PRODUCTION TRANSMISSION medium I extra high voltage high vol tage medium voltage low voltage I I I I I .. .. ... ..... .. . ................ 1 15 k v \f} j{ { if{f to 230 kV 345 kV ·: ·· ··· .... ·:· :· .. '·:·:·:· :' ·: '· to 765 kV I J J I generating transmission stat ions substation I • interconnection substation ___ ..,..-i 69 k v J I I J I • transmission substations substations I J I J:r small industry commerce residences Single-line diagram of a generation, transmission and distribution system. t----spa n j ,., sag ,.. r:-. • ' 1- -- 1 .· . . ···' ··· .. Span and sag of a line. .rl r 8 .. Distributed impedance of a transmission line. .. . . '• :' . "'· ·. .
Transcript
PRODUCTION TRANSMISSION ----'I------DISTRIBUTION----~
medium I extra high voltage high vol tage ~--- 1 lr---~~~~
medium voltage low voltage
I I I I I ................ ................. 1 15 k v
Single-line diagram of a generation, transmission and distribution system.
t----span j ,., sag ,.. r:-. • ' 1-
--t"-~~-=:::::.:::-·-·- ~--~:.-
-- ---------~-------rT'------1
. · . . ···' ··· . .
Span and sag of a line.
.rl r
8
.. Distributed impedance of a transmission line.
. . . . '• :' ~ -~ . "'· ·. .
-7
VOLTAGE CLASSES AS APPLIED
TO INDUSTRIAL AND COMMERCIAL POWER
voltage nominal system voltage class
two-wire three-wire four-wire
low 120 120/240 0
voltage single phase single phase 120/208 0
480 v 0 277/480 0
LV 600 v 347/600
medium 2400
voltage 4 160 0
4 800
MV 6 900
13 800 0 7 200/12 470 0
~~ 23 000 7 620/13 200 0
34 500 7 970/13 800
46 000 14 400/24 940 0
69 000 0 19 920/34 500 D
high 115 000 0
voltage 138 000 0
161 000
HV 230 000 0
extra high 345 000 0
voltage 500 000 0
EHV 735 000 . 765 000 0
All voltages are 3-phase unless indicated otherwise. Voltages designated by the symbol o are preferred voltages.
Note: Voltage class designations were approved for use by IEEE Stanqards Board --., (September4, 1975)
Sectional view of a 69 kV pin-type insulator. BIL: 270 kV; 60Hz flashover voltage under wet conditions: 125 kV ..
... -- ------ .... "'" -.... --·----
....
I · •
' . ,
metal cap
Sectional view of a suspension insulator. Diameter: 254 mm; BIL: 125 kV, 60Hz flashover voltage, under wet conditions: 50 kV ..
Lineman working "bare-handed" on a 735 kV line. He is wearing a special conductive suit so that his body is not subjected to high differences of potential. In the position shown, his potential with respect to ground is about 200 kV . .
neutral , .
£nvironm~ ttd J;((ecfs
Dust, acids, salts, and other pollutants in the atmosphere settle on insulators and reduce their insulating properties. Insulator pollution may pro· duce short-circuits during storms or under momentary overvoltage conditions. Service interruption and the necessity to clean insula_tors periodically is therefore a constant concern, especially where pollution exists.
Towey Gt-o~;~
so lid __ electrical
ground
Transmission-line towers are always solidly con· nected to ground. Great care is taken to ensure that the ground resistance is low. In effect, when lightning hits a line, it creates a sudden voltage rise across the insulators as the lightning current discharges to ground. Such a voltage rise may produce a flashover across the insulators and a conse· quent line outage, as shown by the following example.
..... . ..
solid _ electracal
- ground
{'earth
Flashover produced by a lightning as it flows to ground.
9
· .... · . . ::· .. ··
(
;a
Example .
I
I 'I
I!
A 3-phase 69 kV transmission line with a 81 L of 350 kV is supported on steel towers and protected by a circuit-breaker. The ground resistance at each tower is 20 n whereas the neutral of the transmission line is solidly grounded at the
I I
transformer just ahead of the circuit-breaker. During an electrical. storm, one of the towers is hit by a lightning stroke of 20 kA.
