Power Systems. (Solutions for Volume 1 Class Room Practice Questions) Z pu new = Z pu old. X pu new = X pu = 50. X pu. new = 0.

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2 ower Systems (Solutions for Volume lass Room ractice Questions) ower Systems - j.75 j. j.5. U System, Symmetrical omponents & Fault Analysis. Sol: Given data, Synchronous generator (or) synchronous motor Z sub transient impedance Z transient impedance Z Steady state Generator: pu (new) 5. 4 =.8 Transformer : pu (new).5 4 =.75 Overhead transmission line: pu 5 =. Transformer : pu new Motor: pu new.5 =.5. =.6 5 N. Ans: Select the base MVA as MVA, Base voltage as KV on the Generator side Base voltage on the line side = kv MVA Z pu new = Z pu old MVA Generator:.new old pu new =.5 Transformer: kv kv old new =.5pu pu new =.9 =.9 pu. Transmission line: pu = 5 () Motor : =.4 pu. pu. new =.8 Motor : pu new =.8 Motor : j.86 j.6 =.4958 pu. =.748 pu.

3 : : ostal oaching Solutions pu new =.8 5 =.975 pu. The per unit reactance diagram of the system can given in below..5pu. Ans: (c) Sol: Given data a = 9p.u o 4 5 p.u b c 9 o p.u magnitude of phase current b in p.u =? b? b b b b a = a a a. = b a a a a b a a a a a a a 9 a.9pu b a o pu.4958pu o o 4 5 o p.u.9pu o.748pu.99 p.u. M M M.99 p.u..99 p.u. b c.97 pu a b a = (4) (8 9) = 85p.u b = = Ans: 9.8 a = A; b = A; c = A K = ; K = 4 ositive sequence component = K K a b c [++4 ] = Ans: a =.5 ka a =, b = 5, c =? a + b + c = c = [ a + b ] = [ + 5] = 5 a [a K b K c ] a a =.5 ka 6. Ans: (b) G

4 : 4 : ower Systems T.5 =.5 Z th = Ans: (c) MVA, kv ''. d = j. p.u =.5 refault voltage, E a = p.u Z = j. Z = j. Z = j.5 Solid -G fault Z f = a j. j.5 j.5 j. Z Ea Z Z j E a Z Z Z j n n g n n a a a V a V a V a a MVA n (p.u) n ( ) (kv a base base )..8 p.u j.5 j. j..8 = j.449 p.u f = a = (j.449) = j 4.47 p.u M K base f (ka) = j 4.47 base =.5 ka 8. Ans: (b) MVA b = 5 MVA kv b = kv Z = j.5 Z = j.8 Z = j p.u p.u 5..7 p.u E R f f f 9.4 p.u actual ka

5 : 5 : ostal oaching Solutions 9. Ans: (b) =., =.4, =.5 Fault current = Rated current d p.u =. p.u E R.. ( n ) = n = n =.75 p.u K V b n ( ).75 MVA b.8.75 = 4.8 MVA. Ans: (i) R =9.54kA; (ii) V R =4.kV eq = eq = j. eq = + n + F =.5 +(.5) +(.5) =.5 E R R eq eq n eq...8p.u (i) R kA (ii) V R = R eq = =.65p.u V R.65 4.kV. Ans: (i) Vn 858Volts (ii) Vn 95Volts j. (i) eq j. 5 j. eq j.5 n eq j. ER R R eq eq eq. 5.p.u j. V n = R n = 5.5 =.75p.u 6.6 Vn Volts j. (ii) eq j. 5 j. eq j.5 eq = + n =j. E R. R R.. eq eq eq V n = R n =..5 =.5 p.u 6.6 Vn.5 95Volts. Ans: f =.96 pu. Two identical generators are operate in parallel and positive sequence reactance diagram is given by figure (a).

6 : 6 : ower Systems G j.8 eq.9 jp.u. where G = positive sequence reactance in p.u. of generator () G = positive sequence reactance in p.u. of generator () Negative sequence reactance diagram is given by figure (b). G eq G j.5.75jp.u. Since the star point of the second generator is isolated. ts zero sequence reactance does not comes into picture. The zero sequence reactance diagram is given by figure (c). G R n Fig (b) G Fig.(a) Fig. G eq eq eq eq = j. + (.) =.99 +.j Now all values are in p.u.,then Rpu. pu. For G Fault, Fault current E ( f ) = R = R eq eq eq f j.9 j.75 j..99 (Assume E R =. p.u.) =.99 j.65 =.87 j.756 f =.96 pu.. Ans: (d) Z = j.+j. = j.; Z = j.+ j. = j. Z n =.5 Z Z Z a l Zg Zl Zg Z Ea Z Z Z j. j..4j j.5 For G fault = j. (pu) 6 B (Base urrent) = 6.6 = 75 Amp f (fault current) = ( ac ) B = j A Neutral voltage V. Z N f n n

7 : 7 : ostal oaching Solutions where Z n Z B.5 =.89 V N = = 64. Volts a = b f 4. Ans: 7 ka = = j., f = j.5 a E f = 4pu j. j. j.5 j.5 Figure (a) c Figure (a) a = b c Figure (a) fault 7 ka 6.6 a = a 5. Ans: F = 7.57kA Sol:.5..5 eq eq = x eq =.6 F F eq E R eq kA 6. Ans: V ab =. kv eq =. p.u., eq =. p.u. and Alternator neutral is solidly grounded ( n = ) eq E a From figure (a), b = c From figure (b), a = a ositive sequence current E a a = eq V a eq a eq V a Figure (b) Sequence Network with respective Fig.(a) (assume pre-fault voltage E a = pu.) ositive sequence current j a = = j pu. j. j. Negative sequence current ( a ) = a = j pu.

