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Unformatted text preview: 62
In original positions in the transposition cycle, distance ab = V142 + 3.52 = 14.43 ft
distance a—b’ = V142 + 28.52 = 31.75 ft
distance a—a’ = v 252 + 282 = 37.54 ft D5,, = D; = {/14432 x31752 = 21.04 ft
Dec = {/252 x282 = 26.46 {c
Deq = {/21042 x2546 = 22.71 ft
1
D, = [(#00373 x 37.54)2 «0.0373 x 32]3 = 1.152 ft 22.71
= . 93 ‘7
1.152 5 6 x 10 H/m = 5.963 x 107 x 103 x 1609 = 0.959 mH/mi
XL = 377 x 0.959 x 103 = 0.362 n/mi/phase L = 2x 1071n M (b) r = “08 = 0.0462 ft as in part (a) above, except that r is substituted for D3: Dy; = [(¢0.0462 x 37.54)2 «0.0452 x 32] = 1.282 ft From part (a) above, Deq = 22.71 ft and 22.71
1.282 uln Xc = 2.965x1o‘1n I _ 138,000/Jﬁ
“‘8 " 85,225 = 85, 225 Q  mi/phase to neutral = 0.935 A/mi/phase = 0.467 A/mi/conductor Chapter 6 Problem Solutions 6.1 An 18krn 60—Hz single circuit threephase line is composed of Partridge conduc
tors equilaterally spaced with 1.6 m between centers. The line delivers 2500 kW
at 11 kV to a balanced load. Assume a Wire temperature of 50°C. (a) Determine the per—phase series impedance of the line. (b) What must be the sendingend voltage when the power factor is
(2') 80% lagging (ii) unity (iii) 90% leading? ( c) Determine the percent regulation of the line at the above power factors.
((1) Draw phasor diagrams depicting the operation of the line in each case. Solution: (0) II
II R
From Table A.3, X0 18
0.37
92 x 1.609
0.455 Q/mi 4.242 Q ll 63 and since 1.6 In = (1.6 x 100)/(2.54 x 12): 5.25 ft, Xd = 0.2012 (Table A.4,5’3”) X = 0.465 + 0.2012 = 0.666 Q/rni
0.666
= = '
For 18 km, X 18 x 1.609 451 9
Z = 4.242+j7.451 = 8.575 60.35° 9
(b) For power factor = 1.0, :93;
2500 11.000 I = = 131.2 A ’ = 63'0 V R J3 x 11 M3 0 Vs = 6350 + 131.2 (4.24 + j7.451) = 6906 +j977.6 = 69754 806" sending~end line voltage = J3V5 = \/3 x 6975 = 12,081 V For power factor = 0.8 lagging, I 2500
l "I «5 x 11 x 0.8 V5 = 6350 + 1645—36.87° x 8.574 60.350
7639+j5.60 = 766014.19"
sendingend line voltage = \f3Vs = x/3 x 7660 = 13, 268 V For power factor = 0.9 leading, 2500 I = —————— =145.8A
13' V§x11x09
1c; == 6350+ 145.8;‘25.84° x 8.574 60 35° 6433 + j1247 = 65534 10.97° sendingend line voltage V3 V; = V3 x 6553 = 11, 350 V t ulna.»mn‘M.»I..«L.~4.—‘.huiv5%a‘qmaz:L—LT—ZW,W>~RFKJV~%“Mi¥‘T " ‘ IVsl  lVRl 
———.—— x 1007 MRI 0
7660 — 6350 6350
6975 — 6350 6350
6553 — 6350 6350 ‘70 Regulation at p.f. =0.8 lagging, % Reg. x 100% 20.63 % at unity p.f., ‘70 Reg. x 100% 9.84 % at p.f. = 0.9 leading, % Reg. x 100% 3.20 % (d) For p.f. =0.8 lagging, 64 “7., v5 = 7660 v IRX =1222 V
VR = 6350 V 369° IRR = 696 v In = 131.2 A For unity p.f., Vs = 6975 V
IRX = 978 V IR=131_2A VR=6350V 1RR=557V For p.f. =O.9 leading, Vs = 6553 V . IRX = 1086 V
I}: = 145.8 A 25340 IRR = 618 V VR = 6350 V 6.2 A 100mi, singlecircuit, three—phase transmission line delivers 55 MVA at 0.8
power factor lagging to the load at 132 kV (linetoline). The line is composed
of Drake conductors with ﬂat horizontal spacing of 11.9 ft between adjacent
conductors. Assume a wire temperature of 50°C. Determine (a) the series impedance and the shunt admittance of the line.
