DESIGN OF 130 MW, 200 KM TRANSMISSION LINE

Most Economical Voltage Calculation:

The most economical voltage is given by the following empirical formula:

Economical Voltage (V eco) =5.5*√ Lt1 .6

+ P∗1000cosφ∗Nc∗150

Where,

Lt = length of transmission line =200 Km

P = Power to be transmitted =130 MW

cosØ = Power factor =0.96

For Nc= 1

Then, using the above values

V eco =5.5*√2001.6

+130∗10000 .96∗1∗150

= 177.137 KV

Nearest Standard Voltage= 220 kV

For Nc= 2

Then, using the above values

V eco =5.5*√2001.6

+130∗10000 .96∗2∗150

= 132.587 KV

Nearest Standard Voltage= 132 kV

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Checking technical criterion:

For Nc=1

Surge impedance Loading (SIL) = V2 /Z0

= 2202/400 =121

Multiplying factor (MF) =

P maxSIL = 130/121 = 1.0834

MFlimit for 160 KM from provided standard table = 2.0614

MFcalculated (1.08) < MFlimit (2.0614)

Power transfer capability = MFlimi* SIL

=2.0614*121

=249.4294

For Nc=2

Surge impedance Loading (SIL) = V2 /Z0

= 1322/200 =87.12

Multiplying factor (MF) =

P maxSIL = 130/87.12 = 1.4921

MFlimit for 160 KM from provided standard table = 2.25

MFcalculated (1.1921) < MFlimit (2.0614)

Power transfer capability = MFlimi* SIL

=2.0614*87.12

=179.58

Since the power transfer capability for double circuit is near to the power to be transferred .

s so Nc=2 is taken in this design. Since the voltage level 132Kv is not meet the voltage

regulation for any conductor available. So we have to chose double circuit 220 KV.

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Calculation of Insulation Discs

For all the calculations of number of insulator discs, we considered following value of different factor:

FOWR = Flashover Withstand Ratio =1.15NACF = Non Atmospheric Condition Factor = 1.1FS = Factor of Safety =1.2

1) Number of Insulator Discs Required for the temporary O/V

Temporary O/V = Earth Factor (EF) * Maximum system voltage = 0.8 * (220*√2 * 1.1)

= 273.7917 KVEquivalent Flashover Voltage (Veq= Temporary O/V * FOWR *NACF * FS

Equivalent Voltage (Veq) = 273.7917* *1.15 * 1.1 *1.2 = 415.6158 KV

From standard table number of insulator discs required to withstand above equivalent voltage (Na) = 11

2) Number of Insulator Discs Required to withstand continuous operating voltage:

a) Voltage Level for dry condition = Equiv dry 1 min. voltage*FOWR*NACF * FS

Where,

Equivalent dry 1 min voltage is taken from standard table = 435 KV

Equivalent 1 min dry flashover Voltage = 265 * 1.15 * 1.1 *1.2

= 660.33 kV

From Standard Table number of discs required to withstand above equivalent voltage level (Nb1)

=12

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Voltage Level for given Power Transmission =220 KVNumber of Circuit = 2Power Factor ( cosØ ) = 0.96

Length of Transmission Line (L) = 200Km

b) Voltage Level for Wet condition = Equiv dry 1 min. voltage*FOWR*NACF * FS

Where,

Equivalent wet 1 min voltage is taken from standard table = 395 KV

Equivalent Voltage Level = 395* 1.15 * 1.1 *1.2

= 599.61 KV

From Standard Table number of discs required to withstand the above equivalent voltage level (Nb2) =9

3) Number of Insulator Discs required for switching over voltage:

Voltage Level = switching o/v * Switching to impulse ratio * FOWR *NACF* FS

Where,

Switching to impulse ratio (SIR) = 1.2

SSR = Switching Surge Ratio =2.75

Switching O/V =SSR* (√2 /√3 )*Max system voltage*SIR

So Switching O/V =652.0541KV

Equivalent flashover Voltage Level = 652.0541*1.15*1.1*1.2

= 989.8182 KV

From Standard Table number of discs required to withstand the above voltage level (Nc) = 11

