KOMP Simetris.pdf

SYMMETRICAL COMPONENTS 1 & 2

Presented at the

27th Annual HANDS ON RELAY SCHOOL

March 15 - 19, 2010

Washington State University Pullman, Washington

Stephen E. Marx, P.E. Bonneville Power Administration Malin, Oregon

1

Introduction

The electrical power system normally operates in a balanced three-phase sinusoidal steady-state mode. However, there are certain situations that can cause unbalanced operations. The most severe of these would be a fault or short circuit. Examples may include a tree in contact with a conductor, a lightning strike, or downed power line. In 1918, Dr. C. L. Fortescue wrote a paper entitled Method of Symmetrical Coordinates Applied to the Solution of Polyphase Networks. In the paper Dr. Fortescue described how arbitrary unbalanced 3-phase voltages (or currents) could be transformed into 3 sets of balanced 3-phase components, Fig I.1. He called these components symmetrical components. In the paper it is shown that unbalanced problems can be solved by the resolution of the currents and voltages into certain symmetrical relations.

C

B

C

B

Fig I.1

By the method of symmetrical coordinates, a set of unbalanced voltages (or currents) may be resolved into systems of balanced voltages (or currents) equal in number to the number of phases involved. The symmetrical component method reduces the complexity in solving for electrical quantities during power system disturbances. These sequence components are known as positive, negative and zero sequence components, Fig I.2

Fig I.2

2

The purpose of this paper is to explain symmetrical components and review complex algebra in order to manipulate the components. Knowledge of symmetrical components is important in performing mathematical calculations and understanding system faults. It is also valuable in analyzing faults and how they apply to relay operations. 1. Complex Numbers The method of symmetrical components uses the commonly used mathematical solutions applied in ordinary alternating current problems. A working knowledge of the fundamentals of algebra of complex numbers is essential. Consequently this subject will be reviewed first. Any complex number, such as jba + , may be represented by a single point p, plotted on a Cartesian coordinates, in which a is the abscissa on the x axis of real quantities and b the ordinate on the y axis of imaginary quantities. This is illustrated in Fig. 1.1

Fig. 1.1 Referring to Fig. 1.1, let r represent the length of the line connecting the point p to the origin and the angle measured from the x-axis to the line r. It can be observed that

cos= ra (1.1) sin= rb (1.2)

2. Properties of Phasors A vector is a mathematical quantity that has both a magnitude and direction. Many quantities in the power industry are vector quantities. The term phasor is used within the steady state alternating linear system. It is used to avoid confusion with spatial vectors: the angular position of the phasor represents position in time, not space. In this document, phasors will be used to document various ac voltages, currents and impedances. A phasor quantity or phasor, provides information about not only the magnitude but also the direction or angle of the quantity. When using a compass and giving directions to a house, from a given location, a distance and direction must be provided. For example one could say that a house is 10 miles at an angle of 75 degrees (rotated in a clockwise direction from North) from where I am standing. Just as we dont say the other house is -10 miles away, the magnitude of

3

the phasor is always a positive, or rather the absolute value of the length of the phasor. Therefore giving directions in the opposite direction, one could say that a house is 10 miles at an angle of 255 degrees. The quantity could be a potential, current, watts, etc. Phasors are written in polar form as

= YY (2.1) sincos YjY += (2.2)

where Y is the phasor, Y is the amplitude, magnitude or absolute value and is the phase angle or argument. Polar numbers are written with the magnitude followed by the symbol to indicate angle, followed by the phase angle expressed in degrees. For example oZ 90110= . This would be read as 110 at an angle of 90 degrees. The rectangular form is easily produced by applying Eq. (2.2) The phasor can be represented graphically as we have demonstrated in Fig. 1.1, with the real components coinciding with the x axis. When multiplying two phasors it is best to have the phasor written in the polar form. The magnitudes are multiplied together and the phase angles are added together. Division, which is the inverse of multiplication, can be accomplished in a similar manner. In division the magnitudes are divided and the phase angle in the denominator is subtracted from the phase angle in the numerator. Example 2.1

Multiply BA where oA 355= and oB 453= . Solution ( ) ( )ooooBA 453535453355 +==

o8015= Example 2.2

Solve DC where oC 3515= and oD 503= .

Solution

( )ooooDC 5035315503 3515 == o155 =

4

3. The j and a operator

Recall the operator j. In polar form, oj 901= . Multiplying by j has the effect of rotating a phasor o90 without affecting the magnitude.

Table 3.1 - Properties of the vector j

0.00.11 j+= oj 901=

118012 == oj

jj o == 27013 oj 901 =

Example 3.1

Compute jA where oA 6010= . Solution

)6010(901 oojA = o15010=

Notice that multiplication by the j operator rotated the Phasor A by o90 , but did not change the magnitude. Refer to Fig. 3.1

A

(a) A

jAA

(b) Aj

Fig. 3.1. j effects In a similar manner the a operator is defined as unit vector at an angle of 120o, written as

oa 1201= . The operator a2, is also a unit vector at an angle of 240o, written oa 24012 = .

5

Example 3.2

Compute aA where oA 6010= . Solution

)6010(1201 ooaA = o18010=

A

(a) A

aA

A

(b) Aj

Fig. 3.2. a effects

Table 3.2 - Properties of the vector a

0.00.11 j+= oa 1201=

oa 24012 = ooa 0136013 ==

01 2 =++ aa 12 =+ aa

oa 3031 =+

oa 3031 2 =+ 32 jaa =

32 jaa = oa 3031 =

oa 3031 2 =

4. The three-phase System and the relationship of the 3 In a Wye connected system the voltage measured from line to line equals the square root of three, 3 , times the voltage from line to neutral. See Fig. 4.1 and Eq. (4.1). The line current equals the phase current, see Eq. (4.2)

6

Fig. 4.1

LNLL VV 3= (4.1)

= IIL (4.2) In a Delta connected system the voltage measured from line to line equals the phase voltage. See Fig. 4.2 and Eq. (4.3). The line current will equal the square root of three, 3 , times the phase current, see Eq. (4.4)

VLL

II

IL

Fig. 4.2

=VVLL (4.3) = IIL 3 (4.4)

7

The power equation, for a three phase system, is

LLLIVS 3= (4.5a) cos3 LLLIVP = (4.5b) sin3 LLLIVQ = (4.5c)

where S is the apparent power or complex power in volt-amperes (VA). P is the real power in Watts (W, kW, MW). Q is the reactive power in VARS (Vars, kVars, MVars). 5. The per-unit System In many engineering situations it is useful to scale, or normalize, dimensioned quantities. This is commonly done in power system analysis. The standard method used is referred to as the per-unit system. Historically, this was done to simplify numerical calculations that were made by hand. Although this advantage is eliminated by the calculator, other advantages remain.

Device parameters tend to fall into a relatively narrow range, making erroneous values conspicuous.

Using this method all quantities are expressed as ratios of some base value or values. The per-unit equivalent impedance of any transformer is 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 of the type of winding connections (wye-delta, delta-wye, wye-wye, or delta-delta).

The per-unit method is independent of voltage changes and phase shifts through transformers where the base voltages in the winding are proportional to the number of turns in the windings.

The per-unit system is simply a scaling method. The basic per-unit scaling equation is

valuebasevalueactualunitper

__= (5.1)

The base value always has the same units as the actual value, forcing the per-unit value to be dimensionless. The base value is always a real number, whereas the actual value may be complex. The subscript pu will indicate a per-unit value. The subscript base will indicate a base value, and no subscript will indicate an actual value such as Amperes, Ohms, or Volts. The first step in using per-unit is to select the base(s) for the system. Sbase = power base, in VA. Although in principle Sbase may be selected arbitrarily, in practice it is typically chosen to be 100 MVA.

8

Vbase = voltage base in V. Although in principle Vbase is also arbitrary, in practice Vbase is equal to the nominal line-to-line voltage. The term nominal means the value at which the system was designed to operate under normal balanced conditions. From Eq. (4.5) it follows that the base power equation for a three-phase system is:

basebasebase IVS 33 = (5.2)

Solving for current:

basebase V

SI base

33=

Because S3base can be written as kVA or MVA and voltage is usually expressed in kilo-volts, or kV, current can be written as:

ampereskV

kVAIbase

basebase 3

= (5.3) Solving for base impedance:

base

base

base

basebase S

VIVZ

2

==

ohmskVA

xkVZbase

basebase

10002= (5.4a) or

ohmsMVAkVZ

base

basebase

2

= (5.4b) Given the base values, and the actual values: IZV = , then dividing by the base we are able to calculate the pu values

pupupubasebasebase

ZIVZI

IZVV ==

9

After the base values have been selected or calculated, then the per-unit impedance values for system components can be calculated using Eq. (5.4b)

)()( 2

== Z

kVMVA

ZZZ

base

base

basepu (5.5a)

or

)(1000 2

= ZkVkVAZ

base

basepu (5.5b)

It is also a common practice to express per-unit values as percentages (i.e. 1 pu = 100%). (Transformer impedances are typically given in % at the transformer MVA rating.) The conversion is simple

100_ valuepercentunitper =

Then Eq. (5.5a) can be written as ( ) ( )

22 10100%

base

base

base

base

kVZkVA

kVZMVAZ == (5.6)

It is frequently necessary, particularly for impedance values, to convert from one (old) base to another (new) base. The conversion is accomplished by two successive application of Eq. (5.1), producing:

= new

base

oldbaseold

punewpu Z

ZZZ

Substituting for oldbaseZ and

newbaseZ and re-arranging the new impedance in per-unit equals:

2

= new

base

oldbase

oldbase

newbaseold

punewpu kV

kVkVAkVAZZ (5.7)

In most cases the turns ratio of the transformer is equivalent to the system voltages, and the equipment rated voltages are the same as the system voltages. This means that the voltage-squared ratio is unity. Then Eq. (5.9) reduces to

= old

base

newbaseold

punewpu MVA

MVAZZ (5.8)

10

Example 5.1 A system has Sbase = 100 MVA, calculate the base current for

a) Vbase = 230 kV b) Vbase = 500 kV

Solution

Using Eq. (5.3) ampereskV

kVAIbase

basebase 3

=

a) AamperesIbase 25123031001000 =

=

b) AamperesIbase 5.11550031001000 =

=

Example 5.2 A 900 MVA 525/241.5 autotransformer has a nameplate impedance of 10.14%

a) Determine the impedance in ohms, referenced to the 525 kV side. b) Determine the impedance in ohms, referenced to the 241.5 kV side

Solution First convert from % to pu.