I J
a. Calculate the voltage across each insulator string under normal conditions;
I I I
b. Describe the sequence of events during and after the lightning stroke.
ji . I '
Solution: a . Under normal condit ions, the line-to-neutral
voltage is 69 kV / 1.73 = 40 kV and the current flowing in the tower ground resistance is zero. The steel tower is therefore at the same potent ial as the so lid ground. It follows that the voltage across each insulator string (line to tower) is 40 kV (RMS).
b. When lightning strikes the tower, the voltage across the ground resistance suddenly leaps to 20 kA X 20 n = 400 kV. The voltage between the tower and solid ground is therefore 400 kV, and so the potential difference across all three insulator strings jumps to the same value. Because this impulse exceeds the insulator 81 L of 350 kV, a flashover immediately occurs across the insulators, short-circuiting all three lines to the steel cross-arm. The resulting 3-phase short initiated by the lightning stroke will continue
•' ,, ,. II I
lj ' I I I
I .. ··-·--··-··
1! - --·--
to be fed and sustained by a heavy followthrough current from the 3-phase source. Th is short-circuit current fsc. will trip the circu it breaker, producing a line outage. In view of the many customers affected by such
a load interruption, we try to limit the number of outages by ensuring a low resistance between the towers and ground. In the preceding example, if the tower resistance had been 10 n instead of 20 n, the impulse voltage across the insulators would have risen to 200 kV and no flashover would have occured.
Note that lightning currents of 20 kA are quite frequent, and they last for only a few microseconds
•
fl
RESISTANCE AND AMPACITY OF SOME BARE AERIAL CONDUCTORS
conductor size resistance per conductor ampacity in
at 75°C free air*
cross AWG section copper ACSR copper -ACSR
mm2 rl/km rl/km A A
10 5.3 3.9 6.7 70
7 10.6 2.0 3.3 110
4 21.1 0.91 1.7 180 140
1 42.4 0.50 0.90 270. 200
3/0 85 0.25 0.47 420 300
300 MCM 152 0.14 0.22 600 500
600 MCM 304 0.072 0.11 950 750
. 1000 MCM 507 0.045 0.065 1300 1050
The ampacity indicated is the maximum that may be used without weakening the conductor by
overheating. In practice, the actual line current may be only 25% of the indicated val1,1e.
£'Lecfh'co1 Properties oF trM1fll?is$t'on Lin~ts -- -··--...
1. The voltage should remain as constant as possi· ble over the entire length of the line, from source to load, and for all loads between zero and full load;
TYPICAL IMPEDANCE VALUES PER KILOMETER
2. The losses must be smal(so as to attain a high transmission efficiency;
3. The i 2R losses must not overheat the conductors.
If the line alone cannot satisfy the above re· quirements, supplementary equipment must be added until they are met.
FOR 3-PHASE, SO Hz LINES
type of line
aerial line;
underground cable
0.5
0.1
Xc
n
300 0
30
(
TRANS/r1.Tss:roN LIN£ MODELING
ELECTRIC TRANSMISSION LINE PARAMETERS
Line Resistance
where
pl Rd =-ohms c A
p =resistivity of conductor l= length
A = cross-sectional area
. Resistirity and Temperature Coefficient of Conductor A!aterials
Material
Aluminum Brass Bronze Copper
Hard drawn Annealed
Iron Silver Sodium Steel
Resistidty ( p) at 20°C Temperature Coefficient (a) Micro-ohm em Ohms circular mils per ft at 20°C
2.83 6.4-8.4 13-18
1.77 1.72
10 1.59 4.3
12-88
17.0 38-51 78-108
10.62 10.37 60 9.6
26 72-530
R 2 =R1[1 + a(T2 - T1)j
0.0039 0.002 0.0005
0.00382 0.00393 0.0050 0.0038 0.0044
0.001-0.005
Here R2 is the resistance at temperature T2 , and R 1 is the resistance at
temperature T1•
'k
(
::r
LINE INDUCTANCE
Inductance of a Single-Phase Two-Wire Line
Single-Phase Two-Wire Line Configuration.