3) Number of Insulator Discs required for over voltage due to Lightning:

From standard table Equivalent voltage level for the given system voltage =900 KV

Equivalent flashover Voltage Level = 900* FOWR *NACF* FS

= 900*1.15*1.1*1.2

= 1366.2

From Standard Table number of discs required to withstand the above voltage level (Nd) = 16

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Required Number of Insulator Discs =16

Hence from above table the required number of insulator discs to withstand all types voltage

level in all condition for given system voltage = 16

Air Clearance Calculation

Air clearance parameters calculation for double circuit tower configuration:

a = minimum distance (clearance requirement) from a line conductor to any earthed

object, and is given by the following relation:

a =6.5 inch per 10 KV (rms)

=(6.5*220*1.1)/( √3∗10 )+8

=98.8171 inch

=250.995cm

=2.5099m

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SN

Voltage Level Description Voltage Level Number of

Discs

1)

2)

3)

4)

Temporary O/V appearing across the insulator

Continuous voltage

i) Continuous operating Voltage in Dry condition

ii) Continuous operating Voltage in Wet

condition

Switching Over voltage

O/V due to lightning

415.6158KV

660.33KV

599.61 KV

989.8182KV

1366.2 KV

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12

9

11

16

Air clearance from earthed object (a) = 2.5099 mString Length (l) = 3.5495 mTower Width (b) = 3.7648mCross Arm Length = 5.0198mVertical distance between two adjacent line conductor (y) = 6.3557mHorizontal distance between two adj. line conductor =13.804 mHeight of Earth wire from top cross arm (d) = 5.14504m

Now String Length (l) =√2a =3.5495m

Tower width (b) = 1.5*a = 3.7648m

Cross arm length(CL) = 2*a = 5.0198m

Vertical distance between two adjacent line conductor (y) =

( l+a )

√1−( xy )2

( l+a2a )2

Where 0.25 < x/y< 0.333

=

(3.5495+2 .5099 )

√1−(0 .25 )2( 3 .5495+2 . 50992∗2 . 5099 )

2

= 6.3557m

Horizontal distance between adj. conductor =5.5 a = 5.5*2.5099 =13.804 m

Height of earth wire from top (d) =√3 (CL)−l

= 5.14504m

Conductor and Tower Selection

Conductor and Tower Selection

Line current is calculated as:

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Line current (I) =

P/2√3∗V ll*cos φ =

(130 /2)∗1000

√3∗220∗0 . 96 = 177.688 amp

Comparing this value of the current with the current carrying capacity from the given standard

ASCR conductor table, Conductor “FERRET” is selected.

For Ferret conductor, From ASCR conductor table,

Resistance at 200C (R20) = 0.67950 Ώ/Km

Coefficient of Resistivity (α20) =0.004 /degreec

So Resistance at 650C (R65) = R20 (1 + α20(65-20))

= 0.67950 (1+0.004*45)

= 0.80181Ώ/Km

Transmission Efficiency Criterion

Power Loss for single conductor = I2*R*L

= (177.688)2*0.80181s*200

= 5.063MW/conductor

Transmission Efficiency (ή) = 65 / (65+3*5.063)

= 0.8105

= 81.058%

Transmission Efficiency<94 %. So this conductor is not used used . To get higher efficiency we

proceed in same way and calculate efficiency finally we got the conductor “Wolf” which has

efficiency 94.036%.

Voltage Regulation Criterion

For Conductor Wolf,

This conductor has 37 strands 7Aluminum strands and 1steel strands.

