Electrical Power Systems

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Structure of Power Systems and ew Other Aspects 17 1.3 A generating station has peak demand of 120 MW and its connected load is 200 MW. The energy produced annually is 4 × 10 5 MWhr. Determine (a) load factor (b) demand factor Ans. (a) 0.38 (b) 0.60. 1.4 A power plant has to meet the following load demand: Load A: 100 MW from 8 AM – 6 PM Load B: 150 MW from 6 AM – 10 AM Load C: 50 MW from 6 AM – 10 AM Load D: 20 MW from 10 AM – 6 AM Plot the daily load curve and determine (a) diversity factor (b) load factor (c) daily energy produced. Ans: (a) 1.067 (b) 0.439 (c) 3160 MWhr. 1.5 rom a load duration curve, following data are available: Maximum demand on the system is 40 MW. The load supplied by the two generating units is 28 MW and 20 MW. Unit no. 1 is the base unit and works for all the time. Unit no. 2 is peak load unit and works only for 40% of the time. The energy produced annually by unit 1 is 2 × 10 8 units and that by unit 2 is 15 × 166 units. ind the (a) load factor (b) plant capacity factor (c) plant use factor of both the units. Also (d) determine the load factor of the total plant. Ans: (a) 0.815, 0.356 (b) 0.815, 0.856 (c) 0.815, 0.214 (d) 0.613
18 Electrical Power Systems Resistance and Inductance of Transmission Lines 2.1 INTRODUCTION 2 Power system engineering is that branch of Electrical Engineering which concerns itself with the technology of generation, transmission and distribution of electrical energy. The power system growing into a vast and complex system represents one of the most vital systems in every modern nation. The basic purpose of a transmission network is to transfer electric energy from generating units at various locations to the distribution system which ultimately supplies the load. Transmission line also interconnect neighbouring power utilities which allows not only economic dispatch of electrical power within regions during normal conditions, but also transfer of power between regions during emergencies. An overhead transmission line consists of a group of conductors running parallel to each other and carried on supports which provide insulation between the different conductors and between each conductor and earth. A transmission line has four parameters—resistance, inductance, capacitance and shunt conductance. The shunt conductance accounts for leakage currents flowing across insulators and ionized pathways in the air. The leakage currents are negligible as compared to the current flowing in the transmission lines. The series resistance causes a real power loss in the conductor. The resistance of the conductor is very important in transmission efficiency evaluation and economic studies. The power transmission capacity of the transmission line is mainly governed by the series inductance. The shunt capacitance causes a charging current to flow in the line and assumes importance for medium and long transmission lines. These parameters are uniformly distributed throughout but can be lumped for the purpose of analysis on approximate basis. 2.2 LINE RESISTANCE The dc resistance of a solid round conductor is given by R dc = r×l A where r = resistivity of conductor l = length of conductor A = cross sectional area of conductor ...(2.1)
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Page 6 and 7: To My Wife Shanta Son Debojyoti and
Page 8 and 9: Preface During the last fifty years
Page 10 and 11: Contents Preface vii 1. Structure o
Page 12 and 13: Contents xi 9. Symmetrical Componen
Page 14 and 15: Structure of Power Systems and ew O
Page 16 and 17: 1.2 REASONS OR INTERCONNECTION Stru
Page 18 and 19: Structure of Power Systems and ew O
Page 20 and 21: Case-1: If P 1 = P 2 = P 3 = ... =
Page 22 and 23: Load factor, L = Maximum load = 3 M
Page 24 and 25: \ LD = 1 MW (c) rom eqn.(1.13), coi
Page 26 and 27: Case-2: Very short lasting peak. He
Page 28 and 29: 1.9 DISADVANTAGES O LOW POWER ACTOR
Page 32 and 33: Resistance and Inductance of Transm
Page 34 and 35: 2.4.1 Internal Inductance igure 2.1
Page 38 and 39: Therefore, we can write L 1 = L 11
Page 40 and 41: L a = 0.4605 n log R S| T| Resistan
Page 42 and 43: L = 0.4605 L Resistance and Inducta
Page 44 and 45: =(2Dd 1 · 2Dd 1) 1/4 = (2 Dd 1) 1/
Page 46 and 47: Solution. The distances from strand
Page 50 and 51: Solution: Using eqn. (2.14), Hx = I
Page 52 and 53: = 0.4605 log 6611 . 04 . Resistance
Page 58 and 59: 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 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 72 and 73: Capacitance of Transmission Lines 5
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
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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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Capacitance of Transmission Lines 7
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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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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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