Electrical Power Systems

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Load low Analysis 165 Using eqn. (7.36) \ P21 = –|V2 | 2 |Y12 | cos q12 +|V1 ||V2 ||Y12 | cos (q12 – d2 + d1 ) \ P21 = –(0.98265) 2 × 22.36 cos (116.56°) + 1.05 × 0.98265 × 22.36 cos (116.56° + 3.048° + 0°) \ P 21 = 9.654 – 11.398 = –1.744 pu MW \ P31 = –|V3| 2 |Y13| cos q13 +|V1||V3||Y13| cos (q13 – d3 + d1) \ P31 = –(1.00099) 2 × 31.62 cos (108.4°) + 1.05 × 1.00099 × 31.62 cos (108.4° + 2.68° + 0°) \ P 31 = 10 – 11.953 = –1.95 pu MW P32 = –|V3| 2 |Y23| cos q23 +|V3||V2||Y23| cos (q23 – d3 + d2) \ P32 = –(1.00099) 2 × 35.77 cos (116.6°) + 1.00099 × 0.98265 × 35.77 cos (116.6° + 2.68° – 3.048°) \ P 32 = 16.048 – 15.551 = 0.496 pu MW Real power losses in line 1-2, 1-3 and 2-3, PLoss12 = P12 + P21 = 1.8189 – 1.744 = 0.0749 = 7.49 MW. PLoss13 = P13 + P31 = 2 – 1.95 = 0.05 pu MW = 5 MW. PLoss23 = P23 + P32 = – 0.4903 + 0.496 = 0.0057 pu MW = 0.57 MW. Reactive line flows can be calculated from eqns. (7.35) and (7.37). rom eqn (7.35), we get, Q12 = |V1| 2 |Y12|sinq12 – |V1||V2||Y12|sin(q12 – d1 + d2) \ Q12 = (1.05) 2 × 22.36 sin (116.56°) – 1.05 × 0.98265 × 22.36 sin (116.56° – 3.048°) \ Q 12 = 22.05 – 21.1552 = 0.8948 pu MVAr. \ Q 13 = (1.05) 2 × 31.62 sin (108.4°) – 1.05 × 1.00099 × 31.62 sin (108.4° – 2.68°) \ Q 13 = 33.0788 – 31.9908 = 1.088 pu MVAr. \ Q 23 = (0.98265) 2 × 35.77 × sin (116.6° ) – 0.98265 × 1.00099 × 35.77 sin (116.6° + 3.048° – 2.68° ) \ Q 23 = 30.8836 – 31.3582 \ Q 23 = – 0.4746 \ Q 21 = (0.98265) 2 × 22.36 sin (116.56° ) – 1.05 × 0.98265 × 22.36 sin (116.56° + 3.048° ) \ Q 21 = 19.3122 – 20.0582 = –0.746 pu MVAr. \ Q 31 = (1.00099) 2 × 31.62 sin (108.4° ) – 1.05 × 1.00099 × 31.62 sin (108.4° + 2.68° )
166 Electrical Power Systems \ Q 31 = 30.0629 – 31.0098 = –0.9469 \ Q 23 = (1.00099) 2 × 35.77 sin (116.6° ) – 1.00099 × 0.98265 × 35.77 sin (116.6° + 2.68° – 3.048°) \ Q 23 = 32.0472 – 31.5606 = 0.4866 pu MVAr. Reactive power loss in line 1-2, 1-3 and 2-3. \ QLoss 12 = Q 12 + Q 21 = 0.8948 – 0.746 = 0.1488 = 14.88 MVAr \ QLoss 13 = Q 13 + Q 31 = 1.088 – 0.9469 = 0.1411 = 14.11 MVAr \ QLoss 23 = Q 23 + Q 32 = –0.4746 + 0.4866 = 0.012 = 1.2 MVAr Note that all the results are computed after 2nd iteration and details calculation are given for the purpose of understanding. Example 7.3: Solve problem Ex-7.2 considering bus 2 is P-|V| bus. Details are given in ig. 7.8. Use same line and load data as given in Table 7.3 and 7.2. Solution ig. 7.8: Sample power system. P2 = –2.556, P3 = –1.386, Q3 = –0.452 (0) V1 = (1.05 + j0), V2 = 1.0 + j0.0 Þ P-|V| bus. ( 0) V3 = 1.0 + j0.0 Þ P- Q bus. Bus-2 is a regulated bus where voltage magnitude and real power are specified. or the voltage controlled bus, first the reactive power is computed. Using eqn. (7.16), 3 Q2 = - V2VkY2k q2k - d2+ dk k = 1 å| || | |sin( )
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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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Page 130 and 131: Now we can write, \ |I| cos f = P |
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Page 140 and 141: Eqns (6.15) and (6.17) can be writt
Page 142 and 143: Characteristics and Performance of
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Page 146 and 147: as: \ |V S| = 145.13 kV. \ Sending
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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 174 and 175: 1 1 y13 = y31 = = = ( 10 - j30) Z (
Page 176 and 177: (1) \ V3 = 10011 . - 2. 06° After
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
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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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