1014.0100

% == ZZpu Arranging Eq. (5.5a) and solving for Zactual gives

base

basepu MVA

kVZZ2

)( = ; therefore

a) 9005251014.0

2

525 =kVZ = 05.31

b) 900

5.2411014.02

5.241 =kVZ = 57.6

A check can be made to see if the high-side impedance to the low-side impedance equals the turns ratio squared.

726.457.605.31 = 726.4

5.241525 2 =

11

6. Sequence Networks Refer to the basic three-phase system as shown in Fig. 6.1. There are four conductors to be considered: a, b, c and neutral n.

Network A Network B

a

b

c

nanV bnV cnV

cI

bI

nI

aI

Fig. 6.1

The phase voltages, pV , for the balanced 3 case with a phase sequence abc are

opaan VVV 0== (6.1a)

opbbn VVV 120== (6.1b)

oppccn VVVV 240120

0 =+== (6.1c) The phase-phase voltages, LLV , are written as

oLLbaab VVVV 30== (6.2a)

oLLcbbc VVVV 90== (6.2b)

oLLacca VVVV 150== (6.2c)

Equation (6.1) and (6.2) can be shown in phasor form in Fig. 6.2.

Fig. 6.2

12

There are two balanced configurations of impedance connections within a power system. For the wye case, as shown in Fig. 4.1, and with an impedance connection of Z , the current can be calculated as

== oY

P

Ya Z

VZVI 0 (6.3)

Where is between o90 and + o90 . For greater than zero degrees the load would be inductive ( aI lags aV ). For less than zero degrees the load would be capacitive ( aI leads aV ). The phase currents in the balanced three-phase case are

= opa II 0 (6.4a) = opb II 120 (6.4b) = opc II 240 (6.4c)

See Fig. 6.2. for the phasor representation of the currents. 7. Symmetrical Components Systems The electrical power system operates in a balanced three-phase sinusoidal operation. When a tree contacts a line, a lightning bolt strikes a conductor or two conductors swing into each other we call this a fault, or a fault on the line. When this occurs the system goes from a balanced condition to an unbalanced condition. In order to properly set the protective relays, it is necessary to calculate currents and voltages in the system under such unbalanced operating conditions. In Dr. C. L. Fortescues paper he described how symmetrical components can transform an unbalanced condition into symmetrical components, compute the system response by straight forward circuit analysis on simple circuit models, and transform the results back into original phase variables. When a short circuit fault occurs the result can be a set of unbalanced voltages and currents. The theory of symmetrical components resolves any set of unbalanced voltages or currents into three sets of symmetrical balanced phasors. These are known as positive, negative and zero sequence components. Fig. 7.1 shows balanced and unbalanced systems.

Fig. 7.1

13

Consider the symmetrical system of phasors in Fig. 7.2. Being balanced, the phasors have equal amplitudes and are displaced 120o relative to each other. By the definition of symmetrical components, 1bV always lags 1aV by a fixed angle of 120o and always has the same magnitude as 1aV . Similarly 1cV leads 1aV by 120o. It follows then that

11 aa VV = (7.1a) 1

211 )2401( aa

ob VaVV == (7.1b)

111 )1201( aao

c aVVV == (7.1c)

Where the subscript (1) designates the positive sequence component. The system of phasors is called positive sequence because the order of the sequence of their maxima occur abc. Similarly, in the negative and zero sequence components, we deduce

22 aa VV = (7.2a) 222 )1201( aa

ob aVVV == (7.2b)

22

22 )2401( aao

c VaVV == (7.2c)

00 aa VV = (7.3a) 00 ab VV = (7.3b) 00 ac VV = (7.3c)

Where the subscript (2) designates the negative sequence component and subscript (0) designates zero sequence components. For the negative sequence phasors the order of sequence of the maxima occur cba, which is opposite to that of the positive sequence. The maxima of the instantaneous values for zero sequence occur simultaneously.

Fig.7.2

In all three systems of the symmetrical components, the subscripts denote the components in the different phases. The total voltage of any phase is then equal to the sum of the corresponding components of the different sequences in that phase. It is now possible to write our symmetrical

14

components in terms of three, namely, those referred to the a phase (refer to section 3 for a refresher on the a operator).

210 aaaa VVVV ++= (7.4a) 210 bbbb VVVV ++= (7.4b) 210 cccc VVVV ++= (7.4c)

We may further simplify the notation as follows; define

00 aVV = (7.5a) 11 aVV = (7.5b) 22 aVV = (7.5c)

Substituting their equivalent values

210 VVVVa ++= (7.6a) 21

20 aVVaVVb ++= (7.6b)

22

10 VaaVVVc ++= (7.6c) These equations may be manipulated to solve for 0V , 1V , and 2V in terms of aV , bV , and cV .

( )cba VVVV ++= 31

0 (7.7a)

( )cba VaaVVV 21 31 ++= (7.7b) ( )cba aVVaVV ++= 22 31 (7.7c)

It follows then that the phase current are

210 IIIIa ++= (7.8a) 21

20 aIIaIIb ++= (7.8b)

22

10 IaaIIIc ++= (7.8c) The sequence currents are given by

( )cba IIII ++= 31

0 (7.9a)

( )cba IaaIII 21 31 ++= (7.9b) ( )cba aIIaII ++= 22 31 (7.9c)

15

The unbalanced system is therefore defined in terms of three balanced systems. Eq. (7.6) may be used to convert phase voltages (or currents) to symmetrical component voltages (or currents) and vice versa [Eq. (7.7)]. Example 7.1

Given oaV 535= , obV 1647 = , ocV 1057= , find the symmetrical components. The phase components are shown in the phasor form in Fig. 7.3

Va

Vb

Vc

Unbalanced condition

53o

105o

-164o

Fig. 7.3

Solution Using Eq. (7.7) Solve for the zero sequence component:

( )cbaa VVVV ++= 31

0

( )ooo 1057164753531 ++=

o1225.3 =

From Eq. (7.3b) and (7.3c) o

bV 1225.30 = o

cV 1225.30 =

Solve for the positive sequence component:

( )cbaa VaaVVV 21 31 ++= ( ) ( )( )ooooo 1057240116471201535

31 ++=

o100.5 =


bV 1300.51 = o

cV 1100.51 =

16

Solve for the negative sequence component:

( )cbaa aVVaVV ++= 22 31 ( ) ( )( )ooooo 1057120116472401535

31 ++=

o929.1 =


bV 1489.12 = o

cV 289.12 =

The sequence components can be shown in phasor form in Fig. 7.4.

Negative Sequence

Va2

Vc2Va1

Vc0Vb0

Va0

Zero SequenceVb2

Vb1

Vc1

Positive Sequence

Fig. 7.4 Using Eq. (7.6) the phase voltages can be reconstructed from the sequence components. Example 7.2

Given oV 1225.30 = , oV 100.51 = , oV 929.12 = , find the phase sequence components. Shown in the phasor form in Fig. 7.4

Solution Using Eq. (7.6)

Solve for the A-phase sequence component:

210 VVVVa ++= ooo 929.1100.51225.3 ++=

o530.5 =

17

Solve for the B-phase sequence component:

212

0 aVVaVVb ++= ooo 1489.11300.51225.3 ++=

o1640.7 =

Solve for the C-phase sequence component:

22

10 VaaVVVc ++= ooo 289.11100.51225.3 ++=

o1050.7 =

This returns the original values given in Example 5.2.

This can be shown in phasor form in Fig. 7.5.

Vc2

Vc0

Vc1

Va2

Va1

Va0

Vb0

Vb2

Vb1

Va

Vb

Vc

Fig. 7.5

Notice in Fig. 7.5 that by adding up the phasors from Fig. 7.4, that the original phase, Fig. 7.3 quantities are reconstructed. 8. Balanced and Unbalanced Fault analysis

Lets tie it together. Symmetrical components are used extensively for fault study calculations. In these calculations the positive, negative and zero sequence impedance networks are either given by the manufacturer or are calculated by the user using base voltages and base power for their system. Each of the sequence networks are then connected together in various ways to calculate fault currents and voltages depending upon the type of fault.

18

Given a system, represented in Fig. 8.1, we can construct general sequence equivalent circuits for the system. Such circuits are indicated in Fig. 8.2.

