EP 3026 GENERATION, TRANSMISSION AND DISTRIBUTION 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 , 13.8 – 138 KV leakage reactance 10 % B-C 100,00 KVA, 69 – 138 KV ,leakage reactance 8% If the base 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 . Draw the impedance diagram neglecting magnetizing current transformer resistance and impedance. Determine the voltage regulation if the voltage at the load is 66KV with the assumption that the voltage input to circuit A remain constant .

Solution,

Base voltage for circuit A = 138 x 101

= 13.8 kV

Base voltage for circuit C = 138 x 21

= 69 kV

Base voltage for circuit C = kVABase

1000kV)(Base 2 ×

= 100,00

1000692 × = 476Ω

Per-unit impedance of load in circuit C = 476300

=0.63 p.u

Base impedance of circuit B = 000,10010001382 ×

=1400Ω

Impedance of load referred to circuit B = 300 x 2 2 =1200Ω

Per- unit impedance of load in circuit B =19001200

o.63 p.u

Base impedance of circuit A = 00,10010008.13 2 ×

= 19Ω

Impedance of load referred to circuit A= 300 x 2 2 x 0.1 2 = 12Ω

Per-unit impedance of load in circuit A= 1912

=0.63 p.u

Voltage at load =6966 = 0.957 +jo p.u

Load current = jo0.63jo0.957

++ = 1.52 + jo p.u

Voltage input = (1.52 + jo) (jo .1 +jo.08) + 0.957 = 0.957 + jo.274 =0.995p.u

Voltage regulation = 0.957

0.9570.995− 3.97%

(2) A300,00 KVA 13.8 KV 3Ǿ generator has a sub transient reactance of 15%. The generator supplies two motors over a transmission line having transformer at both ends as shown on the line diagram of fig. The motor have rated inputs of 200,00 and 100,00 KVA, both 12.5KV with X” = 20 % 3Ǿ. Transformer T1 is rated 35,000 KVA, 13.2 Δ – 115 Υ KV with leakage reactance of 10%. Transformer T 2 is composed of 3 single phase transformer each rated 100,00 KVA, 12.5 – 67 KV with leakage reactance of 12%. Series reactance of the transmission line is 80Ω. Draw the reactance diagram with all reactance marked in per-unit. Select the generator rating as base in generator circuit.

Solution, 3∅ rating of T 2 = 3x 100,00 = 300,00 kVA Line to line ration = 12.5 – 67 x √3 = 12.5 – 116 kV A base of 300,00 kVA, 13.8 KV

In transmission line , 13.8 x 2.13

115 = 120 kV

In motor circuit, 120 x 116

5.12 =12.9 kV

X = P.U Z given x ((new)

(given)

kVbasekVbase

) x (given)

(new)

kVAbasekVAbase

Transformer, T1 X = 0.1 x (8.132.13

) 2 x 3500030000

= 0.0784 p.u

T 2 X = 0.1 x (9.125.12

) 2 x 3500030000

=0.094 p.u

The base impedance of transmission line = 30000

10001202 × = 480Ω

Reactance of line = 48080

= 0.167 p.u

Reactance of motor 1 = 0.2 x (9.125.12

) 2 x 2000030000

= 0.282 p.u

Reactance of motor 2 = 0.2 x (9.125.12

) 2 x 1000030000

= 0.563 p.u.

(3) A generator rated 100 MVA , 13.2kV with X” = 20 % is connected through a Δ-Υ transformer to a transmission line whose series reactance is 40Ω . The base chosen for calculation is 200 MVA , 115KV in transmission line . Determine the generator and transformer reactance in-per unit for reactance diagram (a) if the transformer is a 3∅ unit rated 13.8Δ - 120Υ kV , 100MVA with X=8% and (b) 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 base .

Solution,

Base 200MVA , 115kV in transmission line

Base voltage for generator circuit = 115 x 120

8.13 = 13.225 kV

(a) For generator , p.u.z =j0.2x(225.132.13

) 2 x 100200

= j0.4 p.u

For transformer p.u.z (H.T) = j0.08 x (115120

) 2 x 100200

= j0.1742 p.u

For transformer p.u.z (L.T) = j0.08 x (225.13

8.13) 2 x

100200

= j0.1742 p.u (b) 3∅ rating of Δ – Υ Transformer = 3 x 35,000 kVA 13.8Δ = 69 x 3Υ

Base voltage for generator circuit = 115 x 6938.13×

13.28 kV

For generator , p.u z = j0.2 x (28.132.13

) 2 x 100200

= j0.395 p.u

For transformer p.u z (H.T) = jo.08 x (115

693×) 2 x

100200

= j0.1646 p.u

For transformer p.u z (L.T) = jo.08 x (28.138.13

) 2 x 353

200×

=j0.164 p.u

Base impedance for transmission line =kVAbase

100kVbase 2 ×

= 100020010001152

××

= 66.125Ω

p.u z of transformer line = 125.66

40= j0.605 p.u

(4) A 15,000 KVA 8.5 three phase generator has a sub-transient reactance of 20% .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 bank are composed of single phase transformer connected for three phase operation. Each of the three transformers composing each .Bank is rated 6,667 KVA, 10-100 KV with a reactance of 10%. The load represented as impedance is drawing 10,000 KVA at 12.5KV and 80% power factor lagging. Draw a positive sequence diagram showing all impedances in per-unit. Chose a base of 10,000 KVA, 12.5KV in the load circuit. Determine the voltage at the terminals of the generator.

