Electrical Power Systems

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(f ) Vertical sag = dcos q cos q = 297 . 3078 . \ vertical sag = 1.342 × 297 . m 3078 . 15.8 LOCATION O LINE = 1.295 m. (see ig. 15.12) Analysis of Sag and Tension 393 The routing of a transmission or distribution lines requires thorough investigatians and for selecting the most desirable and practical route, following points should be considered: 1. cost of construction 2. cost of easements 3. cost of clearing 4. cost of maintenance. 15.9 SAG TEMPLATE or correct design and economy, the location of structures on the profile with a template is very essential. Sag template is a convenient device used in the design of a transmission line to determine the location and height of stuctures. Sag template can be relied upon to provide the following: 1. Economic layout 2. Minimum errors in design and layout 3. Proper grading of structures 4. Prevention of excessive insulator swing Generally two types of towers are used: 1. The standard or straight run or intermidiate tower. 2. The angle or anchor or tension tower. The straight run towers are used for straight runs and normal conditions. The angle towers are designed to withstand heavy loading as compared to standard towers because angle towers are used at angles, terminals and other points where a large unbalanced pull may be thrown on the supports. or standard towers, for normal or average spans, the sag and the nature of the curve (Catenary or Parabola) that the line conductor will occupy under expected loading conditions is evaluated and plotted on template. Template will also show the required minimum ground clearance by plotting a curve parallel to the conductor shape curve. or the standard tower and same height, the tower footing line can also be plotted on the template. Tower footing line is used for locating the position of towers and minimum ground clearance is maintained throughout. igure 15.13 shows the sag template used for locating towers. In fact there are no clear-cut guide lines for locating the tower positions and sesveral other alternatives may be examined. Ground clearance depends on voltage level and Table-15.1 gives the span length and ground clearance at different voltage levels.
394 Electrical Power Systems ig. 15.13: Sag template for locating towers. Table- 15.1: Span length and ground clearance Voltage span length minimum ground level (m) clearance (m) 0.4 KV 80 4.6 11 KV 100 4.6 33 KV 150-200 5.2 66 KV 200-300 6.3 132 KV 350-360 6.3 220 KV 360-380 7.0 400 KV 400 8.8 Example 15.3: A galvanised steel tower member has original length of 22 cm and cross sectional area 13 cm 2 . With working axial tensile load of 125 KN, the change in length was 0.2 mm. Calculate (a) Stress (b) Strain (c) Modulus of elasticity (d) Percent elongation (e) If ultimate tensile stress is 110000 N/mm 2 , determine the factor of safety. Solution: (a) Stress = (b) Strain = Tensile load Area of cross section \ Stress = 9.615×10 4 N/m 2 Change in length Original length \ Strain = 0.000909. (c) Modulus of elasticity = Stress Strain (d) Elongation = 9. 615 ´ 10 = 0. 000909 = 10577.55 × 10 4 N/m 2 Change in length Original length 125 ´ 10 = -4 13 ´ 10 N/m2 002 . = 22 4 N/m 2 × 100% 3
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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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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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Automatic Generation Control in a R
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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 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
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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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