MECHANICAL DESIGN of TRANSMISSION LINE

Introduction

The mechanical design of this transmission line includes pole structure, wind load

pressure, sag and the tensions on the conductor and also the guy wires. Some details on the

line insulators and the line accessories are further discussed in this part.

Determination of Conductor Sag and Tension in Overhead Lines:

While erecting an overhead line, it is very important that conductors are safe tension. If

the conductors are too much stretched between supports in a bid to save conductor

material, the stress in the conductor may reach unsafe value and in certain cases the

conductor may break due to excessive tension. In order to permit safe tension in the

conductors, they are not fully stretched but are allowed to have a dip or sag.

The difference in level between points of supports and the lowest point on the

conductor is called sag.

Conductor Sag and Tension:

This is an important consideration in the mechanical design of overhead lines. The

conductor sag should be kept to a minimum in order to reduce the conductor material

required and to avoid extra pole height for sufficient clearance above ground level. It is also

desirable that tension in the conductor should be low to avoid mechanical failure of

conductor and to permit the use of less strong supports. However, low conductor tension

and minimum sag are not possible. It is because low sag means a tight wire and high

tension, whereas a low means a loose wire and increased sag. Therefore, in actual pratise, a

compromise in made between the two.

Loadings:

The strength to be designed into a transmission line depends to a large extent on wind

and ice loads that may be imposed on the conductor, overhead ground wire and supporting

structure.

When selecting appropriate design loads, the engineer should evaluate climatic

conditions, previous line operation experience and the importance of the line to the system.

Conservative load assumptions should be made for a transmission line which is the only tie

to important load centers.

Wind load and wind pressure affecting the tension of the line:

Formula:

Wind load = p x [D/12]

Where:

p = wind pressure lb/ft

p = 0.00256 (V)2 KzGRFCf

For Basic Wind Speed 200 km /hr considering the Transmission Line is at zone II

*the value for basic wind speed came from the “New Wind Load Provisions in Philippine

Structural Code” shown in the image below.

Where

Kz = velocity pressure exposure coefficient

Kz = 2.01(h/900)(2/9.5)

Where h = height of the wire at the structure and is between 33 feet and 900 feet

h = 30 + 0.3 (KV) = 30 +0.3(345kv) = 133.5 ft

Kz = 2.01 x 133.5900

2 /9.5

= 1.34

V = basic wind speed

V = 124.27 miles/hr

GRF = gust response factor

GRF= 0.70 (for 750 to 1000 span length in ft.)

Cf = the force coefficient = 1.0 for stranded wires

D = diameter of wire

* The constants found here are all based on the Electrical Engineering Handbook tables

for corresponding data.

Solution:

P = 0.00256 (124.27)2(1.34)(0.70)(1.0)

P = 37.08 lb/ft2

Wind load

WW=PxD12

lb / ft

WW=37.08x 1.21312

lb / ft

WW=3.74 lb / ft

Conductor Load

From Table, The conductor ORTOLAN w/ a cross sectional area of 1033 Kcmil, the nominal

weight is 1165/1000 ft.

WC=11651000

WC=1.165lb/ft

Effective Weight of Conductor Considering the Wind Load

W T=√WC2+WW

2lb / ft

W T=√1.1652+3.742lb / ft

W T=3.91 lb / ft

Calculation of Sag and Tension

In an overhead line, the sag should be adjusted that in the conductors is within safe

limits. The tension is governed by conductor weight, effects of wind, ice loading and

temperature variations. It is standard practice to keep conductor tension less than 50 % of its

ultimate tensile strength i.e. minimum factor of safety in respect to conductor should be 2. We

shall now calculate sag and tension of a conductor when (i) supports are at equal level (ii)

supports are at unequal levels.

(i) When supports are at equal level

Consider a conductor between two equilevel supports A and B with O as the lowest point is

shown in the Fig. It can be proved that the lowest point will be at the mid span.

L = Length of span

W = Weight per unit length of conductor

T = tension in the conductor

Sag, S = w l2

8T

The sag is as result of the tensioning of the line and must not be too low otherwise the

safety clearances may not be met. Also, the sag had to be such that it caters for ice loading in

the winter of temperate climates. If the sag is large, and the line becomes heavily loaded, then

the sag will further increase and branch the safety clearances. Similarity, if the sag is low, then

when the line contracts in the winter, low sag will indicate a high tension, and as a result of this

contraction, the line may snap.

