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Olfl O I. J L5 011' CO: ZE’JTY .

A Thesis Submitted to
The Facult f; of

KICHILBS? 32533 CO "‘33.

Orin D. Dausman

fl

Candidate for the Degree
of

taster of Science.

June 1927.

THESIS

 

I.

II.

111;“

COITENTS.

Page
Introduction
Stateuent of the Eroblem --------------------------- 1
Acknowl edgement s ---------- - ------------------------ 3
Solution of the Problem
General Bethod of Procedure ------------------------ 4
Short Circuit Calculations -
Analytical :Bthod ------------------------------ 8
Short Circuit Calculations by means
of the Calsulatinb Table ----------------------- 26
minimuu short Sircuit Calculations ----------------- 32
Selection of the Kethou of Protection
for the Different Parts of the System ---------- 35
Relay Settings ------------------------------------- 46
Relay Connection Diagrams .......................... 49
Bibliography -------------------------------------------- 52

I.
INTROHJCI‘IOIE.

Statement of Problem.

The purpose of this investigation is to design, as a typical
problem in ProteCtive Engineering, a satisfactory systen of relay
protection for the power network of the Ohio Edison Company. The Ohio
Edison Company, located at Springfield, Ohio, is one of the electrical
Operating companies controlled by the Commonwealth Power Corporation
of Jackson, KiChigan. The general information concerning the power
system, and the data necessary for the design of the protective system
were supplied by the Commonwealth Power Corporation.

The power system, as shown on the accompanying prints, consists
of two generating stations, Had River station.and Rockway station, with
feeders leading from each, a 37.5 kilovolt line from the Tad River
station, and two 12.5 kilovolt connecting circuits between the two.
The Eockway station is in Operation at the prescnt time, while the had
River plant is under construction.

On print No. 1, page 2; may be found the ratings and transient
reactance values of the various units of the power system, and the

locations of oil circuit breakers, instrument transformers, etc. as

provided by the Commonwealth Power Corporation.

 

 

G E. /2,J'oo V.

25,000 Km..879‘.' 3;. #32 Page)?

1500/5.
2:0 r. [35.4%. 125' u a Kw.

’ W

.__{ yea/1,2,0»

1t Area/J.

A

Mad River

 

 

 
   

 

 

 

 

 

 

 

 

 

 

 

  
  
   
 

 

 

 

 

 

r .360", [a
I
I
.
| /2..f or
Woo/,- ; aim/J ‘ “axon/f
H £533.“; L—é iié‘yupu “51.4170” 73PM" 2V?" —/ 00 J15} React
' Power Feeder
I [5.0/5
: #2 Wackwcheggg- 3315/“. Line.
I"'/ Rockwcly a -~'
_Fcca’cr / i l E
a .1 :;
... §: 6. E. 5:000 V. W‘SG. “‘0‘ y ‘4
.9?" *5: 2374' ku- :xzznmt 14 an Inn; -—- 127. Feed: ‘2
:3 K. .97»: "‘ .8 725 .4
Q : ’m/J' ’0‘”- O
. to . l
r ' at
Typtc‘d/ ‘, : To 115': 73 u: v. J”
Feeder a, ' ’m/J' E): .108. 1003/ 511.14“.
I I . 110
u. {2%‘1 1g 1 . 4-7. Pact lg .rx weed. hi i“ ”5‘“
k mAr L “gum. .0 35%... . 0%-
L,J
1 W
6'. .6. “2.303;”? 25:55 :30;;P 31:57:. 23);;
W at. . . 4V 0
2.530%; React WA [07: Rent 0:! 62/0 g geac‘l’
{91-3} 6000 u 7. Read? E law 6000 6217b.
oEl:§V.
’52:.“ 9%... ‘ "" rue/....
800/5. 1%- 840/5. 80%.
I
9‘0ch- FO If
3—4: 30011. c “’1‘!
F. 7 Km. 2?“
7‘ PR/N T N0. 1.
F a’ '
”m" “ "" ONE LINE D/AEPAN M/mv (ONA/ECT/ONS

NHD RIVER 6 POCKWAY STflT/ONS
SHOW/N6 5Y5 TE N REflC 771N655

 

 

 

Aclcn owl edgment s .

At this time, the writer wishes to eXpress his indebtedness to
the Electrical Engineering Department of the Commonwealth Power Corp-
orat ion, not only for submit ting the data for the problem, but also
for the valuable suggestions and for the use of their calculati ng table
in the solution of the problem, and to Prof. L. S. i‘oltz and Prof. A.
Master of the Department of Electrical Engineering at 2.1. S. C. for

their advice and instruction throughout.

II.
SOLUl‘ION OF Tim PROBLEM.

General Kethod of Procedure.

The first step in the solution of the problem was to determine the
short-circuit currents throughout the system in case of a fault at any
one of the various points of the systen. These values of short—circuit
current were first determined by the analytical method, and then, later,
by means of the calculating table. In determining the current values
by the analytical method, the system reactances were reduced to the
ohmic basis, while the percent reactances at a definite kv.a. base were
used for the calculating table work. These two methods of expressing
the reactances were used in order that a better check on the analytical
solution might be obtained.umen the calcuhating table was used. {After
the system reactances had been evaluated, the equivalent reactances for
detennining the short-circuit currents were obtained by a method of
simplifying the circuit diagrams, step by step, as may be seen under
the actual analytical solution. (See pages 10 to 21.)

The analytical method of solution was roughly Checked eXperiment-
ally. An equivalent system, using common laboratory type resistance
boxes, was set up. With.a definite voltage applied on the system, the
current was measured in various parts of the system. The result of
this experimental determination of short-circuit current may be found
on pages£$3~1fih The values of the diort-circuit current obtained by

experiment cheeked very closely with those obtained by the analytical

method.

The determination of the short-circuit currents by means of the
calculating table was carried on in conjunction with I-Ir. C. ‘..’. Meth-
fessel at the Jackson office of the Commonwealth Power Corporation.
The results of these determinations are found on pages :36 —31. It may
be noticed that these short-circuit currents do not check very well
with those obtained by the analytical method. This is not because of
errors in either method of solution, but rather because more accurate
reactance values were obtained for some of the units, two of the Rock-
way tenerators (G1 a 65) were removed from service, and larger trans-
formers were installed between the 12.5 and the 2.5 kilovolt busses of
the Rockway station between the time that the original data was obtained
and the time that the calculating table Was used. These revised values
of reactance are the ones shown on print No. 1, and the ones actually
used in determining the final relay settings.

