Contents Kreatryx. All Rights Reserved.

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2 Contents Manual for K-Notes... Transmission Lines... 3 Underground Cables Overhead Insulators Distribution Systems Per Unit System Load Flow Study Economic Power Generation... Fault Analysis... 1 Power System Stability... 9 Power System Protection Kreatryx. All Rights Reserved. 1

3 Transmission Lines Skin Effect It is tendency of AC current to be concentrated on the surface of conductor. Cause: Non-uniform distribution of magnitude flux linkages Due to skin effect, the effective area of cross section of conductor decrease and hence resistances increases. In case of DC, There is no skin effect so R R DC With increase in frequency, skin effect increases. With increase in, skin effect increases. r AC Inductance of a Transmission line Single Conductor Internal inductance r 8 External inductance from distance d 1 to r d Lex ln d 1 r r d Total inductance ln 8 r r d r d ln ln 1 re 4 r r.7788r = Geometric mean radius (GMR) d Single phase wire line r d Inductance of single wire ln r Total inductance = L1 L d Lsys ln r If radius of both wire is not same, assume radius of 1 st wire d Lsys ln ra r b r.7788r & r.7788r b b a a r a & that of second wire is r b If instead of a single conductor per phase we use multiple conductor, then GMR is replaced by self GND (Geometric Mean Distance) and d by mutual GMD. 3

4 Self GMD self GMD D 11 D 1...D 1n D 1 D...D n... D n1 D n...d nn Where fwd D r.7788r ii i i Self GMD D ' ' D ' '...D i'm'... D ' '...D ' ' Where bwd ii Mutual GMD D r.7788r i i 1 i ml mm i Mutual GMD D ' D '...D '... D ' D '...D ' m n1 n nm Now, with these terms all the inductance expressions change to 1 m 1 mn 1 n Single wire : d ln self GMD mutual GMD 1, wire: ln Self GMD Three phase Transmission line Symmetrical configuration L ph r D ln r r.7788r 4

5 Asymmetrical configurations If conductors are placed horizontally or vertically. 1 3 D D D D L eq ab bc ca ph D r ln r eq In case of bundled conductor, more than one conductor per phase We replace Dab D = mutual GMD between a phase & b phase abeq Similarly, D bc D bc eq D ca D caeq In place of GMR, Self GMD is used Self GMD= Self GMD Self GMD Self GMD a b c Example: Calculate inductance per phase of following circuit? 1 3 Between successive conductors, distance = 3m, Radius of each conductor = 1m Solution 1 4 D D. D. D. D ab eq a1b1 a1b ab1 ab m 1 4 D m bc eq 1 4 D m ca eq D eq = mutual GMD 1 3 Dab.eq Dbc.eq D ca.eq = 5.71m 5

6 1 4 Self GMD rd D r a a1a aa =.341m Self GMD rd D r b b1b =.467m bb1 1 4 Self GMD rd D r c c1c =.158m cc1 1 3 Self GMD Self GMD Self GMD Self GMD c a =.398m GMD L ln GMD Self ln.634mh/ km.398 Remember, Inductance calculated using these formulas is per unit length. b Transposition of Transmission line The position of different lines are changed after regular intervals to reduce radio interference in neighboring communication lines. Capacitance Single Phase Wire System C ab r D ln rr 1 Line to neutral capacitance 6

7 C an r D ln r 1, C bn r D ln r Three phase single conductor system C ph r GMD ln r For bundled conductors C ph GMD ln Self GMD In capacitance calculations, it must always be remembered that there is no concept of r, we simply use radius in calculating self GMD. Performance of Transmission line Classification of lines based on length 1) Short Line l < 8 km or l*f < 4, Where f = frequency ) Medium Line 8 km < l < km 4 < l*f < 1 3) Long Line l > km l*f > 1 Modeling of transmission lines Transmission lines are modeled as port network Vs AVR BIR Under no load Is CVR DIR IR, Vs AVR, Vs VR A 7

8 No Load Voltage Voltage Regulation V s and IR, I s CV R A Vs VR A 1% V R This current is called as line charging current and is responsible for as effect is called as Ferranti Effect. Ferranti Effect Under no-load or light load conditions receiving end voltage becomes more than sending end voltage due to presence of line charging current. Short transmission line V V I R jwl s R R R R V I Z Vs 1 z VR I 1 I s R A = D (symmetrical) AD BC = 1 (reciprocal) Approximate Voltage Regulation For lagging pf For leading pf IR VR= R cos X sin R V R load pf=cos R I R VR RcosR X sin R VR R 8

9 Medium Transmission Line Normal T Model YZ YZ 1 Z 1 V 4 V s I s Y 1 R YZ IR Here all problems are in actual values & not per unit length. Nominal-π-Model YZ 1 Z V V s R I YZ YZ I s Y 1 1 R 4 Long transmission Line Vx VR cosh X IRZc sinh x V R IR IRcoshx sinh x Zc Where V x & I x are voltage and current at distance x from receiving end. Z c R jl = surge impedance G j C In case of long transmission line, we use all promoters per-unit length. For loss less line, R=G= Z c L C For distribution less line, RC = LG For sending end, x = l V V cosh l I Z sinh l s R R c V R I I coshl+ sinh l s R Zc 9

10 cosh l Z sinh l c V V s R 1 I cosh l I s R Z c A = D (symmetric) AD - BC = 1 (reciprocal) Power Transfer Equation Vs Vr A P r cos V cos B B r For Short TL Vs Vr A Q r sin V sin B B B r Z B Z ; A 1 A A 1, = Vs Vr Vr Pr cos cos Z Z Vs Vr Vr Qr sin sin Z Z If resistance of line is neglected Z jx X and 9 s r s r r R V V V V V Pr sin ; Q = cos X X X Remember, the last expression can be applied between any two bases in a power system as long as transmission line connecting them is loss less. Wave Propagation Due to continuous energy transfer between L & C elements of a transmission line we consider energy propagation from sending to receiving end & hence wave propagation. Z c = Surge impedance or characteristic impedance. 1

11 γ = Propagation constant. For loss less line R jlg j C R = G = O j LC j = attenuation constant = phase constant In case of loss less line (no attenuation) LC Velocity of wave 1, Where L & C are per unit length LC Wavelength LC Surge Impedance Loading When load impedance = surge impedance P L Vr(L L) = Surge Impedance Loading Z C If ZL ZC V I Z R R C According to long line V V coshx+v sinhx V e X R R R x yx So no term containing e & hence no reflected wave & hence whenever surge impedance is connected at load, there is no reflection. 11

12 Surge Traversal Theremin equivalent circuit When surge voltage V is induced on the line & line can be represented as Theremin equivalent circuit shown. Z C Z L = Characteristic impedance of line = load impedance or characteristics impedance of second line connected in series to first. L Transmitted voltage V V potential divider Incident Voltage Reflected voltage V1 V V Z ZL ZC V V V [Voltage continuity] Z Z L C V V Z L Z C V V Reflected current, Refracted current, Incident current Z Z C L V Z C Reflection coefficient: Refraction coefficient: V Z Z L V Z Z L L C C V ZL V Z Z C Voltage Control Usually in case of lagging loads, the voltage at receiving end falls below sending end voltage and to boost the receiving end voltage we connect a shunt capacitor at receiving end. Similarly, in case of leading loads, receiving end voltage is higher than sending end voltage so we connect a shunt reactor to avoid over-voltage. Usually in GATE, we need to calculate rating of capacitor for voltage control & it is illustrated through a question shown below: 1

13 Example: A three phase overhead lines has a resistance & reactance of 5 & respectively. The load at receiving end is 3MW,.85 pf lagging at 33kv & we connect a compensating equipment at receiving end to maintain voltage at each end equal to 33 Kv. Find rating of compensating equipment? Solution: Assuming base (MVA) = 3 MVA Base voltage = 33 kv pu power = 3Mw 1pu 3MVA Base impedance = pu impedance = V 36.3 S 5 j VS VR VR Z.568 ; =75.96 PR cos cos Z Z cos cos V V V QR sin sin Z Z cos sin sin S R R Q R Q L = pu 1 PL tan 1 tancos.85 Q =.6197 pu L QR QL QC QC 1.647pu QC = 37.9 MVAR So, we can observe the steps involved Step 1 : Calculate from P L as capacitor does not consume any real power. Step- : Calculate Q R using calculated above Step-3 : QR QL QC than Calculate Q C 13

14 Power Factor Correction Usually, to improve the supply side power factor we connect a capacitor device like capacitor bank or synchronous condenser (synchronous motor under over excited condition). Suppose, initially a load of real power & lagging pf is connected & we want to improve pf to lagging 1 & we connect a capacitive device which consumes real power P & thus net real power after connection. C P P1 PC cos Q P tan ; Q1 P1 tan 1 QC P1 tan1 P tan P 1 cos 1 In case of capacitor bank, PC P P 1 If we wish to calculate capacitance per phase (in both voltage control & pf correction ) Q C 3CV ph Q C 3V Underground Cables Insulation resistance R R ln l r C ph L = length of cable R = Outer radius (sheath radius) R = conductor radius Capacitance Model C C : Core capacitance C : Core to sheath capacitance S 14

