LEVEL 2

Short Circuit ·ABC-Learn It 1n an Hour, Use It Anywhere, Memorize No fofmula

-._-. �_""" .. � __ -_-__ -_-_-_·-. -��---�-- . -:�MO ON H; YUEN, SENIOR. MEMBER, IEEE

Abstract-Short circuit ABC-leun it in iilllour, use it anywhere, memorize no formula. The MYA method for 110lving industrial powel" system short circuits appropriately fits this dexription. Indeed, :;olving short circuit problems with· tile MY A method is u easy u learning the ABC's..

INTRODUCTION

S.HORT CIRCUIT studies are necessary for any power distribution system to determine switchgear rating for

protective relaying, and to determine the voltage drop during starting of large motors. One line diagrams are not complete . unless the short c�rcuit values ii.re solved at various sfrategic points. No �ubstation equipment, motor control centers, breaker panels, etc., can be purchased without knowledge of of the complete short circuit information of the enfire power distribution system.

· · ·

Knowing how .to calculate· short circuit problems is a must for. every electrical engineer. To learn it may be easy for some, difficult for others. However, to db the problems, anywhere in or out of the office where tl:ie references ari: not available

_ _ may not .be an easy task because the conventional mcfthods of solving short circuits involve too many formulas. To memorize them at all times is impractical for the maj6ritY.

WHAT REALLY Is THE MVA METHOD?

--:----=:-Basica!Jy, the. MV A method is a modification of the Ohmic method in which the impedance of a circuit is the rum of the impedances of the various components of the circuit. Since, by definition, admittance is the reciprocal of impedance, it follows that the reciprocai of the system admittance is the sum

-=.- of the reciprocals�_f the admittances of the CO!llponents:-A!so, -.=.::::_:by='definition-;-the:-adm1ttance of a circuit or compo.nent is the

ma.Ximum current or KVXa"t'unit voltage which woiild flow through the circuit or compo�ent to a short circuit or fault

____ when_supplie_d_fr()IIl�-�ource . of infJ!li�Lcapacity.'_-_ Refer to Fig. 1.

.

1 Y=­Zohms

·----:=--- - - -�'�= K�� � �000 X (I0!2 X }_'.".MV� =(KV)2 X Y .

(I)

(2)

(3)

(4)

Paper TOD-73-132, approved by the Petroleum and Chemical Industry Committee of the IEEE Industry Applications Society for presentation al the 1973 Petroleum and Chemical Industry Conference, Howton, Tex., September 17-19. Manuscript released for publication October 31, 1973. -

The allthor ll-with the Bechtel Corporation, San Francisco, Calif.

INFINITE BUS

13.BICV

he • VIZ fil{'JVJISC• VSl./JVJV/Z

Z • 0.01 OHMS 0 13.8 KV

vJE ISC • E2/z VASC E2/z ICVASC MVASC

• 1000 IKVJ· 2 I Z • KV 2/ Z

FAULT

E • Y'JV I SC • SHORT CIRCUIT CURRENT IN AMPERES E •LINE TO LINE VOLTAGE IN VOLTS

V • LINE TO NEUTRAL VOLTAGE IN VOL TS ' Z • Llf!E TO NEUTRAL IMPl!DANCE IN OHMS r.NA SC• SHORT CIRCUIT MVA

Fig. I. One line diagram.

SYSTEM }500 MVA

TRANSFORMER

}llOMVA x • 0.1

50/0.1 •

MOTOR

}

50 MVA x � • o.2

50/0.2 -

MVA 1-2 • MVA1 • MVA7 / MVA1 + MVA2 • 600 • 600/ !500 + l500 • 250

MVA1+J • 250 + 250 • 500 I SC• l500 x 1000 f'/:r x l:LS • 2D900 AMI'S

500

250

13.BICV

Fig. 2. Im�cc diagram. Fig_ 3_ MY A diagram.

y Zo11nu Zpu KV KV� MV� MV�

admittance of a circuit impedance in ohms impedance in per unit line to line voltage $art circuit KV A short circuit MV A (13.8)2 /0.01=19 000 (for Fig. 1).

Practically, the MV A method is used by separating the circuit into components, calculating each component with its own infinite bus as shown in Figs. 2 and 3. Fig. 2 is a typical impedance diagram of a one line diagram. Fig. 3 is an MV A diagram. The conversion fro� a one· line diagram to :in MV A diagram is simple arithmetic. .

Component I, the system, is normally given a short circuit MVA rating. So, one merely writes down 500, which is its system· short circuit MV A. Sometimes, if the system MV A is not available, but its voltage and impedance are given, the short circuit MVA can be calculated with the application of (3).

FUSE TECHNOLOGY COURSE

LEVEL 2

Next, for component 2, use (4). The short cir��t MVA ?f '.ransformer is equal to its own MV A base dmded by its

.,, .. .• per unit impedance. (Use reactance X witli the MVA

method.) .

. .

Next, for component 3, ag� use (4). The short cucmt -Mv A-·contribu"ffonoF-triemotoi-:is-:· eqiial:tc» its ·awn MVA

base -divided by Its own-per unit impedance:-(Use reactance X with the MVA method.)

Now, Jet us examine the MV A diagram, Fig. 3. If a short drcµit is taken at point F, there will be a series flow of MVA1 and MVA2, and their combination will be in parallel with MVA3• The question now is: how do you combine the MV A values in series and in parallel? The answer is again simple arithmetic

series

parallel

(MVA,)X°CMVA2)­MVA1,2 =

(MVA1) + (MV A1)

MV A I+ 2 = MV A I + MV A1 ..

(5)

(6)

From (5) and (6), it can easily be recognized that series MV A combinations are exa·ct1y as resistances computed in parallel. Parallel MV A combinations are exactly as resistances computed in series.

The MV Ase at point F of Fig. 2 then can be calculated as follows:

- 500 x 500 - 250 MVA,,2 - 500 + 500 -

MV Ase = MV AT + MV A3 = 250 + 250 = 500. The tenn with the asterisk is the new MV A 1 value which is the result of combining MVA1 and MVA2• After the opera­tion, the new MVA1 which is 250 MV A, replaces the old MV A1 and MY A2• This scheme of replacing old quantities with new quantities -relates-to computer data memory storage system.

