ebe completely transposed 60 -hz, 25 c three-phase line has flat horizontal phase spacing. the...

11
SI'r"[ ·l ThH"'! -MAL " AYSL.I. ..,J " S.'\JJ". " I l\ " Peperiksaan Senwster Pertama Sidang Akadernilk 1995/96 Oktober - November 1995 EBE 367 - Silltem Penil:UUlttran & Penayihan Kuasa Masa: [3 jam] Sila pastikan bahawa kert:tS peperiksaan ini rnengandungi 11 nluka surat bercetak dan ENAM W soalan sebelum anda Iltlcmulakan peperikflaan ini. Jawab LIMA soalan s8.haja .. Agihan markah bagi soalan diberikan di sut s{:belah kanan soalan berkenaan. Jawab semua soalan di daltam Bahasa Malaysia. .. .2/- 4 1 -'-U

Upload: vanminh

Post on 28-May-2018

223 views

Category:

Documents


0 download

TRANSCRIPT

l.1NIVER~ SI'r"[ ·l ThH"'! -MAL"· " AYSL.I. ..,J " S.'\JJ". ,~, " I l\ "

Peperiksaan Senwster Pertama

Sidang Akadernilk 1995/96

Oktober - November 1995

EBE 367 - Silltem Penil:UUlttran & Penayihan Kuasa

Masa: [3 jam]

Sila pastikan bahawa kert:tS peperiksaan ini rnengandungi 11 nluka surat bercetak dan ENAM

W soalan sebelum anda Iltlcmulakan peperikflaan ini.

Jawab LIMA ~ soalan s8.haja ..

Agihan markah bagi soalan diberikan di sut s{:belah kanan soalan berkenaan.

Jawab semua soalan di daltam Bahasa Malaysia.

.. .2/-

4 1 r:~ -'-U

-2- [EEE 367]

1. Suatu punca 'voita,n seimban:g: ~, sambungan Y, jujukan ,positiC

membekalkan ~voltan talian V,tb·' 480LOo V kepada beban Il seimbang

deagan Z 6. - 30L40o O/fasa. In1pedan taliaml antara punca dan beban

ialah Z T - IL85° C/fasa.

Kira:

A balanced, positive-sequence, Y .. connected voltage source with line voltage

V Ii> - 480 L 00 volts is applied to a balanced A - connected load with

o Z A - 30L40 C/phase. The line iJ'npedance between the 'source and load is

o Z T .. 1 L85 Q/phase.

Calculate:

<a) semua ar'us talian,

the line cu"ents,

(b) semua arus fasa beban ~, dan

the A -/ood cu"ents, and

(c) !Jemua voltan pada terminal beban.

the voitagE!S at the load termirklis"

2 . (a) :Nyatakall TIGA pendeka't:an yang sering digunakan untuk

memudabkan anal isis sistenl kuasa tiga fasa seim bang.

(40%)

(35%)

(25%)

State thre4~ .approaches com17ronly used to simplifY the analysis of a balanced

three-pha~~e power system.

(25%)

... 3/-

41C

···3- [EEE 367]

(b) Rajah S2 Dlellunjukkan 19a:cnbarajab talian tunggal sistem kuasa

tiga fasa mudah dengan empat (4) bas. Penjans tiga fasa berkadar 4160 k.VA, 2.4~ k"V'9 X'~ = 20-;0. Transformer tiga fasa

T 1 mempunyai kadaran 5000 kV A, 2.4/24 kV, X = 5% manakala

bank transformer T] terdiri dari tiga transformer satu fasa

serbasama, dengan kadar setiap satu bersamaan 1500 kV A,

13.8/11 kV, X = 5%.. Behan dianggallikan terdiri dari impedan

tetap ZL =: 10 + j30 O. l-.njang talian penghantaran ialab 200 km

dengao irnpedan siri O.5t"J/knll. Pilib kadaran peDjaDa sebagai nilai

asas rujulkan. Lakarkan m;ar.obarajah reaktan dengao !lemua reaktan

ditandakan daIam per unlit.

Figure S2 shows a single--line diagram of a simple jour-bus three-phase power system. The three-phase generator is rated 4160 kVA, 2.4 kV, X'~ = 2()DA.