15 Capacitance per phase Cph CS 3CC Calculating C S & C C 1) Any of two cores or conductors are connected to sheath & capacitance is measured between remaining core & sheath. C1 CS CC ) All three cores are connected together & capacitance is measured between any core & sheath. C 3C S C C S 3 C C C C 1 3 C C 1 C C 6 3C1 C Cph CS 3CC 6 3) Any one of core is connected to sheath & capacitance is measured between remaining cores. C 3 3CC CS C C ph 3 15

16 Dielectric loss in a UG cable P 3C V tan ph ph where tan loss tangent 1 tan c R ph R = Insulation resistance Overhead Insulators For suspension type string insulator, the model for 3-discs looks like as shown. Let C m C 1 m V V 1 m S V3 V1 1 m 3m The voltage of disc nearest to the conductor is highest. String efficiency No. of discs V V V V 3 string voltage voltage across bottom disc 1% Distribution Systems Sources fed from both ends 1) Assume from ) Calculate I A I A from V A I in I I, I I I & I I I I V V I I r I I I r I I I I r A B A 1 1 A 1 A ) Substitute A & check for sign change. A 1 A 1 A 1 3 4) Node for minimum potential = Node for sign change 5) Calculate minimum potential by KVL Example: Refer Kuestion power systems for that. 16

17 Per Unit System Base value In pu system, energy quantity is expressed as a ratio of some based value. Absolute value or Actual value pu value= Base value Percentage value = pu value x 1% 1 - System S base, V base, I base, Z base Out of these, value must be known, to convert entire system into pu system. S V I base base base S V V I, Z base V I S base base base base base base base Usually, we assume S base & V base as known. 3 - System S base I base V base S base & V base S base 3 V are assumed base = line to line voltage = 3 phase power For start connection V ph V / 3 V Zbase I ph I S base base base base base base For delta connection V ph V 3V Zbase I ph I / 3 S base base base base base base In per unit system, equivalent impedance of transformer referred to primary or secondary in same. 17

18 Change of base If base of system is changed from Vbase old, Sbase old to Vbase new,sbase new pu Z new Z old Load Flow Study Power System Matrices base base pu Vbase new Sbase old V old S new YBUS matrix y1 y1 y1 YBUS y1 y y1 y3 y3 y y y OBSERVATIONS ) The diagonal elements are sum of all admittance connected to that particular bus. ) The off-diagonal elements are negative of admittance connected between two buses. 3) If two buses are not connected to each other than that elements is zero. 4) Matrix is a symmetrical matrix. Y BUS 5) Most of the elements are zero & hence it is a sparse matrix. Total number of zero elements % sparsity = Total number of elements Z matrix BUS Z BUS Y 1 BUS Z BUS matrix used in fault analysis. Suppose a 3 phase SC fault occurs on bus k then fault current V prefault,k If Z Z V, k Pr e fault voltage at bus 'k' prefault Z kk = elements of Z BUS matrix. Z = fault impedance f kk f Due to fault voltage at other buses are also affected. 18

19 V I V I k k.. Z. V Z BUS BUS BUS I f... V I n n. V I Z k f kk V f If Zf Zkk V Z I " j jk f Z Vj Z kk jk V f Post fault voltage at bus j Z jk V V jf j Vf Z kk If there is generator connected to bus j then current supplied by generator. Eg Vjf I " jx d Classification of buses At each bus, there are 4 parameter: V,,P,Q. At any bus, out of these 4 quantities any are specified. 1) Slack Bus / Swing Bus/Reference Bus V, are know quantities. P, Q are unknown quantities. Any extra power needed by the system is supplied by slack bus. ) Generator Bus / PV Bus P, V is specified Q, are unspecified 19

20 3) Load Bus / PQ Bus P & Q are constant as specified V & are unknown Generally, newton Raphson method is used for load flow solution and we form Jacobian matrix, & the order of Jacobian matrix is n m n m N = no. of total buses M = no. of pv buses Economic Power Generation Incremental cost If is cost required to generate an additional unit of energy. = Incremental cost of generator IC i i IC Gi i C P Gi P = Power generated by i th generator C i = cost of th i generator i th Transmission Loss m P P, P i j m PP B L i j ij i1 j1 : Real power injection at i th & j th buses B ij = loss coefficient m : no. of generator units Penalty Factor 1 L i P L 1 P Gi For economic power sharing IC L i i cons tant Constant is called as incremental cost of system. IC = incremental cost i th unit i L i = penalty factor of i th unit

21 From this expression, for m generator we get ( m 1) equation and m i1 P P P Gi D LOSS P D = total power demand For example, refer kuestions on Power systems. th m equation is Fault Analysis Symmetrical Components For an unbalanced 3 phase system, the analysis is done better by means of symmetrical components. Va Zero sequence components Va1 Va Positive sequence components Negative sequence components V a V a 1 Va1 1 Vb 3 V a 1 V c V 1 s A V p = phase voltage = Symmetrical component V p V s A 1 1 ; Where j1 e Power in terms of symmetrical components P 3 V I V I V I a1 a1 a a a3 a Remember, same transformation exist for current also. Sequence Network Alternators Positive Sequence Network Z 1 positive sequence impedance " Z jx 1 d 1

22 Negative Sequence Network V Z Z I Z a a : Negative sequence impedance X d" X q" j Zero Sequence Network V I Z 3Z Z a a n = Zero sequence impedance Z n =Neutral impedance ( Z X l jx l = leakage reactance in case of delta) Transformers Positive Sequence Network Z 1 = Positive sequence impedance Z jx 1 l Negative Sequence Network Z = Negative sequence impedance Z jx l Zero sequence network Depending upon scheme of connection, we close series or shunt connection & method of grounding.

23 Case 1 Shunt connection are closed for delta connection & series connection are closed for star connection with grounded neutral. If primary & secondary are inter changed then circuit becomes mirror image. If neutral is grounded through an impedance Z n Z Z T 3Zn Case Case 3 Case 4 3

24 Case 5 Transmission Lines Positive Sequence Network Negative Sequence Network Zero Sequence Network Z1 Z Zs Zm Z Zs Zm Z s = Self impedance Z m = mutual impedance Remembers, all sequence networks are always drawn in per unit & never in actual values. Fault Analysis The following short circuit faults are considered 1. LG (Single Line to ground fault). LL (Line to line fault) 3. LLG (Line to Line to Ground fault) 4. 3-phase short circuit fault. 4

25 3-phase short circuit fault comes under the category of symmetrical SC fault whereas other 3 faults are called as unsymmetrical SC fault. Order of severity LG < LL < LLG < 3 - SC But if faults occur at terminal of alternator then LG fault is most severe. Occurrence of SC fault LG > LL > LLG > 3 - SC Transient on a Transmission line Equivalent Circuit it it iss V Rt m sin e sin wt Z V L m L R Z 1 tan ; Z R L Maximum momentary current i mm V Z m sin If resistance is neglected, 9 i mm Vm cos Z Short circuit model of a synchronous machine During initial SC period for 1- cycles, current are induced in field & damper winding of machine so reactance at least & called as sub transient reactance X " After initial sub-transient period, current in damper winding in reduced to zero, and this period is called as transient period & reactance of machine is called as Transient Reactance X " d d 5

26 Finally, when current in field winding is also reduced to zero, we enter steady state period & reactance is called as steady state reactance. X d " X d ' Xd Symmetrical Fault Analysis X d We replace alternators by an emf source in series with sub transient reactance and emf source under no-load is usually pu or terminal voltage in pu. Vt Ef pu V base 1 Transformer & Transmission lines are replaced by reactance. The equivalent circuit can be solved either by finding thevenin equivalent across fault or by simple network analysis & fault in SC is calculated. I f Z eq E f Z f Z f = fault impedance. In symmetrical fault analysis, we only consider positive sequence impedance. 1 SC MVA pu Z Z SCMVA eq MVA Z eq f base Z f MVA For example, refer to kuestion on power systems. Selection of Circuit Breakers Usually, circuit breakers are selected on the basis of most severe fault which is 3 phase SC fault. Three ratings of circuit Breakers are important. 1) Rated momentary current Momentary current ( rms ) = 1.6 Isc I is symmetrical SC current which we calculated in previous section. sc ) Making current Making current =.54 Isc 6

27 3) Symmetrical Interrupting Current We need to recalculate by using sub-transient reactance for alternators & transient I sc reactance for synchronous motors. Induction motors & other loads are neglected. Then, we multiply it by a factor to calculate symmetrical interrupting current. This factor depends on speed of circuit breakers which is measured in terms of numbers of cycles it takes to operate. Speed 8 Cycles or slower 5 Cycles 3 Cycles Cycles Factor Unsymmetrical Faults Analysis Line to Ground Fault We first draw equivalent positive, negative & zero sequence networks & calculate thevenin impedance across fault terminals from each network. Assuming equivalent positive, negative & zero sequence reactance are Z 1,Z & Z respectively. Here all sequence currents are equal. Ia1 Ia Ia I a1 Ea Z Z Z 3Z 1 f Fault current I a I a 3I a1 I a 3Ea Z Z Z 3Z 1 f Short Circuit MVA SC MVA 3E a1,i a1 * 7

