1 = 13.8 kv. 1 = 69 kv

Transcription

1 EP 306 GENERATION, TRANSMISSION AN ISTRIBUTION Chapter (1) Representation of Power Systems (1) Three parts of a single phase electric system are designated A,B and C are connected to each other through transformer as shown in fig. The transformer are rate as follows; A-B 100,00 KVA, KV leakage reactance 10 % B-C 100,00 KVA, KV,leakage reactance 8% If the ase in circuit B is chosen as 100,00 KVA, 138 KV, find the per unit impedance of the 300Ω resistive load in circuit C referred to circuit C,B and A. raw the impedance diagram neglecting magnetizing current transformer resistance and impedance. etermine the voltage regulation if the voltage at the load is 66KV with the assumption that the voltage input to circuit A remain constant. Base voltage for circuit A 138 x kv Base voltage for circuit C 138 x 1 69 kv Base voltage for circuit C (BasekV) 1000 Base kva Ω 100, Per-unit impedance of load in circuit C 0.63 p.u Base impedance of circuit B 1400Ω 100,000 Impedance of load referred to circuit B 300 x 100Ω 100 Per- unit impedance of load in circuit B o.63 p.u Base impedance of circuit A 19Ω 100,00 Impedance of load referred to circuit A 300 x x 0.1 1Ω

2 1 Per-unit impedance of load in circuit A 0.63 p.u Voltage at load jo p.u jo Load current jo p.u jo Voltage input (1.5 + jo) (jo.1 +jo.08) jo p.u Voltage regulation 3.97% () A300,00 KVA 13.8 KV 3Ǿ generator has a su transient reactance of 15%. The generator supplies two motors over a transmission line having transformer at oth ends as shown on the line diagram of fig. The motor have rated inputs of 00,00 and 100,00 KVA, oth 1.5KV with X 0 % 3Ǿ. Transformer T 1 is rated 35,000 KVA, 13. Δ 115 Υ KV with leakage reactance of 10%. Transformer T is composed of 3 single phase transformer each rated 100,00 KVA, KV with leakage reactance of 1%. Series reactance of the transmission line is 80Ω. raw the reactance diagram with all reactance marked in per-unit. Select the generator rating as ase in generator circuit. 3 rating of T 3x 100,00 300,00 kva Line to line ration x kv A ase of 300,00 kva, 13.8 KV In transmission line, 13.8 x kv 13.

3 1. 5 In motor circuit, 10 x kv ase kv X P.U Z given x ( ase kv (given) (new) ) x ase kva ase kva (new) (given) Transformer, T 1 X 0.1 x ( ) x p.u T X 0.1 x ( ) x p.u The ase impedance of transmission line 480Ω Reactance of line p.u Reactance of motor 1 0. x ( ) x 0.8 p.u Reactance of motor 0. x ( ) x p.u (3) A generator rated 100 MVA, 13.kV with X 0 % is connected through a Δ-Υ transformer to a transmission line whose series reactance is 40Ω. The ase chosen for calculation is 00 MVA, 115KV in transmission line. etermine the generator and transformer reactance in-per unit for reactance diagram (a) if the transformer is a 3 unit rated 13.8Δ - 10Υ kv, 100MVA with X8% and () if the transformer is composed of three single phase each rated 13.8kV 69kV, 35,000 kva. Also determine the per- unit reactance of the transmission line on the chosen ase.

4 Base 00MVA, 115kV in transmission line Base voltage for generator circuit 115 x 10 (a) For generator, p.u.z j0.x( For transformer p.u.z (H.T) j0.08 x ( ) x j0.174 p.u 13.5 kv 13.8 For transformer p.u.z (L.T) j0.08 x ( 13.5 j0.174 p.u () 3 rating of Δ Υ Transformer 3 x 35,000 kva 00 ) x j0.4 p.u ) x Δ 69 x 3 Υ 13.8 Base voltage for generator circuit 115 x 13.8 kv For generator, p.u z j0. x ( ) x j0.395 p.u For transformer p.u z (H.T) jo.08 x ( ) x j p.u For transformer p.u z (L.T) jo.08 x ( ) x j0.164 p.u ase kv 100 Base impedance for transmission line ase kva p.u z of transformer line Ω j0.605 p.u