the inductance due to internal flux is
Li = Hl0-7) henriesjmeter
The inductance due to external flux linkages is
LP1, P2 = (2 X 10- 7 )ln ( ~:) Substituting D2 = D and D1 = r 1, we get for the external contribution:
L 1 , ·xt = ( 2 X 10-; ) ln { ~ )
/3
/
l ·
The total inductance of the circuit due to the current in conductor 1 only is therefore
L 1 = (2 X 10-')[! + 1n( ~) l henries/meter
Similarly, the inductance due to current in conductor 2 is
L, = (2 X 10-7) [!+In ( ~) l henriesjmeter
Thus L 1 and L~ are the phase inductances. For the complete circuit we have
Lr=Lt+L2
Thus
A more concise form of the inductance expression may be obtained if we observe that
so that
where
1 ln e114 =-
. 4
L 1 = ( 2 X 10- 7) ln ( ~ )
L 2 = ( 2 X 10-7) ln ( ~ )
L1=(4X10-7 )ln({rgD ) r'r.' 1 2
r..' = r..e-11• ' ' = 0.1188ri
140 R r=GM Geo m~lr/c ·t11:-~ n 1Z c.. c/..' :J \
'\
~-
In practice, we deal with the inc:luctive reactance of the line per phase per mile and use the logarithm to the base 10. Performing this conversion, we obtain
where
X= k log ~ ohms per conductor per mile r
k = 4.657 X 10-3/
= 0.2794 at 60Hz
assuming identical line conductors. Expanding the logarithm in the expression ·
1 X = k log D + k log -;
r
The first term is called Xd and the second is Xa . Thus Xd = k log D inductive reactance spacing factor
in ohms per mile
X = k log..!. inductive reactance at 1-ft a r'
spacing in ohms per mile
Factors Xa and Xd may be obtained from tables available in many handbooks.
I.J
Example
Find the inductive reactance per mile per phase for a single-phase line with phase separation of 20ft and conductor radius of 0.06677 ft.
Solution
We first find r ', as follows:
\Ve therefore calculate
r ' = re-11• = (0.06677)(0.7788) = 0.052 ft
. 1 xa = 0.2794log 0.05:2
= 0.35875 Xd= 0.2794log20
= 0.36395 X= X a+ Xd = 0.72247 ohms pet mile
Bundle Conductors
I {,
1. Volf~es a.bave :Z3o kv (E:HV) ,w/fh
e({ed b£comes more e,/XCes.siVe. ,..e_
more Loss ~ well as
,). - · Corona.
1'7
.__ 3. -I h-1! Cr Atl/ e~~,1} lP.A-? -h€ [(ed 1/ u. d
{
Co11Sidera bly by U.slng . rnol"e
fJ,aM one ConducftJr Per fJha~e
4- 7Ju. C,;1d£~cftJr.5 Ore /y, C I ose. pra'XIm,'lj'
(c,.,pared w;~ flu_ 5fAci"c'J b-efw~~l1 phases.
5- fin Imf~rfa1A-1 odvaMI-~Ag(. of bund//'!.1
lS flu. l?.educflor~ ln 1/nl! ;teacftJMUt.s
Bundle Conductor.
General Multiconductor Configurations
n
A MultJconductor Configuration.
VI =jw(Lll/1 +L12/2+ '··+Linin)
V2=jw(L12/1 +L22l2+ ··· +Lznln)
The apparent self- and mutual inductances LJJ and L1k are given by
The above expressions fonn the basis for the evaluation of lin~ inductances in practice.
)8
l?
Inductance of a Multiconductor Single-Phase l.lne
CONDUCTOR· A
CONDUCTOR B
Single-Phase Bundle-Conductor Line.
with
L = (2 X 10-7) ln ( g~~) henriesjmeter
where
(
-- ..
E X A f/1 PLE
Consider a 375-kV, single-phase line wi th bundle conductors as shown in Figure The phase separation D1 is 12.19 m, and the subconductor spacing isS= 45.72 em. The subconductor diameter is 4.577 em. Calculate the line inductance
A 375-kV Single-Phase Bundle-Conductor Line.