At this point the short circuit MVA is solved. To find the. current value, only the voltage value is required. For example, if the voltage is 13.8 k\'.:, t�_:_c_���ntfsc is

��-MV A-X l-000. -500 X 1000 --- . �- --

:;:/3 X KV . ·0 X 13]r�20..900 A.

THE ABC OF THE MV A Up to now, the reader has spent about 15 min in slow read­

ing. He has found that there has been nothing new, and the formulas are no more than good old Ohm's Law arithmetics. Now, he can forget the formulas and start the ABC.

A. Convert to MVA 's ··· Convert all one line components to short circuit MV A's.

Equipment_ such .as generators, motors, ·transformers, etc., are 1orma1Jy given their own MVA and impedance or reactance c.J;,,�s. The short circuit MVA of each is equal to its MVA

divided by its own per unit impedance �r reactance. :;or a feeder where voltage is given and its impedance or re· tctance is known, its short circuit MV A is eq{w to (KV)2 iivided by its impedance or reactance in ohms.

PAGE 4. 7 . 2

A. CONVERT TO MVA'I �1!500 MVA

-� 1

- - v '; ·"" 1.17 OHMS • (!9)2

J.17

6 ) 11'ilVA � -/12KV ,:>.( ==i= x • 0..079

... ' 12 KV 12 ICV

lj) �j 15 'ilVA

-�

f M II Xd • 0..2

{a)

0.2

(b) Fig. 4. The ABC of the MVA. (a) One line diagram. (b) MVA diagram.

B. COMBINE THE MVA'S

1. SERIES: MVA1.2 • MVA1 x MVA2 I MVA1 + MVA7 2. PARALLEL: MVA1 + 2 • MVA1 + MVAz

.3. DELTATDWYE

(a)

��=:: �2 - Y3 • "°J

S • ID 1 x 021 + (0 2 x DJ) + (03 x D1) . . Y • WYE 1

D • DELTA

(b) Fig. 5. The ABC of the MY A. (a) Delta connection. (b) Wye con­

nection.

1.0

.I

.I

.7

.I

..

1 l/A_:_2 ,l 4 5 I 7·1110 15 20 15 lQ <IQ Ml IQ IO 100

Fig. 6. Series combination ratio.

Incoming line short circuit duty in MV A is nonnally given by power companies. Therefore, use the value as given and no conversion is r�quired. However, if impedance or reactance at the terminal is given, find its short circuit MY A by dividing its (KV)2 by its ohms.

As conversion is being made, an MVA diagram is being developed. On� line diagram 4{�) is replaced with MV A diagram 4{b ).

. :·

B. Combine MVA 's I) Series MV A's are combined as resistances in parallel. 2) Parallel MV A's are added arithmetically. Refer to Fig.

· I

I . ,

.; •\

!

FUSE TECHNOLOGY COURSE

LEVEL 2

13.1 KY 300 MVA

:!OMVA )(,i"•0.1

rMllC:. '+ • I • '1

13.8 KV

X • 0.019 OHMll

::. ==�=--===-=_:::_:;...-;_· :::.�=·-�===�·==··_=-··=--::: ___ :::==�-���:=:=- 1-·�--· - . - .

-- ·

- .

1MVA

X•O.OV

X • 0.1>1 I OHMS

20MVA

x. 0.1

x.i· • 11.2 :_:::·-:-_:_----::-_-. :::....:.:_�;:--- _:..�:_,_3,_ _ _.__ __ _

ALL MOTORS

l50-200HP TOTAL• 1 MVA

�"'·.ll.215

X • 0.011 OHMS

.x.i·- 0.211

:!OMVA

X•0.1

Fig. 7. One line diagram.

4(b) for the-following:

MVA = 1500 X 1230 = 675 1'2 1500 + 1230

(this 1s the riew MV A-1) 675X 198 MVA1 3 = 6 = 153 • 75+198

MVA1+4 = MVA1 + MVA4 = 153 + 75 = 228

228 x 1000 f12=../3Xl2 =llOOOA.

-For MVA-1:;-; add MVAi -and MVA;-m series. For MVA1•4, add MV A I and MV � in parallel. I 12 is the short circuit current at 12 kV.

_ _;3) Delta to.- wye conversions are rarely used in ·industrial power distribution systems, but they are again simple arith-

..:.._ITletic.- Re"rer to-Fi� 5(a)and (b).�.::-=- - _-- · · -

--- 4)-ThLQulri;ioint that ru:e.di.JnorCattenti9n is thLJ:enes combination if a slide rule is not available. The attempt here is to be able to solve most short circuit problems with reasonable

-accuracy-without-the use·of-a-slide"rule;·----.:_ __ . -·--With the aid of the curve in Fig. 6, let us analyze the series

combination

·Let A:::; MVA1 ,B =MVA2; and T= total MVA, so that . ·- := A..X_.B_A (B)_..:_- .. - . -

-------,r-----::i: --· . A<.B- -A+B A+B'

.

. , ' + B is platted as a constant on a log-log scale base from 100 which is the ratio of B/A. (Refer to Fig. 6.) For

example, let A = 10, B = 40, and B/A = 4. Read 4, at 11.ori­zontal scale. · From 4 project upward until it intersects the

J 4 I I I I I 15

11

12

C. REDUCE THE MVA DIAGRAM

13.BKV

4,1111CV

Fis. 8. MY A diagram.

curve. Read the lwrizontal value 0.8, which is the result of B/A + B; then

T=A(B/A +B)

=IO X 0.8 = 8.

It is also noted that when combining two quantities in series, the result is always smaller than the smallest of the two. The. example shows the result to be 8 when combining l 0 and �· C. Reduce MVA Diagram

Reducing an MVA diagram takes the same reduction process

required for the per unit impedance diagram, except tlut

MV A quantities are used instead of per unit impecll.nccs or

reactances. Fig. 7 is a typical distrlbution one line diagram including a

delta connected feeder system. Reactanccs only are used for practical purposes. Fig. 8 is an MY A diagram th.at mows all elements in the one line in MY A quantities. Fig. 9{a) shows

FUSE TECHNOLOGY COURSE

LEVEL 2

··- :JOO -

_·f, _-�-.·---10000

.:. _· -_-:-_· �·-,

;·-. .

