Electrical Power Systems

More documents

Recommendations

Info

1 1 y13 = y31 = = = ( 10 - j30) Z (. 001+ j003 . ) 13 1 1 y23 = y32 = = = ( 16 - j32) Z (. 0 0125 + j0. 025) Now Y11 = y12 + y13 + y10 Charging admittance is neglected, i.e. y10 = 0.0 23 \ Y 11 = y 12 + y 13 = (10 – j20) + (10 – j30) = (20 – j50) Y 22 = y 21 + y 23 = y 12 + y 23 = (26 – j52) Y 33 = y 13 + y 23 = (26 – j62) \ Y11 = 53. 85 68. 2 ; \ Y 12 = Y 21 \ Y BUS = Y33 = 67. 23 - 672 . ° . - . ° - ° Y 22 = 5813 634 Y12 = –y12 = – (10 – j20) = –10 + j20 = 22.36 116. 6° Y13 = Y31 = –y13 = – (10 – j30) = 3162 . 108. 4° Y23 = Y32 = –y23 = – (16 – j32) = 35. 77 116. 6° L NM Step-3: Iterative Computation 53. 85 - 68. 2° 22. 36 116. 6° 3162 . 108. 4° 22. 36 116. 6° 58. 13 - 634 . ° 35. 77 116. 6° 3162 . 108. 4° 35. 77 116. 6° 67.23 - 67. 2° V 2 (p + 1) = 1 Y 22 V 3 (p + 1) = 1 Y 33 L NM L NM P2 - jQ2 - Y V - Y V ( p) V * e j 2 P - jQ ( p) V 3 * e j 3 3 Slacks bus voltage V 1 = (1.05 + j0.0) O QP O QP O QP Load low Analysis 161 ( p) 21 1 23 3 ...(i) ( p+1) 31 1 32 2 ...(ii) - Y V - Y V Starting voltage V 2 (o) = (1 + j0); V3 (o) = (1 + j0)
162 Electrical Power Systems Now P - jQ Y 2 2 22 Y Y Y Y Therefore eqn.(i) can be written as: Now 21 22 23 22 = - + 2. 556 j1102 . = 0. 0478 220. 1° 5813 . - 63. 4° = 22 36 116 6 . . ° = 0. 3846 180° = -0. 3846 58. 13 - 634 . ° = 35 77 116 6 . . ° = 0. 6153 180° = -0. 6153 58. 13 - 634 . ° L NM 0. 0478 220. 1 (p + 1) V2 = ( ) 2 * p V P - jQ Y 3 3 33 Y Y Y Y 31 33 32 33 e j Therefore eqn. (ii) can be written as: L NM 0. 0217 229. 2 (p + 1) V3 = ( ) 2 * p V e j Now solve eqns (iii) and (iv) iteratively, 0. 0478 220. 1 (1) V2 = * ( 1 + j0) (1) \ V2 = 0. 98305 - 18 . ° 0 0217 229 2 (1) V3 = * ° + 0. 3846 V + 0. 6153 V = - + 1386 . j0. 452 = 0. 0217 229. 2° 67. 23 - 67. 2° = 3162 108 4 . . ° = 0. 47 175. 6° 67. 23 - 67. 2° = 35 77 116 6 . . ° = 0. 532 183. 8° 67. 23 - 67. 2° ° - 0. 47 175. 6° V - 0. 532 1838 . ° V p = 0 ( p) 1 3 ( p +1) 1 2 ° + 0. 3846 ´ 105 . + 0. 6153 ( 1 + j0) . . ° - 0. 47 175. 6° ´ 105 . - 0532 . 183. 8° ´ 0. 98305 - 18 . ° ( 1 + j0) O QP O QP ...(iii) ...(iv)
Page 3:
This page intentionally left blank
Page 6 and 7:
To My Wife Shanta Son Debojyoti and
Page 8 and 9:
Preface During the last fifty years
Page 10 and 11:
Contents Preface vii 1. Structure o
Page 12 and 13:
Contents xi 9. Symmetrical Componen
Page 14 and 15:
Structure of Power Systems and ew O
Page 16 and 17:
1.2 REASONS OR INTERCONNECTION Stru
Page 18 and 19:
Page 20 and 21:
Case-1: If P 1 = P 2 = P 3 = ... =
Page 22 and 23:
Load factor, L = Maximum load = 3 M
Page 24 and 25:
\ LD = 1 MW (c) rom eqn.(1.13), coi
Page 26 and 27:
Case-2: Very short lasting peak. He
Page 28 and 29:
1.9 DISADVANTAGES O LOW POWER ACTOR
Page 30 and 31:
Page 32 and 33:
Resistance and Inductance of Transm
Page 34 and 35:
2.4.1 Internal Inductance igure 2.1
Page 36 and 37:
Page 38 and 39:
Therefore, we can write L 1 = L 11
Page 40 and 41:
L a = 0.4605 n log R S| T| Resistan
Page 42 and 43:
L = 0.4605 L Resistance and Inducta
Page 44 and 45:
=(2Dd 1 · 2Dd 1) 1/4 = (2 Dd 1) 1/
Page 46 and 47:
Solution. The distances from strand
Page 48 and 49:
Page 50 and 51:
