, MASTER OF SCIENCE · 2 2 2 2A In 0;::61 M •r-I ••·» ::1 I $@3 1 §¤¤° cr uan mg

I

} \A COMPUTER—AIDED FAULT ANALYSTS OF THE. VET ELESTRTC SERVTSE 1>0wER DISTRIBÜTIÖN SYSTEM}

by

George“Hogshead„ M

Thesis submitted to the Graduate Facultyof the Virginia Polytechnic Institute and State University

_ in partial fulfillment of the requirements for the degree of

, MASTER OF SCIENCEin

Electrical Engineering

APPROVED:

gs? ;? * ’L. L. Grigsby, C airman‘ /; g«» i° Ä« ‘

- „QG.

R. Powley 7 M. H. Hopkins

June l977

Blacksburg; VirginiaIII

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Table of Contents

6 EsseAcknowledgements6666 6 6 6 6 6 6 6 6 666666.666 6 6 6 6

iiListof Figures 6 6 6 6 6 6 666666 6 6 6 6 6 6 6 666666 iv

List of Tables666666 6 6666666 6 6 6 6 6 6„. 6 6 6 6 6 iv

I6 INTRODUCTION AND GENERAL DISCUSSION OF THE TOPIC.6666 6 6 lII.”

DESCRIPTION OF THE VPI SYSTEM 6666666. 6 6.66666 2

III. DESCRIPTION OF PROGRAM. 6 6 .....6...6.6..66. 12

3.1 Types of Circuits Programmed 66...... 6 . . 6 . . 12

3.2 Solution Procedure .... 6 .6... 6 . . . 6 . 6 . . 13

3.3 User's Guide .................. 6 6 . . 15

3.4 Data Deck.... . .....6..........6.. 22IV. CONCLUSIONS .... 6 .......... 6 6 .6.... 6 . 30

REFERENCES...6 6 6 . . 6 ........ 6 . . 6 6 . . . 6 . . 32

APPENDIX A........ 6 6 ........... 6 . 6 . . 6 . 33

A. List of Symbols Used in the Program 6.6... 6 6 . 6 . 33

B. Equations Used. 6 ..6. 6 . 6 . 6 .... 6 . . . 6 6 6 36

· ‘ APPENDIX B.6.. 6 . 6 6 6 6 ...... 6 . . 6 6 6 6 .6.. 6 41

A1>1>E1~m1xc.....6..66..666.666 56v1·1tA66666..6..6666.6.66.666..6.6666205A1as'1:RAcr

iii

LIST OF FIGURES

Figure No. _Tit1e Page

1 0ne—Line Diagram of the VPI & SU Substation. 6 30

2 VPI Substation Positive and Negative SequenceDistribution Factors (Percent Impedances_on10 • • • • • 6 • • • • •W-• 6 • • 6 6 5

3 VPI Substation Zero Sequence DistributionFactors (Percent Impedances on 10 MVA Base)"IHfiI°lit€ Bus" 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

4a 8-Ft. Crossarm 6 6 6 6 6 6 6 6 .... 6 6 6 6 8

4b 5-Ft. 7-ln. Crossarm 6 6 6 6 6 6 6 6 6 6 6 6 6 8

4c l0"Ft6 CYOSSBYHI6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 9

4d Bundled Support....666 6 6 6 6 6 6 , , , 9

4e· Steel Conductor Supports 66666.666.6 10

5 Distribution System Fault Study Program Flow6 • • • 6 6 • • • n • • 6 • • • • • 6•6

Illustration of Point Selection Procedure6 6 6 21

LIST OF TABLES

Table No. Title Page

· 1 Positive and Zero Sequence SubstationIIHp€d3IIC€S666666666666666666 7

_ 2 Maximum Three Phase Fault Currents and 5 CycleBreaker Ratings for the 4KV and 12 KV Buses6 6 31

Liv

I. INTRODUCTION AND GENERAL DISCUSSION OF THE TOPICI

The magnitudes of fault currents that can exist on a distribution

circuit are important quantities to the distribution engineer. One

purpose of a fault study of a distribution system is to determine if

circuit breakers can successfully interrupt the fault current without

sustaining equipment damage. Another purpose is to coordinate fuses and

.relay settings so that a fault causes a power outage for the minimumI

number of customers.i

Other studies can be performed using information calculated duringl

the fault study. By knowing the average power at points along the

circuit, the impedances found in the fault study can be used to find the

voltage profile of a particular circuit.i A

Therefore, it can be seen that a fault study is an important tool

to the distribution engineer. Not only does it give him fault currentl

values, the impedance formulation section of the study can be used as al

building block for other important distribution related studies.

In this thesis, the results of an effort to develop a fault analy-

· sis program for distribution systems are presented. Although the

resulting digital computer program is of a general-purpose nature, itl

is particularly suited to the needs of the VPI Electric Service.

A complete report follows which includes a description of the topic,

general information about the VPI system, details of the computer pro- I

gram, a complete fault analysis and a discussion of the results. A I

• detailed user°s guide is also included to aid the reader in using the

program to perform his own fault studies. E

1 ÜEi

E•( i

II. DESCRIPTION OF THE VPI SYSTEM

The VPI Electric Service distribution system supplies power to the

VPI & SU campus and portions of the town of Blacksburg. The maximum _

demand on the system was recorded on January 17, 1977, and was approxi-

mately 29 MW. Under normal operating conditions, the two generators in

the power plant can supply approximately 25% of the demand, while the _

remainder is purchased from Appalachian Power Company.