5 (4) A 15,000 KVA 8.5 three phase generator has a su-transient reactance of 0%.It is connected through a Δ Υ transformer to a high voltage transmission lion having a total series reactance of 70 ohms. At the load circuit of the line is a Υ Υ step down transformer. Both transformers ank are composed of single phase transformer connected for three phase operation. Each of the three transformers composing each.bank is rated 6,667 KVA, KV with a reactance of 10%. The load represented as impedance is drawing 10,000 KVA at 1.5KV and 80% power factor lagging. raw a positive sequence diagram showing all impedances in per-unit. Chose a ase of 10,000 KVA, 1.5KV in the load circuit. etermine the voltage at the terminals of the generator. Choose ase 10,000 kva, 1.5 kv of the load circuit Base kv for line x kw Base kv for generator 15 x kv For generator, p.u z x ( ) x p.u For transformer T 1 p.u z 10 10, x ( ) x p.u For Transformer T p.u z , x ( ) x p.u Basic impedance of transmission line 156.5Ω 100,00 p.u z For transmission line 70 o.o448 p.u 156.5

6 j0.096 j j0.096 j j1 Eg p.u voltage 1 p.u 1.5 p.u z j0.6 1 p.u I 0.8+ jo.6 1 generator terminal voltage j0.6 (j0.37) p.u (5) The one line diagram of an unloaded power system in fig. reactance of the two section of transmission line are shown on the diagram. The generator and transformers are rated as follows; Generator (1) ; 0,000 kva, 6.9 kv, X 0.15 p.u () ; 10,000 kva, 6.9 kv, X 0.15 p.u (3) ; 30,000 kva, 13.8 kv, X 0.15 p.u Transformer (T 1 ) ; 5,000 kva, 6.9 Δ 115 Υ kv, X 10% (T ) ; 1,500 kva, 6.9 Δ 115 Υ kv, X 10% (T 3 ) ; single phase unit each rated 10,000 kva 7.5 kv, X 10% raw the impedance diagram with all reactance marked in per unit and with letters to indicate point corresponding to the one line diagram, choose a ase of 30,000kVA, 6.9 kv in the circuit of generator (1)

7 1 A 0MVA 6.9KV X 0.5 5MVA V X 10% 5MVA 6.9 Y X 10% B J100Ω 100MVA 6.9KV X 0.15 C 115YKV 6.9KV 1MVA X 10% E Y MVA 3 75Y MVA Y13.8KV X 0.15 kv 3 rating of T 3 3x10 MVA 30 MVA Line to line voltage ratio 3 x 75 V 1-3 x 7.5 V 1-8 KV Choose a ase of 30 MVA, 6.9 kv 115 Base kv for transmission line 6.9 x 115 kv Base kv for generator () 115 x 6.9 kv 115 Base kv for generator (3) kv 3 75 For generator (1), p.u z j0.15 x ( ) j0.5 p.u For generator (), p.u z j0.15 x ( ) j0.5 p.u For generator (3), p.u z j0.1 x ( ) j0.16 p.u For Transformer T j0.1 x ( ) j0.1 p.u For Transformer T j0.1 x ( ) j0.5 p.u For Transformer T 3 j0.1 x ( ) j0.18 p.u Base impedance of transmission line ( ase kv ) 1000 ase kva

8 (115) Ω j100 For j100ω line p.u z j 0.7 p.u For j80ω line p.u z j j p.u (6) raw the impedance diagram for the power system shown in fig;. Mark impedances in per- unit- neglect resistance and use a ase of 50,000 kva, 138 kv in the 40 Ω line. The rating of the generators, motors and transformers are, Generator (1), 0,000 kva, 13. kv, x 15% Generator (), 0,000 kva, 13. kv, x 15% Synchronous motor (3) 30,000 kva, 6.9kV, X 0% Three phase Υ Υ transformer 0,000 kva, 13.8V Υ kv, X 10% Three phase Υ Δ transformer 15,000 kva, 6.9 Δ Υ kv, X 10% All transformer are connected to step up the voltage of the generator to transmission line voltage.