Required Clearances:

1. Clearance of conductors passing by buildings

2. Minimum clearances of conductors above ground or rails

3. Crossing clearances of wires carried of wires carried on different supports

4. Horizontal clearances at support between line conductors based on sag.

Working Tension

Given:

Conductor Name = ORTOLAN

Ultimate Strength = 27,700 lb

Safety Factor = 2

L = 250 m 820.21 ft. span length

T = UltimateStrengthSafety factor

T = 277002

= 13850lb

Sag of Conductor

a. Consider conductor load only

Sag=WC x L

2

8 xT

Sag=1.165 x820.212

8 x 13850

Sag=7.07 ft

For bundling

Sag = 7.07 ft x 2

Sag = 14.14 ft. or 4.3 m

b. Consider both conductor and wind load

Sag=W T x L

2

8 xT

Sag=3.91 x820.212

8 x13850

SagT=23.74 ft

This is the slant sag in a direction making an angle θ with the vertical, where value of

θ is given by:

θ = tan-1Ww

W c = tan-1 3.74

1.165 = 72.74o

(ii) When supports are unequal levels

In hilly areas, we generally come across conductors suspended between supports at

unequal levels. Fig. shows a conductor suspended between supports A and B which

at different levels. The lowest point of the conductor is O.

Let

l = Span Length

h = Difference in levels between two supports

x1 = Distance of supports at lower level

x2 = Distance of supports at higher level

T = Tension

S1 = Sag using distance X1

S2 = Sag using distance X2

Calculations:

Span 820.12 ft

X1 + X2 = 820.12;

X1 = 820.12 - X2

For h = 20 ft (to assume that there’s a distance difference level of the 2 towers)

h=W C

2 xT(X22−X1

2 )

20= 1.1652 x13850 (X 2

2−(820.12−X2 )2 )

20 x2 x138501.165

=(X22−820.122+1640 X2−X22)

X2=700 ft

X1=820.12−X 2 ft

X1=820.12−700=120.12 ft

S1=WC x X1

2

2xT=1.165x 120.12

2

2 x13850

s1=6.225 ft

S2=WC x X 2

2

2xT=1.165 x700

2

2 x13850

S2=20.60 ft

LINE SUPPORTS

The supporting structure for overhead line conductors are various type of pole towers

called line supports. In general, the line support should have the following properties:

a. High mechanical strength to withstand the weight of conductor and wind loads etc.

b. Light in weight without the loss of mechanical strength

c. Cheap in cost and economical to maintain.

d. Longer life.

e. Easy accessibility of conductors for maintenance.

The line supports used for transmission and distribution of electric power are of various

types including wooden poles, steel poles, RCC poles, and lattice steel towers. The choice od

supporting structure for a particular case depends upon the line span, cross-sectional area, line

voltage, cost and local conditions.

Steel Poles. The steel poles are often used as a substitute for wooden poles. They possess

greater mechanical strength, longer life and permit longer spans to be used. Such poles are

generally used for distribution purposes in the cities. This type of supports need galvanized or

painted in order to prolong its life. The steel poles are three types., (i)rail poles (ii) tubular poles

and (iii)rolled steel joints.

Spacing and Clearance

1. Vertical Clearance - NESCode 232 Vertical Clearance (Above Ground, Roadway, Rail or

Water Suface)

Ground clearance required for a 345KV transmission voltage is 48 ft.

2. Spacing of Phase Conductors

S = 27.3 ft

3. Space between OHGW and Top Phase Conductor

Standard 1m for Neutral to top phase conductor = 3.28ft

Pole Structure:

Pole Height:

H = 30 + 0.30(Kv)

H = 30 + 0.30(345)

H = 133.5 ft.

