Electrical Power Systems

More documents

Recommendations

Info

6.9 ERRANTI EECT Characteristics and Performance of Transmission Lines 145 During light-load or no-load condition, receiving end voltage is greater than sending end voltage in long transmission line or cables. This happens due to very high line charging current. This phenomenon is known as ferranti effect. A charged open circuit line draws significant amount of current due to capacitive effect of the line. This is more in high-voltage long transmission line. Under no load condition, I R = 0, therefore, eqn. (6.40) can be written as: V S = V R cosh (gl) \ V R = VS cosh ( g l) Now cosh (gl) £ 1, therefore V R is always greater or equal to V S. EXERCISE ...(6.77) 6.1. ind the characteristics of the load at the sending end and the efficiency of a three phase transmission line 160 km long delivering 15 MVA load at 110 kv, 50 Hz and 0.9 power factor (lagging) having inductance 1.356 mH/km per phase, capacitance 0.0085 m/km per phase and resistance 40 ohms. Use nominal P-method Ans. IS = 70.3 20. 8° Amp, VS(L-L) = 117.6 29. 2° kV, power factor = 0.9893, h = 95.3% 6.2. A long three-phase transmission line has resistance 63.5 ohms, reactance per phase 167 ohms, capacitive susceptance to neutral is 1.1 × 10 –3 mho. Determine ABCD constants. Ans. A = D = 0.91 213 . ° , B = 173.3 69. 9° C = 1.067 × 10 –3 90. 7° 6.3. A 50 Hz, three phase, 275 kV, 400 km long transmission line has resistance 0.035 W/km, inductance 1 mH/km and capacitance 0.01 m/km. If the line is supplied at 275 kV, find out the MVA rating of a shunt reactor that would be required to maintain 275 kV at the receiving end, when the line is delivering no load. Use nominal P method. Ans. 47.56 MVAr (lagging) 6.4. The line constants of a transmission line are A = 09301 . 098 . ° and B = 14124 . 8187 . ° ohm. The load at the receiving end is 60 MVA, 50 Hz, 0.8 pf lagging. The supply voltage is 220 kV. Calculate the load voltage. Ans. 202.2 kV 6.5. A three phase 50 Hz transmission line has impedance of (25.3 + j66.5) ohms and a shunt admittance of 4.42 × 10 –4 mho per phase. If it delivers a load of 50 MW at 220 kV at 0.8 power factor lagging, determine the sending end voltage (a) by short line approximation (b) nominal P method (c) exact transmission line equations. Ans. (a) 2338 . 2. 2° kV, (b) 2322 . 233 . ° , (c) 230. 52 2. 50° kV, 6.6. The line constants of a transmission line are A = 0. 986 0. 320° and B = 70. 3 69. 2° ohms. Determine the capacity of a reactor to be installed at the receiving end so that when a load of 50 MVA is delivered at 132 kV and power factor 0.707 lagging the sending end voltage can also be 132 kV. Ans. 48.47 MV Ar.
146 Electrical Power Systems 6.7. A three phase transmission line has A = D = 098 . 15 . ° , B = 755 . 80° and C = 4 × 10 –4 91° . If an impedance (2.64 + j42.3) ohms is connected in series with the line at the sending end, determine the new values of ABCD constants. Ans. Ao = 0963 . 153 . ° ; Bo = 116. 8 82. 8° ohms –4 Co = 4 10 90 ´ ° ; Do = 098 . 15 . ° 6.8. A three phase, 50 Hz, 100 km transmission line has resistance 0.1 ohms/km, inductance 111.7 mH/km and capacitance 0.9954 × 10 –2 m/ph per km. The line is delivering 20 MW at a 0.8 power factor lagging and 66 kV to a balanced load. Determine the efficiency and regulation. Use nominal p method. Ans. h = 93.5%, VR = 17.47% 6.9. A 132 kV, 50 Hz transmission line has the following generalised constants, A = D = 0.9696 049 . ° , -4 B = 52. 88 74. 79° ohms and C = 1177 . ´ 10 90. 15° mho. It is supplying a load of 125 MVA, 0.9 power factor lagging at 132 kV. Calculate the sending end voltage and current. Ans. VS = 166.1 kV, IS = 554 Amp 6.10. A three phase transmission line delivers a load of 5 MW at 0.8 power factor lagging. Resistance of each conductor is 0.5 W/km. Receiving end voltage is 33 kV. If the line loss is not to exceed 10%, determine the length of the line Ans. 27.9 km. 6.11. The ABCD parameters of two transmission lines are A1 , B1 , C1 , D1 and A2 , B2 , C2 , D2 . ind the overall ABCD parameters when (i) the two networks are connected in series (ii) in parallel. 6.12. Two overhead lines are connected in parallel to supply a load of 10 MW at 0.8 power factor lagging at 30 kV. The resistance and reactance of one line (A) are 5.5 W and 13.5 W respectively and for line (B) are 6 ohm and 11 ohm respectively, calculate (a) the KVA supplied by each line and (b) power supplied by each line Ans. (a) 5790, 6730 (b) 4415 kW, 5585 kW. 6.13. A long transmission line has the following generalised constants: A = D = 092 . 53 . ° ; B = 65. 3 81° ohm. Two identical transformers, each of series impedance 100 70° ohms and admittance - 4 2 ´ 10 - 75° mho are connected at the two ends of the line. Determine new values of generalised constants. Ans. Ao = Do = 0843 . 26 . 38° , Bo = 249. 3 76. 85° ohm –4 Co = 34. 38 10 50. 9 ´ °mho
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 118 and 119: Z L (pu) = Z L (ohm) (. 08+ j 03 .)
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 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 174 and 175: 1 1 y13 = y31 = = = ( 10 - j30) Z (
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

Create successful ePaper yourself

Delete template?

Save as template?