Electrical Power Systems

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12.4 ISOCHRONOUS GOVERNOR Automatic Generation Control: Conventional Scenario 309 The “isochronous” means constant speed. An isochronous governor adjusts the turbine valve/ gate to bring the frequency back to the nominal or scheduled value. igure 12.3 shows the schematic diagram of isochronous speed governing system. The measured frequency f (or speed w) is compared with reference frequency f r (or reference speed w r ). The error signal (equal to frequency deviation or speed deviation) is amplified and integrated to produce a control signal DE which actuates the main steam supply valves in the case of a steam turbine, or gates in the case of a hydro turbine. ig. 12.3: Schematic diagram of an isochronous governor. An isochronous governor works very well when a generator is supplying an isolated or when only one generator in a multigenerator system is requireed to respond to changes is load. However, for load sharing between generators connected to the system, droop characteristic or speed regulation must be provided as discussed in next section. 12.5 GOVERNORS WITH SPEED-DROOP CHARACTERISTICS When two or more generating units are connected to the same system, isochronous governors can not be used since each generating unit would have to have precisely the same speed setting. Otherwise, they would fight each other, each will try to control system frequency to its own setting. or stable load division between two or more units operating in parallel, the governors are provided with a characteristic so that the speed drops as the load is increased. The regulation or speed-droop characteristic can be obtained by adding a steady-state feedback loop around the integrator as shown in ig. 12.4, ig. 12.4: Governor with steady-state feedback.
310 Electrical Power Systems The transfer function of the governor of ig. 12.4 reduces to the form as shown in ig. 12.5. This type of governor may be characterized as proportional controller with a gain 1/R. (a) Block diagram of Governor (b) Reduced block diagram ig. 12.5: Block diagram of speed governor with droop Where T g = 1/KR = governor time constant. 12.6 SPEED REGULATION (DROOP) The value of speed regulation parameter R determines the steady-state frequency versus load characteristic of the generating unit as shown in ig. 12.6. The ratio of frequency deviation (Df ) to change in valve/gate position(DE) or power output(DP g) is equal to R. The parameter R is referred to as speed regulation or droop. It can be expressed as: percent frequency change Percent R = ´ 100 ...(12.1) percent power output change or example, a 4% droop or regulation means that a 4% frequency deviation causes 100% change in valve position or power output. ig.12.6: Steady-state characteristics of a governor with speed droop.
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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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Page 274 and 275: Unbalanced ault Analysis 261 Exampl
Page 276 and 277: \ 3 ´ 1.0 ´ j0.4 j0.3 ´ j0.4 + j
Page 278 and 279: Denominator of eqns. (i) and (ii) Z
Page 280 and 281: Hence, rom eqn. (i), V b = V c Unba
Page 282 and 283: ig. 10.18 shows the sequence networ
Page 284 and 285: \ V b = E a = 3752.8 0° 1763 . - 3
Page 286 and 287: Unbalanced ault Analysis 273 ig. 10
Page 288 and 289: Unbalanced ault Analysis 275 10.4 A
Page 290 and 291: where J = moment of inertia of roto
Page 292 and 293: Power System Stability 279 Pi - Pe
Page 294 and 295: Power System Stability 281 (a) ind
Page 296 and 297: On a per-phase basis, power at the
Page 298 and 299: x eq = 0.25 + 0.15 + 02 . ´ 02 . 0
Page 300 and 301: Power System Stability 287 system.
Page 302 and 303: Power System Stability 289 or P i (
Page 304 and 305: \ d1 Pi (dc - d0 ) = z Pmax dc sin
Page 306 and 307: dcr z 1 d0 cr zbPi-Bgdd d dm = C -
Page 308 and 309: We know, K1 = Pmax In ig. 11.14, P
Page 310 and 311: Solution: Prefault operation x A =
Page 312 and 313: \ cosd cr = 0.654 Power System Stab
Page 314 and 315: Similarly, the change in the power
Page 316 and 317: d (2) = d (1) + Dd (2) = 17.855 + 1
Page 318 and 319: ig. 11.19: Sample network of 11.4.
Page 320 and 321: Automatic Generation Control: Conve
Page 328 and 329: Where D = PL = constant. f Theref
Page 332 and 333: \ T p = 32 3 sec. Automatic Generat
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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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