Fig. 8.1

1I+

-

1Vo01

2I+

-

2V

0I+

-

0V

Zero Sequence Network Positvie Sequence Network Negative Sequence Network

0Z 1Z 2Z

Fig. 8.2

Each of the individual sequence may be considered independently. Since each of the sequence networks involves symmetrical currents, voltages and impedances in the three phases, each of the sequence networks may be solved by the single-phase method. After converting the power system to the sequence networks, the next step is to determine the type of fault desired and the connection of the impedance sequence network for that fault. The network connections are listed in Table 8.1

Table 8.1 - Network Connection Three-phase fault - The positive sequence impedance network

is only used in three-phase faults. Fig. 8.3 Single Line-to-Ground fault - The positive, negative and zero

sequence impedance networks are connected in series. Fig. 8.5 Line-to-line fault - The positive and negative sequence

impedance networks are connected in parallel. Fig. 8.7 Double Line-to-Ground fault - All three impedance networks

are connected in parallel. Fig. 8.9 The system shown in Fig. 8.1 and simplified to the sequence network in Fig. 8.2 and will be used throughout this section.

19

Example 8.1 Given puZ o90199.00 = , puZ o90175.01 = ,

puZ o90175.02 = , compute the fault current and voltages for a Three-phase fault. Note that the sequence impedances are in per-unit. This means that the solution for current and voltage will be in per-unit.

Solution The sequence networks are interconnected, as shown in Fig. 8.3

Note that for a three phase fault, there are no negative or zero sequence voltages.

020 ==VV 020 == II

The current 1I is the voltage drop across 1Z

1

11 Z

VI =

175.001

1 jI

o= 71.5j=

The phase current is converted from the sequence value using Eq. (7.8).

pujI oa 9071.5071.50 =+= puajaI ob 15071.5)0()71.5(0

2 =++= puajaI oc 3071.5)0()71.5(0

2 =++=

Calculating the voltage drop, and referring to Fig. 8.3, the sequence voltages are

020 ==VV 111 01 IZV

o = ( ) 0.071.5175.011 == jjV

pu0.0= The phase voltages are converted from the sequence value using Eq. (7.6).

puVa 0.00.00.00.0 =++= puaaVb 0.0)0.0()0.0(0.0

2 =++= puaaVc 0.0)0.0()0.0(0.0

2 =++=

2I+

-

2V

0I+

-

0V

0Z

2Z

1I+

-

1Vo01

1Z

Fig 8.3

Ic

Ia

Ib

VaVbVc

Fig 8.4

20

The per-unit value for the current and voltage would now be converted to actual values using Eq. (5.6b) and knowing the base power and voltage for the given system. The currents and voltages can be shown in phasor form in Fig. 8.4

Example 8.2

Given puZ o90199.00 = , puZ o90175.01 = , puZ o90175.02 = , compute the fault current and voltages for a Single line-to-ground fault. Note that the sequence impedances are in per-unit. This means that the results for current and voltage will be in per-unit.

Solution The sequence networks are interconnected in series, as shown in Fig. 8.5

Because the sequence currents are in series, and using ohms law.

210 III ==

)( 2101

0 ZZZVI ++=

)175.0175.0199.0(01

0 jjjI

o

++=

puj 82.1=

The phase currents are converted from the sequence value using Eq. (7.8). Substituting 210 III == into Eq. (7.8) gives

0000 3IIIIIa =++= 000

20 =++= aIIaIIb

002

00 =++= IaaIIIc Refer to Table 3.2: ( )01 2 =++ aa Note that 03IIa = . This is the quantity that the relay sees for a Single Line-to-Ground fault.

Substituting pujI 82.10 =

)82.1(303 jIIa ==

puj 46.5=

2I+

-

2V

0I+

-

0V

0Z

2Z

1I+

-

1Vo01

1Z

Fig 8.5

21


000 IZV = 111 IZVV =

222 IZV = Substituting in the impedance and current from above

362.0)82.1(199.00 == jjV ( ) 681.082.1175.011 == jjV ( ) 319.082.1175.02 == jjV The phase voltages are converted from the sequence value using Eq. (7.6).

0319.0681.0362.0 =+=aV puaaV ob 238022.1)319.0()681.0(362.0

2 =++= puaaV oc 122022.1)319.0()681.0(362.0

2 =++= The per-unit value for the current and voltage could now be converted to actual values using Eq. (5.6b) and knowing the base power and voltage for the given system. The currents and voltages can be shown in phasor form in Fig. 8.6

Example 8.3

Given puZ o90199.00 = , puZ o90175.01 = , puZ o90175.02 = , compute the fault current and

voltages for a Line-to-Line fault. Note that the sequence impedances are in per-unit. This means that the solution for current and voltage will be in per-unit.


Because the sequence currents sum to one node, it follows that

21 II =

Ia

Va

Vb

Vc

Fig 8.6

2I+

-

2V

0I+

-

0V

0Z

2Z

1I+

-

1Vo01

1Z

Fig 8.7

22

The current 1I is the voltage drop across 1Z in series with 2Z

21

11 ZZ

VI +=

175.0175.001

1 jjI

o

+=

puj 86.2=

pujI 86.22 += 00 =I


pujjIa 086.286.20 =+= pujajaIb 95.4)86.2()86.2(0

2 =++= pujajaIc 95.4)86.2()86.2(0

2 =++=


21 VV = 222 IZV =

)86.2)(75.1( jj= pu5.0=

00 =V The phase voltages are converted from the sequence value using Eq. (7.6).

puVa 0.15.05.00.0 =++= puaaVb 5.0)5.0()5.0(0.0

2 =++= puaaVc 5.0)5.0()5.0(0.0

2 =++=


IcIb

Va

Vb

Vc

Fig 8.8

23

Example 8.4 Given puZ o90199.00 = , puZ o90175.01 = , puZ o90175.02 = , compute the fault current and voltages for a Double Line-to-Ground fault. Note that the sequence impedances are in per-unit. This means that the solution for current and voltage will be in per-unit.


Because the sequence currents sum to one node, it follows that

)( 201 III +=

The current 1I is the voltage drop across 1Z in series with the parallel combination of 0Z and

2Z

++=

20

201

11

ZZZZZ

VI

Substituting in oV 011 = , and 0Z , 1Z , and 2Z , then solving for 1I

pujI 73.31 = 1

20

20 )(

IZZ

ZI += 75.1j+=

120

02 )(

IZZ

ZI += 99.1j+=


pujjjIa 099.173.375.1 =+= pujajajI ob 1.15260.5)99.1()73.3(75.1

2 =++= pujajajI oc 9.2760.5)99.1()73.3(75.1

2 =++=

2I+

-

2V

0I+

-

0V

0Z

2Z

1I+

-

1Vo01

1Z

Fig 8.9

24


210 VVV == 000 IZV =

)199.0)(75.1( jj= pu348.0=

The phase voltages are converted from the sequence value using Eq. (7.6).

puVa 044.1348.0348.0348.0 =++= puaaVb 0)348.0()348.0(348.0

2 =++= puaaVc 0)348.0()348.0(348.0

2 =++= Refer to Table 3.2: ( )01 2 =++ aa


IcIb

Va

Fig 8.10

IR

25

9. Oscillograms and Phasors Attached are four faults that were inputted into a relay and then captured using the relay software. Three-phase fault. Compare to example (8.1)

Fig 9.1a

Fig 9.1b Fig 9.1c

26

Single Line-to-Ground fault. Compare to example (8.2)

Fig 9.2a

Fig 9.2b Fig 9.2c

27

Line-to-Line fault. Compare to example (8.3)

Fig 9.3a

Fig 9.3b Fig 9.3c

28

Double Line-to-Ground fault. Compare to example (8.4)

Fig 9.1a

Fig 9.4b Fig 9.4c

29

10. Symmetrical Components into a Relay Using a directional ground distance relay it will demonstrated how sequential components are used in the line protection. To determine the direction of a fault, a directional relay requires a reference against which the line current can be compared. This reference is known as the polarizing quantity. Zero sequence line current can be referenced to either zero sequence current or zero sequence voltage, or both may be used. The zero sequence line current is obtained by summing the three-phase currents. See Fig. 10.1

From Eq. (7.9) ( ) rcba IIIII ==++ 03 (10.1) This is known as the residual current or simply 03I . The zero sequence voltage at or near the bus can be used for directional polarization. The polarizing zero sequence voltage is obtained from a broken-delta and a grounded wye voltage transformer. See Fig 10.2

03VaV

bV

cV

30

From Eq. (7.7) the zero sequence voltage equals

( )cba VVVV ++= 31

0 (10.2a)

( )cba VVVV ++=03 (10.2a)

Example 10.1

Using the values obtained from example 8.2, calculate 03V . Solution

0=aV puV ob 238022.1 = puV oc 122022.1 =

ooV 122022.1238022.103 0 ++=

puo18025.3 = The zero sequence voltage is puo18025.3 . By connecting the value in the reverse gives 03V which equals puo025.3 . Plotting this, we can show in phasor form what the relay sees, Ia lagging 03V by the line angle. In this case resistance is neglected, therefore Ia lags by 90o. (see Fig 10.3).

