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Solution Manual For Protective Relaying: Principles and Applications, 4th Edition

Solution Manual For Protective Relaying: Principles and Applications, 4th Edition provides the perfect textbook solutions, giving you the help you need to succeed in your studies.

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    Protective Relaying Principles and Applications 4th Edition J. Lewis Blackburn (Deceased) Thomas Domain Solutions Manual Prepared by the following contributors: Sean W. Carr, P.E. Sr. Engineer-Relay & Protection Services Commonwealth Edison Company Sean.W.Carr@ComEd.com James K. Niemira, P.E., ME-EPE Principal Engineer-Power Systems Solutions S&C Electric Company Jim.Niemira@SandC.com John R. Bettler, P.E., MSEE Principal Engineer-Relay & Protection Services Commonwealth Edison Company John.Bettler@ComEd.com William J. Niemira, MSEE Associate-Power Delivery Services Sargent & Lundy LLC wniemira@SargentLundy.com Anthony D. Locatelli, P.E., MSEE Sr. Engineer-Relay & Protection Services Commonwealth Edison Company

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    CHAPTER 2 2.1 A wye-connected ge nerator ha s a namep late rating of 200 MVA , 20 kV, and its subtransient reactance (X ; ') is 1.2 pu. Det ennine its reactance in o hm s. 2.2 The generator of Problem 2.1 is co nn ected in a power system where the base is speci fi ed as 100 MY A , 13 .8 k V. What is the gen era tor reactance (X ; ') in per unit on this system ba se? 2.3 Convert the per-unit an swe r calculated in Prob lem 2 .2 to ohms. Does this match the va lue determined in Pr ob lem 2.1? 2.4 Three 5 MVA si ngle-phase transfo rm ers, each rated 8: 1.39 kY, h ave a leakage impedance of 6%. These can be co nnected in a number of different ways to supply three identical 5 n resisti ve l oa ds. Vari ous transformer and load co nnec ti ons are o ut lin ed in Table P2.4. Co mplete the table column s. Use a three-phase base of 15 MV A. TABLE P2.4 Tota l Z as \'icwcd Case Transformer connection Load conn ce .tion to Line-lo-Iine base kV Load R in from t he high side No . Pr i m ary Se cond al)' secondar y HV LV per un it Per un it Q Wye Wye Wye 2 Wye Wye De lt a 3 Wye Delta Wye 4 Wye Delta Wye 5 De lta Wye Wye 6 Delta Wye Delta 7 De lta Delta Wye 8 De lta Delta Delta 2.5 A three-pha se generator feeds three large synchronous motors over a 16 km, 11 5 kV trans mi ssion line, through a trans former bank, as shown in Figure P 2.S. Draw an e qui valent single-line reactance diagram with all r eac tances indi ca ted in per unit ofa 100 MVA , 13.8 or l i S kV base. Transformer #1 <l do Generator ~ 50MVA 50 MVA 13.8:115 kV 13.2 kV X = 12% X d = 20% Transformer #2 ~ <l M, 115 kV Transmission li n e-- 16 km ~ X = 0.5 Ohm/km Motors 20 MVA 115: 13 . 8kV X = 10 % M, : 10 MVA. 13.8 kV Xd = 20% M 2 : 5 MVA, 13.8 kV X ;; = 20% M 3: 5 MVA, 13.2 kV X ;; = 17%

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    2.6 In the system of Problem 2. 5, it is d es ired to maintain the vo lt age at th e motor bus of 1. L O · per uni t. The three motors are operating at full rating and 90% pf a. Detenn in e the vo lt age required at th e generator temlinals assuming that there is no vo lt age regulating taps or similar equipme nt in this system. b. What is the vo ltage required behind th e subtransient reactance? 2.7 The perce nt imp edance ofa transformer is typically detennined by a short circuit test. In such a test, th e secondary of the transformer is sho rt ed and the voltage on the primary is in creased until rated current fl ows in th e transformer windin gs. The applied voltage that produces rated curre nt di vided by th e rated vo ltage of the transformer is eq ual to the per-unit impedance of th e transformer. A short circuit test on a 150 KVA, 72 00-240 V transformer prov id es th e follo wi ng results: Prima ry vo lt age at 20.8 primary amperes = 2 08. 8 V a. Determ in e th e %Z of th e transformer. b. Calculate the ohmic impedance of th e tran sformer in prima ry and seconda ry tenns. c. H ow mu ch current wo uld fl ow in th e transformer if it s secondary wo uld become sho rt ed during normal operating conditions? (Consider so ur ce impeda nce to be zero.)