Diameter of each strands = 2.59mm

Conductor diameter (D)=18.13mm

Radius of the conductor(R)=9.065mm

GMR for inductance (GMRi) =0.768R

=6.1692smm

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GMR for capacitance (GMRc) = R

=9.065mm

Geometric mean distance(GMD)=3√ y∗y∗2 y

=3√6.3557∗6.3557∗12.7114

=8.00768m

Resistance of the whole line (R) = 43.518 Ώ

Inductance of Whole length (L) =2 * 10-4 ln (GMD/GMRi)*L

=2e-4*ln(8.00768/6.1692∗10−3)*200

=0.2867H

Capacitance of whole Length(C) =

2 Πε

ln ( GMDGMRc )

∗L∗e 3

= 2∗π∗ε

ln (8.007689.065e-3

)

=1.640µF

Now Impedance of the Line (Z) = R + j 314.15 *L

= 43.518+j90.0694

= 100.031<64.211

Susceptance of the Line (Y) = j314.15*C

= j5.1522* e-04

= (5.1522* e-04) <90

A, B, C, D parameters calculation

A = 1 + ZY/2 = 0.9768<0.657

B = Z= 100.031<64.211

C= Y (1+ZY/4)

D = A

Now, Sending end Voltage (Vs) = A*Vr +B*Ir

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= 0.9768<0.657* 220*e3/√3<0 + 100.031<64.211* 177.688<-16.26

= 12.407*e4<0.657+17.774*e3<47.951

= 136.7502<6.1377KV

Now Voltage Regulation = (|Vs|/A -|Vr| )/|Vr|

= 0.1021

= 10.219%

Voltage regulation < 12% so this conductor Wolf can be used.

Corona Inception Voltage Criterion

For wolf conductors

Corona Inception voltage (Vci) =√3*21.1*GMR*m*δ*ln(GMD/GMR)

Where, m = factor of roughness =0.9 δ = relative density of air =.95

Vci = √3∗¿21.1 * 0.9065*0.9*0.95* ln (8.00768/9.065e-3) = 192.152KVHere Vci < Vs so Corona Inception Voltage criterion is met. So we go for next higher conductor. For conductor Panther

GMRc=10.5mmGMD=8.00768mVci= √3∗¿21.1 * 1.05*0.9*0.95* ln (8.00768/10.5e-3) =217.748KVFor conductor LionGMRc=11.13mmGMD=8.00768mVci= √3∗¿21.1 * 1.113*0.9*0.95* ln (8.00768/11.13e-3) = 228.787KVSince Vci>Vs for conductor Lion. So no corona occurs in that conductor.

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Tension Calculation for Different Conductors with Different Span in Different Condition

Four Different conductors below conductor LION in ASCR conductor table is chosen. Hence Tension calculation will be done for conductor “lion, Bear, Goat , Sheep and Dear” with Span length 250 m, 275 m, 300 m, 325 m, and 350 m. Tensions for Toughest, Stringing and Easiest condition are calculated and tabulated below. Sample calculation is also shown.

Sample calculation; Conductor 1: Lion

Area of conductor = 2.94 cm2

Linear expansion coefficient = 1.773 e-5 per 0CModulus of elasticity (E) = 7.87 e+5 kg/cm2

Tension at toughest condition (T1) = (ultimate tensile strength)

( factor of safety ) = 10210

2 = 5105Kg.

Weight of conductor (wc) = 1097Kg/KmWind pressure (Fw) = 100 kg/m2

Weight due to wind force (ww) = Fw*d*(2/3) =100*1000*22.26e-3*2/3 = 1484kgWeight of ice (wice) = 0Weight for toughest condition (w1) = 2√ (wc+wice )2+ww 2 = 1845.444 kg

Weight for the stringing and easiest condition (w2) = 1097kg.

Then, the tension at stringing (T2) and the easiest condition (T3) is calculated using the following equation known as STRINGING EQUATION.

