Transmission and distribution system design

Roll no. PUR072BEL005
Design of a Transmission line whose parameters are:
Selection of Most Economical Voltage and Number of Circuits
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 =95 km
P = Power to be transmitted =145 MW
CosØ = Power factor = 0.96
Nc = Number of circuits
For Nc= 1
Using the above values, we get
V eco = 5.5 * √
100
1.6
+
200∗1000
0.96∗1∗150
= 209.534 kV
∴ Nearest Standard Voltage (V1) = 220kV
For Nc= 2
Then, using the above values, we get
V eco = 5.5*√
100
1.6
+
200∗1000
0.96∗2∗150
= 151.319 kV
∴ Nearest Standard Voltage (V2) = 132 kV
Power to be transmitted (P) =200 MW
Transmission Line Length (Lt) =100 km
Standard Voltage levels
are:
66 kV
132 kV
220 kV
400 kV
500 kV
700 kV
750 kV
765 kV
1000 kV

Checking Technical Criterion
Surge Impedance Loading (SIL):
For Nc =1,
Characteristic Impedance (Zc) =400 Ω
SIL1 =
𝑉12
𝑍𝑐
=
2202
400
= 121 MW
For Nc =2,
Characteristic Impedance (Zc) = 200 Ω
SIL2 =
𝑉22
𝑍𝑐
=
1322
400
= 87.12 MW
Calculation of Multiplying factor (Mf):
For Nc=1,
Mf1 =
𝑃
𝑆𝐼𝐿1
=
200
121
= 1.653
For Nc=2,
Mf2 =
𝑃
𝑆𝐼𝐿2
=
200
87.12
= 2.296
From Table, for 100 Km line length, Mflimit of the line lies between 2.25 and 2.75.
Therefore, using interpolation to find Mflimit for 100km, we have
Mf𝑙𝑖𝑚𝑖𝑡 = 2.75 +
2.25 − 2.75
160 − 80
(95 − 80) = 2.625
Decision:
Here, Mf1<Mf𝑙𝑖𝑚𝑖𝑡 and Mf2<Mf𝑙𝑖𝑚𝑖𝑡 .
Here both are technically feasible. However the mf margin for Nc=1 is 0.972 and for Nc=2 is
0.329. Since, mf margin for Nc=2 is lower, we select double circuit.
Thus,
Line Length
(km)
Mflimit
80 2.75
160 2.25
240 1.75
320 1.35
480 1.0
640 0.75
Voltage Level for given Power Transmission =132 kV
Number of Circuits (Nc) = 2
Power factor (cosφ) = 0.96

Air Clearance and Conductor Spacing Calculation
1) Minimum air clearance required from earthed object is given by
𝒂 =
𝑉𝑟𝑚𝑠 × 1.1
√3
+ 30 𝑐𝑚
=
132 ∗ √2 × 1.1
√3
+ 30
= 148.55 𝑐𝑚 = 1.48 m
2) Maximum String Swing (ɵmax) = 450
3) Length of string or insulator hanging (l) = a Secɵmax= 148.55 * Sec450
= 210.08 cm =2.10 m
∴l=2.10 m
4) Cross arm length (CL) = a (1+tanɵmax) = 148.55 * (1+ tan 450
) = 297.1 cm
∴CL = 297.1 cm = 2.97 m
5) Vertical distance between two adjacent line conductor (y) =
22
1
)(





 








CL
al
y
x
al
Where, 0.25 < x/y< 0.333
Lets take x/y = 0.3
y =
210.08+148.55
√1−(0.3)2(
210.08+148.55
297.1
)
2
= 384.74 𝑐𝑚
∴Distance between two conductors(y) = 3.84 m
6) x= y * 0.3 = 3.84 m * 0.3 = 1.15 m
∴ x= 1.15 m
7) Width of tower (b) = 1.5a = 1.5* 1.48 = 2.22 m
∴ b = 2.22 m
8) Distance between the earth wire and the topmost cross-arm for double circuit:
d= √3 ∗ 𝐶𝑙 − 𝑙 = √3 ∗2.97– 2.10
= 3.04 m
∴ d = 3.04 m
9) Right of way (ROY) = 2*CL + b
= 2* 2.97 + 2.22 = 8.16 m

Therefore, height of tower = h +2 y + d
= Hg + s + 2y + d where, Hg = minimum ground clearance
= 6.1 + s + 2 * 3.84 + 3.04 And s = sag of the conductor
= (16.82 + s) m
Selection criteria for number of earth wire
Voltage levels No. of circuits (Nc) Number of earth wire (Ne)
66 kV *(1/2) 1
132 kV 1
2
1
2
220 kV 1
2
1
2
≥ 400 kV 2
*(1/2)
2
2
From above table, for double circuit of 132 kV, number of earth wire is 2
Air Clearance from Earthed Object (a) = 1.48 m
Length of String (l)= 2.10 m
Cross arm Length (CL) = 2.97 m
Width of Tower (b) = 2.22 m
Vertical Distance between two adjacent line conductors (y) = 3.84 m
Height of Earth Wire from Top Most Cross arm (d) = 3.04 m
Horizontal Distance between Two adj. line conductors
Or Right of Way (2Cl+b) = 8.16 m
Thus, number of earth wire(Ne) =2 is selected.

