new chapter 6 symmetrical components
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CHAPTER 6
SYMMETRICAL COMPONENTS
Chapter 6 Page 1 Symmetrical Components
SYMMETRICAL COMPONENTS
INTRODUCTION
SEQUENCE COMPONENTS
FAULT CONDITION PHASOR DIAGRAMS
SEQUENCE IMPEDANCE FOR TRANSFORMERS
SEQUENCE IMPEDANCE FOR CABLES
SEQUENCE IMPEDANCE FOR MOTORS
SEQUENCE IMPEDANCE FOR GENERATORS
EQUIVALENT CIRCUIT FOR LINE TO GROUND FAULTS
PROBLEM 4
Chapter 6 Page 2 Symmetrical Components
INTRODUCTION The symmetrical components method (mathematical operations) is the foundation for obtaining and understanding ground fault data on three-phase power systems. In short, the method of symmetrical components is one of the relay/coordination engineer’s most powerful technical tools. While the method and mathematics are quite simple, the practical value lies in the ability to think and visualize in symmetrical components. This skill requires practice and experience. The symmetrical components method consists of reducing an unbalanced three-phase system of phasors into three balanced or symmetrical systems: the positive, negative, and zero phase sequence components. This balanced reduction can be performed in terms of current, voltage, and impedance.
BALANCED SYSTEM
A balanced system (i.e., a three-phase fault) consists of three phasors, all equal in magnitude and 120o apart (Figure 6-1). For example:
IA = 1∠0o = 1.0 + j0
IB = 1∠240o = -0.5 - j0.866
IC = 1∠120o = -0.5 + j0.866
IA + IB + IC = 0
|IA| = |IB| = |IC|
Figure 6-1. Example of a Balanced System
Chapter 6 Page 3 Symmetrical Components
Unbalanced System
An unbalanced system (e.g., a line-to-ground fault) consists of three-phasors, not all equal in magnitude or degrees apart (Figure 6-2). For example:
IA = 1∠0o = 1.0 + j0
IB = 2 ∠225o = -1.0 - j1.0
IC = 1∠90o = 0 + j1.0
IA + IB + IC = 0
|IA| = |IC| ≠ |IB|
Figure 6-2. Example of an Unbalanced System
Chapter 6 Page 4 Symmetrical Components
SEQUENCE COMPONENTS The sequence components consist of three sequence sets: positive (+) sequence, negative (-) sequence, and zero (0) sequence.
Positive sequence (+) components consist of three phasors equal in magnitude, displaced from each other by 120o in phase, and having the same phase sequence (abc) as the original unbalanced phasors (abc). The term “positive” derives from the fact that Ib1 is a positive (+) 120o behind Ia1 (Figure 6-3). Note: Subscript 1 identifies the positive sequence component, subscript 2 identifies the negative sequence component, and the subscript 0 identifies the zero sequence component.
Figure 6-3. Positive (+) Sequence Components
Negative sequence (-) components consist of three phasors equal in magnitude, displaced from each other by 120o in phase, and having the phase sequence opposite (acb) to that of the original phasors (abc). The term “negative” derives from the fact that Ib2 is a negative (-) 120o behind Ia2 (Figure 6-4).
Figure 6-4. Negative (-) Sequence Components
Chapter 6 Page 5 Symmetrical Components
Zero sequence (0) components consist of three phasors equal in magnitude and with zero phase displacement (0o) from each other (Figure 6-5).
Figure 6-5. Zero (0) Sequence Components
Chapter 6 Page 6 Symmetrical Components
Operators (J, A)
The “j” operator is a unit phasor with an angle displacement of 90o (Figure 6-6).
j = 1 ∠90o = 0 + j1.0 = j
j2 = 1 ∠180o = -1.0 + j0 = -1.0
j3 = 1 ∠270o = 0 - j1.0 = -j
j4 = 1 ∠360o = 1.0 + j0 = 1.0
-j = 1 ∠270o = 0 - j1.0 = -j = j3
Figure 6-6. The “j” Operator
The “a” operator is a unit phasor with an angle displacement of 120o (Figure 6-7).