Solution '
We have four subconductors; thus
N1 =N2 = .2
The geometric mean distance is therefore
Expanding the products, we have
From the geometry
We thus have
D13 = D-24 = D1 = 12.19 m D
14 = D1 + S = 12.19 + 0.4572 = 12.6472 m
D;!.3 = D1 - S, = 1:2.19-0.4572 = 11.7328 m
Dm = [ ( 12.19)2( 12.6472)(11.7328) ]'
1 4
= 12.1857 m
Observe that
Also
Therefore,
D11 = D22 = r{ = 0.7788r1
= (0.7788) ( 4 ·~77 X 10-2
)
= 1.7823 X 10- 2 m
D12 = D21 = S = 0.4572 m
D.= [ (1.7824 X 10-2 }
2(0.4572)2
] 114
= 9.027 x 10-2 m
As ·a result of the above, we obtain the following result
~-- ---------- - . . - ·- ..
L=(2xto-7)ln( 12.1857 ) 9.027 X 10-2
= 0.981 X 10-6 henriesjmeter
_,
Inductance of a Balanced Three-Phase Single-Circuit Line
2
A Balanced Three-Phase Line.
I 2
8 8 3
0 ~o--··~1·-- 0~
H·Type Line.
-.rr----- ----- -..
Dl2 = D23 = D D13 =2D
2..2._
Z3
Transposition of Line Conductors
a------ c b
b-----""\,. 0 c
C·-----.J b a
.... ~._-I ---~ .. ~ .... ~~---- m ---~ .. ~
Transposed Line.
-------
(
-·
L = (2 X 10- 7) ln ( g~~) henri~jmeter
GMR=r'
1 Xa=klog GMR
Xd=klogGMD
\\ith k = 0.2794 at 60Hz, as before.
(
£XAMPLE
Calculate the inductance per phase of the 345-kV three-phase solid conductor line shown in Figure. Assume that the conductor diameter is 4.475 em and the phase separation D1 is 7.92 m . Assume that the line is transposed.
In this case the geometric mean distance is given by
GMD-:- (DAnDncDcA)l/3
GMR = [ ,§, ( D,;) rN For the case of synunetrical bundle conductors, we have
GMR = [ Nr'(A)N-lr!N
. XL=Xa+Xd
where as before for 60 Hz operation,
EXAmPLE
Xu= 0.2i94log G~H Xd= 0.2794logGMD
An 1100-kV, three-phase line has an eight subconductor-bundle delta arrangement with a 42 in. diameter. The subconductors B.!e ACSR 84/19 (Chukar) with r' = 0.0534 ft. The horizontal phase separation is 72 ft, and the vertical separation is 60ft. Calculate the inductive r:eactance of the line in ohms per mile per phase. ·
(
Thus
From the geometry of the phase arrangement, we have
36 tanB= -
60 8 = 30.96°
D _ 60 AB- cos30.96°
= 69.97 ft
GMD = [(69.97){69.97){72)P13 = 70.64 ft
For Chukar we have r' = 0.0534 ft. The bundle particulars are N = 8 and
A = ( 42 / 2) in. Therefore,
GMR = [s(0.0534)( i! rr Thus
As a result,
= 1.4672 ft
1 Xa = 0.2794log 1.4672
= -0.0465 xd = 0.2794log70.64
= 0.5166
28
(
LINE CAPACITANCE
' IS a
J. PofeKtfiJ differeMCV-J befwe~41 fk Con clue for s
~- fo/e.MftJ dt'(fereMC£ befwee~ LOJ1dllclor..S
CW-d GrcU1\d
~ n~ fime Vp.rt'Al-;al7 of Cltttro~.s
Resu ll.s in CurrfM;ta ( evlled · f-~ ( ( --
Lin-e ,.. ChAr[}}Y\j
~---. - --
L'""\
(
Capacitance of Single-Phase Line
. Single-Phase, Two-Wire Line.