8 : 8 : ower Systems A zero sequence current doesn t exists in - fault because this fault is not associated with the ground a =. n this fault, fault current ( f ) = b = c b = b + b + b = + K a + K a ( a = a ) = (K K) a = [(.5 j.8667) (.5 + j.8667)] a = j.7 a E a b = a = eq eq =.464 p.u. Fault current ( f ) = b = c =.464 pu. Base MVA Base current = Base voltage =.6 pu. But we know that V b = V c V b = V c =.6 ine voltages, V ab = V a V b =. (.6) =.8 p.u. V bc = V b V c = p.u. V ca = V c V a =.6 (.) =.8 p.u.. V ab =.8 =. kv, 7. Ans: f = 4.8 p.u f amp =. ka # G 6 5 = = 9.4A. Fault current in amps, f actual = f pu base = = 787.5A. V a = E a a eq = + j ( j) (j.).4 =.6 p.u. V a = a eq = (j)(.j) =.6pu V a = V a =.6 pu For hase a, V a = V a + V a + V a (V a = ) = V a =.6 =. pu. For hase b, V b = V a + V a + V a = ( k k )V a (V a = V a ) = ( j) + ( j)V a.9 refault voltage. 5. urrent through ground = Fault current f = a a a a eq eq E a eq...().5.(.5.8).. (.5.8) =.4 Substitute a value in equation (). a.4 =.59. (.5.8) f = a =.59 = p.u

9 : 9 : ostal oaching Solutions f amp ka. ka.. 8. Ans: R 6.kA j. j. j.6 eq eq = eq = j.6 eq = + n + = j.5 + (j.5)+j.=j.5 ER R eqeq eq R eq p.u.55 eq 6.kA Ans: (c) Sol: Equivalent reactance seen from the fault point j. j.8 j. j.8 U j. j. j.8 j.8 j. = j.4 Fault level current = / (U) = /j.4 = j8.87. Ans: (c) Sol: S MVA Base MVA eq G. eq.. New.6. 8 eq S MVA. Ans: (b) Sol: G (/9) 9 New on 5 MVA Base 5. [] f f G E. n eq. R 9 MVA.5 p.u 8. p.u kA fg = 8.. = 5.5 n.6 eq =. p.u

10 : : ower Systems 5 fg(actual) = 5 =.9 ka. Ans: f =.4 pu er unit positive sequence reactance diagram of the given system when the breaker closed is shown in fig. = pu.8 Vth Fault current f eq. =.4 pu.875. ower System Stability pu The equivalent reactance with respect to point is [short circuit.u sources] // eq T d l =.4..7 Given prefault voltage (V th ) = pu. Vth Fault current f eq 5 pu = 5(. +.4) =.p.u To find fault level at bus : The equivalent reactance w.r.t. point in reactance diagram is 4.pu T d =. pu eq d T l // = (.+.+.) //. eq Fig. pu l eq pu. Ans:.54k N-m H = 9 kw sec/kva K.E = stored? K.E stroed nertia constant H rating of the machine K.E stored = H S = 9 MVA = 8 MW sec 8 MJ Accelerating torque T a =? a = T a T a a a = s e s = = 998 kw a = = 698 kw 698 T a 5 =.54 kn m. 6. Ans: (c) N s =, f = 6 Hz, S cos 6 MW MVA

11 s H due to moment of nertia, there S is no sudden change in angular velocity 6 NS (88) = 6.5 MJ/MVA SH M 8f 85. Ans: 4 MJ/MVA Generator A n = 4 H eq = 9 4 =6 J/MVA H A New H eq = H Anew + H BNew = = 4 MJ/MVA Ans: f n =.5 Hz Since the system is operating initially under steady state condition, a small perturbation in power will make the rotor oscillate. The natural frequency of oscillation is given by f n 6 5 = 4 dp e d M Generator B n = H eq = 4 = J/MVA H B New 5 = 6 : : ostal oaching Solutions As load increases, load angle () increases, there by Sin increases. sin loading At 6% of loading sin =.6 = 6.86 EV We know that e sin, where E = no-load voltage, V = load voltage de EV cos d. cos SH Moment of inertia M=, f where S = Rating of the machine, f = frequency, nertia constant, H = MW-sec/MVA ( Assume rating of machine pu.) = 5 5 The natural frequency of oscillation at 6% loading, f n de d 5 =.76 M 9.6 = Hz.5Hz / 9.6 rad/sec 5. Ans: (i) KE = 8 MJ; (ii) = 7.5 elec.deg/sec (iii) = 6.75 elec.degree/sec

12 : : ower Systems p = 4, f = 5Hz, G = MVA, H = 8 sec (i) K.E Stored GH= 8 8 MJ (ii) d m p dt p a s a p p = 8 5 = d m (accelecation) dt d dt m e GH 8 M =. 88 8f 8 5 d dt Elec.degree/s = 7.5 mech deg/ s p = =68.7 mech deg /s (t) = t cycles t =. s ec 5 (t) = t (.) = t (.) = 7.5 (.) = 6.74 elec. degree/sec Speed of the motor at end of the cycles. f Before disturbances speed (N s ) = p Speed of the cycle = 5 5 = 4 = mech deg/ s =.94 mech rad/s (iii) cycles- Acceleration maintained constant d mean constant change in angle after dt sec d dt = 7.5 elec. degree/sec d t dt = k Before giving distance, at t= = = () k k = N (t) = N + dn t dt (N =N S ) =5+ dn.... () dt d d dt dt d N dt 6 dn dt 6 d.97 dt () Equation () substitute equation () N(t) = N(t) = rpm 6. Ans: 7 deg

