Transmission Line Design Manual

UNITED STATES DEPARTMENT

OF THE INTERIOR

Water and Power Resources

Service

Denver, Colorado 1980

Transmission Line Design Manual

bY Holland H. Farr

A guide for the investigation,

development,

and design of power transmission

A Water Resources Technical

lines.

Publication

As the Nation’s principal conservation agency, the Department of the Interior has responsibility for most of our nationally owned public lands and natural resources. This includes fostering the wisest use of our land and water resources, protecting our fish and wildlife, preserving the environmental and cultural values of our national parks and historical places, and providing for the enjoyment of life through outdoor recreation. The Department assessesour energy and mineral resources and works to assure that their development is in the best interests of all our people. The Department also has a major responsibility for American Indian reservation communities and for people who live in Island Territories under U.S. administration.

On November 6, 1979, the Bureau of Reclamation was renamed the Water and Power Resources Service in the U.S. Department of the Interior. The new name more closely identifies the agency with its principal functions - supplying water and power. The text of this publication was prepared prior to adoption of the new name; all references to the Bureau of Reclamation or any derivative thereof are to be considered synonymous with the Water and Power Resources Service.

SI METRIC

UNITED

STATES

GOVERNMENT DENVER:

PRINTING

OFFICE

1980

For sale by the Superintendent of Documents, U.S. Government Printing Office, Washington DC 20402, and the Water and Power Resources Service, Engineering and Research Center, Attn D-922, P 0 Box 25007, Denver Federal Center, Denver CO 80225, Stock Number 024-003-00135-O

PREFACE The

purpose

followed

in the

of the line

Interior.

design,

such

of this

manual

is to outline

design

of power

transmission

Numerous

are included

aspects

protection,

spotting.

of the

National

problems

with

as selection clearance

structure

design

patterns, Safety when

the

sixth

some 16 000 circuit to properly distribute

made

other

codes

edition

are made

of NESC

as required.

lightning charts,

of the

are so noted;

sparce

by

and

Interpretations

Some

and

concerning

guying

construction.

considered

voltages engineers

of transmission

insulation,

and

to he

Department

is presented

was current,

while

of lines having power, Bureau

procedures U.S.

tensions,

limitation

to wood-pole

the aspects

Information sags and

structure

are limited

and

on specific

applications.

of NESC. of the Bureau, miles this

of Reclamation,

been

conductors, and

for,

Bureau

conductor

examples

most examples use the 1977 edition The transmission line network encompasses In addition,

have

of their

Code

requirements

by the

of construction,

design

developed

which

galloping

Structure

various

lines

explanations

of type

Electrical

were

studies,

the

some

up to and including have also designed

example however,

standards,

500 kilovolts. and built some

300 substations and switchyards. This total transmission system represents an installed transformer capacity of approximately 22 million kilovolt amperes. In many areas, a Bureau line is the only source of electricity and, if an outage occurs, an area may be completely without power. The vast land area covered

by

Bureau

lines

offers

almost

every

conceivable

type

large percentage of lines are in remote areas-maintenance Therefore, the line designs shown in this manual are more ordinarily be considered. The

Bureau

of Reclamation

recognized

the

need

for

this

of climatic

condition,

and

complete be readily

the manual available

engineers designing new This manual contains

manual

and

consequently

initiated

so that the design expertise gained through years of practical to other organizations as well as being a technical guide

lines and maintaining the engineering tools

many years of transmission line reference and guide for Bureau

design by the Bureau. designers. In keeping

metric units have been shown throughout the There are occasional references to proprietary not be construed in any processes of manufacturers other

facilities. that have

proven

its

of Energy in transmission to have the experience for Bureau

to be successful

over

The manual is not a textbook, but a useful with the Metric Conversion Act of 1975, SI

manual in addition to U.S. customary materials or products in this publication.

way as an endorsement, as we cannot or the services of commercial firms

units. These

must

endorse proprietary products for advertising, publicity, sales,

or or

purposes.

The author, as an electrical contributions Area

the remaining and concepts

a

is both difficult and time consuming. conservative than designs which might

preparation. With the advent of the Western Area Power Administration, Department October of 1977, many of the electrical power features of the Bureau, including most lines, were transferred to the jurisdiction of Energy. However, it was deemed prudent Bureau would

because

Power

Mr. Holland H. Farr, has more than 30 years of transmission line design experience engineer with the Bureau of Reclamation. He gratefully.acknowledges the many to this manual by the personnel of both the Bureau of Reclamation and the Western Administration.

Special recognition to H. J. Kientz for

suggestions, and consultation; R. D. Mohr who provided the technical Bureau of Reclamation, U.S. Department

is given to F. F. Priest his computer treatment

continuity. This of the Interior,

Cdlorado. . ..

111

for his encouragement, of the concepts; and

manual was prepared and Engineering and Research

published Center,

to

by the Denver,

ABBREVIATIONS ACSR

aluminum

conductor,

AIEE Alcoa

American

Institute

Aluminum

AND SYMBOLS steel

of Electrical

Company

ANSI

American

National

,4WG

American

Wire

BIL

basic

impulse

Standards

insulation

level

International extra

IEEE

Institute

K

conductor

loading

LP

low

(distance

MS1 NBS

maximum National

Bureau

NESC

National

Electrical

OGW SAS

overhead ground sum of adjacent

UHV USBR

ultra high voltage U.S. Bureau of Reclamation gigapascal

GPa Hz kcmil

hertz thousand

kPa kV*A

kilopascal kilovolt

kWh MPa

kilowatt

N/m N*m

Institute

Gage

EHV

point

Engineers

of America

CIGRE

high

reinforced

Conference

on Large

Electric

Systems

voltage of Electrical

and

Electronic

Engineers

constant between

low

sag increase of Standards Safety wire spans

circular

mils

ampere hour

megapascal newtons per meter newton meter

iv

Code

points

in adjacent

spans)

CONTENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Preface :\bbrc\

ialions

and

CHAPTER

syn~hols

I. BASIC

Field

data

Safety Cost

6 7 8 9 10

CHAPTER

.....................................

of type Single

wood-pole

(b)

H-frame,

(c) (d)

Single-circuit Double-circuit

(e)

Structures

(f)

Transpositions

structures

...................

4

steel structures steel structures

.................... ...................

s

................... long-span construction ..................... and effective spans ............................. Selection of conductors ................................ Stress-strain curves The parabola and the catenary ........................ Design

instructions

Transmission

data

II. CONDUCTOR

srnnmary

tension

calculations

19 20 21

9 10 14 21

form

23

...................

2s

................................

charts Preparation

........................................ of sag template

Inclined

spans

using

Coppcrwcld

sag calculating 29 32

..........................

38 SO

.................................... ............................... conductors

Galloping Broken

conductors

Insulator

effect

III.

7

SAGS AND TENSIONS

Sag and

Spans

7

................................

line

12

CHAPTER

6 6

Special ruling,

ion

18

6

..................

special conditions ..............................

informat

16 17

.4

structures

wood-pole

for

4

....................... .....................

of construction

(a)

General

1S

2

....................................

11

13 14

1 2

......................................

estimates

(g) Normal,

with

iv

DATA

codes

Selection

5

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

... III

56

................................ on sag and

concentrated

tension

in short

spans

...........

........................

loads

77 99

INSULATION, LIGHTNING PROTECTION, AND CLEARANCE PATTERNS

Insulation

coordination

Lightning Conductor

protection clearance

............................. ............................... ......................... patterns

V

103 106 111

TRANSMISSION

vi

CHAPTER

22

LINE DESIGN MANUAL

IV. STRUCTURE LIMITATION GUYING CHARTS

AND

127 127

General

........................................ .............................. Components of charts ............................... Preparation of charts

23 24

CHAPTER 25

V. ADDITIONAL Stresses Structure

26

in wood-pole spotting

266 266

required .................... ...........................

(c)

Determining

...........................

26%

(d) (e)

Insulator General

........................... ..........................

268 273

Kight-of-way Armor Corona

uplift sideswing instructions

and

building

clearance

sag data

(a)

Sag tables

(b)

Sag and

Transmission

274 282 284

......................

.................................

292

.................................

292 292 300

insulator

line

266

.....................

rods and vibration dampers ........................................

Stringing

Bibliography

213

........................ structures .................................

Data and equipment Process of spotting

28

31

DATA

(a) (b)

27 29 30

127

offset

equations

data

for

inclined

spans

........

...........................

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

303

APPENDIXES A.

A method for computing transmission spans adjacent to a broken conductor

B.

Useful

C.

Conductor

Index

figures

and and

tables

overhead

line

sags and ..................

tensions

307

............................ ground

wire

data

tables

in

............

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..~....

339 441 479

CONTENTS

1

Conductor for

and

USHR

Mathematical

transpositions calculation

form

(metric)

tension

calculation

form

(U.S.

Stress-strain furnished

by

tension

parabolic

and

and and

and curve

catenary showing

length

line data of standard

13 I-I

Typical

sag template

15

Sag and

tension

origin

16

template Sag and

form

for (metric)

form

for

on

example

on

construction

calcldation

span tension

Sag on inclined

span-parameter

of cxampk

% method

problem

on

22

parameter Conductor

23

Conductor

24

problem Overhead

25

example Overhead

26

IIalf-sag

sag

Zmethod sag and

(U.S. customary) tension calculation

galloping sag and

conductors

24 33

......

34 sag 36 problem

on

sag

..............

38

.............. ................

39 44

using 47

span using ..................

form

(metric)

ellipses

on

for

calculation

on

for

,49

example

................

galloping example

form

52

example ........... on galloping conductors (U.S. customary) ground wire sag and tension calculation form for .......... problem on galloping conductors (metric) ground wire sag and tension calculation form for problem

tension

form

problem

an inclined span ........................

21

example

18

36

method method

parameter %method (metric) Results of example problem on an inclirled

on

18

......................

span-equivalent span-average

problem

15 16 17

22

inclined Sag on inclined Restllts

as

...................

form tension

..........................

on

I4

problems ............ curves (U.S. customary) percentage relationship between

summary sag and

12

used

example problems ..................

customary)

Sag

of values

......................... .........................

calculation form for example ................................. (metric) tension calculation form for example (U.S.

11 ....

..................................

Transmission Explanation

template

.......... customary)

for an ACSR, 26/7 conductor ................. Association

curves

calculation

7

........................

equations equations

catenary

tension

span

illustrating

calculations

calculation

and

parabolic Catenary

curves

and creep curves the Almninum

Sag and Sag and

creep

tension

9

19 20

3

tension

and

criteria

..................

sag and

curve curve

18

design

sag and

Parabolic Catenary

17

for

ratenary

Standard

7 8

12

wire

.........................

Standard

in sag and

10

ground lines

solution

Stress-strain

6

overhead

transmission

vii

conductors problem

on

for

(U.S. galloping

customary) conductors

53 54 .... ...

54 55

. .. VIII

TRANSMISSION

LINE DESIGN

MANUAL Page

b'igrrw

Profile

28

Sag and problem

tension calculation form for ................................. (metric)

broken

conductor

29

Sag and

tension

broken

conductor

30 31

of spans

used

for

broken

calculation

form

Curves

for

broken

Sag template

for

33

Conditions condition

for

problem

problem

conductor

60 61

tension

for equilibrium before ......................................

(U.S.

to broken

and

after

68

sohltion

of unbalanced

condition

(metric)

Graphical

solution

of unbalanced

condition

(U.S.

36

Nomenclature

37

tension Sag and

39 40

tension

(U.S.

customary)

calculation

Spans Graphical

rnethod

43

conductor Reduction

required of angle

with

structure

concentrated

height

(metric)

tension

effect

for

insulator effect

problem

problem

(U.S. 94

........................

determining

additional

100 length

of ..........

for concentrated load problem of protection against lightning patterns

according

tension

the

three

types

to

of voltage 112

clearance

pattern

problem

calculation form for .................................

clearance

pattern

problem

form

113

for

side

view

114

of structure

at conductor 121

...................................... structure

49

Clearance

pattern for a 30s ......................................

tangent

structure

so

conductor Clearance

pattern

angle

for

101 110

for

pattern for a 30s tangent ......................................

Clearance conductor

effect

90

insulator

Clearance conductor

51

problem

for

47

conductor

problem

85

form

calculation

(U.S. customary) Assrmled dimensions

18

effect

.......................................

Sag and

elevation

insulator

..................................

16

76

on sag and 78

insulator

loads

for

Superimposed clearance stresses ............................................. Sag and

for

75 ...

81

Tension-temperature curve for customary) .....................................

42

4.5

form

effect

.......... customary)

.................................

41

44

determining insulator .............................. spans

Tension-temperature curve for (metric) ....................................... Sag and

67

unbalanced

Graphical

38

66 .....

conductor

35

(metric)

.......

customary)

due

34

in short tension calculation .......................................

6.5

.............

(metric)

problem

reduced

for

57

...........................

problem (U.S. customary) Curves for broken conductor

32

conductor

..........

27

a 30A

with

single 122

with

duplex 123

structure

with

single 124

...................................... pattern for a 30A angle .......................................

structure

with

duplex 125

CONTENTS

52

Condnctor

sag and

problem 53

on steel

Condnctor

iX

tension

calcnlation

form

for example

strnctnre

limitation

chart

(metric)

tension

calcnlation

form

for example

structure

limitation

chart

(U.S.

sag and

. . . . . . . . .

13s

. . .

136

54

Center

phase

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . of a steel structure limitation chart (metric) . . . . . . . . .

137

5s

angle Example

56

Example

limitation

chart

. .

148

57

Conductor problem

sag and tension on wood-structnre

calculation limitation

form chart

for example (metric) . . . . . . . . .

IS0

58

Conductor

sag and tension on wood-structure

calculation limitation

form chart

for example (U.S. customary)

59

Type

HS

Type

HSB

problem

on steel

of a steel

problem 60 61

for

type

structure

wood-pole

3OS steel

structure

(U.S.

no line

customary)

. .

. . . . . . . . . . . . . . . . . . . . . . . . .

wind force ground wire

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . sag and tension calculation form for

158

example

problem

on

wood-structure

. .

160

Overhead

grourld

wire

example

problem

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . sketch of one pole of a type FIS wood-pole

161

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

161

wood-pole structure sketch of wood pole

sag and

limitation

tension

chart

calculation

on wood-structure

chart

65 66

Single-line

67

structure with X-brace Force triangle showing

68

Force

69

limitation chart (U.S. Force triangle showing

70

Type

3A

71

Type

3AB

72

Type

3TA

73

HalfHalf-

and and

full-sag full-sag

ellipses ellipses

for for

type type

HS wood-pole HSB wood-pole

Half-

and

full-sag

ellipses

for

type

3AC

structure

sketch

limitation

chart

triangle

angle

of top

portion

HS

for (U.S.

wood-pole

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . angle of bias lines for wood-structure

163

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

168

(metric)

showing

of a type

(metric)

form

limitation

customary) Single-line

angle

of bias

for

wood-strncture

. . . . . . . . . . . . . . . . . . . . . . conductor force due to line

168

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

169 177

. . . . . . . . . . . . . . . . . . . . . . . .

178

wood-pole

customary) resnltant

lines

structure

wood-pole

structure

wood-pole

structure

. . . . . . . . . . . . . . . . . . . . . . . .

wood-pole

. . . . . . . . .

189

structure

. . . .

191 193

F&sag ellipses for type 3TA 4267-mm (14-ft) pole spacing

wood-pole structure, tangent, . . . . . . . . . . . . . . . . . . . . . . . .

77

Half-sag

wood-pole

4267-mm Full-sag angle,

ellipses

for

(14-ft) ellipses 11 278-mm

type pole

for

3TA spacing

type (37-ft)

3TA pole

structure,

spacing

structure,

187

tangent,

. . . . . . . . . . . . . . . . . . . . . . . . wood-pole

180

structure structure

76

78

151 154 1ss

compute Overhead

strncture

147

157

63

74 75

with

. . . . . . . . . . . . . . . . . . . . . . . .