Number of Disc Selection
For all the calculations of number of insulator discs of size 254 *154 mm, we considered
following value of different factors:
FOWR= Flashover Withstand Ratio = 1.15
NACF = Non-Standard Atmospheric Condition factor =1.1
FS = Factor of Safety=1.2
Here, System Voltage = 132 kV and Max. System voltage = 145.2 kV
a. 1 minute Dry Test
Equivalent Flashover Voltage = 1 min. dry withstand voltage * FOWR * NACF * FS
Where, 1 min. dry withstand voltage is given in table A-2 for 145.2 kV system
voltage
= 265 kV
∴Equivalent FOV = 265 * 1.15 * 1.1 * 1.2
= 402.27 kV
Nearest higher 1 min. dry FOV voltage (in table A-3) = 435 kV
From table A-3, for 1 minute dry FOV = 435 kV, no. of discs = 7
∴No. of disc = Nd1 = 7
b. 1 minute Wet Test
Equivalent FOV = 1 min. wet withstand voltage * FOWR * NACF * F S
Where, 1 min. wet withstand voltage is given in table A-2 for 145.2 system voltage
= 230 kV
∴Equivalent FOV = 230 * 1.15 * 1.1 * 1.2 = 349.14 kV
Nearest higher 1 min. wet FOV voltage (in table A-3) = 370 kV.
From table A-3, for 1 minute dry FOV = 370 kV, no. of discs = 9
∴ No. of discs = Nd2 = 9
c. Temporary Over Voltage Test
Temporary o/v = EF * maximum system voltage
Where, EF = Earthing Factor = 0.8 (for Nepal)
∴Temporary o/v = 0.8 * 145.2 = 116.16 kV
Equivalent FOV = Temporary o/v * √2 * FOWR * NACF * FS
= 116.16 * √2 * 1.15 * 1.1 * 1.2

= 249.3695 kV
Wet season is the worst condition.
Thus, nearest higher FOV (from table A-3, 1 min. wet FOV) =250 kV
∴No. of discs = Nd3 = 6
d. Switching Over Voltage Test
Switching o/v = Maximum per phase peak voltage * SSR
Where, SSR = Switching Surge Ratio = 2.8
∴Switching o/v = 132 *
√2
√3
* 1.1* 2.8 = 331.49 kV
Equivalent s/w FOV = Switching o/v * SIR * FOWR * NACF * FS
Where, SIR = Switching to Impulse Ratio = 1.2
Equivalent FOV = 331.49 * 1.2 * 1.15 * 1.1 * 1.2
= 603.8 kV
The nearest higher voltage (in table A-3, impulse FOV) = 610 kV
e. Lightening Over Voltage Test
Equivalent impulse withstand o/v = 550 kV (from table A-2) for 145.2 kV
Equivalent impulse FOV = Equivalent impulse withstand voltage * FWR * NAC * FS
= 550 * 1.15 * 1.1 * 1.2
= 834.9 kV
The nearest higher voltage (in table A-3, impulse FOV) = 860 kV
S.N. Test No. of Discs
a. 1 min. Dry Test 7
b. 1 min. Wet Test 9
c. Temporary O/V Test 6
d. Switching O/V Test 6
e. Lightening O/V Test 9
From the table, it is seen that the minimum no. of disc required to withstand all tests is 9.
∴ The No. of discs required for our design (Nd) is 9.

Conductor Selection
I. Continuous Current Carrying Capability/
P = 200 MW, Nc = 2, VL = 132 kV, Cosφ = 0.96
Line current is calculated as:
Line current (IL) =
cos**3
/
llV
NcP
=
200∗10^6/2
√3∗132∗10^3∗0.96
= 455.611 A
Comparing this value of the current with the current carrying capacity from the given
standard ASCR conductor table, the conductor “LYNX” (with current carrying
Capacity 475 A) is selected.
II. Transmission Efficiency Criterion
For LYNX conductor, From ASCR conductor table,
Resistance at 200
C (R20) = 0.15890 Ώ/Km
Coefficient of Resistivity (α20) =0.004 /0
C (For Aluminum)
So Resistance at 650
C (R65) = R20 (1 +α20(65-20))
= 0.15890(1+0.004*45)
= 0.187502 Ώ/Km
Total Resistance of the line for 95 Km = 18.7502 Ώ
∴Total Power Loss (PL) = 3* IL
2
*R65 * Nc
= 3* 455.6112
*18.7502*2
= 23.3532 MW
∴ɳ = 1-
𝑃𝑙
𝑃
= 1-
25.3532
200
= 88.3234 %
This efficiency is <94%. So this conductor cannot be used. To get the higher
efficiency we proceed in the same way and calculate efficiency for SHEEP as shown
below.
For SHEEP conductor, From ASCR conductor table,
Resistance at 200
C (R20) = 0.07771 Ώ/Km
Coefficient of Resistivity (α20) =0.004 /0
C (For Aluminum)
So Resistance at 650
C (R65) = R20 (1 +α20(65-20))