Solution: Using eqn. (2.14), Hx = I
Page 52 and 53:
= 0.4605 log 6611 . 04 . Resistance
Page 54 and 55:
Page 56 and 57:
Page 58 and 59:
Since from symmetry D b1= D b2, l 1
Page 60 and 61:
\ D m = D D D D Resistance and Indu
Page 62 and 63:
dg 1 2 \ Dsa = 0. 7788 ´ 0. 015 ´
Page 64 and 65:
Page 66 and 67:
Capacitance of Transmission Lines 5
Page 68 and 69:
V Ki = 1 pÎ N å q m 2 o m = 1 3.3
Page 70 and 71:
Similarly for the 2nd transposition
Page 72 and 73:
Page 74 and 75:
C an = log R S| T| 00242 . 1 1 1 1
Page 76 and 77:
\ V12 = q ln pÎo \ C 12 = q \ C 12
Page 78 and 79:
I chg = jw C anV LN \ |I chg| = w C
Page 80 and 81:
Line length is 200 km. \ Can (total
Page 82 and 83:
Now applying eqn. (3.7), we have C
Page 84 and 85:
Page 86 and 87:
= 0.008993 m/km. Capacitance of Tra
Page 88 and 89:
Page 90 and 91:
Page 92 and 93:
Synchronous Machine: Steady State a
Page 94 and 95:
Page 96 and 97:
Page 98 and 99:
\ V = 138 Using eqn. (4.18), . 3 =
Page 100 and 101:
om ig. 4.6, Synchronous Machine: St
Page 102 and 103:
Page 104 and 105:
Page 106 and 107:
\ i asy max = from which Synchronou
Page 108 and 109:
Postfault voltage Synchronous Machi
Page 110 and 111:
or the reference phase a, E a = (Z
Page 112 and 113:
5.3 THE PER-UNIT (pu) SYSTEM Power
Page 114 and 115:
Power System Components and Per Uni
Page 116 and 117:
Page 118 and 119:
Z L (pu) = Z L (ohm) (. 08+ j 03 .)
Page 120 and 121:
Page 122 and 123:
Page 124 and 125: Power System Components and Per Uni
Page 126 and 127: \ VL (pu) = 0. 2033 - 76. 68° pu P
Page 128 and 129: Base impedance for the load is ( )
Page 130 and 131: Now we can write, \ |I| cos f = P |
Page 138 and 139: This is a simple series circuit. Th
Page 140 and 141: Eqns (6.15) and (6.17) can be writt
Page 142 and 143: Characteristics and Performance of
Page 144 and 145: om eqn. (6.6), Characteristics and
Page 146 and 147: as: \ |V S| = 145.13 kV. \ Sending
Page 150 and 151: Sending end power factor angle = 8.
Page 152 and 153: \ I R = V R = 208 Characteristics a
Page 154 and 155: 6.6 VOLTAGE WAVES Characteristics a
Page 158 and 159: 6.9 ERRANTI EECT Characteristics an
Page 160 and 161: Load low Analysis 7.1 INTRODUCTION
Page 162 and 163: Applying KCL to the independent nod
Page 164 and 165: y 10 = y 20 = y 30 = y13 ¢ y12 ¢
Page 166 and 167: \ V i = 1 Y ii L NM P - jQ i i * Vi
Page 168 and 169: n 2 i ii ii ik i k ik k i \ Pi - jQ
Page 170 and 171: Load low Analysis 157 ig. 7.6: P-re
Page 172 and 173: Load low Analysis 159 are permitted
Page 176 and 177: (1) \ V3 = 10011 . - 2. 06° After
Page 178 and 179: Load low Analysis 165 Using eqn. (7
Page 180 and 181: Load low Analysis 167 \ Q 2 = -|V 2
Page 182 and 183: |V 1| = 1.05,|V 2| = 1.0,|V 3| (1)
Page 184 and 185: \ L NM e j ( 0) ( 0) ( 0) 1 1 1 2 n
Page 186 and 187: ( p) p The terms DPi and DQi Load l
Page 188 and 189: Load low Analysis 175 P2 = -35.77
Page 190 and 191: (1) P2( cal ) = -2.62 (1) P3( cal )
Page 192 and 193: Load low Analysis 179 Use deoupled
Page 194 and 195: 0 Dd2 \ Dd3 0 ( ) ( ) NM \ 0 Dd2 L
Page 196 and 197: Assuming q ik - d i + d k » d ik,
Page 198 and 199: Table 7.4: Line impedances BUS code
Page 200 and 201: Symmetrical ault 187 is drawing 10