13 : : ostal oaching Solutions E =.pu V =.pu Assuming inertia constant (H) = pu EV sin = j.5+ j.5 = j.pu sin EV j =5.8 GH 4 M. pu f.. a deg/sec t deg Rotor angle = + = =7 deg 7. Ans: cr = 7.6 =, m =.5, m =.5, s =. (rad) =.5 s max 8 sin m 8 sin..5 max = = 8.8 max 8.8 = cos8.8.5cos cos c cos cos.887 = cos [ ].75.4 = cos [.65] = Ans: cr = 55 s =. p.u m =.76.u eq =.7 p.u eq =. p.u eq =. p.u EV m EV m = m r where m EV EV m = m r where r r Substitute these values tot get m & m m m =.45 sin s m = 5.7 =.64 rad

14 : 4 : ower Systems max cr cr 8 sin s m = 6.56 =.8 rad cos s max m m cos p m max m cos cos cos5.7 cos cr = Ans: cr = 88 cr s =.4 m = 6 m m m =.8 m = m.67 s sin.8 = sin (.4) =.578 m 8 sin cos s m.4 m 8 sin = 5.8m s max m m cos m max m cos.4m m cos5.67m cos.578 cos 4.8 m.67m cr = 88. Ans: c = 65 s = e =. e =. sin m =. e =, m = m =.75. =.65 m m c sin s m sin sin 8 sin s m rad 8 cos s m.(.48 cos m m cos m.47).65 cos(4.7) c cos.65 cos Ans: c = 84. s e e.sin m., e m m m. = 7 (rad)=.47 m

15 : 5 : ostal oaching Solutions c c cos(5) cos cos. c = 84. Ans: c = s e e m sin = p.u =, (rad) =.5, e m =. m m m = 8 = 5 (rad)= c c cos.6.5. cos(5) cos 8. m c cos 87.7.(.7.4).4cos(55.4).4 4. Ans:.68 sec S =., H = 5, = 68.5, =, s =. t t c c t c M( c ) s SH ( ) f ( ) s.5(68.5 ) sec 5. Ans: e =.7 pu Sol: E =.5 E =. =. =.5 =.8. Ans: c = 87.7 s =. m e...4.5,.4 m m m. sin m 8 sin EE e sin( ) e eq.5. sin 5. =.7 p.u 6. Ans: ermissible increase = 6.4 S =.5 p.u. max = 5. p.u.

16 : 6 : ower Systems d Before fault, =, a = dt s = e s = max sin = sin max = p.u. max = 4 p.u. max = 8 sin = 8 sin = s max.5 = sin 5 =.5 rad s.5 4 max max = rad.cos cos cos 8 s max o max max max max max o 4.cos4. cos = os c = 6 c = os ( 6 ) 9.4 ermissible increases = c 7. Ans: (d) V =.pu.pu T = 9.4 = 6.4.pu E =.pu.pu when one of the double circuit tripped, then m 5pu. x. 8. Ans: (c) Sol: Before fault Mechanical input to alternator ( s ) = electrical output ( e ) =..u. Given =, V =..u During fault eq = pu.8 E =. p.u, V =..u value cannot change instantaneously. nitial accelerating power ( a ) = s e.. a =. sin.8 a =.56.u. oad Flow Studies. Ans: (a) Y j ; y = Y = j z j. y V=.pu

17 : 7 : ostal oaching Solutions. Ans: (c) Sol: Y = y + y = (j.) + (j.5) = j 7 Y = y + y = (j.5) + (j.5) = j 6 Y = y + y = (j.) + (j.5) = j 9. Ans: (a) Sol: Y = Y = (j.5) + (j.) = j Y = Y = (j.) = j 4. Ans: (b) We know that Y = y + y +y Y = y Y = y From the data, Y = 8, Y =, Y = Y =? 8 = () + y + () y = 8 Shunt Susceptance, y =. 5. Ans: Y = j.8 Sol: Y Bus 4.4 j 5 j.5 j..5.5 j Y Y Y Y 4.4 Y Y = Y = j Y = Y = j.5 Y = Y = j5 Y j4.4 j Y = j...() Similarly Y j.5 j Y = j..() Y Y j = j.4.() Y j. () Y Subtracting () and () Y Y Y j Y Y Y j. 4 j5 j.5 j.4 (4) Solving equation () & (4) we get Y = j.8 6. Ans: (i) (ii) (iii) Y bus Y bus Y bus 4.76 j j 4.88 j 5 Sol: (i) z = j.=j. y = j z = j. = j

18 : 8 : ower Systems y BUS y = j5 y = j.5 = j.5 y = j4 y y y j.6 j.6 j.6 j. j.6 5 j.4 y y Y y y j j5 j.8 j.6 = j 4.76 y y Y y y j j4 j.8 j. = j.7 y y Y y y j5 j4 j.6 j. = j8.64 Y y j Y y j5, Y y 4.76 j 5.7 (ii) z = j.5 j.5 4 y = j y = j.5 = j. y = j z = j.5 5 = j.5 y = j8 y y y j.6 j.6 j.6 j. j.6 5 j.4 j4 y y Y y y = j j + j.8 + j.6 = j9.76 y y Y y y = j j8 + j.6 + j. = j7.64 Y = y = j; Y = y = j; Y = y = j8 y BUS 9.76 j (iii) z =. = j. y = j z = j. = j. y = j5 z = j. 5 = j.5 y = j4 y y y j.8 j.8 j.8 j.6 j.8 5 j. y y Y y y = j j5 + j.4 + j.8 = j4.88 y y Y y y = j j4 + j.4 + j. =.86 y y Y y y = j5 j4 + j.4 + j. = j8.8