_ A schematic diagram of the substation is shown in Fig. 1. The

4 KV bus with the Westinghouse generator tied to it is known as the in-

door bus because it is located inside the power plant. Breaker PP

serves the power plant, and Breakers 1-5, on the indoor bus, connect

campus buildings to the bus via underground cables. There are thirteen

circuits fed by these five breakers and each building has at least two

circuits fed by different breakers. Having multiple circuits in each

building reduces the time the building is without power if something

goes wrong, i.e. circuit breaker failure of a break in the circuit,

because the building can be switched to a healthy circuit.

· Circuits 1A, 1B, ZA and ZB on the 12 KV bus are also underground

circuits feeding campus buildings. Circuits 1A and lB serve a certain

group of buildings, with 1A and 1B both present in each of these build-

ings. The position of the 1A and 1B breakers on opposite sides of theZ

automatic tie breaker (ATB) would allow loss of one—half of the 12 KV

bus without prolonged loss of power to the buildings served by lA and1

1B. The same is true for buildings served by circuits ZA and ZB.

Circuits 1, Z, 4, 5 and 6 on the 4 KV bus are overhead circuits

2

3 IB 12 KV

30 A BUSTO FBNORTH EBLACKSBURG .

69KV OCBB 30 MVA I 1B

4KVTO BU

5 MVA „

—/ 20r I I ZBI

: I 11I

WESTINGHOUSE TIE IGENERATOR _ _ _ _J I

6 I I5 I 5 MVA I 1**

16PP l l 81 2. I A2 _Ü 5 MVA •3 I I 7

4 8 ‘•I

5 2 1

]

_

B _- — _\ WORTHINGTON ‘ I

EAST TIE GENERATOR E‘

Ä Fig. 1 One—-Line Diagram of the VPI & SU SubstationB

I

· I

I

4

serving the town of Blacksburg, as well as circuits 7, 8, ll and 12 on

the 12 KV bus. Switching is also available on these circuits to restore

power to a circuit with a defective breaker from another circuit so that

minimum power outage can be maintained. The general configuration of

the overhead circuits, under normal operation is a radial one.

The source impedances (generator and substation) were calculated

for the positive and zero sequences from Figures 2 and 3. The positive

and zero sequence impedances for the 4 KV and 12 KV buses are given in

Table l. These impedances were found when all three VPI 5 MVA trans-

formers were operating simultaneously. This particular arrangement was

found to give the minimum source impedances which result in maximum

fault currents at the substation. The maximum fault currents at the

buses were desired to properly determine the MVA capacity of the circuit

breakers so that interruption of fault currents is insured. IThe overhead circuits in the VPI system are of five basic geometri-

cal configurations (Figures 4a, 4b, 4c, 4d and 4e). Figure 4a, an

eight-foot crossarm, is the standard being used for the construction of

' new lines. Figures 4b and 4c are remains from the older portion of the

VPI system with 4c being used to carry two circuits on the same cross-

arm. Figure 4d is a bundled configuration which is used only twice in4

the system; Figure 4e is used in only four locations. ' ”

The geometrical configuration of overhead circuits is important in

fault studies due to the mutual impedances that exist between conductors

that are in close proximity. Mutuals also exist between underground I

cables; the concentric neutral cables which exist in the town distribu-

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Table 1. Positive and Zero SequenceSubstation Impedances (10 MVABase)„

‘ uuuuuuuuuuuuuuuuuunuunnnuuuuuuuPositive sequence Zero sequence

im•edance impedance _

4 KV bus j6.8l28Z j2.l000Z

12 KV bus j3.20468Z j2.l747Z

{6-———————— 7.33* -———-9<§--—— 3.67' --——;>

A B C

6.2*5*·•N

Fig. ha 8-Ft. Cressarm

- <-——-—————— Z+.62' ——>1A6- 2-31 —-äß AC .

S, 6.61* 3 · ·

¤

N IFig. Ab 5-Ft. 7-In. Crossarm :

Y 9

<——-———— 3.42°•-ä1.71* A A ÄB c

S!

\ 6.16*NFig. 4c 10-Ft. Crossarm

/ B.78l° „1.18* A C · Yl .66*

N<——-—- .68* —-9 ·Fig. bd Bundled Support

10

B

z.s• I CC A

¢„.:·„a· z.s•I

ll N

2.5° —-%

Fig. be Steel Couductor Supports p

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have the mutuals calculated, but the shielded cables on campus do

not. This is due to the fact that the shields are assumed to contain- the fields produced by currents flowing in the conductors; therefore,

the mutuals for shielded cables are zero; —This should give a good indication of the general layout of the

VPI system and will serve as an introduction to a description of the

computer program used to solve for the fault currents.

III. DESCRIPTION or PROGRAM_ The purpose of this program is to calculate the values of fault

currents for three phase to ground, line to line, line to line to

ground, and line to ground faults on a power distribution system. The

program was written in a general manner so that major circuit types in

use today can be handled.

3.1 Types of Circuits Programmed4

The circuit types included in the program are:

V 1) Three phase Y, 4-wire (three identical phase conductors, one °

multigrounded neutral conductor). I

2) Three phase Y, 4-wire (three identical phase conductors, one

unigrounded neutral conductor).

3) Three phase Y, 3—wire (three identical phase conductors, neu-

tral grounded at substation).

4) Three phase delta (3 identical phase conductors).1

5) Two phase open Y, 3 wire (two identical phase conductors, one

multigrounded neutral conductor).

·I

6) Two phase open Y, 3 wire (two identical phase conductors, oneA

unigrounded neutral conductor).

7) Single phase, 2 wire (two identical phase conductors, neutral4

grounded at substation).

8) Single phase, 2 wire (two identical phase conductors, no

ground).

9) Single phase, 2 wire (one phase conductor, one multigrounded Eneutral conductor). :·12I

13 E

10) Single phase, 2 wire (one phase conductor, one unigrounded

I neutral conductor).

