Electrical Power Systems

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or the reference phase a, E a = (Z g + Z L)I a ...(5.1) igure 5.2 gives the single-phase equivalent of a balanced three-phase network of ig. 5.1. As the system is balanced, the voltage and currents in the other phases have the same magnitude but are shifted in phase by 120°. Now consider the case where a threephase transformer forms part of a three-phase system. Three identical single-phase two winding transformers may be connected to form a three-phase transformer. The windings can be connected in four ways, Y–Y, Y–D, D–Y and D–D. The phase of Y–Y and D–D transformer, there is no phase shift between corresponding quantities on the low and high voltage windings. However, for D–Y or Y–D transformers, there is always a phase shift. igure 5.3 shows a three-phase Y–Y bank. Schematic representation of this transformer is shown in ig. 5.4(a) and ig. 5.4(b) shows the single-phase equivalent of three-phase Y–Y transformer and ig. 5.4(c) shows the single-line diagram. Power System Components and Per Unit System 97 ig. 5.3: Three-phase two-winding Y–Y transformer bank (a) Arrangements of core and coil (b) Single-line diagram. ig. 5.4: (a) Schematic representation of three-phase g-g transformer, (b) Single-phase equivalent of Y–Y transformer, (c) Single-line diagram of transformer.
98 Electrical Power Systems If the transformer is Y–D connected as shown in ig. 5.5 and for obtaining the single-phase equivalent circuit, the delta side has to be replaced by an equivalent star connection as shown in ig. 5.6. rom ig. 5.6, it is also seen that V AN leads V an by 30°. ig. 5.5: Three-phase two-winding Y–D transformer bank. ig. 5.6: Y–D transformer with equivalent star connection. igure 5.7(a) shows the single-phase equivalent of g-D transformer and ig. 5.7(b) shows the single-line diagram. ig. 5.7: (a) Single-phase equivalent of Y-D transformer, (b Single-line diagram.
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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
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: Resistance and Inductance of Transm
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 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 86 and 87: = 0.008993 m/km. Capacitance of Tra
Page 92 and 93: Synchronous Machine: Steady State a
Page 98 and 99: \ V = 138 Using eqn. (4.18), . 3 =
Page 100 and 101: om ig. 4.6, Synchronous Machine: St
Page 106 and 107: \ i asy max = from which Synchronou
Page 108 and 109: Postfault voltage Synchronous Machi
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 158 and 159: 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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Symmetrical ault 217 Solution: Inje
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Hence, new Z = BUS 8.6.3 Type-3 Mod
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or L V 1 V M NM 2 V 0 n O QP = L ol
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Solution: (a) Using equation (8.11)
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Symmetrical ault 225 8.7 ig. 8.31 s
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has the following properties R S| T
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T stands for transpose. Using eqn.
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Using eqn.( 2.46) as given in Chapt
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Also I n = I a + I b + I c Using eq
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Symmetrical Components 235 ig. 9.4:
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Symmetrical Components 237 (b) - co
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Solution: I a + I b + I c = 0, I a
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Also note that I ab1 = I a1 3 30°
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Base voltage of transmission line 1
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Symmetrical Components 245 Example
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Symmetrical Components 247 Example
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9.5 Draw the zero-sequence network
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Unbalanced ault Analysis 251 Assumi
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The symmetrical components of the f
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Unbalanced ault Analysis 255 The bo
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ig. 10.9: Two conductors open. Unba
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Z 1 = j0.691 0.23 ´ 0. 691+ 0. 23
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Unbalanced ault Analysis 261 Exampl
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\ 3 ´ 1.0 ´ j0.4 j0.3 ´ j0.4 + j
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Denominator of eqns. (i) and (ii) Z
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Hence, rom eqn. (i), V b = V c Unba
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ig. 10.18 shows the sequence networ
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\ V b = E a = 3752.8 0° 1763 . - 3
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Unbalanced ault Analysis 273 ig. 10
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Unbalanced ault Analysis 275 10.4 A
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where J = moment of inertia of roto
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Power System Stability 279 Pi - Pe
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Power System Stability 281 (a) ind
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On a per-phase basis, power at the
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x eq = 0.25 + 0.15 + 02 . ´ 02 . 0
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Power System Stability 287 system.
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Power System Stability 289 or P i (
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\ d1 Pi (dc - d0 ) = z Pmax dc sin
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dcr z 1 d0 cr zbPi-Bgdd d dm = C -
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We know, K1 = Pmax In ig. 11.14, P
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Solution: Prefault operation x A =
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\ cosd cr = 0.654 Power System Stab
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Similarly, the change in the power
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d (2) = d (1) + Dd (2) = 17.855 + 1
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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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