= 0.07771 (1+0.004*45)
= 0.0916978 Ώ/Km
Total Resistance of the line for 95 Km = 9.16978 Ώ
∴Total Power Loss (PL) = 3* IL
2
*R65 * Nc
= 3* 333.792
*9.16978*2
= 11.421 MW
∴ɳ = 1-
𝑃𝑙
𝑃
= 1 –
11.421
200
= 94.2895%
Thus, this efficiency is >94% (i.e. 94.21%). So we select the conductor BEAR.
Conductor R20(Ώ/Km) R65 (Ώ/Km) Ploss (MW) ɳ (%) Remarks
LYNX 0.15890 0.187502 23.3532 88.3234 <94%
SHEEP 0.07771 0.0916978 11.421 94.2895 >94%
III. Voltage Regulation Criterion
The SHEEP conductor has 37 strands (30 Aluminum strands and 7 steel strands).
Diameter of conductor (D) =27.93mm
Radius of the conductor(R) =13.965mm
GMR for inductance (r’) =0.768R
=0.768 * 13.965
=10.72512mm = 1.073 cm
GMR for capacitance (r) = R = 13.965 mm.
= 1.3965 cm
Here,
Vertical distance between two conductors (y) = 3.84m
Cross arm length (CL) = 2.97 m
Width of tower (b) = 2.22 m
Horizontal distance between two conductors (2*CL + b) = 8.16 m (i.e. ROY)

Fig. Double Circuit Line Representation
i.e. Dacˈ =8.16 m = Dca’= Dbbˈ = Da’c = Db’b = Dc’a
Dab’= √(𝐷𝑎𝑐′)2 + (𝑦)2= √8.162 + 3.842 = 9.018 m=Dba’=Dbc’=Dcb’
Dca = 2*3.84 = 7.68 m= Dc’a’
Dab=3.84 m= Dbc= Db’a’= Db’c’
Daaˈ=√(𝐷𝑎𝑐′)2 + (2𝑦)2= √8.162 + (2 ∗ 3.84)2 = 11.2 m = Dccˈ
For GMD Calculation:
GMD = (Dab .Dab' . Da'b .Da'b' .Dbc .Dbc'.Db'c .Db'c' .Dca .Dca' . Dc'a .Dc'a')1/12
= ( 3.84*9.018*9.018*3.84*3.84*9.018*9.018*3.84*7.68*8.16*8.16*7.68)1/12
∴GMD=6.496 m
For GMR Calculation:
Here, for GMRL, r’=0.768R = 0.9 cm
Dsa= √𝐷𝑎𝑎 ∗ 𝐷𝑎𝑎′ ∗ 𝐷𝑎′ 𝑎′ ∗ 𝐷𝑎′𝑎
4
=√𝐷 𝑎𝑎′ ∗ 𝑟′ = √11.2 ∗ 9 ∗ 10^(−3) = 0.3174m=31.74 cm
Dsb = √𝐷𝑏𝑏 ∗ 𝐷𝑏𝑏′ ∗ 𝐷𝑏′ 𝑏′ ∗ 𝐷𝑏′𝑏
4
= √𝐷 𝑏𝑏′ ∗ 𝑟′ = √8.16 ∗ 9 ∗ 10^(−3) =0.2709m=27.09 cm
Dsc = √𝐷𝑐𝑐 ∗ 𝐷𝑐𝑐′ ∗ 𝐷𝑐′ 𝑐′ ∗ 𝐷𝑐′𝑐
4
= √𝐷𝑐𝑐′ ∗ 𝑟′ = √11.2 ∗ 9 ∗ 10^(−3) = 0.3174m= 31.74 cm
∴GMRL =√𝐷𝑠𝑎 ∗ 𝐷𝑠𝑏 ∗ 𝐷𝑠𝑐
3
=√31.74 ∗ 27.09 ∗ 31.74
3
= 30.107 cm