62

wood-pole

customary)

strnctnre

. . . . . . . . . . . . . . . . . . . . . . . . showing values needed to

Type 3AC Single-line

64

V-string

90°

194

line

. . . . . . . . . . . . . . . . . .

195

TRANSMISSION

X

LINE DESIGN MANUAL

FigUIV 79

Page Half-sag

ellipses

angle, 80 81

Full-sag angle,

ellipses 4267-mm

Full-sag

ellipses

angle,

8230-mm

82

Half-sag

ellipses

angle,

4267-mm

83

Half-sag

ellipses

84

angle, Full-sag

85 86 87

angle, Half-sag angle,

88

for

11 278-mm

type (37-ft)

for type (14-ft) for

type

(27-ft) for

type

(14-ft) for

3TA

wood-pole

structure,

spacing

..................

pole 3TA pole

90 O line 196

wood-pole structure, spacing ....................

60’

3TA

wood-pole

60°

line

pole

spacing

structure,

60°

line

.................... structure,

60°

line

3TA pole

line 197

structure,

198

....................

wood-pole spacing wood-pole

199

type

3TA

8230-mm ellipses

(27-ft) for type

pole 3TA

spacing .................... wood-pole structure,

45 o line

angle, Half-sag

6096-mm ellipses

(20-ft) for type

pole 3TA

spacing .................... wood-pole structure,

45 o line

angle,

6096-mm

(20-ft)

Full-sag

ellipses

pole

spacing

type

3TA

wood-pole

4572-mm ellipses

(IS-ft) for type

pole 3TA

spacing .................... wood-pole structure,

4572-mm

(15-ft)

pole

spacing

for

200 201

.................... structure,

202 30 O line 203 30°

.................... limitation chart ...... chart (metric) ........ chart (U.S. customary)

89

Instructive Example

example of a wood-structure of a wood-structure limitation

90

Example

of a wood-structure

91

Additional

92 93

Example Example

94 95

Standard guying arrangement for type 3TA structure 29-m type HS 230-kV structure with class 2 Douglas

96

95-ft

97

(one X-brace) 29-m type HSB

98

poles (one X-brace) Free body diagram

99

Free

data

chart

required

limitation for

the

line

wood-structure

customary)

(one

chart chart

crosstie body

HS

structures structures

class

example of pole

2)

(one X-brace) .................................... Free body diagram of pole above

plane

102

Free

customary

customary

example

of pole

2 Douglas

fir

2 Douglas

209

211

poles fir 219

of inflection

and

to the

example between

221

planes

101

diagram

......... fir poleg

.......................... between

with

(U.S.

class

............................... of pole above plane

example 2) ..................................... 95-ft type HSB 230-kV structure

body

.......

217 with

100

crosstie

(metric) (U.S.

214 with

................................... 230-kV structure

diagram

207

210

................................... 230-kV structure

(metric

206 . .

208 wood-pole wood-pole

.....................................

X-brace) type

for for

205

limitation

......................................... guying guying

204

of inflection

(metric 223

class

2 Douglas

fir

poles 232

of inflection

and

to the

2)

.................... planes of inflection

2) ., .............................

234 (U.S. 235

CONTENTS

29-m

type

poles 104 105

HSB

(two

230-kV

3)

107

(two X-braces) Free body diagram

type

crosstie

HSR

(U.S.

230-kV

fir

243

customary

109

customary example Typical sag template

110

Typical

plan

and

111

superimposed Typical plan

and

diagram

profile

114

Average

and

Sag and insulator

example

3)

between

bundles

poles

(II)

snow

of inflection

(U.S. 2.59

spotting conductor

sag template

269 showing

use of sag template 271

284

fair

287 weather

with

different

................................

form

.................. voltages free running stringing

120

insulator offset Profile of spans

121

sag correction ................................... Stationing equation for common survey,

assumption

122

Stationing

equation

123

survey, Station

assumption designations

calculation

form

sheaves

insulator operations

offset and ...............

for

problem ...............

on

problem

on .........

sag correction

Sag and

example (metric)

for

and sag correction for example problem

example

(U.S. customary) on insulator offset

293 297

298 line

common point on a transmission No. 2 ........................... when station back is greater than

line

1

Station

back

designations

point

...........................

301

for

when

station

.........................................

297

and

on a transmission

No.

...

288 290 293

sag

301 station

ahead ......................................... 124

267

......

structures

..........................

119

tension

2S7

....................

with

loss under

for different when using

and

to the

waves in a conductor (A) fair weather, (B) rainfall,

calculation

offset

and

...............

required for calculating data during stringing tension

of inflection

planes

drawing

of corona

Corona loss curves Conductor tensions Dimensions correction

fir

..............................

Schematic of vibration Corona loss curves for

conductor

1 Douglas

255 plane

................................... profile drawing

113

valrles

class

3) .............................. (plastic) used for

uplift

hoarfrost,

with

above

of pole

in determining

(C)

247

structure

of pole

Free

body

24.5

..................................

108

118

1 Douglas

.....................................

95-ft

116 117

class

body diagram of pole above plane of inflection and to the crosstie (metric example 3) .......................... Free body diagram of pole between planes of inflection (metric

106

115

with

..............................

Free

example

112

structure

X-braces)

xi

302 ahead is greater

than

stat&n

302

TRANSMISSION

xii

NESC Functions P curve

conductor P curve conductor 6

H

curve

conductor 7

11 curve conductor

.................. loading constants (K) ......................... sag template of % ...................................

27 37 41

for

computations

for example problem ................................

(metric) computations

example problem .......................... customary)

(U.S. computations

for

computations (U.S.

example problem .......................... customary)

No.

l-broken

I he

computations

for example ................................ for

No.

2-unbalanced

example problem ..........................

No.

2-unbalanced

11

P

curve full-load

computations condition

13

H curve

computations

14

full-load II curve

condition computations

no-load

conditiou

15

H

16

no-load Insulation

conditiou selection

17

Insulatiou

selection

18

Insulation Minimum

curve

(grade

computations

69 No.

2-unbhanced

example problem ..........................

No.

2-unbalanced

for example problem .......................... (metric)

No.

2-unbalanced

problem No. ....................

2-unbalanced

No.

2-unbalanced

problem No. ....................

2-unbalanced

customary)

H

69

problem

for

12

64

problem

10

(U.S.

63

for

condition (U.S. customary) P curve computations for example ................................ condition (metric) condition

63 l-broken

9

computations

l-broken

No.

computations (metric)

curve

No.

................................

Line data condition data

l-broken

problem

(metric)

example

No.

62

for

8

I9

MANUAL

conductor

Calculations

5

LINE DESIGN

for

example

70 70 71

(U.S. customary) for example problem ........................... (metric) for

example

72 73 74 107

(U.S. customary) for 34s kV ........................ ........................ for 230 kV

108

20

Conductor clearance surface-wood-pole

to pole ground wire or crossarm ....................... construction

21

Angular

of suspension

22

USBK Minimurn

23

109

........................ selection for 115 kV factors of safety for wood-pole construction R) .......................................

limitations

wood-pole structures factors of safety for ...................................... California

Conductor clearance surface-wood-pole

insulator

swing

for

129 129 standard 129

.......................... wood-pole

construction

to pole ground wire or crossarm .............. construction in California

in 131 131

. ..

CONTENTS

21-

Stttntttary

of loads

lertgths 2s 26

and

Stttntttary lengths Stttnrttary

of loads and

Sttrntnary Minimum NESC

29

Mirtirttttrtt NESC

irt structure

low-point

of loads

lengths 28

tttetnbers

distartces

and

in slrrtctttre

for for

242

3) . . . . . . . . . .

254

spatt

light,

clearartce

rttedittrn,

and

to buildings-USBR heavy

loading

. . .

266

startdard for . . . . . . . . . . .

275

cxatttple

horizontal clearance to buildings-USBR and heavy loading (rttetric) light, tnedittrn, horizontal

standard

(U.S.

2)

span

various

customary

231

. . .

various

exarttple

tttetnbers (U.S.

span 2) . . . . . . . . . .

for various span cttstorttary exatnple

(metric.

distartces

various

exarttple

rnerttbers

distartces

low-point

for

(rttctric

of loads iii structure rttetttbers artd low-point distartcrs (U.S.

lengths 27

itt structure

low-point

x111

crtstorttary)

3)

for . . . .

275

30

Right-of-way

values-NESC

light

. . . . . . . . . . . ,

276

31 32

Right-of-way Right-of-way

values-NESC values-NE%:

light loading (U.S. crtstotttary) . . . . , rnedittrn loadirtg (metric.) . . . . . . . . .

277 278

33

Right-of-way

values-NESC

ntedirtrtt

. . .

279

34 35

Right-of-way Right-of-way

values-NESC values-NE%:

heavy heavy

(metric) . . . . . . . . . . . (U.S. rttstotnary) . . . .

280

36

Data

frottt

correctiott 37

Data

1% I

problettt

(metric)

front

correction

example example (U.S.

loading

(metric)

loading loadirtg loading

ott insulator

(U.S.

offset

cuslorttary)

and

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . problettt crtstornary)

281

sag 299

on insulator offset artd sag . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . .

340 341

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

342

township showing sectiort rtttrttbering lartd section showing corner and l/l6

. . . . . . . . . . . . . .

B-2

Typical Typical

13-3

Azirttitth

H-4

I~eveloprttertt of forttirtla for rnaxirnttrrt rtiorttertt of resistance ort wood poles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

343

13-s

Grottnd

344

chart

resistivity

in the

Urtited

States

desigrtatiotts

. . . . . . . . . . . . . . . . . . .

TRANSMISSION

xiv

LINE DESIGN

TABLES

IN

MANUAL

APPENDIXES

Pa&?

Table

B-l

Maximum ground

B-2

moment line-USBR

of resistance standard

for pole circumferences .........................

at

moment

of resistance

for pole circumferences ..........................

at

Maximum ground

line-ANSI

standard

B-3 B-4

Pole

circumferences

for

Douglas

fir

Pole

circumferences

for

western

red

B-5

Permanent

set values

for

Alumoweld

B-6

Permanent

set values

for

steel

B-7

Flashover

characteristics

B-9

Flashover Relative

B-10

Barometric

B-l

1

B-13 B-14

Pressure

Permanent

c-2

(metric) Permanent (U.S.

c-3 c-4 c-5

Conductor Conductor Conductor

C-6

Conductor

medium, medium,

pine

385

............... strand ....................

strand

of suspension

insulator

351 419 420

strings

and

air 424

Conductor sag-tension

C-l

southern yellow ................ cedar

.......................... values of air gaps ............... air density and barometric pressure .................... pressure versus elevation

B-12

Equivalent Selected

and

423

Mass per unit species used

B-15

348 ...

........................................

gaps.. B-8

345

volume and relative mass .............................. for poles

temperature computations

area

due

set, and and and and and and

427

to wind

and

for

final

normal 428

velocity

data for standard electrical ........................ conversions

set, creep, and initial ....................................... customary)

of wood

coefficients of expansion ...........................

on a projected metric SI-metric

density

426 426

...........

429 ......

conductors

431 moduhts

values 442

creep, and initial and .................................

final

overhead overhead overhead

.......... data (metric) data (U.S. customary) values for NESC light, ....................

heavy

loading

overhead heavy

ground ground ground

loading

wire wire wire

(metric)

ground (U.S.

430

wire

values

customary)

modulus

for

values

NESC .............

452 462 ....

466 470

light, 474

(2 WI2

Sag, =

Example

ft

26/7

conductor

tension (13- mm

loading =

at no load

14 556 =

ice,

0.19-kPa

wind

at minus

mm

18 638

N

8: T=

SW =

18 638, - (14.556)

18 405/15.9657

=

1152.7839

(15.9657) m

=

18 405

N

18 “C)

CHAPTER x =p/2

Example

= 1/2span,

50 100 150 200 250 300 350 400 450 500

0.043 313 0.086146 0.130 120 0.173 493 0.216 866 0.260240 0.303 613 0.346986 0.390 359 0.433133

curve

795

ruling

kcmil,

10 OOO-lb NESC

ACSR,

heavy

loading =

at no load

=

9.

Design

the Regional by the Denver

may and

office

4-lb/ft2

4190

curve curves .-A

=

(1.0940)

3782.29

at 0 “F)

=

4137.83

lb

ft

Sag=a(c;hz-

z.- 1 cosha

showing

0.000 349 531 0.001 398 367 0.003 147 243 0.005 597738 0.008 750 491 0.012 608 78 1 0.017 174 947 0.022452 180 0.028 444 171 0.035 155 107 0.042589680 0.050753087 0.059 651 036 0.069 289745 0.079 675 954

the

percentage

useful

proportion seven

the technical

further

of the regions. design

in chapter work

Design

of each

between

a clearance

design

l),

1.322 5.289 11.904 21.172 33.097 47.690 64.961 84.921 107.584 132.967 161.087 191.963 225.618 262.074 301.358

relationship

in determining

are discussed

of the Bureau’s to cover

wind

lb

- (47.69)

x a

Instructions

1 084 4 340 9113 17 393 21215 39257 53542 70096 88952 110 144

ft

be particularly

catenary

Directors

in ice,

0.026439 0.052 878 0.079 317 0.105 756 0.132 195 0.158 634 0.185 073 0.211 512 0.237 951 0.264 390 0.290829 0.317 268 0.343707 0.370 146 0.396585

11 is a catenary

Parabolic

0.000940156 0.003 764194 0.008 411558 0.015 087 698 0.023607738 0.034053 971 0.046445 57 1 0.06080607 1 0.077 162 490 0.095 546 05 1

conductor

(l/Z-

47.69

100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500

relationship

- 1),

mm

tension

x = p/2 = l/2 span, ft

Figure

Sag =a(cosh;

1

customary)

26/‘7

H = aw = T- SW = 4190 a = H/w = 4137.83/1.0940

This

21

span

maximum

60 OF sag at no load 60 OF tension

(U.S.

DATA

coshz-

a

1200-ft

Assume:

X

m

4.-Catenary

I-BASIC

at any

point

span

length,

in a span.

II.

on transmission

instructions

transmission

sag and

lines

are issued line

is delegated

to these

and include

directors

the following:

to

TRANSMISSION

I””

0,

lb

LINE DESIGN MANUAL

30

2'0

40

5’0

PERCENT Figure Il.-Catenary

a.

Design

curve showing

percentage

relationship

70

$0

SPAN

80

90

100

LENGTH

between

sag and span length.

104-D-1052.

data.

(1)

Length

of line

(2)

Voltage

of line

(3)

Number

(4)

Type

(5)

Ruling

(6)

Insulators:

(7)

Conductors

and

(8)

Maximum

tension

(9)

Final

of circuits of structures span number,

tension

size, overhead under

at 15.5

and

type

ground loaded

OC (60

wires:

number,

size,

and

type

conditions for conductors and “F) with no wind for conductors

overhead ground wit -es and overhead groul nd

wires (10)

For

steel

towers,

ground

wires

the

horizontal

and

vertical

spacing

between

conductors

and

overhe,

ad

CHAPTER (11)

For

(12)

Final

steel

towers,

sag at

overhead

the

15.5

conductor

OC (60

ground

(14)

wires The annual isoceraunic level This number is calculated

clearance

at 15.5

the

Design

c. d.

Minimum Drawings

loading

e.