The three-phase transformer 'PI is rated 5000 kV A, 2.4124 kV, X = 5% and the

transformer bank T2 is made up of three identical sing/e-phase transformers

each rated 1500 kVA, 13.8//2 kV, X = 5%. The load is ussumed to be

consistingofa constant impedance ojZL = 10 + j30 Q. The transmission line

Je:ngth is 200 km with series inlpet.il.lllCe oJO.5Q/km. Select the generator rating

as reference base value. Draw thl~ reactance diagram with all reactances marked

in per unit"

~: l'ertukaran asas ]lmpledan

~Vote: Change of base of impedance

(75%)

asas kV lama 2 asas kV A bam Per unit Z bam = per unit Z lama ( k ) ( kV A )

asas .V bam asas .. laQla

base kV old 2 base kV A new Per unit Z new = per unit Z old ( --) ( )

base k" base kVA

4 1 rj .l.. ,

new okl

... 4/-

[BEE 367J

Figure S2 {Rajah S2)

3. Talian tila fasa teralih, 60 HZ'I 2:50 C dijaraklkan secara mendatar antara

. pengalir fasa. TaUan J)enghantl.rarll dengan panjang 200 .km terdiri dari dua

berkas pengalir per fasa dan pengalir adalah 795,000 emil ACSR 26/2

pintalan. RuaDI berkas d ialab 40 em ' dengan jarak 10m antara titik tengah

pengalir terberkms, seperti ditu}ljukkan dalam Rajah S3'. Ciri-ciri pengalir

ACSR boleh didapati dalam Jad uaJ. 1.

Kir.:

A completely transposed 60 - Hz, 25°C three-phase line has flat horizontal phase spacing.

The transmission line has a length (if 200 /em consisting of two bundled conductors per

phase and the conductors are 795,000 emil ACSR 26/2 stranding. Bundle spacing is 40

cm with 10 m between adjucent burulle center, as sho»'n in Figure S3. Characteristics of

ACSR conductors are attached in Table 1.

Calculate:

(a) :Reaktan illduktif dalam t;Um per fasa.

Inductive reactance in Wm pel' phase. (25%)

(b) Reaktao kapasitif dalam Q-m per fasa.

Capacitiv~~ reactcmce in Q-m per [.lhase. (25%)

.. . 51 ..

(c) Jumlah rintangan tali an perCasa.

Total line resistance per phar;e.

(d) Jumlah reaktan induktif laUSID per rasa.

Tota/line inductive reactance per phase.

( e ) Jumlab reaktan kapasitir taHlID per fasa.

Tota/line capacitive reactance .Lver phase.

I d

o I o

I dl, d I o () 0 0

a a' b b' C J c'

~ 10m ~-"'''lom---1

Figure S3 (Rajah S3)

4 1 0 .. i. . ..J

[EEE 367]

(10%)

(20010)

(20%)

... 6/-

1"""""1

t" \0 ~

~

\0

J D·d us I

TIbIe A.4 Ch&rllCteristics of aluminum cabla" .... )lBinfofDed (AlumInWl\!~ of America)-ACSR

Code Word

.10M ........... ICMi IIiuIIbIod 0IuQ.

".art '"­,...... MInin "'-­~

Finch {"~~

~ em r-~

CO!ldor

~. M.-.rd CIOW SUIting "-OM"9 Flamingo

RoCIIr; . GroIbut £1IfIIt ~ Sqr.-b 0.-

bole .....,. .... til IM:I!'

u-or QIioIe' 0IIIicft

"'"" ......,.

a-a. Mill ~

, ....... , " 2111000 JI 2,.7000 72

,2Hi6VOO ~ 11711000 84

lMOOOO M 15'0100 ~ 14311100 Sot 1361000 54 unooo 54

,. 11IH.CO . t ~ 1113000 ~

. 1Q.!..1~'" M

Aluminum

---r Sullld ~ {lnchesJ

... J 0.1"9 o.n~

.. 0.1135 4 I 0.l602 ·1· .... 3 0.1116 3 0 .1673 3 0.1628 3 I 0.1582 , , ...... 3 0.'486