Solution, Choose base 10,000 kVA , 12.5 kV of the load circuit

Base kV for line = 12.5 x 3103100

××

= 125 kW

Base kV for generator =125 x 1003

10×

= 712 kV

For generator , p.u z = 0.2 x (2.75.8

) 2 x 1510

= 0.1858 p.u

For transformer T1 p.u z = 0.1 x (2.7

10) 2 x

66673000,10

×

= 0.096 p.u

For Transformer T 2 p.u z =0.1 x (5.12

310×) 2 x

666730000,1×

=0.096 p.u

Basic impedance of transmission line = 00,10010001252 ×

= 1562.5Ω

p.u z For transmission line = 5.1562

70=o.o448 p.u

j0.096 j0.0448

j1

j0.096

j0.1858

Eg+-

p.u voltage = 5.125.12

= 1 p.u

p.u z = 0.8 + j0.6

p.u I = 6.8.0

1jo+

generator terminal voltage = 1+6.08.0

1j+

(j0.237)

= 1.158 9.42 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) ; 20,000 kVA , 6.9 kV , X″ = 0.15 p.u (2) ; 10,000 kVA , 6.9 kV , X″ = 0.15 p.u (3) ; 30,000 kVA , 13.8 kV , X″ = 0.15 p.u Transformer (T1 ) ; 25,000 kVA , 6.9 Δ – 115 Υ kV , X = 10% (T 2 ) ; 12,500 kVA , 6.9 Δ – 115 Υ kV , X =10% (T 3 ) ; single phase unit each rated 10,000 kVA 7.5 kV , X =10%

Draw the impedance diagram with all reactance marked in per unit and with letters to indicate point corresponding to the one – line diagram , choose a base of 30,000kVA , 6.9 kV in the circuit of generator (1)

25MVA6.9 YX” = 10%

20MVA6.9KV

X” = 0.5

100MVA6.9KV

X” = 0.15

115YKV6.9KV

12MVAX = 10%

3 10MVAkV5.73Y753 ×−×

25MVA6.9 -115V

X = 10%D

BA EC31

2

J100Ω

Y 30MVAY13.8KVX” = 0.15

Solution, 3∅ rating of T 3 = 3x10 MVA = 30 MVA

Line to line voltage ratio = 3 x 75 V1 - 3 x 7.5 V1 - 8 KV Choose a base of 30 MVA , 6.9 kV

Base kV for transmission line = 6.9 x 9.6

115 = 115 kV

Base kV for generator (2) = 115 x 115

9.6 = 6.9 kV

Base kV for generator (3) = 115 7535.73

××

= 11.5 kV

For generator (1), p.u z = j0.15 x 2030)

9.69.6( 2 × = j0.225 p.u

For generator (2), p.u z = j0.15 x 1030)

9.69.6( 2 × = j0.225 p.u

For generator (3), p.u z = j0.1 x 3030)

5.118.13( 2 × = j0.216 p.u

For Transformer T1 = j0.1 x 2530)

9.69.6( 2 × = j0.12 p.u

For Transformer T2 = j0.1 x 1230)

9.69.6( 2 × = j0.25 p.u

For Transformer T3 = j0.1 x 3030)

115753( 2 ×

× = j0.128 p.u

Base impedance of transmission line = kVAbase

kVbase 1000)( 2 ×

= 100030

1000)115( 2

××

= 440.83 Ω

For j100Ω line p.u z = 83.440

100j = j 0.227 p.u

For j80Ω line p.u z = 83.440

80j = j 0.181 p.u

(6) Draw the impedance diagram for the power system shown in fig;. Mark impedances

in per- unit- neglect resistance and use a base of 50,000 kVA, 138 kV in the 40 Ω line. The rating of the generators, motors and transformers are,

Generator (1), 20,000 kVA, 13.2 kV, x ′′ = 15% Generator (2), 20,000 kVA, 13.2 kV, x ′′ = 15% Synchronous motor (3) 30,000 kVA, 6.9kV, X ′′ = 20% Three phase Υ – Υ transformer 20,000 kVA, 13.8V1 - 138 Υ kV, X = 10% Three phase Υ – Δ transformer 15,000 kVA, 6.9 Δ - 138 Υ kV, X = 10% All transformer are connected to step up the voltage of the generator to transmission line voltage.

