Electrical Power Systems

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Hence, rom eqn. (i), V b = V c Unbalanced ault Analysis 267 ...(i) Ib = – Ic ...(ii) Va = 0 ...(iii) b 2 V a1 + bV a2 + V a0 = bV a1 + b 2 V a2 + V a0 \ V a1 = V a2 ...(iv) rom eqn. (iii), V a1 + V a2 + V a0 = 0 ...(v) rom eqns. (iv) and (v), we get rom eqn. (ii), we get V a1 = V a2 = – V a0 b 2 I a1 + b I a2 + I a0 = – (b I a1 + b 2 I a2 +I ao ) 2 ...(vi) \ I a1 + I a2 = 2 I a0 ...(vii) Sequence network connection is shown in ig. 10.16. ig. 10.16: Sequence networks of Example 10.6. As V a1 = V a2 , the positive and negative sequence networks are connected in parallel. As I a1 + I a2 = 2I a0, the zero sequence network is connected in series with the parallel combination of positive and negative sequence networks. Example 10.7: ig. 10.17 shows a three phase generator in which phases b and c are sort circuited and connected through an impedance Z f to phase a. Draw the equivalent sequence network.
268 Electrical Power Systems ig. 10.17: Circuit connection of Example 10.7. Solution: As the phases b and c are directly short-circuited, V b = V c \ b 2 V a1 + bV a2 + V a0 = bV a1 + b 2 V a2 + V a0 \ V a1 = V a2 ...(i) Also I a + I b + I c = 0 b g = 0 \ Ia0 = 1 Ia + Ib + Ic 3 Hence Ia = Ia1 + Ia2 + Ia0 = Ia1 + Ia2 Now Va – Vb = ZfIa Substituting \ Va = Va1 + Va2 + Va0 Vb = b 2 Va1 + bVa2 + Va0 Ia = Ia1 + Ia2 \ Va1 + Va2 – b2Va1 – bVa2 = Zf Ia But Va1 = Va2 \ Va1 + Va1 – b 2 Va1 – bVa1 = Zf Ia \ 3 Va1 = Zf Ia = 3 Va2 Z + f \ Va1 = Va2 = Ia1 Ia2 3 ...(ii) ...(iii) b g ...(iv)
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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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Circuit model under fault condition
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Symmetrical ault 209 Example 8.16:
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8.4 SHORT CIRCUIT ANALYSIS OR LARGE
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om equations. (8.9) and (8.10), we
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Using equation (8.14), Z BUS = L NM
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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 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 328 and 329: Where D = PL = constant. f Theref
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Automatic Generation Control: Conve
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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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