31

References Blackburn, J. L., Protective Relaying: Principles and Applications, Mercel Dekker, Inc., New York, 1987 Gross, Charles A., Power System Analysis, John Wiley & Sons, Inc., 1986 ABB, Protective Relaying Theory and Applications, Mercel Dekker, Inc., New York, 2004 Wagner, C. F. and Evans, R. D., Symmetrical Components, Krieger Publishing Company, Florida, 1933 Lantz, Martin J., Fault Calculations for Relay Engineers, Bonneville Power Administration, 1965

2/23/2010

1

HandsHands--On Relay School:On Relay School:Distribution Protection TheoryDistribution Protection Theory

Daniel GoodrichElectrical Engineer, Bonneville Power AdministrationMarch 17th, 2010

AgendaAgenda

y Introductionsy Schedule: y Schedule:

Class 1:00pm to 5:00pm Today Class 7:30am to Noon Tomorrowy Course Format

2/23/2010

2

IntroductionsIntroductionsInstructor: Daniel Goodrich, PEy 1986-2010: Engineer at Salt River Project,

Ph i AZPhoenix, AZy March 2010 to Present: Planning Engineer at

Bonneville Power Administration, Vancouver Washington

y 10 years in System Protection (Transmission, Distribution, and some Generation)

y Active participant in IEEE Power System Relay y Active participant in IEEE Power System Relay Committee, 2005-2009

y And you areName, Utility/Business Affiliation, Experience with Distribution Protection

Course FormatCourse Format

y Theoryy Examplesy Examplesy Questionsy Takeaways (watch for the T) and

reference material

2/23/2010

3

OutlineOutline

y Configurationsy Comparison of Distribution and y Comparison of Distribution and

Transmission Systemsy Class Scopey Load sheddingy Substation feeder breakersy Substation feeder breakersy Reclosingy Fuse saving/sacrificingy Cold Load Pickup

Outline ContinuedOutline Continued

y Line Reclosersy Sectionalizersy Sectionalizersy Distribution Automationy Distributed Generation

2/23/2010

4

Distribution System ConfigurationsDistribution System Configurations

y Radial feeder: A feeder supplying electric energy to a substation or a electric energy to a substation or a feeding point that receives energy by no other means (from IEEE 100)yOpen-Looped: Two distribution circuits

with two active sources that have a normally open tie point between the circuits.

T: What kind of system do you have?

Distribution System ConfigurationsDistribution System Configurations

y Secondary Distribution System Network: The secondaries of the distribution The secondaries of the distribution transformers are connected to a common network for supplying light and power directly to consumers services (IEEE 100)

T: Do you have networks?

2/23/2010

5

Radial FeedsRadial Feeds

y Lowest cost; older systemsy Feeders isolated by geographyy Feeders isolated by geographyy Fault location is very difficult

Substation

Feeder

OpenOpen--Looped SystemLooped Systemy Reduced outage timey Switching flexibility for both planned

Substation A S b t ti B

Normally Closed Switch

Normally Open

Switching flexibility for both planned maintenance and unplanned outages

Substation A Substation BSwitch

2/23/2010

6

OpenOpen--Looped ExampleLooped Example

Secondary Distribution System Secondary Distribution System Network: Network: y Highest level of reliabilityy In locations of high concentrations of load y In locations of high concentrations of load

(downtown areas)

From NREL

2/23/2010

7

Distribution and Transmission Distribution and Transmission System DifferencesSystem DifferencesThe distribution system:y Is more dangerous to the public due to its y Is more dangerous to the public due to its

proximity and common presence y Has many more faults than the

transmission system (20:1 ratio)y Typically has more automatic reclosingyp y g

RELAY RELIABILITYRELAY RELIABILITY

SECURITY

DEPENDABILITY

Takeaway: Are there one or two relays are on each of your distribution substation feeders?

2/23/2010

8

Scope of ClassScope of Class

GeneratingGenerating TransmissionTransmissionStepStep--upup

ReceivingReceivingStationStation

230/69kV230/69kV

Scope of Class (Red Portion)Scope of Class (Red Portion)

GeneratingGeneratingStationStation

SubtransmissionSubtransmission69kV69kV

TransmissionTransmission230kV230kV

StepStep--upupTransformerTransformer

DistributionDistribution12kV12kV

Distribution SubstationDistribution Substation69/12kv69/12kv

Industrial./CommercialIndustrial./CommercialBuildingBuilding

16

2/23/2010

9

Load SheddingLoad Shedding

y For transmission system stabilityy Typically is employed at the distribution y Typically is employed at the distribution

levely Three types, in order of speed:

Under-frequency (0.1 seconds) Under-voltage (2 to 10 seconds)g ( ) Rotating (Rolling) Blackouts (minutes/hours)T: What load shedding do you employ?

UnderUnder--frequency Examplefrequency Example

Coordinated by NERCFrequency % of Total Load

59.5 5%

59.1 1.5%

58.9 5%

58.7 7.5%

58.5 7%

58 3 7%

Restoration is controlled by Operations

58.3 7%

Total 35%

2/23/2010

10

UnderUnder--voltage Load Sheddingvoltage Load Shedding

y Typical purpose is to separate load when substation loses its sourcesubstation loses its sourcey Electromechanical settings, 2.3 Seconds at

Zero voltsy Micro-processor settings, 10 seconds at

63% of nominal

UnderUnder--voltage tripping for Stabilityvoltage tripping for Stability

y Coordinate with other utilitiesy May be coordinated by NERCy May be coordinated by NERCy Phoenix example:

Voltage Time Delay % Load Shed

50% of nominal 2 Seconds 5%

70% 4 5

80% 6 5

90% 10 5

2/23/2010

11

UnderUnder--voltage Restorationvoltage Restoration

y Automaticy In stages; example:y In stages; example:

Feeder 1 (up to 6 MW) instantaneously Feeder 2 after 5 second time delay Feeder 3 after 10 second time delay Feeder 4 after 15 second time delay

Substation Feeder BreakerSubstation Feeder Breaker

y Is typically the dividing point between sub-transmission and distributionsub-transmission and distribution

Breakers

2/23/2010

12

Substation Feeder BreakerSubstation Feeder Breaker

y Traditional protective devices: Instantaneous (50) and time (51) over current Instantaneous (50) and time (51) over-current,

non-directional Ground 50/51 Reclosing relay

Newer configurationsNewer configurations All functions in one relay May add under-voltage under-frequency back May add under voltage, under frequency, back

block/trip, breaker failure

2/23/2010

13

Basic Protection: OverBasic Protection: Over--currentcurrent

y Load current Zero to 600 Ampsy Fault current 500 to 10 000 Ampsy Fault current 500 to 10,000 Ampsy Length 500 feet to 15 miles

Distribution Coordination Distribution Coordination PrinciplesPrinciples1. Give all faults a chance to be temporary2 Lock out a feeder only for permanent 2. Lock out a feeder only for permanent

faults3. Remove only the smallest portion of the

feeder from service

From Electrical DistributionSystem Protection, Cooper Power Systems

2/23/2010

14

Feeder protection basicsFeeder protection basics

y Trip feeder breaker for faults anywhere along the feederalong the feedery Must account for:

Upstream and downstream protective devices Abnormal switching Normally Closed

Switch

Substation A Substation B

Normally Open Switch

Substation A Substation BNormally Closed Switch

Normally Open Switch

New feeder segment

Basic study stepsBasic study steps

1. Determine system parameters1 Source fault duty1. Source fault duty2. Feeder conductor(s) type, length to end of

line

2. Model the system 3. Simulate faults at nearest and furthest

points4. Set relay to trip for any point along the

feeder

2/23/2010

15

Source Fault DutySource Fault Duty

y From system model

COLLIER69.kV 15544

Collier 312.47kV 3 Sample

12.47kV

86 0P 8

0.0P-169

36.9P-2

0.0P107

0.88P-90 8651P93

1153P123 1563P-57

0.00P90

8650P-87

410P122

T: What kind of fault study program do you use?

Conductor length, typeConductor length, typeCONDUCTOR ELECTRICAL CHARACTERISTICS PRIMARY CONDUCTORS

CONDUCTOR DESCRIPTION R (/mile) Xa (/mile) CONDUCTOR LENGTH (mi)

397 A.A. 0.259 0.4587 0.209

477 A.A. 0.216 0.447 0

OV

ERH

EAD

DES

IGN

(P

rim

ary) 795 A.A. 0.131 0.4146 0

266 A.A. (7 strd) 0.386 0.4895 0

3/0 A.A. (7 strd) 0.613 0.5177 0

1/0 A.A. (7strd) 0.973 0.5434 0

2 A.A. (7 strd) 1.55 0.5739 0

2/0 CU. 0.481 0.532 0

1 CU. (7 strd) 0.765 0.56 0

2 CU. (7 strd) 0.964 0.574 0

4 CU. (3 strd) 1.518 0.599 0

6 CU. (1 strd) 2.39 0.637 0

#2 ACSR 1.41 0.665 0

312 A.A.A.C. 0.417 0.479 0

UN

DER

GR

OU

ND

DES

IGN

(P

rim

ary)

500 MCM-AL (DUCT) 0.2472 0.3807 0

500 MCM-AL (D.B.) 0.264 0.2207 0.6354

350 MCM-AL ** 0.2904 0.14784 0

750 MCM-AL (DUCT) 0.1265 0.3965 0

600 CU. 0.11 0.432 0

750 MCM CU. 0.0981 0.3968 0.689

2 AL (D.B.) 1.9628 0.3078 0

4/0 AL (DUCT) 0.6529 0.3458 0

500 CU. 0.15206 0.14098 0

4/0 AL (D.B.) 0.625 0.2434 0

2/23/2010

16

Fault levelsFault levels

(1)(1) Three phase fault in amperes(2) Single line to ground fault in(1)

(2)