Since, in order to determine the prOper relay settings, it is
necessary to know the values of the short-circuit currents under mini-
mum Operating conditions as well as the maximum conditions, the mini-
mum short-circuit current values were next obtained. These may be
found on pages 32-34.

After the short-circuit currents were determined, the method of
protection to be used at the different points in the system were
determined. This was accomplished by making a study of the various
possible schemes of protection for" the types of apparatus used in this
power system, and by selecting the most satisfactory scheme. The
comparative advantages and disadvantages of the schemes considered,

along with the selected schemes, may be found on pages 35-45. The

general method of protection for the entire system is shovm on Print
Ito. 2, page 7.

Follow ing the determination of the proper type of relay protection,
all that remained was to determine the relay settings and to draw the
diagrams. The necessary settings to give proper selectivity under all
short-circuit conditions are given on pages 46 and 47. The time settings
were determined from the curves given on page 113 of the "Relay Hand-
book" (1926 edition), published by the N.E.L.A. These are average
curves for Westinghouse, type CO, overcurrent, inverse and definite
minimum time relay elements. The settings were determined to accomp-
lish the following results:

a. Give prOper selective action under maxim'a, minimum, and
other short-circuit condit ions.

b. Clear maximum short circuit in less than 2.0 seconds.

c. Clear trouble under minimum short-circuit conditions.

d. Carry maximum full load.

9. Clear all trouble with least amount of disturbance.

 

Over/cad Relay .

Po werD/recfi'ana/ Fe/ay.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6’ 4;.
Mad River
A 42.! AW. 3:13 §_0~. h
,. -
I
I
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' 1 1
i I 4: NM
4 IO" yplca
“F4;— ’ I ' 4}" Power Feeder .
o stKvJunc
E :g Roe/(wag Feeder
K
r : a
a:
h
0
k " a
Typica/ ; f 1 ’
N , I
Feeder 7;“ "my .
w-{i—~ T ® r-i f
I -
a {‘3 E? D g“ 53
I I!!!“ .30.! L

 

 

 

 

 

 

 

 

 

 

 

 

NM J NM
N ‘

V1 NOT .

1 Fa any.

E 2500 V. v 3 us

Rockwqg 1&3
% PHI/VT No.2. I
Typica/
Feeder METHOD 0/- REL/9y PROTECT/01
MAD RIVER d ROCK Wfl)’ 5 7277701149

 

 

Short Circuit Calcula ti one.

(Analytical nethod).

Circuit Reactance Reduced to 12.5 Kv. Basis.
Mad River
Gi 25,000 kv.a. 12.5 kv. 13% react. on rating.

Full load current = 25.000 x 1.000, = 1,155 amperes.
12.5 x 1,000 x 53

Voltage to neutnrl 12,500 a 7,225 volts.

Generator react. = 13% = 0.13 x voltaxe
current

= 4.134;;12.500 x 12.5 x:;.000 x V 3 c .8125 ohm.
V"3' x 25,000 x 1,000

T ansformer on 37.5 kv. line.
3 - 1,500 kv.a. 12.5 to 87.5 Xv. 5% react. on rating.

Trans. react. = 5% a .O5_§_voltage to neutrals
full load current (at 12.5 kv.)

- ,05 x 12,599 x 12,5 x 1,000 x V 3 = 1.736 ohms.
V'3‘ x 3 x 1,500 x 1,000

Rockway Feeders.
N0. 1 and No. 2 are similar, each consisting of :5 - No. 4/0
at 2.5' spacing, 2.75 miles long.
Reactance/mile of each conductor a .620 ohm.

React. of each feeder a 2.75 x .620 = 1.705 ohms/wire.

      

Station

   

G1 2,300 V. 6,250 Ev.a. 12$ react. on rating.

Reactance = .12 x voltage to neutnil
full lead current

= ,12 x 2,390 x 2.3 £1,000 3: rs" = .1016 on 22,300 V.
v-s- x 6,°50 x 1,000

Reactance at 12.5 RV. = .1016 X (12,512 = 3,9 ohms,
( 2.5)

GZ'G3 2,300 V. 2.500 kv.a. 12% reactance on rating.

Reactance s ,12 x 2,399 x 2,3 x 1.000 x V 3 = .254 at 2300 V.
V75‘ x 2,500 x 1,000

Reactance at 12.5 RV. = .254 x (12,5)2
( 2.3)

‘ 71L§Q 0317184

G4 6600 v. 10,000 kv.a. sfi react. on rating.

Reactcnce at 6,600 V. = .99 x vgltgge t9 neutnal
full load current

= ,08 x 6,699 x 6.69;»1.000 x V 3 - .3485 ohm.
V_3‘ x 10,000 x 1,000

2
Reactance at 12.5 RV. = .3485 x (12,5) - 1,259 ghms.
( 5-5)

05 5,000 V. 9,375 kvua. Bfi react. on rating.

Reactence at 5,000 V. = 0.08 x 5.000,x 5 x 1.000 x_V'3' s .2134 ohm.
VT 1: 9,375 x 1,000

Reuctonce at 12.5 kv. = .2154 x (12,5)2 . 1,334 ohms,

( 5 )
Ersnsformers to 9, e 959

12.5 kv., 10,000 kv.s. 5% react. on rating.

Reactance a ,95 x 12,509 x 12,5 x 1.000 xiV‘E' a ,782 ghm,
V'3_'x 10,000 x 1,000

 

Transformers between 2,5 & 12.5 kv.93usses.
3 - 1,000 kv.a. 6-1/2% react. on rating.

Reactance = .065 x 12.500 x 12.5 x 1.0009§;V 3 = 3,39 ohms.
V—3_ x 3,000 x 1,000

3 - 2,000 kv.a. 6-1/2fi react. on rating.
Recctcnce - 50% of 1,000 kv.a. size a .5 x 3.39 - 1,695 ohms.
Reactors in 2.5000V. feeders.