T22 [T2+k1] - k2 = 0 ………………………………….. (1)

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Where,

K1={−T 1+α (θ2−θ1 ) AE+W

12 L

2

24T12

AE}and,

K2=

W22 L

2

24AE

.From the above data, the values of K1 and K2 for the span of 250m are given by:K1 = -3210.60; K2 = 7.25e+9;

Using the stringing equation, the value of T2 is found to be, T2 = 3731.26 kg.

Similarly, T3 is calculated by the similar procedure as above. For the calculation of T3 the value of K1 and K2 for 250 m of length are given by:

K1 = -903.54; K2 = 7.25e+9;

Using the stringing equation, we get the value of T3 as:T3 = 2288.303 kg.

In the similar manner, the values of tensions of different conductor for different span length is calculated and presented in the table below:

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Sag and height of tower calculation

We have the relation,Maximum sag (Dmax) = (W L2)/ (8*T3);

Where, W = weight of conductor.L = length of span.T3 = tension at easiest condition.

Sample calculation:

For Lion, W = 1097kg/km. L = 0.25 Km. T3 =2288.303Kg.

Using the above equation, Dmax = 3.74527m.Let the minimum ground clearance (hg) = 7.5m.Height of lower conductor (H1) = hg+ Dmax =11.2452m.Height of middle conductor (H2) = H1+y = 17.60m.Height of top most conductor (H3) = H2+y = 23.956m.Total height of tower (Ht) = H3+l+d = 32.6512m.

Similarly, the maximum sag, H1, H2, H3 and the total height of the tower are calculated and presented in the table below:

Earth wire selection

From earth wire table, earth wire GUINEA is chosen as follows:No of strands = 19; diameter of a strand = 2.92mm; weight of conductor = 590kg/km;Conductor diameter = 14.60mm; conductor area = 127.20mm2;Ultimate tensile strength = 6664 kg;Hence, maximum tension (Te) = 6664/2 = 3332 kg.

Bending moment and tower weight calculation

Sample calculation: for Lion, Ht=32.6512m, 250m span

Taking 80% for tower A; 15% for tower B; and 5% for tower C

Due to Earth wire

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1) Bending moment acting on tower due to E/W considering wind force:BMe1 = Fwe*Ht*Ne

= (100*14.60e-3*250*2/3)*32.6512*2 = 15890.2506 kg-m.

2) Bending moment due to turning of the E/WBMe2=Fte*Ht*Ne

=2∗UTSFs

∗sinα2∗Ht∗Ne

BMe2 = 2*Te*(sin10*0.80 + sin7.50*0.15 + sin150*.05)*Ht*2 =2*3332*(0.046481)*32.6512*2 = 20227.378Kg-m

Due to power conductor1) Bending moment acting on tower due to power conductor considering wind force

BMpw = Fw*(H1+H2+H3)*2 = (100*22.26e-3*250*2/3)*(11.2452+17.60+23.956)*2

= 39178.4904 kg-m2) Bending moment due to turning of the power conductor

BMpt = 2*T1*(sin10*0.80 + sin7.50*0.15 + sin150*.05) *(H1+H2+H3)*2 = 2*5105*0.046481*52.8012*2 =50117.261kg-m

Then the total bending moment is calculated as;BMtotal = BMe1 + BMe2 + BMpw +BMpt

=125413.38 kg-m

Weight of towerWeight of tower (Wt) = 0.0016∗Ht∗√BM∗FS ; Where, Ht is in Ft; BM is in lb-ft.

=0.000631*Ht*√BM∗FS ; where, Ht is in m; Bm is in kg-m.

