Electrical Power Systems

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ig. 10.9: Two conductors open. Unbalanced ault Analysis 257 rom ig. 10.9, we can write V aa¢ = 0 ...(10.43) I b = I c = 0 ...(10.44) In terms of symmetrical components, we can write V aa¢1 + V aa¢2 + V aa¢0 = 0 ...(10.45) Ia1 = Ia2 = Ia0 = 1 3 Ia ...(10.46) Equations (10.45) and (10.46) suggest a series connection of sequence network as shown in ig. 10.10. ig. 10.10: Sequence network for two conductors open. Example 10.1: or the system of example 9.5, the single line diagram is redrawn–in ig. 10.11. The positive, negative and zero sequence networks of the system have already been drawn in igs. 9.10(a), 9.10(b) and 9.10(c). Before the occurrence of a solid L-G fault at line g, the motors were loaded. If the prefault current is neglected, calculate the fault current and subtransient currents in all parts of the system. ig. 10.11: Single-line diagram the system of Example 10.1.
258 Electrical Power Systems Solution: The sequence networks given in igs. 9.10(a), 9.10(b) and 9.10(c) are connected in ig. 10.12 to simulate a solid L-G fault at bus g. Prefault currents are neglected. ig. 10.12: Connection of the sequence networks. E g² = E ² m1 = E ² m2 = V f 0 = 10 11 = 0.909 pu Thevenin equivalent of positive sequence network is shown in ig. 10.13.
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To My Wife Shanta Son Debojyoti and
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Preface During the last fifty years
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Contents Preface vii 1. Structure o
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Contents xi 9. Symmetrical Componen
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Structure of Power Systems and ew O
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1.2 REASONS OR INTERCONNECTION Stru
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Case-1: If P 1 = P 2 = P 3 = ... =
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Load factor, L = Maximum load = 3 M
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\ LD = 1 MW (c) rom eqn.(1.13), coi
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Case-2: Very short lasting peak. He
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1.9 DISADVANTAGES O LOW POWER ACTOR
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Resistance and Inductance of Transm
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2.4.1 Internal Inductance igure 2.1
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Therefore, we can write L 1 = L 11
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L a = 0.4605 n log R S| T| Resistan
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L = 0.4605 L Resistance and Inducta
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=(2Dd 1 · 2Dd 1) 1/4 = (2 Dd 1) 1/
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Solution. The distances from strand
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Solution: Using eqn. (2.14), Hx = I
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= 0.4605 log 6611 . 04 . Resistance
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Since from symmetry D b1= D b2, l 1
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\ D m = D D D D Resistance and Indu
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dg 1 2 \ Dsa = 0. 7788 ´ 0. 015 ´
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Capacitance of Transmission Lines 5
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V Ki = 1 pÎ N å q m 2 o m = 1 3.3
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Similarly for the 2nd transposition
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C an = log R S| T| 00242 . 1 1 1 1
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\ V12 = q ln pÎo \ C 12 = q \ C 12
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I chg = jw C anV LN \ |I chg| = w C
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Line length is 200 km. \ Can (total
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Now applying eqn. (3.7), we have C
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= 0.008993 m/km. Capacitance of Tra
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Synchronous Machine: Steady State a
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\ V = 138 Using eqn. (4.18), . 3 =
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om ig. 4.6, Synchronous Machine: St
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\ i asy max = from which Synchronou
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Postfault voltage Synchronous Machi
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or the reference phase a, E a = (Z
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5.3 THE PER-UNIT (pu) SYSTEM Power
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Power System Components and Per Uni
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Z L (pu) = Z L (ohm) (. 08+ j 03 .)
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\ VL (pu) = 0. 2033 - 76. 68° pu P
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Base impedance for the load is ( )
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Now we can write, \ |I| cos f = P |
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This is a simple series circuit. Th
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Eqns (6.15) and (6.17) can be writt
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Characteristics and Performance of
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om eqn. (6.6), Characteristics and
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as: \ |V S| = 145.13 kV. \ Sending
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Sending end power factor angle = 8.
