Power System Protective Relaying-part Two 4i645s

Power System Protective Relaying-Part Two Wei-Jen Lee, Ph.D., PE Professor of Electrical Engineering Dept. The Univ. of Texas at Arlington

Definition •

Actual Quantity Quantity in per unit = Base Value of Quantity

•

Quantity in percent = (Quantity in per unit)*100

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 threephase 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 IL =

S 3φ 3V LL

o

General Relations Between Circuit Quantities

General Relations Between Circuit Quantities Y − Connection V LN V LL ∠ − 30 o 3V LL VLL2 ∠ − 30 o ZY = = * = IL S 3φ S 3φ 3 ∆ − Connection ID =

I L ∠30 o 3

o VLL S 3φ ∠30 = = ZD 3VLL

2 o V LL 3VLL ∠ − 30 o 3 V 3 V ∠ − 30 LL ZD = = = 3V LL ∠ − 30 o * = LL ID IL S 3φ 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 KVbase ( old ) 2 KVbase ( 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 base ( old )

100 KV = 0.1 * * = 0.2 2 50 KVbase ( new)

Example Two: Base Conversion •

A generator and transformer, as shown below, are to be combined into a single equivalent reactance 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 100 42 Z gen ( new) = 0.25 * * = 1.15 2 25 3.73 2 100 3.9 Z Xfer ( new) = 0.1 * * = 0.364 2 30 3.73 Z eq = Z gen ( new) + 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 (Ipq) 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 Ipq deviates from the maximum torque line.

Relay Polarity •

For ground fault protection, the 60o 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

Ib, Vba

Ic, Vcb

I lags 30o

60o Delta

Fig. 3.7C

Ia-Ib, Vac

Ib-Ic, Vba

Ic-Ia, Vcb

I lags 60o

60o Wye

Fig. 3.7C

Ia, -Vc

Ib, -Va

Ic, -Vb

I lags 60o

90o-45o

Fig. 3.7A

Ia, Vbc

Ib, Vca

Ic, Vab

I lags 45o

Ia, Vbc

Ib, Vca

Ic, Vab

I lags 60o

(max. torque: 45o)

90o-60o

Fig. 3.7A

The 90 -60 Connection for Phase-Fault Protection o

o

The 90 -60 Connection for Phase-Fault Protection o

o

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 I1   I   2 2 a  

 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 Z1 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  2  a a   I b   I   c  2 a a 

 1 I a    I  = 1  b   I c    1 

 1 1 I 0    2 a a   I1  I   2 2 a a  

 Z  0 V0  0   V  = V  − 0  1    V2  0   0  

0 Z1 0

 0  I 0  0   I 1     I 2  Z2   

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, Ib, and Ic are balanced, only I1 appears in the circuit. If there is fault impedance ZF among phases, Z1 should be changed to Z1 + ZF. V-( Z1 + ZF) I1=0

I 1 = I aF

V = Z1

or

I 1 = I aF

V = Z1 + 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).  

Ib = Ic = 0

1  I 0   I  = 1 1  1 3   I 2   1 

1 I a  I a   1 a a 2  0  =  I a   0  3  I   a   2 a a 

1

Va = Z F * 3I 0

Fault Studies for Relay Settings and Coordination •

Sequence Interconnections for Single Phase-to  Ground Faults   Z  0 V0  0   V  = V  − 0  1    V2  0   0  

0

Z1 0

0

   I 1  − Z 0 I 1  0   I 1  = V − Z 1 I 1     I 1  − Z 2 I 1  Z2   

Va = V0 + V1 + V2 = V − ( Z 0 + Z 1 + Z 2 ) I 1 = 3Z F I 1

I1 = I 2 = I 0 =

V Z 1 + Z 2 + Z 0 + (3Z F )

I aF = I 1 + I 2 + I 0 = 3I 1

Fault Studies for Relay Settings and Coordination •

Sequence Interconnections for Phase-to-Phase Faults - It is convenient to show the fault between 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, Vb − Vc = Z F I b , and  1  I 0   I  = 1 1  1 3   I 2   1 

I b = −I c

 1 1   0  0      1 2  2 a a  I b  = aI b − a I b    3 2   a I b − aI b   − I b  2   a a 

Fault Studies for Relay Settings and Coordination •

Sequence Interconnections for Phase-to-Phase   Faults  

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

Z  0 V0  0   V  = V  − 0  1    V2  0   0  

 1 1  V0    2 a a  V1   V   2 2 a a  

0

Z1 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 Vb − Vc = (a 2 − a )V1 + (a − a 2 )V2 = (a 2 − a )(V1 − V2 ) = (a 2 − a)(V − ( Z 1 + Z 2 ) I 1 ) 3I 1 Z F = IbZ F = a − a2 3I 1 Z F (V − ( Z 1 + Z 2 ) I 1 ) = I b Z F = = I1 Z F 2 2 (a − a )(a − a )

Fault Studies for Relay Settings and Coordination •

Sequence Interconnections for Phase-to-Phase Faults I1 = − I 2 =

V Z1 + Z 2 + (Z F )

I aF = I 1 − I 2 = 0 I bF = a 2 I 1 + aI 2 = − j 3I 1 I cF = aI 1 + a 2 I 2 = j 3I 1 •

Assume Z1 = Z2, then I1 = V/2Z1. Just considering the magnitude, 3V V I bF = I cF =