11) Three phase underground (three phase conductors, each sur-

rounded by a concentric neutral conductor).

12) Single phase underground (one phase conductor surrounded by a

concentric neutral conductor).

13) Three phase underground (three phase conductors, each surround-

ed by a shield, and a neutral conductor pulled in the conduits

1 with the phase conductors; neutral and shields multigrounded).

It should be noted here that an approximation was made in circuit type

13. The impedance of the shields are so large compared to the impe-

dance of the neutral conductor that when they are paralleled between

grounding points, the impedance of the shields can be neglected. The

terms rs and xs in the underground shielded equations are simply the

resistance and reactance of the neutral conductor (see Appendix A).

A3.2 Solution Procedure

e The solution procedure of the program is as followsi Before the

study begins, the circuit(s) to be studied is divided into segments

(see useris guide for details). Once the segments have been chosen,

data pertaining to the particular circuit type of the segment is read

· and values of impedance of the segment (self and mutual) are calculated

and the positive and zero sequence impedances of the segment are calcu-

lated in ohms per 1000 feet. This process is repeated for each segment

in the circuit. Z

Once these impedances have been calculated, they are multiplied by

i

14 ’ q_·

the corresponding segment lengths so that the positive and zero sequence

impedances are now in terms of ohms. Having these values, a radial line

search is performed to connect the proper segments together so that the

total positive and zero impedances can be calculated from each point

back to the substation. The radial line search works by starting atthe ending node of the first line segment that is connected to the sub-

station. A search is performed to find a segment whose starting node

is the same as the ending node of the first segment. When one is found,

then the total impedance at the end of this second segment is the total

impedance of the first segment plus the impedance of the second segment.

This process continues until all segments have had total impedances cal-

culated at their ending nodes.

Having these total impedances at each ending node, the three-phase

to ground fault current can be calculated at each point by

IFBPH = (3.2.1) ‘

where VN is the line to ground voltage of the circuit.

· ZSl is the positive sequence impedance of the substation.

ZP1 is the positive sequence impedance from the particular point

to the substation.

ZF is the fault impedance, i.e., not a direct short to ground (a

tree touching a line).

(Note: Subscripts were not used in the thesis text or in the appendicesA

due to the fact that in computer printout, subscripts do not exist.

_ Therefore, all variable names refer to the form in which they appear in

the program.)

15

Currents for the remaining three types of faults are calculated

using symmetrical components. The formulations are given in Appendix

A. The maximum value for each of the four different fault currents atthe ehosen points occur when ZF = O. The minimum fault current is

determined by the value of ZF selected by the user. The program will

automatically calculate the maximum and minimum fault currents at each

y point for each of the four fault conditions.

Similar equations exist for the other types of faults and once the' fault currents are calculated at each point, the MVA capacity of the

circuit breaker on that line is calculated.

Although a great deal more logic was needed to perform the calcula-

tions, this is the general idea of the program. More specific details

about preparing data to be used in the program will be presented in the

user's guide, which follows a flow chart of the program.

3.3 User's Guide

This program is really quite easy to use. First, the individual

circuits to be studied must be selected. The next step is to divide the

circuit into line segments by the following method:

(1) A point must be declared whenever T

a) Conductor sizes change. ·b) Spacing changes, either between conductors and neutral or

between conductors/neutral and ground.

c) The number of phases changes; if the fault currents are

desired beyond these points.

(2) A point must be declared at branches if the fault currents are

A 16

START

READ, DATE, FREQUENCY _

soo . READ, CKTNO, NOSECTes STOPNo

DO 7 I=1, NOSECT

READ, SEGNO(I), CIRTYP(I), LENGTH(I),

‘ o to 116 .No

YES; go to 115N No Yes; go to 150

I· No °

READ PHASE CONDUCTOR DATA

ISY

, _ «CIRTYP=3’h’ Yes, go to 19,

7 or 89 _ NNo

Fig. 5 Distribution System Fault Study Program Flow Chart

17 II

I

READ NEUTRAL CONDUCTOR DATA

19 c0NT1NUE

CALCULATE APPROPRIATE GEOMETRYCONFIGURATION, CONDUCTORl IMEDANCE AND MUTUAL IMEDANCES

WRITE CONDUCTOR DATA

7GO TO 7

115 READ UNDERGROUND CONCENTRIC NEUTRALCONDUCTOR DATA (SINGLE PHASE)

114 CALCULATE UNDERGROUND CABLE IMP.

° WRITE UNDERGROUND CABLE DATA

GO TO 7

· 116 READ UNDERGROUND CONCENTRIC NEUTRAL _CONDUCTOR DATA (THREE PHASE)

GO TO 114

150 READ UNDERGROUND SHIELDEDCONDUCTOR DATA (THREE PHASE)

CALCULATE CABLE IMTEDANCE

(Fig. 5 cont.) l

E 18 Z

WRITE CABLE DATA

7 comrmußZ

IS _ 6 .CIRTYP=ll Yes, go to 140•R 12?

„ No

Yes; 0 to 151N0

CALCULATE APPROPRIATE POSITIVEAND ZERO SEQUENCE IMPEDANCES

92 WRITE SEGNO AND IMPEDANCES(OHMS PER 1000 FEET)

GO TO 89

151 WRITE SEGNO AND SHIELDED CABLEIMPEDANCES (OHMS PER 1000 FEET)

Z GO TO 89

140 CALCULATE IMTEDANCES OF CONCENTRIC _NEUTRAL CABLES (OHMS PER 1000 FEET)

GO TO 92

89 CONTINUE(Fig.

5 cont.) E

19

CALCULATE ALL SEGMENTIMPEDANCES IN OHMS M

WRITE IMPEDANCES (OHMS) _

READ, VN, SBASE, ZSl, ZSO, ZN

CALCULATE FAULT CURRENTSAT SUBSTATION

PERFORM RADIAL SEARCH ROUTINETO FIND TOTAL POSITIVE AND

ZERO SEQUENCE IMPEDANCES FROMEACH POINT TO SUBSTATION

V SOLVE FOR FAULT CURRENTSAT EACH POINT

CALCULATE BREAKER MVAFOR THE CIRCUIT _

GO TO 500

·(Fig‘ 5 finished) ‘

20|

desired at more than one of the branches.