Number

49

23

to tower

steel with

OC (120OF)

’ F) between

the

conductors

and the probable number either per 100 kilometers

for

coefficent

for

the

and

of power outages or per 100 miles

“per-lOO-miles”

conductors

vahle

overhead

ground

due to lighting. of transmission is 1.6 times

locations

of transpositions.

all pertinent data concerning the line Initial entries on the summary form

charts,

so that should

a compact, ready be made when the

steel

tower

are obtained, filled out.

notebook, records-if

along they

sheets for other lines, for easy reference. summary sheet is simple in layout, easy

normally

required,

with summary are kept. The has

room

for

any

and by the time The completed

reference is available. design work is assigned.

entries should be made as data the form should be completely

source.

the

than those given in a. of structures to be used. clearance

A Transmission Line Data Summary Transmission Line Data Summary Form.on figure 12, should be prepared for each transmission line designed. This form should

information

and

conditions.

f. Design data drawings including sag templates, structure limitation diagrams, and conductor height tables for wood-pole structures. 10. shown

no load

value.

clearances, other and characteristics and

o C (60

mmlerical

“per-IOO-kilometers” b.

and

DATA

wires

Midspan

length;

clearances

“F)

(13)

line

I-BASIC

additional

data

that

the transmission form should

might

form,

as

contain

Additional

line is put into service, be placed in a looseleaf

Nothing is better than good to fill out, contains all data

be useful,

and

is an excellent

24

TRANSMISSION

LINE DESIGN

TRANSMISSION LINE

Region: Project: Name of Line: Length: Elevation, min.-max.: NESC loading: Type of

km

MANUAL

DATA SUMMARY

Specifications Voltage: In service: Data by: kPa wind, lb/ft* wind, contractor:

mi

zone,

mm ice, in ice,

construction:

Insulators Size: -. Strength: Number per

mmx

in x

l-Ql( N (

in)

Conductor at 15.5

lb)

No.

+K(O.-), +K(O.-),

and overhead

ground

wire

to ground clearance "C (60 "F)

mm

ft

Overhead

ground

wire

_

-

Name : size:

Type: Stranding: Ultimate strength: Tension limitations 50% us at -"C( OF) initial 33-l/3% US at -"C(eoF) initial 25% US at -"C(OF) final 18% US at 15.5 "Cf 60 "F) final 15% US at 15.5 "C( 60 OF; final Diameter: Area : Temp. coeff. of linear expansion: Modulus of elasticity Final: Initial: NESC Force (weight) per unit length Bare: Iced: Wind: Resultant (with constant): Ellipse resultant: Ruling span: sacs

OC OF

string: Conductor

Conductor

at at

mm* --

kcmil

mm ml*

mm dia.

in dia.

lb

N

lb

lb lb lb lb

N N N

lb lb lb

N

lb in

in

in*

-p&C

__

perOF

2 pergc

----Tn*

per"F

GPa GPa

lb/in* lb/in2

GPa GPa

lb/in2 lb/in2

N/m N/m N/m N/m N/m m

lblft lb/ft lb/ft lb/ft lb/ft

lb/ft lb/ft lb/ft lb/ft lb[ft

ft

N/m N/m N/m N/Ill N/IO In

OF) final: OF) final:

nun Em mm mm mm

ft ft ft ft ft

mm mm mn mm mm

OF) final: OF) final:

N N N N N

lb lb lb lb lb

ft

-

Full load: Cold curve: Ellipse: 15.5 "C (60 49 'C (120 Tensions Full load: Cold curve: Ellipse: 15.5 “C (60 49 'C (120

___ -OC

(

OF)

Key map: Plan-profile drawings: sag template: Stringing sag tables Cond"&r; Overhead ground wire:

Structure

Figure

12.-Transmission

line data summary

Limitation

form.

Chart:

104-D-1053.

ft ft ft ft ft lb lb lb lb lb



CHAPTER

x1 =

a, (M - RS) (l.0005) a1

SP= a, sinh$

II-CONDUCTOR

SAGS AND TENSIONS

= 46.3000 (22.86 - 2.9274) (1.0005) = 1 0855 m 850.6502

= 46.3000 sinh 41$~~~o = 1.0856 m

RSP = RS + SP = 2.9274 + 1.0856 = 4.0130 m 4.0130 X2 = a, sinh- ’ -RSP = 46.3000 sinh- 1 46.3000 = 4.0080 m a2

x, =x2 - x, X=M- x,

= 4.0080 - 1.0855 = 2.9225 m

= 22.86 - 2.9225 = 19.9375 m

L =a, sinhE=

Lu,=L-

850.6502 sinh 8Fi9QJ;b52= 19.9393 m

w12AE (“l I2

_X + sinh _Xco& -x a1

= 19.9393 -

a1

15.688 (850.6502)2 2(33318479)

a1

19.9375 19.9375 850.6502 ‘Osh 850.6502

= 19.9313 m

t, =

Lu,Lu,e - Lu, +t,

19.9342 19.9313-(0.000 19.9342 020 7) +(‘18)=-25030c = ’

Assume T= 12 010 Nm H(no load) H 12 010 a, = - = = 765.5533 m Wl 15.688

H 12 010 = 41.6682 m a2 = i? = 288.2292 x1 =

a2 (M - RS) (1.0005) = 41.6682 (22.86 - 2.9274) (1 .OOOS) = 1 0855 m 765.5533 a1

83

84

TRANSMISSION

SP = a, sinh?=

LINE DESIGN

MANUAL

41.6682 sinh b;T:852 = 1.0856 m

RSP = RS + SP = 2.9274 + 1.0856 = 4.0130 m RSP 4.0130 X, =a, sinh-’ ~ = 41.6682 sinh- ’ 41 .6682 = 4.0068 m a2

X, = X2 - Xl = 4.0068 - 1.0855 = 2.9213 m

X = M - X, = 22.86 - 2.9213 = 19.9387 m X L =a, sinh - = 765.5533 sinh 7F59535y3= 19.9410 m a1

W, Lu2

=L

-

(a,

1’

2*E

x + sinh x cash _x a1 a1 >

al

15.688 (765.5533)’ 2 (33 318 479)

= 19.9410-

19.9387 +sinh 19.9387 cash 19.9387 765.5533 765.5533 765.5533 >

= 19.9338 m

t, =

Lu, - Lu, +t,

Lu,e

Similar

=

19-9338 - 19*9342 + (- 18) = _ 18.97 OC 19.9342 (0.000 020 7)

calculations

temperatures

were

made

for

five

additional

assumed

Assumed T = H (no load), N

The

resulting determine line.

and

the

resulting

Temperature, OC

10 675 9 341 8 007 6 672 6 227

figure, of the

tensions,

were:

temperatures the

tensions

t, t, t, t, t,

are plotted for the

against

desired

the

= - 11.94 = -2.23 = 11.58 = 33.13 = 43.31

assumed

temperatures

and

tensions proceed

on figure in finding

38. Using the

total

this sag

CHAPTER

II-CONDUCTOR

SAGS AND TENSIONS

0 rn z W I-

-23

-13

-5

0

+5

T EM PERATURE, Figure 38.-Tension-temperature

At-18OC.T= H

curve for insulator

11 800N 11 800

a, = - =-z752.1673 15.688 wt

m

H 11 800 a, =~=~~~,~~9~=40.9396

m

+I5

+25

+35

“c effect

problem

(metric).

104-D-1067.

+45

TRANSMISSION

86

x 1

LINE DESIGN MANUAL

= a, (M - RS) (1.0005) = 40.9396 (22.86 - 2.9274) (1.0005) = 1 0855 m 752.1673 a1

Xl

SP = a, sinh a, = ‘40.9396 sinh 40;0983~6= 1.0856 m

RSP = RS + SP = 2.9274 + 1.0856 = 4.013 m RSP

X, = a, sinh- ’ -

4.013 = 40.9396 sinh- 1 40.9396 = 4.0066 m

a2

X, = X2 - X, = 4.0066 - 1.0855 = 2.9211 m

X = M - X, = 22.86 - 2.9211 = 19.9389 m D, =a2 (.osh:-

1) =40.9396(cosh4s6-

1) =O.O144m=

14mm

D, =a, (coshz-

1) =40.9396(cosh~~~f~6-

1) =O.l962m=

196mm

D, =o,(coshc-

l) =752.1673(cosh7!~~q368~~3-

l) =0.2643m=264mm

D, = D, + D, - D, = 264 + 196 - 14 = 446 mm

At-

1 OC, T=9220N

H

9220

a, = - = = 587.7104 m w, 15.688

H a2 = k=

9220 = 31.9884m 288.2292

x1 = a, (M - RS) (1.0005) = 3 1.9884 (22.86 - 2.9274) (1.0005) = 1 0855 m a1 587.7104

SP = a, sinh 2

= 3 1.9884 bnh 31{yii4

= 1.0857 m

CHAPTER

RSP=RS+SP=2.9274+ RSP

X, = a2 sinh- ’ -

a2

II-CONDUCTOR

87

SAGS AND TENSIONS

1.0857=4.0131rn

= 3 1.9884 sinh- ’ 34.y1814 = 4.0026 m

X, = X, - X, = 4.0026 - 1.0855 = 2.9176 m

X = M - X, = 22.86 - 2.9176 = 19.9424 m D, =a,(cosh:-

1) =31.9884

D, =a2 (coshz-

1) = 31.9884(cosh~~~~~4-

D, =a1 (cash:-

D, =D,

+D,

(cosh31~~~~~4- 1) =O.O184m=

1) =0.2507m=251

l) =587.7104(cosh:89;~7412~4-1)

-D,

=338+251

- 18=571

18mm

mm

=0.3384m=338mm

mm

At 15.5 OC, T = 7740 N

H = ~7740 = 493.3707 m

a’ = w,

15.688

H

7740 = 26.8536 m a2 = w = 288.2292 =

x 1

a2

(M - RS) (1.0005) = 26.8536 (22.86 - 2.9274) (1 .OOOS)= 1 0855 m 493.3707 al

Xl

SP=a? sinh -

a2

= 26.8536 sinh :ey5y6

= 1.0858 m

RSP=RS+SP= 2.9274+ 1.0858=4.0132m RSP

X, = a, sinh- ’ -

a2

= 26.8536 sinh- l 2:08:3;?6 = 3.9984 m

88

TRANSMISSION X,

=X2

X=M

- Xl

= 3.9984

LINE DESIGN MANUAL

- 1.0855 = 2.9129 m

- X, = 22.86 - 2.9129 = 19.9471 m

D,=a,(cosh$-

1) =26.8536(~osh:6p88;;5~-

1) =O.O219m=22mm

D, =a,(cosh$-

l) =26.8536(co~h~~~~~~~-

l) =0.2982m=298mm

D, =a, (,osh~-

1) =493.3707(cosh499;83477d7-

D, =D, +D, -D,

At32

l) =0.4033m=403mm

= 403 + 298 - 22 = 679 mm

OC. T=6760N

H

6760 = ~ = 430.9026 m a1 = w, 15.688

H a’ =w=

x1 =

a2

6760 = 23.4536 m 288.2292 (M - RS) (1.0005) = 23.4536 (22.86 - 2.9274) (1 .OOOS)= 1 0855 m

430.9026

a1

SP = a, sinh 2 = 23.4536 sinh

1.0855 = 1.0859 m 23.4536

RSP = RS + SP = 2.9274 + 1.0859 = 4.0133 m

X2 =a, sinh-’ !?!f

= 23.4536 sinh- ’ ~~~,j3~6 = 3.9940 m

a2

X, = X2 - Xl = 3.9940 - 1.0855 = 2.9085 m

X=M-

X, =22.86-

2.9085 = 19.9515 m

CHAPTER

II-CONDUCTOR

SAGS AND TENSIONS

D, =a, (yxh$--

l) = 23.4536 (cash $jp)&T;6 - $ = 0.0251 m = 25 mm

D, =a2 (c~sh$-

1) = 23.4536 (cash :;:23p6‘

1) =0.3409m =341 mm

D, =a, (cosha$-

1) =430.9026(cosh~~!~~256-

l) =0.4620m=462mm

D, =D,

+D,

- D, =462+341-

25=778mm

At43 OC, T=6260N

H a1 =w 1

6260 = -= 15.688

399.0311 m

H

6260 = 21.7188 m az = ii = 288.2292

x, =

a,@f - RS) (1.0005) = 21.7188 (22.86 - 2.9274) (1.0005) = 1 0855 m 399.03 11 a1

SP=a, sinh;

Xl

1.0855 = 21.7188 sinh 21.7188 = 1.0860 m

RSP=RS+SP=2.9274+

1.0860=4.0134m

X, = a, sinh-’ RSP = 2 1.7 188 sinh- l ~~~~~8 = 3.9909 m a2

X, = X2 - X, = 3.9909 - 1.0855 = 2.9054 m

X =M - X, = 22.86 - 2.9054 = 19.9546 m

D, =a, bosh:-

1) =21.7188

(cosh211f!!f~8-

1) =0.0271 m=27mm

90

TRANSMISSION

D, =a, (&$

1) =21.7188

LINE DESIGN

(cosh;;y;;;8-

MANUAL

1) =0.3677rn=368rnm

D3=al(cosh~-I) =399.0311 (coih;g-f,.,~l -I) =0.4990m=499rnm D,

=D,

+D,

=499+368-27=840mm

-D,

U. S. Customary Figure

DC-576

39 shows

the

U.S.

customary

sag and

tension

computations.

(6-76)

&,tL

SAG CALCULATIONS

LOADINGWeight Factws: Dead Welpht

(W’)

1, n 750 St.1

0. Initial.- &m°Fd.2d Fenal. -&?.e.?FX Loaded. OF Final. &OF

-5e -

Computed by -

4

3/7q

lb

SC,‘&!?&% @&!-lb

Resultant:

lb

0.000 o//

Ib/ft2 Wind(W”‘)/

[

90 .*n

I

Figure 39.-Sag

H = T - W,

H

a2 =r=H

t

8

I

0

60

1

and tension

T>/

7

Permanent Set 0.009

IbItt

Creep Total

Ib/ft

2.330

T

calculation

form

Modulus. (E) Final-x initial/,.56x

1 SAGFACTOR 1 SAG,ft FEET

I 0.09d9

lLuf&LLn//

I

I

I

I

3gg5*53g3 = 202 *3058 19.75

f-t

ft

78

I / I I I

effect

(sag) = 4000 - 2.5306 (1.78) = 3995.5393 lb 3gg5-53g3 = 1578 8901 2.5306 *

106 lb/i,? 106 lb/in2

Final AE .m lnttial AE s

150

for insulator

151

o.oo--O.ooa

Ib/ft

per “F

SPAE;LENGTH(S)

tnch Ice.

w

(W’“)

Ib/ft

Area (A) ti in2 Temp. Coeff. of Linear Exp.:

oF * UNSTRESSED 1TEMP. LENGTH

No Ice. No Wind (W’)

a,=-=

lb

Date

LOADING .km

Kfe

Ad.2

o/n

problem

tb lb

1 SW,Ib

1 TENSION,Ib

I.? 7959

I

I I

(U.S. customary).

I

JImlo

n;d

CHAPTER

x 1

II-CONDUCTOR

SAGS AND TENSIONS

= a, (M - RS) (1.0005) = 202.3013 (75 - 9.6) (1 .OOOS)= 8 3838 ft 1578.8901 4

Xl= 202.3058 SP=a2 sinhz

8.3838 sinh 202.3058 = 8.3862 ft

RSP = RS + SP = 9.6 + 8.3862 = 17.9862 ft

X2 = a, sinh- ’ E

= 202.3058 sinh- ’ :a’;~~5!8

= 17.9626 ft

a2

X, =X2 - Xl = 17.9626 - 8.3838 = 9.5788 ft

= 75 - 9.5788 = 65.4212 ft

X=M-X,

X sinh -=

L=a,

1578.8901 sinh l;;;p;;;1

= 65.4399 ft

4

Lu,=L-

w,

b,

2AE

= 65.4399 -

I2

--X+sinh&c& a1 i 4

a1 )

2.5306 ( 1578.8901)2 2 (7 490 285)

65.4212 65.4212 65.4212 1578.8901 + sinh 1578.8901 ‘Osh 1578.8901

= 65.4050 ft Temperature = 0 OF = t, Assume T = 3000 lb w H (no load) H = == 1 1.075 al=w H 3000 = -= a, = w 19.75 x 1

=

2790.6977 ft

151.8987 ft

a, (M - RSI (1 .OOOS) _ 15 1.8987 (75 - 9.60) (1 .OOOS)

SP=a2 sinh:=

Ql

2790.6977 151.8987 sinh 1;.l5;;;7 .