3 0 .1436 1 ! O. ~ 384 ~ ! G. U29 ! ~ooo l!'.& I ::=, ~ f ~ 0.12S1

0 .1173 ~Ooo ' ~ U. I~I"

?tSOOO 28 2 0.174' 7H000 30 Z 0.1128 715500 ~ 3 0.11M 115100 26 2 0 .lB59 115&00 JO 2 01~. .... ~ SoC 0 ,1111

1138000 &4 3 o 1011~ 531000 26 2 O.lW 131000 30 2 0.1454 805000 54 a 0.1059 805000 til 2 O.1~Z5

MtIOOI H 2 O.l4aa

561500130 0.1362 4170001 28 0.13&6 U?OQO 30 0.1251

,.,1001" , 0.1236

317100 30 Z 0.1151

_4OO?6 Z 0.1138

""TT O.lOU 300000 16 2 0.1074 3OlOWO 30 2 0 .1000 L2 .. ~12t2 0 ,101)

" 111 7

19

19 19 19 19 19

"

St ...

Shnd ~ (InctleaJ

0 .084' 0.0114 0.1157 O.<m! 0.08"4

0.1030 0.1004 00177 O.0t48 0 .0821 0 .0892

I 0 .011'2 O. 1~4 I~ I 0 .1 325 7 i

; I ~:g~~ v.n14

7 O.UIO 19 0.0177

7 0.1151 1 0.1290

19 0 .0926 7 !U111

O. I{I8!! 7 0.1218

1. 0.0874 7 0.1058 ., 0.1186 J I ". I i~ I 0 .1382

0.1054 0.1211 I O.osel O.lIIH Ig: 0.0835

, Q.HlOO 0.07ea

~a..ct on ~ 1'"'-~ 61 '!I. c:onducr.",IY.

IF

0u\IidI . ~ (incNl,

1.110 tlOZ l .m U~l 1.802

1.545 1.501 1.4&6 1.424 1.312 1.331

1293 1.24$ 1.Ii.

~; x. ........ ... ~ ~ c....-.. (0hIM 1* RMI:uMcI

'. Resis1Mct (Ohms PI' ~ PIll Mile) ConduI:Iar PI' (.....,.... .. Mile at 1 Ft . ConductDr

~ Wclight w.. CunnI I --I I r-,- 260(; (7rF) Smell CunentJ we (122"1') Cunen.l ~ s.cq poet .......

a.- ~ (Poundl AMI. .~ 7",~ Mc-.) lft~) Mila qI SIIw9h per .. IOHir c..atvt A.W.G. ( ....... , Mile) ('-) . (~) I It::

11100, o.otl1 $7_ 0." 41100 O.OUO 10300 O.OSII

'_1°·' 1340 dOll. 1300 0.0Gi2 )l5O !lOU' 1:00 O-C1J4 1180 0.0713

ilOOO O.~

1 0000lI0 11000 to 711 o.QIIq tIIDGf.IO 53.200 10237 ~ lO000O ti0400 I.. o.otW 150000 41100 "10 o.QUI iQOOOO 44100 ",, o.~ 7W000 Cll00 1082 .lCMillt'

I25HzI~HzI80tt.r

0.0Me O.OUO 0.0591 G.OIl!1 0.0121 O.08S 0.0153 0.0155 0.0651 0.0St2 00et4 0.0&85 o.t)73S 0 .0737 0.07. 0.0714 0.01. 00718

do I 25tt.r i 50Hz 11OH!