— (3) A point may be declared anywhere the fault currents are desired.

(4) Point zero is always the substation.·

For example, in Fig. 6, if the currents are desired at point A, then .

point B must also be declared a point for two reasons; (1) the number

of phases has changed; (2) the conductor size and geometry has changed.

l If the currents at A are not desired, then point B does not have to be

declared a point in order to find the currents farther down the line be-

cause geometries, wire sizes and phases are the same from point 0 to

point C (the spacing is assumed to be the same from 0 to C).

If the currents at D are wanted, point C must be declared for the

same reasons as above. If the currents at D are not required, C need

not be declared. However, if the currents at point H are desired and

currents at D, E, F and G are not required, point C does not have to be

declared because no changes have taken place from point 0 to H. The

remaining segments in Fig. 6 can be reasoned out in the same manner.

Once the points have been selected, according to the above rules,

. they must be numbered. However, they need not be numbered in any par-

ticular order; but in this study, for convenience, the points have been

numbered consecutively in the overhead circuits. Furthermore, the seg-

ments should be numbered consecutively, beginning at the·substation and

numbering sequentially toward the end of the line, so that the last seg-

ment number in the circuit is equal to the total number of segments in |

the circuit. E

21

3 — #2 ALG D„¢

·

. 4 — 2 —~ 0 4-2/OALB 4-2/OAL /0ALG4 #2ALE .3¢ AA c 3¢ 3¢

2‘#2^L 2-//2ALl¢ Fl¢

A A — 2/0 AL H

Fig. 6 Illustration of Point Selection Procedure

22

3.4 The Data Deck

To begin the data deck, the first card required contains the

following data: the month (DATE1), the day (DATE2), the year (DATE3)

and the system frequency (f). The data card for June 5, 1977, with a

system frequency of 60 HZ would be H

6 5 77 60 (A)

It should be noted here that all data is read in an unformatted manner

so that data elements do not have to appear in a certain location on the

data card. Individual values need to be separated by a space or a

comma in order for them to be read.

The next data card begins the data set required for each circuit

studied. This card contains the circuit number (CKTNO) and the number

of line segments that the line has been divided into (NOSECT). For a

circuit number 5 with 18 segments the data card would be

5 18 (B)

· The next card required contains the segment number (SEGNO), circuit type V

' (CIRTYP), length of the segment (LENGTH), starting node of the segment

(SB), and the ending node of the segment (EN). The CIRTYP refers to the

list of conductor configurations presented in section 3.1. For example,

let segment number 15 be a three phase multigrounded configuration

(CIRTYP = 1), 300 feet long, starting at point 6 and ending at point 7.

The data card describing this data would be

15 1 300 6 7 .

23 II

Next, the properties of conductors in the segment are read• First, a

list must be defined for conductor sizes and conductor materials.

NSIZE(l) = 750 MCM MATRL(l) = AL

NSIZE(2) = 350 MCM MATRL(2) = CU

NS1ZE(3) = 336 MCM MATRL(3) = AL—UGU

N3IZE(4) = 6/0 MATRL(4) = CU-UGI

NSIZE(5) = 2/0 MATRL(5) = C-CN —J

NSIZE(6) = l/0 MATRL(6) = CSNI

NSIZE(7) = #1 MArRL(7) = c a csu

NSIZE(8) = #2

Ns1zE(9) = #6

NSIZE(l0) = #6

NSIZE(ll) - #10

NSIZE(12) = #12

NSIZE(l3) = 3/O I

The conductor sizes are self·explanatory1 The conductor material

abbreviations are defined in this manner:‘ AL — Aluminum

CU - CopperAL—UG

~ Aluminum underground

CU—UG — Copper underground‘

C-CN - Copper, concentric neutral

CSN — Copper shielded neutral I

C & CSN — Copper conductor and a copper shielded neutral I

The next data card read is for the phase conductors. The information I

I

I

24 :

Icontained is: conductor type (CNDTYP), material (MAT), resistance (R), A

geometrie mean radius (GMR), diameter (D), height of phase A above the

ground (HC), B above the ground (HB), C above the ground (HC), distance

from Aeto B (HHAB), distance from A to C (HHAC), and RHO (RHO), the

resistivity of the earth. For a 336 MCM aluminum phase conductor, R =.306 ohms/mi, GMR = .0255 ft., D = .741 in., HA = 30 ft., HB = 30 ft.,

HC * 30 ft., HHAB = 3.67 ft., HHAC = 7.33 ft., RHO = 100 meter-ohms,

the data card would be q3 l .306 .0255 .741 30 30 30 3.67 7.33 100 (D)

where the 3 and the 1 come from the list of the NSIZEs and MATRLS. A

If the same conductor had been used except two phase, the data card

would be

3 1 .306 .0255 .741 30 30 0 7.33 0 100A

since there is no phase C, HC = 0 and HHAC = 0.

