Transmission Line designed on basis of data available for a given Hydropower system. Looking this document you can yourself design the Transmission Line system.

1. 1 | P a g e
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*
150**cos
1000*
6.1 Nc
PLt


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*
150*1*96.0
1000*130
6.1
200

= 177.137 KV
Nearest Standard Voltage= 220 kV
For Nc= 2
Then, using the above values
V eco =5.5*
150*2*96.0
1000*130
6.1
200

= 132.587 KV
Nearest Standard Voltage= 132 kV

2. 2 | P a g e
Checking technical criterion:
For Nc=1
Surge impedance Loading (SIL) = V2
/Z0
= 2202
/400 =121
Multiplying factor (MF) =
SIL
P max
= 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) =
SIL
P max
= 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.

3. 3 | P a g e
Calculation of Insulation Discs
For all the calculations of number of insulator discs, we considered following value of different
factor:
FOWR = Flashover Withstand Ratio =1.15
NACF = Non Atmospheric Condition Factor = 1.1
FS = 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 KV
Equivalent 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
Voltage Level for given Power Transmission =220 KV
Number of Circuit = 2
Power Factor ( cosØ ) = 0.96
Length of Transmission Line (L) = 200Km

4. 4 | P a g e
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

5. 5 | P a g e
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)/( 10*3 )+8
=98.8171 inch
=250.995cm
=2.5099m
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
11
12
9
11
16
Required Number of Insulator Discs =16

6. 6 | P a g e
Now String Length (l) = 2 a =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) =
22
2
1
)(





 








a
al
y
x
al
Where 0.25 < x/y< 0.333
=
 
2
2
5099.2*2
5099.25495.3
25.01
)5099.25495.3(





 


= 6.3557m
Horizontal distance between adj. conductor =5.5 a = 5.5*2.5099 =13.804 m
Height of earth wire from top (d) =
= 5.14504m
Conductor and Tower Selection
Conductor and Tower Selection
Line current is calculated as:
Air clearance from earthed object (a) = 2.5099 m
String Length (l) = 3.5495 m
Tower Width (b) = 3.7648m
Cross Arm Length = 5.0198m
Vertical distance between two adjacent line conductor (y) = 6.3557m
Horizontal distance between two adj. line conductor =13.804 m
Height of Earth wire from top cross arm (d) = 5.14504m

7. 7 | P a g e
Line current (I) =
cos**3
2/
llV
P
=
96.0*220*3
1000*)2/130(
= 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 200
C (R20) = 0.67950 Ώ/Km
Coefficient of Resistivity (α20) =
So Resistance at 650
C (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

8. 8 | P a g e
GMR for capacitance (GMRc) = R
=9.065mm
Geometric mean distance(GMD)=
=
=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/ )*200
=0.2867H
Capacitance of whole Length(C) = 3**
ln
2
eL
GMRc
GMD







=
=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
= 0.9768<0.657* 220*e3/√3<0 + 100.031<64.211* 177.688<-16.26

9. 9 | P a g e
= 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) = *21.1*GMR*m*δ*ln(GMD/GMR)
Where, m = factor of roughness =0.9
δ = relative density of air =.95
Vci = 21.1 * 0.9065*0.9*0.95* ln (8.00768/9.065e-3)
= 192.152KV
Here Vci < Vs so Corona Inception Voltage criterion is met. So we go for next higher conductor.
For conductor Panther
GMRc=10.5mm
GMD=8.00768m
Vci= 21.1 * 1.05*0.9*0.95* ln (8.00768/10.5e-3)
=217.748KV
For conductor Lion
GMRc=11.13mm
GMD=8.00768m
Vci= 21.1 * 1.113*0.9*0.95* ln (8.00768/11.13e-3)
= 228.787KV
Since Vci>Vs for conductor Lion. So no corona occurs in that conductor.

10. 10 | P a g e
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 0
C
Modulus of elasticity (E) = 7.87 e+5 kg/cm2
Tension at toughest condition (T1) = = = 5105Kg.
Weight of conductor (wc) = 1097Kg/Km
Wind pressure (Fw) = 100 kg/m2
Weight due to wind force (ww) = Fw*d*(2/3)
=100*1000*22.26e-3*2/3 = 1484kg
Weight of ice (wice) = 0
Weight for toughest condition (w1) =
= 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.
T2
2
[T2+k1] - k2 = 0 ………………………………….. (1)

11. 11 | P a g e
Where,
 








 AE
T
LW
AETK 2
1
22
1
1
24
121  and,
K2= AE
LW
24
22
2
.
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:

12. 12 | P a g e
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
1) Bending moment acting on tower due to E/W considering wind force:

13. 13 | P a g e
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/W
BMe2=Fte*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 conductor
1) 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-m
2) 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 tower
Weight of tower (Wt) = ; Where, Ht is in Ft; BM is in lb-ft.
=0.000631*Ht* ; 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:

14. 14 | P a g e

15. 15
Cost per unit length calculation
Assumptions: Cost of the steel used in tower = Rs.150000 per tonne
Number of towers (Nt) = ((total length)/(length of span))+1
Cost 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 nos
Cost 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.20150
Cost 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.13278.85
Cost of steel per km = Rs 150000*.438 = Rs. 65700
Total 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

16. 16
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=
Where i=10% and n=25 years
This gives annual capital cost (A) = Rs. 703873.348
Power 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. 538831.98
Total annual cost = Rs. 703873.348+ Rs. 538831.98
=Rs. 1242705.328
The 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.

17. 17
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) =13.965mm
Resistance of the whole line per phase(R) = 18.3395Ώ
Inductance of Whole length (L) =0.264H
Capacitance of whole Length(C) = 1.7517 μF
Now Impedance of the Line (Z) = 84.9414<77.531 Ώ
Susceptance of the Line (Y) =j5.5031*e-04 siemen
= 5.5031*e-04<90

18. 18
A, B, C, D parameters calculation
A = 1 + ZY/2 = 0.979<0.2711
B = Z= 84.9414<77.531
C = 5.440*e-04<90.146
D = 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 = 21.1 * 1.3965*0.9*0.95* ln (8.00768/13.965e-3)
= 277.161KV

19. 19
Mechanical characteristics
Length of span =350m.
Tension at toughest condition = 7955kg
Tension at stringing condition = 5902.87kg
Tension at easiest condition = 3923.8231kg
Heights:
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 tonne
Tower Cost per unit length=Rs.6187763.968 /Km
Total annual cost =Rs. 1106280.389/km