Solution, Choose a base of 50MVA, 138kV in the 40Ω line

base kV for the generator (1) and (2) = 138

8.13138× = 13.8kV

base kV for synchronous motor = 138

9.6138× = 6.9kV

base kV for the Y-Y transformer = 138

8.13138× = 13.8kV

base kV for the Y-Δ transformer = 138

9.6138× = 6.9kV

for the generator (1) and (2) p.uz = J 0.15 × 2

8.132.13⎟⎠⎞

⎜⎝⎛ ×

2050

= J0.343 p.u

for synchronous motor p.uz = J 0.2 × 2

9.69.6⎟⎠⎞

⎜⎝⎛ ×

3050

= J0.333 p.u

for 3∅, Y-Y transformer p.uz = J 0.1 × 2

8.138.13⎟⎠⎞

⎜⎝⎛ ×

2050

= J0.25 p.u

for 3∅, Y-Δ transformer p.uz = J 0.1 × 2

9.69.6⎟⎠⎞

⎜⎝⎛ ×

1550

= J0.333 p.u

for 40 Ω line p.uz = ( )2138

5040×J = J 0.105 p.u

for 20 Ω line p.uz = ( )2138

5020×J = J 0.0525 p.u

Chapter (2) Series impedance of transmission lines

(7) One circuit of a single phase transmission line is composed of three solid 0.25 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).

Solution,

s

mx D

DL ln102 7−×=

23cbcaccbcbabbacabaas DDDDDDDDDD =

Daa = Dbb = Dcc = 0.7788r = 0.7788 x 0.25 x 10-2 Dab = Dbc = Dcb = Dba = 6m Dac = Dca = 12m ∴Ds = 0.481 m

mncebeaecdbdadm DDDDDDD =

Dad = Dbe = 4m Dcd = 15m Dae = Dbd = Dce = 177 m

Dm = 1779177151779 ××××× = 11.915 m

Lx = 2 x 10-7 ln 481.0915.11

= 6.419 x 10-7 H/m

For cord Y,

22eddeeedds DDDDD =

Dde = Dee = 0.7788r Dde = Ded = 6m

∴Ds = 0.153 m Dm = 11.915

Ly = 2 x 10-7 ln s

m

DD

= 8.71 x 10-7 H/m

L = Lx + Ly

= 15.129 x 10-7 H/m

(8) A three phase double – circuit line is composed of 300,000 (mil 26/7 ACSR ostrich conductors arranged as shown in Fig. Find the 60 Hz inductive reactance in ohms per mile per phase. ( Ds = 0.0229′)

Phase A (a -a )

Phase C (c -c )

Phase B (b -b ) Solution,

s

meq

DD

L ln102 7−×= H/m

Phase A, a-a′ B, b-b′ C, c-c′ m.n

bababaabPAB DDDDD ′′′′=

Dab = 10.1 ft = Ddb′ Dab′ = 21.9′ = Da′b

22 22 9.211.10× ×=PABD

= 14.8′

mncbcbcbbc

PBC DDDDD .

′′′′= Dbc = 10.1 ft = Db′c′

Dbc′ = Db′c = 21.9′ 4 22 9.211.10 ×=P

BCD = 14.8′ mn

cdcacacaPCA DDDDD .

′′′= Dca = Dc′a′ = 20′

Dac′ = Da′c = 18′ 718.9DP

CA ′= 4 P

CAPBC

PAB

Pmeq DDDD ××=

= 16.1′ For phase A, aas

bSA dDD ′−=

Ds = 0.0229′ da-a′ = 26.9′ ∴ b

SAD = 0.785′ For phase B, bbs

bSB dDD ′−=

db-b′ = 21′ ∴ b

SBD = 0.693′ For phase B, dc-c′ = 26.9′ ∴ b

SCD = 0.785′ 3 2b

Seq 0.6930.785D ×= = 0.753′

bSeq

Pmeq

D

DL ln102 7−×= H/m

= 753.0

1.16ln102 7−×

= 6.13 x 10-7 H/m/P Xl = 2πfl = 2π x 60 x 6.13 x 10-7 x 1609 = 0.372 Ω/mi/P

(9) A single phase 60 Hz power line is supported on a horizontal cross arm. Spacing between conductors is 2.5m. A telephone line is supported on a horizontal cross arm 1.8m directly power line with a spacing of 1.0m between the centers of its conductors. Find the mutual inductance between 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.

bφaφ

Solution,

Due to Ia, flux leakage ac

adacd D

DI ln102 7−×=ψ

Dad = Dbc = 22 )5.025.1(8.1 ++ = 2.51m

Dac = Dbd = 22 )5.025.1(8.1 −+ = 1.95m

Due to Ia, 95.151.2ln102 7

acd I−×+=ψ

Due to Ib, 95.151.2ln102 7

abd I−×−=ψ

Since Ia and Ib are 180° out of phase, the total flux (due to both Ia and Ib)

95.151.2ln104 7

)( atotalcd I−×=ψ

Mutual inductance, M = a

cd

Iψ

= 95.151.2ln104 7

aI−×

= 1.0098 x 10-7 H/m 60 Hz telephone circuit V/km for 150A Vcd = 2πf MIa = 2π x 60 x 1.0098 x 10-7 x 150 x 103 = 5.71 V/km

(10) Find the GRM of each of the unconventional conductors shown in Fig: in terms of the radius of the individual strands.