= 3.5618 ft

= 3.5615 ft

91

‘12

TRANSMISSION

RSP=RStSP=9.6+3.5618=

LINE DESIGN

MANUAL

13.1618ft

x2

RSP = a, sinh-’ = 151.8987 sinh-’

x,

=x2

~~~~91g87= 13.1454 ft

a2

- x,

= 13.1454 - 3.5615 = 9.5839 ft

X =M - X, = 75 - 9.5839 = 65.4161 ft

L = a, sinh:

65.4161 = 2790.6977 sinh 2790.6977 = 65.4221 ft

wl(al)2 LU, =L - 2AE

= 65.4221

x ( c+sinh-f

1

cosht

1.075 (2790.6977)2 2 (7 490 285)

-

1

65.4161 65.4161 65.4161 2790.6977 + sinh 2790.6977 ‘Osh 2790.6977

= 65.3959 ft

Lu, = Lu, + Lu,e(t,

t, =

- to)

Lu, - Lu, 65.3959 - 65.4050 + o = _ 12.10 OF Lu, e + to = 65.4050 (0.000 011 5)

Assume T = 2700 lb = H (no load) H 2700 a, = - = 1075=2511.6279ft WI * H 2700 a, = p = 19.75 = 136.7089 ft x 1

=a,@4 - RS) (1.0005) _ 136.7089 (75 - 9.6) (1.0005) = 3.5615 ft 2511.6279 a1

SP = a, sinh:

RSP=RS+SP=

= 136.7089 sinh

3.5615 = 3.5619 ft 136.7089

9.6 + 3.5619 = 13.1619 ft

CHAPTER X2 =a,

sinh-’

@E=

II-CONDUCTOR

93

SAGS AND TENSIONS

136.7089 sinh- l 1:; y;899 =

13.1417

ft

a2

X,

=X2

- X,

= 13.1417

- 3.5615

= 9.5802

ft

X = M - X, = 75 - 9.5802 = 65.4198 ft

L

=

a,

sinh

Lu, =L-

5

= 25 11.6279

sinh

65.4198

25 11.6279

F!&&?

= 65.4273 ft

cash -x a1

1.075 (25 11 .6279)2 2 (7 490 285)

= 65.4273 -

65.4198 65.4198 65.4198 2511.6279 + sinh 25 11.6279 ‘Osh 25 11.6279

= 65.4037 ft

t, =

Lu, - Lu, Lu,e

Similar temperatures

+t,

=

65.4037 - 65.4050 65.4050 (0.000 011 5) = -1*73 OF

calculations were:

Assumed

were

made

T-- H (no load), lb 2400 2100 1800 1500 1400

At

The figure,

resulting determine

of the

line.

0 OF,

temperatures the tensions

w,

1.075

five

additional

assumed

tensions,

= = = = =

the

resulting

11.30 28.72 53.58 92.53 110.88

are plotted against the assumed tensions for the desired temperatures and proceed

1 ft

and

Temperature, OF t, t, t, t, t,

T = 2645 lb

H 2645 = 2460.465 a, = - = -

for

on figure 40. Using in finding the total

this sag

94

TRANSMISSION

LINE DESIGN MANUAL

2600

2400.

g

2200

; 0v) z E

2000

1800.

1600

-20

0

+20

l 40

T EM PERATURE, Figure 40.-Tension-temperature

curve for insulator

+80 -

+60

effect

+I00

OF problem

(U.S. customary).

104-D-1068.

+I20

CHAPTER a2

=-=

x

H w

2645 -= 19.75

II-CONDUCTOR

SAGS AND TENSIONS

133.9241 ft

- RS) (1.0005) 133.9241 (75 = 9.60) (1.0005) = 3.5615 ft a1 2460.465 1

=a,(M 1

SP=a2 sinh$

= 133.9241 sinh 3.5615 = 3.5619 ft 133.9241

RSP=RS+SP=9.6+3.5619=

X2 = a, sinh- 1 Q

13.1619ft

= 133.9241 sinh- 1 l:‘; kyll

= 13.1408 ft

a2

X, =X2 - X, = 13.1408 - 3.5615 = 9.5793 ft

X = M - X, = 75 - 9.5793 = 65.4207 ft

D,=a,(cosh:-

D, =D,

l) =133.9241(khll~~~~~)481-

+D,

1)=0.6452ft

- D, = 0.8698 + 0.6452 - 0.0474 = 1.4676 ft

At 30 OF, T = 2075 lb H 2075 a, = - = = 1930.2326 ft w, 1.075 H _ 2075 a2 = w- = 105.0633 ft 19.75 x

=a,(M-

1

RS) Ql

(1.0005) = 105.0433 (75 - 9.60) (1.0005) = 3 5615 ft 1930.2326

95

TRANSMISSION

SP = a, sinh 2 = 105.0633 sinh RSP=RS+SP=9.60+3.5622 RSP

X, = a, sinh-’ -

LINE DESIGN MANUAL

3.5615 = 3.5622 ft 105.0633

= 13.1622 ft

= 105.0633 sinh”

1~~~6~3 = 13.1280 ft

a2

X, =X2 - Xl = 13.1280 - 3.5615 = 9.5665 ft

X=M-

X, =75- 9.5665 =65.4335 ft

D, =a,(cosh$

1) =105.0633(cosh1;;;;;3-

l)=O.O604ft

D, =a, (ah:

- 1) = 105.0633 (cash ll;;;;;;3

- 1) = 0.8213 ft

D, =a1 (co&:

- 1) = 1930.2326(cash

D, =D,

1;;;;;;6-

1) = 1.1092 ft

+D, - D, = 1.1092 + 0.8213 - 0.0604 = 1.8701 ft

At 60 OF, T = 1733 lb --

H 1733 a, = - =x= Wl .

1612.0930 ft

a, = /f=F5=87.7468 x 1- _ a,

ft

04 - RS) (1 .OOOS) 87.7468 = (75 9.60) (1 .OOOS) = 3.5615 ft a1 1612.0930

SP = a, sinh:

= 87.7468 sinh

RSP=RS+SP=9.60+3.5625

3.5615 = 3.5625 ft 87.7468

= 13.1625 ft

CHAPTER

II-CONDUCTOR

SAGS AND TENSIONS

13.1625 X, = a, sinh-’ RSP = 87.7468 sinh-’ 87.7468 = 13.1136 ft a2

X, =X2 - X, = 13.1136 - 3.5615 = 9.5521 ft

X = M - X, = 75 - 9.5521 = 65.4479 ft

D, =a,

(cosh$-

l) =87.7468(cosh$~~~~8

D, =a2(ysh2-

$ =87.7468(cosh~~:~~~~-

D, =aI(cosht-

- l) =O.O723ft

I) =0.9817ft

l) =1612.0930(xsh1~~~~~~O-

$ =1.3287ft

D, = D, +D, -- D, = 1.3287 + 0.9817 - 0.0723 = 2.2381 ft

At90°F,T=15131b

H

a, = - = g$ WI *

= 1407.4419 ft

H

1513 = = 76.6076 ft a2 =w 19.75 x

1

=

Q-2

(M - RS) (l-0005) a1

= 76.6076 (75 - 9.60) (1 .OOOS) = 3 56 15 ft 1407.4419

SP = a, sinh x, = 76.6076 sinh 7iyo1756 = 3.5628 ft a2

RSP=RS+SP=9.60+3.5628 RSP

X2 =a2 sinh-’ -

= 13.1628ft

= 76.6076 sinh- l :i’i$i

a2

X, =X2 - X, = 13.0989 - 3.5615’= 9.5374 ft

= 13.0989 ft

98

TRANSMISSION

X=M-

X, =75-

9.5374=65.4626

LINE DESIGN MANUAL

ft

D, =a, kosh$

1) =76.6076(cosh;;~;;6-

1) =O.O828ft

D, =a+sh$-

1) =76.6076(cosh;~:~;;~-

l)

D, =a1 (cosh$

D, =D,

1) = 1407.4419 kosh lg4;;;g

+D,-

= 1.1306ft

- 1) =1.5227 ft

D, = 1.5227 + 1.1306 - 0.0828 = 2.5705 ft

At 110 OF:, T = 1405 lb

H a, = - = E Wl . H a2 = w

x

= 1306.9767 ft

1405 == 71.1392 ft 19.75

=a2(M-

= 71.1392 (75 - 9.60) (1.0005) = 3 5615 ft

Ra(1.0005)

1

1306.9767

a1

SP =a2 sinhs

= 71.1392 sinh ;i5fi12

= 3.5630 ft

a2

RSP=RS+SP=9.60+3.5630= X2 =a, sinh- l g

13.163Oft

= 71.1392 sinh-’

~~‘:~~~ = 13.0890 ft

a2

X, =X2 - Xl = 13.0890 - 3.5615 = 9.5275 ft

X=M-

X, =75-

D, =a,(cosh~-

9.5275 =65.4725 ft

1) = 71i1392 (cash ~~~1631~2 - 1) = 0.0892 ft

CHAPTER

II-CONDUCTOR

SAGS AND TENSIONS

99

D, =a,(cosh$ 1) =71.1392(cosh;;~;~;;-1) = 1.2075ft 65.4725

-

1306.9767

D,

18.

=D,

- D 1 = 1.6403

+D,

Spans

With

Concentrated

infrequent

and

arrangements in addition figure 3.1 .

are used. to the dead

are confined Such force

to the

span

1. 2. string

Assume

a desired

Calculate that will

relating or switchyard

problems applied.

are complicated A method which

ft

ft

to spans spans

with concentrated in which

loads

taps

are

or tie-down

by the elastic effects of the tap or tie-down adequately treats this problem is shown on

to this problem than the method shown on figure 41 would be to sag normal sag for a given temperature and then add a calculated length for

the

force

may be determined F. F. Priest: spring

tension

of the

by the

at some

tie-down,

following

given

see figure

procedure

42. The

which

required

was developed

temperature.

the angle that will be formed by a vertical result from the horizontal tension in the

to the tie-down after installation 3. By multiplying the length reflected length of the insulator The difference in the originally

= 1.6403

= 2.7586

to substation

to compensate

additional length of conductor by a former Bureau engineer,

- 0.0892

Loads.-Problems

mainly

Probably a better approach the conductors to the calculated of conductor

+ 1.2075

1

(0 = tan-’ H/P). of the insulator string string is obtained (i,

line and the position of the insulator conductor and the vertical force due by the sine of this = isin 0).

angle,

the horizontal

between the length of the insulator string as it will lay in the near horizontal sagged span and its calculated horizontal reflected length after the tie-down

indicates the additional amount of conductor same characteristics as the originally sagged

required to give the final span without the tie-down.

tied-down

span

position is made, about

the

Example Conductor: Span length Force Spring Length

Calculate might short the

242 mm2 (477 kcmil), = 45.7 m (150 ft)

of hardware on tie-down tension at 15.5 ‘C (60 of insulator

sags and

string

tensions

be applicable during span, such as in the insulator

effect

=

ASCR

24/7

= 444.8 N (100 lb) OF) = 889.6 N (200 1829

for the

mm

conductor,

lb)

(6 ft)

without

tie-down,

for a range

of temperatures

that

installation. If the insulator force will be appreciable in a comparatively example used here, the original sags should be determined by considering

(see sec.

16).

LOO

TRANSMISSION

LINE DESIGN

MANUAL

P

LeveI' Span

Inclined

Span

s = 2PL + wL2

8H CONCENTRATED LOAD AT

CENTER OF SPAN

H H-TeH

t

P

Level Span

Inclined S=

Span

L, L, (2P + WL) 2LH

LOAD AT ANY POINT ON

SPAN

=Horizontal span length between conductor support points, m (ft) = Horizontal tension in conductor, N (lb) = Sag, from line of supports at concentrated load, m (ft) =Concentrated load, N (lb) = Linear force factor (weight) of conductor, N/m (Ib/ft) ;,L*= Horizontal distance from concentrated load to points of support, m (ft)

L H S P

Figure 41.Spans

with

concentrated

loads. 104-D-1069.

CHAPTER

II-CONDUCTOR

SAGS AND TENSIONS

i = Length of insulator string, mm (ft) length of insulator string, in - Horizontal reflected ni = i-iH, mm (ft) H - Horizontal tension in conductor, N (lb) P - Vertical force added by tie-down (hardware tension) , N (lb) Figure 42.-Graphical method for determining concentrated load problem. 104-D-1070.

Assume the following

length

‘C

(OF)

-18 -1 15.5

(0) (30) (60)

mm 625 780 917 1039 1149

For 15.5 OC:

mm

required

by previous

for

calculations:

Tension,

SW, w

N

(2.05) (2.56) (3.01) (3.41) (3.77)

3750 3015 2571 2268 2050

(lb)

(843) (678)

(578)

(510) (461)

For 60 OF: 2571/1334.4

e = tan-’

578/300

= tan- l 1.926 70

= tan- ’ 1.926 70

= 62.57O

= 62.57O

ih = 1829

(ft)

+ spring

of conductor

sag and tension values have been obtained Temperature,

e = tan-’

additional

101

Sin

8 = 1829 (0.887 57)

= 1623.37 mm Ai = 1829 - 1623.37 = 205.63 mm

_

ih = 6 sin 8 = 6 (0.887 57) = 5.33 ft

Ai = 6 - 5.33 = 0.67 ft = 8 in

TRANSMISSION

102 The

Ai vahle

is the additional

Considering 8 constant can be made:

for setting

Temperature, OC (OFI -18 -1 15.5 32 49

(0) (30)

(60) (90)

(120)

amount

of conductor

the spring

Horizontal N 3750 3015 2571 2269 2050

LINE DESIGN MANUAL

tension

to be added at other

to the span after

temperatures,

the

the initial following

tension, (lb)

Hardware force, N (lb)

Spring tension, N (lb)

(843) (678) (578)

444.8 444.8 444.8 444.8 444.8

1501 1120 890 733 620

(510) (461)

(100) (100) (100) (100) (100)

(337.5) (251.9) (200) (164.7) (139.3)

sagging. tabulation



107.58

700-ft spans, 1400-ft LP t$,,=

1.5(2616)

+ 1131 + 6.002(963)

+ 4.865(635)

+

= 2013 lb/in2

107.58

800-ft spans, 1600-ft LP

s,= 1.5(2989)

+ 1293 + 6.002(1100) 107.58

+ 4.865(725)+

18.11(1100)

+ 12.02(725) 12 159.66 x T 1 = 2300 lb/in2

900-ft spans, 1800-ft LP s,=

1.5(3363)

+ 1454 + 6.002(1238)

+ 4.865 (816)

107.58

+

18.11(1238)

+ 12.02(816)

159.66

12 x> 1

= 2589 lb/i2

lOOO-ft spans, 2000-ft LP sN=

1.5(3737)+

1616 + 6.002(1375)+4.865(907)+, 107.58

= 2876 lb/in’

242

TRANSMISSION

LINE DESIGN MANUAL

1 lOO-ft spans, 2200-ft LP s,=

1.5(4110)+

1200-ft

spans, lS(4484)

sN=

1777 + 6.002(1513)+4.865(997) 107.58

18.11(1513)

+

+ 12.02(997) 12 159.66 >( 1 > = 3163 lb/in*

2400-ft LP + 1939 + 6.002(1650)

+ 4.865(1088)+

= 345 1 lb/in*

107.58

1300-ft spans, 2600-ft LP 1.5 (4857) + 2101+ s,=

6.002 (1788) + 4.865 (1179)

+

18.11(1788)

107.58

+ 12 12.02(1179) >( > = 3739 lb/in* 159.66 r

1400-ft spans, 2800-ft LP lS(5231)

sN=

Table point

+ 2262 + 6.002(1926)

18.11(1926)

+ 4.865(1269)+

107.58

25 shows

a summary

+ 12.02(1269) 159.66

of loads

in the

structure

members

12 )( r >

for

various

= 4027 lb/in*

span

lengths

and

low

distances.

Table 25.~Summary of loads in structure members for various spans lengths and low-point distances (U.S. customary example 2) SAS/Z, ft Member

Position

600

700

800

900

1000

1100

1200

1300

1400

2 400

2 600

2 800

8 348 7 637 3 142 8 395 10 075 9 889 3 501 3 739

8 991 8 223 3 384 9 042 10 851 10 650 3 770 4 027

LP, ft 1200 AG&EF GC&FC GF AB&DE BC&CD KN&LM L N

Adjustable braces, lb Nonadjustable braces, lb Crosstie, lb Crossann (compressive), lb Crossarm (compressive), lb X-brace, lb Pole, lb/in* Pole, lb/in*

Example and

double

3.-Stress

analysis

1400

3854 3524 1451 3 875 4650 4 563 1 615 1725

4496 4113 1693 4521 5426 5 326 1 886 2013

for a 29-m

(95-ft)

1600

1800

5137 4 698 1934 5166 6 200 6083 2 154 2300

type

5781 5 287 2 176 5813 6 976 6 846 2424 2589

HSB

2000

2 200

6423 5 874 2418 6459 7751 7 606 2 693 2876

7064 6 460 2659 7 104 8525 8 366 2961 3 163

230-kV

structure

--

7 707 7 049 2 901 7 749 9 299 9 126 3 231 3451

with

class

1 wood

poles

X-brace:

Metric Figure

103