28 3 E I a1 Z Z Z 3Z 1 f In pu SCMVA 3 Z Z Z 3Z 1 f pu 3 MVA base Z Z Z 3Z 1 f MVA Line to Line Fault Here, we calculate equivalent positive & negative sequence impedance Z 1 & Z respectively. I a1 Ea Z Z Z 1 f Fault current I b j 3 E Z Z Z 1 a f Short Circuit MVA SC MVA 3 Z Z Z 1 f pu 3 MVA base Z Z Z 1 f MVA Line to Line to Ground Fault I I a1 a Ea Z Z Z 3Z Ia1 Z Z Z 1 f Fault Current = 3 Iao 8

29 Short Circuit MVA SC MVA 3 Z Z 1 3Z Z 1 f pu 3 MVA base Z Z Z 3Z 1 f MVA Remember, all fault analysis will be done in pu system. Power System Stability Two types of stability are studied: 1) Steady State Stability ) Transient Stability Steady State Stability For Steady State Stability dp d And for this condition to be true. P P e max If power demand is greater than maximum demand than machine goes out of synchronous. EV For a loss less machine, P max X Transient Stability Swing Equation Md P P m e dt M = inertia constant ( MJ-S / elect - rad) P m= mechanical input (MW) = electrical output (MW) P e = rotor angle Another Form Hd P P m e f dt H = inertia constant ( MJ / MVA) P & P e both are in pu m S 9

30 GH M 18f GH M f (MJ S / elect - deg) (MJ S / elect - rad) G = machine rating (MVA) If two alternators are swinging coherently. Then they can be replaced by a single alternator having M M M eq 1 But H cannot be added directly, they must first be on same base. If machines are not swinging coherently, then MM 1 Meq M M Accelerating Power, P P P a m e 1 In steady state P m P e In transient, P m P so rotor accelerate or decelerate. e Equal area criterion For system to possess transient stability P d a There are basically 3 stages in stability analysis Before Fault We say maximum power transferrable is P max,1 & e P P sin max,1 During fault We say maximum power transferrable is P max, P P sin e max, After Fault We say maximum power transferrable is P max,3 P P sin e max,3 3

31 Critical clearing angle It is the maximum value of beyond which if the fault is cleared system will be unstable. The time instant corresponding to this angle is called as critical clearing time assuming fault occurs at t =. Case-1 : Fault occurs on TL near to bus Pmax, P max,3 Cr P max,1 clearing angle By equal area criteria P P sin d m max,1 P 1 m sin P max,1 For critical clearing max t Cr max m H Cr = Critical Clearing Time f P Case- : Fault occurs on one of parallel lines close to bus Before Fault P max,1 EV X X X During Fault g 1 EV Pmax, X After Fault P max,3 eq EV X X g 1 31

32 P sin P 1 m max,1 1 Pm max sin P max,3 For transient stability P d a c m m max,3 P d P P sin d c For critical Clearing max t Cr H f Cr P m Case-3 : Fault occurs in middle of one of parallel lines The equivalent reactance during the fault is highest and thus P max, is lowest Pmax,1 Pmax,3 Pmax, P d a c For critical clearing, m max, max,3 m P P sin d P sin P d max c 1 Pm sin P max,3 3

33 P P cos P cos cr cos1 Pmax3 Pmax m max max,3 max max This is a generic formula and can be applied to other two cases as well after substituting value of P max,1, P max, & P max,3. But t cr before. can only be calculated from cr in previous two cases using expression written Power System Protection In our current protection, normally a current transformer is connected between protected elements and the relay. Plug setting multiplier (PSM) Fault current T ratio Pick up current Usually pick up current = Relay setting x Rated secondary current of CT Pick-up current is minimum current above which a relay operates. Differential Relays The current through operating coil k I I 1 If this current is greater than pick-up current then it operates, else it does not operates. K I I I Trip 1 pick up 1 pick up K I I I Block We usually provide a restraining coil to avoid relay mal-operation. Relay operates if Nr I1 I K I I K I N 1 pu I pu = pick up current N r = Number of turns in restraining coil N = Number of turns in operating coil The ratio operating coil current & restraining coil current is called as bias of differential relay. 33

34 Protection of Transmission line 1) Mho relay is at least affected by power surges& thus it is used for protection of long transmission lines. It is inherently directional. ) Impedance relay is used for protection of medium transmission lines. 3) Reactance relay is unaffected by ground resistance & hence used for earth fault protection & also for short transmission Lines. These relays are collectively called as distance relays. Protection of Transformers Differential relays are used for protection of large transformers and CT are always connected in configuration opposite to power transformer. Example: 1) If power transformer is Y then CT is Y : ) If power transformer is then CT is Y Y Buccholz relay used to prevent any incipient fault below oil level in a transformer of small KVA. 34

35 Kuestion Power Systems

36 Contents Manual for Kuestion... Type 1: Capacitance and inductance... 3 Type : SIL loading... 4 Type 3: ABCD parameter... 5 Type 4: Surges... 7 Type 5: Voltage control... 8 Type 6: Power factor correction... 1 Type 7: UG cable Type 8: Overhead Insulators... 1 Type 9: Distribution Systems Type 1: Economic Dispatch Type 11: Per unit system Type 1: Symmetrical Components... Type 13: Fault Analysis... Type 14: Thevenin Impedance... 4 Type 15: Power System Stability... 5 Type 16: Swing equation... 7 Type 17: Power System Matrices... 8 Type 18: Load flow Solution Type 19: Plug setting Multiplier Answer Key Kreatryx. All Rights Reserved. 1

37 Type 1: Capacitance and inductance For Concept, refer to Power Systems K-Notes, Transmission Lines Common Mistake: While calculating capacitance we do not use GMR but rather the radius of conductor, so this is the basic difference in Inductance and Capacitance calculation. Sample problem 1: The conductor of a 1km long, single phase, two wire line are separated by a distance of 1.5m. The diameter of an each conducting is 1cm. If the conductors are of copper, the induction of the circuit is? (A)5.mH (B)45.3mH (C)3.8mH (D) 19.6mH Solution: (C) is correct option Inductor of a single phase circuit with copper conductors d L L.ln( ) ab a r Given d=1.5m, r=.5cm, r =.7788 L L ab ab 1.5.4ln mh / km mH / km For 1km length of the line L ab =3.8 mh m Unsolved Problems: Q.1 Single phase concentric cable 5 km long has a capacitance of. F per km, the relative permeability of the dielectric being 3.5. the diameter of the inner conductor is 1.5 cm and the supply voltage is 66 kv at 5 Hz. Calculate the inner diameter of outer conductor (A) 3.9 cm (B) 1.95 cm (C) 3. cm (D) 1.6 cm Q. A double circuit 3-phse transmission line is shown below. The conductor a,a I ; b,b I and c,c I belong to the same phase respectively. The radius of each conductor is 1.5 cm. Find the inductance of the double circuit line in mh /km /phase. (A).191 (B).455 (C).755 (D).95 3

38 Q.3 A 3 phase double circuit line is shown in the figure. The diameter of each conductor is cm. Determine the capacitance per km length of the line, assume that the line is transposed. (A).11 F (B).15 (C).19 F (D).175 F F Q.4. Fig. shows the spacing s of a double circuit 3-phase overhead line. The phase sequence is ABC and the line is completely transposed. The conductor radius is.75 cm each. Inductance per phase per km is (A).46 mh (B).737 mh (C).63 mh (D).589 mh Q.5 A single phase 6 Hz transmission line and a telephone line both are supported on a horizontal cross-arm in the same horizontal plane. The spacing between transmission line conductors is.5m and conductors of the telephone line are of solid copper spaced.6m between centres. The distance between the nearest conductors of the two lines is cm. A current of 15 A is flowing over the power line. The value of mutual inductance between the circuit is? (A).1 mh/km (B).67 mh/km (C).46 mh/km (D).33 mh/km Type : SIL loading For Concept, refer to Power Systems K-Notes, Transmission Lines Common Mistake: While calculating SIL if we are using standard formula then we need to consider Line to Line Voltage and if we use phase voltage we need to multiply the formula by a factor of 3. Sample problem : A loss less transmission line having Surge Impedance Loading (SIL) of 8 MW is provided with a uniformly distributed series capacitive compensation of 3%. Then, SIL of the compensated transmission line will be (A) 1835 MW (B) 8 MW (C) 75 MW (D) 357 MW Solution: (C) is correct option 4

39 Let characteristic impedance Z impedance / km Z 1 sc c admittance / km Z 1 oc 1 pu Given that for line 3% series capacitive compensation is provided. Hence the series impedance of line is.7 or 7% of original value..7 Z.836 pu new 1. V Surge imedance loading (SIL)= Z SIL 1 (SIL) Z Z (SIL) Z c1 c 1 c 1. (SIL) 8M 75 MW.836 Unsolved Problems: c Q.1 The L/C ratio for 13KV and 4KV lines are typically 16*1 3 and 6.5*1 3 respectively. The neutral 3-phase loading for the two line will be respectively? (A) 44 MW and 5 MW (C) 64 MW and 44 MW (B) 44 MW and 64 MW (D) 18.9 MW and 56 MW Type 3: ABCD parameter For Concept, refer to Power Systems K-Notes, Transmission Lines Common Mistake: While using Transmission Parameters the voltages used should be per phase and not line to line voltages. Sample problem 3: The generalized circuit constants of a 3-phase, kv rated voltage, medium length transmission line are A D.936 j B 35.5 j C ( 5.18 j914) 1 6 If the load at the receiving end is 5 MW at kv with a power factor of.9 lagging, then magnitude of line to line sending end voltage should be (A) kv (B). kv (C) kv (D) 46.3kV Solution: (C) is correct option 5