Page 202 and 203: \ X TH = ig. 8.4: Thevenin equivale
Page 204 and 205: ig. 8.5: Circuit diagram of Example
Page 206 and 207: = | Vo| | Vo| | Z| ´ × (MVA) Base
Page 208 and 209: \ 1 - V 1f = j0.14 × (-j4.165) \ V
Page 210 and 211: Base current I B = Per unit reactan
Page 212 and 213: Symmetrical ault 199 Example 8.9: T
Page 214 and 215: Solution: Let Base MVA = 12 Base Vo
Page 216 and 217: Solution: Let Base MVA = 100 Base V
Page 218 and 219: Therefore, current to be interrupte
Page 220 and 221: Circuit model under fault condition
Page 222 and 223: Symmetrical ault 209 Example 8.16:
Page 224 and 225:
8.4 SHORT CIRCUIT ANALYSIS OR LARGE
Page 226 and 227:
om equations. (8.9) and (8.10), we
Page 228 and 229:
Using equation (8.14), Z BUS = L NM
Page 230 and 231:
Symmetrical ault 217 Solution: Inje
Page 232 and 233:
Hence, new Z = BUS 8.6.3 Type-3 Mod
Page 234 and 235:
or L V 1 V M NM 2 V 0 n O QP = L ol
Page 236 and 237:
Solution: (a) Using equation (8.11)
Page 238 and 239:
Symmetrical ault 225 8.7 ig. 8.31 s
Page 240 and 241:
has the following properties R S| T
Page 242 and 243:
T stands for transpose. Using eqn.
Page 244 and 245:
Using eqn.( 2.46) as given in Chapt
Page 246 and 247:
Also I n = I a + I b + I c Using eq
Page 248 and 249:
Symmetrical Components 235 ig. 9.4:
Page 250 and 251:
Symmetrical Components 237 (b) - co
Page 252 and 253:
Solution: I a + I b + I c = 0, I a
Page 254 and 255:
Also note that I ab1 = I a1 3 30°
Page 256 and 257:
Base voltage of transmission line 1
Page 258 and 259:
Symmetrical Components 245 Example
Page 260 and 261:
Symmetrical Components 247 Example
Page 262 and 263:
9.5 Draw the zero-sequence network
Page 264 and 265:
Unbalanced ault Analysis 251 Assumi
Page 266 and 267:
The symmetrical components of the f
Page 268 and 269:
Unbalanced ault Analysis 255 The bo
Page 270 and 271:
ig. 10.9: Two conductors open. Unba
Page 272 and 273:
Z 1 = j0.691 0.23 ´ 0. 691+ 0. 23
Page 274 and 275:
Unbalanced ault Analysis 261 Exampl
Page 276 and 277:
\ 3 ´ 1.0 ´ j0.4 j0.3 ´ j0.4 + j
Page 278 and 279:
Denominator of eqns. (i) and (ii) Z
Page 280 and 281:
Hence, rom eqn. (i), V b = V c Unba
Page 282 and 283:
ig. 10.18 shows the sequence networ
Page 284 and 285:
\ V b = E a = 3752.8 0° 1763 . - 3
Page 286 and 287:
Unbalanced ault Analysis 273 ig. 10
Page 288 and 289:
Unbalanced ault Analysis 275 10.4 A
Page 290 and 291:
where J = moment of inertia of roto
Page 292 and 293:
Power System Stability 279 Pi - Pe
Page 294 and 295:
Power System Stability 281 (a) ind
Page 296 and 297:
On a per-phase basis, power at the
Page 298 and 299:
x eq = 0.25 + 0.15 + 02 . ´ 02 . 0
Page 300 and 301:
Power System Stability 287 system.
Page 302 and 303:
Power System Stability 289 or P i (
Page 304 and 305:
\ d1 Pi (dc - d0 ) = z Pmax dc sin
Page 306 and 307:
dcr z 1 d0 cr zbPi-Bgdd d dm = C -
Page 308 and 309:
We know, K1 = Pmax In ig. 11.14, P
Page 310 and 311:
Solution: Prefault operation x A =
Page 312 and 313:
\ cosd cr = 0.654 Power System Stab
Page 314 and 315:
Similarly, the change in the power
Page 316 and 317:
d (2) = d (1) + Dd (2) = 17.855 + 1
Page 318 and 319:
ig. 11.19: Sample network of 11.4.