Solution; (a)

Daa = Dbb =Dcc = Ddd = 0.7788r

Dab = Dad = Dbc = Dcd = Dda = Ddc = Dba = Dcb = 2r Dac = Dca = Dbd = Ddb = 8 r

∴ Ds = GMR = 16 484 r)8((2r)(0.7788r) = 1.7228r (b)

Daa = Dbb =Dcc = Ddd = 0.7788r Dab = Dad = Dbc = Dcd = Dda = Ddc = Dba = Dcb = 2r Dac = Dca = 2 222 rr − = 2 3 r

∴ Ds = GMR = 16 42104 r)3((2r)(0.7788r) = 1.692r (c)

Daa = Dbb =Dcc = 0.7788r Dab = Dbc = Dba = Dcb = 2r Dac = Dca = 4r ∴ GMR = 9 243 )4()2()7788.0( rrr = 1.7037r

(d)

b

a

c

de

f

Daa = Dbb =Dcc = Ddd = Dee = Dff = 0.7788r Dab = Dba = Dbc = Dcb = Dcd = Ddc = Dde = Ded = Def = Dfe = Dfa = Daf = Dfb = Dbf = Dbd = Ddb = Ddf = Dfd = 2r Dac = Dca = Dae = Dea = Dce = Dec = 4r Dad = Dda = rrr 12)2()4( 22 =−

Dbc = Dcb = Dcf = Dfc = 2 222 rr − = 2 3 r

∴ GMR = 36 4226186 r)3(r)12((4r)(2r)(0.7788r) = 2.1r

(11) A three phase 60 Hz line has flat horizontal spacing. The conductors has a GRM of 0.0133 with 10m between adjacent conductors. Determine the inductive reactance per phase in ohms per kilometer. What is the name of this conductor? Solution,

Ds = 0.0133 Dm= 3

132312 DDD D12 = D23 = 10m ∴ Dm = 12.6m

Inductance per phase, L = 2 x 10-7 ln s

m

DD

= 2 x 10-7 ln 0133.0

6.12

= 1.37 x 10-6 H/m = 1.37 x 10-3 H/km Inductive reactance = 2πfl = 2π x 60 x 1.37 x 10-3 = 0.5167 Ω/km

(12) Calculate the inductive reactance in ohm per kilometer of a bundle 60 Hz three phase line having three ACSR Rail conductors per bundle with 45cm between conductors of the bundle. The spacing between bundle center is 9, 9 and 18 (Ds=0.0386′) Solution,

D= 45cm, Ds = 0.0386′ 45.03048.00386.0 ××== sd

bs DD

= 0.0728m Dm = 3 1899 ×× = 11.34m

∴ XL = 2π x 2 x 10-7 ln 0728.0

34.11

= 3.8 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 between adjacent conductors. Compare the inductive reactance in ohm per kilometer per phase of this line that of a line using a two conductor bundle of ASCR 26/7 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 bundles. The spacing between conductors in the bundle is 40cm.(Ds = 0.0415′) Solution,

Ds = 0.0415′ For 30 lines, Deq = 3 221111 ×× = 13.86m

Inductive reactance, XL = 2π x 60 x 2 x 10-7 ln 3048.00415.0

86.13×

= 5.27 x 10-4 Ω/m = 0.527 Ω/km For 2 bundle, 10 conductor, 4.03048.00415.0 ××=b

sD = 0.07113m Dm = 11m

XL = 2π x 60 x 2 x 10-7 ln 07113.011

= 3.8 x 10-4 Ω/m = 0.38 Ω/km

Compare with two condition, =38.0

527.0

20 horizontal conductor = 1.39 times of two bundle conductors. (14) Six conductors of ASCR Drake constitute a 60Hz double circuit three phase line arranged in Fig. The vertical spacing however is 14′ the longer horizontal distance is 32′ and the short horizontal distance are 25′. Find the inductance per phase per mile and the six inductive reactance in ohm per mile. ( Ds = 0.0373′)