0.0410 OJM12 0.05" 0.0I0Ii ·C.~i

t 0.0Me 0.0IM 0.0875 0.0Ie4-G.OIIO D.OIIO 0.0710 0.0720 0.0711 0.0721 0.01'" 0.07tO 0.0711 0.0771 0.07t2 O.a.» 0.01l0I 0.011. O~ 0.., Q,oea 0.0871 0.01" 0.0l0I

1'00000 40200 -; 544 I oodi' ,i ~iJOO a1 100 -; 019 I G.G4~ I

1HO 1 I)l10 1010

C0I3t I 0.0840 /. 0.0842 1 008441 o.ot2.1 0.01351 0.0157 I,O.OMI

I~·~I~·~ ~·~?l~~ 0.004 0.1~ 0.10~~,O!035

10Hz 10Hz

o.s37 0.0155 o.J42 0.07f1 UtI 0.0771 0.3&4 0.0774 c.~! 0.0Il02

0.368 0.01-14 0.312 0.0121 0." 0.0830 0.3U 0._ 0.312 0.0141 0.37. 0.0157

0.380 0.OM7 0 .• 6.0178 o.m (I.QStO

' .013

iiUOtiilO 134'lGD I ;;., $ I {i.;~ i UI2 see. ooo '2300 I 4HI 2 I O.038i:.1 U4fI 550000 ]1400 5-..0 O.a. .

iliv 910

,~O!!:'" i ,-,.~ ... ~ i ~.~ .. ~. i , .. ~-!~ Ilo.llP.8 !I).!~ IJ 11." !O .• 121 .

10.104 10.\04 I(U C4 O.l O-t o.t14Si !) H!5 1 (un~l()u~1 ~.!~~ • ~. !~! ' . ~!~~ '?!~ I~·!!!! ~.~!: ~.:!!: I~' :~~:

a.l 0 .•

O.OIM O.OIIO!

.. -~., i -.---. . -.. . ~

I ~;;; F;;; I;';; j 0 ,, ; ~:,~~I~:'~~ i 1.108 500000 31200 5770 0.0371. 100 0.1298 0.12. 0 .• I 0.0112 ,,40 .,000 311400 I'll O.03Q 910 ;g; ~g~ I g.g~ gi ~~ o.12I11JO.12M 0.12M 0.12. 0 .• 3 0.0lI04 I.OH 450000 21300 485S O.~· 830 0.1"' 0 .1452 0.1472' 0.141 0.407 0.0132 l .ot;l 450000 28100 5193 O~ 840 0.131 0.131 10.,31 0.131 0.1442 0.1442 0.1442 0.1442 0.405 0.0121 1.081 450000 34100 1">865 00372 840 0.131 0.131 0.131 0.131 0.1442 0.'442 0.1442 0.1442 0.319 0.0120 1.000 .,tooo 24500 4!i27 0.0»1 100 0.140 0.140 10'''' C 141 0.1 .. ' 0 .1671 0.1st\ 0.1101 u.412 Q,Qt43

0.977 4COOOO 23100 43191 0.032t 770 0.141 0.147 0.1" 0148 0.1'1' 0.1631 0 .• 111 0.1'" o.4U

I n.09!iO

ono 400000 :tIHlOO 46111 C.033f 780 0.147 0,147 0.141 0141 0.&111 0.1'" 0 .111' 0.111' 0.412 OOt4l

1.01' ~ 31100 521l I 0.0311 ?80 0,141 0.147 10.147 0 1.7 0.111. O.II!S OJ6HI 0,1$11 0 .• 0.0137 0.853 3IO- 22100 4101 0.012'( 1M 0.1" 0.155 0.116 01SS D.l. 0.1715 0.1156 O.n?1i 0.411 0.CJI$7 0.9H *Il00 2.UOO UtI O.03a7' 1to 0.1S<1 0.1&4 11:1.154 OJ!>, OJ7OCI O.17M 0.1120 0.1120 0.411 0.0153 ii.iH · 360= 22a 4:=~lo.~1 no 1),'" .0.'" 0.188 0.1&8 .u .... 0.18se 01 .... 0,'_ a.no I O,0H5

I I I. ... . __ -.~L18&· ft ... ~ .. I:w .... 1 0.t53 3&0000 27200 4111 0.03Z8 730 OJ. 0.11,. r·1H IlI.OII wt u. • r ._-, . --I !!.~!! 0.0.1;7 0 .... 300000 11'30 3412 O.D2ICt 670 0.'" 0.," O.ltt 0198 D .• 0.0Ne 0.813 3DIHlOO 2IJOO U33 O.OlOf· 110 0.111 0"" O,ltt 019' 0.211 0.424 0.0lI0