Similarly, if the same conductor had been used, except single phase,

the data card would be

3 l .306 .0255 .741 30 0 0 0 0 100

since now HB, HC, HHAB, HHAC = 0.·

If the CIRTYP equals 3, 4, 7 or 8, this is all the data that is

required for that segment since no neutral wire is present. However, Efor CIRTYP's 1, 2, 5, 6, 9 and 10, a data card for a neutral conductor Ais required. For a neutral conductor, the data card contains the follow- A

1ng information: conductor type of the neutral conductor (CNDTYN), A

25

hneutral conductor material (MATN), resistance of the neutral (RN), GMR

of the neutral (GMRN), diameter of the neutral (DN), height of the neu—

tral (HN), distance from phase A to the neutral (HHAN). A neutral con-

duetor of 4/0 copper, RN f .303 ohms/mi, GMRN = .01579 ft., DN = .522W

in., HN - 25 ft. and HHAN = 5.67 ft. would have the data card _

· 4 2 .303 .01579 .522 25 5.67 (E)l

This completes the data for a segment of CIRTYP 1, 2, 5, 6, 9 and 10.

If the CIRTYP = 11, 12 or 13, then the data for an underground lcable must be read. For a CIRTYP 11 (three phase UG), the following

information is read: conductor type (CNDTYP), conductor material (MAT),

· resistance (R), geometric mean radius (GMR), diameter (D), distance

from center of A to center of B (DAB), from B to C (DBC) and from C to

A (DCA), and RHO. Therefore, if the conductor is a 350 MCM AL—UG cable,

R ¤ .295 ohms/mi., GMR = .01983 ft., D = .611 in., DAB = .1133 ft., DBC

- .1133 ft., DCA = .1133 ft. and RHO = 100 meter—ohms, the data card

( would beI

· 2 3 .295 .01983 .611 .1133 .1133 .1133 100.0 (F)

If CIRTYP = 12 (single phase UG), the data is very similar: CNDTYP,

MAT, T, GMR, D, the diameter of the entire cable, including concentric

neutral (S), and RHO. A single phase underground concentric neutral

cable with the following data--2/0 AL, R = .771 ohms/mi., GMR = .01317

ft., D = .406 in., S = .0883 ft., RH0 = 100 meter—ohms-—wou1d havethefollowingdata card:

5 3 .771 .01317 .406 .0883 100 (G)

l I

26l

The data card for the concentric neutral of both CIRTYP ll and 12 is the

same. The data contained is conductor size of one strand of the con-

centric neutral (CNDTYN), conductor material (MATN), resistance of one

strand of the concentric neutral (RN), GMRN of one strand of the con-

centric neutral, diameter of the concentric neutral and number ofstrands in the concentric neutral (NN). For a concentric neutral of

#12 copper, RN = 8.849 ohms/mi., GMN ¤ .00262 ft., DEE = 1.06 in., NN

. ¤ 13 strands, the data card would be

12 5 8.849 .00262 1.06 13 (H)

For a CIRTYP 13 (three phase shielded underground), the following

information is required: conductor type (CNDTYP), conductor material

(MAT), resistance (R), geometric mean radius (GMR), diameter (D) and

RH0 (RHO). Therefore, for a 350 MCM shielded aluminum conductor, R =

.295 ohms/mi., GMR = .01983 ft., D = .611 ft., RH0 = 100 meter-ohms,

the data card would be

2 3 .295 .01983 .611 100 (I)I

lThe data read for the neutral conductor pulled in the conduit with the

three phase conductors contains the following information: conductor

size (CNDTYN), conductor material (MATN), resistance (RN), geometric

mean radius (GMRN), and diameter (DN). Therefore, the neutral conductor

' accompanying a shielded underground cable, having neutral conductor

data 4/0 copper, RN = .303 ohms/mi., GMRN = .01519 ft., DN = .522 in.,

would have a data card as follows:

2 7 .303 .01579 .522 (J)

27

The MATN = 7 comes from the fact that the neutral is copper pulled in

with shielded cable having a copper shield (C & CSN).

So, in conclusion, one card of type (A) is needed to start the

data deck. Then, for each line studied, one card of type (B) is next.

· For each line segment, the following cards are_needed:

Type (C) card

1£ c1RTYP = 1 ~ 10Type (D) card

If CIRTYP = 1, 2, 5, 6, 9, 10

Type (E) card

or _

1f c1RTYP = lll Type (F) card

Type (H) card

or

‘ If CIRTYP = 12

Type (G) card

Type (H) card

or

(IfCIRTYP_= 13 .

Type (I) card

Type (J) card VB

28 .

After an entire group of line segments has been read, one more card

must be read. The information contained is: line to neutral voltage”

I of the circuit (VN), MVA base (SBASE), positive sequence source impe-

dance (ZS1), zero sequence source impedance (ZSO) and the fault impe-

I dance (ZN). For a VN = 2400 volts, SBASE = 10 MVA, ZS1 = j.11773, ZS0

= j.03629, ZN = 10 ohms, the data card isA

2400 10 (0.0, 0.11773) (0.0, 0.03629) 10

After all lines have been studied, a final data card is read. This

card contains the data3

0 0

This card is simply a control card which signals the program that all

_ lines have been studied and the program stops.A

A simple example deck to illustrate the data deck follows:·” U

6 5 77 60 A

11 2

1 1 1 980 0 1

· 3 1 .306 .0255 .741 30 30 30 3.67 7.33 100

3 1 .306 .0255 .741 26.33 5.17 A

2 12 100 1 2 _

6 3 .973 .01175 .362 .085 100 gr.”

12 5 8.849 .00262 1.02 10 if2400 10 (0.0, 0.11773) (0.0, 0.03629) 10 I0

OAThisis a complete data deck for circuit number ll containing two line

{Ölg·lt

I

29 '

segments, the first a three phase overhead line, the second a single

phase underground cable with concentric neutral. The data does not

appear in well defined columns due to the fact that unformatted read F

is used. However, data cards read in an unformatted manner are much _

easier to type and the above example deck is in the form of an actual

data deck.