Solution; Phase A, a-a′ B, b-b′ C, c-c′ For P

ABD

Dab = D ba ′′ = 22 5.314 + = 14.43′

Dab′ = D ba′ = 22 5.2814 + = 31.753′ 4 22 753.3143.14 ×=P

ABD = 21.4′ For P

BCD

Dbc = D cb ′′ = 22 5.314 + = 14.43′

Dbc′ = Db′c = 22 5.2814 + = 31.753′ =P

BCD 21.4′ For P

CAD Dca = Dc′a′ = 28′

Dca′ = Dc′a = 25′ 4 22 2528 ×=P

CAD = 26.46′ 4 P

CAPBC

PAB

Pmeq DDDD ××=

= 22.97′ For phase A, aas

bSA dDD ′−=

da-a′ = 22 2528 × = 37.536′ ∴ b

SAD = 1.183′ For phase B, bbs

bSB dDD ′−=

db-b′ = 32′ ∴ b

SBD = 1.092′ For phase C, ccs

bSC dDD ′−=

dc-c′ = da-a′ = 37.536′ ∴ b

SCD = 1.183′ 3 2 092.1183.1 ×=b

SeqD = 1.152′

bSeq

Pmeq

D

DL ln102 7−×= H/m

= 152.1

97.22ln102 7−×

= 5.99 x 10-7 H/m XL= 2πfl = 2π x 60 x 5.99 x 10-7 Ω/m/Ph = 2.256 x 10-4 x 1609 = 0.363 Ω/m/phase

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 220kV, 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 = 0.554″)

20 20

38 Solution,

Deq = 3 382020 ×× = 24.8′, r = 12554.0

= 0.0462′

Cn = )ln(

2

rDkeq

π F/m to neutral

= )

0462.08.24ln(

1085.82 12−××π = 8.8466 x 10-12 F/m

= 8.8466 x 10-12 x 1609 F/mi

XC = fcπ2

1=

1609 x 12-10 x 8.84661

= 0.1864 x 106 Ω/mi Xa′ = 0.0912 x 106 Xd′ = 0.06831 log d = 0.06831 log 24.8 = 0.0953 x 106 Xc = Xa′+ Xd′= 0.1865 x 106 Ω/mi to neutral

For 175 mi, Xc = 175

101865.0 6×= 1066Ω to neutral

Ichg = 2πf CanVan

= 2π x 60 x 310220 3×

x 8.8466 x 10-12 x 1069

= 0.681 A/mi For 175 mi, Ichg = 0.681 x 175 = 119A Reactive power, Q = 3 Vl Ichg

= 3 x 220 x 103 x 119 = 45.3 Mvars

Xa’ = 0.912 × 106

Xd’ = 0.06831 log d

(16) Find the 60Hz capacitive susceptance to neutral per mile per phase of the double circuit line in ostrich. ( r = 0.0283′)

Solution, r = 0.0283′ DAB = bababaab DDDD ′′′′

Dab = baD ′′ = 22 5.110 × = 10.11

Dab′ = baD ′ = 22 5.1910 × = 21.9 DAB = 14.8′ DBC = DAB = 14.8′ DCA = 22 1810 × = 18.97′ Deq = CABCAB DDD = 16.1′ aa

bSA rdD ′−=

da-a′ = 22 2018 × = 26.9 ∴ b

SAD = 9.260283.0 × = 0.873′ bb

bSB rdD ′−= = 210283.0 × = 0.77′

cc

bSC rdD ′−= = 9.260283.0 × =0.873′

3 b

SCbSB

bSA

bSeq DDDD ××=

= 0.837′

Cn = )ln(

2

s

eq

DDkπ

= )

837.01.16ln(

1085.82 12−××π

= 1.8807 x 10-11 F/m = 1.8807 x 10-11 x 1609 F/mi

BC = CX

1=

fcπ211

= 2πfc

= 2π x 60 x 1.8807 x 10-11 x 1609 = 1.1407 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 between conductors are 25′ and third is 42′. The conductors are ASCR . Osprey. Determine 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. ( D = 0.879 in) k = 8.85 x 10-12

Solution,

r = 122

879.0×

= 0.0366

Can =

rD

kmeqln

2πF/m

Dmeq = 3 422525 ×× = 29.72′

Cn=

0366.072.29ln

1085.82 12−××π= 1.335 x 10-8 F/mi to neutral

Xc = nfCπ2

1

= 8

3

10335.1602101

−

−

××××

π

= 198.69 kΩ - mi to neutral For 150 mi, Cn = 1.335 x 10-8 x 150 = 2.0025 μF to neutral XCn= 1324.166 Ω

(18) Calculate the capacitive reactance in ohm – kilometer of a bundle 60 Hz, 30 line having three ACSR Rail conductors per bundle with 45cm between conductors of the bundle. The spacing between bundle centers is 9,9 and 18m.(D = 1.165in)k = 8.85 x 10-12