X - E -‘ 29 . .0967 ohm at 2,500 V.
I 300
Reactance at 12.5 kv. a .0967 x (12,5)2 -_2.4l75 ohms.
( 2.5)

10

Determination of Lhort-Circuit Current in Case of a
Fault on the 37.5 Kv. Line from the mad hiver Station.

(All switches Closed)

The complete simplified diagram of the Ohio Edison System, showing the
circuit reactances reduced to the 12.5 kv. (Ohmic) basis, is shown in Fig. l
on the following page. The succeeding figures show the method of simplifi-
cation of this diagran in order to obtain the resulting single equivalent
reactance for the system.

Fig. 2 is a.more simple diagram of the system, showing only those cir-
cuits along whioh power may flow in case of a fault at "X", and the equiv-
alent reactances for the different portions of these circuits.

Fig. 3 represents Fig. 2 converted into a simple network. n F g. 4,
the parallel paths of Fig. 3 have been replaced by their equivalent circuits.
The method of combining these reactances is the simple method used in can-
bining parallel resistances in D.C. circuits, since the pocerfactor is being
neglected in these computations. The equivalent reactances of the generators
are also shoxn in Fig. 4.

Fig. 5 is a reproduction of Fig. 4 with the generators replaced by their
equivalent reactances and.connected at the common point "0". The successive
figures from Fig. 5 to 10 show the steps in the simplification of the network
to the final euuivalait reactance of 2.275 ohms.

Short-Circuit Current, ISC = 12.500 = 5.165 amperes.
V75 x 2.275

Current delivered by T‘.’ad River aei erator,

1G1 = X x Isc = ,539 x 3,165 = 2,100 angeres
K01 .8125

 

 

 

 

 

0.8m!
Mad River
[.75 c.
Tgp/ed/
Feeder 37.5. Irv.
Linc

 

/. 704'
l. 705’

e Typt'c‘d/
I 334 Feeder

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2.4175"

Typ/ca/
FEedcr

 

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6’ .802:

l. 73(-

I. 73‘

rill“.

Trans .

 

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VVVVVVV

 

 

 

 

 

 

 

 

14

Current delivered by Ziocicvay Station,

Iiiockway g -:—- 3 13¢ = 52.2.. X 0,105 = 1.065 adheres
1‘11 ockway l ' C 10
Current delivered by each Eoclczay Feeder,

. . I ~ I. ‘ C: ‘7 -. -,_ ,
IRock-Teodor " Iiioc‘may :- d ‘ 1065 ' d = “3"‘5 “mares

All the above currents are found on the 12,500 volt basis.

The actual current in the transm'm si on line is

 

._. 9 RQQ - '7 g 5:.- r. ,0 9
ISO at 37.5 m. rztu u )5. (1,165 L0...) @431. S
07,500

this Case is the total reactance as fbund in

15

Determination of short-Circuit Current in Care of a Fault at the

i JOAV

rad River Bus or any Circuit Lirectll Connected to the Bus.

 

(All switches closed.) (all based on 12.5 Xv.)

The total reactance ir this ca-e is the total reactance as found in

(.1

('f

h case for the 37.5 kv. line

(3

‘minus the reactance 0f the 37.5 kv. transformers. Referring back to Fifi. 10,

Xsc a .539 ohms.

Short-Circuit Current,
ISc = 12,500 = 13,380 amperes.
V U K .539

Current delivered by Had River Generator,

1., = Xsc x Isc a ,539 x 13,380 = 8,889,
“'1 X3. .8125
1
Current delivered by Rockmay Station,

Iliock'qasr g XSC X 150 z .539 X 13,380 3 4.5g!)
Y \
“hockway 1'610

Current delivered'by each hockwqy Feeder,

O 9 4 :1 rs :.. 0') p:
‘ * = "t - ‘1 Co : ~
IRockmay leeder {nockway . ~ = uOO/ .21.:9

Current delivered to 12,5 h . Bus by G4 & G5,

 

04+? = X Rockwav Ctation x IRocha a ,758 x 4,500 = 3,2 5.
‘5 .' 13.5 kv. “ 5’ 1.038

Current delivered to 12,5 kv. Bus from 2.5 kv. bus,

 

12 5 tv c figfiockway_3t. x I Rockway = ,758 x 4,500 = 1,295.
' ‘ ' x2.5 kv. 2.798

r“ ‘ v '7 t) _ e‘ r
1G4 ‘ M 3* IGI, a; 35 " _:____l 0:? x 39"95 - __L_1 L50
K94 + trans x trans. 2.002

135 ‘ K 12.5w, >1 10,4 “.35 -= 1&0 x 3,..95 = 1,030

 

 

V VVVVVVUVV

 

XG5 + trans & trans. 2,116 Based
on
I2000 kya. 59$ trans. x I2.5 kv. = ,139 x 1,205 = 89§ 12.5 kV-
trLIIS. x2000 KVB. trans. 1.695
IlOOOkva, = guano. x 12.5 M. .. 1.150 x 1,205 = 4 2
x1000 kva. trans. 5°39

16

Detenninat ion of Short—Circuit Current in Case of g

1 5 Irv Bus at Rockne. Stations

         

(111 Switches Closed.) (111 Based on 12,500 v.)

By referring back to figure 2 to 8 and comparing figure 8 with figure
12, it will be seen that figure 12 represents the system with a fault on the
12.5 kv. bus at "X".

The final equivalent short-circuit reactance is .520 ohms as represented
by Fig. 14.

Short-Ci rc‘uit Cur rent ,

Isc = 12,5Q 2- 13,850 am‘geres,
T3 1 .520

Current delivered by Had River Generator,

10' s Xsc x Isc = ,529 x 13,850 = 4,325.
1 K ’20 :Iad River 1.665

Current delivered by each iioclcway Feeder,

I q .2. 2 = 4 7'75 2 3 2 162.
IRockway Feeder Iva. ' ’0“ / ’
Current delivered by Rocmvar] Station,
‘ :.- 3 a Z,
IRoclmay - 150 x Isc ,529 x 1 ,850 9,5 9,
X R0 clctvay .758

Current delivered to Bus by G4 8: G5.