19 : 9 : ostal oaching Solutions Y y j; Y = y = j5; Y = y = j4 Y BUS 4.88 j ower Systems -. Transmission & Distribution. Basic oncepts &. Transmission ine onstants: 7. Ans: (b) y = y = j5 y = y = j5 Y = y + y = j5 Y = y + y = j5 Y = Y = Y = Y = y = j5 Y = Y = y = j5 Y = y + y = j5 j5 = j. 8. Ans: 5 (5 to5) Number of Buses (N) = Number of non- zero elements = 8 =N+N (N =Number of transmission lines) + N = 8 N = 5 Minimum number of transmission lines and transformers = 5. Ans: n For same length, same material, same power loss and same power transfer f the voltage is increased by n times, what will happen to area of cross section of conductor. oss = oss R oss V cos oss oss V cos oss R V av V cos R av oss av = constant oss = onstant a a V V V n given V R

20 : : ower Systems a n a n this efficiency is constant since same power loss.. Ans: (b) We know that = Vcos () ( V cos) ower loss = R a = a () Substitute eq () in eq. () Vcos a K a V cos a V cos volume (V cos) R a ( volume area) R y B n V 4, R.5 R 4 V V n R 9 V 9 V V V R V V cos V cos R V sin V sin... (i) R = 7.9 mh = 9. F f suppose on phase B, on phase R V A V V + B. Ans: (b) R B y R V n A V V + B V R V V R cos cos R

21 : : ostal oaching Solutions sin sin st condition never be zero, because all the positive parts never becomes zero 4. Ans: (b) Self-inductance of a long cylindrical conductor due to its internal flux linkages is kh/m. r r r a ln ln 8 r d self mutual int self 8 due to r int r mutualdue toext ext self ln r r due to ln d Ans: K H/m ( st term is independent of diameter) 5. Ans:.6% n =. mh/km increased 5% n d. n mh / km r.mh / km d.. n r.. d n r d. n mh / km r ext d 5.5 = n r 5.5 e d r r = d d..5. n r d.55. n r.55. d e r.4r = d d d d.4r 44.69r 44.69r =.65 =.6% 6. Ans: (b) d = 4; (i) n After Transposition GMD (ii) n After Transposition GMD 4 48 GMD < GMD < n > n Resistances R = R Z 5. m

22 : : ower Systems n /.5.5. r x r x r x Z Z S V Z S n V Z r y / 7. Ans: (b) The impedance of a Transmission line Z =.5+.5/phase/km Spacing is doubled d =d ; R=.5 radius is doubled r = r =.5 /phase/km GMD l ln GMR l remain constant.5 f =.5 = f B let R. ;R A r R r R R R r R (z ) new =.5+j.5/km. 8. Ans: (c) r x =.m r y =.4m GMD system = GMD a. GMD b r y GMD d / 6 d d d d d a = ( ) /6 =.89m GMD b = GMD a =.89 GMD system = =.89 m. GMD GMD (Self GMD) system ( self GMDof system a) self GMD b self GMD a r.5 r.5.5 r.5 x x = (.7788 (.) (.5) ) /9 =.76m r Self GMD b / 4 y. ry =.96m SelfGMD =.6m GMD.ln mh / km GMR.89.4ln.6 =.9 7 H/m 9. Ans: d =.74 m r =.5cm =. mh/km 6 a H / m d d GMD d.599 d. n d =.74 m x b /9

23 : : ostal oaching Solutions. Ans:.5 nf/km f = 5Hz, d =.4m, r =.m v = kv r GMD n GMR = 6 n. = 9.75nF/km nterline capacitance =.5nF / km. Ans:.94 Self GMD = kr Self GMD = R RR 9.75 =.7788R RR = R.7788 kr =.94 R k =.94. Steady state performance analysis Of Transmission lines. Ans: (c) A = D =.96 + j.6 =.96.98, B =.5 + j8 = , = ( j94) 6, V r = 5 MW, p.f =.9 lag, V S (-) =? V S ph V r ph r r ph V Sph A V r ph B kv r V cos r r ph 5 M = 45.7 A k cos (.9) = k (.96.98) (476.4)( ) =.47.7 kv V S ( ).4.6kV V R Vs A kV. Ans: (c) oad delivered at nominal rating V rl = kv VS Vr A % V.R % V r 4.94 % = 6%. Ans: (c) A = D =.95.7 ; B = =.69 ; V S = V r = 8 kv

24 : 4 : ower Systems R, Y are neglected VS Vr max n nominal- B = Z Z = = +j9 = 9 88 max.6 MW 9 4. Ans: 8.4 kw A = B = V = kv AD-B = AD B V ` = S = V r +B r = V cos cos = 8.4kW 5. Ans: (b) omplex power delivered by load: S = V = (6) (5) = = j 5 VA omplex power absorbed by load S load = j 5 VA Ans: (b) i.e., load absorbs both real and reactive power. 6. Ans:.96 lag Short transmission line having impedance = + j5 cos Vs Vr cos B Source 6kV os(68.8 ) =.675 Vs Vr Q sin B =5 +j5 AV B r cos 6 6kV O A D oad MW cos 68. =.9 AV B V=6 cos(68.) r sin MVAR