9 Choose a ase of 50MVA, 138kV in the 40Ω line 13.8 ase kv for the generator (1) and () kV ase kv for synchronous motor kV ase kv for the Y-Y transformer kV ase kv for the Y-Δ transformer kV for the generator (1) and () p.uz J for synchronous motor p.uz J for 3, Y-Y transformer p.uz J J0.343 p.u 0 50 J0.333 p.u J0.5 p.u for 3, Y-Δ transformer p.uz J J0.333 p.u 15 for 40 Ω line p.uz J J p.u ( 138) for 0 Ω line p.uz J 0 50 J p.u ( 138)

10 Chapter () Series impedance of transmission lines (7) One circuit of a single phase transmission line is composed of three solid 0.5 cm radius wires. The return circuit is composed of two 0.5 cm radius wires. The arrangement of conductors is shown in fig. Find the inductance of the complete line in henrys per meter (and in milli- henrys per mile). L x s 10 3 aa 7 ln a ac m s aa cc r x 0.5 x 10 - a c c a 6m ac ca 1m s m a c cc ca c m m n ad d cd ad e 4m cd 15m ae d ce 177 m m m L x x 10-7 ln ae e ce x 10-7 H/m For cord Y, s dd eede de ee r de ed 6m ed

11 s m m L y x 10-7 ln s L L x + L y x 10-7 H/m m 8.71 x 10-7 H/m (8) A three phase doule circuit line is composed of 300,000 (mil 6/7 ACSR ostrich conductors arranged as shown in Fig. Find the 60 Hz inductive reactance in ohms per mile per phase. ( s 0.09 ) Phase C (c -c ) Phase A (a -a ) Phase B ( - ) L 7 meq 10 ln H/m s Phase A, a-a B, - C, c-c P n.m AB aa a a a 10.1 ft d a 1.9 a 10.1 P AB

12 P n. m BC c c c 10.1 ft c c c 1.9 c P 4 BC P n m CA. caac a c ca c a 0 ac a c 18 P CA c dc P meq 16.1 For phase A, SA sd a s 0.09 d a-a P P P AB BC CA a SA For phase B, SB sd d SB For phase B, d c-c SC 3 Seq L 10 ln H/m P meq Seq ln x 10-7 H/m/P X l πfl π x 60 x 6.13 x 10-7 x Ω/mi/P

13 (9) A single phase 60 Hz power line is supported on a horizontal cross arm. Spacing etween conductors is.5m. A telephone line is supported on a horizontal cross arm 1.8m directly power line with a spacing of 1.0m etween the centers of its conductors. Find the mutual inductance etween the power line and telephone circuits and the 60 Hz voltage per kilometer induced in the telephone line if the current in the power line is 150A. φa φ ue to I a, flux leakage ψ ad c ac d ue to I a, ue to I, ψ ψ cd ( ) 7 I a ln +.51m + ( ) 1.95m cd I a 10 7 d I a.51 ln ln 1.95 Since I a and I are 180 out of phase, the total flux (due to oth I a and I ) 7.51 ψ cd ( total ) 4 10 I a ln 1.95 ψ cd Mutual inductance, M I a ln I a x 10-7 H/m 60 Hz telephone circuit V/km for 150A V cd πf MI a π x 60 x x 10-7 x 150 x V/km ad ac

14 (10) Find the GRM of each of the unconventional conductors shown in Fig: in terms of the radius of the individual strands. Solution; (a) aa cc dd r a ad c cd da dc a c r ac ca d d 8 r s GMR (0.7788r) (r) ( 8r) 1.78r () (c) aa cc dd r a ad c cd da dc a c r ac ca r r 3 r s GMR (0.7788r) (r) ( 3r) 1.69r aa cc r a c a c r ac ca 4r GMR ( r ) (r) (4r) r

15 (d) a f c d e aa cc dd ee ff r a a c c cd dc de ed ef fe fa af f f d d df fd r ac ca ae ea ce ec 4r ad da (4r) (r) 1r c c cf fc r r 3 r GMR (0.7788r) (r) (4r) ( 1r) ( 3r).1r (11) A three phase 60 Hz line has flat horizontal spacing. The conductors has a GRM of with 10m etween adjacent conductors. etermine the inductive reactance per phase in ohms per kilometer. What is the name of this conductor? s m m m 1.6m Inductance per phase, L x 10-7 ln x ln x 10-6 H/m 1.37 x 10-3 H/km Inductive reactance πfl π x 60 x 1.37 x Ω/km m s

16 (1) Calculate the inductive reactance in ohm per kilometer of a undle 60 Hz three phase line having three ACSR Rail conductors per undle with 45cm etween conductors of the undle. The spacing etween undle center is 9, 9 and 18 ( s ) 45cm, s s sd m m m XL π x x ln x 10-4 Ω/m 0.38 Ω/km (13) A 60Hz three phase line composed of an ASCR Bluejay conductor per phase has flat horizontal spacing of 11m etween adjacent conductors. Compare the inductive reactance in ohm per kilometer per phase of this line that of a line using a two conductor undle of ASCR 6/7 conductors having the same total cross sectional area of aluminium as the single phase conductor line and 11m spacing measured from the center of the undles. The spacing etween conductors in the undle is 40cm.( s ) s For 30 lines, eq m Inductive reactance, XL π x 60 x x 10-7 ln x 10-4 Ω/m 0.57 Ω/km For undle, 10 conductor, s m m 11m