(b) the ABCD constants of the line. (c) the sendingend voltage, current real and reactive powers and the power
factor. (d) the percent regulation of the line. Solution: (0) Decl
series impedance Z €/"11.9—x"11‘.9"x"2—x "1T.9 = 15 a
100 x (0.1284 + j0.399 + 30.3286) 12.84 +j72.76 = 73.88180° Q .100 106 _4 O
3—2 (O.9012+0.0803) ‘ 2915 “0 £5 5.83 x 10'4 ; 90° 5 Z
2
shunt admittance Y 65 Z
A = D=1+3}: 73.88 x 5.83 x 10—4 4119: = 0.979 0.2190 H
,_1
+ 2
B = Z = 73.88/__8_C£Q
ZY
C = Y 1+——)
4
— —4
= 583x 10‘4 (1+73'88x°f3x 10 (170°) 5 = 5.768x10‘4490.108° 5 (Check: AD — BC = 1 must be satisﬁed) 55 000
I = ———’——— 0.8 — '0.6 = 192.4 — 144.3 A Current in series arm: _ .132, 000 _ _
I...”ies = 1924—11443“ ﬂ x 2.915x 10 4
= 192.4— 3122.1 = 227.9é—32.40°
Vs = 133;)00 +'227.94—32.40° x 73.884 80°
= 87,563 + 312, 434 = 88,44148.08° v to neutral
1V5} = J3 x88,441 = 153.2 kV, linetoline
I5 = 192.4 — 3122.4 + 32.915 x 104 x (87, 563 + 3'12, 434)
= 188.8—j96.9 = 2125—27.?
1151 = 212 A
as = 8.08° — (47.2") = 35.28°
P5 = (J3 x 153.2 x 212) cos 35.28° kW = 45.92 MW
Q5 = (x/z'a x 153.2 x 212) sin 35.28° kvar = 32.49 Mvar (sendingend) p.f. cos 35.28° = 0.816 lagging ll Meg. = “00% IVR. FLI (1532/0979) — 132 ‘ = ————— — 8.55 132 x100% 1 % V 6.3 Find the ABC'D constants of a 7r circuit having a 6009 resistor for the shunt branch at the sending end, a l—kQ resistor for the shunt branch at the receiving
end and an 809 resistor for the series branch. 66 Solution:
80 Q
600 Q 1000 Q
VR
V5 = VR+ IR+ m x 80 = VR+80IR +0.0sz
= 1.08VR + SOIR
15 = 13 + Vi + w = 0.001VR + 0.0018VR + IR + 0.13313 1000 600
= 0.0028VR + 1.1331}; The ABC D constants are A B 1.08 C
80 D D 0.0028 S
1.133 6.4 The ABCD constants of a threephase transmission line are A = D 0.936 + 30.016 = O.93640.98°
B = 33.5+j138 = 142476.4°o
C = (—5.18+j914)x10"6 s The load at the receiving end is 50 MW at 220 kV with a power factor of
0.9 lagging. Find the magnitude of the sendingend voltage and the voltage
regulation. Assume the magnitude of the sendingend voltage remains constant. Solution:
50 000
I = __7__ —25.84° = 145.8 —25.84° A
R «3 x 220 x 0.9;— 4—
VR = 220’000 = 127,00040°v J1?
Vs = 0.93640.98° x 127,000&°+142¢76.4° x 14585—2584?
= 118, 855 + 32033 + 13,153 + 3'15, 990 = 133.234 777° kV With lineto—line sendingend voltage IVs] = J3 x 133.23 =‘230.8 kV, 230.8 V = —— = _' I RNLI 0.936 2460kV
246.5 — 220 6.5 A 70 mi, singlecircuit, three—phase line composed of Ostrich conductors is ar
ranged in ﬂat horizontal spacing with 15 ft between adjacent conductors. The
line delivers a load of 60 MW at 230 kV with 0.8 power factor lagging. 68 Q; = 1.086 x 0.742 x sin (2.125° — (—36.54°)) = 0.503 per unit
= 100 x 0.503 = 50.3 Mvar A = bra = 1+l(0.1166 675° XOJQQLE) = 099040.248"
2 2
V — V
‘7cReg_ = LSW x 100%
RJ‘L
= Whom = 9.73% 6.6 A singlecircuit, threephase transmission line is composed of Parakeet con
ductors with ﬂat horizontal spacing of 19.85 ft between adjacent conductors. Determine the characteristic impedance and the propagation constant of the
line at 60 Hz and 50°C temperature. Solution:
At 50°C and 60 Hz, from Table A.3, for Parakeet conductors, r = 0.1832 Q/mi X. = 0.423 Q/mi
Deq = {719.853 x 2 ft = 25 ft
At 25 ft, Xd(inductive) = 0.3906 Q/mi Therefore, 2 = 0.1832 + j (0.423 + 0.3906) Q/mi
0.834; 77.31° n/mi 0.0969 x 106 n . mi 0.0955 x 106 0  mi 10"6 4 9 °
X; + Xd 0.0969 + 0.0955 5.1975 x 10290" S/mi H X;
X 4(capacitive) 11 Characteristic impedance: _ z __ 0.834(7731‘3 _ o
Zc  \/; — m 9 — 40061—6345 9 Propagation constant: 7 = M = 0.334 x 5.1975 x 106577.31° + 90° mrl = 2.08 x10’3 mi—1 6.7 Using Eqs. (6.23) and (6.24) show that, if the receivingend of a line is ter