-.-. -

��

·.--_.,:..-;·· 12

(a)

- , -- -- v1�

�

-

. . ' 19'! B TN

-

-

--

-

--

- .

-.- ;..._ ----=--- - _. - .

s J.llG • 10• "

3IMI Y3 " o; " lOOOO

s J.9tl x 10• v, " 0. " --, ..

-- " 20191

200

13.BKV

, .. •• 181CV

s J.IMI .�o·· Ye" De " --, .. - " 20;;'• yp..._ i./)' ··r\) 6 .\,Q "t"Y" [") � . ··

.

S • (Y.3-f .. XJY,)�+ IY3l X 1.Yel • !Y,l X IYel • (100001 x (191) + (100001 • (1911 + (1Nl • (191)

• J.N • 1oe

(b)

,,

(c)

Fig. 9. MVA reduction.

=- ---- --· -· : . .:.--;-1e first step reduction sequence. Note that there are three .U!tS:i6 be.C:i.lcwated, F1--; F2-; and F ;:-=-�-The7"fii-st� �ep r�Cfllctffin"cornbines the series -�d p�allel >mponents so that the simplest diagram can be accomplished, id that any fault can be solved in random fashion. Fig. 9(a) ows..that..iterns-4-<uld-5-have been-combined to-make a new items 6 and 7 have been combined to make a new 6; items

-10 have been combined to make a new 8. Fig. 9(b) con­rted a delta to a wye configuration. Fig. 9(c) is further duced_to Fig. JO(a), indicating items 2 and 4 have been ·mb·iii.;i �-��friake·-a-new �1:-:- Figure 1 O(a) then ill the simplest igram. th.at-would allow . the solving of any of the three .tlts in·-random".selection. Figs. IO(a)-{c) show the reduction �-Ss..� solving.::faulta.:F1::;.Fn:Uid.F3.J-respectively.

. WHY TIIE MY A METHOD? . . .1 ••• .: are many reasons why. the MVA method is recom­:nded for industrial power short circuit calculations. ) It does not require a common MV A base as required by : per unit method. ·

PAGE 4. 7 . 4

Fl

(a)

•.lBKV

(b)

(c)

Fig. IO. MVA reduction.

4.18KV

l

1 j

. . � 1 t

·t � !

FUSE TECHNOLOGY COURSE

LEVEL 2

F 2.4 KV

SYSTEM

1J.B KV X • 0.151 OOms

5000 ICVA X • O.OGli

2500 KVA x � -o.1e

Fig. 11. One line diagram for comparison of methods.

2) It is not necessary to convert impedances from one voltage to another as required by the Ohmic method.

3) The conversion formulas as used for both the Ohmic and the per unit methods are complex and not easy to . memorize.

4) Both the Ohmic and the per unit methods usually end up with. small decimals resulting from converting impedances from one voltage to another or from converting impedances to .the same common base. Therefore, one can make mistakes

-in the decimals, with resulting wrong answers. 5) The MV A method utilizes large whole numbers denoting

MVA quantities. With a little practice, one can estimate the mlt by looking at the combination. For example, 10 and 10

-ill .series become 5; 10 and 100 in series become 9.1; and 10 and 10 000 in series give 10. A small number combined with too large a number, 100 times larger or more, will have no ef­fect on the small number.

In order to further prove the preceding points, it is necessary --to give the following comparison of methods that are utilized

- in solving industrial power system short circuits.

COMPARISON OF METHODS A one line diagram, Fig. 11, is shown. Solve the three-phase

- --fault at point F' with-and without motor· contribution. �. - ·.- -=_NgJe -that r��i::Jfillc'es only_ a.Ie. considered-T�jhe tlu:e��cases

being compared .. It is felt that using impedances would give the same result, but would complicate the calculations. It is

--·also-widely recognized-and - a_cceptable-by -industries to use reactances only in calculating industrial power system short circuits, in that it would result in a higher short circuit value, perhaps by- 0-� percent in most cases. Reference [I]

_exemplifies the use of reactances rather than impedances. c-:-·Figs. 12(a)-{c) tabulate the conversion calculations for the three methods. Figs. 13-15 show the three methods utilized for comparison. Fig. 16 tabulates the results of the three

-=-:.mettrod;-:-:. . ·- --

CAN PHASE-GROUND FAULT BE SOLVED? The answer, of course, is yes. Solving phase-ground fault is

as easy as solving three-phase fault. Refer to Figs. 4{a) and (b ). This problem is taken from the

California State Professional Engineer Registration Examina-

sYSTEM

1.

ll.SKV FEEDEll

2.

TRAHSl-OflMEll

J..

MOTOll

4.

SYSTEM

1.

1J..SKV FEEOEll

2.

TRANSFORMER

J.

MOTOR.

4.

SYSTEM

1J..IXV FEEDER

2.

TRAHSFORMER

MOTOR

�-

PAGE 4.7.5

�IC MlTHOO

k • 10CXl •..;,,KVt2 KV � 1000. (11.8) 2 .

llOllCOO • 11.:11..Jlli!!J

X • Q 1�] QHM;i

X• 10CXJ • IX �.U.) • !KV) 2 KVAb

• 1000. 0.0'5!5. 12.412 ISOOO

x . 1000 • IXP.U.w IKV)2 KV Ab

- 1000 • 0. , •• (2.111)2 2500

(a)

PEA UNIT METHOD (500000 KVA RASE)

x BASE KVA • 111 P.U.• KVA S.C. 500000. 1 .

500000

XP.U.• !OHMS)• !BASE KVAI

• O.OllJ OHMS

• O..Je'! OHMS

•• 1.000

H>OO • IKV1 2 ( 0.1511•1500000) . (13.1)2. 1000

x IX P.U.I •(BASE KVAI P.U. • KVA.T

x

ID.055l • 15000001 . 5000

IXl'.U) •!BASE KVAI �.U.• KVAm

ID. 181 • 15000001 . 2SOO

(b) ' MVA METHOD

MVA1 • 500

!KV) 2 (13.BI 2 MVAi• -- •--

X OHMS 0.151

5 MVAT MV"J • -- • --XP.U. 0.05li

MVAm 2.5 MVA4• -- • --XP.U. 0.15

(c)

• 12'!0 -

• 91 -

• 15.8 -

-�

• 5.500

• 32.000 --

.

rig. 12. (a) Ohmic conversion. (b) Per unit conversion. (c) MVA �version.

ti on of August, 1965. As noted, the three-phase fault has been solved to be 228 MV A at the 12-kV bus. Since the positive sequence fault is equal to the negative sequence fault, therefore,

MVAx1 = MV An = 228.