I = X 110010751 1 I Rockway == ,758 x 9,520 = 6,939,

G4 + 65
384 + G5 trans. 1.038

Current delivered from 2.5 k'V. Bus.

 

12,5 kv. = Xfiggkmax x Ifiockway = ,758 x 9,520 = 2,599.
e 2.5 kv.+ trans. 2.798
104 5 £04 + (,5 5, mm x In +05 -= 499.: “g x 6.930 = 5....953 .
x04 a trans. " "
105 - Eua+es , ,n,n, x IG4 , G5 - ,998 x 6,930 = 3,499,

x55 a trans. 2'116

17

IZOOO kva. ‘ X trans, I 12 5 kV '3 .139 X 2,590 = 1,737,
trans. X2000 kva ' ° 1.695

11000 ha. ‘ X trans, X I 2.5 kv. ‘ 13 X 2,590 3 863.
tnans. X1000 kva. 3.39

‘51 ' GJ+Ga+§a x12.5 kv. =.%1%%§ ‘ 2.590 = lessen.
1

I " XI -. $1568 x2590=575
G2 XGI+GZ+G5 2.5 a. m— , ———L

X62

I - +0 +0 1 I. - 1,668 x 2,590 = 575,
G3 M ‘05 1w. 7.50

18

Determination of Short-Circuit Current in Case of 8.

Fault on the 2.500 V- Bus at Rocka Stati on,

 

(All Switches Closed.) (All based on 12,500 V.)

By referring to Figs. 2 to 6 and centering with No. 15, it will be
8$811 that Fig. 15 represents the system with a fault on the 2,500 V. bus.
Figs. 15 to 19 show the simplification of the system until the short circuit
reactance of .858 ohms is obtained.

Short -Circuit Curran t ,

Isc - 12,599 = .499 aggres.
V3- 1 .858

Current delivered by 61' G2, G5,

12,500 V 3' £30 1: ISO . .858 x 8,400 g 4.529.
X2500 V 1.668

Current to fault fran 12.5 kv. Bus,

 

 

112.5 kv. - Xsc x Isc - .858 x 8,400 - __,_Q8_Q.
X13.5 kv. 1.768
1G1 a x2590 I x 12500 V.’ éf)? x 4,320 = 4,493.
1 O
IG2 - 1:500 1 x 12500 ‘7. . #238 x 4,320 . 969.
z 0
1G - 259° n x 12500 VL - 1.668 x 4,320 = 969,
3 XGS 7.50

Current flowing between 12.5 kv. on 2500 V. Bussos.

12900 kva. . g trans. x I12.5 kv. - LISQ x 4,080 a 2,749,

““18- x2300 kva. 1.695
rans. ‘
Ilooo kva, ' W1 112.5 h. - .139 x 4,080 . 1 559 ;
tmns' x1 00 m. 5.39
9mm.

Current supplied to 12.5 kv. Bus by 85 6; G4.

185+G4 " “trans. 1‘ 112.5 kv. ‘ ...-.———1 gig I 4,080 = 4—43 515
5+G4 6° '
trans.

 

0

  

Feed: A:

 

 

 

 

. 638

"""

 

 

.858 2.4/76"
5 Feeder
gs‘rem Reactor

Fig. 20.

  

.658
XL:

F12. IQ.

X

 

1.66!

‘v-vv'

 

 

 

 

 

 

 

 

 

 

20

18 = XE9+G§ trans x IG4+G5 = l,Q§8 x 2,515 = 1,299,
4 m 2032
§u4 & trans. °

IG5 II énlfijls 3; trans! I IG4+G5 '5 “.925 X 2,515 = 1.335.
‘5 a trans. ““115

Current suyplied from red River Station,

1G1 3 la 5 .15! jEEQS X 112.5 Irv. II .638 X 4,080 = 1,565.
X Mad River a: 1°6°5
Rodkway Feeders

Current delivered'by each Rockway Feeder,

130cm” Feeder = 1G1 1 2 = 1,555 -;- 2 = 782.5

21

Deteminati on of Short-Circuit Current in Case of;

Fault on a 2 5 Reactence Protected Feeder

     

(211 Switdhes Closed.) (111 based on 12,500 v.)

The short-circuit reactance for the 2,500 V. feeder is equal to the
short-circuit reactance for the 2500 V. bus plus the reuctance of the curr-
ent limiting reactence, as illustrated by figures 20 and 21.

Isc = 42.500 2 2,299 amber-es,
V3- 2: 3.2755

 

This value is 26.2,? of the short-circuit value on the 2500 V. bus.
Therefore, all the other values 2.1 11 be 26.273 of their corresponding bus
short-circuit v21 ue 3.

Current delivered by 81, 62, 2 G5 a 26.2% of 4,520 a 1,152,

Current to fault fnom 12.5 ks. bus: 26.2% of 4,080 = 1,979,

IGl ‘ a 26.2% of 2,400 a 629,
1G2 = 26.2% of 960 = 251,
165 . 26.2fl of 960 s 251,

12000 kva. trans. 26.2% of 2,720 = 751

—-—-L
r: : :7 __
11000 ha. trans. 3 56.3/9 Of 1,360 - 3§6I

154 + G5 = 26.2; of 2,515 a .222.
IG4 ' = 26.2% of 1,290 = §@§,,
1G5 = 26.2% of 1,225 = .521,
19.1 = 26.2% of 1,565 a .210,

Current delivered by each Rockway Feeder

. 26.2% of 702.5 = 205,

 

 

 

 

 

 

 

 

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C'x'rcul'f' Curran/3.

 

Svstem Reactances

For Experimental Determination of Short-Circuit Currents.

Unit, Kva, Capacity 2 React, st Rating $‘Beact, at 45,999 kva.

oi 25,000 13% 25.4%

C1 6,250 . 12% 86.4%

Ga 2,500 22 21.6%

G4 10,000 8% 56.9%

G5 9,575 6% 56.4%

Trans. 04 10,000 5% 22.5;

Trans. c5 10,000 5% 22.5;

Trans. 37.5 kv. 4,500 5% 50.05

Trans. 12.5-2.5 3,000 6-1/2% 97.5%
1000

Trans. 12.5-2.5 6,000 6-1/2fl 48.7%
2000

Rockney Feeder 1.705 ohms at 12.5 kv. 49.1%

Reactance Protected Feeder 2.4175 ohms at 12.5 kv. 69.6%

Values ofzresistance used in the experiment and indicated on the dia-
gram , Fig. 22, are equal to 10 x fl reactance based on 45,000 kv.a. as shown

above.