It should be noted here that for overhead circuits on crossarm‘ types pictured in Figs. 4a, 4b and 4c, the conductor heights were

assumed to be 30 feet, with the neutral height 25 feet. For Fig. hd,

the neutral height is 30 feet with the other heights calculated

_ by the dimensions given in the figure. In Fig. he, phase B is 30 feet

high with the other conductor heights also calculated from the dimensions

given in the figure. All three phase underground circuits were assumed

to be spaced so that the three conductors formed an equilateral triangler'

It should be noted at this point that conductor data (resistances,

GMRs, etc.) were obtained from Reference 7 and from cable manufacturers°

catalogs.

· This concludes the user°s guide for this program. Results of

fault current calculations for the entire VPI Electric Service distri-”

bution system follow.

l IV. CONCLUSIONS

The program was tested by running the first eight segments of the

system which had been previously studied by Wylie, Auer and Ender [2]

and the results of the two studies differed by a maximum of .003%. This

would lead one to believe that the program written for this thesis is

accurate.

Some important results of the study are contained in Table 2.A

Since the Wiley and Wilson study was performed in 1972, one of the old'A

generators has been replaced, causing a new value of source impedance.

Also, Wiley and Wilson performed their study with only one of the VPI

5 MVA transformers operating, which is not the condition for maximum

fault current. As a result, the values of the fault currents at the

substation are considerably different from those obtained in the Wiley

8 Wilson study. However, the fault currents occurring on a circuit do

conform to the same pattern in both studies.

It is felt that the contribution of this thesis is in two areas.

First, an up-to—date system description has been provided for the VPI

° Electric Service distribution network. Secondly, a digital computer

program has been developed to perform general-purpose fault studies.

The program can be used for several additional distribution system

studies. Specifically, system voltage profiles can be performed readily.

By adding the storage of fuse characteristics to the program, a fuse-

relay coordination study is possible. Additionally, this program should

be useful in future system planning.A

ä

30

31

Table 2. Maximum Three Phase Fault Currentsand 5 Cycle Breaker Ratings for the 4 KV and 12 KV Buses

4 KV Bus 12 KV Bus

Maximum three phase current (amps) 20385.62 14448.26

Breaker MVA (5 cycle) 161.45 343.29

I

REFERENCES1. Clarke, Edith, Circuit Analysis of AC Power Systems, Vol. L & II,

_ John Wiley & Sons, New York, N.Y., 1943 and 1950.

2. Wylie, R. A., G. G. Auer and R. C. Ender, "Digital Calculation ofSequence Impedances and Fault Currents for Radial Primary Distri-

_ bution Circuits", AIEE Transactions, PT. III, Power Apparatus andSystems, Vol. 79, February 1961, pp. 1264-1277.

3. Electrical Transmission and Distribution Reference Book, WestinghouseElectric Corporation, Pittsburgh, PA., 4th Ed., 1964.

E4. Webb, R. L. and O. W. Manz, Jr., "Impedance Measurements on Under-

ground Cables", AIEE Transactions, Pt. III, Power Apparatus andSystems,‘Apri1, 1936, p. 359.

5. Wiseman, R. J., E. H. Salter and G. B. Shanklin, "Resistance and. Reactance of 3-conductor Cables", AIEE Transactions, Pt. III,

Power Apparatus and Systems, Vol. 53, December 1934, pp. 1581-1589.

6. Lewis, W. A. and P. D. Tuttle, "The Resistance and Reactance of’ - Aluminum Conductors, Steel Reinforced", AIEE Transactions, Pt. III,Power Apparatus and Systems, Vol. 77, pp. 1189-1215. .

7. Silent Sentinels, Westinghouse Protective Relays, Westinghouse Elec-. tric Corporation, Newark, N.J., 1949. °

8. Distribution System Protection Manual, McGraw-Edison Co., PowerSystems Div., Canonsburg, PA., Bulletin No. 71022, pp. 17-48.

9. Calabrese, G. O., Symmetrical Components, Ronald Press Co., New York,N.Y., 1950.

' 32

. " 1D 1· 1

y APPENDIX AV

A. List of Symbols Used in the ProgramAt

The following variable names were used in the computer program:

DATE 1 - Two~digit number representing the month of today's datel

(i.e. January = l) V

DATE 2 - Two-digit number representing the day of today°s date

„ (i.e. 27th = 27)

DATE 3 — Two—digit number representing the year of today°s date

(i.e. 1977 = 77)

f — System frequency (HZ)A

CKTN0 - Circuit number under study

DNOSECT — Number of segments in the circuit under study

SEGNO - Segment number

SB — Starting node of a particular segment

EN — Ending node of a particular segmentD

LENGTH — Length of a segment (ft)

LENTHO - Length of a segment (thousands of feet) 1 ·

· ZHOLDl — Positive sequence impedance being held at a pointa

ZHOLD0 - Zero sequence impedance being held at a point

NUMP — Dummy variable

•CIRTYP - Circuit type ‘

CNDTYP — Phase conductor type

MAT - Phase conductor material

R — Phase conductor resistance (ohms/mi)

GMP — Phase conductor geometric mean radius (ft)33

34 '

D — Phase conductor diameter (in)

HA - Height of phase above the ground (ft)

· HB - Height of phase B above the ground (ft)

HC - Height of phase C above the ground (ft)

HHAB - Distance from phase A to phase B (ft) gHHAC — Distance from phase A to phase C (ft)

RH0 - Resistivity of the earth (meter—ohms)

CNDTYN - Conductor type of neutral

MATN — Material of neutral conductor

RN — Resistance of neutral conductor (ohms/mi)

GMRN — Geometrie mean radius of neutral conductor (ft)

DN - Diameter of neutral conductor (in)

HN - Height of neutral conductor above the ground (ft)

HHAN — Distance from neutral conductor to phase A (ft)