The zero sequence fault MVA, however, must be calculated, and its MVA value then is combined with the positive and negative MVA values. .

Refer to Fig. 4{a) again. During a fault on the 12-kV bus, only the transformer and the motor contribute to zero sequence MVA's. The delta primary of the transformer blocks

FUSE TECHNOLOGY COURSE

LEVEL 2

SYSTEM

,., . ____ 1

x1 • 11.3*1 OHMS• 13.IKV . x2 • Q.151 OHMS• 13.IXV

_ •• Xl,2 • 0631 OHMS• 13.aKV

x1.2 • IO.IO'I) • I0.031•2.4KV (••) • 0.011 OHMS• 2.4KV

CABLE XJ" 0.00!3 OHMS• 2.4KV

=-�;���=-�- �.!!2���-��::::._::::.._�::::':· ::::.c...Xf.� �OHMS• �V ----....

-------'1-c-=-------

-----�llA.i<-• IKV) 2 • 12.412 • 72.1

MOTOR

SYSTEM

CABLE

. TRANSF

.. _. ____ OR

2.41(11

F 2.4Kll

xF a.on x4 • O.:IM OHMS • 2.4Kll XF+M • (0.0791 • I0.36ill • � OHMS

(Q.0791 • i<LJ&a)

Mll:�+M • (KV) 2 • (2.412 • 1111 x;- � -

•• OHMS 2.4 •OHMS lJ.l > 12.4/13.I) 2

• OHMS l3.B • (D.03)

PER UNIT METHOD BASE Ml/A• 500

X1•1.000

x2 • 0.39'5

X3 • 5.500

XF 1.8911 Ml/AF,• �/8.11941 • �

(15.-) • (32.000)

WITHOUT MOTOR CONTRIBUTION

X F+M • (15.8911) + (32.0001 • 15.67

WITH MOTOR CONTRIBUTION

Fig. 14. Reactancc diagram.

Ml/A METHOD

500

15.1 15.1 15.1

MVA 1+4 ·�

-- . ---�:...:·--

-----------------------·

- .::=--...._ -· :..At£rHCllS- :....OHMIC- -PER UNIT FAULT CONDIT� METHOD METHOD

FAULT 0 2.4KV BUS 72.8 Ml/A 72.1 MllA , .. -yovr MOTOR CONTRl8UTION

_T • 2.4Kll BUS 118MVA -.. ... I WITH MOTOR CONTRIBUTION

Fig. 16. Result of comparoon of methods.

--Ml/A METHOD

72.1 MVA

&8.2MVA

PAGE 4.7.6

J.l'HA£E FAULT Ml/A• 221

1119 Ml/AX OT THEREFORE:

MllAx 1•MVAx 2·m 12KV Ml/AX 0 •Ml/AX OT+ MVA XOM

• 191 + 150

150 -�

15 .

1!50 •T.f-MOTOR ZERO SEOUENCE REACTANCE

Fii;. 17. Zero 5equcnce fault power.

Ml/A 1,2 • �2" ill MVA1,J • 114 X 348/114 +348 • !!f MVAfO • J X 16 • i1!§ 1Fo

.•258X 1000/ i/j• 12• �

POSITIVE SEQUENCE POWER

NEGATIVE SEQUENCE POWER ZERO SEQUENCE POWER

Fig. 18. Phase-ground fault Qf MY A circuit.

any zero sequence power flowing from the system and across the transformer. Therefore, Fig. 17 shows the zero sequence power circuit

MV AxoT = MV Ax1 = MV An = 198

(the transformer zero sequence reactance is equal to its posi· tive and negative reactances)

15 MVAxoM=-= 150MVA 0.1

(since the zero sequence reactance of the motor is about t of its positive sequence reactance). The total zero sequence fault power then is equal to the sum, which is

MVAxoT + MVAxoM = 198 + 150 = 348.

The phase-ground fault power is obtained with-the use of Fig. 18. Since these are three branches in parallel, the s.hnplest approach is to take one branch out of the circuit and solve its MVA value, then multiply. the value by 3, which gives the final answer

MVA1,2=228/2=114 114 x 348 .

MVA - -86 \ '3 - 114 + 348 -

MV AFo = 3 X 86 = 258 258 x 1000

IFO = VJX 12 = 12 400A.

The problem, as shown in Fig. 4(a), is also solved with the per unit method as Appendix I. This gives further comparison of

ii ::1

i I ',I

I ' j

! ;[ r

I [ i I

FUSE TECHNOLOGY COURSE

LEVEL 2

8'!1 •...., -=- ����-IM+4a0�·221 MVA

- . - 221 • 1000 I• --- • 10800 Yla12 =

IKVJ2 11212 ·--·-· '4&0

.J .J

'•1"'·2·1..,. ------ · l.O -�-0."91 X1 + x2 + x0 t JXt 0.456 + 0.556 + 0.428 + 0.3 2.0ll -

-· _ 11pu.: �1,1

_� J • 0."92 • 1_.•n � _

1, · 1An x 7230• �

Fig. 19: Phase to ground fault with added fault react2Ilce Xr-

,. �� I "' �.r. '-------v_,/ '1 -: c

ZERO

POS.

NEG.

lo O/'EAATORS

• � •••• -0.500 + J 0.11811 12 �D• O°b • -0.500 • J 0.11511

•3• 1.0 ••••

CONN[CTIONS BETWEEN SEQUENCE NETWORKS

Fig. 20. Two lines ta ground fault.

-----Fig.-.Z 1: Two-phase-to ground.fault by per unit-method.

-

·-,�:- t--· - -�y� � -f ====- MVAX-- -MVA:e----1 1

• 183.5 • --1:!!._ �·221

�-- . �VAF•JMVAXa •Jalt8•�

IF•� Fig. 22. Two-phase to ground fault by MVA me.thod.

PAGE 4. 7. 7

�-��r��1� SYSTEM SYMETRICAL

SHOAT CIRCUIT· 500 MVA VII• SM'VA --------��tMI

·:�'7_: .. ��-.-:�·-·

SM'VA + MVA SC 1 • -:· . : - ...