minated by its characteristic impedance ZC, then the impedance seen at the
sending end of the line is also Zc regardless of line length. 69 Solution:
If ZR = ZC, then 13 = VR/ZC, and VR  IRZC = 0. VR ‘l' IRZC 7L
——6
2
l’rR + 6.7L
2Zc From Eq. (6.23) V; From Eq. (6.24) Is H where L is the length of the line. Finally, Z, = Vs/Is = 2,; (which is independent of L) 6.8 A 200—mi transmission line has the follOwing parameters at 60 Hz 0.21 Q/mi per phase
0.78 Q/mi per phase
5.42 x 10‘6 S/mi per phase resistance T
series reactance ac
shunt susceptance b H is
a 1:23.17; ﬂan—gvﬁm 7‘ sick“ (a) Determine the attenuation constant a, wavelength A and the velocity of
propagation of the line at 60 Hz. 3; (b) If the line is open circuited at the receiving end and the receiving—end
voltage is maintained at 100 kV lineto—line use Eqs. (6.26) and (6.27) to
determine the incident and reflected components of the sendingend voltage ‘
and current. (c) Hence determine the sendingend voltage and current of the line. Solution: (0) 0.21 Q/mi z, = 0.78 Q/mi (0.21+j0.78) Q/mi = 0.8081773? Q/mi
5.42 x 10'3 5 7731" S/rni (/5 = 2.092 x 10"3 4 82.47° mi" 0; +113 = (2.744 x 104 +j2.074 x 10*) mi—1
2.744 x 10"1 nepers/mi g: 27r x103 ' Q‘Q N ‘i
ll Attenuationconstant a W velen h A = = —————— ‘ = '
a gt 6 2074 mi 3030 m1
3
Velocity of propagation Af = 3%]: = % rni/s = 181770 mi/s (5) Characteristic impedance: Zc = \/E = 386055—753" f2
3/ 70 When the receiving end is open circuited, 13 = 0. Then, V . V .
from Eq. (6.26) Vs = 356L635!“ + €5°L€JBL
V . V .
from Eq. (6.27) 15 = led?“ — i€°L€_JBL
2Z6 2Zc
W w
incident reﬂected
where L = 200 mi = length of the line
e°‘L = 1.0564 5’“ = 0.9466
BL = 2.074 x 10*3 x 200 x 1:3 deg = 23.7"!" Hence, at the sending end (taking the receivingend line voltage as reference), the line
toline voltages and currents are 0 incident voltage 12, = 13— LCﬁ x 1.0564; 2377" IN = 52.82 23.77° kV
reﬂected voltage 11,. = 1—30 L0° x 0.9466 (—23.77° RV = 47.33 (—23.77" kV
incident current I, = 4.9L x kA 2 x 386.05 £—7.53° «55 30° = 78.991 13° A reﬂected current I, = 100$ 0'9466d3'i kA ‘2 x 386055—753" x .5530:
= 70.784133.76° A (The 30° angle in the denominator of the second fraction of the current equations above
represents a phase/line V conversion.) (6) V5 = VI; + V, = 52.82( 23.77° +47.334—23.77° kV = 91.685 1.38° kV
Is = I; +I, = 78.99413" +7078; 133.76° A = 608370.78; 604° A where all angles are expressed with respect to receivingend line voltage. 6.9 Evaluate cosh0 and sinh 6 for 0 = 0.51 82°. Solution:
0.5/8_2° = 0.0696 + j0.4951
0.4951 radian = 28.37°
cosho = % (Eo.m9sﬂ+ €—0.0696 _28_37o)
= % (0.9433 + 30.5094 + 0.8207 — 30.4432) = 0.8820 +j0.0331
sinh9 = % (0.9433 + 30.5094 — 0.8207 +j0.4432) = 0.0613 +j0.4763 71 6.10 Using Eqs. (6.1)., (6.2), (6.10) and (6.37) show that the generalized circuit con 6.11 stants of all three transmission line models satisfy the condition that AD  BC = 1
Solution:
Shortline model (from Eq. (6.1) and (6.2)):
A = D = 1 B = Z C = O
AD—BC = l—ZXO =1
Mediumlength line model (from Eq. (6.10)):
ZY
A = D=<1+£2z> B=Z C=Y(1+T>
2 2 2 2 2
AD—BC = (1+ZZ)—ZXY(1+Z—Y = 1+ZY+Z Y _Zy_ ZY =1
2 4 4 4
Longline model (from Eq. (6.37)):
A = D = cosh'yl B = chinh'yl
_ sinh'yl
C — ZC
AD — BC = cosh2 'yl  Zc sinh 71 (513171)
C
2 2
_ 2 I 2 _ 6’11 + 6‘71 6'1! __ 6—71
— cosh 'yl smh 'yl — ( 2 ———2
62'” + 2 + 6’27! 62'” — 2 + 6“?“
r —T“— ‘ “7—”  1 The sendingend voltage, current and power factor of the line described in Exam—
ple 6.3 are found to be 260 kV (lineto—line), 300 A and 0.9 lagging, respectively.