For GMRc,
Here, for GMRC, r=R=11.725 mm=11.725*10-3
m
Dsa= √𝐷𝑎𝑎 ∗ 𝐷𝑎𝑎′ ∗ 𝐷𝑎′ 𝑎′ ∗ 𝐷𝑎′𝑎
4
=√ 𝐷 𝑎𝑎′ ∗ 𝑟
= √11.2 ∗ 11.725 ∗ 10^(−3)
=0.3623 m =36.23cm
Dsb = √𝐷𝑏𝑏 ∗ 𝐷𝑏𝑏′ ∗ 𝐷𝑏′ 𝑏′ ∗ 𝐷𝑏′𝑏
4
= √ 𝐷 𝑏𝑏′ ∗ 𝑟
= √8.16 ∗ 11.725 ∗ 10−3
=0.3093m =30.93 cm
Dsc = √𝐷𝑐𝑐 ∗ 𝐷𝑐𝑐′ ∗ 𝐷𝑐′ 𝑐′ ∗ 𝐷𝑐′𝑐
4
= √ 𝐷𝑐𝑐′ ∗ 𝑟
= √11.2 ∗ 11.725 ∗ 10^(−3)
=0.3623m=36.23cm
∴ GMRC=√ 𝐷𝑠𝑎 ∗ 𝐷𝑠𝑏 ∗ 𝐷𝑠𝑐
3
=√36.23 ∗ 30.93 ∗ 36.23
3
= 34.36 cm
Now, Inductance per unit length (L) = 2 ∗ 10−7
∗ ln (
𝐺𝑀𝐷
𝐺𝑀𝑅 𝐿
) H/m
= 2 ∗ 10−7
∗ ln (
6.496∗100
30.107
)
= 6.143*10-7
H/m = 6.143*10-7
*103
*103
=0.6143 mH/km
∴Capacitance per unit Length(C) =
2𝜋𝜀
ln(
𝐺𝑀𝐷
𝐺𝑀𝑅𝑐
)
∗ 1000 𝐹/𝑘𝑚 [Ɛ0 = 8.85*10-12
F/m]
=
2𝜋∗8.85∗10−12
𝑙𝑛(
6.496∗100
34.36
)
∗ 1000 = 1.891*10-8
F/km
∴Capacitance per unit length = 1.891*10−8
F/km
Resistance of whole length = 12.255 Ώ
∴ Resistance per unit length = 0.12255 Ώ/Km
Impedance of the line (Z) = R65 + j XL =R+j(2π*f*L )
= (0.12255+j*(2π*50*6.143*10-4
) ) [f=50 Hz]
= (0.12255 +j 0.1929) Ω/km
=(0.12255 +j 0.1929) * 100 Ω
=12.255+j 18.333 Ω

= 22.051∠𝟓𝟔. 𝟐𝟑 0
Ω
Susceptance of the line (Y) = j w C = j*2π*50*1.891 *10-8
*95= j 5.643 *10-4
Siemens
= 5.643*10-4
∠900
Siemens
Calculation of ABCD parameters
Since 100 km line length lies on medium
Transmission line (i.e. 50 –200 km),
Calculation of parameters is done
Using π-model.
Fig: Nominal π- model of T.L.
𝐀 = 𝐃 = 1 +
YZ
2
= 1 +
5.643 ∗ 10−4
∠900
∗ 22.051∠56.230
2
= 𝟎. 𝟗𝟗𝟒∠𝟎. 𝟏𝟗𝟗° = 𝟎. 𝟗𝟗𝟒 + 𝐣𝟑. 𝟒𝟓𝟖
𝐁 = Z =22.051 ∠ 56.23
𝐂 = Y (1 +
YZ
4
) = 5.643 ∗ 10−4
∠900
(1 +
5.643 ∗ 10−4
∠900
∗ 22.051∠56.230
4
)
= −𝟗. 𝟕𝟓𝟕 ∗ 𝟏𝟎−𝟕
+ 𝐣𝟓. 𝟔𝟐𝟖 ∗ 𝟏𝟎−𝟒
= 𝟓. 𝟔𝟐𝟖 ∗ 𝟏𝟎−𝟒
∠𝟗𝟎. 𝟎𝟗°
∴| 𝐼 𝑅| @ full load = 333.79 A
Cos Φ = 0.95(lag)
∴Φ = -18.190
Then, IR =333.79∠-18.19 A
∴VR per phase @ full load =
132
√3
∗ 𝟏𝟎𝟎𝟎 ∠ 00
= ( 𝟕𝟔𝟐𝟏𝟎∠𝟎°)V
∴|VR| per phase @ full load = 76210 V = 76.21 kV
Therefore, Sending end voltage is given by
∴ VS (per phase) @ full load = A VR + B IR
=(0.994∠0.199°)*( 76210∠0°) +(22.051 ∠ 56.23)*333.79∠−18.190
=
𝟖𝟏𝟓𝟒𝟗. 𝟏𝟗 + 𝒋 𝟒𝟕𝟗𝟖. 𝟔𝟔
VS VR
IR
Y/2
ZIS
Y/2
2