E = volts applied between Generator“ and Failt at A, B, C, etc.

I = milliamperes in circuits as recorded.

Location of Circuit E. 19,0. S.C. Current
Fault Considered. ID C x 164

A A 12.65 19.0 3,120

Gi 12.65 12.5 2,050

Rockway Feeders 12.65 6.5 1,068

Location of
Fault

Circuit
Considered.

B B
G!
l
Hockney Feeders
G4
G5
5-2000 kva. trans.
3-1000 £' "
C C
Rockway Feeders
G4
Gs

341000 trans.

3-1000 trans.

3-2000 trans .
G4
G5

Hockney Feeders

ID.c.

10.0
5.0

8.25

49.0
14.0

6.0

8.0
16.0
7.5
7.0

9.5

8.0. Current
x 164

I

3.0.
12,800

0,540
4,270
1,640
1,475

758

369

15,150

4,025
5,450
5,290

1,645

8,060
2,610
986

986

(x 164.5)

 

 

 

Location of Circuit E II) C 8.0. Current
Fault Considered. ‘ ‘ ID 0 x 154.5
E n 12.5 15.0 2,455

Determination of multiplier for converting D.C.,milliem0eres as read in
the experimental emivalent system to short-circuit current.

At the base of 45,000 1:11.24. and 12.5 kv. the full load current :71 11 be

45,999 x 1,999 = 2,075 mg.
12,500 x V“:-

Since 1003’; system reactance is represented by 1,000 o'mns in the ex-
perimental circuit, the constant of the set-up at 12,65 volts corresponding

to 100% reactence is

I - 12.65 = .01265 array” or 12.1.65 millimperes,
1,000
.‘. One millie-mpe re represents 2,075 = 164‘. map,
12.65 '
and likewise when E a 12.6, one 3.1.5.. represents 2.975 a 164.5 amp.
12.6

Short Circuit Calculations by teens

of the Calculating Table,

 

System Reactances for Short-Circuit Calculations.

 

Unit. Kva, 05951513 ineact. at Rating Sgfiezct. at 100.000 Eva.
Gi 25,000 12 48
Gl(not run- 6,250 10 160
ins)
02 2,500 10 400
G3 2,500 10 400
G4 10,000 12 1230
G t 1 - 9 375 12 “5
5(noingyn ’ . 1“
Trans. 04 10,000 5 50
Trans. 05 10,000 5 50
0mm. 57.5 kv. 4,500 5-1/2 4 122
Trans. 12.5-2.5 kv. 6,000 6.5 108
each
Rockway Feeder 1.705 ohms at 12.5 kv. 100

React. Protected Feeder 1,300 2.01 155

Z7

&eet NO. 1.
COWONWEALTH £01113 CORPORATION OF MI CHIGAN

Electrical Engineering Department
Jackson. Michigan.

SHORT CIRCUIT ggnwgylon

 

 

 

 

 

OPERATING 00- Umncai W0 ' 8.007410%
STATION 6.4.5221 Cycles_ 44 Made bycuLz-anDate I wax-.17. ' '
BASIS or QALCUIMION References

( ) Present ( ) Ultimate (V) 1127 RNA Base me me

 

PURPOSE OF CALCULATION -
(.4 0033 Application ( ) Bus Stresses (V) Relay Settings
( ) Reactor " ( ) LT. Resistors( )Trouble Report

 

REGOWDATIONS

 

SUMMARY OF CALCULATIONS
' KV V
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28
Sheet No. 3 -
COMMONWEALTH POWER WRPORATION OF MLCHIGAN

Electrical Engineering Department
Jackson. Michigan.

SHORT CIRCUIT GALGJLATION

 

OPERATING 00. file fizl'fl‘. (:1. W0 S.C.Calc. No. 36.
STATION (11: fl,.tga Cycles ‘4 Made by alum-cm Date hag-‘1
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32
Minimum Short-Circuit Calculations.

The conditions for minimum short-circuit values are assumed with
only G2 and G4 at the Hockney station in.0peration. fhe resulting

current values are detennined.as follows.

Fault on the 37.5 kv. Line from.xad River.

The sequence of figures from Fig. 2.3 to Fig. 26; on the following
page, shows the reduction of the system to a single equivalent reactance
of 29633 on a basis of 100,000 lama.

% Reactance = 296 on 100,000 kV.a. basis.

Short-Circuit kv.a. at fault = _1_Q_9_,QQ_Q = 33,800
2.96

Short-Circuit Current at fault - 33,800 = 1,569 amps,
W x 12.5

Current in each Rockway Feeder = 789 amgsI

"I

Current delivered by G4 = 4
6

‘4 x 1560 = 1 135 amvs

(

 

on

.p.

Current ddlivered'by G2 a 170 x 1560 = 425 amos, at 12.5 kv.
624

Fault on zed River ginsI
‘70 Reactance = 296 - 122 . 174 (See Fig. 23 a. 25.)

Short-Circuit lama. at fault a lOOIQQQ a 57,400 Iona.
1.74

Short-Circuit Current Lt fault a 57.400_ = 3,555_Q__a_p_..m 5
V3 1 12.5

Durrent in each Rooney Feeder = 1,325 amps,

Current delivered by G4 = 454 x 2,550 a 1,939 amps.
624

Current delivered by G2 . 179 x 2,650 a 720 amps. at 12.5 KV.
624

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Fig.

gives a re

34

Fault on 300.1 W853! 1245 75v;— 3118.

% Reectance a 295 - 172 . 124 (see Fig. 24, 26 e 27).

Short—Cir uit Kv.a. at cult = 100.000 = 80,600 kv.&.

1.84

Short-Circuit Curran at fault = 80,600 = 3.720 amps.

VET x 13.5

 

Current delivered‘by G4 = 454 x 3,720 = 2.710 amps. ,
6&4

Current delivered'by 02 - 170 x 3,720 = 1.010 &E)S. at 12.5 kv.