8700 A9300 A

3000 A2000 A

(2) Single line to ground fault in Amperes

End of lineSubstation

Over-current pickup < 3000 AmpsGround over-current pickup

2/23/2010

17

Transformer Damage and Fuse Transformer Damage and Fuse CurvesCurves

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7

5070

100

200300400500700

1000

5070100

2003004005007001000

1

A. Transf. damage curve. 22.00 MVA. Category 3Base I=555.00 A. Z= 7.0 percent.

SECONDS

23457

10

20304050

2345710

20304050

3.4.5.71

3.4.5.71

1. Pole Riser Fuse S&C 167-62-85Total clear.

A

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7CURRENT (A)

.01

.02

.03

.04

.05

.07.1

.2

.3

.01

.02

.03

.04

.05

.07

.1

.2

.3

TIME-CURRENT CURVES @ Voltage By

F N

Fuse Blowing/Damage RegionsFuse Blowing/Damage Regions10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7

100

200300400500700

1000

100

2003004005007001000

1


Fuse blows

SECONDS

23457

10

2030405070

2345710

2030405070

4.5.71

4.5.71

1. Pole Riser Fuse S&C 167-62-85Total clear.

A

Fuse blows

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7CURRENT (A)

.01

.02

.03

.04

.05

.07.1

.2

.3

.4

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4


F N

Fuse does not blow

2/23/2010

18

Fuse Curve ChangesFuse Curve Changes

A fuse will operate at a faster time depending on:depending on:y Pre-fault loadingy Ambient temperaturey Material used

Fuse Minimum Melt vs. Maximum Fuse Minimum Melt vs. Maximum ClearClear

At 2,000 Amps:Maximum clear is about 0.04 secondsMinimum melt is about 0.02 seconds

2/23/2010

19

Transformer Damage CurveTransformer Damage Curve

y Set by ANSI Standard

Electrical Damage curveMechanical Damage curve, depends on transformer impedance

Transformer Damage Curve POTENTIAL

DAMAGE AREA

38

2/23/2010

20

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7

2030405070

100

200300400500700

1000

2030405070100

2003004005007001000

1 1. Pole Riser Fuse S&C 167-62-85Total clear.

2 2. Main Normal OC Ph IEEE_EI TD=3.500CTR=240 Pickup=6.5A No inst. TP@5=0.6484s


Area of Coor-dination

SECONDS

23457

10

20

2345710

20

.3

.4

.5

.71

.3

.4

.5

.71

A

Downstream device

Upstream device

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7CURRENT (A)

.01

.02

.03

.04

.05

.07.1

.2

.01

.02

.03

.04

.05

.07

.1

.2


device

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7

2030405070

100

200300400500700

1000

2030405070100

2003004005007001000

1 1. Pole Riser Fuse S&C 167-62-85Total clear.

2 2. Main Normal OC Ph IEEE_EI TD=3.500CTR=240 Pickup=6.5A No inst. TP@5=0.6484s

3


SECONDS

23457

10

20

2345710

20

2.3.4.5.71

2.3.4.5.71

3. FDR Normal OC PH IEEE_EI TD=4.500CTR=240 Pickup=2.5A No inst. TP@5=0.8336s

A

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7CURRENT (A)

.01

.02

.03

.04

.05

.07.1

.2

.01

.02

.03

.04

.05

.07

.1

.2


F N

2/23/2010

21

Coordination:Fuse operates before feeder breaker curve, before main breaker curve, before transformer damage curveFault at end of

line

Fault at substation

Coordination Coordination

Pickup

Curve Curve ComparisonComparisonof Time Dialsof Time Dials

Time Dial 15

Time Dial 0 55Time Dial 0.55

42

2/23/2010

22

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7

2030405070

100

200300400500700

1000

2030405070100

2003004005007001000

1

1. Sample EI IEEE_EI TD=1.000CTR=240 Pickup=1.A (Dir) No inst. TP@5=0.1852s

3. Sample VI IEEE_VI TD=1.000CTR=240 Pickup=1.A (Dir) No inst. TP@5=0.1869s

Coordination Coordination SECONDS

23457

10

2345710

.2

.3

.4

.5

.71

.2

.3

.4

.5

.71

1

2

3

Curve Curve ComparisonComparison

of curve shapeof curve shapeModerately Inverse

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7CURRENT (A)

.01

.02

.03

.04

.05

.07.1

.01

.02

.03

.04

.05

.07

.1


F N

2. Sample MI IEEE_MI TD=1.000CTR=240 Pickup=1.A (Dir) No inst. TP@5=0.2413sExtremely Inverse

y

43

Very Inverse

Relay to Fuse CoordinationRelay to Fuse Coordination

FeederBreaker

Fuse

Load

Fault

51

Relay

Fuse maximum melt should be 0.2 to 0.3 seconds faster than relay, for any pointseconds faster than relay, for any point downstream of the fuse.The challenge is typically at the maximum fault values.

T: Where do you employ fusing?

2/23/2010

23

Fused SingleFused Single--Phase TransformerPhase Transformer

Fused ThreeFused Three--Phase TransformerPhase Transformer

2/23/2010

24

Fusing CubicleFusing Cubicle

Current limiting fuse operatedCurrent limiting fuse operated

2/23/2010

25

Successful Successful CoordinationCoordination

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7

30405070

100

200300400500700

1000

30405070100

2003004005007001000

1

1. FDR Normal OC PH IEEE_EI TD=4.500CTR=240 Pickup=2.5A No inst. TP@5=0.8336sIa= 3006.8A (12.5 sec A) T= 0.83s

2

SECONDS

23457

10

20

2345710

20

.3

.4

.5

.71

.3

.4

.5

.71

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7CURRENT (A)

.01

.02

.03

.04

.05

.07.1

.2

.01

.02

.03

.04

.05

.07

.1

.2


2. Xfmr fuse Lo BAY-0-NET065Total clear.Ia= 3006.8A T= 0.18s

FAULT DESCRIPTION:Bus Fault on: 0 Sample 12.47kV 3LG

Automatic ReclosingAutomatic Reclosing

yWhy? ~70% of faults are temporaryy Improved customer service and reliabilityy Improved customer service and reliabilityy Survey Results:Utility First Reclosure Second Reclosure

1 Instantaneous 20 seconds

2 5 25

3 15 N/A3 15 N/A

4 5 N/A

5 2 N/A

6 Instantaneous 30

7 Instantaneous 15

2/23/2010

26

Reclosing ContinuedReclosing Continued

y A wide variety of practicesy The presence of underground feeders y The presence of underground feeders,

distributed generation and SCADA tend to reduce reclosing

T: What is your reclosing sequence?y g q

Fuse Sacrificing/SavingFuse Sacrificing/Saving

y Fuse sacrificing allows the fuse to blow before an upstream device for fault before an upstream device for fault conditionsy Fuse saving allows the upstream device to

operate once before the fuse. If the fault is temporary, the fuse might be saved.

T: Does your company employ fuse saving or fuse sacrificing?

2/23/2010

27

Fuse Fuse SacrificingSacrificing

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7

30405070

100

200300400500700

1000

30405070100

2003004005007001000

1

1. FDR Normal OC PH IEEE_EI TD=4.500CTR=240 Pickup=2.5A No inst. TP@5=0.8336sIa= 3006.8A (12.5 sec A) T= 0.83s

2

SECONDS

23457

10

20

2345710

20

.3

.4

.5

.71

.3

.4

.5

.71

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7CURRENT (A)

.01

.02

.03

.04

.05

.07.1

.2

.01

.02

.03

.04

.05

.07

.1

.2




Fuse SavingFuse Saving

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7

2030405070

100

200300400500700

1000

2030405070100

2003004005007001000

1

1. FDR Normal OC PH IEEE_EI TD=4.500CTR=240 Pickup=2.5A Inst=1500A TP@5=0.8336Ia= 3006.8A (12.5 sec A) T= 0.83s

2

SECONDS

23457

10

20

2345710

20

.2

.3

.4

.5

.71

.2

.3

.4

.5

.71Addition of

instantaneous element on relay

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7CURRENT (A)

.01

.02

.03

.04

.05

.07.1

.01

.02

.03

.04

.05

.07

.1


F N



2/23/2010

28

Fuse Saving SchemeFuse Saving SchemeDC +

The instantaneous (50) element is supervised by the reclosing relay (79)(79)

Trip Bus

Fuse Saving in Microprocessor RelayFuse Saving in Microprocessor Relay

79 (O->I)Reset Time (Seconds) 60#1 Pickup51P (3I>) NA50P-1 (3I>>1) Enable50P-2 (3I>>2) Lockout50P-3 (3I>>3) NA#1 Open time (Seconds) 0.1

#2 Pickupp51P (3I>) NA50P-1 (3I>>1) Disable50P-2 (3I>>2) Lockout50P-3 (3I>>3) NA#2 Open time (Seconds) 15.0

2/23/2010

29

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7

2030405070

100

200300400500700

1000

2030405070100

2003004005007001000

1

1. Xfmr fuse 150 kVA Lo BAY-0-NET015Total clear.Ia= 3006.8A T= 0.01s

2

2. FDR Normal OC PH IEEE_EI TD=4.500CTR=240 Pickup=2 5A Inst=1500A TP@5=0

Fuse Saving Fuse Saving Coordination Coordination ProblemProblem

SECONDS

23457

10

20

2345710

20

.3

.4

.5

.71

.3

.4

.5

.71

CTR 240 Pickup 2.5A Inst 1500A TP@5 0Ia= 3006.8A (12.5 sec A) T= 0.83s

FAULT DESCRIPTION:Bus Fault on: 0 Sample 12.47kV 3

Cannot predict which element will operate first

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7CURRENT (A)

.01

.02

.03

.04

.05

.07.1

.2

.01

.02

.03

.04

.05

.07

.1

.2


Instantaneous vs. InstantaneousInstantaneous vs. Instantaneous

2/23/2010

30

Fuse to Fuse CoordinationFuse to Fuse Coordinationy Common in overhead distribution and in

industrial applicationsppy Using the same fuse type (K, N) of different

sizes helps with coordination because of similar curve shapesy Tables are available for coordinationy The general principle is that the maximum y The general principle is that the maximum

clearing time of the downstream (protecting) fuse is less than 75% of the minimum melting time of the upstream (protected) fuse.