Find the corresponding receivingend voltage, current and power factor. Solution: >
Horn Example 6.3, A = D = cosh'yl = 089041134" >
B = chinh'yl = 406.41—548" x0.4597g84.93° Q = 1868217945" 0 sinh 'yl 0.45971 8493"
= ————— S = . 319 . ° 5
Zc 406.4 _5.480 1 131 X 10 041 [‘3 ﬁl'lfl = [—3 ’illrvil 0
ll f’_"—I 9:5 L__——J
n V1.2 = DVsBIs 13 = —CV5+AIs
260 V “‘  0° kV = 150.11 0° V 5 £4. ‘/—k 72 Is = 3005—cos’10.9A = 3005—2584‘ A
VR = 0.8904M x 150.11 0° — 186'824L451230300ﬂ RV
= 108355—2276" kV
[VB] = \/§x 108.85 IN = 188.5 kV lineto—line
13 = —1.131 x 104490.41o x 150.11 x 10'3/_Q_°_+ 0.531904% x 3005—25.84° A
= 372.05—48.95° A
IIR = 372 A The receivingend power factor is then pi. = cos(—22.76°+48.95°) = 0.897 lagging 6.12 A 60 Hz threephase transmission is 175 mi long. It has a total series impedance
of 35 + j 140 Q and a shunt admittance of 930 x 10‘6L9m S. It delivers 40 MW
at 220 kV, with 90% power factor lagging. Find the voltage at the sending end
by (a) the shortline approximation, (b) the nomial7r approximation and
(c) the longline equation. Solution:
1 = 175mi
Z _ 35+j40 = 144.3475.cho n
Y 930 x 106 s
40 000
I = ~_...’____ = 116.6 25.84° A
R «5 x 220 x 0.9 g— (a) Using the shortline approximation, VS = 127,017+116.65—25.84° x144.3g75.96° = 127,017+10,788 + 312, 912
= 138,4oszﬂv I
[Vsl = «ﬁx 138,408 = 239.73 kV (b) Using the nominal7r approximation and Eq. (6.5), 0.1342 c o c,
Vs = 127,017( 2 4165.96 +1)+144.3475.96 x116.65—25.84 127, 017(0.935+ j0.0163) + 10, 788 +j12,912 = 129, 549+ 3'14, 982
130, 412%
\/§ x 130,412 = 225.88 kV lel (c) Using the longline equation, Zc 144.3175.96° I o
(930 x 10—6 1905) “ 394ﬂ 72 = y/144.3x930x 10—54165.96° = 0.3663483.0° = 0.0448 + 30.364
8044890364 = 1.0458420.86° = 0.9773+j0.3724 — 73 (00448510364 095625—2086" = 08935—303405 cosh ~12 = (0.9773 + 30.3724 + 0.8935  j0.3405) /2 = 0.9354 + 30.0160
sinh 71 = (0.9773 + j0.3724 — 0.8935 + j0.3405) /2 = 0.0419 + 30.3565
VS = 127,017(0.9354 +j0.0160)+116.6525.84° x 394 (—7.020 (0.0419 + j0.3565)
= 118, 812 +32, 032 + 10, 563 + 312, 715 = 129, 315 + 3'14, 747
= 130,153L6i: v
[V51 = \/§ x 130,153 = 225.4 kV 6.13 Determine the voltage regulation for the line described in Prob. 6.12. Assume
that the sendingend voltage remains constant. Solution: By Problem 6.12, voltto—neutral results,
V5 = 130.15 kV VR = 127.02 kV For IR = 0, V5 2 V3 cosh '71,
130.15 'VR’NLI 10.9354 + j0.0161 13 12 w
139.12 — 127.02 6.14 A three—phase 60Hz transmission line is 250 mi long. The voltage at the sending
end is 220 kV. The parameters of the line are R = 0.2 Q/mi, X = 0.8 (2/1111 and Y = 5.3 118 /mi. Find the sendingend current when there is no load on the
line. Solution: 2 (0.2+ 30.8) x 250 = 206147596"
Y = 250 x 5.3 x 10‘6 = 1.325 x10‘3490° II 71 = x/ZY = «206.1 x 1.325x 103/165.96° = 05220582930
= 0.0639+j0.5187
206.1 '7’.96°
Zc = x/Z/Y = ———&——— = 3945—7020 12 1.325 x 10—34 90°
By Eq. (6.39) for I3 = 0, sinh'yl
I = V
61 = 0.5187 rad = 29.72°
ee’ei‘“ = 0.9258+j0.5285
e_°lejm = 0.8147— 30.4651
1
cosh‘yl = (0.9258+0.8147+ j0.5285— 30.4651) = 087095208: 2 74 sinh 71 = %[0.9258 — 0.8147 + 3' (0.5285 + 04651)] = 0499958361"
I 220, 000/\/§ 0.4999; 83.61°
S = * —————— = 185.0 8.’°A
_394¢—7.02° x 0.870942.086° $1. 6.15 If the load on the line described in Prob. 6.14 is 80 MW at 220 kV, with
unity power factor, calculate the current, voltage and power at the sending end.