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    • 2.1 Find impedan ce in o hm s from per - unit. MV A n = 200, kV n = 20. X -d = Z P\J = 1. 2 Per Eq . (2. 17 ) Zn = 2.4 Impedance X "d = 2.4 o hm s kV ll 2X Z ru MV A B 200

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    • 2.2 - Method 1 Convert per-unit impedance from one base to ano th er. MVA . = 200. M VA2 = 100. kV . = 20. kV~= 13.8. X "J = Z I PU = 1.2 P cr Eq . (2.33) Z MVA, kV ,' = lPU x 1.1V A x -- , I ~ I kV 2 - loa 20 2 1.2 )( - .- - 200 13 . 8 2 1. 26024 Per- unit impedance in new base Z2PU = 1.2 6 02~ • 2.2 - Method 2 Convon impedance from ohms (Result 2. 1) to per- unit. MVA H= 100. kV B = 13.8. x~ ,, =Zn=2.4 P cr Eq. ( 2. t5 ) 1. 26024 Per- unit impedance Zp u = 1.2 602 ~ ZP U = Zn = MVA . x Zn Zs kv i 100 )( 2 . 4 13 . 8 2

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    • 2.3 Conve rt im pe dance f ro m pe r· unit (Result 2.2) bac k to ohms. M VA. = 1 00 . kV H = t3.8. X" d = Z P\J = 1.26024 P cr Eq . (2. 17 ) 2.4 Zn = 13 . 8 2 Ie 1 . 2602 4 1 00 Im pe dance X "d = 2.~ ohms. This matches Res ult 2. 1.

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    J-~ 'V.1iI f ' ..... HV '-v ~t:£J~ V e. ~ ~IA.'" P4. bfV I-iv , I..V 'I fUIIJ!... ~R-- Y ~y '{ y '( I ~ .'alt 2·<I'cf' ~ -'e Jl. -.1' t)." b S- ~~:'J&I' " r ~ y y A i3.8S1 ~.'1of jo.IJ" t.l. (, 1 • 'J • ;, y .6. Y /'$.iS1 /, Ho d 0.01. r ~ . t{ y A A n.es;, I, ~c;oc jO" ,.'67 S- ~ Y ~I 8 I ;,<ff)f .J P. Ob ~ . C- D. Y A 8' ,2. <fOg .J P, 04 /.1.17 1 A 6. Y g- 1, 39() JO.D1- S t' I> A ~ &' I.YiD V f). 02- /. "67 . • J ~ ?;kV¥ f3 ~ 15.j56 Jd :e- & ; VU ;-& Wp- :: /2 V~ r+-v J IS'MVIf ""'"'- /J ~il4 SdZe_~lr£",~d- r ~.