a = 1 ∠120o = - 0.5 + j0.866
-a = 1 ∠300o = + 0.5 - j0.866
a2 = 1 ∠240o = - 0.5 - j0.866
-a2 = 1 ∠60o = + 0.5 + j0.866
a3 = 1 ∠360o = + 1.0 + j0 = 1.0
Chapter 6 Page 7 Symmetrical Components
-a3 = 1 ∠180o = - 1.0 + j0 = - 1.0
Figure 6-7. The “a” Operator
Chapter 6 Page 8 Symmetrical Components
FAULT CONDITION PHASOR DIAGRAMS Sequence Currents
Figure 6-8 shows and the following characteristics apply to the current sequence component sets for three-phase faults, line-to-line faults, and line-to-line-to ground faults.
o No negative or zero sequence currents flow for three-phase faults; only positive sequence currents flow.
o Only positive and negative sequence currents flow for line-to-line faults.
o Positive, negative, and zero sequence currents flow for faults involving ground.
Sequence Voltages
Figure 6-9 shows and the following characteristics apply to the voltage sequence component sets for three-phase faults and line-to-line faults.
o No negative or zero sequence voltages exist for a three-phase fault and the positive sequence voltage collapses to zero at the point of the fault.
o No zero sequence voltages exist for line-to-line faults.
o Positive, negative, and zero sequence voltages exist for faults involving ground.
Chapter 6 Page 9 Symmetrical Components
Sequence Currents
Figure 6-8. Sequence Current Components
Chapter 6 Page 10 Symmetrical Components
Sequence Voltages
Figure 6-9. Sequence Voltage Components
Chapter 6 Page 11 Symmetrical Components
SEQUENCE IMPEDANCE FOR TRANSFORMERS The zero sequence equivalent circuits of three-phase transformers deserve special attention because of the different combinations of connections (e.g., delta-wye, wye-delta, etc.). Figure 6-10 shows the various transformer connection combinations and the corresponding zero sequence current flow equivalent diagram.
Referring to Figure 6-10, the following observations are noted:
o If either one of the neutrals of a wye-wye (Y-Y) transformer bank is ungrounded, zero sequence current (I0) cannot flow in either winding.
o Where both neutrals of a wye-wye (Y-Y) transformer bank are grounded, zero sequence current (I0) flows in both windings.
o In delta-wye (Δ-Y) or wye-delta (Y-Δ) transformer grounded banks, zero sequence currents (I0) have a path only through the wye (Y) winding.
o No zero sequence currents (I0) flow in a delta-delta (Δ-Δ) transformer bank.
o If the connection from neutral to ground contains an impedance (ZN), the zero sequence equivalent circuit model must be modeled as an impedance of 3ZN.
Chapter 6 Page 12 Symmetrical Components
Zero Sequence Transformer Models
Figure 6-10. Transformer Zero Sequence Model
Chapter 6 Page 13 Symmetrical Components
SEQUENCE IMPEDANCE FOR CABLES Positive Sequence Impedance
Usually given in tables
Negative Sequence Impedance
Z2 = Z1
Zero Sequence Impedance
Z0 > Z1
SEQUENCE IMPEDANCE FOR MOTORS Positive Sequence Impedance
X1 = Xd’’
Negative Sequence Impedance
X2 = X1
Zero Sequence Impedance
Since motors are ungrounded, they have no zero sequence impedance.
Chapter 6 Page 14 Symmetrical Components
SEQUENCE IMPEDANCE FOR GENERATORS Positive Sequence Impedance
X1 = Xd’’
Negative Sequence Impedance
X2 is usually 20% higher than X1
Zero Sequence Impedance
X0 is usually much smaller than X1
EQUIVALENT CIRCUIT FOR LINE TO GROUND FAULTS
Ilg = 3E
Z1 + Z2 + Z0 + 3Zn
Chapter 6 Page 15 Symmetrical Components
PROBLEM 4
SCC = 1200 MVA
10 MVA 6 %
13.8 kV
Cable (0.1 Ohms)
2.5 MVA 5.5 %
6 Ohms
4.16 kV
200 Ground Fault HP
Calculate the line to ground fault current in the 4.16kV system using the following methods: A. Approximate method (only grounding resistor)
B. Ignoring motor contribution.
C. Including motor contribution. Note: For the utility, transformers, cable, and motor: Assume that the total impedance is reactive.