7.5 MVA TRANSF SMVA. •STARTING llfVA MVA SC• SHOAT CJRCUIT ti/VA

�-ooHM

X • O.OUJ l!ll/13.IKV V d • INSTANTANEOUS VOLTA.CU DAOP

13.8 KV

X • 0.19 OHM

(71 MVAI F2 (SHAT CIRCU1n

4000 HP INDUCTION MOTOR J.S MVA STARTING MVA • 21 AT 13.8 KV

· (19.3 MVA AT 13.2 KV)

v d • 21121 + 71 • 21/92. 11.2211 V TlJ.; = 1-0.22ii • 0.772

v Tl3.2 • o.m • 1�1 2 13 .2

• O.!IOll OR l!0.51' V TlJ..ll • MOTOR TERMINAL

VOLTAGE• 13..l KV V T1:L2 • MOTOR TERMINAL

VOLTA.GE• 112 KV

Fig. 23. Motor starting voltage drop calculation.

the amount of calculation involved between the MV A and the per unit methods.

Phase-ground fault with an added fault neutral reactance also can be calculated with the MV A method. Fig. 19 il­lustrates _the preceding problem with an added fault neutral reactaiice ){1. Note that using both the MV A and the per unit methods obtain the same result except that the MV A method shows much less calculation.

Two-PHASE TO GROUND FAULT

Can two-phase to ground fault be solved with the MVA method? The answer is again, yes. Fig. 20 shows a two-phase to ground fault diagram and connections between sequence networks. As indicated, 11 is the fault current between phases C,B, and ground.

In order to develop an MV A equation for two-phase to ground fault calculations, the classical symmetrical component equations are utilized as basis. Relationships between phase and sequence quantities are expressed by the following:

·

Va = Vo + V1 + V2 (7)

Vb= V0 +a2V1 +aV2 (8)

Ve= V0 +aV1 +a2 V2 (9)

Ia =Io +11 +12 (IO)

lb =10 +a211 +all (11)

le =10 +al1 +a212. (12)

From the preceding equations, the following relationships are obtained:

Vo = i CVa + vb + Ve) . (13)

V1 = i CVa +a Vb +al Ve) (14)

V2 = i CVa +al vb+ a Ve) (15)

lo =j(Ia+Ib +le) (16)

11 = j- (Ia +alb +al le) (17)

12 = i (Ia + a2 h +ale)- (18)

FUSE TECHNOLOGY COURSE

LEVEL 2

PAGE 4 . 7 .8

120 �----.----------------.. 115 1101-

------<'---------------;

100P....,.,,....--�--------------1

-::-·-:::::;-· - - ·-- . . . . .

. ___ ...:....�-. · .

.23 ==re::: ·- . ..., -

.::::=--T�� . .'-··...:....

EQUATION fOR LINE -C-, · HW

y•MX•-- X I.IV".

"" - aD

1 2 10

l 4 20

y

o,.� . .,..,,, o.,

..,

SHORT ClllCUIT MVA

5 I 30

7 I 40

(a)

(SMVA, 1.1141

A

9 10 11 50

12 13 14 IQ 70

15 11 aD

:OLVE FOi'! X ft'ITH -C­

• • y (SMVAJ

suasTITUTE X INTO "A" y 1.04

y UM .___. ____ --"l�x MVA,.· y (SMVAJ MVA,. -M·-· --

SllVA

EQUATION FOR LINE "A".

� • Y2·Y1

• . • , •2 .• ,

y - 0 1.04 - 0 ---·--x · MVA,. O· MVA,..

. y 1.114 OR--·--MVA,.·X MV"..

i-v.t.,.,. 01

(b)

yjMVA,.) • 1.04[MVA,. • y(SMVAI] y • UW MV A,,jMV A,.+ SMV A

SMVA •STARTING MVA

MVA,. • SHORT CIRCUIT MVA V1 • Y •STARTING VOLTAGE

1.GI • 71 71 + 21

- 0.801

• 80.51' Al'rf!OX.

Fig. 24. (a) Voltage drop by graphic solution. (b) Graphic solution proven by analytic geometry.

Connections between sequence networks can be found from From the preceding analysis, (22)-(25) can be derived for the application of both the per unit method 13).::{18):=.·;::_--:-_-::::�� --

- .

- Io ·+ Ii .tt-I2 =O

___ ___ ·::-_-_:-· ro _= Vi '.".. Vi·_··.--.--=-.-=_:_--- - · ---'---'---� ------.---�

(19)

(20)

E Ii=

-------x (X2)(Xo )

1 +----­(X:z) + (Xo )

(22)

-JlquatiOfl& (l 9Tanl-(20) �e sa:t�fie<l-if-the-sequence net- (23) 1orlcs are connected as follows:

-----·-·---·-==- ··------� !F-5-f ,,__tf�----� ld by addition of (11) and (12)

Ib+Ic=·2fo +(a2 +a)(I1 +I2). ..

ncE__a2 .:!: a ��_:::-__ l, th�r�fore, _.;;-- _ .. --:-___ _

----�Ib_:±_f.c.·:::./i'= 210:. (Ii_-U2). __

nee --· -- ·--. --- - --- -- ·-

ere fore

----�- -

Io+ Ii + I2 = 0 Ii+I1=-Io .

and the MY A method

_ MYAiX.(MYA2 + MY.Ao) MVAx, -MYA1+(MYA2 + MY.Ao)

(24)

(25)

Fig. 21 shows the use of (22) and (23) for solving the two­ph.a.se to ground fault of the same problem, the per unit method. Fig. 22 indicates the utilization of (24) and (25) for solving the two-pha:se to ground fault of the same problem, the MV A method. It is obvious as shown, that the use of the

(21) MV A method is simpler.

I 1

� I I I l

] � " 1 i 1

I

I I

FUSE TECHNOLOGY COURSE

LEVEL 2

PAGE 4 . 7 . 9

120 -1���������+-����������������������������������� 115 --- --------------------- ------ - - -·---- -----�----

110-1���������+-�����������������������������������

w Cl <(

BO

� 60 0 > ;rl!

