LEVEL 2 Short Circuit ·ABC-Learn It 1n an Hour, Use It Anywher e, Memorize No fofmula _ -. �_ " � _ _ -_- _ _ -_-_ -_ ·-. -��---�--. -:�MO ON H; N, SENIOR MEMBER, EE Abstct-Short it «C-leun it o, u it anywhe, memore formula. e MYA met for nlving indu w system ort its appptely fits this dtn. Ind, lving short cit p blems with· tile A met u u ling the «s. INTRODUCTION S . HORT CIRCUIT studies are neces 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 e not complete . unless the short c�rcuit values re solved at various sfratec points. No �ubstation equipment, motor control centers, breaker panels, etc., can be purchad 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 r. every electrical enneer. To learn it may be easy for me, difficult r others. However, to db the problems, anywhere in or out of the office where tle references ar not available _ _ may not . be an easy task because the conventional mhods of solving short circuits involve too many foulas. To memoe them at all times is impractical for the maj6ritY. WHAT REALLY Is THE MVA METHOD? ---Basiy, 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 vious components of the circuit. Since, by definition, admittance is the reciprocal of impedance, it llows that the reciprocai of the system admittance is the sum - of the reciprocals� the admittances of the COlponents:-so, -bdefinitionthe:dmtance of a circuit or compo . nent is the maum current or unit voltage wch wod flow through the circuit or compo�ent to a short rcuit or ult _ _ __ when_supplie_d_fr�-�ource . of ini�Lcapacity_-_ Refer to Fig. 1. . 1 Y=- Z ohms ·- - - - -�'�= K �� � � 000 X ( 2 X . MV� =(KV)2 X Y . (I) (2) (3) (4) Paper TOD-73-132, approved by the Petroleum and Chemil Industry Committee of the IEEE Industry Applitions Society for presentation al the 1973 Petroleum and Chemil Industry Conferen, Howton, Tex., September 17-19. Manusipt relead r publition October 31, 1973. - The ahor lwith the Bechtel Corporation, San Francis, . INFINITE B 13BV he • VIZ filVJI• VSlVJV/Z Z • 0.01 OHMS 0 13.8 KV E ISC • E2/z VA E2/z VA MVA • 10 IKVJ· 2 I Z • KV 2/ Z FAULT E • V I • SHORT CIRCUIT CURRENT IN AMPERES E •LINE LINE VOLTAGE IN VOLTS V • LINE TO NEUTRAL VOLTAGE IN VOL ' Z • LlE TO NEUTRAL IMPANCE IN OHMS rA • SHORT CIRCUIT MVA Fig. I. One line diaam. SYEM } 20 MVA TRANSFORMER } MVA x • 0.1 2.1 • MOTOR } 2 MVA x � • o 2.2 - MVA 1- 2 • MVA 1 • MVA7 / MVA1 + MVA2 • <0 • <0/ 0 + 0 • 2 2 MVA1+J • ú + ú • 20 I • 0 x 10 x lS • ⶐ0 20 ú 13.BV Fig. 2. Im� diaam. Fig_ 3_ MYA diam. y Zo Zpu KV KV� MV� MV� admittance of a circuit impedance in os 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). Practicly, the MV A method is used by separating the circuit into components, calculang 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 diagr. Fig. 3 is an MV A diagram. The conversion fro� a one· line diagram to MV A diagram is simple arithmetic. . Component I, e system, is normally ven a short circuit MVA rating. So, one merely writes down 500, which is its stem· short circuit MV A. Sometimes, if the system MV A is not available, but its voltage and impedance are ven, the short circuit MVA can be calculated with the application of (3).
12
Embed
Short Circuit ·ABC-Learn It 1n an Hour, Use It Anywhere, Memorize ...
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
LEVEL 2
Short Circuit ·ABC-Learn It 1n an Hour, Use It Anywhere, Memorize No fofmula
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
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.
--- 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
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
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
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):=.·;::_--:-_-::::�� --
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
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
·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.
-��:·:;:::::=-�·::;::-�·�-::-:::·:��':_ ) .. : ���:�� .. :.�_:-:�:�: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.