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    Problem 2.4 Col A Col B Col C Col D Col E Col F Col G Col H Col I Col J Col K Col L Col M PU leakage reactance equivalent Y ‐ connected resistive load (ohms on LV side) PU equivalent wye ‐ connected load resistance Case HV LV Load connection HV LV HV LV equivalent in line (Col J / Col H) per unit (Col K + j*Col i) ohms (Col L * Col G) 1 wye wye wye 13.856 2.407 12.799 0.386 0.060 5.000 12.940 12.940 + j0.060 165.623 + j0.768 2 wye wye delta 13.856 2.407 12.799 0.386 0.060 1.667 4.313 4.313 + j0.060 55.208 + j0.768 3 wye delta wye 13.856 1.390 12.799 0.129 0.060 5.000 38.818 38.818 + j0.060 496.840 + j0.768 4 wye delta delta 13.856 1.390 12.799 0.129 0.060 1.667 12.939 12.939 + j0.060 165.613 + j0.768 5 delta wye wye 8.000 2.407 4.267 0.386 0.060 5.000 12.940 12.940 + j0.060 55.211 + j0.256 6 delta wye delta 8.000 2.407 4.267 0.386 0.060 1.667 4.313 4.313 + j0.060 18.404 + j0.256 7 delta delta wye 8.000 1.390 4.267 0.129 0.020 5.000 38.818 38.818 + j0.020 165.623 + j0.085 8 delta delta delta 8.000 1.390 4.267 0.129 0.020 1.667 12.939 12.939 + j0.020 55.208 + j0.085 Base power (3 ‐ phase MVA) HV winding voltage (kV) HV delta LL voltage (kV) HV wye LL voltage (kV) Zbase for HV delta Zbase for HV wye 15.000 8.000 8.000 13.856 4.267 12.799 Base power (3 ‐ phase MVA) LV winding voltage (kV) LV delta LL voltage (kV) LV wye LL voltage (kV) Zbase for LV delta Zbase for LV wye 15.000 1.390 1.390 2.407 0.129 0.386 Winding connection LL Voltage Base (kV) Base Impedance (ohms) Intermediate calculations Total Z viewed from HV side

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    c.k;yj 13~: / Mv'lttJE..t) (kVoL-P Y ~fu · tJW ;::: ~O'-VV-w~OLD k\[~~J ~thr "1/':; 2D1.(~)(~~:;tc ~6."1. I o.~{,£/"< 1-1 X:: I;;<'/; C ~) =- Zrn.~ O.~Yf"'-< -r-~ )c = Iv %' ~;:)= o,j 10 I 0.>0 f'''- ;t( / Xd "" Jo1, (p.) ~ ;Wll %/ ~d) r- IA1 Z- y./' . 2tJ ~ C~) '" foz> 7. I ~ a r tA13 'f.J" -;; /11, t~ ) (',~ . } y ~ 3111.) '3. II f>lA UV\12. ;"1r.(l~ : . ~f<'-- '" ~~ -z. - v"r5"lc.v)" V' b ...... ItS"k." - / ez, ,.,.jit- =- /:) z. . '2-'s- ...t::'l..- , , f ble"ro.t /W:- ( e'~ ~. ) (II. 'M ~ '" 0 to 05' I r$ Z , l- r -IV • f) ~

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    'lVtt

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    JKN 2~ 1~1 4 6 MVA E 10 ·vo lt ·amp Problem 2.6 PU E 1 Prob 2.6.a Voltage at generator bus to keep 1 PU voltage at mot or bus Ymotor := 1.0 · PU Load cu rrent · mo t ors at rated power and 0.9 PF . Problem does not say explicitly, but we assume that the PF is lagging as this is the most probabl e. Total motor bus power in per unit Snt Olo r := IO·MVA + 5·MVA + 5·MVA . /acos(O . 9) = (0.18 + 0.08 7") PU • lOO . MVA J ISmotorl = 0.2 PU ar g(S motor) = 25 .842 dcg Cu rrent is found from the relation t ha t complex power is equal to voltage times complex conjugate of current. ( Smolo, ) Im otor := = ( 0.1 8- O . 0 87~ PU V motor p moto rl = 0. 2 PU arg( lmotor) = -25 .842dcg Une reactance in PU (from prob. 2.5) Z Lin c := j ·0.0605 PU T1 reactance in PU (f rom prob. 2.5) Z n: = j·O .2 4P U T2 re actance in PU (f rom prob. 2.5) ZT2 := j·O.50 PU Gener a tor bus voltage is motor bus voltage plus vollage drop in T1, line , and T 2. I VOcu. busl = 1.0 79 PU arg(VOcn.b us) = 7.67 1dc g Prob 2.6 b - Generator internal voltage behind subtransient reactance: add intemal vo lt age drop to the generator tenninal vo ltage Generator subtransient reactance in PU (from prob. 2.5) ZGcn := j ·0.366 VOen. inl ema l := VOen. bu s + Im otor·ZOcn = (I 102 + 0.21] PU IVOe ll .illtem a ll = I.122PU arg(V Ge n.i nl emal) = 10 .7 9 Id eg