H - Average height of a group of conductors (ft)

_ HH,HHN — Average horizontal spacing of a group of conductors (ft) _

$S,SSN — Average spacing between a conductor and the image of

A another (ft)

S,SN — Geometric mean spacing between a group of conducusrs

THETA — The angle formed between lines drawn from a conductor to

its own image and the image of another conductor (degrees)

THETAR — Theta in radians

X — Reactance of conductor (ohms/mi) A

RR — Real component of self impedance with earth return extern-

al to the conductor (ohms/mi)

35

RM - Real component of the mutual impedance (ohms/mi) _

XX - Imaginary component of self impedance (ohms/mi)ft

XM - lmaginary component of mutual impedance (ohms/mi)

ZZ - Complex self impedance (ohms/1000 ft)

ZM - Complex mutual impedanee (ohms/1000 ft);

S - Outside diameter of a single phase, concentric neutralS

cable (feet)l

DEE - Diameter of concentric neutral (in.)

„ NN - Number of strands in the concentric neutral

lZll — Positive sequence self impedance (ohms/1000 ft)

ZOOl - Zero sequence self impedance without neutral wire but with

presence of earth taken into account (ohms/1000 ft)

ZO02 - Zero sequence self impedance of a circuit with a uniground-

ed neutral (ohms/1000 ft)

Z003 — Zero sequence self impedance for a circuit with a multi-

grounded neutral

ZllT - Zll in ohms

· Z00lT - Z0Ol in ohms

ZOOZT - ZOOZ in ohms

Z003T - Z0O3 in ohms

VN - Line to neutral voltage (volts) ‘

· SBASE - MVA base (MVA)

zs1 — Positive sequence source impedance (ohms)

ZS0 - Zero sequence source impedance (ohms)

ZN - Real part of fault impedance (ohms)

I36 I

IV — Line to line voltage (volts)

ZBASE — Impedance base (ohms)I

. „ZF — Complex fault impedance (ohms)

ZS2 · Negative sequence source impedance (ohms)

IFBPHO — Three phase fault current at substation (amps)

IFLLG — Line to line fault current at substation (amps)

IFLLG0 - Line to line to ground fault current at substation (amps)

IFLG0 — Line to ground fault current at substation (amps)

ZTl - Positive sequence impedance at point I (ohms)

ZT0 — Zero sequence impedance at point I (ohms)

IF3PH - Three phase fault current (amps)

IFLL — Line to line fault current (amps)

IFLLG — Line to line to ground fault current (amps)

IFLG — Line to ground fault current (amps)

SCC - Short circuit capacity of breaker (MVA)

RAT2 — Rating of a 2-cycle breaker (MVA)

RAT4 - Rating of a 4—cycle breaker (MVA)

. RAT5 — Rating of a 5—cycle breaker (MVA)

‘ _B. Eguations Used

Overhead lines (reference 2)

Circuit type Phase conductorsl Neutral conductor1,2 S = (SAB*SAC*SBC)§ SN = (SAN*SBN*SCN)§

SS = %(SSAB+SSAC+SSBC) SSN=%(SSAN+SSBN+SSCN) IH =·%(HA+HB+HC)

1HN = HN I

ilrmcul) I

37

_Circgit tyge Phase conductors_ Neutral conductoru 3,4 same as 1,2 no neutral conductoä

1 5,6 S - SAB SN · (SAN*SBN7§SS = SSAB SSN =·%(SSAN+SSBN)

H =·%(HA+HB) Sw HN =~HN

. HH = IHHAHI HHH =-%(1HHAH|+|HHHH|)7,8 u same as 5,6 V no neutral conductor9,10 H = HA su = SAN

SSN = SSAN_ . HN = HNHHN = |HHAH|

HHBC = |HHAC—HHAB1

HHBN = 1HHAN—HHAB1 H

HHCN = 1HHAN—HHAC1

.1Sxy = [HHxy2 + (Hx — Hy)2]2

1SSxy = [HHxy2 + (Hx + Hy)2]2

lwhere xy = ab, ac, bc, an, bn or cn.

X = •2794(f/60) loglo EzzgüäyTHETA = sin-l(HH/SS) .KN S NN(l/(NN—l))

lRR = 0MEGA(2.s28 x 10-4 - 2.599 x 1O_7H(f/RH?)2 + 2.717 x

10-1OH2§f/RHO) x (3.36 + loglo ä(RHO/f)? + 5.084 x 10—l4H3(E/RH0)7”)

1

38

1R = 0MEGA(2.528 x 10*4 — 1.299 x 10—7(f/RH0)2SS cos THETAR +

6.785 x 10‘1l(f/RH0)SS2 eos 2 THETAR(3.66l + 1ogl0(é)(RHO/f)) ·

+ 2.951 X310'll(£/Ruo)s2TuETAR(s1X 2 THETAR) + 6.355 X 10*5x (f/RH0)2S3 cos 3 THETAR)

1XX — ·3 ~4 1 -2¤ omega(3.4944 x 10

I7.4113 x 10 1ogl0(EQ(RH0/f) +

2.599 x 10-7H(§/RH0)§ — 9.271 x 10—1lH2(f/RH0) + 5.084 x

1XM ¤ omega(2.4715lx 10-3 + 7.4113 x 10—4logl0(é)(RH0/f)2 + 1.299 x

10-7(f/RH0)E-S cos THETAR - 2.318 x 10_ll(f/RHO)S; cos 2

THETAR) ’

Each of the above four equations is used for the phase conductors and

for the neutral conductor.

