Electrical Power Systems

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Case-1: If P 1 = P 2 = P 3 = ... = P n = P Then C = n å i i=1 P ´ C P ´ n Structure of Power Systems and ew Other Aspects 7 That is, the coincident factor is equal to the average contribution factor. Case-2: If C 1 = C 2 = C 3 = ... = C n = C, Then C = C ´ å Pi = n å i=1 n C i ...(1.17) n = C ...(1.18) P å i=1 n i=1 i That is, coincidence factor is equal to the contribution factor. Load actor: It is the ratio of the average load over a designated period of time to the peak load occurring on that period. Therefore, the load factor (L) is defined as: or L = L = Average load Peak load Average load ´ T Peak load ´ T ... (1.19) Energy served \ L = ...(1.20) Peak load ´ T where T = time, in days, weeks, months or years. If T is large, L is small. The reason for this is that for the same maximum demand, the energy consumption covers a larger time period and results in a smaller average load. Load factor is less than or equal to unity. Annual load factor is defined as: Annual Load actor = Total annual energy Annual peak load ´ 8760 ...(1.21) Loss actor: It is the ratio of the average power loss to the peak-load power loss during a specified period of time. Therefore, the loss factor (LL) is defined as: LL = Average power loss Power loss at peak load ...(1.22) Equation (1.22) is applicable for the copper losses of the system but not for iron losses.
8 Electrical Power Systems Example 1.1: A power station supplies the load as tabulated below: Time Load (hours) (MW) 6 AM – 8 AM 1.2 8 AM – 9 AM 2.0 9 AM – 12 Noon 3.0 12 Noon – 2 PM 1.50 2 PM – 6 PM 2.50 6 PM – 8 PM 1.80 8 PM – 9 PM 2.0 9 PM – 11 PM 1.0 11 PM – 5 AM 0.50 5 AM – 6 AM 0.80 (a) Plot the load curve and find out the load factor. (b) Determine the proper number and size of generating units to supply this load. (c) ind the reserve capacity of the plant and plant factor. (d) ind out the operating schedule of the generating units selected. Solution: (a) igure 1.4 show the plot of load curve ig. 1.4: Load curve of Ex–1.1. Units generated during 24 hours = (2 × 1.2 + 1 × 2 + 3 × 3 + 2 × 1.5 + 4 × 2.5 + 2 × 1.8 + 1 × 2 + 2 × 1 + 6 × 0.5 + 1 × 0.8) MWhr. = 37.80 MWhr Average load = \ Average load = Units generated Time in hours 37. 80 24 = 1.575 MW.
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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 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
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Similarly for the 2nd transposition
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Capacitance of Transmission Lines 5
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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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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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