17 X L π x 60 x x ln x 10-4 Ω/m 0.38 Ω/km 0.57 Compare with two condition, horizontal conductor 1.39 times of two undle conductors. (14) Six conductors of ASCR rake constitute a 60Hz doule circuit three phase line arranged in Fig. The vertical spacing however is 14 the longer horizontal distance is 3 and the short horizontal distance are 5. Find the inductance per phase per mile and the six inductive reactance in ohm per mile. ( s ) Solution; Phase A, a-a B, - C, c-c For P AB a a a a P AB For P BC c c c c 1.4 P BC P CA For ca c a 8

18 ca c a P CA P meq.97 For phase A, d 4 P P P AB BC CA SA s a a d a-a SA For phase B, SB sd d SB For phase C, SC sdc c d c-c d a-a SC Seq L 10 ln H/m P meq Seq ln x 10-7 H/m XL πfl π x 60 x 5.99 x 10-7 Ω/m/Ph.56 x 10-4 x Ω/m/phase

19 Chapter (3) Capacitance of Transmission Line (15) Find the capacitive reactance for 1 mi of the line as shown in Fig. If the length of the line is 175 mi and the normal operating voltage is 0kV, find the capacitive reactance to neutral for the entire length of the line, the charging current per mi and the total charging megavolt-amperes. (r ) eq , r X a X d log d πk Cn F/m to neutral eq ln( ) r 1 π x 10-1 F/m 4.8 ln( ) x 10-1 x 1609 F/mi 1 1 XC πfc x 10-1 x x 10 6 Ω/mi Xa x 10 6 Xd log d log x 10 6 Xc Xa + Xd x 10 6 Ω/mi to neutral For 175 mi, Xc 1066Ω to neutral 175 I chg πf C an V an π x 60 x x x 10-1 x A/mi For 175 mi, I chg x A Reactive power, Q 3 Vl I chg 3 x 0 x 10 3 x Mvars

20 (16) Find the 60Hz capacitive susceptance to neutral per mile per phase of the doule circuit line in ostrich. ( r ) r AB a a aa a a a 1.5 a AB 14.8 BC AB 14.8 CA eq 16.1 AB BC CA SA rd a a d a-a SA SB rd SC rd c c Seq 3 SA SB SC πk C n eq ln( ) s π ln( )

21 x F/m x x 1609 F/mi 1 1 B C πfc X C 1 πfc π x 60 x x x x 10-5 mho/mi to neutral (17) A three phase 60 Hz transmission line has its conductor arrange in a triangular formation so that two of the distances etween conductors are 5 and third is 4. The conductors are ASCR. Osprey. etermine the capacitance to neutral in microfarad per mile and the capacitive reactance to neutral in ohm-mile. If the line is 150mi long find the capacitance to neutral and capacitive reactance of the line. ( in) k 8.85 x r πk C an F/m meq ln r meq π C n x 10-8 F/mi to neutral 9.7 ln Xc πfc n π kω - mi to neutral For 150 mi, C n x 10-8 x μf to neutral X Cn Ω

22 (18) Calculate the capacitive reactance in ohm kilometer of a undle 60 Hz, 30 line having three ACSR Rail conductors per undle with 45cm etween conductors of the undle. The spacing etween undle centers is 9,9 and 18m.( 1.165in)k 8.85 x cm 45cm 45cm 9m 9m 1.165in r m 4 πk C an meq ln S meq m rd m s 1 π C an ln x 10-1 F/m 1 Xc πfc n π x 10 6 Ω km/p (19) A 60 Hz 30 line composed of one ACSR Bluejay conductor per phase has flat horizontal spacing of11m etween adjacent conductors. Compare the capacitive reactance in ohm kilometer per phase of this line with that of a line using a two conductor undle of ACSR 6/7 conductors having the same total cross sectional area of aluminum as the single conductor line and the 11m spacing measured etween undles. The spacing etween conductors in the undle is 40cm. ( 1.59 in) k 8.85 x in

23 r 0.016m 1 d 40cm 0.4m eq m rd m s 1 πk π C n eq ln ln 0.08 S x F/m to neutral x F/km to neutral 1 X cn πfc n 46 l kω- km to neutral For same total cross-sectional area Finch, r 1 1 πk π C n eq ln ln r x 10-9 F/km to neutral 1 X cn kω- km to neutral πfc n X cn (single conductor) > X cn (undled conductor) (0) Six conductors of ACSR rake constitute a 60 Hz doule circuit three phase line arranged as shown in Fig. The vertical spacing, however, is 14ft the longer horizontal distance is 3 Hz and the short horizontal distance are 5 ft. Find the capacitive reactance to neutral in ohm-miles and the charging current per mile per phase and per conductor at 138 kv. ( ) X c?, I chg?