shows

the

structure

outline

and

other

data.

Using

the

nomenclature

from

example

1,

CHAPTER V-ADDITIONAL

DATA

243

Conductor:

403 mm : ACSR, 45/7 27 mm 0.38-kPo wind on iced (l3-mm radial) conductor - 20.07 N/m Vertical force with l3-mm radial ice 27.26 N/m OGW: IO mm, H.S. Steel, 7-wire Diometer - 9 mm 0.38- kPo wind on iced (l3-mm radial) OGW= 13.23 N/m Vertical force with l3-mm radial ice II.79 N/m DiOmeter:

I---

1 1

Pole Circumference, mm

Position 8 or K or M or R or

Pr, N*m

771 857 1247 1401

13 L N s

74 208 IO1 754 312 751 444 314

Douglas Fir Working Stress - 51.02 MPa 6,706m tan

- +j&

= 0.7727

sin a - 0.6114 cos a - 0.7913 a - 37O41’ Figure

103.-29-m

type HSB 230.kV

V,

=

H, = Load

in adjustable

braces

structure

(27.26)(LP)

vg

(20.07)(SAS/2)

Hg =

AC

and

load

=

fir poles (two X-braces).

(11.79)(LP) (13.23)(SAS/2)

EF:

LAG’ = L,’ Compression

with class 1 Douglas

= V,/sin a = 1.635Vc

in crossarm: )-

LAB

- L,’

= - Vc/tan a = - 1.294Vc

104-D-1 111.

244

TRANSMISSION

Load

in nonadjustable

braces

Compressive

force

in crossarm

between

I- -

L, loads U,’ For

3 V, and = U,’

transverse

B and

C and

‘=-V,/tana

‘=L,,

D

between

and

C :

=-1.294V,

GF:

in crosstie

Vertical

= 0.5 V,/sin a = 0.8 18 V,

= L,’

LBC Load

FC:

CC and

L,,’

LINE DESIGN MANUAL

COS

2 Vg are shared

= UK’

loads

L,,’

H,

= u,

equally

’ = U,’

Hg , a

and

a - L,,’

= UN’

plane

a = 0.647Y,

COS

by two = Up’

of inflection

poles:

= Ue’

= u,

’ = Us’

HJexists.

The

= 1.5 v,

location

+ vg

of this

plane

parts

and

is found

by:

X(PrB)

3.048(74 208) +P,B = 101 754+74208

xo =& Xl

A plane

exists.

Its

location

Y C&M 1

considered

moment

Axial

444314+312751

is known,

points

wind

forces

of zero

’

the structure

may

be separated

into

each

part

on conductors

reaction by

the

and

overhead

ground

wires

are resisted

equally

by each

pole

moment: RI;’

dividing

= 2 266 m

separately.

Horizontal at the

of zero

by:

= 5.486 - 2.266 = 3.220 m

Yl =y-Yo position

is found

5.486(312 751)

y” =PpR +PrM=

When

’

= 3.048 - 1.285 = 1.763 m

=x-x,

PQ also

of inflection

= 1 285 m

at Jcaused moment

‘J UJ”

arm

by (pole

CR, f) --R/z horizontal

Rd’=-1.5H,-H, wind

force

is found

by

,, _ (3H,) (1.285) + Cur,> (4.181) = 0.575H, 6.706 = q/

taking

moments

spacing):

+ 1.247H,

about

Hand

CHAPTER Taking

moments

B

about

in the

pole

V-ADDITIONAL

above

the

plane

DATA

245

of inflection

(fig.

104)

gives

force

&“:

Figure 104.-Free body diagram of pole above plane of inflection and to the crosstie (metric example 3).

-F/ -1.5

Hc + Hg 1.285(1.5H,

FG8)---

+I$)+

1

2.896H,

_

- -0.744H,

2.591

[

- 1.614H,

FF” = FG” The

outside

( fi’ . carry

carry

on the

FGRand FH” CF is:

10 percent

of

braces

CG and

inner

cos a

L,, load

EF,

and

Load

,, _ 0.9FG" 0.9(-0.744H, --=

L,

The

AG

braces,

90 percent.

- 1.614H,)

0.7913

while

the

inside

braces,

=-0.846H, - 1.836H,

--L,"

8) -

in the

outer

L*(y=

braces

-O.lF&’ cos a

AG

=

and

EFis:

-O.l(-0.744H, - 1.614H,) = O.O94OH,+0.204H, 0.7913

L,, e --L,," The

load

in the

crossarm

LBc)) =(-L,") = l.l69H, LBC

II -

--

LCD”

portions

BC

and

CD

is:

cos.a + O.SH, =-(-0.846H, + 1.453H,

- 1.836H,)(O.7913)+

O.SH,

CG and

246 The

load

in the

crossarm

TRANSMISSION

LINE DESIGN MANUAL

portions

DE

AB

and

is:

LAB " = -(LAG ” cos a + H,) = - (O.O94H, + 0.204H,)

L, The

(0.7913) -H, = -

- O.l61H,

l.O74H,

rt -- -L& moment

B

at

D is

and

given

by:

MD “=-xo(l.5Hc

+Hg) = - 1.928H, - 1.285H, N-m

MB” = MD” For

the

portion

of pole

between

the

MK ” =x1 (lSH,

planes

of inflection,

the

moment

at

K

and

L

is:

+ Hg) = 2.645H, + 1.763H, Nom

ML” = MK” The

area

of the

pole

at

K

and

L,

excluding

the

23.8.mm-diameter

hole

for mounting

the

X-brace

is:

AK=x-

nD2

23.80 =a (272.8)2 - 23.8(272.8) = 58 449 - 6493 = 51 956 mm2

A, =A, The

section

zK=x-

modulus

at

K

and

L

is:

71D3 23.8D2 = 5 (272.8)3 (j

_ 23’8(;72’8)2

= 1993 118- 295 198= 1 697920mm3

z, =z, The

horizontal

reaction

in the

poles

RP The

axial

reaction

‘Q” = UP

in the

poles

at P and

Q is:

11 -

-RQ ” = - 1 .5Hc - Hg

at P and

Q is:

3H, + 2Hg (17.898) + 0.575H, + 1.247H, = 8.581H, + 6.585H, 6.706

” = - UQ"

The

force

at K can

he found

CHAPTER

V-ADDITIONAL

by

moments

+H

taking

247

DATA

about

point

M (fig.

105):

9

F lgure 105.-Free body diagram of pole between (metric example 3).

+H

Q

15.632(1.5H,

" =-

FK

planes of inflection

+I$)+

2.266(1.5&

+Hg)

1

= - 4.003H, - 2.669H,

6.706

FK u --FM” Since the

the

division

installation,

VN,

WM

and

of load

assume

L,,

net

area

A,=4A, The

all load

X-braces

KUand

is taken

by

LTand

one

X-braces

set of braces.

The

VNand force

WMdepends in X-braces

of the

nD2

4.003H, + 2.669Hg

Of-

--

sin 45O

M- - L,,

pole

(less

M- L,”

the

X-brace

= -5.662H,

- 3.775H,

)

= -L,,”

mounting

hole)

23.80 = 2 (396.80)2 - 23.8(396.80)

at

Mand

N is:

= 123 661- 9444 = 114 217 mm2

=A,

section

modulus

nD3 ZM=F-

at

Mand

N is:

23.8D2 6 = & (396.80)3 - ‘9

= 5 509 039 mm3

z,

upon

KU, LT,

is:

=KU

The

between

that

= z,

(396.80)2 = 6 133 592 -- 624 553

TRANSMISSION

248 Taking

about point

moments

u MA4

the

By superposition, and horizontal

loading

by its respective Stress At

in the point

total poles

M(fig.

-

- - 2.266

values

LINE DESIGN

105): (1

.5Hc + HE) = - 3.399H, - 2.266H,

of the forces

and

can be combined

for total

load

factors

and

safety

MANUAL

bending

moments

loading.

The

strength

computed of each

separately member

for

tabulated.

is:

L :

SL =-

UL

+-

AL

ML ZL

where : UL” = UJ” + 0.707LL,” UL

and UL = UL’ + UL”

” = 0.575H, + 1.247H, + 0.707 (5.662H, + 3.775H,) = 4.578H, + 3.9 16Hg

U,’ = 1.5v, + vg u, = U,’ + UL” = 1.5V, + Vg + 4.578H, + 3.916H, A, =51 956mm’ ML” = 2.645H, + 1.763H, N-m

Z, = 1 697 920 mm3 s

lSV, L

At

point

+ Vg +4.578H, 51 956 mm*

+ 3.916H,

+

N:

‘N SN=ANfZN

MN

vertical

can be divided

CHAPTER

V-ADDITIONAL

DATA

249

where : UN” = Ue” and UN = UN’ + UN” U,‘=

1.5vc + vg

UN” = 8.58lH,

+ 6.585H,

UN = UN) + UN)) = 1.5 v, + vg + 8.581H, + 6.585H, A,

= 114217mm2

MN

” = - 3.399H, - 2.266H,

zN = 5 509 039

+ vg + 8.581H, + 6.585H, 114 217 mm2

s, = Adiustable

mm3

braces

AG

and

1000,,,,,, + 3.399Hc+2.266Hg 5 509039 mm3 ) ()](lOO~zmm’)

El;: L AG’ = 1.635V,

183-m Spans, LP = 366 m

+ 1000 = kPa

LA/'

O.O94OH,+0.204H,

LAG=LAG'+LAGA

16 312

345.26+493.88= 839

17 151 N

18 987

401.85 + 574.87 = 977

19 964 N

21 750

460.32 + 658.51=

22869N

24424

516.91+

V,=9977N V =4315N 4~3673~

iTI"=

N

213-m Spans, LP = 426 m

V,= 11 613N V = 5023N d= 4275N Hg= 2818N 244-m Spans, LP = 488 m

1119

V,=13303N 2: ii;;; H;= 3228N 274-m Spans, LP = 548 m

Vc= 14938N v-

6461 N

f(= 5499N Hg= 3625N

739.50 = 1256

-

25680N

250

TRANSMISSION

Adjustable

braces

AG and

LINE DESIGN

MANUAL

IS-Continued

LAG’ = 1.635 Vc 305-m Spans, LP = 610 m

LAG))= O.O94OH,+0.204Hg

LAG=L*Gt+L*G)

27 188

575.37 + 823.14 = 1399

28587N

29862

631.96 + 904.13 = 1536

31 398 N

32625

690.52 + 987.77 = 1678

34 303N

35 300

747.11+

1068.76 = 1816

37 116 N

38063

805.58+ 1152.40 = 1958

40 021 N

V,=16629N V = 7 192 N $= 6121N

Hg= 4035N 335-m Spans, LP = 670 m

V,=18264N V = 7899N l(= 6723N I+= 4432N 366-m Spans, LP = 732 m F= = 1gg54N 8630N l(= 7346N Hg= 4842N 396-m Spans, LP = 792 m

V,=21590N V = 9338N ig= 7948N fig= 5239N 427-m Spans, LP = 854 m

V,=23280N V = 10 069 N f(= 8570N Hg= 5649N

Nonadjustable

spans, m 183 213

244 274 305 335 366 396 427

braces

LP, m

366 426 488 548 610

670 732 792 854

GC and

FC :

Lee' = O.S18V,, N

N

8 161 9 10 12 13 14 16 17 19

L Gc"= 0.846H,+ 1.836Hg,

499 882 219 603 940 322 661 043

3107 + 4 445 = 7 552 3617+ 5174= 8791 4128+ 5927=10055 4652+ 6656=11308 5178+ 7408=12586 5688+ 8137=13825 6215 + 8 890 = 15 105 6724+ 9619=16343 7250+10372=17622

LGC = L&

+ L&

15 713 18 290

20937 23572 26 189 28765 31427

34004 36665

CHAPTER Crosstie

V-ADDITIONAL

DATA

251

GF : spans, m

LP, m -

183 213

366 426 488 548 610 670 732 192 854

244 214 305 335 366 396 421

Crossarm

183 213

LP, m

-

11 12 13 15

817 910 969 062

L m” = -l.O74H,

- O.l61H& N

-12 910

-3945-390=

-4335

-15 021

-4591-454=

-5045

-17 214

-5259-520= -5906-584= -6574-650= -1221-114= -1890-780= -8536-843=

-5 719 -6490 -7224 -7935 -8670 -9319

-19 330 -21518 -23634 -25 820 -21937 -30124

-9204-909=-10113

LAB = LAB’ + LAB” -. N ’ -17245 -20012 -22993 -25 820 -28142 -31569 -34490 -31316 -40231

BC and CD (compressive):

spans, m

LP, m -

183

366 426 488 548 610 610 132 792 854

244 274 305 335 366 396 427

LAB’ = -1.294 vc, N

366 426 488 548 610 670 732 192 854

244 214 305 335 366 396 421

213

6455 1514 8607 9665 10 159

AB and DE (compressive):

spans, m

Crossarm

L ,;=0.64lv,, N

L&

= -1.294 vc, N -12 910 -15 027

L &=

-l.l69H,

- 1.453Hg, N

-4294-3518= -4991-40951

-1812 -9092

-5725-4690=-10415 -6428-5267=-11695 -7155-5863=-13018 -7859-6440=-14299 -8587-7035=-15622 -9291-7612=-16903 -10018-8208=-18226

-17 214

-19 330 -21518 -23634 -25 820 -21937 -30124

X-brace: spans, m 183 213 244 214 305 335 366 396 427

m,

L,;=-5.662#-

366 426 488 548 610 670 132 192 854

-207969139=-29935 -24205-10638=-34843 -27721-12186=-39913 -31 135 - 13 684 = -44 819 -34 651- 15 232 = -49 889 -38066-16731=-54191 -41593-18219=-59812

3.175Hg,

-m

-45002-19171=-64779 -48523-21325=-69848

LCD = ‘SD’

+ LCD:

-20722 -24 119 -21629 -31025 -34536 -37933 -41442 -44 840 -48350

252

TRANSMISSION

Poles

(at

point

L):

1.5 Vc + Vg + 4.578H, s,

LINE DESIGN MANUAL

+ 3.916Hg

=

+

51956

mm2

183-m spans, 366-m LP s

=

1.5(9977)

+ 4315 + 4.578(3673)

L

+ 3.916(2421)+

2.645(3673)

+ 1.763(2421) (1000) = 9113 kPa

1697.92

51 956

I

2 13-m spans, 426-m LP SL =

C

1.5(11 613) + 5023 + 4.578(4275) 51956

+ 3.916(2818)

+ 2.645(4275)

1

+ 1.763(2818)

(1000) = 10 607 kPa

1697.92

244-m spans, 488-m LP s

= L

1.5 (13 303) + 5754 + 4.578(4897) 51956

+ 3.916(3228)

+ 2.645 (4897) + 1.763 (3228) 1697.92

1

(1000) = 12 150 kPa

274-m spans, 548-m LP s

= L

1.5(14 938) + 6461 + 4.578(5499) 51956

+ 3.916(3625)+2.645(5499)

+ 1.763(3625) 1697.92

1

(1000) = 13 644 kPa

305-m spans, 610-m LP s

L

4.578(6121)

=

+ 3.916(4035)

51956

+ 2.645(6121)

+ 1.763(4035) 1697.92

1

(1000) = 15 187 kPa

335-m spans, 670-m LP s = 1.5(18 264) + 7899 + 4.578(6723) L

+ 3.916(4432)

51956

+ 2.645 (6723) + 1.763(4432) 1697.92

1

+ 2.645 (7346) + 1.763(4842) 1697.92

1

(1000) = 16 681 kPa

366-m spans, 732-m LP (19 954) + 8630 + 4.578(7346) 51956

+ 3.916(4842)

(1000) = 18 225 kPa

396-m spans, 792-m LP SL =

1.5 (21 590) + 9338 + 4.578(7948) 51’956

+ 3.916(5239)

+ 2.645(7948) + 1.763(5239) 1697.92

1

(1000) = 19 719 kPa

CHAPTER

427-m spans, 854-m

l’oles .-

(at point

253

+ 3.916(5649)

+ 2.645 (8570) + 1.763(5649) 1697.92

1

(1000) = 21262

kPa

N):

I .5 V, + Vg + 8.581H,

+ 6.585Hg

&TN=

3.399H, + 2.266Hg +

[

DATA

LP

(23 280) + 10 069 + 4.578(8570) 51956

s, =

V-ADDITIONAL

114 217 nun2

5 509 039 mm3

1ooo mm )(~](looo~~m2)

+lOOO= kPa

183-m spans, 366-m LP S,=

1.5(9977)

+ 4315 + 8.581(3673) 114 217

+ 6.585(2421)

+ 3.399(3673) + 2.266(2421) 5509.039

1

(1000) = 3846 kPa

213-m spans, 426-m LP S,=

1.5(11 613) + 5023 + 8.581(4275) 114 217

+ 6.585(2818)

+ 3.399(4275) + 2.266(2818) 5509.039

1

(1000) = 4477 kPa

244-m spans, 488-m LP sN=

1.5 (13 303) + 5754 + 8.581(4897)

+ 6.585(3228)

+ 3.399(4897)

+ 2.266(3228)

1

(1000) = 5128 kPa

114 217

5509.039

274-m spans, 548-m LP shl=

1.5(14 938) + 6461 + 8.581(5499) 114 217

+ 6.585(3625)

+ 3.399(5499) + 2.266(3625) 5509.039

1

(1000) = 5759 kPa

305-m spans, 610-m LP slv=

1.5 (16 629) + 7192 + 8.581(6121) 114 217

+ 6.585 (4035) + 3.399(6121)

+ 2.266(4035)

1

(1000) = 6410 kPa

5509.039

335-m spans, 670-m LP s,=

1.5(18 264) + 7899 + 8.581(6723) 114 217

+ 3.399(6723) + 2.266(4432) 5509.039

1

+ 6.585 (4842) + 3.399(7346) + 2.266(4842) 5509.039

1

+ 6.585(4432)

(iOO0) = 7040 kPa

366-m spans, 732-m LP SN’