If r1 = r2 = r, we have
'TTEo
CAn= (D) In -r
Converting to microfarads (J.LF) per mile and changing the base of the logarithmic term, we have
C _ 0.0388 F .1 A B - Jj J.L per m1 e
2log( r) The capacitance to neutral for conductor A is defined as
q 2we0
CAN= VA = ( D ) In - ·
rl
Like\\;se, observing that the charge on conductor B is - q, we have
_ -q _ 2we0
CsN- Vs- {D) ln -
r2
(
Observe that
In terms of p.F per mile, we have
C -0.0388
AN- D log-
r
p.F per mile to neutral
A c.e e ~------~1~(------0
A CAN CeN e e----~1 r-( -·· -~1.-( --tle
N
Capacitance between Lines as Series Combination of Capaci· tances to Neutral. ·
The capacitive reactance Xc is given by
where
X = - 1- = k' log D ohms · mile to neutral
c 2TT/C r
k' = 4.1 X 106
I
~xpanding the logarithm, we have
1 X c = k' log D + k' log r
The first term is called Xd., the capacitive reactance spacing factor, and the second is called X
0., the capacitive reactance at 1-ft spacing.
~----- ---- -· , __
.:.)\
Example
Xd. = k'log D 1
xa.=k'logr
Xc=~d.+Xa'
Find the capacitive reactance in ohms· mile per phase for a single-phase line with phase separation of 20 ft and conductor radius of 0.06677 ft for 60-Hz operation.
Solution
Note that this line is the same as that of Example .. We have for f= 60Hz:
\Ve calculate
As a result,
6 .
k' = 4·1 ~ 10 = 0.06833 X 106
xd. = k'log2q
= 88.904 X 101
xa. = k'log 0.0~677 = 80.32 X 103
Xc=Xd.+Xa.
Xc = 169.224 X 103 ohms· mile to neutral
32.
Including the Effect of Earth
* s Jt 0 u I d b .e. Ttt k.cn .tJ'I to a CC.Ol.U1 t iF
Conductors ~ /1of l}l'tJh ~ov<7h a. hove
GroUMd .
-..f,---- • Q Physical choroe
H
l H
__.!,____ • -q lmooe choroe
Image Charge Concepl
/////
· General Multiconductor Configurations
p
a . c .~ . -.<.e .. __ - -----·:--------
H/2
H/2
~·
Capacitance of a Single-Phase Line Considering the Effect of Ground · ·
,~ D
-q
0 B
A' e'
Single.Phase Line and Its Image.
C - '1TEo
:AB- ( D H ) ln -·
r HAB'
The capacitance to neutral is obtained using
C - q AN-
VA 2'1Tt0 -----=--- farads per meter
Observe that again
zn(D ·_!!_) r HAB'
C -CAN AB- 2
35
(
...
Example
Find the capacitance to neutral for a single·phase line with phase separation of 20ft and conductor radius of 0.06677 ft. Assume the height of the conductor above ground is 80 ft.
Solution
We have
D= 20ft r = 0.06677 ft H= 160ft
As a result,
Therefore we have
C - 2'77'£o . AN,- ( 20 · 160 )
In 0.06677 · 161 .2452 _ 21it0
5.6945 farads per meter
If we neglect earth effect, we have
C - 2'77'£o AN2- ( 20 )
In 0.06677
2 '77'£0 - 5.7022 farads per meter
The relative error involved if we neglect earth effect is: c -c ,
Find the capacitance to neutral for the single-circuit, three-phase, 345-kV line with conductors having an outside diameter of 1.063 in. with phase configuration as shown in Figure
Solution
GMD = [(23.5)(23.5)(47)f13
= 29.61 ft 1.063
r = (2)(
12) = 0.0443 ft
2rrE0
·
CAs= In( G~D) = 8.5404 X 10- 12 farads per meter
~ 8 • ~23ft Gin-J--2311 Gin-1
Now, :rncfucl€.. Earfh EFFect
H =so {t
H1 = H2 = H3 = 2 X 50= 100 ft 4o
( H$ ) _ l (100)(100){100) ]
113
ln Hm -ln (102.72)(102.72)(110.49) ·
= -0.0512
I 2 3
la_23.5fl ~ 23.51! 1 50ft ~ __,
50ft
Le 2'
0 3'
e
Thus
(
1 CAN= (18 X 109 )(6.505- 0.0512)
= 8.6082 X 10-12 farads per meter
(
TWO-PORT NETWORKS
r ~·J = r.. y., 1 r v.l Jr Yrs Yrr V,.