= 81690.25∠3.36 V = 81.69 ∠ 3.36 kV
Hence, |VR| per phase @ no load = |
𝑉𝑠 @ 𝑓𝑢𝑙𝑙 𝑙𝑜𝑎𝑑
𝐴
| = |
81.69
0.994
| = 82.18 kV
∴Voltage Regulation (V.R.) =
| 𝑉𝑟@ 𝑛𝑜 𝑙𝑜𝑎𝑑|−|𝑉𝑟 @ 𝑓𝑢𝑙𝑙 𝑙𝑜𝑎𝑑|
|𝑉𝑟 @ 𝑓𝑢𝑙𝑙 𝑙𝑜𝑎𝑑|
=
82.18−76.21
76.21
= 7.83 %
Since calculated voltage regulation < 12 %, the conductor SHEEP satisfies voltage regulation
criterion.
IV. Corona Inception Voltage Criterion
For BEER conductor,
Maximum system voltage = 132 * 1.1 = 145.2 kV (rms)
Corona inception voltage (Vci) = √3 ∗ Air dielectric strength * GMRC * m * δ * 𝑙𝑛 (
𝐺𝑀𝐷
𝐺𝑀𝑅𝑐
)
Where, Air dielectric strength = 21.21 kV/ cm (rms)
GMRc = 34.36 cm
GMD = 649.6 cm
m = Roughness factor = 0.9
δ = Relative density of air = 0.95
∴Vci= √3 ∗ 21.21 * 34.36 * 0.9 * 0.95 * 𝑙𝑛 (
649.6
34.36
)= 3170.909 kV
Since Vci> Maximum system voltage (145.2 kV), there is no corona effect on BEER
conductor.
So, Corona Inception Voltage criterion is satisfied and all the technical criteria is met by
BEER conductor.
Hence the best five conductors which satisfy all the criteria are:
Thus, the conductor SHEEP can be used for our design.
1. SHEEP 2. DEER 3. ZEBRA
4. ELK 5. MOOSE

Tension Calculation for Different Conductors
1. Toughest condition -T1 tension and Sag is minimum (Dmin).
- Wt. of conductor (w1)
2. Normal Operating Condition (Stringing Condition) – T2 tension and S2 sag
3. Easiest condition – T3 tension and Sag is maximum (Dmax).
Let wc = weight of conductor per unit length
ww = weight per unit length due to wind
wice= weight per unit length due to ice
∴ Weight during toughest condition =
w1 = √(𝑤𝑐 + 𝑤𝑖𝑐𝑒)2 + 𝑤 𝑤
2
Calculation of Tension @ toughest condition (T1)
T1≤
𝑈𝑇𝑆
𝐹𝑆
where, UTS = Ultimate Tensile Strength of the conductor
FS = Factor of safety = 2
Calculation of Tension at Normal condition (T2)
T2 is given by stringing equation
T2
2
(T2 + k1) – k2 = 0
Where,
k1 = -𝑇1 + α (θ2 – θ1) A ϵ +
𝑤1
2 𝑙2
24 𝑇1
2 Aϵ
k2 =
𝑤2
2 𝑙2
24
Aϵ
ϵ= Modulus of Elasticity
α= Coefficient of linear expansion
A = Cross-section area of conductor
θ2 = Temperature at normal condition = 270
C
θ1 = Temperature at toughest condition = 00
C
w1 = per unit length conductor weight @ toughest condition
w2 = per unit length conductor weight @ stringing condition
Calculation of Tension @ Easiest condition (T3)
wc + wice
ww
w1

