Power System Protective Relaying-Part Two
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Power System Protective Relaying-Pa Relay ing-Part rt Two Two Wei-Jen Lee, Ph.D., PE Professor of Electrical Engineering Dept. The Univ Un iv.. of Texas at Arlington
Definition Actual Quantity •
Quantity in per unit =
•
Quantity in percent = (Quantity in per unit)*100
Base Value of Quantity
Advantages • •
•
•
More meaningful when comparing different voltage levels The per unit equivalent impedance of the transformer remains the same when referred to either the primary or the secondary side The per unit impedance of a transformer in a three phase system is the same, regardless the winding connection The per unit method is independent of voltage changes and phase shifts through transformers
Advantages •
Manufacturers usually specify the impedance of the equipment in per unit or percent on the base of its nameplate ratings
•
The per unit impedance values of various ratings of equipment lie in a narrow range
General Relations Between Circuit Quantities S 3φ = 3V LL I L V LL = 3V LN ∠30 I L =
S 3φ 3V LL
o
General Relations Between Circuit Quantities
General Relations Between Circuit Quantities Y − Connection V LN
Z Y =
I L
=
V LL ∠ − 30 o 3
*
3V LL
S 3φ
=
V LL2 ∠ − 30 o S 3φ
∆ − Connection I D =
I L ∠30
Z D =
3
V LL I D
=
o
=
V LL Z D
=
S 3φ ∠30 o 3V LL
3V LL ∠ − 30 o
I L
= 3V LL ∠ − 30 o *
3V LL
S 3φ
=
3V LL2 ∠ − 30 o
S 3φ
Base Quantity Selections •
VA, V, I, and Z are four power quantities
•
One has to select two base quantities and derive the other two
Base Conversion Z pu ( new)
= Z pu ( old )
*
MVAbase ( new) MVAbase ( old )
*
2 base ( old )
KV
2 KV base ( new )
Example One: Base Conversion •
A 50-MVA, 34.5:161 kV transformer with 10% reactance is connected to a power system where all the other impedance values are on a 100 MVA, 34.5 or 161 kV base. The reactance of the transformer under new base is:
Z pu ( new)
2 KV 100 base ( old ) = 0.1 * * 2 50 KV base ( new )
=
0.2
Example Two: Base Conversion •
A generator and transformer, as shown s hown below, are to be combined into a single equivalent reactance r eactance on a 100 MVA, 110 kV (high voltage side) base.
Example Two: Base Conversion • • •
The transformer is operated at 3.9 kV tap. New base voltage at high side is 110 kV. The base voltage at low side is: 110*3.9/115 = 3.73 kV Z gen ( new) = 0.25 * Z Xfer ( new) = 0.1 * Z eq = Z gen ( new)
100 25
100
*
*
42 3.73 3.9
2 2
2
= 1.15 = 0.364
30 3.73 + Z Xfer ( new) = 1.15 + 0.364 = 1.514
Transformer Polarity
Transformer Polarity •
The ANSI/IEEE standard for transformers states that the high voltage should lead the low voltage by 30o with Y-∆ or ∆ -Y banks.
Relay Polarity •
Relays that sense the direction of current (or power) flow at a specific location and, thereby, indicate the direction of the fault, provide a good example of relay polarity.
•
A directional-sensing unit requires a reference quantity that is reasonably constant against which the current in the protected circuit can be compared.
Relay Polarity •
Definition of maximum torque line and zero torque line
•
Solid state units can have adjustments for (1) the maximum torque angle and (2) the angle limits of the operating zone
Relay Polarity
Relay Polarity •
•
In Fig. (A), the maximum operating torque or energy occurs when the current flow from polarity to non-polarity (I pq) leads by 30o the voltage drop from polarity to non-polarity (Vrs). The minimum pick up of the directional unit is specified at the maximum torque. Higher current will be required when I pq deviates from the maximum torque line.
Relay Polarity •
For ground fault protection, the 60 o unit of Fig. (B) is used with a 3Vo reference and the zero unit of Fig. (C) with a 3Io current reference.
