. a = 1∠120 = −0.5 + j 0.866 a 2 = 1∠240 = −0.5 − j 0.866 a 3 = 1∠360 = 1 From which we can say that 1 + a + a2 = 0. Symmetrical components : Let us consider a set of three unbalanced phasors of a three phase system. According to Fortescue’s theorem, these three phasors can be resolved into the following three sets of balanced phasors as 1. Positive sequence components consisting of three phasors with equal magnitudes, equally displaced from one another by 120° and phase sequence is same as that of original phasors. 2. Negative sequence components consisting of three phasors with equal magnitudes, equally displaced from one another by 120° and phase sequene is opposite to that of original phasors. 3. Zero sequence components consisting of three phasors with equal magnitudes and in 0° phase displacement as shown in fig. below:

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Fig. 1 Symmetrical components of three unbalanced phasors. These three sets of balanced phasors are called the symmetrical components of the original unbalanced phasors. Suppose that the three phases are represented by a, b and c such that the phase sequence is abc. Thus the phase sequence of the positive sequence components is abc and and the phase sequence of the negative sequence components is acb. It is assumed that the subscripts 0, 1, 2 refer to zero sequence, positive sequence and negative sequence respectively. If Va, Vb and Vc represent an unbalanced set of voltage phasors, the three sets of balanced phasors are written as 1. Positive sequence components – Va1, Vb1, Vc1. 2. Negative sequence components – Va2, Vb2, Vc2. 3. Zero sequence components – Va0, Vb0, Vc0. The phase voltages Va, Vb, Vc can be written as the sum of the symmetrical components as : Va = Va 0 + Va1 + Va 2 Vb = Vb 0 + Vb1 + Vb 2 Vc = Vc 0 + Vc1 + Vc 2 In a three phase balanced system, zero and negative sequence components are zero. Expressing the component quantities of ‘b’ and ‘c’ phases in terms of ‘a’ phase and using the ‘a’ operator where a = 1∠120° Va = Va 0 + Va1 + Va 2 Vb = Va 0 + a 2Va1 + aVa 2

We get,

Vc = Va 0 + aVa1 + a 2Va 2 In matrix form,

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

1  Va 0  a  Va1  a 2  Va 2 

We can write the above equation as V abc = AV 012

where

1 1 A = 1 a 2 1 a

1 a  a 2 

is called as the symmetrical component transformation matrix 3

Va  Va 0    V abc = Vb  and V 012 = Va1  Vc  Va 2  Using the inverse of the symmetrical component transformation matrix A-1, the above matrix equation can be written as V 012 = A −1V abc 1 1 1  1 1 a a 2  A −1 = where  3  1 a 2 a  Upon writing the matrix equation,

Va 0  1 1 1 V  = 1 a  a1  3  Va 2  1 a 2

1  Va  a 2  Vb  a  Vc 

From which we can write, 1 (Va + Vb + Vc ) 3 1 Va1 = Va + aVb + a 2Vc 3 1 Va 2 = Va + a 2Vb + aVc 3 I a0  1 1 1   I a   I  = 1 1 a a 2   I   a1  3   b  2  I a 2  1 a a   I c  Va 0 =

Similarly for current,

(

)

(

)

From which we can write, 1 (Va + Vb + Vc ) 3 1 Va1 = Va + aVb + a 2Vc 3 1 Va 2 = Va + a 2Vb + aVc 3 Thus any unbalanced quantity can be resolved into their sequence components. Note : From the expression of zero sequence current 1 I a0 = [ I a + I b + I c ] 3 Ia + Ib + Ic = In Va 0 =

(

)

(

)

1 In 3 ⇒ I n = 3I a 0 If neutral current flows in a system it indicates unbalanced condition, then definitely zero sequence current exists in the system. 2. In a three wire delta connected and also star connected system, In = 0 1. In a star connected, 4 wire system, ∴ I a 0 =

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∴ Ia = 0 If neutral current is zero, then zero sequence current is zero. --------------------------------------------------------------------------------------------------------Solved problems : 1. The line to ground voltages on the high voltage side of a step up transformer are 100 kV, 33 kV and 38 kV on phases a, b and c respectively. The voltage of phase ‘a’ leads that of phase ‘b’ by 100° and lags that of phase ‘c’ by 176.5°. Determine the symmetrical components of voltage. Solution : Va = 100∠0° Vb = 33∠ − 100° Vc = 38∠176.5° Va 0 =

1 [Va + Vb + Vc ] 3

1 [100 + j 0 − j32.4986 − 5.7303 − 37.9291 + j 2.3198] 3 1 = [ 56.34 − j 30.18] = 18.78 − j10.0595 = 21.3045∠ − 28.1757°kVolts 3 1 Va1 = Va + aVb + a 2Vc 3 1 = [100∠0° + 1∠120° x33∠ − 100° + 1∠240° x38∠176.5] 3 1 = [100 + j 0 + 31.01 + j11.2866 + 20.97 + j 31.69] 3 1 = [151.98 + j 42.9766] = 52.66 + j14.9766 = 52.646∠15.7897°kVolts 3 1 Va 2 = Va + a 2Vb + aVc 3 1 = [100∠0° + 1∠240° x33∠ − 100° + 1∠120° x38∠176.5] 3 1 = [100 + j 0 − 25.279 + j 21.212 + 16.9555 − j 34.0075] 3 1 = [ 91.6765 − j12.947] = 30.55 − j 4.265 = 30.846∠ − 7.947°kVolts 3 2. The unsymmetrical components of phase fault current in a 3-phase unbalanced system are I a 0 = 350∠90° A, I a1 = 600∠ − 90° A and I a 2 = 250∠90° A. Determine the phase currents Ia, Ib and Ic. Solution : The currents Ia, Ib and Ic are given by the following matrix equations. =

[

]

[

]

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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. 10

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 13

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 14

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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