ch6-soln - 62 In original positions in the transposition...

Info iconThis preview shows pages 1–23. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 12
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 14
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 16
Background image of page 17

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 18
Background image of page 19

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 20
Background image of page 21

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 22
Background image of page 23
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 62 In original positions in the transposition cycle, distance a-b = 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 10-7 x 103 x 1609 = 0.959 mH/mi XL = 377 x 0.959 x 10-3 = 0.362 n/mi/phase L = 2x 10-71n 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 18-krn 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 sending-end 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" sending-end 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° sending-end 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 100-mi, single-circuit, three—phase transmission line delivers 55 MVA at 0.8 power factor lagging to the load at 132 kV (line-to-line). 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 sending-end 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 10-6 _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, line-to-line I5 = 192.4 — 3122.4 + 32.915 x 10-4 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 (sending-end) 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 600-9 resistor for the shunt branch at the sending end, a l—kQ resistor for the shunt branch at the receiving end and an 80-9 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 three-phase 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 sending-end voltage and the voltage regulation. Assume the magnitude of the sending-end 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 line-to—line sending-end 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, single-circuit, 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 single-circuit, three-phase 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 10-6 n -. mi 0.0955 x 10-6 0 - mi 10"6 4 9 ° X; + Xd 0.0969 + 0.0955 5.1975 x 10-290" 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 10-6577.31° + 90° mrl = 2.08 x10’3 mi—1 6.7 Using Eqs. (6.23) and (6.24) show that, if the receiving-end 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 line-to—line use Eqs. (6.26) and (6.27) to determine the incident and reflected components of the sending-end voltage ‘ and current. (c) Hence determine the sending-end 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 10-4 +j2.074 x 10*) mi—1 2.744 x 10"1 nepers/mi g: 27r x103 ' Q‘Q N ‘i ll Attenuation-constant 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 receiving-end line voltage as reference), the line- to-line 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 receiving-end 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: Short-line model (from Eq. (6.1) and (6.2)): A = D = 1 B = Z C = O AD—BC = l—ZXO =1 Medium-length 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+-Z-Z)—ZXY(1+-Z—Y = 1+ZY+Z Y _Zy_ ZY =1 2 4 4 4 Long-line 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 sending-end voltage, current and power factor of the line described in Exam— ple 6.3 are found to be 260 kV (line-to—line), 300 A and 0.9 lagging, respectively. Find the corresponding receiving-end 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 = DVs-BIs 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 receiving-end power factor is then pi. = cos(—-22.76°+48.95°) = 0.897 lagging 6.12 A 60 Hz three-phase 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 short-line approximation, (b) the nomial-7r approximation and (c) the long-line equation. Solution: 1 = 175mi Z _ 35+j40 = 144.3475.cho n Y 930 x 10-6 s 40 000 I = ~_...’____ = 116.6 -25.84° A R «5 x 220 x 0.9 g— (a) Using the short-line 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 nominal-7r 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 long-line 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)+116.65-25.84° 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 sending-end voltage remains constant. Solution: By Problem 6.12, volt-to—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 60-Hz 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 sending-end 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 10-3/165.96° = 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_°le-jm = 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 sending-end 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 three-phase 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 sending-end line-to-neutral voltage, the sending-end current and the percent voltage drop at full load. (b) Determine the receiving—end line-to-neutral 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 equivalent-7r 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—39-52 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: Fi-om 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 138-kV 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 two-conductor bundle 40 cm from the original conductor and a center-to—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 = 1-041 (41% Increase) (b) D, = «0.0217 x (GA/0.3048) = 0.1688 ft 20.67 ‘ X = 0.07541n 0-1688 — 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 receiving-end power-circle 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 receiving-end voltage fixed at 220 kV. Using the power-circle diagram constructed for Prob. 6.21, determine the sending-end 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 BOO-mi 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 line-capacitor-line. (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 300-mi 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, 345-kV shunt reactor whose admittance is 0.0021 g—90° S is con- nected to the receiving end of the 300-mi 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 sending-end 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° = 1.17775-O.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 one-fourth 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 open-circuited, 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 receiving-end voltage at t = 0, 2T, 4T and 6T, and hence at t = 2nT where n is any non-negative 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 \ 123-12 Rs-Rc 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:—l-M—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 ...
View Full Document

{[ snackBarMessage ]}

Page1 / 23

ch6-soln - 62 In original positions in the transposition...

This preview shows document pages 1 - 23. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online