24 πk C n ln Phase A, a-a B, - C, c-c eq S For phase A, Radius, r rd SA a a d a a SA SA For phase B, d SC 3 SB S For P AB SA SB SC a a a P BC P AB 1.4 ca 8ft c a ca 5 ft c a P 4 CA P eq 4 P P P AB BC CA.97 1 π C n 19.7 x 10-1 F/m.97 ln X c πfc n π x 10 6 Ω-mi to neutral

25 I chg πf V an C an π x 60 x A/mi 3 x 19.7 x 10-1 x 1609 Chapter (4) Sag Calculations and Some Features of Lighters of Lightning Protections (1) A transmission line conductor at a river crossing is supported from two towers at heights of 150 and 300 aove water level. The horizontal distance etween the towers is If the tension in the conductor is 460 l and the conductor weights l/ft, find the clearance etween the conductor and the water at a point midway etween the towers. Assume Paraolic configuration. y 1 150ft, y 300ft, l 1100ft, T 460 l, W c l/ft h y - y ft Th x 1 l- Wl ft Th x l+ Wl

26 ft x 1 + x x ft Wx d 1 ½ T ft Wx ½ T ft WX d T 460 From point B, Δd d d Clearance y - Δd () An overhead line has the following data. Span length 160m, conductor diameter 0.95cm, weight per unit length of the conductor 0.65 kg/m. Ultimate stress 450 kg/cm, wind pressure 40 kg/cm of projected area. Factor of safety is 5. Calculate the sag. Solution 1 160cm, diameter 0.95cm, Wc 0.65Kg /m, muximun stress 450Kg/cm, W wind 40 Kg / cm, S.F 5, Sag, d? max imum stress S.F working stress 450 Working stress 850 Kg / cm 5 πd π 0.95 Working tension kg 4 4 W wind 40kg/cm 0.95 cm 38kg/cm 0.38kg/m W Wv + Wn kg / m 1 WL d 3.998m T 60.5 (3) A transmission line conductor having a diameter of 19.5mm weights 0.85kg/m. The span is 75m. The wind pressure is 39 kg/m of projected area with ice coating of 13mm. The ultimate strength of the conductor is 8000kg. Calculate the maximum sag if the factor of safety is and ice weights 910 kg/m 3.

27 Solution iameter 195mm, Wc 0.85 kg/m, L 75m, wind Pressure 39mm Mun strength 8,00kg, Max sag d?, p.f, ice weight 910kg/m 3 overall diameter dia + L cm Projected area per wire length ( )m 1m m wind load 39kg /m 10 - m 1.775kg/m The area of section of ice [ π ( r + t) π r ] m Ice load 910kg/ m m kg/ m W ( Wc + W ) + W Working stress Max sag, d ice wind ( ) + (1.775).717kg / m max imumstree s.f WL T 6.41m.717(75/ ) kg (4) Consider an insulator made up of three porcelain disks as in Fig. Let the capacitance from the insulator hardware to the line conductor e considered negligile. Let the capacitive susceptances at a given frequency have the values shown in Fig. The lines in Fig represent metal. Find the proportion of voltage across disk.

28 Let V 1-4 E and V 1- V I 1-9Bv I 3Bv I -3 I 1- + I 9Bv + 3Bv 18V I 3 1Bv 4 V -3 V 9B 9B V 1-3 V 1- + V 3 v + V V I 3 V 1-3 B V B BV 3 3 But, I 3-4 I -3 + I 3 1BV + I 3 V 3-4 9B V 1-4 E BV BV BV V V 3 91 B V E V E 0.39E 7 V 1- V 0.39E V V E 0.319E V 3-4 V 0.39E 0.44E 7 7 V 1-4 E