25 : 5 : ostal oaching Solutions sin sin MW.4 + = Q c Q c =.754 MW cos =.959 lag.96 lag 7. Ans: (a) f = 5 Hz Surge impedance Z = Velocity of wave 5 V f =.9 y = [fc] l = 5 4 = Ans: (b) V s = V r =, =.5, V V S r Real power r sin.. sin.5 sin (.5) Reactive power ( VS )( Vr ) ( V ) Qr os ( ).. cos But Q r + Q = 9. Ans: (c) Sol: V =. +jq G G Q = Q r =.68 p.u = Active power sent by bus () V V sin = G Q G Q = Active power received by bus () j. Q loss V =. Q G G 5+jQ G load + jq load =5+j5 load + jq load = +j Q G

26 : 6 : ower Systems V V = sin Q = Reactive power sent by bus () V = V V cos Q = Reactive power received by bus () V cos V = V Active power balance at bus (): Active power balance at bus : G = + load + G = load = = = 5 = 5 VV sin 5. sin =.5 = V Q V V cos Q sin 5 V V cos V. cos cos. =.4 pu =.4 pu o o =.68 pu Q loss =.68 pu Reactive power balance at bus (): Reactive power balance at bus (): Q G = Q + Q load Q + Q G = Q load Q G = Q G = (.4) Q G = 6.4 pu Q G =.4 pu Q G =6.4pu, Q G =.4pu, Q loss =.68 pu.4. Transient Analysis &.5. Wave Traveling Analysis. Ans: (c) et l be the total length of line Total reactance of line =.45p.u. = f.45 Total inductance of line = 5 Total susceptance of line =.p.u = f Total capacitance of line = 5.45 nductance/km = 5 l. apacitance/km = 5 l Velocity wave propagation (V) = km km Q line = Q loss = Q Q =.4 (.4)

27 : 7 : ostal oaching Solutions V = l 5 l l 5 = ength of the line (l) = km. Ans: (c) Sol: Since load impedance is equal to surge impedance, the voltage & current wave forms are not going to experience any reflection. Hence reflection coefficient is zero. V reflection = i reflection =.. Ans: (c) Sol: V. pu A Reactor The transmitted (or) refracted voltage V = V Z Z Z Here indicates that the voltage V is calculating in transient condition V = 5 4 V = kv 5. Ans: (b) cable =.85 mh/km cable =.85 F/km ine =.4 mh ine =.87 F/km Z (able) Z (ine) Z s The Reactor is initially open circuit V = V + V =. +. =. p.u V = reflected voltage V = Switched voltage 4. Ans: (b) V = 5 kv, Z =, Z = 4,.4.87 V Z V Z Z kv = 8.7 kv 6. Ans: (d) Sol: A short length of cable is connected between dead-end tower and sub-station at the end of

28 : 8 : ower Systems a transmission line. This of the following will decrease, when voltage wave is entering from overhead to cable is (i) Velocity of propagation of voltage wave. (ii) Steepness of voltage wave. (iii) Magnitude of voltage wave. Dead end Velocity of propagation V (ine) = 8 V (able) = V able > V (OH line) 7. Ans:.9 kv Sol: kv V r 8 m / s J J A B Z A = 5 Z B = 7 V V V V 4 able Surge absorbers (or) rotect T/F from Travelling wave. S/S Z = 6 V 5 V V ZB Z Z B 7 k 75 V = 4.9 kv V (Re flection of V ) V V 4 6 V 4 A Z Z Z Z B B kv 6 7 Z Z A A Z Z B B K.9 kv Ans: (d) Sol: Given data V 6 =.9 5 V7 V4 57 = 6.8 kv 6 V9 V V 67 V 7 V 6 V V 8 D (or) step voltage (line is of infinite length) V 9.6. Voltage ontrol. Ans: (a) A = D =.9 B = 9 =.95 9

29 : 9 : ostal oaching Solutions V S Q reactive =? Without shunt reactor VS Vro A By adding shunt reactor V ro = V S R = (no load) Q R = Q reactor VS Vro A sin( ) Vro sin( ) B B Vr Q r At V ro = V S A sin( ) sin( ) B B To get at ( V ro = V S ) VS r B A cos( ) V B = cos( ) A cos ( ) = cos(9 ).9 cos (9 ) cos (9 ) = sin =, = sin(9 A ) = or k B D S.9 sin(9. Ans: (d) = Q = tan (6.86) = (.749)=499.46kW R + j Q R cos( ) ) V ro R(S) s-motor = j S Total = S S m s m = ( + j499.46) + ( j) = + j cos %.986lag 4.9. Ans: (a) M = 4 V, 5 Hz, pf =.6 lag, i/p = 4.5 kva p.f =.6 load total supply =? M 4V, 5Hz S V ; 4.5 kva Q Sh (-) = (tan tan ) = Real power drawn by M = M = S M cos M = kw =.7 kw Q sh (-) =.7[tan(cos.6) tan(cos.8)] =.575 kvar.575 Q S/ ph kvar.55 kvar V Reactive power supplied S 55 (4) (5) = 55 =. F