1.5(19 954) + 8630 + 8.581(7346) 114 217

(1000) = 7693 kPa

254

TRANSMISSION

LINE DESIGN

MANUAL

396-m spans, 792-m LP (21 590) + 9338 + 8.581(7948)

S,=

+ 6.585 (5239) + 3.399(7948) + 2.266(5239) 5509.039

114 217

1

(1000) = 8323 kPa

427-m spans, 854-m LP 1.5 (23 280) + 10 069 + 8.581(8570) 114 217

SN’

Table point

26 shows

a summary

of loads

+ 6.585 (5649) + 3.399(8570) + 2.266(5649) 5509.039

in the

structure

members

for

various

1

(1000) = 8974 kPa

span

lengths

and

low

distances.

26.~Summary of loads in structure members for various span lengths and low-point distances (metric example 3)

Table

SAS/Z, m 213

183 Member

244

274

335

366

396

427

670

732

792

854

34303 31427 12910 34 490 41442 59872 18225 7 693

37116 34004 13969 37 316 44840 64779 19719 8 323

40021 36665 15062 40 237 48350 69848 21262 8 974

LP, m

Position

Adjustable braces, N Nonadjustable braces, N Crosstie, N Crossarm (compressive), N Crossarm (compressive), N X-brace, N Pole, kPa Pole, kPa

305

AG&EF GC&FC %&DE BC & CD KN&LM L N

366

426

488

548

610

17151 15 713 6 455 17 245 20722 29935 9113 3 846

19964 18 290

22869 20937 8 607 22 993 27629 39913 12150 5 128

25680 23572 9665 25 820 31025 44819 13644 5 759

28587 26 189 10759 28 742 34536 49889 15187 6 410

2: iii 24119 34843 10607 4 477

31398 28765 11817 31569 37933 54797 16681 7040

U. S. Customary Figure

Load

106

shows

in adjustable

the

structure

outline

and

other

data.

V,

=

(1.8682)(LP)

Vg =

(0.8079)(LP)

H,

= (1.3754)(SAS/2)

Hg =

(0.9066)(SAS/2)

braces

AG and

EF:

LAG’ = LEF) = &/sin a = 1.635Vc Compression

load

in crossarm:

tLAB

Load

in nonadjustable

braces

- LDE

CC and

L,.’

= L,’

‘=-

K/tana=-1.294V,

FC: = 0.5 V, /sin a = 0.8 18 V,

CHAPTER

L

V-ADDITIONAL

Conductor: 759 kcmii, ACSR, 45/7 Diameter - 1.063 in 8- lb/f t ’ wind on iced (+-in radial) conductor - I.3754 lb/ft Vertical force with i-in radial ice - 1.8682 Ib/ft oGw:i-in, H.S. steel, 7-wire Diameter - 0.360 in 8-Ib/ft’ wind on iced (i-in radial) OGW- 0.9066 lb/f t Vertical force with t-in radial ice = 0.8079 Ib/ft

I

x X 1

255

DATA

I

Position

Pole Circumference, in

8 or D KWL

30.37 33.74

M or

49.08

N

ROTS

Pr, Ib=ft 54 75 230 328

55. I5

730 047 976 655

Douglas Fir Working stress = 7400 lb/in* .

tan

a - v

sin

a -

cos

a = 0.7913

- 0.7727

37O 41’

106.-95-ft

Compressive

I

0.6114

a Figure

22 ft

type HSB 230.kV

structure

force in crossarm between LBC

Load in crosstie

‘=L,,

with class 1 Douglas

B and ‘=-

fir poles (two X-braces).

C and between

V,/tana=-1.294

D and C :

V,

GF: LcF) = LAG’ 00s a - L,,’

cos a = 0.647 V,

104-D-1112.

256

TRANSMISSION

Vertical

loads

3

&I ’ = For

transverse

V, u;

and

2

Vg are

= UK’ = u,’

H,

loads

shared

equally

by

= (j-,’

= UN’

= (J,’

Hg,

and

LINE DESIGN

a plane

two

MANUAL

poles:

= Ue’

of inflection

= U,’

HJexists.

= Us’ = 1.5 v,

The

location

+ vg

of this

plane

parts

and

is found

by:

x (prB1

lO(54

xo =PrK +-Prs = Xl =x-x() A plane

of inflection

PQ also

75 047

= lo-

exists.

Its

When

position

considered Horizontal at the

of zero

separately. wind forces

points

of zero

- Y,

moment

Axial

is known,

on conductors

J

reaction

at

by the

moment

moments

= 4*22

ft

655

- 7.43

by:

976) + 230

976

= 10.57

ft

the structure

and overhead

may ground

= 7’43

ft

be separated wires

into

are resisted

equally

each

part

by each

pole

moment:

caused arm

about

B

&“=m

Rd’=R$=-l.SH,-

by horizontal (pole

u ,, = (3H,) J

Taking

328

= 18.0

Rt;‘=R;=

dividing

is found 18(230

Y(PrM)

=Y

+ 54 730

4.22=5.78ft

location

y” =PrR +PrM =

Yl

730)

(4.22)

+ (*H,)

is found

(13.77)

22

pole

(1

above

the

plane

.5Hc + Hg) + 9.5H, 8.5

FF u -- FG”

force

by

taking

moments

about

107)

force

Hand

spacing):

in the

4.22

wind

Hg

I

= 0.575H, +

1.252H,

of inflection

(fig.

= - 0.745H,

- 1.614H,

gives

Fc”:

CHAPTER

-0 4 1

-In a6

The

CF,

HC+Hg AG and E&

braces,

Load

90 percent.

LCG

load

in the

load

LB,”

L, The

load

braces,

of

F,

CG and

” and

CF,

FH”

while

the

inside

braces,

is:

- 1.6 14H,)

= - 0.847H,

- 1.836H,

0.7913

- LCG”

outer

-

in the

10 percent

0.9 (- 0.745H,

=

AG

braces

,, _- 0. I&”

The

inner

cos a

LAG

L,y,

carry

on the

0.9FG”

rr=-

L CF 1)-The

257

Figure 107.-Free body diagram of pole above plane of inflection and to the crosstie (U.S. customary example 3).

Fe”

-I.5

outside

carry

DATA

;n#? 9 -

-z dI

V-ADDITIONAL

-0.1

COSa

and

EF

is:

(-0.745H,

=

-

1.614H,) = 0.0941Hc

0.7913

0.204H,

+

M -

- - LAG”

crossarm

BC

portions

and

CD

is:

= (-a&“)

cos a + 0.5H, = - (- 0.847H, -

= l.l7OH,

+ 1.453H,

1.836H,)

(0.79

13) +

0.204H,)

(0.7913)-

OSH,

rt -

- - L,”

in the

LAB

crossarm

portions

” =- (LAG” = - l.O74H,

LABu --

L,,”

cos

AB

and

a + H,) = -

- O.l61H,

DE

is:

(O.O941H,

+

H,

CG and

TRANSMISSION

258 The

moment

B

at

D

and

is given

For

the

portion

of pole

MANUAL

by:

MD ” = - x,, (1 SH, MB”

LINE DESIGN

+ I$)

= -

6.33H,

- 4.22H,

lb* ft

= MD”

hetween

the

planes

of inflection,

the

moment

at

K and

“=x,(1.5Hc+Hg)=5.78(1.5Hc+Hg)=8.67Hc+5.78Hg

MK

L

is:

lb.0

ML” = MK” The

area of the pole

L

at Kand

, excluding

the 15/16-inch-diameter

hole

for mounting

the X-brace

is:

.rrD2

A,=TAL The

‘K

section

=

=t (10.74)2 - E (10.74) = 90.59 - 10.07 = 80.52 in2

=A, modulus

TD3 -32

15 ED

at

K

15D2/16 6

L

and

is:

= &(10.74)3

- 0.15625(10.74)2

= 121.62-

18.02 = 103.6in3

z, =z, The

horizontal

reaction

in the

poles

P

at

and

Q is:

Rp” = Rp” = - 1.5Hc - Hg The

axial

reaction

in the

poles

u*” = (34;2fJ

at

P

and

Q is:

g ) (58.71) + 0.575H, + 1.252Hg = 8.581H, + 6.589Hg

‘P” = - ‘Q ” The

force

at

FK

FK

K

can be found

”=-

by

51.28(1.5H,

” = - F,”

taking

moments

about

+Hg> + 7.43(1.5H, 22

point

+Hg)

M (fig.

1

108):

= -4.003H,

- 2.669H,

CHAPTER

V-ADDITIONAL

DATA

259

Figure 108.-Free body diagram of pole between (U.S. customary example 3).

Since

the division

installation,

assume

of load that

between

all load

X-braces

is taken

KUand

LTand

by one set of braces.

X-braces The

planes of inflection

VNand

force

WMdepends

in X-braces

KU,

upon

LT, VN,

WMis:

and

4.003H, + 2.669H,

II LKU -L KU ” = The

net

area

of the

pole

The

section

>

n -- - LWM”

X-brace

= $(1~62)~

= - 5.662H, - 3.775H,

mounting

- E

(15.62)

hole)

at

Mand

= 191.62

-

N is:

14.64

= 176.88

= 373.48

- 38.21

in2

=A, modulus

nD3

at Mand

15D2

/16

6

‘hf=x-

Taking

w -L, -

LLT

(less the

ED

A,

sin 4S”

(

moments

about

N is:

- 0.15625(15.62)2

= 0.098(15.62)3

point

MM“=-

M (fig. 7.43(1.5H,

MM n --MN”

108):

+Hg)=-

ll.l4H,

-

7.43H,

-

= 335.36

in3

260

TRANSMISSION

By superposition, and horizontal by

its respective Stress At

the

loading total

in the

poles

values

of the

LINE DESIGN MANUAL

forces

and

can be combined

for total

load

factors

and

safety

bending

moments

loading.

The

strength

computed of each

separately member

for vertic:tl can be divicl~~ll

tabulated.

is:

L:

point

s, =- UL &-ML AL

ZL

where : UL

)) = UJ” + 0.707L,,"

and UL = UL' + UL"

UL" = O.S75H, + 1.252H, +0.707(5.662H,

+3.775H,)=

U,’ = l.SV, + vg U, = UL'+ UL" = 1.5Vc + Vg + 4.578H, + 3.921H, A, = 80.52 in2 ML"

=

8.67H,

+

5.78H,

lb*ft

ZL = 103.60 in3 s, =

= Poles

(at

(%;LuL”

+(~)(ni)

1.5Vc + VK + 4.578H, + 3.921H,

+

80.52 in2 point

N):

‘N SN=ANfZN

where : UN

“= UQ" and UN = UN'+ UN"

MN

4.578H, + 3.921H,

CHAPTER

V-ADDITIONAL

DATA

261

UN’ = 1.5vc + vg UN” = 8.581H, + 6.589H, UN = UN’ + UNf’ = 1.5 V, + Vg + 8.58 lH, + 6.589H, AN = 176.98 in2 MN

“=-1l.l4H,

- 7.43H,

ZN = 335.36 in3 s, = Adjustable

1.5V, + V’ + 8.581H, + 6.589H,

braces

176.98 in2 AC

and

600-ft Spans, LP = 1200 ft

+

11.14Hc + 7.43Hg 335.36 in3

EF: L AG' = 1.635V,

LAGn = O.O94OH,+ 0.204Hg

LAG = LAG' + LA/

3666

77.63 + 110.98 = 189

3855 lb

4277

90.62 + 129.54= 220

4497 lb

4881

103.51+

147.90 = 251

5138 lb

5499

116.50 + 166.46 = 283

5782 lb

6110

129.39 + 185.03 = 315

6425 lb

Vc = 2242 lb V = 9701b f( = 825 lb Hg=

544Ib

700-ft Spans, LP = 1400 ft

V, = 2616 lb V = 11311b g=

9631b

Hg= 635 lb 800-ft Spans, LP = 1600 ft

V, = 2989 lb V = 12931b l$=llOOlb

Hg = 725 lb 900-ft Spans, LP = 1800 ft

Vc = 3363 lb v I(

= 1454 lb = 1238 lb Hg= 8161b lOOO-ft Spans, LP = 2000 ft

vc = 3737 lb v = 1616 lb g=1375lb Hg= 907 lb

262

TRANSMISSION

LINE DESIGN

MANUAL

Adjustable braces AG and E&Continued ’ + LAG”

L AG’ = 1.635 V,

LAG” = O.O94OH, + 0.204Hg

6720

142.37 + 203.39 = 346

7066 lb

7331

155.27 + 221.95 = 377

7708 lb

1300-ft Spans, LP = 2600 ft

7924

168.25 + 240.43 = 409

8351 lb

1400-ft Spans, LP = 2800 ft

8553

181.20 + 258.92 = 440

8993 lb

1100-ft Spans, LP = 2200 ft

LAG

= LAG

Vc = 4110 lb v = 1777lb 4=15131b Hg = 997 lb 1200-ft Spans, LP = 2400 ft

Vc = 4484 lb 2

: ;g

;

H; = 1088 lb

Vc = 5231 lb V = 2262 lb I$ = 1926 lb

Hg = 1269 lb

Nonadiustable Spa% ft 600 700 800 900 1000 1100 1200 1300 1400

Crosstie

braces

GC and

FC :

LP, ft

L Gc’= 0.818V,, lb

lb

1200 1400 1600 1800 2000 2200 2400 2600 2800

1834 2140 2445 2751 3057 3362 3668 3973 4279

699 + 999 = 1698 816+ 1165 = 1981 932+1332=2264 1049 + 1498 = 2547 1165 + 1665 = 2830 1282+1831=3113 1398+1998=3396 1515+2164=3679 1631+2330=3961

-

L cc” = 0.847H,

+ 1.836Hg,

GF : Spans, ft

LP,

600 700 800 900 1000 1100 1200 1300 1400

1200 1400 1600 1800 2000 2200 2400 2600 2800

ft

L ‘$

= 0.647 Vc, lb 1451 1693 1934 2176 2418 2659 2901 3142 3384

3532 4121 4709 5298 5887 6475 7064 7652 8240

CHAPTER

AB

Crossarm

and

DE

LP, ft

L AB’ = -1.294Vc, lb

600 700 800 900 1000 1100 1200 1300 1400

1200 1400 1600 1800 2000 2200 2400 2600 2800

-2901 -3385 -3868 -4352 -4836 -5318 -5802 -6285 -6769

BC

and

263

DATA

(compressive):

Spans, ft

Crossarm

V-ADDITIONAL

L ABn = -l.O74H, lb

- 0.161Hg,

-88688= -974 -1034-102=-1136 -1181-117=-1298 -1330-131=-1461 -1477-146=-1623 -1625-161=-1786 -1772-175=-1947 -1920-190=-2110 -2069-204=-2273

-3875 -4521 -5166 -5813 -6459 -7104 -7749 -8395 -9042

CD (compressive):

spans, ft

LP, ft

L cD’= -1.294Vc, lb

600 700 800 900 1000 1100 1200 1300 1400

1200 1400 1600 1800 2000 2200 2400 2600 2800

-2901 -3385 -3868 -4352 -4836 -5318 -5802 -6285 -6769

L &=

-l.l7OH, lb

- 1.453H

8.

-966791=-1757 -1127922=-2049 -1287-1054=-2341 -1448 - 1186 = -2634 -1609-1317=-2926 -1770-1449=-3219 -1931-1581=-3512 -2092-1713=-3805 -2253-1844=-4097

L CD = “7” -4 -5 -6 -6 -7 -8 -9 -10 -10

+ bD: 658 434 209 986 762 537 314 090 866

X-brace:

Poles

(at

point

spans, ft

LP,

600 700 800 900 1000 1100 1200 1300 1400

1200 1400 1600 1800 2000 2200 2400 2600 2800

ft

L KG = -5.662Hc lb

- 3.775Hg,

-4673-2054= -6727 -5451-2396= -7847 -6230-2738= -8968 -7009-3080=-10089 -7788-3422=-11210 -8566-3765=-12331 -9345-41Oi=-13452 -10124-4449=-14573 -10903-4791=-15694

L):

1.5 V, + Vg + 4.578Hc + 3.921Hg SL = 80.52 ina

600-ft spans, 1200-ft LP s, =

1.5(2242)

+ 970 + 4.578(825) 80.52

+ 3.921(544)

+

= 1320 lb/in2

264

TRANSMISSION

LINE DESIGN

MANUAL

700-ft spans, 1400-ft LP 1.5(2616)+

S,=

1131+4.578(963)

+ 3.921(635)

+ = 1541 lb/in2

8052

800-ft spans, 1600-ft LP SL =

1.5(2989)+

1293+4.578(1100)

+ 3.921(725)

+

8.67(1100)

80.52

+ 5.78(725) 103.60

12 >(

T

= 1760 lb/in2 >

900-ft spans, 1800-ft LP SL =

1.5(3363)

+ 1454 + 4.578(1238)

+ 3.921(816)

+

8.67(1238)

80.52

+ 5.78(816)

103.60

12 I( -7 >

= 1980 lb/in2

lOOO-ft spans, 2000-ft LP s, =

1.5(3737)

+ 1616 + 4.578(1375)

+ 3.921(907)

80.52

8.67(1375)

+ (

+ 5.78(907)

103.60

12 )O T

= 2200 lb/in2

1 lOO-ft spans, 2200-ft LP s, =

1.5(4110)+

1777 +4.578(1513)+

3.921(997)

+

8.67(1513)+

80.52

5.78(997)

103.60

12 )O r

= 2420 lb/i2

1200-ft spans, 2400-ft LP 1.5 (4484) + 1939 + 4.578(1650) s,

=

+ 3.921(1088)

8.67 (1650) + 5.78(1088) +

80.52

103.60

-12 JO 1

= 2640 lb/in2

1300-ft spans, 2600-ft LP s

= 1.5(4857)+

2101 +4.578(1788)

L

+ 3.921(1179)

+

= 2861 lb/in2

80.52

1400-ft spans, 2800-ft LP SL =

Poles