At Tension = T = 13,850 lbs

Where:

H1 = height starting from the ground to the 1st phase conductor

H2 = height starting from the ground to the 2nd phase conductor

H3 = height starting from the ground to the 3rd phase conductor

H4 = height starting from the ground to the OHGW

H1 = 133.5 – 27.3 – 27.3 – 3.28 = 75.62 ft

H2 = 75.62 ft + 27.3 = 102.92 ft

H3 = 75.62 ft + 54.6 = 130.22 ft

H4 = 75.62 ft + 57.68 = 133.5 ft

Mass:

Where:

M = T1H1 + T2H2 + T3H3 + [(10% x T)(H4)]

M = [(13850) (75.62+ 102.92 + 130.22)] + (0.1) (13,850)(133.5)

M = 4,465,378 lb-ft = 53,584,542 lb-in

Circumference (c) bottom:

Where s = ultimate fiber stress using 46,350.36 lb/ in2

C=3√ M0.00026386∗S

= 3√ 53,584,5420.00026386∗46,350.36

C=163.63∈¿

Butt Diameter:

Butt Diameter=Cπ

=163.63π

=54.54∈¿

Butt Diameter = 54.54 in x 1 ft12∈¿¿ x

1m3.28 ft

= 1.38 m

Top Diameter:

Ratio of the top to bottom is 0.60

Top Diameter = 0.6 x 54.54 in.

Top Diameter= 32.72 in

Top Diameter = 32.72 in x 1 ft12∈¿¿ x

1m3.28 ft

= 0.83 m

Pole Circumference, Top:

C = top diameter x π

C = 32.72 x π

C = 102.8 in.

Taper

Taper=Cbottom−C top

60

Taper=163.63−102.860

=1.01∈¿

Projected Area:

A=(D¿¿ top+Dbutt )x2 x LPole x 12¿

A=(54.54+32.72 )(2)(133.5)(12)

A=279,581.04 ft2

POLE SETTING DEPTHS

It is indicated that the 10% of the total length of the pole above the ground is depth of

the portion of the pole to be on the ground.

The total length of pole above the ground is 133.5 ft. so the 10% of the 133.5 ft pole is

13.35 ft pole.

The additional 13.35 ft will be the depth of the pole to be under the ground. These

giving the total length of the pole to:

H = 133.5 + 13.35

H = 146.85 ft.

CORNER POLE:

The Angle 30˚ & 40˚ of two cables

Since T1 = T2 = 13,850 Kg is already computed

∑ Fv=0

Vertical Force:

Fv = T1 Sin 30 – T2 Sin 40

Fv=13,850 Sin 30 – 13,850 Sin 40

Fv= -1977.60

Horizontal Force:

Fh = T1 Cos 30 – T2 Cos 40

Fh = 13,850 Cos 30 - 13,850 Cos 40

Fh = 1384.74

Resultant Force:

R=√F h2+Fv2

R=√(−1977.6)2+(1384.74)2

R=2,414.20

Angle:

θ=tan−1 FvFh

θ=tan−1 1384.741977.6

θ=35O

GUY WIRES

The various grades of guy strand are almost universally furnished in accordance with

ASTM specifications. The ultimate strength for each size and grade is given. The so-called

double galvanized is commonly used. In transmission construction a factor of safety of 2 is

general for guys, although this may be somewhat reduced.

GUY REQUIREMENT: IEEE C2-1997 (EE Handbook pp(18-66))

Distance dig from the pole not less than ¼ or more than 1 to 1 ½ of the height of the guy

attachment.

GUY WIRE TENSIONS

D = 0.551 x H

D = 0.551 (146.85)

D= 80.91 ft

Length of guy

L1=√D2+H2

L1=√80.912+75.822

L1=110.88 ft

L2=√D2+H 2

L2=√80.912+103.02 L2=130.99 ft

L3=√D2+H2

L3=√80.912+130.222

L3=153.30 ft

L4=√D2+H2

L4=√80.912+133.5L4=156.10 ft

DESIGN DATA FOR GUYS

According to the guy requirements, for dead ends, the allowable stressed must be less

than 66.67% of the ultimate strength of the guy used.

Guy at Phase Conductors

Siemens-Martin grade Guy

Plow Steel

Diameter of 3/4 inch

26,200 lbs Nominal Breaking Strength

Guy at Static Wire

Siemens-Martin grade Guy

Mild Plow Steel

Diameter of 3/4

35,900 lbs Nominal Breaking Strength