Assume that the sendingend voltage is held constant and calculate the voltage
regulation of the line for the load speciﬁed above. Solution: 220 80,000
V =—=127kV I =——=209.95A
R \/§ R \/§x220 With values of cosh '71 and sinh 71 from Problem 6.14, V5 = 127, 017(0.8703 + 10.0317) + 209.95 x 394 (—7.020 x 0.4999ﬁ 83.61°
110, 528 +j4, 026 + 9, 592 +340, 232 = 128, 0144 20.23° V to neutral [V5] = «5 x 128,014 = 221.7 IN
I — 209 95 (0 8703 + ‘0 0317) + “127’000 x o 4999 83 61°
5 ‘ ' ' J ' 394 5—7020 ' L‘—
= 182.72 + j6.66——1.77+j161.13 = 246844284" A
P3 = J5 x 221.7 x 246.8cos(20.3° — 4284") = 87,486 kW (or 87.5 MW)
At [3 = 0,
127 000
V = * = V
l R] 08709 145, 826 to neutral
145.8  127 6.16 A threephase tranrnission line is 300 mi long and serves a load of 400 MVA,
0.8 lagging power factor at 345 kV. The ABCD constants of the line are = D = 0.818011.3°
172.2% 9
0.001933% S A
B
C H (a) Determine the sendingend linetoneutral voltage, the sendingend current
and the percent voltage drop at full load. (b) Determine the receiving—end linetoneutral voltage at no load, the sending
end current at no load and the voltage regulation. Solution:
345. 000 400 000
VR = = 199,18640o V I = ——'— = 669.45—36.87° A
\/§ R J5. x 345 75 V5 = 0.81804130 x 199,186/_Q:+172.2¢84.2° x 669.4 (—36.87‘
256, 738% v
0.0019334 904° x 199,1864ﬁ+ 0.818011.3° x 669.4 (—36.870 447.7 A
256, 738 — 199,186
256, 738 ll 67'
I x100 = 22.4% Voltage drop =
256,738L2_0_1_53 = = 3 861 18.8’° V
VRJ” 0.8180;1.3° 13’ L—i"
I5,“ = 0.001933490.4° x313,861418.85° = 606.7/109.25° 313,861  199,186
199,186 % Reg. = = 57.6% 6.17 Justify Eq. (6.50) by substituting for the hyperbolic functions the equivalent
exponential expressions. Solution: , l sinh 71/2 671/2 — 6‘71/2
left hand side .— tenh1§ — W — m _ l 71 —vl 1 '71_ 'yl
right hand side M  2 (6 +6 ) iii—— ll sinh’yl _ %(67‘ —— 6'7‘) 57’ — 5'7‘
(611/2 _ {vi/2)? €41/2 _ E—yz/2
= W = W Therefore, left hand side = right hand side 6.18 Determine the equivalent7r circuit for the line of Prob. 6.12. Solution:
By Eq. (6.46) and Problem 6.12, sinh '7l = 0.0419 +j0.3565 = 0.359( 833° 0.359 833°
2’ = 144.3 7596" _——————/— = . 75. °
 L—x 0.3663183.0° 141 41—3952
By Eq. (6.49) and Problem 6.12,
Y’ _ 1 O.9354+j0.016—1 6 D
2 ‘ 3945—79? x 0.359483.3° ' 471x10 LEE—s 76 6.19 Use Eqs. (6.1) and (6.2) to simplify Eqs. (6.57) and (6.58) for the short transmis
sion line with (a) series reactance X and resistance R and (1)) series reactance
X and negligible resistance. Solution:
Fiom Eq. (6.1) and (6.2), it follows that, for a short line ‘ A=D=1 B=Z=R+jXélZlds C=O
(a) V V V 2
FromEq. (6.57): P3 = '5” Rlcos(¢—6)—I 3' cosqb
IZl lZl
V V V 2
From Eq. (6.58): QR = l 3” Rlsin(¢—6)' RI sin¢
lZI lZl
(b) IfR=0,thenB=Z=X¢90° and
P3 = IV‘glleRlsincS
_ lellVRl _ WEI?