T3 is given by Stringing equation
T3
2
(T3 + k1’) – k2’= 0
Where,
k1’ = -T2 + α (θ3 – θ2) A ϵ +
𝑤1
2 𝑙2
24 𝑇2
2 Aϵ
k2’ =
𝑤2
2 𝑙2
24
Aϵ = k2
θ3 = Temperature @ easiest condition = 650
C
Now, we perform tension calculation for conductors “SHEEP,DEER,ZEBRA,ELK,MOOSE”
with Span length 250 m, 275 m, 300 m, 325 m, and 350 m. The Tensions for Toughest, Stringing
(Normal) and Easiest condition are calculated and tabulated below.
Sample Calculation
For SHEEP Conductor
Area of conductor (A) = 462.60 mm2
Coefficient of linear expansion (α) = 17.73*10-6
/0
C
Modulus of Elasticity (ε) = 0.789*106
kg/cm2
Ultimate Tensile Strength (UTS) = 15910 kg
Wt. of conductor per unit length (wc) = 1726 kg/km
Wind Pressure (wp) = 100 kg/m2
Conductor diameter (d) = 27.93 mm
Thickness of ice (t) = 10 mm = 0.01 m
‫؞‬Wt due to wind (ww) per km = (wp*1000)*(d*2/3) kg/km
= 100*1000*27.93*10-3
*2/3
= 1862kg/km
Wt due to ice loading (wi) = π t (d+t) ρice *1000 kg/km
= π * 0.01 (0.02345+0.01)* 950 *1000
= 998.31 kg/km
‫؞‬Wt. @ Toughest condition (w1)= √(𝑤𝑐 + 𝑤𝑖)2 + 𝑤 𝑤
2
= √(17262 + 18262
= 25838.921 kg/km
‫؞‬ Wt. @ Stringing Condition (w2) = wc = 1726 kg/km
‫؞‬Wt. @ Easiest Condition (w3) = wc = 1726 kg/ km

Temperature @ Toughest condition (ɵ1) = 00
C
Temperature @ Normal Condition (ɵ2) = 270
C
Temperature @ Easiest Condition (ɵ3) = 650
C
Calculation of T1
T1 =
𝑈𝑇𝑆
𝐹𝑆
=
15910
2
= 7955kg ( since, T1=110.9kN=11316.3kg)
Calculation of T2
k1 = -T1 + α (θ2 – θ1) A ϵ +
𝑤12 𝑙2
24 𝑇12 Aϵ
k2 =
𝑤22 𝑙2
24
Aϵ
From stringing equation,
T2
2
(T2 + k1) – k2 = 0
or, T2
3
+ k1 T2
2
– k2 = 0
or, T2
3
-5239.54 T2
2
– 2.831*1010
= 0
∴T2= 6020.697 kg
Calculation of T3
k1’ = -T2 + α (θ3 – θ2) A ϵ +
𝑤12 𝑙2
24 𝑇22 Aϵ
= -1871.33
k2’ = k2 = 2.8316*1010

From String’s equation, T3
2
(T3 + k1’) – k2’ = 0
or,T3
3
+k1’T3
2
– k2’ = 0
or,T3
3
– 2.831*1010
=0
or, T3 = 3815.932 kg
Hence, five best conductors have been chosen. The tension due to these five different conductors
for different span length from 250m to 350m (step of 25m) are shown in the following tables :

Sag and Tower Height Calculation
The maximum sag between two towers is given by
Dmax =
𝑤3𝑙𝑠 𝑝2
8𝑇3
Where, w3= weight of conductor per unit length @ easiest condition.
Lsp = Span length
T3 = Tension @ easiest condition.
Minimum ground clearance = hg =
( 𝑉𝑠𝑚𝑎𝑥−33)
33
+ 17 𝑓𝑒𝑒𝑡
Where, Vsmax= Maximum system voltage.
Now,
Height of lowest conductor = h1 = hg + Dmax
Height of middle conductor = h2 = h1 + y [For Nc=2]
Height of topmost conductor = h3 = h2 + y
Height of tower = ht = h3 + d’
From air clearance section, we have
d’ =distance between earth wire and topmost power conductor = 3.04 m
y= vertical distance between two conductors = 3.84m

Vsmax= 132*1.1 =145.2 kV
hg=
(145.2−33)
33
+ 17 𝑓𝑒𝑒𝑡 = 20.4 ft. = 6.21792 m
Sample Calculation
For SHEEP Conductor
w3 = wt. per unit length @ easiest condition = 1219 kg/km. (=w2 in table 1.1)
lsp= 250m = 0.25 km (assume)
T3 = 1684.78 kg (from table 1.1)
= 3.533m
∴ h1 = hg + Dmax = 6.21792+3.533= 9.75092 m
h2 = h1 + y = 9.75092+2.00967 = 11.76059m
h3 = h2 + y = 13.77026 m
ht = h3+d’ = 19.2687 m
Similarly, the maximum sag, h1, h2, h3 and the total height of the tower for different span lengths
for five best conductors are calculated and presented in the table shown here:

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
Diameter of earth wire (de) = 14.60mm
Area of Conductor = 127.20mm2
Ultimate tensile strength = 6664 kg
Hence, maximum tension (T1e) = 3332 kg