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    JK N 2-1-14 Pr o bl em 2.7 Transformer ra t ed power and rated voltages JKN 2-1-1 4 page 1 of 2 kVA =: lOOO ·vo h ·a mp 6 MVA =: 10 ·vo lt ·amp PU .: I Srlltcd := 150 -kVA Vruled.prima ry: = 72 ()() ·volt Vru tcd.sccondary := 240· volt T ra nsformer rated current [ 150. kVA = 20.8 33A ra ted.primary := 7. 2. kV I IS OOOO·vo lt ·a mp = 625 A ra ted.sccondarv := ' J 240· volt This confi rm s that the sho rt circuit test was in fact perfo rm ed at ra ted current . The impedance vo ltage is there fo re eq ua l to the voltage measured during the test , or 208.8 volts. U&.. The transformer per cent impedance voltage (% IZ or %Z) is equal to the measured impedance vol tage expressed as per cent of the rated voltage or in this case : ,, __ V ~ " ~ S ~ I __ , = O.029 I' U V ru ted.primary 208.8 ·volt = 0 .029 PU 72()() ·vo lt x ,. = 2. 9 % UJL The eq uivalent ohmic impedance referr ed to the pri mary or seco nd a ry is equal to the co rresponding base im ped an ce times the pe r un it impedan ce . , V rul ed.primary- Zbasc.pr im ary :- , (7.2·kV )- = 345 Ml O. IS·MVA 345 Ml XT.ohms.prima ry := Xr Zb asc_ pri mary = 10 .022 n XT.ohms .s econd, II Y := Xr Z ba sc _ sccondmy = 0. 01 11 fl V 2 raled. sc condary = 0 .384 n Sratcd 2 (O.24-k V) = 0.3 84 n O. I SO ·MVA

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    2. 7.c There are a few ways to arrive at the short circuit current. JKN 2-1-14 page2of2 One way is to divide the rated voltage by the equivalent ohmic impedance on either primary or secondary. Alternatively, 1.0 PU voltage can be divided by the PU impedance to determi ne the short circuit c ur rent in PU of rated current. The PU short circuit current can then be multiplied by the base c ur rent (i .e. , rated current ) on either primary or secondary as desi red . This method can be further simplif ied to "divide rated current by per unit impedance." Both methods are i ll ustrated here: v d . rate .pnmary 7 18.391 A XT .o hm s.prima ry v rated.secondary -====~ = 2 1 55L724A XT.ohms.sccondary noo · vo lt IO .022·ohm 240·vo lt .Ol ll· ohm 718.419 A 2162 1.6 22 A (NOT E: this result can also be used to illustrate the concept of significant digits. The result is 21.6 kA, athough the computer intemally carries out calculations to greater precision than is significant. ) Another method to arrive at the same result: I ISc.pU := - = 34.483 PU Xl 'SC.Primary:= ' Sc.PU· 'ratcd.primary = 71 8.3 91 A 34.5·20.8·amp = 717.6 A 'SC.secondary := ' Sc.PU· ' ratcd.secondary = 21.552 kA And finally, another method, often the simplest: lrat ed .primary Xl 718.39 ' A 'rate d. seconda ry = 2 1 55 1.724 A X T 34. 5· 6 25 ·amp = 2 1562.5 A 20 .8·amp = 7 1 7.241 A 0 29 625·a mp = 2155 1. n4A 0.029

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    C HAPT ER 3 3.1 FOllr boxes represent an AC generator, reactor, resistor, and capacitor and are connected to a source bus XY as shown in Figure P 3. 1. From the ci rc ui t and phasor diagrams, id enti fy e ac h box. y I, I I, ! 13 t IJ x 1 2 3 4 I, 1 3 I, ~ ------1r---~--- ~ V XY 3.2 Two transformer banks are connected to a common bus as shown in Figure P3.2. What are the phase relations between th e voltages V AN and V A'N'; V llN and V 13 'N', V CN and VC'N'? T sformer '"" Baok 2 ~ No.1 ,*~. ----- A r r*-----il----- B & _+- _______ C ~ ~ 'N - Tran sformer No.2 Baok 2 3 r-:: ~ A' ~. B' ~' C' ~' N' _L ~

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