0.183 210000 11110 2115 0.0211:' NO UH :: Slmetsdc

0.44' 0.101' 0.801 210000 1IteO 32n 0.021( 100 0.235 ~ .. de \'U3& 0.100II

0.721 4JO 14010 2442 0.02441 .,0 0.271 I ~ I I 0.411 0.10" 0.7.' ./0 170.t0 U1. O.OUt I uo 0.271 0.445 0.1032 0-180 181100 12660 l2171 0.0230 I 410 0.311 I O.<t!9 O.'~7

O.'JOO , .. 100 li430 2473 0 .OZ4', i 100 0.311 i 0.,.2 I I O.4t2 0.!04' 0.142 310 112!O tl36 O.Gt17 4tO 0.350 0._ ; !u. 0.107.

tfor COfIducrof It ?~-C . .... It 25·C. wind 1.4 t'I>iIn per hour (2 flIsec:). ~ - 10Hz. n:urftflr A,porox. 7S11o~" is 75'!1. of the .. Aq;wot&. C_ CMJying ~ in~" ~ .............. !he QIIIIMt wIIidI will pIOduce 5O"C ~uctCW twnp. (2S·C tiM) wid'! n'c iii' ....., .. wind '.4 .... PI' 1Iour,

....!. r---

Q C\l ~

4. (a> 0)

[BEE 367]

Dalam memodelb.n taUao penghaDtaran, talian boleh

dirnodelkan bergant.ung kepada jaraknya. Nyatakan TIGA

kategori tRlian pen:lthulltaran dan jarak masing .. masing dalam

kilometero

In modeling the translnission lines. lines can be modeled in terms of

lellgth. State THREl~ categories of transmission lines and their

corresponding length in kilometer.

(ii) JeJ.askan seeara ringkJli51 konsep beban impedan pusuan (SIL)

yang digunakan d.,.hun sistem kuasa.

Explain briefly the c01J!cept of surge impedance loading (SIL) used in power

sYJ,,'tems.

(20%)

(b) Paramet.~r .ABCD suatu tali:IlD tiga fasa teralib jarak sederhana, 50

··Hz 34S-kV, diberikan !;eperti berikut:

The ABel) parameters of a ,':ompiete/y transposed, three-phase 50-Hz, 345-kV

medium length line are as jolkJWs:

A - 0.9706 L 0.1590 per unit

B - 70.29 L 84.780 0

-4 0 C - 8.277 xlO L 90.08 S D - 0.9706 LO .159° per unit

... 8/-

(c)

··8·· [EEE 367]

Taliao ini membekalkan StRatu beban penuh 700 MW beroperasi

dengan faktor kuasa 0.9: ,~ Illtendahuiu pada voltan talian 950/0 dari

voltan terk.adar.. Hitungkllo:

This line is supplying a.fuIlI04~!d (~l700 MW at 0.99 leading power factor at 95% of

rated voltage. Calculate the following:

(i) Voltan talian V" a~·us talian I., dan kuasa nyata p. bujung

baJlltaran.

Send,ng-end line voltage. VSt line current lSI and real power P s.

(U) Pelr81tusan regulasi voltaD.

Percent voltage regulation

(iii) Efisiensi talian pen;~hflntaran pada beban penuh.

(i)

Transmission-line efficiency at full load.

(70%)

Da,lam konfigurarli sistem kuasa umum, konsep

fUlldamentai ialab pembahagian sistem kelt.da zoo peirlinduDgan. T t 'rangkan seeara ringkas kOBsep zon

perlindungan ini.

In a more general pol'-J,er system configuration, a fundamental concept is the

division. of a system into .protective zones or zones of protection. State this

concept in short.

(ii) TllkrifkaDpbrasa dste:m perlindungan utama dan sistem perlinduDgan kedua.

Define the terms primtuy (md backup protection systems.