Is I,
I I + + ___,.. ___. v, v,
A Two-Terminal Pair or Two-Port Network.
J. ·nnJ,·'d 7-k wrrWs /, rerms of bol-h
V~/faJ'S Of" h'nd''':J V~/f~«-S Tn f-t:.rm.S
0 r Bofl, Curr€NJrs c Trat\>fer Pr~blem.;
~ . R n d i ~';} V d I~..._~ ClN\.cl ~ rre.....J o.:t c=' 11<
__ 'P~h·-- --"( fern"'nA.I.r in fer!Y15 of fk orher tfJoJr ·· ( /ro.MSnu'.SSloll Probfe.m)
_..--- ---· .... . . . . ·-· ~ -·
/I
f /""'
(
..........
-..-
( 0) n - network
(b) v-network
Zs
( c) T- network
I I I -
,....-
Yso Yro
Zr
(
-
T~AJ\/SMt:ssroN ~oSLEM
~=A~+Bir
Is = C~+Dir
to represent the two-port network. In matrix form, we thus have
Z/2 v Z/2
+ l M ..... .....
I ... Is ...
Ir
v s y
..... .....
A Symmetrical T -Network.
the middle point of the T be ~,1 ; then we obtain
V, = Is(;)+ VM
Is= ~\1¥ + Ir
~~=~+Ir(;) From these eq~ations, we can write
V, = ( 1 + Z2Y) ~ + z( 1 + z4y) Ir ( zy.)
Is = Y~+ 1 +T Ir
-"" +
"
v r
--
A= 1 + ZY 2
B=Z(1+~y) C=Y
D= 1 +ZY ·2 .
z + o----.-----i t-----e---<> +
A r.-Network.
A= ( 1 + Z2Y)
B=Z
C= Y(1 + z:) D=A
(
- ·-
v s v .. v r Is
A, a, I .. A2 82
Ir
c, o, c2 02
A Cascade of Two two-Port Networks.
From which, eliminating (V:\t• 1.\1 ), we obtain
Thus the equivalent ABCD parameters of the cascade are
A =A1A 2 + B1C2
B = A 1 B2 + B 1 D2
C= C1A2 + D1C2
D = C1B2 + D1D2
{
TRANSMISSION LINE MODELS
AssUm[Jfions I· -r7iL line i.s o perafi11) lhlder
5 inUSoi cla r I bo lanced I Sfe~AdyS t crt-~ Ce>n d i t\o)t~· ·
It is practical to introduce the complex variable(} in the definition of the ABCD parameters. Vole define
8 = vl= /ZY
Also we have
As a result, A= cosh8 B = Zcsinh8
C= _!_sinh8 zc D=A
Observe that the total line series impedance and admittance are given by
Z=zl · Y=yl
Evaluating ABCD Parameters
e9 + e-9
A=-- -2
= ~ ( eetl!1+ e-etf -8z)
e9 - e-e sinh8 = 2
= ~ ( e9tl!1- e-9tj 82)
cosh 8 =cosh 81cos 82 +}sinh 81sjn 82
sinh 8 =sinh 81cos 82 + jcosh81sin 82
(
Example
Find the exact ABCD parameters for a 235.92-mile long, 735-kV, bundle-conductor line with four subconductors per phase with subconductor resistance of 0.1004 ohms per mile. Assume that the series inductive reactance per phase is 0.5541 ohms per mile and shunt capacitive susceptance of 7.4722 X 10-6 siemens per mile to neutral. Neglect shunt conductance.
Solution
The resistance per phase is
r = 0·1~04 = 0.0251 ohmsjmile
Thus the series impedance in ohms per mile is
z = 0.0251 + j0.5541 ohmsjmile
The shunt admittance is
y = j7 .4722 X 10-6 siemensjmile
For the line length,
Z = zl = (0.0251 + j0.5541)(235.92) = 130.86/87.41 o
Y= yl= j(7.4722 X 10-6 ){235.92) = 1.7628 X 10-3j90°
\\7e calculate 8 as
Thus,
8=Jzy
= r ( 130.86j87 .41° H 1.7628 x l0-3j90° r12
= 0.0109 + j0.4802
81 = 0.0109 82 = 0.4802
We change 82 to degrees. Therefore,
82= (0.4802)(1:0) =27.5117°
·- ·~·.