40 Power received by load =5MW Current at receiving end = r 3V I cos r r I A r I cos =.9 = V AV BI KV s r r 3 Line to line sending end voltage = KV KV Magnitude of line to line sending end voltage =3.78KV Unsolved Problems: Q.1 A km 3-phase 5Hz transmission line has the following data A D B / phase C.19 /phase The sending end voltage is 3KV. The receiving end voltage is maintained as kv. The maximum power that can be transmitted and the correspond reactive power. (A) 33.8Mw, 311 MVAR (C) 311Mw, 33.8MVAR (B) 33.8Mw, MVAR (D) 311Mw 33.8 MVAR Q. A 5Hz, 3-phase transmission line is km long has total series impedance of (35 + j 14) ohms / phase and a shunt admittance of 93 x Mhos / phase. It delivers a load of 4Mw at kv with.9 p.f. lag. Find the magnitude of sending voltage for line to line in kv. (A) 3.4 (B) 9.8 (C) 7.6 (D) 36.5 Q.3 The generalized circuit constants of a nominal circuit representing a 3-phase transmission line are A = D =.98.3, B =.5 76 ohm, C =.5 9 mho, The two terminal voltages are held constant at 11 kv. If shunt admittance and the series resistance are zero, steady state stability limit is(mw) (A) (B) (C) (D) Q.4 A short 3 phase transmission line with an impedance of (6 + j8) Ohms per phase has sending and receiving end voltages of 1 kv and 11 kv respectively for some receiving end load at a point of.9 lagging. The sending end power factor is (A).88 lag (B).8 lag (C).84 lag (D).78 lag 6

41 Q.5 Two identical 3-phase transmission lines are connected in parallel to supply a load of 1MW at 13 KV and.8 pf lagging at the receiving-end. The constants of each transmission line are as follows; A.981,B 175 / phase What are the values of constant A and B for the combined network? (A) (C) A 1.961,B 75 (B) A 11,B.9875 (D) A.981,B 575 A.981,B 175 Type 4: Surges For Concept, refer to Power Systems K-Notes, Transmission Lines Common Mistake: If a line is terminated in characteristic impedance then there is no reflection so you need not solve the entire problem if you encounter such a case. Sample problem 4: A surge of kv magnitude travels along a lossless cable towards its junction with two identical lossless overhead transmission lines. The inductance and the capacitance of the cable are.4 mh and.5 μf per km. The inductance and capacitance of the overhead transmission lines are 1.5 mh and.15 μf per km. The magnitude of the voltage at the junction due to surge is (A) 36.7 kv (B) kv (C) 6.7 kv (D) kv Solution: (A) is correct option Surge impedance of cable L Z L=.4 mh/km, C=.5F/km 1 C surge impedance of overhead transmission line L Z Z ; L=1.5 mh/km, C=.15F/km 3 C Now the magnitude of voltage at junction due to surge is being given by as 3 V Z V 36.7KV Z Z

42 Unsolved Problems: Q.1 A 13 kv transmission line AB is connected to a cable BC. The characteristic impedances of the overhead line and the cable are 4 and 8 respectively. Assume that these are purely resistive A 5 kv switching surge travels from A to B. The value of the reflected component of this surge when two first reflection reaches A (A) kv (B) kv (C) kv (D) kv Q. The ends of two long transmission lines A and C are connected by a cable B. The surge impedances of A,B and C are 5, 7 and 6 ohms respectively. A rectangular voltage wave of KV magnitude and of infinite length is initiated in A travels to C. Second impressed voltage on C is (A) 14.1 KV (B) 8.8 KV (C) 3.5 KV (D) 5.3 KV Q.3 A surge f 1KV travelling in a line of natural impedance 6Ω arrives at a junction with two lines of impedance 8Ω and Ω respectively. What is the value of surge voltage? (A)4.9KV (B)43.76KV (C)45.98KV (D)4.4KV Q.4 What is the first impressed surge voltage and currents into line-3 from following network (A)38.96 KV, A (B) KV, 63.4 A (C)77.9 KV, A (D) KV, A Type 5: Voltage control For Concept, refer to Power Systems K-Notes, Transmission Lines Common Mistake: It is always easier to solve these type of problems in pu system rather than actual values. Sample problem 5: A 3-phase 11 kv generator feeds power to a constant power unity power factor load of 1MW through a 3-phase transmission line. The line-to line voltage at the terminals of the machine is maintained constant at 11 kv. The per unit positive sequence impedance of the line based on 1 MVA and 11 kv is j.. The line to line voltage at the load terminals is measured to be less than 11 kv. The total reactive power to be injected at the terminals of the load to increase the line-to-line voltage at the load terminals to 11 kv is (MVAR) (A) 1 (B) 1.1 (C) 1 (D) 1.1 Solution: (D) is correct option 8

43 Given Load Power = 1 MW VS = VR = 11 kv p.u. (KV) j. (11) imedance of line Z j.4 L MV 1 So as to make the sending end voltage and receiving end voltage equal, the net reactive power demand at load must be zero. Let QC be the reactive power injected at load side and Qr be the reactive power received from supply. Q Q Q r C d Q Q Q Q r C C r In this case β=line impedance angle =9.4, =, A 1. V V A s r Pr cos( ) V cos( ) r B B cos(9 ) 11 cos(9 ) sin V V A sin( ) V sin( ) B s r Qr r Q Q r r B MVAR Unsolved Problems: sin( ) 11 sin(9) Q.1 The receiving end voltage of a km long are head transmission line is to be kept constant at 1 KV. Its Voltage regulation is % with a load. It a capacitor is connected in parallel which the load to reduce the regulation to 1% the sending end voltage is (A) 9KV (B) 1KV (C) 11 KV (D) 1KV Q. A 75Kv transmission line has A =.855, B = 75 what is the ratio of compensating equipment required in MVAR at the receiving end to maintain the sending and receiving end voltages at 75kV. The load at the receiving end is 15MW at.8 power factor lagging (A) 7.56 (B) (C) (D) Q.3 Three supply points A, B and C are connected to bus bar M. Supply point A is maintained at a nominal 75 KV and is connected to M through a 75/13 KV transformer (.1p.u.reactance) and a 13 KV line of reactance 5. Supply point B is nominally at 13 KV and is connected to M through a 13 KV line of 5 reactance. Supply point C is nominally at 75 KV and is connected to M by a 75/13 KV transformer (.1p.u reactance) and 13 KV line of 5 reactance. If at a particular point, the line voltage M falls below its nominal value 9

44 by 5KV, calculate the magnitude of reactive volt ampere injection required at M to reestablish the original voltage. The p.u values are expressed on a 5 MVA base and resistance may neglect throughout. (A) 7.6MVAR (B) 15.7 MVAR (C) 3 MVAR (D) 38 MVAR Q.4 A 3-phase, 5 Hz overhead transmission line is feeding a.8 power factor lagging load with both the sending end and receiving end line voltage held at 11KV. The sending-end voltage leads the receiving end voltage by 15 and the line constants are as following A.961,B 183 If a compensating device is required to met the demand of load, then VAR rating (in MVAR) of device should be? (A)4.71 (B).84 (C)19.3 (D) 1.87 Type 6: Power factor correction For Concept, refer to Power Systems K-Notes, Transmission Lines Common Mistake: While using formula for rating of compensating device the real power is always total three phase power. Also, we already consider Q as negative and hence answer comes out as positive for capacitor. Sample problem 6: At an industrial sub-station with a 4 MW load, a capacitor of MVAR is installed to maintain the load power factor at.97 lagging. If the capacitor goes out of service, the load power factor becomes (A).85 lag (B) 1. lag (C).8 lag (D).9 lag Solution: (C) is correct option Let the initial power factor angle =φ1 1 Given cos P tan tan KVAR supplied by capacitor 4 1 (tan tan14.7 ) cos.8 lag Hence if the capacitor goes out of the service the load power factor becomes.8 lag. 1

45 Unsolved Problems: Q.1 An induction motor operating at.8 pf lag consuming 3 KW. A zero power consuming synchronous motor is connected across the induction motor to raise the pf to.9 lagging. The reactive power drawn by the synchronous motor is? (A)97. KVAR lag (C)36 KVAR lag (B)97. KVAR lead (D)36 KVAR lead Q. A 4 V, 5 Hz, three phase balanced source supplies power to a star connected load whose rating is 1 3 kva,.8 pf (lag). The rating (in kvar) of the delta connected (capacitive) reactive power bank necessary to bring the pf to unity is? (A) (B) 1.6 (C) 16.6 (D) 1.47 Type 7: UG cable For Concept, refer to Power Systems K-Notes, Underground Cables. Common Mistake: While connecting core to core and core to sheath make sure that you combine capacitors correctly in series or parallel. Sample problem 7: Consider a three-core, three-phase, 5 Hz, 11 kv cable whose conductors are denoted as R,Y and B in the figure. The inter-phase capacitance(c1) between each line conductor and the sheath is.4 μf. The per-phase charging current is (A). A (B).4 A (C).7 A (D) 3.5 A Solution: (A) is correct option Given 3-Ф, 5 Hz, 11 kv cable C1 =. μf C =.4 μf Charging current IC per phase =? Capacitance Per Phase C = 3C1+ C C = 3*. +.4 = 1 μf ω = πf =