ZZ = (R + RR) + j(X + XX)

ZM = RM + jXM

Zll = ZZ — ZM

Z001 = ZZ + 2ZM

Z002 = Z001 + 3ZZN — 6ZAN I

ZOO3 = Z001 — 3ZAN2/ZZN

1 ·

I

IFBPH 'Z l]:*ZS+ZFI2= .,.(§l_.V!L....IFLL lZ11+Z22+ZS+ZF

IFLG ‘Z0O3+Z11+Z22+ZS+3ZF‘ .

1

I

12 . . _Z003+3ZF+aZ..Z2 „ . -IFLLG IG) VN Z11Z22+ZS+(Z11+Z22)(2003+3ZF)I

a = 1/120

UNDERGROUND CABLE EQUATIONS (Referencé 8)

ZZAA== (11+4.788 x 10-5*2*PI*F) + j2*PI*f(4.681 x 10-4+6.096 x 10-51

1og€(1/GM11)(RH0/f)2) . (A1)

ZMAB = (4.788 x 10-5*2*PI*f) + j2*PI*f1

(4.681 x 10-a + 6.096 x 10-5 1oge(1/S) (RH0/f)2) (A2)

zzNN == ((RN/3NN)+4.788x1

(:..661 X 10’°"+6.096 X 10-5 10g€(RH0/f)2 +1E 6.096 X 10-5 16g€(1/(s2)3)+6.096 X 10-5

(1/3NN)(1oge(1/G11RN)+(NN-1) 1og€(l/KN(DEE/2))) (A3)

ZMAN == (4.788 x 10—5*2*PI*f) + j2*PI*f(4.681 x 10-4 +1 1' . 9 -6 . 2 E 26.096 x 10 1oge(1/((DEE/2)S ) )(RHO/f) ) (A4)

ZZNN = (RN/NN+4.788 x 10-5*2*PI*f) +

jgj-?—y¢—§—(4.681 x 10-4+6.096 x 10-5 log (1/CMRN)NN 1 E(RHO/f)2)+(NN-—1)4.681 X 10-4+(NN-1)

16.096 X 10-5 16ge((2/KN*0EE)*(R110/£)2) (A5)

40ZMAN= (4.788 x 10éS*2*PI*f) + j*2*PI*f _A 1

(4.681 x 10äA+ 6.096 x l0_51oge(2/DEE)(RH0/f)2) (A6)

ZZNN = RN/NN + j2*PI*f*6.096 x 10_S((NN-1)/NN '

1oge(2/(KN*DEE)) + (1/N)loge(1/GMRN) — 1og€(1/S)) (A7)

zMAN = j2*PI*f*6.096 x 10é5(1og€(2/DEE) — 1og€(1/S)) (A8)

_ 2Z113_phaS€ — ZZAA ZMAB — ZMAN /zzNN (A9)

s 2Z0033_phaSe ZZAA + ZZMAB — (3 ZMAN /ZZNN) (A10)

· Z1ll_phaS€ = ZZAA — ZMAB (A11)

2Z003l_phaSe ZZAA + 2 ZMAB — (3 ZMAN /ZZNN)

(from reference 7)

Z113_phaSe = R + jX q (A12)

Z0033_phaS€ = R + RN + j(X - XN) (A13)' For circuit type 11 use equations A1, A2, A3, A4, A7, A8, A9, A10.

For circuit type 12 use equations A1, A2, A5, A6, A11, A12.l

For circuit type 13 use equations A13, A14.

44 4 APPENDIX B

1·R0cRAM L1sTIuc

41

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APPENDIX C — PROGRAM RESULTS

Table of Contents

Results of Circuit Number gage1. . . . . . . . . . ............ . . . • 58

2.... . . . . . ..... . .... . . . . . . . 634.......................... 725.......................... 776.......................... 847.......................... 938.......................... 104ll.......................... 11312.......................... 126

110. . . . . . . . .... . . . . . . . . . . . . . . 143

140................. . . . ...... 140

210.... . ...... . . . . . . . . . . . . . . . 153

220. . . . . . . .......... . . . ...... 161

240. . . . . . . . . . . . . . . . . . . . . . .... 166

· 320. . . . . . . . . . . . . . . . . . . . . . .... 169

340, , , . . . . . . .... . . . . . ..... . . . 174

410.......................... 177420...... . ................· . . . 181510. . . . . . . . . . . . . . . . . . . . . . . . . . 186

520. . . . . . . . .... . . . . . . . . . . . . . . 190

1A. . . . . . . . . . . ............... 195

ZB. . . ........... . .......... . 202

56

S7

— In the pages that follow, the point numbers correspond to points on

system maps provided by the author for the Electric Service.

58

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1A COMPUTER—AIDED FAULT ANALYSIS OF THE

. VPI ELECTRIC SERVICE POWER DISTRIBUTION SYSTEM

George L. Hogshead

(ABSTRACT)

In this thesis, a general—purpose digital computer program to

Péffßfm fault 8¤ölYS€S is developed and discussed in detail. The

program is utilized to perform a complete fault study of the distribu-

tion system of the VPI Electric Service, as an example test case.

Equations are gathered and a computer program is written to calculate

the fault current magnitudes on a distribution system when three phase,

phase to phase, phase to phase to ground, or phase to ground faults

occur.

The program is written to include basic circuit types that are in

use in the utility industry today. The program provides results that

are in tabular form for easy use.

The VPI Electric Service system is discussed and results are

“ included for the major distribution circuits in the system. System

maps have been prepared for the circuits and are available at the

Electric Service.

A user's guide is included so that the program can be used whenever

system changes demand a new study.

Documents

, MASTER OF SCIENCE · 2 2 2 2A In 0;::61 M •r-I ••·» ::*1* I $@3 1 §¤¤° cr uan mg

, MASTER OF SCIENCE · 2 2 2 2A In 0;::61 M •r-I ••·» ::1 I $@3 1 §¤¤° cr uan mg