QR —— X 0056 X 6.20. Rights of way for transmission circuits are difﬁcult to obtain in urban areas
and existing lines are often upgraded by reconductoring the line with larger
conductors or by reinsulating the line for operation at higher voltage. Ther
mal considerations and maximum power which the line can transmit are the
important considerations. A 138kV line is 50 km long and is composed of
Partridge conductors with flat horizontal spacing of 5 In between adjacent con
ductors. Neglect resistance and ﬁnd the percent increase in power which can be
transmitted for constant [Vsl and WE) while 6 is limited to 45° (a) if the Partridge conductor is replaced by Osprey which has more than twice
the area of aluminum in square millimeters, (b) if a second Partridge conductor is placed in a twoconductor bundle 40 cm from the original conductor and a centerto—center distance between bundles
of 5 In and (c) r if the voltage of the original line is raised to 230 kV with increased conductor
spacing of 8 m. Solution: Length of 50 km is a short line and with resistance neglected the generalized circuit constants are A = IQ and B = X 4 90°. Then, since resistance is neglected conductor heating is
disregarded; and from Eq. (6.57), lVSllVRl
X P3 = cos 45° 77 or, inversely proportional to X if we assume constant [V51 and IVE}. Additionally,
Deq = 3 5 x 5 x 10 '= 6.30 m, or 6.30/0.3048 = 20.67 ft
(0) . _, 20.67 _
For Partrzdge. X — 0.0/o4ln 0.0217 — 0.5172 Q/km
F 0 ‘X — 007’4ln 20'67 — 04969 Q/km
or spray. — . 5 0.0284 — . Ratio of PR (new/old): 0.5172 0.4969 = 1041 (41% Increase) (b) D, = «0.0217 x (GA/0.3048) = 0.1688 ft
20.67 ‘
X = 0.07541n 01688 — 0.3625 Q/km = 1.427 (42.7% increase)
(c) PR increases by factor of (gig)2 = 278 due to increased V. PR decreases due to increase
of X.
Deg ; v38X8X15 = 33.07ft
33.07
= 0. 7 4 = 
X 0 5 ln 00217 05526 km
D ase factor  05172
we ‘ 0.5526
0.5172
' = .7 = .
Resultant factor of increase 2 8 x 0.5526 2 602
Increase = 160.3% However, in addition to the increase in conductor spacing and insulation, larger conduc tors will probably be required since current will increase by a factor of about 230/138 and NPR loss in the line by a factor of about 2.78 for the increase in load at the same
power factor. 6.21 Construct a receivingend powercircle diagram similar to Fig. 6.11 for the line
of Prob. 6.12. Locate the point corresponding to the load of Prob. 6.12 and
locate the center of circles for various values of IVS] if [VRI = 220 kV. Draw
the circle passing through the load point. From the measured radius of the latter circle determine V3 and compare this value with the values calculated
for Prob. 6.12. 78 Solution:
Use scale of 1” = 50 MVA. By comparing the work in Problem 6.12(c) with the equation
V5 = AVR + 81;; we ﬁnd
A = 0.9354+j0.0160 = 0.9364135:
B = 3941—702"(0.0419+j0.3565) = 141447628" 9
ﬁ—a = 7628" O.98° = 75.3" 2 2
[A] IVRI = 0.9354 x 220 = 320.2 MVA
B 141.4
Use above data to construct load line through origin at cos’1 0.9 = 25.8° in the ﬁrst quadrant.
Draw a vertical line at 40 MW. The load point is at the intersection of this line and the load line. The radius of the circle through the load point is 7.05”. 7.05 X 50 = 352.5
M! anl ,
—— = 302.5
lBl
352.5 x 141.4
[V5] 220 226 5 k 6.22 A synchronous condenser is connected in parallel with the load described in
Prob. 6.12 to improve the overall power factor at the receiving end. The sending
end voltage is always adjusted so as to maintain the receivingend voltage ﬁxed
at 220 kV. Using the powercircle diagram constructed for Prob. 6.21, determine
the sendingend voltage and the reactive power supplied by the synchronous condenser when the overall power factor at the receiving end is (a) unity
(b) 0.9 leading. Solution: ' On the diagram for Problem 6.21 draw a new load line in the fourth quadrant at cos‘1 0.9
with the horizontal axis. Draw power circles at radii IVSIIVRl/IBI = 311, 327, 342, 358, 373
and 389 MVA for [Vsl = 200, 210, 220, 230, 240 and 250 kV, respectively. This provides the
power circle diagram that we can use for parts (a) and (b). For p.f.= 1.0 read V5 = 214 kV at 40 MW on the horizontal axis. The vertical distance
between the horizontal axis and the load line in the ﬁrst quadrant respresents the kvar of the
capacitors needed. The value is 19.3 kvar. For p.f. = 0.9 leading, read IV5 = 202 kV where the vertical line through 40 MW intersects
the load line in the fourth quadrant. The vertical distance between the two load lines at
40 MW represents the kvar of capacitors needed. The value is 38.6 kvar. 6.23 A series capacitor bank having a reactance of 146.6 52 is to be installed at the
midpoint of the BOOmi line of Prob. 6.16. The ABCD constants for each 150 mi portion of line are D = 0.953410.3°
90.334841O Q
0.001014190.1° S A
B
C '79 (0) Determine the equivalent ABCD constants of the cascade combination of
the linecapacitorline. (See Table A.6 in the Appendix.) ( b) Solve Prob. 6.16 using these equivalent ABC'D constants. Note to Instructor: This problem is somewhat long, but the solution
is interesting to show that the ABCD constants of networks in series
as given in Table A.6 can be calculated by matrix multiplication.