A!

Current thru each 12.5-2.5 kv. Trans. = 505 emns. at 12.5 kv.

 

 

 

 

Fault on 2.5 1w.l‘.oc1’:.vejj Bus.
29 is obtained fnam Fig. 25 and reiuoes to Fig. 31, which
cuctance of léup based on 100,000 kv.a.
Short-Circuit kv.e. at fault = 100,009 = 69, 400 kv.u. .
1.44 i
Short-Circuit Current at failt a 69,003 = 3.209.amps. at ,
V3 x 15.5 1;.5 xv.
Current delivered by 02 = 2;4 x 3,400 = 1.150 amps. at 12.5 kv.
6L4
Current delivered by C4 = 420 x 3,200 = 2,959 unis.
624
Current thru each 12.5-2.5 kv. Grins. = 1.025 emos.
I
1
t
E

L

avg ‘nLL'r; IL- ‘.._1 _.‘_‘

-. a; g

4.11

if '1' J “...-‘1"

 

Selection of Xethods of Protection for the

Different Parts of the System.

Types of apparatus to be Protected.
1. Generators.
2. Transformers.
3. Rockway Circuits.
4 "ypical Feeders.

. i
I. .fienengtors.
A. gossible Schemes for Protection.
1. Simple Differential Scheme.
2. Self-Balanced Schemes.

3. Power—Directional Schemes.

B. édvm %1 es and Di sadvantg -es of Di fferent Schemes-

 

1. Simple Qifferential Scheme.

Advantages.

Simplicity.

Ease of Application.

Sensitive settings per-
mitted, resulting in quick

action.
No Special apparatus required.

Protects Against

Phase-to-phase short circuits.

Grounds (If neutral is
grounded solidly on through a low
resistance.)

2. Self-galanced Scheme.

Advantages.

Eliminates the normal unbal-
ance due to two current trans-
fonners having dissimilar

Disadvantages.

Current transformers
must have identical char-
acteristics, if sensitive
sett ing s- are required.

Does Not Protect Against

High resistance grounds.
Open circuits.
Short-circuited turns.
Faults in leads from
generator to bus.

Di sadvan tages .

Carrying two leads (hav-
ing a large potential differ-
ence) through the same

-‘.._.—-..-_...

‘mnvLsc- 11.111 -

-.~

 

36

characteristics. This makes transformer Opening.

possible a very low relay Difficulties in machine

setting so that the generator construction and installs-

can be disconnected for low tion.

fault current, with small Special current trans-

resulting damage. formers are required.
Protects Against Does Not Protect Against
Faults actually inside the Faults not detected by the

machine, as with the simple simple differential scheme.

differential scheme. Faults in leads from

generator to bus.

3. Power-Qirectignal Scheme.

Advantages Disadvantages
Comparatively simple and Cannot be given a sensitive
not very expensive. setting, on account of cir-
culating and other transient
currents.

Requires the use of po-
tential transformers.

Protects Against Does net Protect Against

Severe phase-to~phase Short-circuitei turns.
short-circuits. Open circuits.

Severe ground faults Any fault until it is
(under some conditions.) serious.

C. SelectedpProtective Schemes for seneratorgi,

It leirecmmnended that the Simple Differential SCheme be used for
the protection of the Had River Generator and of the 10,000 kv.a. gen-
erator (G4) on the 12.5 kv. Rockway Bus.

This scheme has all the advantages listed above fortill of the
three schemes which might be applied to these generators. The neutrals
are solidly grounded. Therefore, there will be no necessity for extra-
sensitive relay settings and slight dissimilarity of instrument trans-
former characteristics will not be troublesame. Since the leads are
all brought out in both of these machines, the adaptation of this

scheme will be of relatively low cost.

 

 

37

It is recommended that no protection be provided for the 2,500

kv.a. generators (G2 and G?) on the 2500 volt Rockwqy Bus. The leads

are not brought out on both sides in the manner necessary for differ—

ential protection, and the advantages offered by any type of overcurrent

or power-directional protection are not sufficient to warrant the in—

stallation of the necessary equipment.
II. Tnansfgrvers,
A“ Possible Sch mes for Protection.
1. Overcurrent.
2. Current Differential.
3. Power Differential.

4. Directional Overcurrent.

 

B. Advantages and Disadvantages of Different Schemes.

l. Overcurrent.
Advantages

Simplicity.

Protects Against

Short-circuits and grounds
taking excessive currents
whether in the tnansformer or
in the secondary distribution
systen.

2. Current-Differential Scheme.

Advantages

There is no interruption in
case of a through short cir-
cuit; thus the fault is con-
fined to the anallest possible
area.

An internal fault gives a
large differential current, so

Disadvantages

Is not selective; Oper—
ates on faults outside the
transformer.

Not Suitable For

Protecting against open
circuit.

Protecting against an
incipient or other fault
taking a “all current.

Protecting transformers
connected in parallel.

Disadvantages
Exact balance of normal
currents is required, and
calls for additional appa-

ratus and rafiier complicated

connections.
Not very sensitive to
grounds on ungrounded

that a sturdily-built relay
can produce a strong, sensi-
tive action.

The relay is simple, in-
expensive, and not liable to
get out of order.

There are no volt age con-
nections requiring voltage
transformers.

Protects.sgeinst

Phase-to—phase short cir-
cuits, short—circuited turns,
and low resistance grounds.

3.<Eower;Qifferentiel Geheme.

Advantages

Same as for the current-
differential scheme.

38

systems and systems with
high resistence between
ground and neutral.

Does Not Protect Against

Open circuit.
High resistance ground.

Disadvantages

Kore complicated, more ex-
pensive, more difficult of
adjustme nt than curren t -
differential scheme.

Requires voltage connections.

4. Di recti anal {yer cur rent Scheme ,

Advantages

Accurate current balanciig
is unnecessary.

Application.nnd adjustnert
are simple.

Protects Against

Phase-to-phese short cir-
cuits.
Low-resistance grounds.

Disadvanteges

Does not discriminate be-
tween line and transformer
faults, unless an added step
is introduced in the definite-
minimum time setting.

Requires pot ntiel trans-
formers.

Reguires high current
settings on account of poss-
ible transient conditions.