45cm 45cm

9m

45cm

9m Solution, D = 1.165in

r = 24

3048.0165.1 ×= 0.0148m

Can =

bS

meq

DD

k

ln

2π

Dmeq = 3 1899 ×× = 11.34m D b

s = rd = 45.00148.0 × = 0.0816m

Can =

0816.034.11ln

1085.82 12−××π

= 11.27 x 10-12F/m

Xc = nfCπ2

1

= 27.11602

1010 312

××× −

π

= 0.2354 x 106 Ω km/P (19) A 60 Hz 30 line composed of one ACSR Bluejay conductor per phase has flat horizontal spacing of11m between adjacent conductors. Compare the capacitive reactance in ohm kilometer per phase of this line with that of a line using a two conductor bundle of ACSR 26/7 conductors having the same total cross sectional area of aluminum as the single conductor line and the 11m spacing measured between bundles. The spacing between conductors in the bundle is 40cm. (D = 1.259 in) k = 8.85 x 10-12 Solution,

D = 1.259 in

r = 122

3048.0259.1××

= 0.016m

d = 40cm = 0.4m Deq = 3 221111 ×× = 13.86 m D b

s = rd = 4.0016.0 × = 0.08m

Cn =

bS

eq

DDk

ln

2π=

08.086.13ln

1085.82 12−××π

= 1.078 x 10-11 F/m to neutral = 1.078 x 10-11 F/km to neutral

Xcn = nfcπ2

1

= 246 l kΩ- km to neutral For same total cross-sectional area

Finch, r = 122

3048.0293.1××

Cn =

rDkeqln

2π=

01642.086.13ln

1085.82 12−××π

= 8.2523 x 10-9 F/km to neutral

Xcn = nfcπ2

1 = 321.435 kΩ- km to neutral

Xcn (single conductor) > Xcn(bundled conductor) (20) Six conductors of ACSR Drake constitute a 60 Hz double circuit three phase line arranged as shown in Fig. The vertical spacing , however, is 14ft the longer horizontal distance is 32 Hz and the short horizontal distance are 25 ft. Find the capacitive reactance to neutral in ohm-miles and the charging current per mile per phase and per conductor at 138 kV. ( D = 1.108″)

Solution, Xc = ?, Ichg = ?

Cn =

bS

eq

DDk

ln

2π

Phase A, a-a′ B, b-b′ C, c-c′ For phase A,

Radius, r = 122

108.1×

= 0.04617′

aabSA rdD ′−=

aad ′− = 22 2825 + = 37.54′ 631.1 ′=b

SAD bSAD = b

SCD = 631.1 ′

For phase B, bbd ′− = 32′

3204617.0 ×=bSBD = 1.2155′ bSD = 3 b

SCbSB

bSA DDD = 228.1 ′

For P

ABD

Dab = D ba ′′ = 22 5.314 + = 14.43′

Da′b = 22 5.2814 + = 31.753′ =P

BCD PABD = 21.4′

Dca = 28ft = Dc′a′

Dca′ = 25 ft = Dc′a 4 22 2528 ×=P

CAD = 26.46′ 4 P

CAPBC

PAB

Peq DDDD ××=

= 22.97′

Cn =

282.197.22ln

1085.82 12−××π= 19.27 x 10-12 F/m

Xc = nfcπ2

1=

160927.196021012

×××π

= 0.815 x 106 Ω-mi to neutral

Ichg = 2πf VanCan

= 2π x 60 x 310138 3×

x 19.27 x 10-12 x 1609

= 0.0978 A/mi

Chapter (4) Sag Calculations and Some Features of Lighters of Lightning Protections

(21) A transmission line conductor at a river crossing is supported from two towers at heights of 150′ and 300′ above water level. The horizontal distance between the towers is 1100′. If the tension in the conductor is 4260 lb and the conductor weights 0.594 lb/ft, find the clearance between the conductor and the water at a point midway between the towers. Assume Parabolic configuration. Solution,

y1= 150ft, y2= 300ft, 2l = 1100ft, T = 4260 lb, Wc = 0.594 lb/ft h = y2 - y1 = 300-150 = 150ft

x1 = l-WlTh2

= 550 - 594.01101504260

××

= -428 ft

x2 = l+WlTh2

= 550 + 594.01101504260

××

= 1528 ft

x = 2

21 xx +=

21528428+

=978 ft

d1 = ½ T

Wx21 =

42602428594.0 2

××

= 12.77 ft

D2 = ½ T

Wx22 =

426021528594.0 2

××

= 162.77 ft

d = T

WX2

2

= 42602

978594.0 2

×× = 66.684′

From point B,

Δd = d2 – d = 162.77 – 66.684 = 96.093′ Clearance = y2 - Δd = 300 – 96.093 = 203.907′

(22) 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 4250 kg/cm2, wind pressure 40 kg/cm2 of projected area. Factor of safety is 5. Calculate the sag. Solution

21 =160cm, diameter = 0.95cm, Wc = 0.65Kg /m, muximun stress 4250Kg/cm2,

Wwind = 40 Kg / cm2, S.F = 5, Sag, d = ?