Fuse to Fuse CoordinationFuse to Fuse Coordination

With the 75% reduction, the two fuses coordination i d d f 2 500is reduced from 2,500 Amps to 2,000 Amps

2/23/2010

31

Sample Fuse Coordination TableSample Fuse Coordination Table

From ee.Washington.edu

Cold Load PickupCold Load Pickup

yWhen a protective device operates due to the energization of a cold feeder to the energization of a cold feeder, caused by inrush of distribution transformers

2/23/2010

32

Transformer Transformer EnergizationEnergization CurrentCurrent

Feeder Feeder EnergizationEnergization CurrentCurrent

3 000 Amps ~600 Amps~3,000 Amps 600 Amps

2/23/2010

33

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7

200300400500700

1000

2003004005007001000

2

Cold Load Pickup SolutionCold Load Pickup Solution----PhasePhase

SECONDS

23457

10

2030405070

100

2345710

2030405070100

.71

.71

1

1. FDR Normal OC PH IEEE_EI TD=4.500CTR=240 Pickup=2.5A Inst=1500A TP@5=0.8336s

2 FDR OC Cold Load PH IEEE EI TD=12 000

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7CURRENT (A)

.01

.02

.03

.04

.05

.07.1

.2

.3

.4

.5

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5


F N

2. FDR OC Cold Load PH IEEE_EI TD=12.000CTR=240 Pickup=2.5A Inst=3000A TP@5=2.223s

Cold Load Pickup SolutionCold Load Pickup Solution----GroundGround10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7

200300400500700

1000

2003004005007001000

SECONDS

23457

10

2030405070

100

2345710

2030405070100

.71

.71

11. FDR Normal Gnd OC IEEE_VI TD=5.00CTR=240 Pickup=0.75A Inst=1560A TP@5=

2

2. FDR Cold Load Gnd IEEE_VI TD=10.00CTR=240 Pickup=1.5A Inst=3000A TP@5=1

Q: Does your utility use Cold Load settings, and if so, how is it accomplished?

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7CURRENT (A)

.01

.02

.03

.04

.05

.07.1

.2

.3

.4

.5

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5


F N

2/23/2010

34

Line Line ReclosersReclosers

y Is a breaker on a stick: It can sense and interrupt faultsinterrupt faultsy Typically has 3 or 4 automatic reclosuresy Commonly employs fuse savingy Can be single or three-phasey Are rated by:y Are rated by:

Continuous and interrupting rating in Amperes

Interrupting medium (Vacuum or Oil) Hydraulic or electronic control

Line Line ReclosersReclosers

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Effective Placement of Effective Placement of RecloserRecloserOn the edge of a residential development

RecloserRecloser Application FactorsApplication Factors

1. System Voltage2 Maximum fault current at recloser 2. Maximum fault current at recloser

location3. Maximum load current4. Minimum fault current in the zone of

protectionp5. Coordination with upstream and

downstream devices6. Ground fault Sensing

From Electrical DistributionSystem Protection, Cooper Power Systems

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36

RecloserRecloser--Fuse Coordination Simple Fuse Coordination Simple ExampleExample

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7

200300400500700

1000

2003004005007001000

23

1. First trip: Instantaneous2. Wait 2 seconds, reclose3 S d t i F t C

SECONDS

23457

10

2030405070

100

2345710

2030405070100

.4

.5

.71

.4

.5

.71

1

2. Recloser Slow Phase IEEE_EI TD=10.000CTR=1000 Pickup=0.4A No inst. TP@5=1.8525sIa= 3006.8A (3.0 sec A) T= 0.90s

3. Xfmr fuse 1500 kVA BAY-0-NET140Total clear.Ia= 3006.8A T= 0.55s FAULT DESCRIPTION:

Bus Fault on: 0 Sample 12.47kV 3LG

3. Second trip: Fast Curve4. Wait 5 seconds, reclose5. Third and fourth trip: Slow

curve6. Lockout after fourth trip

Employs fuse saving

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7CURRENT (A)

.01

.02

.03

.04

.05

.07.1

.2

.3

.01

.02

.03

.04

.05

.07

.1

.2

.3


For No

1. Recloser Fast Phase IEEE_EI TD=1.000CTR=1000 Pickup=0.4A Inst=1000A TP@5=0.1852sIa= 3006.8A (3.0 sec A) T= 0.09s

Employs fuse saving Wont go to lockout for

temporary faults

RecloserRecloser Coordination with Coordination with Upstream BreakerUpstream Breakery Time curves are the same as for circuit

breakersbreakersy Instantaneous element of the upstream

breaker must not reach past the recloser FaultLevels:

Phase2,100Amps

Gnd 1,420 Amps Fault Levels:

Bkr1245R

Gnd1,420Amps FaultLevels:

Phase970Amps

Gnd620Amps

3.45miles 5.54miles

Substation Line Recloser

End of line

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RecloserRecloser Coordination with Coordination with Upstream BreakerUpstream Breaker

For a ground fault downstream of the recloser the time element

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7

200300400500700

1000

2003004005007001000

1. Substation Breaker IEEE_VI TD=5.000CTR=240 Pickup=0.75A Inst=3000A

recloser, the time element of the substation breaker coordinates well with the instantaneous element of the recloser.

For a fault at the end of the line, the time elements coordinate.

SECONDS

23457

10

2030405070

100

2345710

2030405070100

5.71

5.71

End of line fault

Fault just downstream of recloser

12

2. Recloser IEEE_VI TD=3.600CTR=1000 Pickup=0.08A Inst=900A

See table in TTY Window.

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7CURRENT (A)

.01

.02

.03

.04

.05

.07.1

.2

.3

.4

.5

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5


F N

Fault just downstream of recloser

RecloserRecloser--RecloserRecloser--Breaker Breaker CoordinationCoordination

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RecloserRecloser ChallengesChallengesy Changes in load current over timey Feeder length too long; fault duty below load y Feeder length too long; fault duty below load

levelsy Coordinating 3 or more curves (breaker,

recloser, fuse)y Getting SCADA to remote locationsy Battery maintenancey Battery maintenancey Voltage input from both sides needed for

reversabilityy If recloser is too close to substation, cannot

coordinate due to high fault levels

SectionalizerSectionalizerFrom IEEE 100:A self-contained circuit-opening device that A self contained circuit opening device that

automatically opens the main electrical circuit through it after sensing and responding to a predetermined number of successive main current impulses equal or greater to a predetermined magnitude or greater to a predetermined magnitude. It opens while the main electrical circuit is de-energized. It may also have a provision to be manually operated to interrupt loads.

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From Hubbell/Chance 10D

SecitionalizerSecitionalizer ApplicationsApplications

yWhere fault duty is very high, and time curves wont coordinate (typically curves won t coordinate (typically between breaker or recloser curves and fuse curves.y Between two protective devices with

time curves that are too close together.

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10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7

2030405070

100

200300400500700

1000

2030405070100

2003004005007001000

1

1. FDR Normal OC PH IEEE_EI TD=4.500CTR=240 Pickup=2.5A Inst=1500A TP@5=0.8336Ia= 3006.8A (12.5 sec A) T= 0.83s

2

SECONDS

23457

10

20

2345710

20

.2

.3

.4

.5

.71

.2

.3

.4

.5

.71

Area of difficult coordination

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 100002 3 4 5 7CURRENT (A)

.01

.02

.03

.04

.05

.07.1

.01

.02

.03

.04

.05

.07

.1


F N



Distribution AutomationDistribution Automation

Technique used to limit outage duration and restore service to customer through and restore service to customer through fault location identification and automatic switching. Reconfiguring a feeder can also be done to reduce losses, or eliminate temporary overloading

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41

Basic Example

Fiber

Substation

Distribution Operations Center

Substation

Fiber

OPENAutomatic Switch

MID-POINTAutomatic Switch


CLOSED OPEN CLOSED

Basic Example

Fiber

Substation

Distribution Operations Center

Substation

Fiber

FAULT

OPENAutomatic Switch



OPEN CLOSE

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Automation Effects on RelayingAutomation Effects on Relaying

y Abnormal feeder configurationsy Circulating currents when feeders are y Circulating currents when feeders are

connected to two substations simultaneouslymay need directionalityy May lose fuse saving and fault locating

ability

IEEE Power System Relaying Committee Working Group D11 Effect of Distribution Automation on Protective Relaying

Distributed GenerationDistributed Generation

Distributed Resource (DR): A source of electric power that is not directly of electric power that is not directly connected to the bulk power transmission system. DR includes both generators and energy storage technologies

Source: IEEE Standard 1547 for Interconnecting Distributed Resources with Electric Power Systems

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43

Example: Diesel Generators (2,000 Example: Diesel Generators (2,000 kVAkVA))

Sample ControlsSample Controls

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Hydroelectric (750 Hydroelectric (750 kVAkVA, 480V), 480V)

Solar Concentrator (25 kV Solar Concentrator (25 kV StirlingStirling engine)engine)

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Residential PhotovoltaicResidential Photovoltaic

Typical 3 kW Residential PV SiteTypical 3 kW Residential PV Site

Solar S

Utility AC DisconnectBilling Meter SystemBilling Meter

and Breaker Panel

DC DisconnectInverter

PV non billing Meter

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Bicycle Power!Bicycle Power!