The bending moment of different conductors at different length of span and weight of tower are calculated and shown in the tabulated form below:

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Cost per unit length calculation

Assumptions: Cost of the steel used in tower = Rs.150000 per tonneNumber of towers (Nt) = ((total length)/(length of span))+1Cost of tower per unit length = (cost per tower*Nt)/length of transmission

Sample calculation:

For Lion; 250m; weight of tower= 10.30955 tonne

Number of tower = (200/0.25 ) +1= 801 nosCost per tower = cost per tonne* weight of tower

= Rs.150000*10.3185= Rs. 1547775

Cost of tower per unit length = (1547775*801)/200 = Rs.6198838.875

Similarly the cost of tower per unit length of different conductor and different span are shown in table below:

Most economical span and conductor selection

Data available;Cost of aluminum per tonne = Rs.20150Cost of steel per tonne = Rs. 150000

Sample calculation:

For Lion; span =325m; aluminum weight per Km = 659 kg; steel weight per km = 438 kg

Cost of power conductor

Cost of aluminum per km = Rs. 20150*0.659 = Rs.13278.85Cost of steel per km = Rs 150000*.438 = Rs. 65700Total Cost of power conductor = (Rs. 13278.85 + Rs. 65700)*6 = Rs.473873.1/km

From table above,Tower cost per unit length = Rs. 5915213.448

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Capital cost/length = conductor cost/km + tower cost/km = Rs.473873.1+ Rs. 5915213.448 (P) =Rs. 6389086.548

Annual capital cost (A) is calculated as:

A= (1+i)n∗i(1+i)n−1

∗PV

Where i=10% and n=25 yearsThis gives annual capital cost (A) = Rs. 703873.348Power loss per Km PL=27.338 Kw;

load loss factor (LLf)=k1*LF+K2*LF^2 = (0.2*0.5)+(0.8*0.5*0.5) =0.3

Cost of Energy loss/Km = PL *LLf * time* rate per Kwh =27.338*0.3*365*24*7.5

=Rs. 538831.98Total annual cost = Rs. 703873.348+ Rs. 538831.98

=Rs. 1242705.328The total annual cost calculation is shown in table below:

From the above table,The minimum total annual cost is acquired by the conductor Sheep. Hence we go

for the conductor Bear for the transmission line of 130 Mw and 200Km.

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Transmission line characteristics of the conductor Sheep

Electrical characteristics

67 strands with 7 Steel strands and 32Aluminum strands.Diameter of each strands = 3.99 mm

GMR for inductance (GMRi) = 10.7251mm GMR for capacitance (GMRc) =13.965mmResistance of the whole line per phase(R) = 18.3395Ώ Inductance of Whole length (L) =0.264HCapacitance of whole Length(C) = 1.7517 μFNow Impedance of the Line (Z) = 84.9414<77.531 ΏSusceptance of the Line (Y) =j5.5031*e-04 siemen

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A, B, C, D parameters calculation

A = 1 + ZY/2 = 0.979<0.2711B = Z= 84.9414<77.531C = 5.440*e-04<90.146D = A=0.979<0.2711

Now, Sending end Voltage (Vs) = A*Vr +B*Ir

= 0.979<0.2711* 220/√3<0 + 84.9414<77.056* 177.688*e-03<-16.26 = 132.4296<5.9652KV

Sending end current(Is) = C*Vr +D*Ir

= 168.563<7.2185

Now Voltage Regulation = (|Vs|/A -|Vr| )/|Vr| = 0.0649 = 6.497%

Corona Inception Voltage Criterion

Corona Inception voltage (Vci) =21.1*GMR*m*δ*ln(GMD/GMR)

Where, m = factor of roughness =0.9 δ = relative density of air =0.95

Vci = √3∗¿21.1 * 1.3965*0.9*0.95* ln (8.00768/13.965e-3) = 277.161KV

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Mechanical characteristics

Length of span =350m.Tension at toughest condition = 7955kgTension at stringing condition = 5902.87kgTension at easiest condition = 3923.8231kgHeights:Maximum sag =6.73561 m; H1=14.2356m; H2=20.591m; H3=26.947m; Ht=35.6415m

Bending moment =178830.9kg-m

Tower weight=14.3985 tonneTower Cost per unit length=Rs.6187763.968 /KmTotal annual cost =Rs. 1106280.389/km

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