(at

1.5 (5231) + 2262 + 4.578(1926) 80.52

point

N):

l.5Vc+Hg+8.581Hc+6.589Hg s,=

176.98 in2

+ 3.921(1269)

+

8.67(1926)

+ 5.78(1269) 12 103.60 )O T

= 3081 lb/in2

CHAPTER

V-ADDITIONAL

DATA

265

600-ft spans, 1200-ft LP S,=

1.50242)

+ 970 + 8.581(825)

+ 6.589(544)

+ 11.14(825)

176.98

+ 7.43(544)

335.36

(

)O

12 = 558 lb/in2 1

700-ft spans, 1400-ft LP S,-=

1.5(2616)+

1131+ 8.581(963)

+ 6.589(635)

+

176.98

800-ft spans, 1600-ft LP S,=

1.5(2989)

+ 1293 + 8.581(1100)

+ 6.589(725)

+

176.98

11.14(1100) (

+ 7.43(725)

335.36

12 )O

r

= 744 lb/in2

900-ft spans, 1800-ft LP S,=

1.5(3363)+

1454 +8.581(1238)

+ 6.589(816)+

176.98

lOOO-ft spans, 2000-ft LP s,=

1.5(3737)

+ 1616 + 8.581(1375)

+ 6.589(907)

+

176.98

1 loo-ft spans, 2200-ft LP s,=

1.5(4110)+

1777+8.581(1513)+6.589(997)+ (

176.98

11.14(1513)+7.43(997) 335.36

12 )O - = 1023 lb/in2 1

1200-ft spans, 2400-ft LP s,=

1.5 (4484) + 1939 + 8.581(1650)

+ 6.589(1088)

+

176.98

1300-ft spans, 2600-ft LP SN’

1.5(4857)

+ 2101+ 8.581(1788)+

6.589(1179)

176.98

)O

1400-ft spans, 2800-ft LP sN=

1.5 (5231) + 2262 + 8.581(1926) 176.98

+ 6.589(1269)

+

12 = 1209 lb/in’ 1

TRANSMISSION

266 Table point

27 shows

a summary

of loads

LINE DESIGN MANUAL

in the

structure

members

for

various

span

lengths

and

low

distances.

Table 27.-Summary

of loads in structure members for various span lengths and low-point distances (U.S. customary example 3) SAS/Z, ft 600

Member

700

800

900

1000

1100

1200

1 300

1400

2000

2200

2400

2600

2800

6425 5881 2418 6459 7762 11 210 2200 930

7066 6475 2659 7104 8537 12 331 2420 1023

7708 7064 2 901 7749 9 314 13452 2640 1116

8351 7 652 3142 8 395 10090 14573 2861 1209

8993 8240 3 384 9042 10 866 15 694 3081 1 302

Position LP, ft

Adjustable braces, lb Nonadjustable braces, lb Crosstie, lb Crossarm (compressive), lb Crossarm (compressive), lb X-brace, lb Pole, lb/in2 Pole, lb/in2

26.

Structure

location, wood-pole

Data

(a)

l

necessary

1800

3855 3532 1451 3875 4658 6727 1320 558

4497 4121 1693 4521 5434 7841 1541 651

5138 5782 4709 5298 1934 2176 5166 5813 6209 6986 8968 10089 1760 1980 744 837

spatting is a term line structures and bracing

Required.-The

following

to the

structure

plan-profile

limitation

scales and

for

guying

for

the For

and

line: These the

process

equipment

drawings

are

required

are prepared

specified

conductor,

and

a conductor

charts,

by the

field

span,

and

ruling height

table

Process ofSpotting.-Figure

Code

109 shows

or the

the details

plan and profile drawing with the sag template spotting structures. Figure 110 also shows the

applicable

State

for the various

or

clearances over ground, railroads, highways, communication circuits, and These clearances should be calculated in accordance with the latest edition Safety

at the structure

for

Reqnired conductor other power lines.

Electrical

ground

the

the conductor

National

above

of determining

on the plan and profile drawings. is also determined for each location.

template showing types and heights.

of the

height

used

data

of structures on a transmission drawings of the transmission line.

Plan

l

(b)

and Equipment the locations and profile

forces. A sag template made loading conditions. The

1600

type of transmission the amount of guying

l

l

1400

Structure

Spotting.-

height, and structures,

determining

AG&EF GCLFC GF AB & DE BC & CD KU< L N

1200

or municipal

of the sag template,

and figure

structure

code. 110 is a typical

superimposed showing the method of using it for method of using the 15,5 OC (60 “F) curve of the

template to determine the proper conductor and structure heights. The curve labeled “15.5 ‘C (60 o F) Final” represents the conductor position. The lower two curves, marked “8.2-m (27-ft) are exactly the same curves as the 15.5 ’ C final curve, but clearance” and “8.8-m (29-ft) clearance” displaced vertically the corresponding the

8.8-m

clearance

8.2 and 8.8 m, respectively. Therefore, point on the 8,2-m clearance curve curve.

Referring

again

to figure

110,

any point on the final curve is 8.2 m above or 8.8 m above the corresponding point on the

8.8-m

clearance

curve

just

touches

the

CHAPTER ground the

line

8.8-m

of the

profile.

clearance

line

Therefore, touches

the

the

V-ADDITIONAL conductor

ground

DATA

267

is 8.8 m above

the

ground

at the

point

where

line.

Ccnductw: 201 md (387.5 kcmil), ACM, 2W7 Ruling Span =213.4 m (700 f t ) Max. Tension = 32 472 N (7300 lb), 45% Ult. NESC Heavy Loading: IS-mm (IL&in) ice with a 0.38-kPa (8-lb& wind at -18 OC (0 OFI

cut out to prrmit drawing curve on the plan-profile

/

the I sheetr.

TYPE

low point span.

Figure

The

process

109.-Typical

of spotting

usually

2083 + 50 on figure 110, and the position described above. in U.S. customary span to the right is selected, either the various

types

sag template

(plastic)

progresses

the spans to the Please note that

used for spotting

from left the

left

to right

structures.

on the

of it are spotted station numbering

ter of ES the in each

21.3 IO.3 15.2 12.2

Ill m Ill Ill

HS

@Ott)@Oft& QofO{ &oft)-

GROUND -

104-D-1113.

profile.

The

structure

at Sta.

before the template is placed in referred to in this section are

units. After the required position of the conductor has been determined for the of the structure at Sta. 2083+50, the location and height of the next structure by scaling or by use of a pole template. For convenience, the pole template for of structures

is marked

on the margin

of the template.

For

the span

under

discussion,

the structure location selected is at Sta. 2090 +20, the structure is a type HS with 18.3-m (60-ft) poles, and the span length is 204 m (670 ft). This information should be recorded on the drawing. The template is then moved to the right and the next span and structure located by repeating the process.

TRANSMISSION Although

the process

the profile

of spotting

for several

spans

or railroad

crossings,

powerline

which

require

will

special

structures

ahead

because

ahead

there

to one of the fixed

is a choice

to determine that

of structure

of equal

heights.

The

points at each of the as much as possible. transmission

(c)

may

example,

and

high

it is best to examine angle

or low

of the structure.

points,

points Such

and

work

backward.

be desirable

desirable ruling

profile

to make

layout

In the

highway

in the profile

conditions

more

is to have

span,

a smooth

is a sign

of good

sections

often

than

one

of line layout

spans of nearly

conductor design.

The

where

in order

uniform

profile,

and

length

structures

conductor

structures should lie in a smooth flowing curve to equalize This is called grading the line and is an important part

UplifLUplift, in a rough profile

occur

refer

to the three

conductor sag is drawn conductor will contract 60 o F), the conductor template. supports

conductor

crossings,

as line

attachment

structure of the

loading design

of a

line.

Determining

Uplift

the

left to right, such

and it is usually a matter of determining the most these fixed locations. Sometimes it is desirable to

it may most

less than

smooth

line

locations

The

from

be conditions

the location

structure, between

locations,

to or slightly

progresses

may

and affect

structure

the best arrangement.

are equal

usually there

or communication

consideration

fix the location of a transmission line desirable arrangement of the structures move

LINE DESIGN MANUAL

Therefore, of alternate

or upstrain, where the

structures

is a condition which conductor supports

at Sta. 2105+35,2112+40,

should be avoided, if possible. are at different elevations. For and

2121+70

on figure

for a temperature of 15.5 ‘C (60 ’ F), but as the temperature and the sag will decrease. When the temperature reaches minus assumes the position indicated by the minus 51 ‘C cold curve

minus 51 OC curve on the template between the conductor 2 105 + 35 and 2121+70), it can be determined whether the support of the intermediate structure (Sta. 2112+40) is above or below the cold curve. 21.3-m (70-ft) structure at Sta. 2112 +40, the conductor support is approximately on the

conductor For the

by placing structures

lll.‘The

decreases, the 51 ‘C (minus shown on the

the (Sta.

cold curve. Suppose, however, that the 21.3-m (70-ft) structure is replaced by a 19.8-m (65-ft) structure. The conductor support would then be below the cold curve and the conductor would exert an upward pull on the structure-this upward pull is the uplift or upstrain. Uplift at a structure will cause the conductor to pull the insulators cause the conductor to pull away from crossarm. Uplift may possibly be avoided

up into the crossarm, and with pin-type insulators it might the insulator and possibly pull the insulator pin out of the by adjusting structure locations on the plan-profile drawing,

to take advantage of terrain, by using a higher structure at the point of uplift or by attaching weights to the conductor. If these methods fail, then the conductor must be dead-ended. Structures should not be located at uplift points if it can be avoided because the only function of such a structure is to hold the conductors conductors during hot

Insulator

(d)

that

tends

from

to swing

the

swinging force

of the

to the distance of the

adjacent

wind

pressure

Sideswing.-Suspension

pressure. Conductor to limit the sideswing on the conductor

against weather.

clearance in order an insulator

spans

suspended adjacent

to the

insulator between

fall

of the

The

rapidly

away

vertical

conductor

low

the

to sideswing by insulator insulation.

is equal force

structure,

supported the

caused

length

of the

by horizontal

wind

sideswing, so it is necessary The horizontal wind pressure

tends

by the

of the adjacent

a short

to one-half

that

supported

of conductor points

from

to support

are subject is reduced conductor

The

length

conductor

sometimes

on a structure

spans.

force

string. the

insulators

to the structure to maintain proper

in the two is equal

and

the

to keep insulator

by the spans. conductor

total string

insulator

On rough low

wind

pressure

the insulator

terrain points,

plus string

string one-half is equal

where

each

as indicated

CHAPTER by the

conductor

template,

the low points from

is still

swinging.

the

Too

structure.

distance

To

the

checked

the point

much

insulator

falls

sideswing

might

(e) the

instructions

allowable,

low-point

distance,

of structure

could

for

extra

proper

angle

be used In

for

by a broken

conductor

policy and

of 230

kV

and

When strainextra clearance on both

sides

adjacent

span

or pin-type for broken is not

the

to a special

maximum

the

be added

the

this

If

value

area,

Structure at the

value

chart.

be used,

outside

the

heights

bottom

lengths

and

structure

heights

used

for

pole

lengths

over.

strength

structure

of 13.7

Class

1 poles

of the

are given

is needed

limitation

number

of guys,

as shown

major

highways,

major

for

any

chart, on the

m (45 ft) are used

in

or less; for extra

reason.

should guying

be used charts,

at a

should

adjacent

to the crossing span. Other states broken (1977), d o not require

required

clearance for

broken

to maintain

over

the

circuits,

clearance

edition

lines

communication

sufficient

latest

NESC

structures conductors.

of a crossing,

Whenever the under the outside The

This

railroads,

conductor

clearance

major

conditions

and

required

are governed conductor

highways,

major

on transmission

lines

above.

the

enough

it is necessary

be reduced structure.

with

communication

or lake crossings

one-half

may

falls

is necessary.

could

m (50 ft) and

additional

railroads,

of the spans in

to provide major

When sagged

over

which,

River than

correction

the

limitation

type

point

or

limits,

be used.

by the

correct

be provided in either

rules

powerlines,

or where