•
The Fig. (C) is also used for power or var applications.
Relay Polarity Connection
Unit Type
Phase A
Phase B
Phase C
Maximum torque occurs when
30o
Fig. 3.7C
Ia, Vac
I b, V ba
Ic, Vcb
I lags 30o
60o Delta
Fig. 3.7C
Ia-I b, Vac
I b-Ic, V ba
Ic-Ia, Vcb
I lags 60o
60o Wye
Fig. 3.7C
Ia, -Vc
I b, -Va
Ic, -V b
I lags 60o
90o-45o
Fig. 3.7A
Ia, V bc
I b, Vca
Ic, Vab
I lags 45o
Ia, V bc
I b, Vca
Ic, Vab
I lags 60o
(max. torque: 45o)
90o-60o
Fig. 3.7A
o
o
The 90 -60 Connection for Phase-Fault Protection
o
o
The 90 -60 Connection for Phase-Fault Protection
Directional Sensing for Ground Faults: Voltage Polarization
Directional Sensing for Ground Faults: Voltage Polarization
Directional Sensing for Ground Faults: Current Polarization
Symmetrical Components 1 I a I = 1 b I c 1
1 a2 a
1 I 0 a I 1 I 2 a2
1 I 0 I = 1 1 1 3 I 2 1
1 a a2
1 I a 2 a I b I c a
Zero Sequence Current and Voltage for Ground Fault Protection
Sequence Networks •
Single-Line Diagram
Sequence Networks •
Positive Sequence Network
Sequence Networks •
Negative Sequence Network
Sequence Networks •
Zero Sequence Network
Sequence Network Reduction •
Consider faults at bus H for the positive sequence network of the sample system. The Z 1 is equal to the parallel of (Xd”+XTG+X1GH) and (Z1S+XHM)
Fault Studies for Relay Settings and Coordination • • • •
Three-phase fault studies for applying and setting phase relays Single-phase-to-ground fault studies for applying and setting ground protection relays Fault Impedance: Faults are seldom solid, but involve varying amount of resistance. It is generally ignored in protective relaying and faults studies for high voltage transmission or subtransmission system.
Fault Studies for Relay Settings and Coordination •
•
In distribution systems, very large or basically infinite impedance can exist. High impedance fault detection relay may be necessary for distribution system. For arcing fault, the arc resistance varies a lot. However, a commonly accepted value for currents between 70 and 20,000 A has been an arc drop of 440 V per foot, essentially independent of current magnitude. Therefore, 440l Z arc =
I
Ω
Fault Studies for Relay Settings and Coordination •
•
In low voltage (480 V) switchboard-type enclosures, typical arc voltages of about 150 V can be experienced. Substation and Tower-Footing Impedance is another highly variable factor. Several technical papers have been written and computer programs have been developed in this area with many variables and assumptions. The general practice is to neglect these in most fault studies and relay applications and settings.
Fault Studies for Relay Settings and Coordination 1 I 0 I = 1 1 1 3 I 2 1
1 1 I a a a 2 I b I c 2 a a
1 I a I = 1 b I c 1
1 1 I 0 a 2 a I 1 I 2 2 a a
Z 0 V 0 0 V = V − 0 1 V 2 0 0
0 0 I 0 Z 1 0 I 1 I 2 0 Z 2
Fault Studies for Relay Settings and Coordination •
Sequence Interconnections for Three-Phase Faults
Fault Studies for Relay Settings and Coordination •
Sequence Interconnections for Three-Phase Faults – Three-Phase faults are assumed to be symmetrical. – The positive-sequence network can be used to calculate the fault current. – Since Ia, I b, and Ic are balanced, only I1 appears in the circuit. If there is fault impedance Z F among phases, Z1 should be changed to Z 1 + ZF. V-( Z1 + ZF) I1=0
I 1 = I aF =
V Z 1
or
I 1 = I aF =
V Z 1 + Z F
Fault Studies for Relay Settings and Coordination •
Sequence Interconnections for Single Phase-toGround Faults
Fault Studies for Relay Settings and Coordination •
Sequence Interconnections for Single Phase-toGround Faults – A phase-a-to-ground fault is represented by connecting the three sequence networks together (either with or without fault impedance).
