5 Ruling Span or Equivalent Span (Spans of Unequal Length)

388



Electrical Power Systems



Le = Lavg +



2

(Lmax – Lavg)

3



...(15.68)



where

Lavg = average span in line



1

=

n



n



å Li

i =1



...(15.69)



Lmax = maximum span in line

...(15.70)

= max[L1, L2, L3, ..., Ln]

The tension of line T can be calculated using this equivalent span length and expression for

sag is detined as

d=



wLe2

8T



...(15.71)



15.6 EFFECT OF ICE

The sag is determined for the span design at which the transmission line is constructed so that

accumulations of snow or ice and excessive temperature changes will not stress the conductor

beyond its elastic limit, may cause permanent stretch or fatigue failures from continued

vibrations. In mountaneons areas, the thickness of the ice formed on the conductor is very

significant. Accumulations of ice on the line conductor has the following effects on the line

design:

1. Increase the dead weight per meter of the line

2. Increase the projected surface of the line subject to wind pressure.

More likely configuration of a conductor with ice coating is shown in Fig. 15.7. However, for

the sake of simplicity, it can be assumed that ice coating is uniform over the surface of the

conductor, as shown in Fig. 15.8.



Fig. 15.7: More likely configuration of cross-sectional area of ice covered conductor.



Analysis of Sag and Tension 389



Fig. 15.8: Assumed configuration of cross-sectional area of ice covered

conductor (uniform ice coating).



where

ti = thickness of ice (cm), uniform over the surface of conductor

dc = diameter of conductor (cm).

The cross-sectional area of the ice is

Ai =



LNMb



1

p dc + 2ti

4



g



2



OQP



- dc2 cm2



\

Ai = pti(dc + ti) cm2

\

Ai = pti(dc + ti) × 10–4 m2

Volume of the ice per meter is

Vice = 1 × Ai m3/m

\

Vice = pti(dc + ti) × 10–4 m3m

Let the weight of the ice is wc(Kg/m3 ), so that the weight of ice per meter is,

wi = wcpti(dc + ti) × 10–4 kg/m

Therefore, total vertical load on the couductor per merter length is

wT = w + wi

where

wT = total vertical load on conductor per meter length.



...(15.72)



...(15.72a)

...(15.73)

...(15.74)



w = weight of conductor per meter length.

wi = weight of ice per meter length.



15.7 EFFECT OF WIND

We assume that wind blows uniformly and horizontally across the projected area of the conductor

covered without ice and with ice. Fig. 15.9 shows the force of wind on conductor covered without

ice and Fig. 15.10 shows force of wind on conductor covered with ice.



390



Electrical Power Systems



Fig. 15.9: Wind force on conductor without ice.



Fig. 15.10: Wind force on conductor covered with ice.



The projected area per meter length of the conductor with no ice is

...(15.75)

Sni = Ani.l

where

Sni = projected area of conductor covered without ice in square meter per meter

length.

Ani = cross-sectional area of conductor covered without ice in square meter.

l = length of conductor (meter)

For 1-meter length of conductor with no ice

Sni =



dc

´ 1 m2/m

100



...(15.76)



with ice covered

where



Swi = Awil



...(15.77)



Swi = projected area of conductor covered with ice in square meter per meter

length.

Awi = cross-sectioral area of conductor covered with ice in square meter.

l = length of conductor (meter)

For 1-meter length of conductor,



bd + 2t g ´ 1 m /m

c



i



2

...(15.78)

100

The horizontal force exerted on the line as a result of the pressure of wind without ice (Fig.

15.9) is

F = Sni.P

...(15.79)



Swi =



Analysis of Sag and Tension 391



For 1-meter length of conductor,

F=



dc

. p kg/m

100



...(15.80)



where

F = horizontal force due to wind pressure exerted on line (kg/m)

p = wind pressure (Kg/m2)

with ice covered (Fig. 15.10), it is

F = Swi.p

For 1-meter length of conductor,

F=



bd + 2t g . p kg/m

c



i



100

The effective load acting on the conductor is

we =



b



F 2 + w + wi



g



2



...kg/m



...(15.81)



...(15.82)



...(15.83)



Fig. 15.11 shows the force triangle



Fig. 15.11: Force triangle.



Therefore sag can be calculated as

d=



weL2

kg.