30 : : ower Systems 4. Ans: (c) Sol: Given data A =.85 5 = 5 B = 75 = 75 ower demand by the load = 5 MW at upf D = R = 5 MW Q D = ower at receiving end V V A R B B s R R = cos V cos cos = 8.46 o cos7 V V A R B s R So Q R = sin V cos sin sin 7 = 7.56 MVAR n order to maintain 75 kv at receiving end Q R = 7.56 MVAR must be drawn along with the real power. So Q = Q = 7.56 MVAR So compensation equipment must be feed in to 7.56 MVAR to the line. 5. Ans: (a) th =.5 pu ; 5 MVA, kv V th =kv th To boost the voltage 4 kv shunt capacitor is used. V V Q sh ap S Q sh ap V VS (kvbase) pu MVA base ( ) kk Q sh ap 8.8 kvar 48.4 To reduce voltage by kv, shunt reactor is used. V Qsh nd V Q sh nd S kk 9.9 MVAr Ans: (d) V =. V F =.9f Reactive power absorbed by reactor V = V Q MVAr f Then reactive power absorbed V V Q f Q Q V V.V V f f f.9f

31 : : ostal oaching Solutions..9 = Q. = 4.4 MVAr.9 7. Ans: (c) et characteristic impedance (Z c ) = = Z Y sc oc =.. impedance / km admittance / km = p.u. Given that for a given line series capacitive compensation is provided. Hence the series impedance of line is.7 or (7) of original value. Z new =.7. =.86 p.u. Surge impedance loading (S) = S S S S = Z c Z = Z c c = 75 6 = 75MW. 6 V Z 8. Ans: (b) Sol: phase, kv, 5Hz, kw load, at power factor =.8 kvar demand of oad (Q ) = sincos.8.8 Q = 5 kvar c kvar demand of load at upf = So as to operate the load at upf, we have to supply the 5 kvar by using capacitor bank. kvar rating of - connected V ph capacitor bank = = 5 kvar ph ph ph 5 = 4 4 f = 54 =.5 F.6 F 9. Ans: (c) Sol: Given Data: et the initial power factor angle = After connecting a capacitor, the power factor angle = Given = cos.97 = 4.7 (tan tan ) = kvar supplied by capacitor 4 6 (tan tan4.7) = 6 = 6.89 cos =.8 lag Hence if the capacitor goes out of service the load power factor becomes.8 lag

32 : : ower Systems. Ans: (d) Sol: The appearance will inject leading VArs into the system is induction generator, under excited synchronous generator, under excited synchronous motor and induction motor..7. Under ground cables. = 5 km =. F/km E r =.5 core d =.5 cm V = 66 kv, 5Hz = f D =? E r(rms) =? c(rms) =? (a) oncentric cable: core a placed exactly of the center of the cable h = r F / M ln(d / d) =. 6.. = D ln( ) d D ln d. = 9.7 D ln. 97 d D =.977cm (b) E r(rms) = V R r ln r =.75 E rms = 9.4 kv/cm R D r d ln.5 (c) At charging current = l = =.7A. Ans: (b) V = kv; =.6 F; =.96 F (i) =.6 F (given) From network = S + S + =.6 F.. () D d.97 e.97 D = d e =.5 e.97

33 : : ostal oaching Solutions (ii) Sheath S S S S S Sheath S =.96 F (given) From network = S.96 F S =. F From (). + =.6 S Sheath =.4 F Effective capacitance from core to neutral /ph = S + eq (s ) (s ) s s s =. +.4 =.74 F. Ans: (b) c =.5 F s =. F eq s c.5..f eq S S S 4. Ans: 8.kW Sol: Given data = 4 km -core ground cable =.77kVAr/km F = 5Hz Dielectric material is.5 cos =.5 = cos - (.5)

34 : 4 : ower Systems = Q tan.77 4 Tan = 8. kw. Ans: (b) V = 7.5 kv = /8 ' 8 V 5. Ans: (a) =. 6 F =.4 6 F f = 5 Hz V = kv /ph = + = = 6 = F. erphase charging current = V ph ph = 6 5 = A..8. Overhead line nsulators V + V = V V = ( + K) V V 7.5 kv K 8 V V = 5.55 kv V = V + V =.5 kv. Ans: (b) V = kv f = 5 Hz ine V = 7.5kV. Ans: (d) n = ; - ; 4 kv ; = 8% string Vph nv 4k /.8 V V 5 kv string V V V V V kv, 5Hz - V ( K) V V ( K)

35 : 5 : ostal oaching Solutions K () 4 75% 4. Ans: (b) f = 5 Hz V = kv apacitance of insulators is 5 times the shunt capacitance between the link and the ground. V p R S..5 A A A 5A V q. Q et V D be the drop of voltage in line Applying KV, V V D V Q = V V Q = V D 5 e V D = V p V Q = V e = e ( + K) e + e = 5 e But V D = ( ).+( ).5 +( 6). = = = 45.55A.45 V D = Here we have to take magnitude only K = = =. 5 5 V D = 8.77 V = = 8.7V e ( + K) + e = V Q = V = 5.7V. e ( + K) = e = kv e = e ( + K) = =.46 kv... Distribution Systems. Ans: (a). Ans: (d) All the loads are at unity factor. et us take current in 4 m section as such that currents in remaining sections are shown. Assume that loop resistance feeder rω /m (reactance is neglected). S 4m m m m S 4V 5Hz A A A 4V 5Hz