46 V Charging current I V( C) c X c = = Amp 6 Q.1 The charging current drawn by a cable with 3 cores and protected by a metal sheet when switched on to a 11 KV, 5Hz supply. The capacitance between two cores with the third core connected to the sheath is measured to be 3.7.F (A) 9.3 A (B) A (C) 16.3 A (D) A Q. In a 3 core cable capacitance between any two cores is.5µf and capacitance between any one of the core to sheath is.3 µf. What is capacitance measured between any two cores connected together and third core? (A)1.3µF (B) 1.µF (C) 1.1µF (D) 1.5µF Type 8: Overhead Insulators For Concept, refer to Power Systems K-Notes, Overhead Insulators. Common Mistake: Voltage Distribution can be obtained by using Voltage division and also by using KCL. Sample problem 8: Consider a three-phase, 5 Hz, 11 kv distribution system. Each of the conductors is suspended by an insulator string having two identical porcelain insulators. The self-capacitance of the insulator is 5 times the shunt capacitance between the link and the ground, as shown in the figure. The voltages across the two insulators are (A) e 3.74 kv,e.61 kv 1 1 (B) e 3.46 kv,e.89 kv (C) e 6. kv, e 4.3 kv 1 (D) e 5.5 kv, e 5.5 kv 1 Solution: (B) is correct option At A point I I I 1 3 C e 5C e 5C e 1 3 5e 6e......(1) 1 1

47 11k e e 6.35kV.....() 1 3 From equation (1) and () we can get e =.886kV and e =3.46kV 1 Unsolved Problems: Q.1 In a four disc string insulator capacitance grading is employed in order to get 1% string efficiency. The capacitance of shunt capacitor is.5 F and the capacitance of a disc near to cross arm is.1 F. The capacitance of a disc near to power conductor is (A).35 F (B).4 F (C).5 F (D).15 F Q.. The equivalent capacitor arrangement of the two string Insulator is shown below. The maximum voltage that each unit can with stand should not exceed 17.5 kv. The line to line voltage of the system is? (A) 17.5 kv (B) 33 kv (C) kv (D) 3.3 kv Q.3 In a 3 unit insulator string, voltage across the lowest unit is 17.5 KV and string efficiency is 84.8%. The total voltage across the string will be equal to? (A)8.85KV (B)44.5KV (C)88.5KV (D)44.5KV Q.4 If the voltage across the string of a string insulator assembly is 38KV, number of insulators discs are 4 and voltage across the lowest disc is 1KV, string efficiency is? (A)79.1% (B)7% (C)5% (D)1% Q.5 It is required to grade a string consisting of three suspension insulators. Determine the line to pin capacitance that would give the same voltage across each insulator of the string if the pin to earth capacitance are all equal to.5c? (A) C, C, C (C) C, C, C 5 3 (B) C, C, C 16 8 (D) 5C, 8C, 1C 13

48 Type 9: Distribution Systems For Concept, refer to Power Systems K-Notes, Distribution Systems. Sample problem 9: A distribution feeder of 1 km length having resistance, but negligible reactance, is fed from both the ends by 4V, 5 Hz balanced sources. Both voltage sources S1 and S are in phase. The feeder supplies concentrated loads of unity power factor as shown in the figure. The contributions of S1 and S in 1 A current supplied at location P respectively, are (A) 75 A and 5 A (C) 5 A and 75 A Solution: (D) is correct option Assume point of minimum voltage is A Assume resistance of entire feeder = R V 4 I.4R.(i) A 1 V 4 I.R I.R I 3.R A = 4.6 R I I 3 I 1 1 I = 5.(iii) I + 1R..(ii) From (i) & (ii) 4.4I1R 4.6IR 1R.6I.4I1 1 (iv).4i.4i. (v) 1 Contribution of I to I I 1A Contribution of I to I P 1 P A (B) 5 A and 5 A (D) A and 1 A Unsolved Problems: Q.1. A two line d.c. distributor fed from both ends is shown below. The Minimum Voltage on the distributor is (A) (B).45 (C) (D)

49 Q. A single phase two wire feeder 15m long is shown below what is the voltage at the sending end. The impedance A at the feeder is (.6 +.1) ohms / km (A) (B) 5.56 (C) (D) 45.1 Q.3 A dc wire ring main distributor is shown below. Find the total currents supplied by two sources. The resistance of each conductor.1 ohms/km. The source voltages are equal (A) 9.78 A, 19.8 A (B) 1. A, A (C) A, 94. A (D) 115. A, A Q.4 A wire DC distributor ABCDEA in the form of a ring main is fed at point A at V and is loaded as shown in the figure. The minimum potential on the distributor is (A) V (B)16.15 V (C)16.45 V (D)18.8 V Type 1: Economic Dispatch For Concept, refer to Power Systems K-Notes, Economic Power Generation. Point to Remember: Just remember the basic formulas for factors related to plant and no need to memorize all factor just a few important ones are fine. 15

50 Sample problem 1: The figure shows a two-generator system applying a load of PD = 4MW, connected at bus. The fuel cost of generators G1 and G are : C1(PG1) = 1 Rs/MWh and C(PG) = 15 Rs/MWh and the loss in the line is Ploss(pu)=.5P G1(pu), where the loss coefficient is specified in pu on a 1MVA base. The most economic power generation schedule in MW is (A) PG1 =,PG = (B) PG1 =,PG = (C) PG1 =,PG = (D) PG1 =,PG = 4 Solution: (A) is correct option 1 Let penalty factor of plant G1 L 1 ; PL.5P G1 PL 1 P P P L G1 So,L.5 P P P G1 G1 G1 1 Penalty factor of plant G is L =1 ; PL 1 P For economic power generation C L C L 1 1 where C1 and C are the incremental fuel cost of plant G1 and G 1 So, P G P P pu It is an 1 MVA, so G G1 G1 G P 1 MW G1 5 P P L G Loss P.5 pu 1 MW L

51 Total power, P P P P L G1 G L 4 P P MW G Unsolved Problems: G Q.1 Two power plants interconnected by a tie line as shown in the above figure have loss formula coefficient b11 = 1-3 mw -1. Power is being dispatched economically with plant as 1Mw and plant as 15 Mw. The penalty factor for the plants 1 and are respectively. (A) 1 and 1.5 (B) 1.5 and 1 (C) 1 and zero (D) zero and 1 Q. A two bus system is shown below If 1 Mw is transmitted from plant 1 to the load, a transmission loss of 1 Mw is incurred. Find the required generation for each plant when the system x is Rs.5/mwhr. The incremental fuel costs are dc dp dc. P 16 Rs / Mwhr ;.4 P Rs / Mwhr 1 G1 dp G1 G (A) 14.5 MW and 15 MW (C) MW and 15 MW Q.3 The incremented fuel cost of two plants are given as.1p 15 Rs/ Mwhr IC 1 1 I C Rs/ Mwhr G (B) 15 MW and MW (D) 135. MW and 15 MW The minimum and maximum generation limits are 5 Pi 1, (i 1,) The load on the system is 18 MW. The generation limits are (A) 9 MW and 9 MW (C) 8 MW and 1 MW (B) 1 MW and 8 MW (D) 6 MW and 1 MW Q.4 The power generated by two plants are P1 = 5 mw and P = 4 mw the loss coefficients are B11 =.1, B =.5 and B1 = -.5. The demand on the system will be (A) 9 MW (B) 84.5 MW (C) 79.5 MW (D) 74.5 MW 17

52 Q.5 For the system whose line diagram is shown in figure. I1 = 1. and I =.8 per unit. If the voltage at bus 3 is 1. per unit. Find the loss coefficients B1 line impedances are (.4+j.16) per unit, (.3+j.1) per unit and (.+j.8) per unit for sections a, b and c respectively (A).554p.u. (B).188p.u. (C).477p.u. (D) 1.5p.u. Type 11: Per unit system For Concept, refer to Power Systems K-Notes, Per Unit System. Common Mistake: The voltage ratings of Transformers need not be necessarily be the base values, you need to transform the voltage bases as per transformer voltage ratios. Sample problem 11: Two generator units G1 and G are connected by 15 kv line with a bus at the midpoint as shown below G1 = 5MVA, 15 kv, positive sequence reactance XG1 =5% on its own base G = 1MVA, 15 kv, positive sequence reactance XG =1% on its own base L1 and L = 1km, positive sequence reactance XL =.5 Ω/km For the above system, the positive sequence diagram with the pu values on the 1 MVA common is? 18