S.F = stressworkingstressimummax

Working stress = 5

4250 = 850 Kg / cm2

Working tension = kg5.6024

95.08504d850

22

=×π

×=π

×

Wwind = 40kg/cm2 × 0.95 cm = 38kg/cm = 0.38kg/m W = m/kg7529.038.065.0WWv 222

n2 =+=+

d = m998.35.6022

807529.0T

WL21 22

=×

×=

(23) A transmission line conductor having a diameter of 19.5mm weights 0.85kg/m. The span is 275m. The wind pressure is 39 kg/m2 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 2 and ice weights 910 kg/m3.

Solution Diameter = 195mm, Wc = 0.85 kg/m, 2L = 275m, wind Pressure = 39mm Mun strength = 8,00kg, Max sag d = ?, p.f = 2, ice weight = 910kg/m3 overall diameter

= dia + 2L = 1.95 + 2.6 = 4.55cm

Projected area per wire length = (4.55 × 10-2)m × 1m = 4.55× 10-2 m2 wind load = 39kg /m2 ×10-2 m2 = 1.775kg/m The area of section of ice = [ ]22 r)tr( π−+π = 1.327 × 10-3m2 Ice load = 910kg/ m2 × 1.327 × 10-3m2 = 1.20757 kg/ m W = 2

wind2

ice W)WWc( ++

= m/kg717.2)775.1()20757.185.0( 22 =++

Working stress = f.s

imumstreemax = 2

8000 = 4000 kg

Max sag, d = T2

WL2 = 40002

)2/275(717.2 2

×

= 6.421m (24) Consider an insulator made up of three porcelain disks as in Fig. Let the capacitance from the insulator hardware to the line conductor be considered negligible . 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.

Solution, Let V1-4 = E and V1-2 = V

I1-2 = 9Bv I2 = 3Bv I2-3 = I1-2+ I2 = 9Bv + 3Bv = 128V

V2-3 = B9

I 32− = B9Bv12 =

34 V

V1-3 = V1-2 + V2 – 3 = v + 34 V=

37 V

I3 = V1-3 × 2B = 37 V × 2B =

314 BV

I3-4 = I2-3 + I3 = 12BV + 3

14 BV = 3

50 BV

V3-4 = B9

I 32− = 3

50 BV × B91 =

2750 V =

27113 V

But, V1-4 = E

27113 V = E

∴ V= 27

113 E = 0.239E

∴ V1-2= V= 0.239E

V2-3= 34 V =

34 × 0.239E = 0.319E

V3-4= 2750 V =

2750 × 0.239E = 0.442E

V1-4= E

(25) Find the proportion of voltage on each disk of the 3 disk suspension insulator shown in Fig.

Solution, Let V1-4 = E and V1-2 = V I1-2 = 9 BV I2 = 3 BV I2

’ = B × V2-4 = B (V1-4 - V1-2) = B (E-V) = BE – BV

∴ I2-3 = I1-2 + I2 – I2’

= 9BV + 3 BV – (BE + BV) = 13BV – BE

V2-3 = B8

BEBV13 − = 9

EV13 −

V1-3 = V1-2 + V2-3 = V + 9

EV13 − = 9

EV22 −

I3 = 2BV1-3 = 2B × 9

EV22 − = 9

BE2BV44 −

I3′ = B × V3-4 = B × (V1-4 – V1-3)

= B × (E - 9

EV22 − )

= 9

EV22BV10 −

I3-4= I2-3 + I3 – I3′

=(13VV – BE) + (9

BE2BV44 − ) - (9

EV22BV10 − )

= 9

EV21BV183 −

V3-4 = B9

I 43− = 9

EV21BV183 − × B91 =

81E21V183 − = E

∴ 81E + 30E = 381V

E=111381 V = 3.432 V

41

21

VV

−

− = V432.3

V = 29.14 %

41

32

VV

−

− = V432.93

V432.3V13 − = 30.9 %

41

43

VV

−

− = V432.381

)V432.3(21V183×− = 39.9 %

(26) Find the unstressed length at -20°F a 400,000 cmil copper conductor in a span between two supports 400ft apart and of equal height. The maximum tension in conductor is to be 6500lb. The weight of the conductor is 1.25lb per ft and of its ice load 0.75lb per ft . The wind load, with ice present, is 1.0lb per ft. Take the modulus of elasticity to be 15,000,000 lb per sq-in use short parabolic formula. Solution, Unstressed length, Lu = ? (at – 20° F)

A= 400,000 mil = 400,000 × 4π × 10- 6

in2

2L= 400ft T = 6500 lb We= 1.25 lb/ft Wice=0.75 lb/ft Wwind=1.0 lb/ft E= 15,000,000 lb/in2