36 : 6 : ower Systems KV From S and S is given as V S V S = ( 4 r) + ( ) ( r) + ( )(r) + ( 5) (r) = = A as = A, ontribution to load at point from source S is A from source S is A. Ans: Sol: Given Data: V r = s = = V s = V r + V V = (8 6.86) (.5+j.) + ( ) (.5+j.) = V s = = F.= cos (angle between V s and sc ) = os (4.7) =.74 lag ower Systems -. ircuit Breakers f r = 9 khz 5.. Ans: (b) = A, =. 6 F, = H i V i = V V i = kv.. Ans: (a) Maximum voltage across circuit breakers contacts at current zero point = Maximum value of Restriking voltage (V max ) V rmax = ARV ARV = K K K V max sin K = No Armature reaction K = Assuming fault as grounded fault K = ARV/phase 7. V max 6 6. Ans: (a) = 5 H =. 6 F v i B F Fault F os = Sin =

37 V r max 7. = 8.8kV : 7 : ostal oaching Solutions A.R.V =.5 sin = 96.7kV 4. Ans: (d) Sol: Making current =.55 B.55 = 44.5 ka 5 5. Ans: (a) Sol: For -, breaking current MVA 5kV = 8 ka Making current =.55[8 ka] = 4 ka 6. Ans: (c) Sol: R.5 5mH.5 5.5H 7. Ans: (c) Sol: A.R.V = K K V m sin K first pole clearing factor K =.5 ( fault) K Due to armature reaction K = (Armature reaction not given) - p.f angle of the fault cos =.8 = 6.86 o V m = maximum value of phase voltage of the system kv V m = 8. Ans: (b) Sol: AR is initiated at the instant of contact separation due to high field gradient (or) field ionization properties of the Arc is column of ionized gases. 9. Ans: (b) Sol: High resistance method of Arc interruption, it is resistance is increased as to reduce the current to a value insufficient to maintain the arc. When current is interrupted the energy associated with its magnetic field appears in the form of electrostatic energy. A high voltage appears across the contact of circuit breaker. f this voltage is very high and more than with standing capacity of the gap between the contacts, the Arc will strike again.. Ans: (d) Sol: When interrupting a low inductive current (shunt reactor (or) magnetizing current of Transformer) the current become abruptly zero well before natural zero instant this phenomenon known as current chopping. A current chopping phenomenon is very severe during the interruption of low magnetizing current.

38 : 8 : ower Systems. rotective Relays operating time of relay must be less than zone operating time.( TZ R TZ R ). Ans: (d) Sol: Relay current setting = 5% primary current (fault current) SM relay current setting T ratio Ans: (c) Sol: The minimum value of current required for relay operation is the plug setting value of current. Minimum value of negative sequence urrent required for relay operation 5 =. A But for a line to line fault R R And fault current f R.7A Minimum fault current required =.7 A.. Ans: (a) Sol: From figure, it is clear that zone of relay and relay are overlapped. f there is a fault in overlapped section (line), the fault should be clear by relay. Hence zone 4. Ans: (b) 4 Sol: ; i 66 6 i i 5 : Ans: (b) Sol: The active power restrained over current relay will have characteristics in R- plane. Operates 6. Ans: (b) Sol: T ratio = 4/5 = 8 Relay current setting = 5% of 5A =.5 5A =.5A r imary current (fault current) SM Relay current setting T ratio 5.58 The operating time from given table at SM 5 is.4 the operating time for TMS of.5 will be.5.4 =.7 sec R

39 : 9 : ostal oaching Solutions 8. Ans: (c) Sol: Mho relay is selected for long Transmission line should be less affected due to power swings. mpedance of long line is very high effect of AR resistance. 9. Ans: (a) Sol: The angle between voltage coil voltage and voltage coil current is adjusted with the help of phase shifting network so it is possible to adjust the maximum torque angle. V = 45, maximum torque angle = 45, the relay operated torque is 7.7% of maximum torque.. Ans: (a) Sol: NO O O NO NO O O NO R R= The operation of relay depends only on reactance seen by the relay. Reactance relay is not affected due to Arc resistance, occupies more space on R diagram. ower Systems - 4. Fundamentals of power economics. Ans:.(i) = MW, = MW, =8.4 Rs/MWhr (ii) = 5 MW, =5 MW, =.5 Rs/MWhr (iii) = 6.66 MW, =86.64 MW, = 6.6 Rs/MWhr (iv) 4 MW; (v) 7 MW;(vi) 5 MW (vii) 5 MW Sol: (i) =? =? G G D = 4 MW As per equality constraints + = 4...() From the coordination equation c = c. + = = 4. () Solving () & () =.66 MW = 6.66 MW G violated its min power limit = min = MW = 4 = MW =. + = Rs/MWhr =. + 6 = 8.4 Rs/MWhr always decided by the unit (or) group of units which are participated in the economic dispatch. = 8.4 Rs/MWhr

40 : 4 : ower Systems (ii) D = 5 MW + = 5 MW. (). + = = 4. (4) Solving () and (4) =8. MW, =.8 MW G is violating its maximum power limit. = 5 MW = 5 5 = 5 MW =. 5 + =.5 Rs/MWhr = = Rs/MWhr is always decided by the unit or group of units which are participated in the economic dispatch. =.5 Rs/MWhr (iii) D = 5 MW + = 5. (5). + = = 4 (6) Solving (5) and (6) = 6.6 MW = 86.6 MW =. + = 6.6 Rs/MWhr =. + 6 = 6.6 Rs/MWhr = 6.6 Rs/MWhr (iv) D min = (min) + (min).. N (min) D (min) = 4 MW (v) D(min) economic dispatch D (min) ED: t is the minimum demand on two generators, such that both generators operate at the economic dispatch. To solve for D min ED Solve for D = D min alculate, and, max = max {, } = max =? = max =? D min ED: D = 4, = ; = =, = 8.4 (max) = Rs/MWhr = max. + = = MW. + 6 = = 5 MW D min ED = + 5 = 7 MW Region c = c G G 5 5 vi) D max = max + max = = 5 MW vii) To solve for D max ED Solve for D = D max alculate, and, min = min {, } = min =? = min =? D max ED = + D max ED D = D max = 5 MW = 5 MW ; = 5 MW