Reasons for Distributed GenerationReasons for Distributed Generation

y For reliability: Data Centers and Water treatment plantstreatment plantsy Economics: Photovoltaics with incentivesy Environment: Renewables (Wind, Solar)

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Distributed Generation IssuesDistributed Generation Issues

y Public and Utility worker safety is the major concernmajor concerny Each Utility should have its own

Interconnection Guidelines Documenty Getting all of the correct information

takes several meeting iterations: One line and three line diagram Intended operation (fast transition, soft

unload, peaking, load reduction) Protective devices, coordination study and

relay settings

Distributed Generation Issues ContinuedDistributed Generation Issues Continued

y Is islanding a possibility?y Cannot accept customer owned and y Cannot accept customer owned and

maintained equipment as a substitute for utility protection.y Operation of generation during hold tags,

also called contact tagsyWitness testing of generators at

commissioning and for re-verification

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Types of ParallelingTypes of Paralleling

y Open-transitiononly needs to meet NEC requirementsNEC requirementsy Fast-transitiongenerators parallel for

0.1 seconds or less Not within the scope of IEEE 1547too fast

for a need for relaying Requirements can include:x Signed inter-connection agreementx Switch meets UL1008x Utility witnesses a demonstrationx Back-tripping in case switch gets stuck

Types of Paralleling ContinuedTypes of Paralleling Continued

y Soft unloading For larger loads where fast transition will not For larger loads where fast transition will not

work Generator typically parallels for 2 to 20

seconds

y Continuous (solar, co-generation)

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IslandIslandA condition in which a portion of a area electric

power system (the utility) is energized solely by p y ( y) g y yone or more local electric power systems (the distributed generator)y Is a possibility when the generator size is equal

to or greater than the minimum feeder load. At SRP, this value is around 500 kWy IEEE 1547: Unintended islandsgenerator

must be off line within 2 secondsy Transfer tripping may be required, or voltage

sensing at the substation

Ground fault simulator (no grounding banks)Ground fault simulator (no grounding banks)

Why a special ground fault detection scheme?

12.47 kV ground fault detected at SRP sub, but not at distributed generator

Because primary side f di ib i

SRP

CustomerSwitchboard

of distribution transformer is delta connected, there is no return path to customer generator for ground fault current

SRPTransformer

CustomerGenerator

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Ground fault simulator (with grounding bank)Ground fault simulator (with grounding bank)

Grounding bank provides a return path for ground fault current.

CustomerSwitchboard

CT

CustomerGenerator

GFR

Theory of operation Fault on delta side of xfmr hasno path to ground, relative to DG Grounding bank (GB) providesa path to ground CT on neutral of GB provides anoperating quantity for DG GFP (Iop) Or, use open delta VT connectionin conjunction with GB}To SES

Grounding Banks forGrounding Banks forDG Ground Fault ProtectionDG Ground Fault Protection

SLGF

Ground fault current pathw

w~Transfer switch Distributed

Generator (DG)Delta-wye xfmr

Generator GFP

Grounding bank(3 single phase xfmrsconnected grounded wyeprimary, delta secondary with CTon neutral of primary side connection)

Generator GFP

m

Alternateconfiguration(Open delta VTconnectedto delta primary -still requires GB, but VT provides relayoperating quantity)

mw

w

Iop

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Sympathetic TrippingSympathetic TrippingTransient pickup of relays or other healthy

feeders due to the conditions created by ya fault on another feeder of the system. (Othman, Mamdoh and Karnal)

Example: Fault on transmission system causes voltage to temporarily sag on distribution system with a lot of low distribution system with a lot of low inertia induction motors (air conditioners), and the feeder trips

Solution is to raise the pickup of the feeder relays.

ReferencesReferences

The Art and Science of Protective Relaying, C. Russell MasonRussell Mason

Electrical Distribution System Protection, Third Edition, Cooper Power Systems

Protective Relaying Principles and Applications, J. Lewis Blackburn

IEEE Standard for Interconnecting Distributed Resources with Electric Power Systems

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52

Disturbance photosDisturbance photos

Failed Capacitor BankFailed Capacitor Bank

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Failed SwitchFailed Switch

Open CTOpen CT

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4 kV Energized at 12 kV4 kV Energized at 12 kV

Distribution Breaker FailureDistribution Breaker Failure

2/23/2010

55

PT FirePT Fire

Junction Box Under WaterJunction Box Under Water

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TrampolineTrampoline

1DG Interconnection ProtectionDG Interconnection Protection

ProtectingUtility Distribution with Interconnecting

Distributed Generation Green Power, MicroGrids and Smart Grids

Technical & Industry Practice Background What IEEE Standard 1547 Doesnt Tell You

Chuck Mozina -- is a Consultant, Protection and Protection Systems for Beckwith Electric and resides in Palm Harbor (near Tampa), Florida. His consulting practice involves projects relating to protective relay applications, protection system design and coordination. He specializes in generator and power plant protection.

Chuck is an active 25-year member of the IEEE Power System Relay Committee (PSRC) and is the past chairman of the Rotating Machinery Subcommittee. He is active in the IEEE IAS I&CPS, PCIC and PPIC committees, which address industrial system protection. He is a former U.S. representative to the CIGRE Study Committee 34 ( now B-5) on System Protection and has chaired a CIGRE working group on generator protection. He also chaired the IEEE task force that produced the tutorial The Protection of Synchronous Generators, which won the PSRCs 1997 Outstanding Working Group Award. Chuck is the 1993 recipient of the Power System Relay Committees Career Service Award and he recently received the 2002 IAS I&CPS Ralph Lee Prize Paper Award. His papers have been republished in the IAS Industrial Applications Magazine.

Chuck has a Bachelor of Science in Electrical Engineering from Purdue University andChuck has a Bachelor of Science in Electrical Engineering from Purdue University and is a graduate of the eight month GE Power System Engineering Course. He has authored a number of papers and magazine articles on protective relaying. He has over 25 years of experience as a protection engineer at Centerior Energy( now FirstEnergy), a major investor-owned utility in Ohio where he was the Manager of the System Protection Section. He is also a former instructor in the Graduate School of Electrical Engineering at Cleveland State University as well as a registered Professional Engineer in the state of Ohio.

2Presentation Objectives

DG Interconnection Protection

Distributed Generation Overview- What is distributed generation (DG)?

Explore Types of DGs

n

- Explore Types of DGs- Explore the Future - Green Power, MicroGrids and Smart Grids- Define interconnection protection- Tell you what IEEE 1547 Doesnt

Protection Variables & Standards Generator types: Synchronous, Induction, Asynchronous Interconnection Transformer grounding makes a big difference! Protection to meet interconnection challenges Comparison of various power company and standards Comparison of various power company and standards

including IEEE 1547 Distribution protection coordination issues IEEE Distribution Survey Data - DG issues and impact on distribution systems

Introduction to DG Interconnection Protective Relays

Protection that allows the DG to operate in parallel to the utility

What is DG Interconnection Protection?


Large non-utility generators do not require specific interconnection protection

- Integrated into transmission system

- Breaker(s) are tripped by transmission line/bus/transformer protection.

Smaller DGs do require specific interconnection protection

3To Utility SystemTo Utility System

Generator vs. Interconnection Protection


n Utility SystemDG System

InterconnectionRelay

InterconnectionTransformer

Utility System

DG System

Local Loads Local Loads

Generator Interconnection

Dispersed Generator Interconnection Protection Areas Detection of loss of parallel operation with utility Fault backfeed detection

Outline


Detection of damaging system conditions Abnormal power flow Restoration

Tripping, Automatic Restoration and Automatic Reclosing on Utility System Trip of generator or main incoming breaker to separate from utilityhow to

decide Auto restoration strategies Utility auto reclosinghow to protect your generator

Use of Digital Technology for DG Interconnection/Generator Protection Advantages of the technology User-selectable functionality Self-diagnostics Communications capability Oscillographic capability Software demo

4DG Interconnection Protection

Impact of Smart Grids on DG

Enhance use of communication networks from protection and pdata acquisition

Peer to Peer as well as DG site to central location communications.

MircoGrids made possible by enhance communication network.Green Power and it ascending role in DG projection that are interconnected to the utility at the distribution level.y

Until Public Utility Regulatory Policies Act (PURPA) in 1978, U.S. utilities were not required to interconnect with small generators.

- Started DGB k ith t i t th i t ti t ti b i

Brief DG History


n

- Beckwith gets into the interconnection protection business- Hot until late 1980s when tax incentive terminated

Late 1990s DG again hot- Driven by high utility rates and de-regulation- DGs can generator cheaper at source of consumption

+ Peak Shaving and Load Following- Hot until early 2000s when natural gas prices increased

Late 2000s Green Power drives resurgence of DGs- Regulates require utilities to generator a portion of their

power from green sources.