Chapter 6 Page 16 Symmetrical Components
PROBLEM 6 SOLUTION Utility Contribution SCC = 1200 MVA Z(p.u. on 100 MVA base) = 100 SCC (MVA) Z(p.u. on 100 MVA base) = 100 1200 = .0833 P.U. 10 MVA Transformer (Branch 100) Z% = 6.0 % Z(p.u. new) = Z(p.u. old) * MVA (base-new) MVA (base-old)
Z(p.u. new) = 6.0 * MVA (base-new) 100 MVA (base-old)
Z(p.u. new) = 0.060 * 100 10
Z(p.u. new) = 0.6000 P.U.
Chapter 6 Page 17 Symmetrical Components
2.5 MVA Transformer Z% = 5.50 % Z(p.u. new) = Z(p.u. old) * MVA (base-new) MVA (base-old)
Z(p.u. new) = 5.50 * MVA (base-new) 100 MVA (base-old)
Z(p.u. new) = 0.0550 * 100 2.50
Z(p.u. new) = 2.200 P.U.
0.1 ohm Cable Z (branch) = X(branch) = 0.1 ohms X (p.u.) = X (ohms) = 0.10 = 0.0525 P.U. Z (base) 1.90440
6 ohm grounding resistor Z (branch) = R(branch) = 6.0 ohms R (p.u.) = R (ohms) = 6.0 = 34.671 P.U. R (base) 17.3056
Chapter 6 Page 18 Symmetrical Components
2000 HP Motor X(d’’) = 0.17 (assumed) X(p.u.) = X(d’’ old) * MVA (base-new) MVA (base-old)
X(p.u.) = 0.17 * MVA (base-new) MVA (base-old)
X(p.u.) = 0.17 * 100 2.00
X(p.u.) = 8.5 P.U. Ibase = 100,000 = 13,879.019 amps √3 * 4.16
1) Using Approximate Method Ilg = (4160/ √3) = 2400 6 6 = 400 amps
Chapter 6 Page 19 Symmetrical Components
Chapter 6 Page 20 Symmetrical Components
2) Ignoring Motor Contribution
Z1 = Z2 = j (.0833 + 0.6 + 0.05250 + 2.2) = j (2.9358) Z0 + 3 Zn = j (2.2) + 3 (34.671) Ilg = 3E Z1 + Z2 + Z0 + 3Zn
= 3E j (2.9358) + j (2.9358) + j (2.2) + 3 (34.671)
= 3 j(8.0716) + 104.13
= 3 = .028756 P.U. 104.3257 Ifault = IP.U. * Ibase
Utility
10 MVA
TXMR
Cable
2.5 MVA
TXMR
j 0.0833
j 0.6
j 0.0525
j 2.2
Positive & Negative Sequence
Per Unit Impedances
2.5 MVA
TXMRj 2.2
3 * (34.671)
Grounding
Zero Sequence
Per Unit Impedances
= .028756 *13,879.019 = 399.105 amps (for all practical purposes, it is equal to 400 amps)
Chapter 6 Page 21 Symmetrical Components
3) If Motor Contribution was to be Added
j 0.0833
XP.U. = (2.9358) * (8.5) (2.9358) + (8.5) = 2.18212 P.U. Ilg = 3E Z1 + Z2 + Z0 + 3Zn
= 3E j (2.18212) + j (2.18212) + j (2.2) + 3 (34.671) = 3 j(6.56424) + 104.13
= 3 = .02875 P.U. 104.2199 Ifault = IP.U. * Ibase
Utility
10 MVA
Cable
2.5 MVA
j 0.6
j 0.0525
j 2.2
Positive & Negative Sequence
2.5 MVA j 2.2
3 * (34.671)
Groundin
Zero Sequence
200 HP
j
j 8.5
Chapter 6 Page 22 Symmetrical Components
Chapter 6 Page 23 Symmetrical Components
= .02875 *13,879.019 = 399.5114 amps which is basically 400 amps (not worth all the effort)
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