I b = I c = 0
1 I 0 I = 1 1 1 3 I 2 1
1
1
I a I a 1 a a 2 0 = I a 0 3 I a 2 a a
V a = Z F * 3I 0
Fault Studies for Relay Settings and Coordination •
Sequence Interconnections for Single Phase-to Ground Faults Z
0 V 0 0 V = V − 0 1 V 2 0 0
0
Z 1 0
0
I 1 − Z 0 I 1 0 I 1 = V − Z 1 I 1 I 1 − Z 2 I 1 Z 2
V a = V 0 + V 1 + V 2 = V − ( Z 0 + Z 1 + Z 2 ) I 1 = 3 Z F I 1
I 1 = I 2 = I 0 =
V Z 1 + Z 2 + Z 0 + (3 Z F )
I aF = I 1 + I 2 + I 0 = 3 I 1
Fault Studies for Relay Settings and Coordination •
Sequence Interconnections for Phase-to-Phase Faults - It is con conven venien ientt to show show the fault fault bet betwee ween n phases phases b and c with fault impedance of ZF.
Fault Studies for Relay Settings and Coordination •
Sequence Interconnections for Phase-to-Phase Faults I a = 0, V b − V c = Z F I b , and I b = − I c 1 I 0 I = 1 1 1 3 I 2 1
1 1 0 0 1 a a 2 I b = aI b − a 2 I b 3 2 a I b − aI b − I a 2 a b
Fault Studies for Relay Settings and Coordination •
Sequence Interconnections for Phase-to-Phase Faults
1 V a V = 1 b V c 1
Z 0 V 0 0 V = V − 0 1 V 2 0 0
1 1 V 0 2 a a V 1 V 2 2 a a
0
Z 1 0
0
0 0 0 I 1 = V − Z 1 I 1 − I Z 2 I 1 Z 2 1
Fault Studies for Relay Settings and Coordination •
Sequence Interconnections for Phase-to-Phase Faults V b − V c = (a 2 − a )V 1 + (a − a 2 )V 2
= (a 2 − a )(V 1 − V 2 ) = (a 2 − a)(V − ( Z 1 + Z 2 ) I 1 ) = I b Z F =
3 I 1 Z F
a − a2
(V − ( Z 1 + Z 2 ) I 1 ) = I b Z F =
3 I 1 Z F (a − a )(a − a ) 2
2
= I 1 Z F
Fault Studies for Relay Settings and Coordination •
Sequence Interconnections for Phase-to-Phase Faults I 1 = − I 2 =
V Z 1 + Z 2 + ( Z F )
I aF = I 1 − I 2 = 0 I bF = a 2 I 1 + aI 2 = − j 3 I 1 I cF = aI 1 + a 2 I 2 = j 3 I 1 •
Assume Z1 = Z2, then I1 = V/2Z1. Just considering the magnitude, V 3V I bF = I cF =
2 Z 1
= 0.866
Z 1
= 0.866 I 3φ
Fault Studies for Relay Settings and Coordination •
Sequence Interconnections for Double Phase-toGround Faults • The connection for this type of fault is similar to the phase-to-phase fault with the addition of the zero sequence impedance in parallel with the negative sequence impedance.