Bending Moment and Tower Weight Calculation
For design purpose, we consider 80% of the towers are of class A, 15% of the towers are of class
B and 5% of the towers are of class C.
The bending moment acting on the tower are due to the followings:
 wind force on power conductor (BMPw)
 wind force on earth wire (BMEw)
 Turning of power conductor (BMPT)
 Turning of earth conductor (BMET).
Sample Calculation
For SHEEP and span length (lsp) = 250m
a. BM due to power conductor
BM due to wind force (BMPw)
= Fwp * (h1+h2+h3)*Nc
Where,
Fwp = Wind force =wp * dp * lsp * 2/3 wp =
Wind pressure = 100kg/m2
dp = diameter of
power conductor = 27.93 mm = 27.93*10-3
m
h1 = height of bottom most conductor =
9.75162m (table 2.1) h2 = height of middle
conductor = 11.76129m (table 2.1) h3 =
height of top most conductor = 13.77096 m
(table 2.1) Nc = No. of circuits = 1
lsp = span length = 250m
∴ BMPw = wp* dp*lsp* 2/3* (h1+h2+h3)*Nc
= 100*27.93*10-3
*250*2/3*(9.75162 +11.76129 +13.77096)*2
= 32847.32787 kgm

BM due to turning (BMPt)
= 2*T1*(0.8 sin10
+0.15 sin7.50
+ 0.05 sin 150
)*(h1+h2+h3)*Nc
Where,
T1 = Tension @ toughest condition = 7955kg (table 1.1)
∴ BMPt = 2*7955*(0.8 sin10
+0.15 sin7.50
+ 0.05 sin 150
)* (9.75162
+11.76129 +13.77096)*2
= 15910* 0.046482* 35.28177*2 = 52183.7574 kgm.
b. BM due to earth wire
BM due to wind force (BMEw)
= Fwe * ht * Ne
= wp*de*lsp*2/3*ht*Ne
Where,
Fwe = Wind force on earth conductor Wp = wind pressure =
100kg/m2
de = Diameter of earth conductor = 14.60 mm =
14.60*10-3
m (for GUINEA) lsp = Span length = 250m
ht = tower height = 19.26933m (from table 2.1)
Ne = No. of earth wire = 1
∴ BMEw = 100*14.60* 10-3
*250*2/3*19.26933*2
= 93774.434 kgm
BM due to turning (BMEt)
BMEt = 2 T1e*(0.8 sin10
+0.15 sin7.50
+ 0.05 sin 150
)*ht*Ne
Where T1e = Tension on earth conductor @ toughest condition
T1e = UTS of GUINEA /2 = 6664/2 = 3332 kg
∴ BMEt = 2*3332*(0.8 sin10
+0.15 sin7.50
+ 0.05 sin 150
)*19.26933*2
= 11937.193 kgm
∴ Total Bending Moment (TBM) = BMPw+BMPt+BMEw+BMEt
= 106345.7123 kgm ∴ Tower weight (TW)
Where FS = Factor of Safety = 2

∴ TW = 0.000631*19.26933*√𝟏𝟎𝟔𝟑𝟒𝟓. 𝟕𝟏𝟐𝟑 ∗ 𝟐 = 5.60734 tonnes
Similarly, we can calculate the total bending moment and tower weight for different
conductors at different span length:

Tower Cost per Unit Length Calculation
Assumptions:
Cost of steel = Rs 1, 50,000 per tonnes
No. of towers = + 1 = Nt [Lt = Total Length and lsp = Span length]
∴Cost of tower per unit length = Cost per tower * Nt /Lt
Sample Calculation
For SHEEP and for span length = 250m lsp = 250m
Lt = 100 km
Tower weight (TW) = 5.60734 tonnes (from table 3.1)
Cost per tower = cost per tonne ∗ weight of tower
= 1, 50,000 *5.60734
=Rs. 841101
Cost of tower per unit length = Cost per tower * Nt / Lt
= 841101*401/100
= Rs. 3372815.01/km
Similarly we can find tower cost/km for 5 different conductors for their different span length. The
tower cost per unit length is shown in the following table:

Hence, the tower cost per unit length is minimum for SHEEP conductor with span
length 275 m.

Most Economical Span and Conductor Selection:
Assumption: Cost of Al/tonnes= Rs 20105
Cost of Steel/ tonnes = Rs 150000
Load Factor (LF) = 0.5
Loss of Load Factor (LLF) = k1*(LF)+k2*(LF)2
k1 = 0.2
And k2=0.8 ∴ LLF = 0.5 * 0.2 + 0.52
* 0.8 = 0.3 [Note:
k1+k2=1]
Per unit energy cost = Rs 7.50 /-
Life span (n) = 25 years Now, Rate of interest (i) = 10%
∴ Annuity factor (
Annual Capital cost = ϒ*capital cost per km
Capital cost per km = Tower cost per km (from table 4) + power conductor
cost per km
Total cost of power conductor per km = (cost of Al/km + cost of steel/km)
* Total no. of conductor
Cost of Al/km = Weight of Al/km * cost of Al/tons
Cost of Steel/km = Weight of Steel/km * cost of Steel / tons
Cost of energy loss/km = PL*LLF*time*Rate of cost
PL = Power loss = IL
2
* r65 * Total no. of power conductor
Total annual cost per km = Annual energy loss cost + Annual capital cost
Sample Calculation:
For SHEEP conductor
From table (4.1), the economical span length = 275 m
From ACSR table, Weight of Aluminum = 1036 kg/km
Weight of Steel = 690 kg/km
Total cost of power conductor per km = No. of conductors*cost of power conductor
per km/conductor
= 6* (20105*1036+150000*690)*10-3
= 745972.68
From above calculation,
Tower cost per km = Rs. 3343801.241 (for span length of 275m from table 4.1)