41 : 4 : ostal oaching Solutions =.5 = min = =. + = = MW = = 5 MW D (max) ED = + 5 = 5 MW ( ) MW 5 as per Dmin as per Dmin E.D as per Dmax & Dmax E.D ( ) as per 5 Dmax E.D as per Dmin & Dmin ED. as per Dmax MW. Ans: G =.44 MW G = 56.5 MW G =.98 MW df Sol:.5 6 d df d df d = 5 MW Optimal generation schedule = = = = 6.; = 8.68 G =.5(8.68) +6(8.68) =.44 MW G = (8.68) +4(8.68)4 = 56.5 MW G =.5(8.68) +5(8.68)9 =.98 MW G = + High cost generators violates their min. power limits All the generators satisfies their min. &max. power limits 4MW 7MW 5MW Dmin Dmin ED Dmax E.D ower cost generators violates their max. power limit 5MW D MW. Ans: = 6.6 MW = 7. MW Sol: (G ).6G 8G 5 MW G 65 MW (G ).9G 7 G 5MW G 5 MW D = 6 MW d d. G 8 4

42 : 4 : ower Systems d d..8 G.8 7 [8.88] = = Ans: (c) MW 7.MW Ans: = 8 MW = 5 MW Sol:. Rs / hr 8 F F.5 4 5Rs/ hr 4 df.4 Rs/MWhr d df d Rs/MWhr = 85 = MW MW.5 Note: Here Generator is violating upper limit which cannot be allowed instead it is fix to generate 8 MW and remaining rest of the load is shared by unit D = + = 8 + = 5 MW = 8 MW, = 5 MW 6. Ans: 7976 Rs/annum Sol: Given: Alternators capacity = MW oad = MW df. RS/MWhr d df. 5 RS/MWhr d When the load is economically divided between two generators df df d d.. = 5.. () + = () Solving () and () = 4.9 MW and = 59.9 MW..5 F x x Rs/hr F. 5 y.6 5 y Rs/hr Substitute and values in the above equation F =.5(4.9) x Rs/hr F =.6(589.9) y Rs/hr (F +F ) Economic = (8.98+x+94.9+y) = ( x + y) Rs/hr When the load is equally shared = 5 MW, = 5 MW

43 : 4 : ostal oaching Solutions Substitute and values F and F F =.5(5) x Rs/hr F =.6(5) y Rs/hr (F +F ) equals = (775+x+y) Rs/hr saving = (F + F ) equal (F + F ) economic = 9. Rs/hr savings in fuel cost per annum = Rs/annum = 7976 Rs/annum. 7. Ans: (c) Sol: ncremental fuel cost of generator A for maximum power generation = 6 Rs/ MWhr ncremental fuel cost of generator B for minimum power generation = 65 Rs / MWhr As the incremental fuel cost for maximum generation of generator A is less than t the incremental fuel cost for minimum generation of generator B is Hence we can operate the generator A at its maximum output of 45 MW and the remaining will be generated by generator B. 8. Ans: (c) Sol: df = b + RS/MWhr d df = b + 4 RS/MWhr d For most economic generation df df = d d = Given + = + = = MW, = MW. 9.Ans: = MW, = 5 MW = 9 MW Sol: ~ B B. + = d = 6 B 6 ~ d = 6MW. 6.. () Assuming lossless problem d d. 6.4 d d 4+. = =.5. () Substitute () in () +.5. = = = = MW =.5 = 5 MW =. = 9 MW b + = b + 4

44 : 44 : ower Systems. Ans: =. MW Sol: = MW d = 5.5 MW # ~ ~ # = 5 df df d d = 5 59 MW.6 df d (. d 7) d d (. +7) = 5( B ) B = B () B = = (. +7) = 5(. ). + (.5 ) = = 5 7 =. MW ower received = d = + =. + (.)(.) = 5.5 MW. Ans: (b) Sol: - - ~ ~ B = B = B =, B ( B ).5. Ans: (c) Sol: = ~ ~ D = 5 MW = 8 MW = 8 5 = MW p /8.75 enalty factor =. 6. Ans: =.565, =.5 dc Sol:.5 5 Rs/MWhr dp dc dp.5 MW d. d 75 Rs/MWhr =

45 : 45 : ostal oaching Solutions =. =.5 D R.5 AFR 4 MW / Hz 4 MW / Hz c = c = = Ans: (b) 5. Ans: (c) 6. Ans: (a) 7. Ans: (a) Sol: Unit commitment is optimally out of the available generating sources to meet the expected load and provide a specified margin of operating reserve over a specified period of time. 8. Ans: (d). oad frequency control. Ans: (c) Nominal frequency is 6 Hz, Regulation is.. When load of 5 MW, The regulation =. 6 6 Hz / MW 5 5. Ans: (a) D =, R =.5, We know that hange in load D D f, R where f = change in frequency. Ans: (b) f = 5 Hz, generator rating = MVA Generator frequency decreases. f f MW 4. Ans: (c) The energy stored at no load = 5 = 5 MJ Before the steam valves open the energy lost by the rotor = 5. 6 = 5 MJ As a result of this there is reduction in speed of the rotor and, reduction in frequency f new 5. Ans: (c) = 49.4Hz 5 Sol: % regulation = f f p p = 4% 5