8T



...(15.84)



Example 15.2: A stress-crossing overhead transmission line has a span of 150 m over the

stream. Horizontal wind pressure is 20 kg/m2 and the thickness of ice is 1.25 cm. Diameter of

conductor is 2.80 cm and weight is 1520 kg/km, and an ultimate strength of 12900 kg. Use a

factor of safety of 2 and 912 kg/m3 for the weight of ice. Using the parabolic method, determine

the following:

(a)

(b)

(c)

(d)

(e)

(f)



Weight of ice in kg per meter

Total vertical load on conductor in kg/m

Horizontal windforce exerted on line in kg/m

Effective load acting on conductor in kg/m

Sag in m

Vertical sag in meter



392



Electrical Power Systems



Solution:

(a) Using eqn. (15.73)

wi =

wc =

ti =

\ wi =

\ wi =

(b) Using eqn. (15.74)

wT =

w=

\ wT =

(c) From eqn. (15.82)

F=



wcp ti(dc + ti) × 10–4 kg/m

912 kg/m3

1.25 cm, dc = 2.80 cm



912 × p × 1.25 (2.80 +1.25) × 10–4 kg/m

1.45 kg/m

w + wi

1520 kg/km, wi = 1.45 kg/m

(1.520 + 1.45) = 2.97 kg/m



bd



c



+ 2ti



100



g p kg/m



dc = 2.80 cm, ti = 1.25 cm,

p = 20 kg/m2



\F =



b2.80 + 1.25g × 20 kg/m

100



\ F = 0.81 kg/m

(d) Using eqn.(15.83),



we =



\we =



b



F 2 + w + wi



g



2



F 2 + wT2



F = 0.81 kg/m, wT = 2.97 kg/m

\ we =



(e)



T=



b0.81g + b2.97g

2



2



=3.078 kg/m



Ultimate strength

Factor of safety



Ultimate strength = 12900 kg

Factor of safety = 2.0

\T=



12900

= 6450 kg

2



Using eqn. (15.84),

d=

\d=



w e L2

8T



b g



3.078 ´ 150

8 ´ 6450



2



m = 1.342 m.



Fig. 15.12



Analysis of Sag and Tension 393



(f ) Vertical sag = dcos q

cos q =



\



2.97

3.078



vertical sag = 1.342 ×



(see Fig. 15.12)

2.97

m

3.078



= 1.295 m.



15.8 LOCATION OF LINE

The routing of a transmission or distribution lines requires thorough investigatians and for

selecting the most desirable and practical route, following points should be considered:

1.

2.

3.

4.



cost

cost

cost

cost



of

of

of

of



construction

easements

clearing

maintenance.



15.9 SAG TEMPLATE

For correct design and economy, the location of structures on the profile with a template is very

essential. Sag template is a convenient device used in the design of a transmission line to

determine the location and height of stuctures. Sag template can be relied upon to provide the

following:

1.

2.

3.

4.



Economic layout

Minimum errors in design and layout

Proper grading of structures

Prevention of excessive insulator swing



Generally two types of towers are used:

1. The standard or straight run or intermidiate tower.

2. The angle or anchor or tension tower.

The straight run towers are used for straight runs and normal conditions. The angle towers

are designed to withstand heavy loading as compared to standard towers because angle towers

are used at angles, terminals and other points where a large unbalanced pull may be thrown on

the supports.

For standard towers, for normal or average spans, the sag and the nature of the curve

(Catenary or Parabola) that the line conductor will occupy under expected loading conditions is

evaluated and plotted on template. Template will also show the required minimum ground

clearance by plotting a curve parallel to the conductor shape curve. For the standard tower and

same height, the tower footing line can also be plotted on the template. Tower footing line is

used for locating the position of towers and minimum ground clearance is maintained throughout.

Figure 15.13 shows the sag template used for locating towers. In fact there are no clear-cut

guide lines for locating the tower positions and sesveral other alternatives may be examined.

Ground clearance depends on voltage level and Table-15.1 gives the span length and ground

clearance at different voltage levels.



394



Electrical Power Systems



Fig. 15.13: Sag template for locating towers.

Table- 15.1: Span length and ground clearance

Voltage

level



span length

(m)



minimum ground

clearance (m)



0.4 KV

11 KV

33 KV

66 KV

132 KV

220 KV

400 KV



80

100

150-200

200-300

350-360

360-380

400



4.6

4.6

5.2

6.3

6.3

7.0

8.8



Example 15.3: A galvanised steel tower member has original length of 22 cm and cross

sectional area 13 cm2. With working axial tensile load of 125 KN, the change in length was 0.2

mm. Calculate

(a) Stress (b) Strain (c) Modulus of elasticity (d) Percent elongation (e) If ultimate tensile

stress is 110000 N/mm2, determine the factor of safety.

Solution:

(a)



Stress =



\

(b)



Stress = 9.615×104 N/m2

Strain =



\



125 ´ 10 3

Tensile load

=

N/m2

Area of cross section

13 ´ 10 -4



Change in length

0.02

=

Original length

22



Strain = 0.000909.