∴ Capital cost per km = Tower cost per km + Power conductor cost per km
= Rs (3343801.241+ 745972.68)
= Rs 4089773.921
Annual capital cost = ϒ * Capital cost per km
= 0.110168 * 4089773.921
= Rs 450562.2133per km
Power loss per Km (PL) = = IL
2
R65' * Total number of conductors
IL = 455.611 A [From conductor selection section]
R65 = 0.0916978 Ώ/Km [From Voltage Regulation section in conductor selection]
Total number of conductors= 6
∴ PL = 455.611 2
*0.0916978 *6 = 114208.537 W/km = 114.2085 kW/km
Annual Cost of energy loss per km = PL * LLF * time * cost per unit energy
= 114.2085 * 0.3*(365*24)*7.50
= Rs. 2251049.535/-
Total annual cost per km = Annual cost of energy loss per km + Annual capital cost
= Rs 2251049.535+ Rs 450562.2133
= Rs 2701611.748/-
Similarly,
We can calculate the Total annual cost per km for each conductor with their
respective economic span length. The tabulated form of the calculation is shown
below:
Hence,
From above table, it can be seen that MOOSE is the most economical
conductor with span length of 275m.

Transmission line Characteristics of the conductor MOOSE
A. Electrical Characteristics
The MOOSE conductor has 61 strands with 7 Steel strands and 54 Aluminum
strands.
Diameter of each strands = 3.53mm
Diameter of conductor (d) = 31.77 mm = 3.177 cm
Radius of conductor = 15.885 mm = 1.588 cm
GMR for inductance (GMRL) = 33.9041 cm
GMR for capacitance (GMRc) = 38.6856 cm
GMD for Double circuit = 686.5811 cm
Resistance of the whole length(R) = 9.16978 Ω@ 650
C
Inductance of Whole length (L) = 0.060164 H
Capacitance of whole Length(C) = 1.9333μF
Impedance of the Line (Z) = 9.1678 +j 18.901Ω
Susceptance of the Line (Y) = j0.6074*10-3
Siemens
Calculation of A, B, C, D parameters
A = 1+YZ/2 = 0.9943 < 0.16045°
B = Z (1+YZ/4) =21 . 0071 < 64.125°
C = Y = 0.000606 < 90.08°
D = A = 0.9943 < 0.16045°
Sending end Voltage (Vs) = A*Vr+B*Ir
= 82.5209<5.08190
kV (per phase)
∴ Voltage Regulation = (|Vs|/A -|Vr|)/|Vr| = ((82.5209/0.9943)- 76.21)/76.21
= 8.9%<12%

Corona Inception Voltage Criterion
Corona Inception voltage (Vci) =21.21*GMR*m*δ*ln(GMD/GMR)
∴Vci =√3 * 21.21 * 38.6856*0.9*0.95* ln (686.5811
/38.6856)
Vci = 3494.9778 kV>Vsmax
B. Mechanical characteristics:
Length of span =275 m
Tension at toughest condition = T1 = 8125 kg
Tension at stringing condition = T2= 5855.13 kg
Tension at easiest condition =T3= 3616.56 kg
Tower Heights:
H1 = 11.45084 m
H2 = 13.46051 m
H3 = 15.47018 m
Ht = 20.9685 m
Maximum sag (Dmax) = 8.74 m
Bending Moment on Earth wire due to Wind Force (BMEw) = 12797.919 kgm
Bending Moment on Earth wire due to Turning (BMET) = 12990.2765 kgm
Bending Moment on Power Conductor due to wind force (BMPw) = 47040.4545
kgm
Bending Moment on Power Conductor due to turning (BMPT) = 61002.9773 kgm
Total Bending Moment (TBM) = 133831.63 kgm
Tower Weight = 6.8453 tonnes
Tower Cost = 6.8453*(Rs 150000)= Rs 1026795
No. of Towers (Nt) = 365
Capital Cost per km = Rs. 3747801.75
Total Annual cost per km = Rs. 2092785.581

Transmission and distribution system design

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Transmission and distribution system design

Similar to Transmission and distribution system design (20)

Recently uploaded

Recently uploaded (20)

Transmission and distribution system design