(10%)

... 9/-

422~

-9 .. - [EEE 367]

5. (a) Talian penghantaran atas sa~tlll fasft, tanpa hilang mernpunyai impedan

c:iri ZA ::: 400 Q, keiajtta.tl jg,elombang 'UA= 3 ,x 108 mIs, dan paojang

. ~ = 30 k:m disambungl~~all kepada kabel·satu fasa, tanpa hilang

deugan impedan ciri ZB, = 100 Q, kelajuan gelombang VB = 2 I loB

mIs, dan panjang R.. B == 20 kOl. Pads hujllng bantaran talian A, satu

\'oltao unit langkab dengJlln J/Qagnitud Evolt dikenakan dari punca de

dengan r'intangan diri dulallLlan Z. = ZA- Hujung terimaan talian B

ditamatkan dengan ZR::: 2 ,ZB = 200 O. Lakarkan gambarajab kekisi

(paotulan) untuk 0:5 t :5 6 T .'

4. single.~phCls.e lossless overhead i'ine with characteristic impedance olZA = 4000,

vel~ity o/wave VA = 3 x 1 rft mlf! and length ~ = 30 km is connected· to a single

phas,e !ossless cable with characteristic impedance ZB = 100 Q,wove ~elocity VB =

2 xl OS ml.s, and length t B =:: 20 km. At the sending end of line 'A, · '(1 step voltage

of magnitude E volts is apI,Jied and Zs = ZA. At the receiving 'end of line B.

ZR = 2 ZB = 200 Q. Draw thE! lattice diagram for 0 ~ t ~ 6 T .

(50%)

(b) .Andai kata sebuah motor iuduksi tiga fasa, sambunganY, ~OO hp,

50 ... Hz, 4 .. 16;,.kV mempunyai efisiensi beban peuuh 88%, faktor kuasa

0.75 menyusul. Motor j)j[j dL~ambunlkan kepada bas penyuap. Anda

dikebendalki melnbetulkalu f'a,ktor kuasa beban motor ini menjadi 0.90

menyusu~ Rlenggunakan bank kapasitor disambungkan pada terminal

beban. JHitungkan:

Assume that a three-phase 5()O .. ·hp. 50-Hz, 4. 16-kV Y-connected induction motor

has a full-load efficiency of 8891it a lagging puwer factor of 0.75. and is connected

to a feeder" If is' desired to correct the power factor oj the load to a lagging power

factor of 0.9 by connecting thrde capacitors at the load Calcukzte:

••• 10/-

, J.O~ [BEE 367]

(i) Kadaralt bank kap,uit:(JII" dalam kV ARS.

Tht~ rating of the capacitor hank in kV ARS.

(ii) Kapasitan dalam luikr'ofarad setiap unit (fasa) jika bank

kapasitor disambul1lgkan dalam delta (A).

The capacitance in microftrads of each tout if the ca~itor...v are connected in

delta (A).

(50%)

6 • Tig. transformer satu rasa meDllbentlik suatu bank transformer peDlagibao

Y - A uotuk membekalkan, kuasa kepada beban tiga fasa seimbang

sambuagan Yya.ng menyerap 2;00 kVA kuasa kompleks pada faktor kuasa

0.8 menyusul dan beroperasi padavoltan bllian terkadar sisi ketuaran (sekunder) transformer. AnJr:gal)kan ketiga-tiga transformer satu fasa

mempunyai kadaran serbasalna. idan setiap satu berkadar 75 kV A,

12,470/415 V.

Three sitJgie-phas'e transformers are connected in Y - A. to supply power to a balanced

three-phase Y -cormected 200.JcVA /Otfl}w:ith a 0.8 Iaggmg power factor operating at rated

line voltage of output side transf(jnn!ij~r, Assume that the three single-phase transformers

have equal ratings and ecICh is rated (':ft 75 kVA, 12s470/415 J-~

... 11/-

424

, N 11.; [EEE 367]

<a) I~akarkan. lambarajah Skf~nUl!l:ik menunjukkan sambungan bank transformer.

Draw the schematic diagram showing' the connection.

(25%)

(b) Tentukan Jrdlai·,nilai voltaJrit d~lln arus fasa dan talian daJam kedua-dua

belitao ballk transfornaea6 (si!ui utama dan sisi sekunder).

Determine the values of phase an(1'/ine voltage and (''U"ent in both primary and

secondary"side of the transfonner bank.

(75%)