29 (5) Find the proportion of voltage on each disk of the 3 disk suspension insulator shown in Fig. Let V 1-4 E and V 1- V I 1-9 BV I 3 BV I B V -4 B (V V 1- ) B (E-V) BE BV I -3 I 1- + I I 9BV + 3 BV (BE + BV) 13BV BE 13BV BE 13V E V -3 8B 9 13V E V E V 1-3 V 1- + V -3 V V E 44BV BE I 3 BV 1-3 B 9 9 I 3 B V 3-4 B (V 1-4 V 1-3 ) V E B (E - ) 9 10BV EV 9

30 I 3-4 I -3 + I 3 I 3 44BV BE 10BV EV (13VV BE) + ( ) - ( ) BV 1EV 9 I BV 1EV 1 183V 1E V 3-4 E 9B 9 9B 81 81E + 30E 381V 381 E V 3.43 V 111 V V V V V V V 9.14 % 3.43V 13V 3.43V 30.9 % 93.43V 183V 1(3.43V) 39.9 % V (6) Find the unstressed length at -0 F a 400,000 cmil copper conductor in a span etween two supports 400ft apart and of equal height. The maximum tension in conductor is to e 6500l. The weight of the conductor is 1.5l per ft and of its ice load 0.75l per ft. The wind load, with ice present, is 1.0l per ft. Take the modulus of elasticity to e 15,000,000 l per sq-in use short paraolic formula. Unstressed length, L u? (at 0 F) A 400,000 mil 400,000 4 π 10-6 in L 400ft T 6500 l W e 1.5 l/ft W ice 0.75 l/ft W wind 1.0 l/ft E 15,000,000 l/in T Stretch L ( ) AE W ( π 400, h v W + W ( ).361 l/ft 6 )

31 1 WL P L [1 + ( ) ] 6 T L u P Stretch 400, (at -0 F) 7. The unstressed length at 3 F of a 450,000 cmil copper conductor is 600 ft. The supports are at equal height and are 600 ft apart. Find the temperature at which the conductor has a sag of 14 ft, with no ice and no wind. The weight of conductor is 39 l/ft. Take the characteristics of copper cale to e as W c 1.39 l/ft, E l/in, α 0.000,0095 use the paraolic formula. Solution; L u1 600 (3 F) A450,000 cmil 450,000 4 π 10-6 in L W c 1.39 l/ft, t? E l/in 0.000, Wl d T L u P stretch lt Stretch AE 1 WL P L [1+ ( ) ] 6 T d T T l P 600 [1 + ( ) ] stretch π , L u L u L u1 [1 + (t t 1 )] [ ,0095 (t -3)] t 96.1 F

32 8. At a river crossing an overhead transmission line has a span of 560 m with the two supports of lowest conductor at 15 m and 95 m aove the water level. The weight of the conductor is kg/m. If the tension is adjusted to 100 kg, determine the clearance of the conductors aove the water level at a point 15 m the ase of the higher tower. Solution; p d p p 1 d y d 1 y 1 x 1 x x 15 m 560 m L 560 m, y 95 m, y 1 15 m, w0.394 kg/m,t100kg h y -y m Th x 1 L m wl Th x L m wl x x m Wx 0.394(715) d m T 100 Wx d m T 100 Δd d d m clearance y - Δd m

33 9. An overhead transmission line at a river crossing is supported from two towers at heights of 40 m and 90 m aove water level the horizontal distance etween the towers eing 400 m. If the maximum allowale tension is 000 kg, find clearance etween the conductor and water level at a point midway etween the towers. Weight of the conductor is 1 kg/m. Solution; B d p y d A y 1 d 1 x 1 x d x 15 m L y 1 40m, y 90 m, u400 m, T 000 kg, w1 kg/m h y y m Th x m wl Th x l m wl x1 + x x 50 m wx d m T 000 Δd d d m clearance y - Δd m

34 30. In fig: let there e 4 disks, the capacitive susceptance of the two connectors across each disk eing 10 B. The susceptance of each metal connector to ground is 4 B. Find the ratio of the voltage across the ottom disk to the voltage from line to ground. V4 5? V1 5 V 1- V I 1-10 BV I 1 4 BV I -3 I 1- + I 14BV I 3 7 V -3 V 10B 5 1 V 1-3 V 1- + V -3 V 5 I V 13 4B 5 48 V I 3-4 V+ 14BV V BV 118 V 3-4 V V 1-4 V + V

35 38 4 I 3 V BV I 4-5 I I 3 + BV BV 50 V 4-5 V V BV 13 V V V Where V1 5 V1 4 + V4 5 38V V 500 By TU(MawLaMyaing) yanaungoo@gmail.com