5 Major Reason Cost of utility power is very high in many parts of the U.S. due to:

Deregulation problems (CA)


Why is DG Becoming Popular Again?

n

Deregulation problems (CA) Settlement of stranded investment (East Coast-Midwest)

DGs Can Generate Power Cheaper Than Utility Power From A Central Plant Use of waste heat (CHP) enhances DG economics

New Technologies Are Being Tried Microturbines Diesels powered from synthetic fuel Methane (from landfills) powered gas turbines Fuel Cells

PV PV Wind

Federal and State Government Push for Renewable Resources(Green Power)

Many Utilites are Receptive to DGs That Can Provide Power During System Peak Load

UTILITY CONCERNS:

Protection of the system fromDG OWNER CONCERNS:

Protection of the generator from

Protection Objectives


Protection of the system from damage

System Short Circuit Damage:loss of coordination, customer outages

Transient Overvoltage:during islanding, transformer connections, ferroresonance

Protection of the generator from damage Automated Reclosing:

shaft to torque on generator

Transient Overvoltage:also affects generator

Power Quality:during islanding voltage and harmonics

Power Quality:affects load served from generator bus

6Projection: A significant amount new generating capacity maybe be supplied by none utiltiy Generation in the next decade and beyond.This new capacity will take two forms:


n

This new capacity will take two forms:

Large capacity plants owned & operated by non-utility entities (IPPs and NUGs)- Gas turbines & combined cycle plants- Typically greater than 10MW- Connected into utilitys transmission and subtransmission systems Protection integrated into transmission and subtransmission systems

Small dispersed generating units - Connected into utilitys distribution system- Typically less than 10MW Installations require DG interconnection protection

IEEE PSRC WG Report- Intertie Protection of Consumer-Owned Sources of Generation 3

MVA L

DG Standards History


MVA or Less

First guideline addressing DG

Published in 1985

Good work and full of application information

7 Std 1001 from IEEE SCC23- IEEE Guide for Interfacing Dispersed Storage and Generation

DG Standards History


Facilities with Utility Systems

First standard addressing DG

Published in 1988

Withdrawn and replaced by IEEE 1547 series of standards, still a good work and full of application informationstill a good work and full of application information

Standard 929- Focus: Small inverter based systems sourced from

Recent History & Today


- Focus: Small inverter based systems sourced fromPV, Fuel Cells, Microturbines

IEEE 1547 and its Follow-up Standards- Focus:Produce universalDG interconnection

protection document to be used as a minimumtechnical requirement

Can be purchased from IEEE

8IEEE 1547 Says DGs Shall :

Not Cause Overvoltages or Loss of Utility Relay Coordination


Disconnect When No Longer Operating in Parallel With the Utility.

+ Only Discusses 81O/U and 27, 59

Not Energize the Utility when it is De-energized Not Create an Unintentional Islands Use Utility Grade Relays Not Cause Objectionable Harmonics Not Cause Loss of Synchronism That Results in

Objectionable Flicker

Impact of size- Intended to cover up to 10MW

IEEE 1547: Addressed Areas


Impact of Islanding- Creation of unintentional islands must be detected and

eliminated as fast as possible

Local Disturbances- Quality of service on the utility system should not be degraded

9 1547.1: Standard for Conformance Test Procedures forEquipment Interconnecting DR with EPS

More IEEE 1547


- Scope: Specifies the Type, Production and Commissioningtests that shall be performed to demonstrate thatinterconnection functions and equipment of a DR conform toIEEE 1547

1547.2: Application Guide for IEEE Std. 1547 forInterconnecting DR with EPSg

- Scope: Provides technical background and applicationdetails to support understanding for IEEE 1547 Standard forInterconnecting DR with EPS

IEEE 1547.3: Draft Guide for Monitoring, InformationExchange and Control of DR Interconnected with EPS

More IEEE 1547


Other 1547.4: Address DG Installation on Low Votage Networks both Spot and Street Neworks

- Scope: Facilities interoperability of one or more DRsinterconnected with EPS

10

Complete List of IEEE 1547 Standards


IEEE Std 1547TM Standard for Distributed Resources Interconnected with Electric Power Systems

IEEE Std 1547 1TM Standard for Conformance Test Procedures for EquipmentIEEE Std 1547.1TM Standard for Conformance Test Procedures for Equipment Interconnecting

Distributed Resources with Electric Power Systems

IEEE 1547.2TM Application Guide for IEEE Standard 1547, Interconnecting Distributed Resources

with Electric Power Systems

IEEE 1547.3TM Guide for Monitoring, Information Exchange, and Control of Distributed Resources

Interconnected with Electric Power Systemsy

IEEE P1547.4 Draft Guide for Design, Operation, and Integration of Distributed Resource Island

Systems with Electric Power Systems

IEEE P1547.5 Draft Technical Guidelines for Interconnection of Electric Power Sources Greater Than

10 MVA to the Power Transmission Grid

IEEE P1547.6 Draft Recommended Practice for Interconnecting Distributed Resources with Electric

Low-cost natural gas

Factors that will accelerate DG installationsare:


- That has changing for the worse and de-accelerated DG in early 2000s ---- Now prices have gone down.

High electricity rates

The advance of CHP as driver

New technology : microturbines and fuel cells. Microturbines are pretty much a dead technology.

Unreliability of utility supply to critical process and loads

The Northeast blackout of August 2003 and Calif. Rotating Blackout.

Government Mandated Green Power High Buy Back and/or Mandating a Percent of Future Power be Green.

11

Consumer Shaving Strategies


n

Peak ShavingNo electrical export

Load Following


n

Load

Generation

Morning Midday Night

12

Federal and State Governments Push for Renewable Resources. Green Power Is In.

Why Green PowerDG Interconnection Protection

n

Two Basic Strategies:1. PUC Mandate That a Percentage of Generation is Green by a Given Date. This Typically Fosters Installation of Large Blocks of Green Energy Installation such as Wind Farms .

2. Increase the Buy Back Rate and Let Market Forces (TypicallyIPPs) Install Green Generation This Typically Fosters SmallerIPPs) Install Green Generation. This Typically Fosters Smaller

Generators Installed on Distribution Systems.

Technological Advances have Reduced Green Power Costs.

Not Green- Burn conventional fuel

DG: Green or Not Green


Gas Diesel, oil, gasoline

Green- Use renewable sources to reduce reliance of fossil fuels: Solar Methane (from decomposition) Wind Hydro Diesels Powered by Syn Fuel Biomass (burn it) Tidal

13

Loads Loads LoadsUtility Substation

Islanded Operation of DG with Utility Loadis Generally Not Allowed


n

Loads LoadsLoads Loads Loads


LoadsFeeder Island

Greatly complicates restoration- Requires synchronizing at utility substation- Inhibits automatic reclosing

Power quality issue- DG may not be able to maintain voltage, frequency and harmonics within

acceptable levels (load generation; no harmonic sink)

LoadsDG

If DG creates a feeder island,reclosing requires synchronizing at the uti l i tysubstation

Loads Loads LoadsUtility Substation

DG Facility Islanding to the Utility is Allowed


n



Feeder deenergizes when utility opens feeder Restoration responsibility on the DG

- Requires synchronizing to utility- Inhibits automatic reclosing

Loads

LoadsDG

DG can create its own island,and synchronize to the utility

DG Island

14

Smart Grids


Full Integration of all Components of the Distribution System Through By-Directional Communication

n

Through By-Directional Communication.

+ Peer to Peer Relay Communication

+ Adaptive Relaying

+ Full Control of Load

+ Energy Storage

+ MircoGrid Operation System During Contingencies

Smart Grids --- MircoGrids


15

Smart Grids --- MircoGrids


DG Interconnection ProtectionHigh Penetration of DG on Distribution SystemsWill Require Smart Grid Technology to Control

System Voltage

n

Voltage control with high levels of DG require some type ofof DG require some type of adaptive watt/VAR control

16

Seamless integration of DGs into the utility protection system despite:

Challenges for the Protection EngineerApplying Todays Technology


- Too many cooks in the kitchen Owner, consultant, packager, utility

- Ownership boundaries

- Conflicting objectives of DG owners vs. Utility Want everything for nothing Want to pay for nothing

Making sure protection is correct and operational over the life of the installation- Settings are properly developed

- If system or installation changes, assess the impact on the existing protection

Loss of parallel operation (anti-islanding)- Voltage and frequency (27, 59, 81-U, 81-O) ; power (32F, 32R-U);

Interconnection Protection: Core Elements


instantaneous overvoltage (59I) , rate-of-frequency change(81R),Transfer Trip

- Based on load (real and reactive) not equaling generation Except 59I, which is based on self-excitation of resonant

induction and synchronous generators with pole top capacitors

Fault backfeed detection- Ground over/under voltage (27N, 27N/59N), ungrounded systems- Phase and ground overcurrent (51V, 51N), grounded systems

Directional overcurrent (67, 67N) and impedance (21) may beused

- Negative sequence overcurrent (46), detects ground faults- All based on sensing abnormally high current or

abnormally low/high voltage as a result of faults

17

Detection of damaging system conditions- Open phase condition or load imbalance (46, 47), negative

Interconnection Protection: Core Elements


sequence current and voltage- Phase sequence reversal (47), negative sequence voltage- Loss of synchronism (78)- Based on severe quantity imbalance (including reverse phase

rotation) or power system and DG going out of step

Abnormal power flow detection- DG Export (32F)

L DG I t (32R U)- Low DG Import (32R-U)- Transformer magnetizing (32F, very sensitive)- These strategies are used when the DG is sized so it cannot

physically or contractually export (peak shaver)

Facilitat

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