40

20

.2 3 4

2

5 6 7 8 9 10

3 4 5

� l\OTDft STARTING VOLTA!;E OkOI'

1. S•t o,,.r.tor 1'A1•on ]H'\VA of short circuit sc•ll� and'""'"' it 10 lu Mirllne falh on 10Lt"L on the 'l -wolU!Je scale.

2. S•t Operator "I" on 21 surting -,..y,. on short circuit suilc.

J. S•lng Operator "'C" so IU h.alrlirw fa th on ttw: lnt•rucalon of Operator ''B" and 10't't "'°luge.

i.. "-••d 80.Sl wlt..gc on croulf19 point 1Ud• by O�rat ions '"-" and "t" ..

5. If It Is d•slrad to find �t surtln9 voltag• c•n be obtained with Inc.rosing bus voltagcs to I07X .and 109X. (corrnpondll't9 to 2}l and Sl hp uttln91 abov. no,....I trantfor.r volt.ag•).

11

A. SwlftCJ 0p.9:ntor ''C" so Its hairline hits on Intersection of Opcntor "111 and lOTX. volt•CJC·

' kead 82.Sl on crot.tlng point IMdc by O�ntors ,.._11 and ''C".

I. Swing 0SM"r•tor ''C11 10 Its h•lrllne hlh on lntmrHctlOf'I of Opentor ''111 aM 109lo volt•ire-

'-••d 85l. on c�• Ing point .. ci. by Openton ''A" •nd ''C'".

12

6

13 14

7

15 16

8

17 18

9

19 20

10

SHORT CIRCUIT MVA,

Fig. 25. Slide rule model.

MVAMETHOD FOR INSTANTANEOUS VOLTAGE EsT™ATE

Large motors are frequently connected to power systems �-:-�_.consisting of complicated-networks of lines and cables for . which a calculation of the voltage drop would be diffic�t.

Yet, it may be critical to know approximately what the __ voltag�.at certain bus must be. . Titis is because the voltage

affects the motor torque in a square function; i.e., motor -:-torque -varies as· the_square _of the voltage for a 10-percent =��OltBge drop== · �--=�-.-- :-· .-· ---�� · -- - ---·

------· ·

torque er (E) 2

_T= (0.9]2 =�.81 or 81 p�rcent.

The torque loss is 19 percent. The voltage drop may be estimated with reasonable ac­

curacy, however, if the short circuit MVA is known at the � -. paint. of po�er delivery .. -\V!len motor starting MV A is dra\vn

_from - a system, the voltage drop in per. unit of the initial -· voltage irapproximately equal to the motor starting MY A

_ divid�d. by the su� of this MY A and the short circuit MY A

V = MVA1 pu MY A3 + MV Ase. (26)

• • g. 23 shows an example applying the MY A method in estimating the voltage at the 13.8-kY bus when a large motor is started.

Fig. 24(a) shows a graphic solution of the problem. Figure 24(b) illustrates the validity of the graphic solution by analytic geometry. Fig. 25 shows a slide rule made for the sole purpose of solving instantaneous voltage drop in starting large motors. The instruction for the use of the slide rule as shown is self-explanatory.

Fig. 26 is a compilation of standard industrial nominal voltages and motor terminal voltages. Note that the unique relationship between the nominal and terminal voltages is 4 percent different. 1hls unique relationship aids the slide rule operation in solving instantaneous voltage drop during motor starting.

THE S™PLE COMPUTER TIME SHARE PROGRAM Why is it necessary to develop a computer program for such

an easy method?. The answer is simply economics. True, the MY A method enables the engineer to quickly calculate the faults on a power system, but how about documentation? After the engineer finishes his calculation, the result needs typing, proofreading, etc. From rr{;inua} calculation to print· ing, the estimated cost for a problem as shown in Appendix II, having 12 components and three faults, is approximately S 100 including· the engineer's pay. But the time share program solution from input to print-out costs approximately S24 (S 12 for engineering time and $12 for computer and terminal

FUSE TECHNOLOGY COURSE

LEVEL 2

A• TRANSFORMER SECOHoARY VOLTAGE

B •MOTOR TERMINAL VOLTAGE ·

/lJ8 . U0/"60. 2400/2300. 41!!0/<4000. 12000/11600 •. • 13800/13200 • 22'00/22000 . 104%

WHEN ONE 2 112 TRANSFORMER TAP IS USED:

- • A/B • 490/4'50 • 24a0/Zl00 • 42ll0/4000 • 1Z300/11500

• 1•100/13200 • 23500/22000 • 107"'

·2 - 2 112 TAPS ABOVE WILL GIVE· A/B • U2 ------ -- -- -· -· ·-- -· -ig. 26: Percent ·motor termiral voltaze..J:Clated to transformer sec­

ondary voltage.

line). As the- problem involves more components and more mlts, the cost differential increases noticeably.

Appendix II is a time share computer solution of Fig. 7. 'he program itself is a conversational type. The user of the rogram can_input the data_by a.prepared tape or by tYPirig 1e data as the computer.is readytoreceive llie data. The input data are separated into two sections. Section one to use lines 200 through 399 for MV A items as shown with

1e use of the MVA method. For example, item I is 300 IV A, item 2 is 200 MV A, etc. Section two is to use lines 00 through 999 for ·command sequence. For example, line 00 data I, 4, 5 instructs the· computer to combine items 4 1d 5 in series; the first number, 1, is for a series operation. t line 400 data, there is a 2, 8, and 9, which instructs the �mputer to combine items 8 and 9 in parallel; the first 11mber, 2, i�for a parallel operation. -A 3, 3, 4, 6 command lStructs the computer to convert a delta to wye operation of

..,,4,and6. �--; to Appendix ll. The fault 1 result is 533.4 MVA,

hich is close to the manual solution result of 533 MVA. ate the computer asked for a kilovolt input. The user 1tered the voltage. The computer then asked whether the :er prefers an interrupting duty or a momentary duty compu­tiol.f_, __ As sho'!VJl, fault 1 requires an interrupting calculation td the. computer gave a series of output selections to meet NSI Standard latest requireme�t of multipliers. The com-1ter solution sequence is exactly as shown on Fig. IO(a) for ult I, manual solution.--For fault 2, line 410 data are replaced with new commands $.own::.Notk� ihat...thtu;f;g__ue_nJ,:ej..Qllows_the MVA diagram,