53 Solution: (A) is correct option Positive sequence diagram of the above system 1 15 X.5.1 G X.1.1 G X.51 j1. L X.51 j1. L 15 5 Unsolved Problems: Q.1 Assuming S buses of MVA and 4MVA G1 G Tr 3MVA 5MVA 6MVA 11KV 11KV 11KV ( ) : 66KV (Y) X =. pu X =.5 pu X=.1 pu From the figure find the actual impedance in ohm for given Sbase as 5 MVA? (A)j.8874 (B)j.6853 (C)j.63 (D)j.93 Q. For the power system shown in the figure below, the specifications of the components are the following: G1: 5 kv, 1 MVA, X = 9% G: 5 kv, 1 MVA, X = 9% T1: 5 kv/ kv, 9 MVA, X = 1% T: kv/5 kv, 9 MVA, X = 1% 19

54 Line 1: kv, X = 15 ohms Choose 5 kv as the base voltage at the generator G1, and MVA as the MVA base. The impedance diagram is? Type 1: Symmetrical Components For Concept, refer to Power Systems K-Notes, Fault Analysis. Point to Remember: Try to understand the zero sequence network for Transformer using Switch Model. Sample Problem 1: A three-phase alternator generating unbalanced voltages is connected to an unbalanced load through a 3-phase transmission line as shown in figure. The neutral of the alternator and the star point of the load are solidly grounded. The phase voltages of the alternator are E 1 V,E 1 9 V,E 1 1 V. a b c The positive-sequence component of the load current is (A) A (B).33-1 A (C) A (D) A Solution: (D) is correct option Both sides are grounded So, E 1 a I 5 9 a Z j a E 19 b I b Z 3j b

55 E 11 c I.5 3 c Z 4 j c 1 We know that I I I I ; where =1 1 =1 4 a1 a b c 3 1 I ( 18 1 ).5 (4 3 ) a1 3 1 I a1 3 1 I 5j j.883.5j a1 3 Unsolved Problems: Q.1 A single phase load of 1kv is connected across lines b and c of a 3-phase supply of 3.3kv. Determine the positive sequence component of line a is (A) 3.3A (B) 17.5A (C) A (D) 48.5A Q. In a unbalanced 3 phase power system, the currents are measured as Ia = zero, Ib = 66 and Ic = 6-1. The corresponding sequence currents are Ia Ia1 Ia (A) zero 3-j3-3+j3 (B) zero -3+j3 3-j3 (C) zero -9+j33 9-j33 (D) zero 9-j33-9+j33 Q.3 When a 5 MVA, 11 KV 3-phase generator is subjected to a 3-phase fault, the fault current is j 5 p.u. When it is subjected to a line-to-line fault, the positive sequence current is j 4p.u. The positive and negative sequence reactance are respectively (A) j. and j.5 p.u (C) j.5 and j.5 p.u (B) j. and j.5 p.u (D) j.5 and j.5 p.u Q.4 A 3 - line supplies a delta connected load. The line current line1 is 1 A, taking the current in line 1 is as reference, current in line is Positive symmetrical component of line, if phase 3 is only switched off (A) (B) (C) (D)

56 Type 13: Fault Analysis For Concept, refer to Power Systems K-Notes, Fault Analysis. Common Mistake: While calculating any reactance please make sure that unit is correct as answer may be in pu or in ohms. Sample problem 13: A -MVA, 6.6-kV, 3-phase alternator is connected to a 3-phase transmission line. The per unit positive-sequence, negative-sequence and zero-sequence impedances of the alternator are j.1, j.1 and j.4 respectively. The neutral of the alternator is connected to ground through an inductive reactor of j.5 p.u. The per unit positive-, negative- and zerosequence impedances of transmission line are j.1, j.1 and j.3, respectively. All per unit values are based on the machine ratings. A solid ground fault occurs at one phase of the far end of the transmission line. The voltage of the alternator neutral with respect to ground during the fault is? (A)513.8 V (B)889.9 V (C)111. V (D) 64. V Solution: (D) is correct option Total zero sequence impedance, +ve sequence impedance and ve sequence impedances Z = (Z) Line+ (Z) Generator = j.4 + j.3 = j.34 pu Z1 = (Z1) Line+ (Z1) Generator = j.1 + j.1 = j. pu Z = (Z) Line+ (Z) Generator = j.1 + j.1 = j. pu Zn = j.5 pu for L-G fault I I E.1 a a1 Z Z Z 3Z j. j. j.34 j.15 1 n 6 generator MVA A 3 generator KV Base 3 Fault current I 3I I 3 j j A f a Base n n 6.6 Neutral Voltage V =I Z and Z =Z Z n f n n B pu V = V = 64. V j1.1 pu

57 Unsolved Problems: Q.1 A 3-phase 5 Hz generator is rated at 5 MVA, kv with Xd =.p.u.it supplies purely resistive load of 4 Mw at kv. The load is connected directly across the terminal of the generator. If all the three phases of the load are short circuited simultaneously through a fault reactance of.1 p.u, find the initial symmetrical r.m.s. current in the generator in ka in the generator on a base of 5 MVA, kv. (A) 7.17 (B) 7.75 (C) (D) 56.8 Q. When a 5 MVA, 11 KV 3 phase generator is subjected to a 3 phase fault, the fault current is j5 PU. When it is subjected a line to line fault, the positive sequence current is j4pu. The positive and negative sequence reactance s are respectively (A) j., j.5 (B) j., j.5 (C) j.5, j.5 (D) j.5, j.5 Q.3 Two 11 KV, MVA, three phase, star connected generators operate in parallel. The positive, negative and zero sequence reactance s of each being respectively j.18, j.15 and j.1 P.U. The star point of one of the generator is isolated and the other is earthed through a resistor. The potential of the neutral for a line to ground fault in P.U. (A).33 (B).96 (C).89 (D) 1.44 Q.4 In the power system shown, the values are marked are the per unit reactance s taking MVA and 11 KV as base values in the generator circuit. Both transformers are rated for 11/11 KV. A three phase to ground fault with a fault impedance of j.88 pu occurs at bus. The fault current supplied by generator is (A) 154.8A (B) 64.4A (C) 19.6A (D) 17.8A Q.5 A 1 MVA, 13.8 kv alternator has positive, negative and zero sequence reactance of 3%, 4% and 5 % respectively. What reactance must be put in the generator neutral so that the fault current for a line to ground fault of zero fault impedance will not exceed the rated line current. (A) 13.8 (B) 14.8 (C) 15.5 (D)

58 Type 14: Thevenin Impedance For Concept, refer to Power Systems K-Notes, Fault Analysis. Common Mistake: While calculating transfer reactance, shunt branches of pi-network are neglected. Sample problem 14: A generator is connected to a transformer which feeds another transformer through a short feeder. The zero sequence impedance values are expressed In pu on a common base and voltage are indicate in figure. The Thevenin equivalent zero sequence impedance at point B is? (A).8+j.6 (B).75+j. (C).75+j.5 (D) 1.5+j.5 Solution: (B) is correct option Per unit zero sequence reactance diagram of the given single line diagram is shown below. Thevenin equivalent impedance, Zth at B is Zth=j.1+j.5+j.7+.75=.75+j. Unsolved Problems: Q.1 For the network shown, the zero sequence reactance in PU are indicated. The zero sequence driving point reactance of the node 3 is (A).1 PU (B).3 PU (C).1 P.U (D). PU 4

59 Q. For the diagram shown below, the steady state symmetrical fault current for a 3 phase fault on the 11 kv feeder is (A) 1 p.u (B) 15 p.u (C) p.u (D) 5 p.u Q.3 A 3-phase fault occurs at the middle point F on the transmission line as shown in the figure. The transfer reactance appearing between the generator and the infinite bus is? (A)j.9 pu (B) j.575 pu (C)j.6 pu (D)j.65 pu Q.4 For figure shown below calculate the max power transfer (during the fault ), when the system is healthy, fault occurs at the middle of the line and fault is cleared by the breaker? (A)1.567 pu (B).7 pu (C).6 pu (D).618 pu Type 15: Power System Stability For Concept, refer to Power Systems K-Notes, Power System Stability. Point to Remember: If you are not able to remember the entire expression for critical clearing angle then better understand the method and you can compute using integration. Sample problem 15: A cylinder rotor generator delivers.5 pu power in the steady-state to an infinite bus through a transmission line of reactance.5 pu. The generator no-load voltage is 1.5 pu and the infinite bus voltage is 1 pu. The inertia constant of the generator is 5MW-s/MVA and the generator reactance is 1 pu. The critical clearing angle, in degrees, for a three-phase dead short circuit fault at the generator terminal is (A) 53.5 (B) 6. (C) 7.8 (D)