Does Not Protect Against

Open circuits.

High resistance grounds.

Short-circuited turns, ex-
cept serious short-circuits.

Applicable only where there
is a source of back feed to
the transformer.

C. Selected Protective Schemes for Transformers.

1. For Transformer on 37.5 kv. line.

39

It is recommended that a canbinati on of the current-differ-
ential and the overcurrent schemes be used at this location.
A comparison of the advantages and disadvantages of the

schemes outlined above shows that the current-differential
scheme gives the most complete protect ion without excessively
high cost. Satisfactory Operation may be expected even though
bushing type transformers must be balanced against standard
type current transformers, since it will not be necessary to
make sensitive relay settings.

The overcurrent scheme of protection sould be added in
order to give protection to the system in case of a fault on
the transmission line, rather than for the protection of the
transformer itself.

2. 6,000 kv.a. Transformers between 12.5 kv. a 2,500 V.
Roch-ray Busses.

Same as above. (For same reasons).

3. 10,000 kv.a. Transformer between 10,000 Iona. Generator
and 12.5 kv. Hockney Bus.

It is recommended that the current differential scheme be
used to protect this transformer. The advantages of this
scheme for this unit are the same as those listed for (1) 3c
(2) above.

NOTE: The application of the currmt differential schemes
as recommended above will give a uniform system of protection
for the generators and transformers. rIhis will be of great

advantage in test and maintenance work after the system is

4O

put in operation. It should also be observed that the nec-
essary current transformers for this type of protection are

already in place throughout the system.

III. Rockway Circuits.

A. Possible Schemes for Protection,

1. Schemes Commonly Applied to Parallel Feeders.
a. Overcurrent Protection with Directional Selectivity.
b. Overcurrent Inverse Time Protection.
0. Current Differential Protection by Current Balancing.
d. Current Differential Protection using Selective-Differ-
ential-Current Relays.
e. Current Differential Protection with Directional Selec-
tivity.
2. Sdhenes.Applicable to LOOp Systems or Hetworks.
a. Overcurrent Time and Directional Scheme.
b. Pilot Sire Schemes.
B. édvdntagegzgnd.Disadvantages of Different Schemes.
l. Schemes Commonlyaépplied to Parallel_Peede;g;between
generating Stations and Substatigns,
a. Overcurrent Protection with Ddrectional Selectivity.

(Overcurreut rehays at source, directional at receiver.)

.Advantages Disadvantages

Simple and reliable method Requires voltage trans-
of protecting any number of former connections.
parallel feeders. Requires time operation

Not affected by removal of of overcurrent relays.
part of the feeders from Requires overcurrent
service. elements in conjunction

When only one feeder re- with directional elements
mains in service it has over- if normal flow of power is
current protection. to be allowed in reverse

direction.

Protects Against

Short-circuits on lines
and buses, except bus at
gene rat ing sta ti on .

Low resistance grounds, on

systems with neutrals grounded
solidly or through low resist~

ance.

41

Does Not Protect Against

High resistance grounds.
Single grounds on systems
not solidly grounded.

b. Overcurrent-Inverse Time Protection.

Applicable only vhere there are alwavs at least three
parallel feeders in service and all are of about equal impedance.

c. Current-Differential Protection by Current Balancing.

Advantages

It is extremely simple.

Easy to apply.

Employs standard inex-
pensive apparatus.

Eliminates use of voltage
connections.

Does not require fine
adjustment of relay.

Is not affected by through
short circuits.

Operation at generating
end can be made practically
instantaneous.

Protects.Against

Line shor t -c ircuit s.

Disadvantages

Does not discriminate
between good and faulty
lines that are paired to-
gether.

Not applicable except
in pairs.

Does Not Protect Against

nus short-circuits.

d. Current-Differential Protection using Selective-Differ-

ential-Current Relays.

Advan taxes

Ho poteitial transformers
are required.

Protection is practically
instantaneous.

Di sadvan tages

Protection depends on

the correct functioning

of three relays with their
contacts in series if
protection of the remain-
ing feeders is to be con-
tinued.

Protects Against

Currents in faulty line
exceeding those in good line.
Overcurrent in the last

line in service, if relays
are suitably arranged a set.

42

Does Not Protect Against

Currents in faulty line
equal to or less than
currents in good line.

Bus short-circuits.

e. Current-Dif7erential Protection, with Directional

Selectivity. (Cross—Connected Power Directional Scheme.)

Advantafes

always selects the faulty
line.

Practically instantaneous
in operation.

It can is set to Operate
on etrrents smaller than the
full-load current of each
feeder.

It may be applied to any
system, no matter how com—
plex, if the feeders are run
parallel between the switch-

“. Schemes Applicable to Loop

' 'c- '~'- 1““- ~43
D1 sauaa..bar,b8

hill not clear the
trouble then the buses or
all the feeder: are involved.

There may be trouble in
cutting feeders in and out
of service under heavy loads.

The wiring is complex.

It is necessary to pro-
vide a.ceans for changing
the 0.9. c nnections when
the circuit breaker Operates
if the remaining feeders
are to be protected.

‘tens or he works.

a. Overcurrent Zime and Directional Schemes.

(Inverse definite-minimum time overcurrent and direction%
a1 relays are put on the loop there it enters and leaves each
station and are made Operative on power flow in the direc ion

from the station bus.)
Advantages

Least expensive of all
satisfactory schemes.

Simple to install, check
and maintain.

Applicable to ground pro-
tection with little or no
additional SQJanBRt.

Ecually satisfactory for
protecting short and long
sections of the 100p.

Applicable To

LOOps having four sta-
tions or less.

Disadvantages

Not instantane ous .

Operates only on over-
current values of fault
Current .

Kot Applicable

By itself to loOpS hav-
ing more than four stations.

LOOpS having any number
of stations, if used in
conjunction with another
scheme.

b. Pilot Hire achemes.
Advantages
Pot affected by throuji

0311at1r‘ on fault currents
that are small compared with
load currents.

Operating almost instantan-
eously, and not interferring
with successive time settings
emgloyed in other parts of the
system.

Discrim.inating eslscially
well on short lines.

"ot reauirinn additional
e1uipr ent for protecti ion
against ground currents.