.g...-1 O(b �-The:::f�f-fault-2-is-261.9-MVA which was anually solved to be 262 MV A, Fig. IO(b ). The computer ain asked for a kilovolt input and a 4.16 was given. The-next-question -again was-for!fiteirupting duty ·or omentary duty. The answer was momentary so the com-1ter gave two answers that are in accordance with [2] . For fault 3, line 410 data are replaced with new commands at f<:>.!_lmv _the seql!ence as shown on Fig. 10( c ). The computer k-ed for a kilovolt input, and 0.48 was given. Because 0.48 r-is ·a-low-voltage··systei:n,. the -computer automatically inte� _ _9!1t_fiv�_ansvii:_r�_J_o suit the user's choice. The multi­ers area:.IJ-lii"accordance with theIEEFRed Book #141, ".tinn IV ...

the MV A method is mastered in about an hour's lie, 1t will take no more than fifteen minutes to learn to use � computer program. Appendix III is a pre-made graph for ick estimate of instantaneous voltage in starting large motors.

PAGE 4.7.10

CONCLUSION

The paper described a unique easy to learn and easy to remember method for solving industrial power distribution system short circuit problems. The examples given proved its effectiveness in terms of speed, accuracy, and .economy over other conventional Ohmic and per unit methods. The writer has been using it for the past twenty years for many projects, small and large, and found it most effective because it seldom required one to memorize formulas as with other methods.

The MV A method also has been taught in various evening schools and corporation sponsored 1eminars, including the University of California Extension, ITC College, Bechtel Corporation, Pacific Gas and Electric Co., etc., in the San Francisco Bay area for the past seven years.

REFERENCES

[1) Electrical Powu Distribution for Industrial Plants, IEEE Standard 141, 1969, ch. IV.

[2) IEEE Red Book,iEEE Standard 141, 1969, sect. IV.

APPENDIX I The following problem is taken from the California �ate

P.E. Registration Examination of August, 1965. The problem will be solved with the per unit method.

UTILITY 5Y'ST£H . tt.

- � G_Lr; .... · --=-62"'•_,_.,v �---"'__3 . "'--6 '8-�:........; x•i.&1 a""' j r"\ Fo11ult � ISOOHVA <J Q - .

15HVA .,.. I SHVA 69/ I 2KV I 2KV .-0.075 .d ... 0.200

A. Solve for )•phu.e f•ult B. Solve for sinqle llna to ground f•ult A. )-PHASE FAULT

ICVAbau • I SDHVA

KVbua • 69 .11nd 12 respectively

xb 69. (KV)2• 1000. (69)2• 1000. Jl.8 Ohms ue KVAb•s• 150000

x • b•u 12 • (12)2 x 1000 • Q,96 Ot.m. 150000 I 50000 • 1250 A

� I 50000 • 72)0 A

VTXT2 Xutlllty p.u. • 1.0 x ��:bu• ·,��ggg • O.IOOp.u .

s.c.

Xtr•nsf. p.u. • tr•ud p.u. :�:b•�• • 0.07511 lf��g� • 0.750 p.u. r•t•d

x1 J • Xr•ted Ohms • J.87 • 0.121 p.u . .,,. p.u. xbue JCT xmotor di". 0.200 x 17��gg. 2.00 p.u.

.u. (•

x. • xut. + x11,,. + xtr•nsf

• 0.100 +.0.121 • o. 750 • a. 971 ·

xb • 0.200

x • 0.971 x 2 • 0.655 •9 �

x.do11 • 0.100

I fault • f-• Hss • 1.525 A ag

1r.ult J ph (ii 121tY • 1f•ult p,u. x 1b•H 12 J(V • 1 . 525 x 7210. �

I j � { l f i I

J l l -� } ;

FUSE TECHNOLOGY COURSE

LEVEL 2

------ --··- -·--

r •• 1.0

x • o.655 •9

·�-----'�-- --- -···

z.ro �nc.9

XO nt • 3xn -.otor • 0

X0 tr• D.750

x n Jl'IOtor

XO nt • ��1"1) T�=-�..:u�:.':� of

XO tr • !:��ir:�o=Unc•

1motor dO .. • �c!=::• t�b�::�lent Xmotor dO" • t x ..at.or dJl1 • 1.000

Xmotor di"• Posltlw S.qu11nca Subtran1l•n1: k.aacu� of the Hotar

E • 1.0

----· ·-·-· _. -·-==-===-·---·-

x. • o.1so :11__ 1 - � . o.1i2e 9 � l.];O --

_____ .. _. --=---!..1 .. -:....!.2-=.1.a - r. ________ xi� x2 -� xo + 3x" + jxrau1 t -

-='-==-....:..;..;..-'-:::.:...--··: -· . . _ .. : �Z::f...:: ... :-· -==-- ·-

·. - :-:"7 ___ __ _

'.1 • '•2 - I • I •0 0.655 • 0.655 + o.428 • I • 0.575

-r:m-fault p.u. • (�+l.2+-1�--� J"(0.575)

• 1.n 1hult • I fault p.u. x lb.es• 12 KV

· -1._n x nJo -12,4ilo A

--- · ·--··-=-·-·-- .- .

,·-·· :.:..

Jl[ADY 'TM'( .11£AOY

PAGE 4 . 7.11

200 o.tiTA JOO, 200, 10000, 10000, 200, 10000 210 D.f.TA 200, 5. 10, 24, 15, 4

400 DATA I, 4, 5, I, 60 7, 2, 8, 9, 2, 8, 10, ), ) . . ._ .(, _,O DATA I. 11, 12, 2, !, 11,· 1, J. 8, 1, 2, 4, 2,;. ), 1, 2, 6, 2, 1, 2

"""