60 Solution: (D) is correct option Total reactance X= j1+j.5 =j1.5 pu Ps=Pe1=.5 pu EV Before fault P 1. pu m1 X 1.5 During fault Pm= pu After the fault Pm3=1. pu 1 Ps sin 3 Pm1 3 (radians).5 rad 18 1 P s sin 18 sin 15 max P 1 m3 15 (radians).168 rad max 18 1 P ( ) P cos s max m3 max 1.5(.168.5) 1. cos15 Critical clearing angle cos cos c P 1. m Unsolved Problems: c Q.1 A synchronous generator is connected to an infinite bus through a loss less double circuit transmission line. The generator is delivering 1. P.U power at a load angle at 3. When a sudden fault reduces are peak power that can transmitted to.5 P.U. After clearance of fault the peak power that can be transmitted becomes 1.5 per limit. Find the critical clearing angle (A) (B) (C) 6.3 (D) 56.6 Q. A 5 Hz 3-phase generator is supplying 6% of man power to an infinite bus though a reactive network. A fault occurs which will increase the reactance of the network between the generator to the infinite bus to 4% of the original value. When the fault is cleared, the maximum power that can be delivered is 8% of the original maximum value. Determine the critical clearing angle for the condition described. (A) 56.6 (B) 6.4 (C) 66.5 (D)

61 Q.3 The single line diagram of the power system is having a three phase fault at the point P. The generator is delivering 1. P.U. power at the instant preceding the fault. The Maximum power transfer before, during and fault in P.U. are (A).3,.8,.3 (B) 1.5,.8,.3 (C).3,,.3 (D).3,, 1.5 Q.4 A 5 Hz, 6 pole generator with H=4. pu delivers 1. pu to an infinite bus bar through a network in which resistance is negligible, and is supplying 45% of its peak power capacity as shown. A 3-phase fault occurs at the point F of the outgoing radial line. Faulted line is cleared by the opening of the line circuit breaker. Critical clearing time is (A).34 sec (B).78 sec (C).37 sec (D).6 sec Type 16: Swing equation For Concept, refer to Power Systems K-Notes, Power System Stability. Common Mistake: Both M and H are inertia constants and the only way to distinguish between them is through their units. Sample problem 16: A 5 MW, 1 kv, 5 Hz, 3-phase, -pole synchronous generator having a rated p.f=.9, has a moment of inertia of 7.5 X 1 3 # kg-m.the inertia constant (H ) will be (A).44 s (B).71 s (C) 4.88 s (D) 5.4 s Solution: (A) is correct option Given Synchronous generator of 5 MW, 1 kv, 5 Hz, 3-φ, -pole, P.F =.9, Moment of inertia M = 7.5 *1 3 kg-m Inertia constant H =? P 5 MW Generator rating in MVA G MVA cos.9 1 f 1 5 N 3 rpm pole 7

62 1 1 N stored K.E. = M M 7.51 MJ MJ 6 6 Stored Kinetic Energy inertia costant (H)= Rating of generator (MVA) H=.44 sec Q.1 A 5Hz two pole turbo attenuator rated 5 MVA, 13. kv has an inertia constant H = 5 MJ / MVA. The input less the rotational losses is 65, HP and the electrical power developed is 4 Mw. Determine the angular acceleration in electrical deg/sec is (A) (B) 9.45 (C) (D) 3.4 Q. A pole, 5 Hz, 11KV turbo-generator has a rating of 6 MW, power factor.85 lagging. Its rotor has a moment of inertia of 88 kg-m. The inertia constant in MJ per MVA and its momentum in MJ-s/electrical degree respectively are? (A)8.5149,.1447 (B)7.376,.485 (C)6.15,.185 (D)6.15,.1447 Q.3 A synchronous generator of reactance 1.pu is connected to an infinite bus-bar through a transformer and a line of total reactance of.6pu. The infinite bus-bar voltage is 1.pu and the generator no load voltage is 1.pu. The inertia constant is H=4 Mw-sec/MVA. The resistance and machine damping may be assumed negligible. The system frequency is 5Hz. The frequency of oscillations in Hz if the alternator is loaded to 5% of maximum power limit. (A).63 (B).758 (C).48 (D).54 Q.4 A 1 MVA synchronous Generator operates on full load at a frequency of 5Hz. The load is suddenly reduced to 5 Mw. Due to time lag in governor system the Steam value begins to close after.4 seconds. Determine the change in frequency that occurs in this time. Given H = 5 kw S/kVA. (A) 5.5 Hz (B) 5 Hz (C) 51.5 Hz (D) 5 Hz Q.5 A power station consists of two synchronous generators A and B at rating 5MVA and 5MVA with inertia constant at 1.6 p.u and 1 p.u respectively on their own base MVA ratings. The equivalent p.u inertia constant in the system on 1 MVA common base will be (A).6 p.u (B).615 p.u (C) 1.65 p.u (D) 9 p.u Type 17: Power System Matrices For Concept, refer to Power Systems K-Notes, Load Flow Study. Common Mistake: Number of non-zero off-diagonal elements in bus admittance matrix are double the number of transmission lines in a power system network. 8

63 Sample problem 17: The network shown in the given figure has impedances in p.u. as indicated. The diagonal element Y of the bus admittance matrix YBUS of the network is (A) j19.8 (B) +j. (C) +j. (D) j19.95 Solution: (D) is correct option I = V Y + (V - V )Y =.5V - j1(v - V ) =- j9.95v + j1v I = (V - V )Y + (V - V )Y = j1v - j9.9v - j.1v 1 3 Y = Y + Y + Y 11 3 =- j j j =- j19.95 Fault analysis using Z bus Sample problem 18: For a power system the admittance and impedance matrices for the fault studies are as follows. j8.75 j1.5 j.5 j.16 j.8 j.1 Y j1.5 j6.5 j.5 bus ; Z bus j.8 j.4 j.16 j.5 j.5 j5. j.1 j.16 j.34 The pre-fault voltages are 1. pu. at all the buses. The system was unloaded prior to the fault. A solid 3-phase fault takes place at bus. The post fault voltages at buses 1 and 3 in per unit respectively are (A).4,.63 (B).31,.76 (C).33,.67 (D).67,.33 Solution: (D) is correct option The post fault voltage at bus 1 and 3 are. Pre fault voltage. V 1 1 bus V 1 3 V V 1 At bus solid fault occurs Z(f) =, r = 9

64 Fault current I f Z f V V r Z Z Z Z rr f f 1 4 j j.4 V(f) = V () - Z I(f), i i ir i 1 i 1 f f 3 ;V = Prefault voltage V (f) = V - Z I = 1 -j.8(-j4) = V (f) =.68 pu V (f) = V - Z I = 1 -j.16(-j4) = V (f) =.36 pu Unsolved Problems: Q.1 A single line diagram of a power system is shown. The diagonal elements of the Y bus matrix is (A) j.7 pu, j.75 pu, j.45pu (B) -j.7pu, -j.75pu, -j.45pu (C) -j7.pu, -j6pu, -j9.pu (D) j7. pu, j6.pu, j9.pu Q. The Reactance s between the various buses in a power system are given in the table below. Y and Y33 of YBUS are Buses Reactance in p.u (A) -j.85, -j 1.15 (B) -j 19, -j 3 (C) -j.65, -j 1.15 (D) -j 16, -j 8.5 3

65 Q.3 The power system network shown in figure, where bus numbers impedances are marked, assuming equal R/X of impedances, find the bus impedance matrix element z. (A). (B).5 (C) 3.5 (D) 4.5 Q.4 The Ybus matrix of a 1-bus interconnected system is 9% sparse. Hence the number of transmission lines in the system must be? (A)45 (B)8 (C)1 (D)6 Type 18: Load flow Solution For Concept, refer to Power Systems K-Notes, Load Flow Study. Common Mistake: These type of questions are based on basic power angle equation and performance equations of short Transmission Line. Sample problem 19: In the following network, the voltage magnitudes at all buses are equal to 1 pu, the voltage phase angles are very small, and the line resistances are negligible. All the line reactance s are equal to j1ω The voltage phase angles in rad at buses and 3 are (A) ϴ =-.1, ϴ3 =-. (B) ϴ =, ϴ3 =-.1 (C) ϴ =.1, ϴ3 =.1 (D) ϴ =.1, ϴ3 =. Solution: (B) is correct option.1pu power must flow from slack bus to Bus3 so that demand at Bus 3 is satisfied and therefore, VV 1 3 1*1 Power P sin( ) sin( ) X 1 Therefore, 3 13 sin 3 3.1rad Since, no power flows from Bus1 to Bus so both must have same angles Therefore, rad 31

66 Unsolved Problems: Q.1 Figure shows a 3-phase system supplied at 11 KV at A. The load currents are balanced and the pf s (all lagging) are with respect to supply voltage at A. The impedances shown are per phase values. Voltage at load point C is (A) 9.93 KV (B) 1.3 KV (C) 1.66 KV (D) 9.7 KV Q. A simple power system has been shown in fig. all quantities are phasors VA = 1pv, ZAB = j.5 pv, SDA =. pu, SDB =. pu. If QGB is made as a zero, voltage VB is (A).7 3 (B).5 4 (C) (D) Q.3 The station loads are equalized by the flow of power in the cable. The Generator can generate a maximum of. pu real power. Reactive power flow in the cable (A).638 pu (B).638 pu (C) 1.76 pu (D) Q.4 For a 15 bus power system with 3 voltage controlled bus, the size of Jocobian Matrix is (A) 11 x 11 (B) 1 x 1 (C) 4 x 4 (D) 5 x 5 Q.5 A 5Hz transmission line 3km long has a total series impedance of 4+j15 ohms and a total shunt admittance of 1-3 mho. The receiving end load is 5MW at kv with.8 lagging power factor. Find the sending power with the short line approximating (A) 5.15MW (B) 53.MW (C) 5.58MW (D) 51.15MW 3