The problem also shows the large reduction in voltage accomplished by series capacitors in the middle of the line. Compare results of
Problems 6.16 and 6.23. Solution: Let A 0.95344 03" 90.33; 84.1°
0.001014 1 901° 0.9534 4 03° 0 10$ 0.9534 4 03° 50.915—78.65° x 0.9534 4 0.30 90.33; 84.1°
0.0010144 901° 1.10224 027° 00010141901o 0.95344 0.3° _ 0.9597; 1.18° 42.304 645°
_ . 0.0020845 904° 0.95971 1.18° _ AB
— CDnew Ax[1.o& 1.465—90 PA (b) For VR and In from Problem 6.16,
Vs 0.959741.18° x 199, 18GLQi+ 42,30 4 645° x 669.4 ("36.87°
216, 87014.5° 00020841904o x 199,186L0‘: + 0.959711.18° x 669.4 5—36.87° 520.4; 444° ' 216,870 — 199,186
‘ 216,870 (Compare this voltage drop with that of Problem 6.16)
216, 870145" Is it Voltage drop x100 = 8.15% V = = ' . °
R'NL 0.959741.18° 22°’ 977$; V
15,,” = 00020845904" x225,97753.32° = 4709493.? A
225, 977 —— 199, 6
‘70 Reg. = 18 x 100 = 13.45% 199, 186
(without capacitors 57.6%) 6.24 The shunt admittance of a 300mi transmission line is y. = O+j6.87><10’6 S/mi — 80 Determine the ABCD constants of a shunt reactor that will compensate for
60% of the total shunt admittance. Solution:
Capacitive susceptance: BC = 6.87 x 10“6 x 300 = 0.002061 S
Inductive susceptance: BL = 0.6 x 0.002061 = 0.001237 S ’
For the shunt reactor, A = D 2 1.0g): 
B = 0
C = —j0.001237S 6.25 A 250—Mvar, 345kV shunt reactor whose admittance is 0.0021 g—90° S is con
nected to the receiving end of the 300mi line of Prob. 6.16 at no load. (a) Determine the equivalent ABCD constants of the line in series with the
shunt reactor. (See Table A6 in the Appendix.) (b) Rework part (b) of Prob. 6.16 using these equivalent ABCD constants and
the sendingend volage found in Prob. 6.16. Solution: For the shunt reactor A'=D=1.0 B ll
0 C = —j0.0021 S Aeq = 0.818;1.3°+172.2484.2°x0.00215—90° = 1.17775O.88°
Becl = 17224842" Deq = 0.818413“
Ceq = 0.001933490.4°+0.00214—90" x0.81841.3° = (100021728325o (b) From Problem 6.16, V5 = 256, 7384 20.15°. So, 256,738;20.15° _ 0
Va,“ — mo— — 21199942103 v 15,)” = 0.000217% x 217, 999% = 47.34104.28° A
Recall that the shunt reactor is in the circuit only at no load. So, from Problem 6.16,
V1;pr = 199,186&° V
and 217, 999 — 199, 186
199,186 (compare'with Problems 6.16 and 6.24) ‘70 Reg. = = 9.45 ‘76 6.26 Draw the lattice diagram for current and plot current versus time at the sending end of the line of Example 6.8 for the line terminated in (a) an open circuit
(b) a short circuit. 81 Solution: m Arena (a) 23 = 00, and for current ,, z _§2_Zc=1%:=_1
R ZR'l’Zc O—Zc 3
= — = 1
p5 0+Zc +
120
Initially F = ——3 = 4 A P5 :1 pg: 0 0 :7
4 z T 4A f
“ t 2T 0 ):
3’!” —4A ‘ 4T 0 ‘ 4 .‘ 4A (b) Z R = 0, and for current _ _o—zc __1
p“ " O+Zc 
_ 0—2c _ 1
ps — 0+Zc
4, 120
Initially i‘ = W — 4A 1 P5=1 PR:
0
I 4
4A
4
8A
4
12A .
4
l'6A
4
20A 0 2T 4T 67' 6.27 Plot voltage versus time for the line of Example 6.8 at a point distant from the
sending end equal to one—fourth of the length of the line if the line is terminated
in a resistance of 10 Q. w“: ‘ 82
Solution:
Imagine a vertical line on the diagram of Fig. 6.15(b) at onefourth the line length from the sending end toward the receiving end. Intersections of this line and the slant lines occur at
T = 0.25T, 1.75T, 2.251", 3.75T, etc. Changes in voltage occur at these times. The sum of
the incident and reﬂected voltages are shown between slanted lines and determine the values plotted below.