IS THIS THC FIAST FAULT Tt I( .lll»4? l•YES, O-Nll, ?1

MID[ NII 1

""' )00.0

2 J 4 5 6 7 ft � .

10 11 12

1 2 J 4· s

6 7 ft 9

10 11 12

200,0 10000.0 10000.0

200.0 10000.0

SE"' ES SERI ES

200.0 s .o

10.0 24.o 15 .0

i..o

PAMLLEL PAIV.LLEL DELTA IN[ SEklES PAMLLEL StklES 5Elt1ES PAMLLEL SERI ES PAMLLEL

INPUT kV 713.B

a 10000.D 200.D 10000.0 200.0

5 .o 10.0 15 .o 24.0

10000.0 196.1 J9".0 201%.1

15.0 4.0 )9.0 J.2

J9".0 42.2 200.0 20196.1 198.0 JB.1 2)6.1 20196.1 300.0 2JJ.4

A AHO 15 196.1 196. 1

15.0 ]9.0

196.1 20196.1

J.2 42.2 38.1

198.0 2J6. 1

� 4

ENTER I fU. IMTE�ftUPTINC DUTY, 2 ru Mi1H£HTAAY DUTY ?1

IHTEIUWPTINC. DUTY X/R MTlt Xlk ltJLTIPLIEk HVA AT FAULT I SYl'V1ETllltAL

READY ,.,TAPE READY

0-25 1.0

5Jl.4 22316.J

26 ..... 1.1

586.8 2"547 .9

41-6<1 1.2

640.1 26779.6

410 DATA 11 II, 12, 2, 81 ti, I, 1, 6, 1, 2. 4, 2, I. 2,

::l\UH

IS THIS THE FIRST f'AULT Tl B( kUN7 .T•YES, O• ..... ?O INPUT Tiii L IHE ""S \IHEkE PklHTIUT STAkTS ?12 .5 NfDE HO MYA

6 7 B 9

10 11 12

12 4.o

DELTA \/YE S£k1ES PAMLL£L SEklES SEklES PAMLLEL SEklES PAAALLEL

INl'llT KV 74.16

A a 10000.0 196.1

. J96.D 20196.1 15,0 i..o

. ]9.0 J.2 J00.0 20196.1 200.0 20196. I 295.6 198.0 49].6 )96.o 219. 7 42.2

A AND B 19".I

20196. I J.2

42.2 295,6 198.0 49J.6

� .

EHTU. I f'lll IHT(IUtUP'TINi: DUTY, 2 FIR 11t1HEHTAAY Dl!TY 72

""1£HTMY DUTY X/lt !ltATll 0-10 IVE• 10 X/ft MULTIPLIER 1.5 1.6 l'WA AT FAULT 392.8 419.0 1 SY"'1t'.TIUCAl 54520.2 5815<..9

""DES 4 5 6 7 B 9 ft 10 J 4

11 12 B 11 J B 2 .. 2 J 2 6

2

IVEk 60 1.3

693 ... 29011.2

1, I. JI 2, 1. I

""DES J

11 12 8 11

' 4

e.

FUSE TECHNOLOGY COURSE

LEVEL 2

PAGE 4 . 7 . 1 2

APPENDIX III

'·P11M

A READY -MADE GRAPH FO R QUICK ESTIMATE OF

INSTANTANEOUS VO LTAGE I N STARTING LARGE MOTORS .

r S · T H I � THE F I J\�.r -rAULT Td BE f\U�? l • Y E S ,- o.;-H •• ---

?O . -- - -

I I I

I NPUT rw1 .uHE NQS \IHCRE P l't l NTillIT STAATS ... . .. ...

I I '? 1 2 , 6

•

littoE-- Hfl l"\VA .---:-.- -- :-• .:::· - - --·-1 2 � � ·::.:_ � . o -=�-: -= _ .... -;� ·

I I I H .... ?i "'

I "' ...

·" - · �-·--. : . . -

• A a . . A AHO I HIDES 6 SrA I ES · JDD . D -- ·20 1 96 . 1 295 . 6 6 7 S E R I ES -. -· 200 . 0 "2D 1 9 6 . I 1 9 B . D 4

-C· PARALLEL . . 295 . 6 1 96 , D 493.6 2

_,.

"' - r--9 SER I ES 493 . 6 3 9 6 . D 2 1 9 . 7 3 I D PARALLEL 2 1 9 . 7 3 9 . D 258 . ) B ...

c::i � U -.s..ER I E.5..--- - · -25B. 7 _ 1 5 , D 1 • . 2 1 1 1 4 , 2 . . -=� .. �-· .. 4 . D @:IJ '2" PARALLEL - - . ---_ •. 1 2

0 .>• >

NPUT KV ? . 48

1ULT I P L I ER 1 . 0 I . 2 5 1 . • H I GH SPEED

FUSE reR FEEDER

(D.OOu SEC

1 . 6 H I GH SPEED

FUSE FUR HllTIR ST ARTER ( 0 . 004 SEC

SHORT CUCUIT !WA LIV HllLDED

CASE C I RCU I T BREAKER '1 R LIV FUSES

l/V MOTOR STARTER W I TH

• HOLDED C I RCU I T BREAKER

I

VA AT FAULT SYMMET R I CAL

1 8 . 2 2 1864 . 8

UR FUSE

22. 7 2733 1 . 0

. UR LESS)

25 .4 306 1 0 . 7

- � , - - · · - - - -"- -- - ­··· - - ··· ···

IR LESS)

-

=-==-:.. - ==-=-=------ ·--· · _ ___ ,

-��:·:;:::::=-�·::;::-�·�-::-:::·:��':_ ) .. : ���:�� .. :.�_:-:�:�:i·�� Moon H. Yuen (M'54-SM'66) received the B .S .E.E. degree in 1 94 8 fro m Heald Engineering Col­

lege, San Francisco, Calif., and the B M C from the University of California Extension, Berkeley, in 1 970.

Currently, he is the Head of the Electrical Group of Scientific Development Facilities Engi­;�ii:Pt��e::-:---neering, Bechtel Corporation, San Francisco, Calif. Since 1 96 5, he has also been the Organizer Ii and Teacher of electrical engineering and design courses at for the University of California Ex­=���"/!;.���;. -�=· ·,:.:-=-;.::··�· Jtm}eri�on and ..Bechtel Manpower Development Programs , and is the author of 12 technical papers

• .. �.,,;i;1 and various class texts on these subjects. In 1 97 3 , he was granted a U.S. Patent in "Solid State · ··r.�1�! Control-Low Voltage Heating of Motors." � Mr. Yuen is a Registered Professional Engineer in the States of California, Louisiana, Michl-. gan, New York, South Carolina, Oregon , Hawaii ; Illiilois, Florida , Texas , Washington, and Wash­mgton, D.C. He is a member of the National Council of Engineering Examiners. He is also registered in the Province of Alberta, Canada, and a member of the Instrument Society of America.

' ., 4