New Chapter 6 Symmetrical Components

CHAPTER 6

SYMMETRICAL COMPONENTS

Chapter 6 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


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


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


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


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


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


-a3 = 1 ∠180o = - 1.0 + j0 = - 1.0

Figure 6-7. The “a” Operator


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.


Sequence Currents

Figure 6-8. Sequence Current Components


Sequence Voltages

Figure 6-9. Sequence Voltage 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.


Zero Sequence Transformer Models

Figure 6-10. Transformer Zero Sequence Model


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’’


X2 = X1


Since motors are ungrounded, they have no zero sequence impedance.


SEQUENCE IMPEDANCE FOR GENERATORS Positive Sequence Impedance

X1 = Xd’’


X2 is usually 20% higher than X1


X0 is usually much smaller than X1

EQUIVALENT CIRCUIT FOR LINE TO GROUND FAULTS

Ilg = 3E

Z1 + Z2 + Z0 + 3Zn


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.


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.


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


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



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)


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



= .02875 *13,879.019 = 399.5114 amps which is basically 400 amps (not worth all the effort)

New Chapter 6 Symmetrical Components

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