Stretch= 2L (AET )

= 400 (66 101510

4000,400

6500

×××π

× −)

= 0.5517

W= 2v

2h WW + = 22 )75.025.1(1 ++ = 2.2361 lb/ft

2P= 2L [1 + 2)T

WL(61 ]

= 400.3156 ′ ∴ Lu = 2P – Stretch = 400,3156 – 0.5517 = 399.7639 ′ (at -20° F) 27. The unstressed length at 32˚ 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 lb/ft. Take the characteristics of copper cable to be as Wc = 1.39 lb/ft, E = 15 × 106 lb/in2 , α = 0.000,0095 use the parabolic formula. Solution; Lu1 = 600′ (32° F)

A=450,000 cmil = 450,000 × 4π × 10- 6

in2

2L= 600′ D=14′ Wc = 1.39 lb/ft, t2= ? E= 15 × 106 lb/in2

∝ = 0.000,0095

d= 21

TWl2

Lu= 2P – stretch

Stretch= AE

lt2

2P= 2L [1+ 2)T

WL(61 ]

d= 21×

T30039.1 2×

T=28

30039.1 2× = 4467.8571 lb

2P= 600 [1 + 2)8571.446730039.1(

61 × ]

= 600.8711

stretch = 66 101510

40000,45

8571.44673002

×××π

×

××−

= 0.5057

Lu2 = 600.8711 – 0.5057 = 600.3654′ Lu2 = Lu1 [1 + ∝ (t2 – t1)]

600.3654= 600[1 + 0.000,0095 (t2 -32)] t2 = 96.1° F

28. 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 above the water level. The weight of the conductor is 0.394 kg/m. If the tension is adjusted to 1200 kg, determine the clearance of the conductors above the water level at a point 215 m the base of the higher tower. Solution;

y2

d2

p2

p

d

215 mxx2

560 m

y1

p1

d1

x1

2L = 560 m, y2= 95 m, y1=15 m, w=0.394 kg/m,T=1200kg h = y2-y1=95 – 15= 80 m

x1 = L - wL2Th =

2560 -

560394.0801200

×× = - 155 m

x2 = L + wL2Th = 280 + 435 =715 m

x = x2 – 215 = 715 – 215 = 500m

d2 = T2Wx 2

2 = 12002

)715(394.0 2

×= 83.926 m

d = T2

Wx2=

12002500394.0 2

×× = 41.042 m

Δd = d2 – d= 83.926 – 41.042 = 42.884 m clearance = y2 - Δd = 95 – 42.884 = 52.116 m

29. An overhead transmission line at a river crossing is supported from two towers at heights of 40 m and 90 m above water level the horizontal distance between the towers being 400 m. If the maximum allowable tension is 2000 kg, find clearance between the conductor and water level at a point midway between the towers. Weight of the conductor is 1 kg/m. Solution;

d2

y2

B

p

215 mxx2

2L

y1

d1

x1

d

A

d

y1=40m, y2= 90 m, u=400 m, T= 2000 kg, w=1 kg/m h= y2 – y1 = 90 – 40 = 50 m

x1 = 1 - wL2Th = 200 -

4001502000

×× = - 50m

x2 = l + wL2Th = 200 +

4001502000

×× = 450 m

x = 2

xx 21 + = 245050 + = 250 m

d2= T2

wx2=

200024501 2

×× = 15.625 m

Δd = d2 – d= 50.625 – 15.625 = 35 m clearance =y2 - Δd = 90 – 35 = 55 m

30. In fig: let there be 4 disks, the capacitive susceptance of the two connectors across each disk being 10 B. The susceptance of each metal connector to ground is 4 B. Find the ratio of the voltage across the bottom disk to the voltage from line to ground.

Solution,

51

54

VV

−

− = ?

V1-2 = V I1-2 = 10 BV I1 = 4 BV I2-3 = I1-2 + I2 = 14BV

V2-3 = B10I 32− =

57 V

V1-3 = V1-2 + V2-3 = 5

12 V

I2 =V13 × 4B = 548 V

I3-4 = 548 V+ 14BV =

5118 V

V3-4 = 5118 ×

108BV =

50118 V

V1-4 = 512 V +

50118 =

50238 V

I3 =V1-4 × 48 = 50

4238 × BV

I4-5 = I3-4 + I3 = BV5

11850

4238⎟⎟⎠

⎞⎜⎜⎝

⎛+

× = 50

2132 BV

V4-5 = 50

2132 × 108BV =

5002123 V

51

54

VV

−

− = V4512500

500V2132×

× = 0.4725

⎥⎦

⎤⎢⎣

⎡=+=+= −−− 500

V45125002123

50V238VVVWhere 544151

By TU(MawLaMyaing)

yanaungoo@gmail.com

057-27630