67 Type 19: Plug setting Multiplier For Concept, refer to Power Systems K-Notes, Power System Protection. Common Mistake: Remember to consider the relay setting while calculating Pick-up Current, it is not necessarily equal to rated secondary current of Current Transformer. Sample problem : The overcurrent relays for the line protection and loads connected at the buses are shown in the figure. The relays are IDMT in nature having the characteristic t.14 Time Multiplier Setting (Plug setting multiplier) 1 op. The maximum and minimum fault currents at bus B are A and 5 A respectively. Assuming the time multiplier setting and plug setting for relay RB to be.1 and 5A respectively, the operating time of RB (in seconds) is (A).67s (B).356s (C).145s (D).148s Solution: (A) is correct option CT ratio = : 1 PSM = Maximum fault current CT primary current 1 Top = =.67 s Unsolved Problems: Q.1 A fault current of A is passing on the primary side of a 4/5 CT. on the secondary side of the CT an inverse-time over current relay is connected whose plug setting is set at 5%. The plug setting multiplier will be (A) 5 (B) 5 (C) 1 (D) 3 33

68 Q. The rated secondary current of a current transformer (CT) is 5A. The plug setting of relay connected to the secondary is.5 A. If the effective VA burden on the CT is 1 VA, then the VA burden of the relay at.5 A plug setting is (A).75 A (B) 3 A (C) 6 A (D) 1 A Q.3 An over-current relay, having a current setting of 1.5% is connected to a supply circuit through a current transformer of ratio 4/5. The pick-up current value is? (A).65A (B)1A (C)1.5A (D)15A 34

69 Answer Key Type 1 A B C C B Type B Type 3 B B B A B Type 4 C A D B Type 5 C B D A Type 6 B D Type 7 B B Type 8 B C B A A Type 9 A A B B Type 1 B C C B B Type 11 B B Type 1 B B B D Type 13 B A B C B Type 14 C B A D Type 15 B C D D Type 16 A C B B D Type 17 C D C A Type 18 C D C D B Type 19 C C C 35

70 Kreatryx Subject Test Power Systems

71 KST- General Instructions during Examination 1. Total Duration of KST is 6 minutes.. The question paper consists of parts. Questions 1-1 carry one mark each and Question 11- carry marks each. 3. The question paper may consist of questions of Multiple Choice Type (MCQ) and Numerical Answer Type. 4. Multiple choice type questions will have four choices against A, B, C, D, out of which only ONE is the correct answer. 5. All questions that are not attempted will result in zero marks. However, wrong answers for multiple choice type questions (MCQ) will result in NEGATIVE marks. For all MCQ questions a wrong answer will result in deduction of 1/3 marks for a 1-mark question and /3 marks for a -mark question. 6. There is NO NEGATIVE MARKING for questions of NUMERICAL ANSWER TYPE. 7. Non-programmable type Calculator is allowed. Charts, graph sheets, and mathematical tables are NOT allowed during the Examination. 1

72 Q.1 A generated station has a maximum demand of 3 MW, a load factor of 6% and plant capacity factor of 5%. The reserve capacity of the plant is (A) 5 MW (B) 4 MW (C) 6 MW (D) 1 MW Q. Consider a step voltage wave of magnitude 1 pu travelling along a loss less transmission line that terminates in a reactor. The voltage magnitude across reactor at the instant the travelling wave reaches the reactor is (A) 1 pu (B) 1pu (C) pu (D) 3 pu Q.3 Consider two buses connected by an impedance of j 5. The bus 1 voltage is 13, and bus voltage is 1. The real and reactive supplied by bus 1, respectively are (A) 1W, 67.9 VAr (C) 76.9 W, 56.7Var (B) 1W, 134 Var (D) 76.9 W, 56.7 Var Q.4 Two power plants interconnected by a tie-line as shown in the above figure have loss formula coefficient B11 = 1 3 MW 1. Power is being dispatched economically with plant 1 as 1 MW and plant as 15 MW. The penalty factory for plants 1 and are respectively. (A) 1 and 1.5 (B) 1.5 and 1 (C) 1 and zero (D) zero and 1 Q.5 A relay is connected to 4/5 CT and set at 15% with primary current of 4 A, the plug setting multiplier is Q.6 In a 14 bus power system network, there are 5 voltage controlled buses. The size of the Jacobian matrix useful for power flow analysis will be: (A) 16 x 16 (B) 3 x 3 (C) 1 x 1 (D) 8 x 8

73 Q.7 Two alternators each having 4% speed regulation are working in parallel. Alternator 1 is rated for 1 MW and alternator is rated for 8 MW when the total load is 1 MW the Loads shared by alternators 1 and would be respectively (A) 4 MW and 6 MW (B) 6 MW and 4 MW (C) 5 MW and 5 MW (D) 1 MW and zero Q.8 A generator is connected to a transformer which feeds another transformer through a short feeder (see. Fig.) The zero sequence impedance value are expressed in pu on a common base and are indicated in fig. The venin equivalent zero sequence impedance at point B is (A).8 + j.6 (B).75 + j. (C).75 + j.5 (D) j.5 Q.9 In the system shown in figure a three phase static capacitive reactor of reactance 1 P.U. per Phase is connected through a switch at motor bus bar. The limit of steady state power, if the switch is closed is (A).49 P.U (B).965 P.U (C) 1.44 P.U. (D).99 P.U Q.1 A synchronous generator is connected to an infinite bus through a lossless double circuit transmission line. The generator is delivering 1. PU power at a load angle of 3 º when a sudden fault reduces peak power that can be transmitted to.5 P.U after clearance of fault, the peck power that can be transmitted becomes 1.5 P.U. The critical clearing angle is (A).35 radians (C) 1.1 radians (B) 1.35 radians (D) 1.1 radians 3

74 Q.11 A three phase star-connected alternator is rated 3MVA, 13.8 KV and has following sequence reactance values: X1.5P.U, X.35P.U, X.1P.U The neutral of alternator is solidly grounded. The alternator line currents when a double lineground fault occurs on its terminals (Assume Alternator is unloaded) will be (A) 3.5 P.U (B).37 P.U (C).68 P.U (D) 4.81 P.U Q.1 For a single circuit transmission line delivering a load of 45 MVA at 13kV and power factor.8 lag. Has transmission line parameters as A = D =.99.3 º, C = º, B = 7 69º sending end line voltage is (A) KV (B) 76.1 KV (C) KV (D) KV Q.13 A 1 MVA, 138 KV alternator has positive, negative and zero sequence reactance of 3%, 4% and 5% respectively. What reactance must be put in the generator so that the fault current for a line to ground fault of zero fault impedance will not exceed the rated line current? (A) 14.8 Ω (B) 1.38 Ω (C) 16.1 Ω (D) 1.8 Ω Q.14 A 3 unsymmetrical spaced transmission line configuration is shown below. What are the values of interline capacitances conductor is 1 cm (A).11,.11, 1.9 (B) 8.44, 8.44, 7.66 (C) 4., 4., 3.83 (D) none of the above C, C and ab bc C ac respectively in nf/km Radius of each Q.15 A 3 bus power system is shown below. The inductance is having P.U. Reactance and capacitance having P.U. susceptance. The Total self-admittances are (A) J 3.8, j 5.7, j 8.5 (B)-j14, -j 1.5, j4.5 (C) j15.5, -15, -6.5 (D)-18.5, ,

75 Q.16 Synchronous generator rated as 6 MW,.8 pf lag 5 Hz, 4-pole has moment of Inertia of 3, kg-. The Angular momentum in MJ-sec/Mech rad is. m (A).8 (B).41 (C)4.71 (D).35 Q.17 1 MVA, 11KV, 3-phase synchronous Generator is working at No load, rated voltage. The positive sequence Sub-transient current of symmetrical fault on general terminal is 5. P.U. The negative sequence reactance is 9% of positive sequence reactance and zero sequence reactance is % of positive sequence reactance. The neutral of the general is solidity grounded. The short circuit MVA of line to ground fault is. (A)5 (B)38 (C)456 (D)715 Q.18 Two conductor cables 1km long loaded distribution system as shown in figure. Both ends are 5v, at end A current is 9A. If minimum allowable voltage to the consumer is 45 and 1.7 cm then the cross section of each conductor in is. cm Q.19 At a 33kv, 5Hz bus a load of 3 Mw at.8 lagging power factor is connected to improve the power factor to unity a synchronous phase modifier is connected across load which is taking a real power of 5 MW. If the synchronous phase modifier is represented as a -connected lossy capacitor bank where each lossy capacitor represented as an RC parallel network then what is R, C values respectively. (A) 17.8,.639F (B) 653.4,.13F (C) 653.4,.639F (D) 17.8,.13F Q. A 6 Hz generator connected directly to an infinite bus operating at a voltage of 1 p.u. has a synchronous reactance of 1.35p.u. The generator no load voltage is 1.1p.u. and its interia constant H is 4 MJ/MVA. The generator is suddenly loaded to 6% its maximum power limit. The frequency of the resulting oscillation of the generator rotor is (A).44 Hz (B).55 Hz (C).88 Hz (D)5.5 Hz 5