Fault Studies for Relay Settings and Coordination •
Sequence Interconnections for Double Phase-toGround Faults V b = V c = Z F ( I b + I c ) I a = I 0 + I 1 + I 2 = 0 V b = V 0 + a 2V 1 + aV 2 V c = V 0 + aV 1 + a 2V 2 V 1 = V 2 V b = ( I b + I c ) Z F = 3 I 0 Z F
Z 0 V 0 0 V = V − 0 1 V 2 0 0 V 0 = − Z 0 I 0 V 1 = V − Z 1 I 1 V 2 = − Z 2 I 2
0
Z 1 0
0 I 0 0 I 1 I 2 Z 2
Fault Studies for Relay Settings and Coordination •
Sequence Interconnections for Double Phase-toGround Faults V b = 3 I 0 Z F = V 0 + (a 2 + a )V 1 = V 0 − V 1 3 I 0 Z F = V 0 − V 1 = − Z 0 I 0 − V + Z 1 I 1
I 0 = − I 2 = −
V − Z 1 I 1 Z 0 + 3 Z F V − Z 1 I 1 Z 2
I 1 = −( I 2 + I 0 )
Fault Studies for Relay Settings and Coordination •
Sequence Interconnections for Double Phase-toGround Faults (Solid faults) I 1 = Z 1 + I 2 = − I 1 I 0 = − I 1
V Z 2 Z 0 Z 2 + Z 0 Z 0
Z 2 + Z 0 Z 2 Z 2 + Z 0
Fault Studies for Relay Settings and Coordination •
Sequence Interconnections for Double Phase-toGround Faults (Line-to-line fault impedance: Z F and phase-to-ground fault impedance: Z FG) I 1 = ( Z 1 +
I 2 = − I 1 I 0 = − I 1
Z F 2
)+
V ( Z 2 + ( Z F / 2))( Z 0 + ( Z F / 2) + 3 Z FG ) Z 2 + Z 0 + Z F + 3 Z FG
( Z 0 + ( Z F / 2) + 3 Z FG )
Z 2 + Z 0 + Z F + 3 Z FG ( Z 2 + ( Z F / 2))
Z 2 + Z 0 + Z F + 3 Z FG
Example: Fault Calculations on a Sample System
Example: Fault Calculations on a Sample System •
Step One: Transfer all the constants to a common base - VG: X d " = X 2G = 0.16 *
100 80
= 0.2 pu
- VS: No change
100
- Tw Two o win windi ding ng Tr Tran ansf sfor orme mer: r: X TG = 0.11 * 80 = 0.1375
Example: Fault Calculations on a Sample System •
Step One: Transfer all the constants to a common base 100 - Th Thre reee win windi ding ng tra trans nsfo form rmer er::
X HM = 0.055 *
= 0.03667 pu
X HL = 0.360 *
= 0.2400 pu
X ML = 0.280 *
X H = X M =
1 2 1
(0.03667 + 0.2400 − 0.18667) = 0.0450 pu (0.03667 − 0.2400 + 0.18667) = −0.00833 pu
2 1 X = (−0 03667 + 0 2400 + 0 18667) = 0 1950 pu
150 100 150 100 150
= 0.18667 pu
Example: Fault Calculations on a Sample System •
Step One: Transfer all the constants to a common base Z base =
115 2 * 10 6 6
= 132.25Ω
100 *10 - Line Im Impedance: 82 X 0 = = 0.6200 pu 132.25 X 1 = X 2 = 0.18147 pu
Example: Fault Calculations on a Sample System •
Develop sequence network for different fault conditions (Fault at point “G”). Positive & negative sequence network
Example: Fault Calculations on a Sample System •
Develop sequence network for different fault conditions (Fault at point “G”). Zero sequence network
Example: Fault Calculations on a Sample System •
For the fault at G, the right r ight side impedance (j0.18147+j0.03667+j0.03=j0.2481) is parallel with left side impedance (j0.20+j0.1375=j0.3375). Z 1 = Z 2 =
j 0.3375 * j 0.2482 j 0.3375 + j 0.2481
= j 0.1430
Example: Fault Calculations on a Sample System •
The right side network is reduced for a fault at bus G by first paralleling X0S + ZH with ZL and then adding ZM and X0GH (The equivalent impedance is equal to j0.6709). Paralleling with the left side impedance (XTG = j0.1375), the zero sequence impedance X0 = j0.1141.