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\ I R = V R = 208 Characteristics a
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6.6 VOLTAGE WAVES Characteristics a
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6.9 ERRANTI EECT Characteristics an
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Load low Analysis 7.1 INTRODUCTION
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Applying KCL to the independent nod
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y 10 = y 20 = y 30 = y13 ¢ y12 ¢
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\ V i = 1 Y ii L NM P - jQ i i * Vi
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n 2 i ii ii ik i k ik k i \ Pi - jQ
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Load low Analysis 157 ig. 7.6: P-re
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Load low Analysis 159 are permitted
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1 1 y13 = y31 = = = ( 10 - j30) Z (
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(1) \ V3 = 10011 . - 2. 06° After
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Load low Analysis 165 Using eqn. (7
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Load low Analysis 167 \ Q 2 = -|V 2
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|V 1| = 1.05,|V 2| = 1.0,|V 3| (1)
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\ L NM e j ( 0) ( 0) ( 0) 1 1 1 2 n
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( p) p The terms DPi and DQi Load l
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Load low Analysis 175 P2 = -35.77
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(1) P2( cal ) = -2.62 (1) P3( cal )
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Load low Analysis 179 Use deoupled
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0 Dd2 \ Dd3 0 ( ) ( ) NM \ 0 Dd2 L
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Assuming q ik - d i + d k » d ik,
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Table 7.4: Line impedances BUS code
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Symmetrical ault 187 is drawing 10
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\ X TH = ig. 8.4: Thevenin equivale
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ig. 8.5: Circuit diagram of Example
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= | Vo| | Vo| | Z| ´ × (MVA) Base
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\ 1 - V 1f = j0.14 × (-j4.165) \ V
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Base current I B = Per unit reactan
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Symmetrical ault 199 Example 8.9: T
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Solution: Let Base MVA = 12 Base Vo
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Solution: Let Base MVA = 100 Base V
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Therefore, current to be interrupte
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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 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.
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Automatic Generation Control: Conve
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12.4 ISOCHRONOUS GOVERNOR Automatic
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Where D = PL = constant. f Theref
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\ T p = 32 3 sec. Automatic Generat
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om ig. 12.20, state-variable equati
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The area control error for area-1 a
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Automatic Generation Control in a R
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Corona 357 air around the conductor
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14.4 POTENTIAL GRADIENT OR THREE-PH
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Similarly, G b = G c = r r V bn ln
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Corona 363 In the above expressions
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P c = 2.1f HG log 10 Vn DI HG r K
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Solution: rom eqn. (14.38), air den
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\ V 0 = 110.15 KV(rms) V n = 220 3
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\ W = 55 ´ b69. 28 - 57g 2 635 . -
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Analysis of Sag and Tension 15.1 IN
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where Dl = l0. DT MA DT = T1 - T0 T
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when x = L l , s = 2 2 , \ l 2 = H
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or approximately, 2 d = wL 8H ig. 1
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Squaring eqn. (15.37), we get, s 2
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under the action of T, H and wx. or
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\ T max = 572.59 kg (d) rom eqn. (1
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ig. 15.6: Case of negative x 1. Ana
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where Analysis of Sag and Tension 3
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or 1-meter length of conductor, Ana
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(f ) Vertical sag = dcos q cos q =
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Analysis of Sag and Tension 395 = 0
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Analysis of Sag and Tension 397 Ass
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Analysis of Sag and Tension 399 Dis
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Total weight of the conductor = wl
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Analysis of Sag and Tension 403 Whe
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Optimal System Operation 16.1 INTRO
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Optimal System Operation 407 where
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Optimal System Operation 409 2. The
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Complete algorithm is given below:
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DP g1 = Î 2 DP g2 = Î 2 Using Tay
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Optimal System Operation 415 Equati
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\ 2 C2 = 0.18 Pg2 + 32 Pg2 + K2 Whe
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Optimal System Operation 419 Equati
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DP L = rom eqns (16.36) and (16.39)
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Optimal System Operation 423 Now we
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Optimal System Operation 425 The pr
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Example 16.6: Consider Ex-16.5, wit
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\ DP g (l) = 264.1657 MW rom eqn. (
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\ (4) 2 2 P = 0.00005 × (188.95) +
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Optimal System Operation 433 The te
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\ Also d P 3 d 2 P 3 3 =- G 32 si
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Let the current in line K be I K1.
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+ HG I K Optimal System Operation
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Now, \ V1 = 1.198 14. 5º, \d1 = 14
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Optimal System Operation 443 B 11 =
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Z 1 = (0.03 + j0.12)pu; I 1 = (1.3
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Objective Questions Objective Quest
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Objective Questions 449 26. or a de
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Objective Questions 451 56. Peak lo
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82. Inductance of a conductor due t
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Objective Questions 455 108. In ter
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Objective Questions 457 137. In a p
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Objective Questions 459 (c) are var
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203. A system is said to be effecti
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Answers of Objective Questions 1. (
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Bibliography Bibliography 465 1. O.
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Index accelerating (or decelerating
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methods for voltage control 115 mul
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