2Z1

= 0.866

Z1

= 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 Vb = Vc = Z F ( I b + I c ) I a = I 0 + I1 + I 2 = 0 Vb = V0 + a 2V1 + aV2 Vc = V0 + aV1 + a 2V2 V1 = V2 Vb = ( I b + I c ) Z F = 3 I 0 Z F

 Z  0 V0  0   V  = V  − 0  1    V2  0   0   V0 = − Z 0 I 0 V1 = V − Z 1 I 1 V2 = − Z 2 I 2

0 Z1 0

 0  I 0  0   I 1     I 2  Z2   

Fault Studies for Relay Settings and Coordination •

Sequence Interconnections for Double Phase-toGround Faults Vb = 3I 0 Z F = V0 + (a 2 + a )V1 = V0 − V1 3I 0 Z F = V0 − V1 = − Z 0 I 0 − V + Z 1 I 1 I0 = −

V − Z1 I1 Z 0 + 3Z F

V − Z1 I1 I2 = − Z2 I 1 = −( I 2 + I 0 )

Fault Studies for Relay Settings and Coordination •

Sequence Interconnections for Double Phase-toGround Faults (Solid faults) V I1 = Z Z Z1 + 2 0 Z2 + Z0 Z0 I 2 = − I1 Z2 + Z0 Z2 I 0 = − I1 Z2 + Z0

Fault Studies for Relay Settings and Coordination •

Sequence Interconnections for Double Phase-toGround Faults (Line-to-line fault impedance: ZF and phase-to-ground fault impedance: ZFG) V I1 = ( Z 2 + ( Z F / 2))( Z 0 + ( Z F / 2) + 3Z FG ) ZF (Z1 + )+ 2 Z 2 + Z 0 + Z F + 3Z FG ( Z 0 + ( Z F / 2) + 3Z FG ) I 2 = − I1 Z 2 + Z 0 + Z F + 3Z FG ( Z 2 + ( Z F / 2)) I 0 = − I1 Z 2 + Z 0 + Z F + 3Z 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 = X 2G " d

100 = 0.16 * = 0.2 pu 80

- VS: No change - Two winding Transformer: X TG

100 = 0.11 * = 0.1375 80

Example: Fault Calculations on a Sample System •

Step One: Transfer all the constants to a common base 100 X HM = 0.055 *

- Three winding transformer:

X HL X ML

= 0.03667 pu

150 100 = 0.360 * = 0.2400 pu 150 100 = 0.280 * = 0.18667 pu 150

1 X H = (0.03667 + 0.2400 − 0.18667) = 0.0450 pu 2 1 X M = (0.03667 − 0.2400 + 0.18667) = −0.00833 pu 2 1 X L = (−0.03667 + 0.2400 + 0.18667) = 0.1950 pu 2

Example: Fault Calculations on a Sample System •

Step One: Transfer all the constants to a common base 115 2 * 10 6 Z base = = 132.25Ω 6 100 * 10 - Line Impedance: 82 X0 = = 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 side impedance (j0.18147+j0.03667+j0.03=j0.2481) is parallel with left side impedance (j0.20+j0.1375=j0.3375). j 0.3375 * j 0.2482 Z1 = Z 2 = = j 0.1430 j 0.3375 + j 0.2481

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 =

•

1 100,000 = j 6.993 pu = j 6.993 = 3510.8 A at 115kV j 0.143 3 * 115

The division of current from the left (IaG) 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 1 .0 = j 2.5 pu j (0.143 + 0.143 + 0.1141) = 3I 1 = j 7.5 pu = 3764.4 A at 115kV

I1 = I 2 = I 0 = I aF

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 that a singlephase-to-ground fault occurs at the H or 34.5 kV terminal

Example: Fault Calculations for Autotransformers

Example: Fault Calculations for Autotransformers 100 X HM = 0.08 * = 0.05333 pu 150 100 X HL = 0.34 * = 0.68 pu 50 100 X ML = 0.216 * = 0.54 pu 40

1 (0.0533 + 0.68 − 0.54) = 0.09667 pu 2 1 = (0.0533 + 0.54 − 0.68) = −0.04334 pu 2 1 = (0.68 + 0.54 − 0.0533) = 0.58334 pu 2

XH = XM XH

Example: Fault Calculations for Autotransformers •

The sequence networks are shown below:

Example: Fault Calculations for Autotransformers •

When the fault happens at Bus H, the equivalent sequence impedance for the networks are: • Positive & negative sequence network X1 = X 2 =

(0.057 − 0.0433 + 0.0967) * (0.08) = 0.04637 pu (0.057 − 0.0433 + 0.0967 + 0.08)

• Zero sequence network (0.032 − 0.0433) * 0.583 + 0.0967 = 0.085177 pu (0.032 − 0.0433 + 0.583) 0.085177 * 0.28 X0 = = 0.06527 pu 0.085177 + 0.28 X left =

Example: Fault Calculations for Autotransformers •

Single-phase-to-ground fault at H: I 0 = I1 = I 2 =

1.0 = 6.3287 pu (0.04637 + 0.04637 + 0.06527)

I 0 = I 1 = I 2 = 6.3287 * I aF

100 * 10 6 3

= 1059.1A at 345kV

3 * 345 * 10 = 3I 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.