120 v
I 90 v
. 60 v 120 V 0 T 2T 3T 41' 6.28 Solve Example 6.8 is a resistance of 54 Q is in series with the source. Solution:
For voltage,
__ 54—30 _ g
’0’ ’ 54+30 _ 7
_ 90—30 ‘ i
p“ ‘ 90+30 _ 2
Initial voltage impressed on line:
30
120 = 42.86 V
30 + 54 X
Final value:
120 X 90 — 75 V
90+54 '
42.86 V 75 11—“ 75 V
r 42.86
21.43 v
2* 54.29
6.12 V
3T 70.41
3.05
4r 73.47
0.87 v
74.34 a 7‘ 2T 37 4T 5T 6.29 Voltage from a dc source is applied to an overhead transmission line by closing
a switch. The end of the overhead line is connected to an underground cable.
Assume both the line and the cable are lossless and that the initial voltage along
the line is 11+. If the characteristic impedances of the line and cable are 400 Q and 50 9, respectively, and the end of the cable is opencircuited, ﬁnd in terms
of 12+ ' 83 (a) the voltage at the junction of the line and cable immediately after the
arrival of the incident wave and (b) the voltage at the open end of the cable immediately after arrival of the
ﬁrst voltage wave. Solution: (a) The initial wave of voltage 123’ arriving at the juction with the ‘cable “sees” the Zc of
the cable. So, at the end of the overhead line: 50  400 = —0.777
50 + 400 PR =
and the voltage at the juction is
(1 — 0.777) 12+ = 0.2231); which is the refracted voltage wave travelling along the cable.
(6) At the end of the cable p3 = 1.0 and m = (0.223+0.223)v+ = 0.4462); 6.30 A dc source of voltage V; and internal resistance R5 is connected through
a switch to a lossless line having characteristic impedance Re. The line is
terminated in a resistance R3. The travelling time of a voltage wave across the
line is T. The switch closes at t = 0. ((1) Draw a lattice diagram showing the voltage of the line during the period
t = O to t = 7T. Indicate the voltage components in terms of V5 and the
reflection coefﬁcients p3 and p,. (b) Determine the receivingend voltage at t = 0, 2T, 4T and 6T, and hence
at t = 2nT where n is any nonnegative integer. (c) Hence determine the steady state voltage at the receiving end of the line
in terms of V5, R5, RR and RC. (d) Verify the result in Part (c) by analyzing the system as a simple dc circuit
in the steady state. (Note that the line is lossless and remember how inductances and capacitances behave as short circuits and open circuits to
dc.) Solution: (0) R \ 12312 RsRc
V: c =~——C 5
’ VS(RC+RS) ”“ RR+RC " Rs+Rc 84 t = 0’ =>
15 = 2T => VR(2T)
t = 4T => VR(4T) VR(0) = o VR(O) + V] + PRVJ = (1+ PR)Vf
VR(2T) + PapRVf + PJPQRVI (1+ PR)VI + P5PR(1+ PR)Vf (1 + PR)(1+ PaPR)Vf Vn(4T) + 931%"! + 93/032"; (1 + pR)(1 + PsPR)Vf + P3P§2(1+ rm)";
= (1 + pa) [1 + PsPR + (papaﬂ V! t = 6T => VR(6T) II Hence at any given t = 2nT, n—l n
e 1
MM) = (1+ PR) 2003012)] V; = (1+ PR) Vf
j:  PJPR
(c) At the steady state, 12 —+ co. If R5 or RR # 0, p,pRI < 1 and (p,pR)" —+ 0 as n —> 00.
Hence,
V
moo) = (1+ PR)1 ’
 PsPR
. RR — Rc R5 — RC
5 = —
Ince, pg RR + RC and p, R3 + RC
. 2RR
1 + pg _ R}; + Rc
1 = (RS+RC)(RR+Rc)
1 ~ PsPR 2Rc (R5 + RR)
R
V = ___C.__ .
’ R5 + Rc V5
2BR (Rs + Ra) (RR + Re) Re RR
3‘”) RR+RC x 2Rc(Rs+RR> " Rs+Rc S Rs+RR VS (d) 85 ‘27 5 9:: 5:1,. , _ (line) Chapter 7 Problem Solutions 7.1 Using the building—block procedure described in Sec. 7.1, determine Ybus for
the circuit of Fig. 7.18. Assume there is no mutual coupling between any of the branches. Solution: First the voltage sources are converted to current sources. Then, the building blocks are given
as follows: we.” e:—lM—H~(«~w€»n my: a. (D o (D c)(—j1.0) 3L} '11](—j2.o) 8H 'ﬂejzs) @© ©© @@ .._..;A bignuawu.‘ a. Combining these together yields CD ® © © 6) (D —j5.5 32.5 3‘2 3'0 30
® j2.5 j11.5 3'4 jO » 3'5
@ 3'2 3'4 j14 3'8 3'0
GD .7'0 3‘0 3'8 j10 3'2
@ 3‘0 3‘5 3'0 j2 —j7.8 ...
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 Spring '09
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