Hot rcauirings hort- -circ it
calcukitions to determine the
on rent that will flow.

mmple constrzmc ion.

Not reguiring potential

connections.

C. Selected Protective Scheme for Rock:

43

general)

Disadvantages

Ha

Co 0 silot'uire.

C s 01‘ current tians-
forner- izauing the reuuired
cha'ac eristics.

Diffic ultj oi‘ r041111g
current trexsformers of
similar characteristics at
all currents.

Possibility of an unde-
tected break in the pilot
wire.

Possibility of false
Operation due to "olta :9
in nilot wires iniuced by
short- -circuit current in
neighboring circuit.

, a
\.’l

O
Qfld‘

(+C

'irCuits.

It is recommended that the sche1e of overcurrent protection with direc-

tional selectivity be usei at the Fock.1ny end

and that the cross-connected

power directional in conjunction with plain overload be used at the :ad

River and of the Rocrwa;' circuits.

Examination of the advantages and disadvanta es of the iif ercnt pro-

tective schemes listed above shows that th

cross-connecte1 po."er directiona

scherne, the overcurrent time and directional scheme, and the pilot wire

scheme are all that discriminate between the good

and the faulty circuit

unless the curre11t in the faulty circuit is appreciably greater than the

current in the good circuit. The pilot vire msch

we is ex :ensive to install,

44

and at the same time no excess current protection is provided, in case of a
bus short or other fault outside the actual circuits protected, without the
addition of the common overcurrent protection.

A consideration of the problem shows that it is possible to provide
adequate protection as long as the Kai River generator is in Operation by
installing over current-power directional relays at the Rockway end and
excess current relays at the Had River and of the circuits. If the had
River generator is not in Operation, this scheme still gives over current
protection, but it will not give discrimination in case of a fault on one
of the feeders.

An installation using overcurrent-power directional relays at both ends,
as in the loop protective scheme, will give discrimination when the had
River generator is not running as well as when it is running, provided the
time settings of all the relays are pr0penly changed each time the generator
is put on or taken off the bus. Obviously, it would not be desirable to use
this method in order to obtain discrimination.

The cross connected power directional systen of protection gives ade-
quate protection and discrimination in case of a phase to phase short cir-
cuit on either feeder. However, in order to leave the remaining circuit
protected, in case one circuit has been removed due to a fault or for any
other reason, and in order to make the scheme extaidable when another feeder
is added some time in the future, it is necessary to use a complicated
system of wiring. Difficulties will also be encountered when trying to re-
move one circuit from service, or replace a circuit in service during periods
of heavy load unless the protective system is rendered inactive during the

switching period.

.5
U1

By using the recommended scheme of protection, which is a combimtion
of the loop protective scheme and the cross-connected power directional
shheme, it is possible to obtain good discrimination in case of a fault on
either feeder, whether or not the Had River generator is running. The use
of the overcurrent relays at the had River end of the circuits provides for
protection against sustained through short-circuits and bus faults at Rock-
Way. At the same time they furnish more positive ground fault protection
than can'be depended upon When the directional elements are used. The con-
nect ions for the recommended system are more simple than if the cross con-
nected system were used throughout. Thereiwill be no difficulty in putting
the second feeder in parallel with the first even when heavily loaded, pro-
vided the Rockway end is closed before the Ka' River end is closed. Like-
wise, thereivill be no danger of destroying the continuity of service when
removing one of the feeders provided the Had River end is Opened first. The
wiring of the reccmnended system is less complicated than that of the cross?
connected system and may also be more easily extended when a third feeder
is installed.

IV. .Ixnical Feeders.

It is recommended that overcurrent, inverse-time relays be used on all
the feeder circuits. It is further recommended that induCtion type relays,
rather than plunger type relays, be used because of their greater accuracy,
more permanent adjustments, and greater reliability,‘and because they place

In all cases, in order to provide

less burden on the current transformers.

adequate protection from ground faults, three relays instead of only two

should be used.

Form ”0—423-26—lw-D'e May's

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

. _ 46
COMMONWEALTH POWER CORPORATION 3““ "°' '
OF NICHlflAN .
ELECTRIOAL ENGINEERING DEPARTMENT
J J_ R E L A Y D A T A J 7' J___J__J __ J
'9 OPERATING co. @110 Edison Company- _W.0-, V x. s. J
' STATION mam Cycles 59 m_._o.n.n.___ MEL-..
Minimum Short Circuit __ _ V f Imp. Date
a 1 '1: .19!”- Pgtawuw ,_.
swag 7, App. Dan:
J APPTLICTRICAL [MUNICH mtc J
i snfililfialgcltigillCAL CHOIR.“
. References J__ __ _____J_J__J I By 252° _ _ J
SUMMARY or SETI'INGS _ w M:
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COMMONWEALTH POWER CORPORATION

OF MICHIGAN
ELECTRICAL ENGINEERING DEPARTMENT

RELAY DATA

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III

BIBLIOGRAPR

Relay Handbook
Protective Relays

Silent Sentinels - Protective Relays
for A.C. and D.C. Systems

Report of Committee on Protective
Devices

Report of Committee on Protective
Devices

Relay Practice
Trend in Relay Protection
Schemes of Relay Protection

Present Day Practice in Relay
Protection

System Connections and Relay Schemes
in Use by the Niagara Falls
Power Company.

Installing Eirectional Relays

Typical Relay Connections

Application and Operation of Relay
Systems

Protective Relays€Reverse-Power Type

52

Published by N.E.L.A., 1936.

v. H. TOdd.
Westinghouse Bulletin, S.P. 1666.
J1. of.A.I.E.E., Hov., 1926.

J1. of A.I.E.E., Nov., 1925.
Electric Journal, 1924, pg. 319.
Electrical World, 1922, pg. 66.

J1. of A.I.E.E.. 1931, pg. 690.
Electrical Jorld, 1920, pg. 133.
Electric Light a Power, Rov.,'25.

Electrical Korld, l9L5, pg. 457.

Electric Journal, 1921, pg. 29.

Electric Journal, 1921, pg. 497.

Power, 1920, pg. 987.

‘INIILI‘I‘! all" It

' ROOM use ONLY

   

O 8972

1 3030

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