(c) Modulus of elasticity =



Stress 9.615 ´ 10 4

=

N/m2

Strain

0.000909



= 10577.55 × 104 N/m2

(d)



Elongation =



Change in length

× 100%

Original length



Analysis of Sag and Tension 395



0.02

× 100%

22

= 0.0909%



=



(e)



Ultimate Stress

Working Stress

110000

=

= 1.14

9.615 ´ 10 4



Factor of safety =



Example 15.4: An overhead transmission line at a river crossing is supported from two towers

of heighs 40 m and 80 m above water level with a span of 250 m. Weight of the conductor is 1.16

kg/m and the working tension is 1800 kg. Determine the clearance between the conductor and

the water level midway between the towers.

Solution:

Using eqn. (15.65)

L

hT

–

2

wL

L = 250 m, h = (80 – 40) m = 40 m, T = 1800 kg

w = 1.16 kg/m

250 40 ´ 1800

\

x1 =

–

= – 123.27 m

2

116 ´ 250

.

x1 is negative means, both the towers are on the same side of the point of maximum sag

cansidering parabolic configuration. Fig. 15.14 shows this condition.



x1 =



Fig. 15.14



Horizontal distance of mid point, P from O (OP´)

L

250

=

– x1 =

– (–123.27)

2

2

= 248.27 m



396



Electrical Power Systems



Horizontal distance of point, B from O (OB¢ )

= L – x1 = 250 – (–123.27)

= 373.27 m

Therefore, height of mid point P above O,



w



FG L - x IJ

H2 K



2



1



116 ´ 2.48.27

.

=

= 19.86 m

2 ´ 1800

2T

Similarly, height of point B abve O,



d1 =



d2 =



b



w L - x1



g



2



=



b



116 ´ 373.27

.



g



2



= 44.9 m

2 ´ 1800

2T

Hence mid point P is (d2 – d1) = (44.9 – 19.86)

= 25.04 m below point B.

Height of the mid point P with respect to A

= (19.86 – 4.9) m = 14.96 m

Therefore, clearance between the conductor and the water level mid–way between the

towers will be

s = (40 + 14.96) m = 54.96 m

or

s = (80 – 25.04) m = 54.96 m

Example 15.5: An overhead transmission line at a river crossing is supported from two towers

at heights of 30 m and 70 m above the water level. The horizontal distance between the towers

is 250 m. If the required clearance between the conductors and the water midway between the

towers is 45 m and if both the towers are on the same side of the point of maximum sag, find the

tension in the conductor. The weight of the conductor is 0.80 kg/m.

Solution:



Fig. 15.15



Analysis of Sag and Tension 397



Assuming parabolic configuration as shown in Fig. 15.15.

L = 250 m, w = 0.80 kg/m.

Difference in level between the two supports

h = 70 – 30 = 40 m.

Note that both the towers are on the same side of the point of maximum sag. Hence x1 is

negative.

Using eqn. (15.61),

h=

As x1 is negative, x2 = L – x1



\



j



{bL - x g

w

h=

bL - 2 x g

2T

h=



\



e



w

2

2

x 2 - x1

2T



w

2T



1



2



2

- x1



1



}

...(i)



For points A and B, h = 40 m

\



b



g



0.8 ´ 250

250 - 2 x1 = 40

2T



250 - 2 x1

= 0.40

T

For points A and P, h = 45 – 30 = 15 m,

Horizontal distance between A and P

\



=



...(ii)



250

= 125 m.

2



Using eqn. (i)



0.8 ´ 125

(125 – 2x1) = 15

2T

125 - 2 x1

= 0.3

T

Dividing eqn. (ii) by eqn. (iii), we get

\



...(iii)



250 - 2 x1

0.4

4

125 - 2 x1 = 0.3 = 3

\

x1 = – 125 m

substituting x1 = – 125 in eqn. (ii), we get



b g = 0.40



250 - 2 -125

T

\



T = 1250 kg.



Ans.



398



Electrical Power Systems



Example 15.6: An overhead line is supported on two towers 300 m apart having a difference in

level of 10 m. The conductor radius is 1 cm and weighs 2.3 kg/m. Determine the sag at the lower

support when the line is subjected to wind pressure of 55 kg/m2 of projected area. The maximum

tensile strength of copper is 422 × 105 kg/m2. Factor of safety is 2.3.

Solution:

Span length, L = 300 m

Weight of the conductor, w = 2.3 kg/m

Radius of conductor = 1 cm

Diameter of conductor, dc = 2 cm.

Using eqn. (15.80)

F=



dc

p kg/m

100



p = 55 kg/m2

\



F=



2

× 55 kg/m = 1.1 kg/m.

100



Using eqn. (15.83), In this case

we =



\



F 2 + w2 =



.

b11g + b2.3g

2



2



we = 2.55 kg/m.



Fig. 15.16



Cross-sectional area of conductor Ac =



bg



p2



pdc 2

4



2



\



Ac =



\



Ac = 3.142 × 10–4 m2



4



cm2 = 3.142 cm2



\ Allowable Tension



422 ´ 10 5 ´ 3.142 ´ 10 -4

kg

2.3

T = 5764.88 kg.

T=



\