Example: Fault Calculations on a Sample System •
Three-phase fault at Bus G I 1 = I aF =
•
1
j 0.143
= j 6.993 pu = j 6.993
100,000 3 * 115
= 3510.8 A at 115kV
The division of current from the left (I aG) and the right (IaH) are: I aG = 0.4237 * 6.993 = 2.963 pu I aH = 0.5763 * 6.993 = 4.030 pu
Example: Fault Calculations on a Sample System •
Single-phase-to-ground fault at Bus G I 1 = I 2 = I 0 =
1.0 j (0.143 + 0.143 + 0.1141)
= j 2.5 pu
I aF = 3 I 1 = j 7.5 pu = 3764.4 A at 115kV
Example: Fault Calculations for Autotransformers •
Autotransformers have become very common in recent years.
•
Consider a typical autotransformer in a system, as shown in the figure, and assume assu me that a single phase-to-ground fault occurs at the H or 34.5 kV terminal
Example: Fault Calculations for Autotransformers
Example: Fault Calculations for Autotransformers X HM = 0.08 * X HL = 0.34 * X ML
100
150 100
= 0.05333 pu = 0.68 pu
50 100 = 0.216 * = 0.54 pu 40
X H = X M = X H =
1 2 1 2 1 2
(0.0533 + 0.68 − 0.54) = 0.09667 pu (0.0533 + 0.54 − 0.68) = −0.04334 pu (0.68 + 0.54 − 0.0533) = 0.58334 pu
Example: Fault Calculations for Autotransformers •
The sequence networks are shown below:
Example: Fault Calculations for Autotransformers •
When the fault happens happens at Bus H, the equivalent equivalent sequence impedance for the networks are: • Positive & negative sequence network X 1 = X 2 =
(0.057 − 0.0433 + 0.0967) * (0.08) (0.057 − 0.0433 + 0.0967 + 0.08)
= 0.04637 pu
• Zero sequence network X left = X 0 =
(0.032 − 0.0433) * 0.583 (0.032 − 0.0433 + 0.583)
0.085177 * 0.28 0.085177 + 0.28
+ 0.0967 = 0.085177 pu
= 0.06527 pu
Example: Fault Calculations for Autotransformers •
Single-phase-to-ground fault at H: I 0 = I 1 = I 2 =
1.0 (0.04637 + 0.04637 + 0.06527)
I 0 = I 1 = I 2 = 6.3287 *
100 * 10 6 3 * 345 * 10
3
= 6.3287 pu
= 1059.1 A at 345kV
I aF = 3 I 0 = 3 * 6.3287 = 18.986 pu = 3177.29 A at 345kV
Example: Fault Calculations for Autotransformers •
Fault current distribution
Example: Open-Phase Conductor •
•
A blown fuse or broken conductor that opens one of the three phases results in a serious unbalance that has to be detected and resolved as soon as possible. The sample system (Assume phase a open at “H”)
Example: Open-Phase Conductor •
The positive sequence network (X-Y indicates the fault location)
Example: Open-Phase Conductor •
The negative sequence network (X-Y indicates the fault location)
Example: Open-Phase Conductor •
The zero sequence network (X-Y indicates the fault location)
Example: Open-Phase Conductor •
It is necessary to consider the load current in this case. (This is similar to two-phase-to-ground fault calculation)
Example: Open-Phase Falling to Ground on One Side •
Same sample system as before.
•
Assume phase a conductor on the line at bus H opens and falls to ground on the H side (right side).
Example: Open-Phase Falling to Ground on One Side • •
Same sequence network Since this is a simultaneous fault (an open phase fault and a phase-to-ground fault), we can insert three ideal transformers at H to isolate the open phase fault and phase-to-ground fault. (Load current can be ignored)
Example: Open-Phase Falling to Ground on One Side
Example: Open-Phase Falling to Ground on One Side
Example: Open-Phase Falling to Ground on One Side •
The other possibility is that the open conductor falls to ground on the line side. We can use the same approach as before. However, the load must be considered in the fault current calculation.
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