._)V
(
co~h 8 = ~ ( e0·0109j27 .5117° + e - 0
·0109
/ 27.5117°)
= 0.8870j0.32_42° ·--· -- - ---
From the above,
D=A = 0.8870j0.3242°
We now calculate sinh 8 as
\Ve have
As a result,
(;X AMPLE
sinh 8 =! ( e 0·0109j27 .5117°- e-om09j-21 .5117°)
= 0.4621/88.8033°
. /Z ( 130.86j87 .41° ) 112
zc = v y = 1.7628 X 10-3j90°
= ( 74234.17/-2.59 f 12
= 272.46/-1.295°
C=_!_sinh8 zc
B =Zcsinh8
= 125.904/87.508°
= 1.696 X 10-3/90.098°
Fi~ cJ \{, , T_s tMtd I; ewvJ effic!'enc;t
5 j
oF TrOM.5 m t'55t'tJn , G/ ven . M vAR :.1-?"0o fti rA k VR = 7oo lc v oJ; tl, o .. 9. 5 p. F.. L'rj~ ""!J
Solution
We have the apparent power given by
Sr= 1500 X 106 VA
The voltage to neutral is
Therefore,
V. = 700 X 103
V r /3
Ir= '1500X106 j-cos-10.95
a( 100v~ to') = 1237.18/-18.19° A
~=A~+Blr
= ( 0.8870j0.3242°) ( 700 ~ 10
:)) fS
+ ( 125.904/87.508°) ( 1237.18/-18.19°)
= 439.10151/19.66° kV
The line-to-line value is obtained by multiplying the above value by .f3, giving
'VsL = 760 ~546 kV
The sending-end current is obtained as
I.= C~+Dlr
= ( 1.696 X 10-3/90.098°) ( 700 ~ 10:3)
t/3
+ ( 0.887/0.3242) ( 1237.18/-18.19°)
= 1100.04/18.48° A
The sending-end power factor is
cos 4>, = cos(19.66- 18.48) = 0.99979
....... .
52
r
--
As a result, the sending-end ~ower is
Ps = 3( 439.10151 X 103 }(1100.04)(0.99979)
= 1448.78 X 106 MW
The efficiency is
~ Tl=-
~ 1500 X 106 X 0.95 -
1448.78 X 106
= 0.9835
Lumped Parameter Transmissi~n Line Models
+
1 v,
-ls
V: =A V,. + Blr Js = CV,. + DJr
z,= e
-1,
A-1 r,=a~
+
1 v,
Equivalent 'IT Model.of a Transmission Line.
Z =B fl
A-1 Y.= B
53
';
i.
I
.. · . .r-··· : ---· ·
Approximations to the ABCD Parameters of Transmission Lines
Consider the series expansion of the hyperbolic functions defining the A, B, C, and D parameters given by
62 64 66
A = 1 + 2 + 24 + 720 + .. .
( 62 64 6(i )
B=Z 1 +6+ 120 + 5040 + ... .
C= }"(1 + 62 +~ +L+ ... ) 6 120 5040
D=A
Using 6 = /ZY, we obtain the following:
zy z:.~yz z:!y:! A = 1 + 2 + 24 + 720 + ...
( zy Z 2Y 2 Z 3Y:3 )
B=Z 1 +6+120+ 5040 + ...
( zy Z 2Y 2 z:!ya )
c = y 1 + 6 + 120 + 5040 + .. .
I=Or- Less Tha.N\
Lih.e . we. c~
ZY A=D=I+-
2
·( ZY) B=Z 1+T .
C= Y(1 + ~y)
or
+ o----....-----4 1---...... ---o+
r~ Y/2 Y/2
Nominal ., Model of a Medium Transmission Line.
If only the first tenn of the expansions is used, then
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