Page 320 and 321:
Automatic Generation Control: Conve
Page 322 and 323:
12.4 ISOCHRONOUS GOVERNOR Automatic
Page 324 and 325:
Page 326 and 327:
Page 328 and 329:
Where D = PL = constant. f Theref
Page 330 and 331:
Page 332 and 333:
\ T p = 32 3 sec. Automatic Generat
Page 334 and 335:
Page 336 and 337:
Page 338 and 339:
Page 340 and 341:
om ig. 12.20, state-variable equati
Page 342 and 343:
The area control error for area-1 a
Page 344 and 345:
Page 346 and 347:
Page 348 and 349:
Page 350 and 351:
Page 352 and 353:
Automatic Generation Control in a R
Page 354 and 355:
Page 356 and 357:
Page 358 and 359:
Page 360 and 361:
Page 362 and 363:
Page 364 and 365:
Page 366 and 367:
Page 368 and 369:
Page 370 and 371:
Corona 357 air around the conductor
Page 372 and 373:
14.4 POTENTIAL GRADIENT OR THREE-PH
Page 374 and 375:
Similarly, G b = G c = r r V bn ln
Page 376 and 377:
Corona 363 In the above expressions
Page 378 and 379:
P c = 2.1f HG log 10 Vn DI HG r K
Page 380 and 381:
Solution: rom eqn. (14.38), air den
Page 382 and 383:
\ V 0 = 110.15 KV(rms) V n = 220 3
Page 384 and 385:
\ W = 55 ´ b69. 28 - 57g 2 635 . -
Page 386 and 387:
Analysis of Sag and Tension 15.1 IN
Page 388 and 389:
where Dl = l0. DT MA DT = T1 - T0 T
Page 390 and 391:
when x = L l , s = 2 2 , \ l 2 = H
Page 392 and 393:
or approximately, 2 d = wL 8H ig. 1
Page 394 and 395:
Squaring eqn. (15.37), we get, s 2
Page 396 and 397:
under the action of T, H and wx. or
Page 398 and 399:
\ T max = 572.59 kg (d) rom eqn. (1
Page 400 and 401:
ig. 15.6: Case of negative x 1. Ana
Page 402 and 403:
where Analysis of Sag and Tension 3
Page 404 and 405:
or 1-meter length of conductor, Ana
Page 406 and 407:
(f ) Vertical sag = dcos q cos q =
Page 408 and 409:
Analysis of Sag and Tension 395 = 0
Page 410 and 411:
Analysis of Sag and Tension 397 Ass
Page 412 and 413:
Analysis of Sag and Tension 399 Dis
Page 414 and 415:
Total weight of the conductor = wl
Page 416 and 417:
Analysis of Sag and Tension 403 Whe
Page 418 and 419:
Optimal System Operation 16.1 INTRO
Page 420 and 421:
Optimal System Operation 407 where
Page 422 and 423:
Optimal System Operation 409 2. The
Page 424 and 425:
Complete algorithm is given below:
Page 426 and 427:
DP g1 = Î 2 DP g2 = Î 2 Using Tay
Page 428 and 429:
Optimal System Operation 415 Equati
Page 430 and 431:
\ 2 C2 = 0.18 Pg2 + 32 Pg2 + K2 Whe
Page 432 and 433:
Optimal System Operation 419 Equati
Page 434 and 435:
DP L = rom eqns (16.36) and (16.39)
Page 436 and 437:
Optimal System Operation 423 Now we
Page 438 and 439:
Optimal System Operation 425 The pr
Page 440 and 441:
Example 16.6: Consider Ex-16.5, wit
Page 442 and 443:
\ DP g (l) = 264.1657 MW rom eqn. (
Page 444 and 445:
\ (4) 2 2 P = 0.00005 × (188.95) +
Page 446 and 447:
Optimal System Operation 433 The te
Page 448 and 449:
\ Also d P 3 d 2 P 3 3 =- G 32 si
Page 450 and 451:
Let the current in line K be I K1.
Page 452 and 453:
+ HG I K Optimal System Operation
Page 454 and 455:
Now, \ V1 = 1.198 14. 5º, \d1 = 14
Page 456 and 457:
Optimal System Operation 443 B 11 =
Page 458 and 459:
Z 1 = (0.03 + j0.12)pu; I 1 = (1.3
Page 460 and 461:
Objective Questions Objective Quest
Page 462 and 463:
Objective Questions 449 26. or a de
Page 464 and 465:
Objective Questions 451 56. Peak lo
Page 466 and 467:
82. Inductance of a conductor due t
Page 468 and 469:
Objective Questions 455 108. In ter
Page 470 and 471:
Objective Questions 457 137. In a p
Page 472 and 473:
Objective Questions 459 (c) are var
Page 474 and 475:
203. A system is said to be effecti
Page 476 and 477:
Answers of Objective Questions 1. (
Page 478 and 479:
Bibliography Bibliography 465 1. O.
Page 480 and 481:
Index accelerating (or decelerating
Page 482 and 483:
methods for voltage control 115 mul
show all

Electrical Power Systems

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?