The

the

hardware,

plan-profile.

structure

structure

weights

the allowable

structures.

should

considerations. It is our

of 15.2

as indicated

all crossings

NESC

are normally

line.

some

span

3 poles

structures,

on the

between

the insulator

line.

for lengths

of structure,

wood-pole

powerlines

the

class

in a transmission

California,

major

tall

type

transmission

used

and

insulators,

suspension If

distance

to hold

is within

is measured

limits.

the

vertically

in the

the specified

prescribed

However,

insulators

on the

regarding

each

lines,

are normally

spans,

The

by

the

more

spans.

a failure

spans

spans

273

as acting

of the

area in which

than

type

for

On all wood-pole class 2 poles

line

be greater

or another

can cause

adjacent

General Instructions.-Instructions

design

long

the

be within

adjacent

sideswing

of adjacent

within

to provide

strings,

the

of the

sum

will will

be adjusted

insulator

points the

of the

DATA

to be considered

distance

whether

low

sideswing

outside

low-point

against

so defined

of insulator

fall

of conductor

determine

between

is then

may

the length

V-ADDITIONAL

ruling

are used on both sides of a crossing, it is not necessary For lower voltages, when suspension-type structures

increased

involving span,

decrease

special

to use spans

ruling

sag in the

to seriously

the

crossing the

structures

longer

than

conductors

or long spans approximately

should

tension

due

to a broken

in most

cases.

are to be handled 1.7 times

be dead-ended

conductor as special

the

ruling

span

at both

ends

of the

in an studies.

or shorter span

and

span.

terrain slopes across the right-of-way, conductor on the high side to meet

approximately Other policies

span

clearance

to allow are used

in conductors

and

overhead

50 percent in the regarding substations

1. It is Bureau policy a substation or switchyard.

sufficient clearance all requirements. ground

span terminating and switchyards

to install self-supporting In general, this means

structures that the

wires on are:

under the

should full

load

substation

(no guys) within structure adjacent

be maintained should

normally

or switchyard 183 m (600 ft) of to the substation

274

TRANSMISSION

or switchyard conductor 2.

will and

When

is not

the

reduced,

should

he a steel

overhead

may

be varied

and

any special

to meet

The

method

that

is designing

of approach the

tension

requirements

between

span

structures

before

angle

in the transmission

and design

roads, should

proceeding

where

with

the

and

overhead ground

final

tension

ground

wires

wire

substation

power

tensions

or switchyard

or communication

be discussed the

in the

conductor

overhead of the

and

tensions

yard.

conductors

structural

as railroads,

unbalanced

the

in a span

the

or switchyard

steel

the into

of conductor

of the such

to the substation

slack

is reduced

of reduction

requirements

of accepting

to the

clearance

amount

the

crossing

due

wire

midspan The

capable

wire

ground

sufficient

be maintained.

structure

ground

overhead

LINE DESIGN MANUAL

with

design

the

of the

lines.

design

group

transmission

line. 3.

The

be made deflection

deflection

as small as possible. angle reduces the

or switchyard

structure.

On wood-pole

lines

all guyed

structures

27.

Right-of-Way

transmission

line

strings require wide enough clearance Sufficient

should

sandy

that imposes

soil or other

have

a separate

and Building

design.

Today’s

soil with

poor

anchorplate

Clearance

higher

this angle additional

voltages,

obstruction is essential

that may to avoid

for

each

.-Right-of-way wider phase

adjacent to the right-of-way. are hanging in their no-wind

It is legally possible for someone right-of-way, and occasionally this

be at the flashover

or switchyard

characteristics

guy

clearance

should

is encountered,

strand.

is a very spacings,

important consideration in and unrestrained insulator

ever before. A right-of-way in a high-wind situation,

edge of the right-of-way to trees, buildings, pole

Some of these position.

structure

be less than loo because a larger transverse load on the substation

bearing

a wider right-of-way and greater clearances than to give adequate clearance between conductors

from any clearance

obstruction conductors

where

line at the substation

It is preferred clearance and

hazards

on private lines, and

are not

obvious

to erect a structure, such as a building, at the is done. The only way we can protect ourselves

very and

must be and also property. any other when

the

edge of our others is to

make our right-of-way wide enough to provide a minimum electrical clearance between the outer conductor, at a maximum wind condition of 0.43 kPa (9 lb/ft2), and an imaginary building with a wall on the edge of the right-of-way. Tables 28 and 29 show the horizontal distance required as clearance between Tables 30 through ruling spans.

a conductor and a building for various line voltages 35 show the required right-of-way for transmission

Sometimes there is a tendency to reduce the right-of-way require shorter spans (to keep the conductors safely within would

be more

expensive

than

initially

because

of the

and elevations above sea level. lines of different voltages and

width to keep costs down, but this would the right-of-way) and the line probably

additional

structures

required.

CHAPTER

V-ADDITIONAL

DATA

275

Table 28 .-Minimum

horizontal clearance to buildings- USBR standard for NESC light, medium, and heavy loading (metric)

kV

Ruling span, m

Conductor

69

84 mm2 ACSR 6/l

115

135 mm2 ACSR 26/7

138

242 mm2 ACSR 2417

161

242 mm2 ACSR 2417

230

483 mm2 ACSR 4517

345

483 mm2 ACSR 4517 duplex

213 305 213 305 213 305 213 305 305 366 427 305 366 427

Basic clearance, m

3.048 3.048 3.048 3.048 3.048 3.048 3.048 3.048 3.048 3.048 3.048 3.048 3.048 3.048

Increase for voltage,’ m

0.2003 .2003 .3420 .3420 .4836 .4836 .9086 .9086 .9086 1.6170 1.6170 1.6170

Increase for elevation, m

Minimum horizontal clearance to buildings,2 m

3 percent of ‘increase for voltage” for each 305 m of elevation over 1006 m

3.048 3.048 3.249 3.249 3.389 3.389 3.532 3.532 3.956 3.956 3.956 4.666 4.666 4.666

r The increase for voltage is: ’ At 1006-m elevation and a

Table 29.-Minimum

horizontal clearance to buildings-USBR standard for NESC light, medium, and heavy loading (U.S. customary)

kV

Conductor

69

No. 4/O AWG ACSR 611

115

266.8 kcmil ACSR 2617

138

477 kcmil ACSR 2417

161

477 kcmil ACSR 2417

230

954 kcmil ACSR 4517

345

954 kcmil ACSR 4517 duplex

Ruling spa ft

Basic clearance, ft

700 1000 700 1000 700 1000 700 1000 1000 1200 1400 1000 1200 1400

10 10 10 10 10 10 :: 10 10 10 :8 10

l The increase for voltage is: 2 At 3300-ft elevation and at 60 “F with a 9-lb/ft’

wind.

Increase for voltage,’ ft

0.66 0.66 1.12 1.12 1.59 1.59 2.98 2.98 2.98 5.31 5.31 5.31

Increase for elevation, ft

3 percent of ‘increase for voltage” for each 1000 ft of elevation over 3300 ft

Minimum horizontal clearance to buildings,2 ft 10.00 10.00 10.66 10.66 11.12 11.12 11.59 11.59 12.98 12.98 12.98 15.31 15.31 15.31

Table 30.-Right-of-way Maximum

kV’

69

Conductor

Ruling span, m

conductor tension,2 newtons per conductor

84 mm2 ACSR 6/l

213

a12 900

115 138

135 mm2 ACSR 26/7 242 mm2 ACSR 2417

213 213 305

a24900 a16400

161

242 mm2 ACSR 24/l

213

230

483 mm2 ACSR 4517

305 305

a23 100 a24900 a23 100 a31100

366 427 305 366

a30200 a29 300 a31 100 a30200

427

a29 300

305

305

345

483 mm2 ACSR 4517

duplex

b16 900

Insulator string length, mm

Conductor swing 0.43kPa wind l/3 low point Degrees m

869 869

65O19’ 65O19’

1219 1372 1219

3746 7 705 8 964

3745 7576 3460 73746 116

a12900

r 69 through 161 kV are H-frame wood-pole construction; 2 Maximum conductor tensions are limited by:

Conductor sag at 15.5 oc, mm

values-NESC Iigh t loading (metric)

Right-of-way,4 m

2 %

63OO2' 57OO9’ 63OO2'

3.048 3.048 3.658 4.267 3.658

3.048 3.048 3.249 3.389 3.249

21 28 23 24 29

g

1372 1676 1676 2286

57OO9’ 57009’ 57OO9’ 50°38'

7.6255 4.5550 7.8809 8.6982

4.267 5.182 5.182 7.620

3.389 3.532 3.532 3.956

;; 34 41

Iz m u

12851 17676 8 964 12851

2286 2286 3658 3658

50°38' 50°38' 50°38' 50°38'

11.7036 15.4341 9.7590 12.7644

7.620 7.620

9.144 9.144

3.956 3.956 4.666 4.666

47 55 48 54

17676

3658

5OO38'

16.4949

9.144

4.666

61

230 and 345 kV are steel tower construction.

3 At 1006-m elevation, and at 15.5 OC with a 0.43kPa wind. 4 At 1006-m elevation, and rounded off to next highest meter.

Minimum horizontal clearance to buildings,3 m

7.6736 4.1705 4.2996 7.4292

7 705

a 18 percent ultimate strength at 15.5 oC fmal, no load. b 25 percent ultimate strength at -18 OC final, no load.

4.1925

Outside phase to structure centerline, m

$

E n z 5

z ?

Table 3 1.-Right-of-way

kV’

69

Conductor

No. 4/O AWG ACSR 6/l

115

266.8 kcmil ACSR 2617

138

477 kcmil ACSR 24/l

Ruling span, ft

411 kcmil ACSR 2417

230

954 kcmil ACSR 4517

345

954 kcmil ACSR 45/l duplex

Conductor sag at 60 OF, ft

Insulator string length, ft

12.28 24.86 11.34 23.27 12.28 25.25 12.28 25.25 29.31 42.09 57.89 29.37 42.09 57.89

2.5 2.5 4.0 4.0 4.5 4.5 5.5 5.5 7.5 7.5 7.5 12.0 12.0 12.0

700

a2900

a29oo

700

b3800 a3700 a5600 a5200 a5600 a5200 a7000 86800 a66oo a7000 86800 a66oo

700

1000 161

Maximum conductor tension,* pounds per conductor

1000 1000 700

1000 1000 1200 1400 1000 1200 1400

values-NESC light loading (US. customary)

r 69 through 161 kV are H-frame wood-pole construction; * Maximum conductor tensions are limited by:

Conductor swing 9-lb/ft* wind l/3 low point Degrees Et 65O19’ 65O19’ 63OO2' 63OO2' 57009’ 57OO9’ 57009’ 57009’ 50°38' 50°38' 50°38' 50°38' 50°38' 50'38'

230 and 345 kV are steel tower construction.

a 18 percent ultimate strength at 60 OF fmal, no load. b 25 percent ultimate strength at 0 OF fmal, no load. ’ At 3300-ft elevation, and at 60 OF with a 9-lb/ft* wind. 4 At 3300-ft elevation. and rounded off to next highest 5 feet.

Outside phase to structure centerline, ft

Minimum horizontal clearance to buildings,3 ft

Right-of-way,4 ft

10 10

10.00 10.00

70

24.86 13.67 24.31 14.10 24.99

90

s

12 12 14 14

10.66 10.66

75 95 80

!z

14.94

17

25.83 28.51 38.34 50.56

:5’ 25 25

13.43

31.99 41.82 54.04

i8 30

11.12 11.12 11.59 11.59 12.98 12.98 12.98 15.31

105 90 110 135 155 180 155

15.31 15.31

175 200

7

$ 0 =i 5 z :

2

Table 32.-Right-of-way

kV’

Conductor

69 115 138

84 mm2 ACSR 6/l

213

135 mm2 ACSR 26/7

213

242 mm2 ACSR 24/l

161

242 mm2 ACSR 24/l

230

483 mm2 ACSR 45/l

345

Ruling span, m

483 mm2 ACSR 4517

duplex

Maximum conductor tension,’ newtons per conductor

Conductor sag at 15.5 oc, mm

a15 500

values-NE,!%7 medium loading (metric) Insulator string length, mm

Conductor swing 0.43-kPa wind l/3 low point Degrees m

305

bll300

a19 100

4033 7521 3111

305

b21300 a26700

6947 4 111

869 869 1219 1219 1372

305

305 305 366 427 305 366

b28500 a267OO b28500 b37400 b36900 b36400 b37400 b36900

1655 4 111 7655 8948 12 821 17525 8948 12 821

1372 1676 1676 2286 2286 2286 3658 3658

50038' 50°38' 50°38' 50°38' 50°38'

11.6850 15.3174

427

b36400

17525

3658

50°38'

16.3782

213 213

r 69 through 161 kV are H-frame wood-pole construction; 2 Maximum conductor tensions are limited by:

65O19’ 65O19’

3 At 1006-m elevation, and at 15.5 OC with a 0.43-kPa wind. 4 At 1006-m elevation, and rounded off to next highest meter.

3.048 3.048

4.4531

7.6291

3.048 3.048 3.658

63OO2'

1.2185 4.6062

3.658 4.267

7.5835 4.8616

4.267 5.182 5.182 7.620 7.620 7.620

230 and 345 kV are steel tower construction.

a 25 percent ultimate strength at -29 OC final, no load. b 18 percent ultimate strength at 15.5 OC fmal, no load.

Minimum horizontal clearance to buildings,3 m

63OO2'

57009’ 57009’ 57009’ 57009’

4.4542

Outside phase to structure centerline, m

7.8389 8.6859

9.7467 12.7458

9.144 9.144 9.144

3.249 3.249 3.389 3.389

Right-of-way ,4 m

z %

22 28

g

5;

ul

i:

3.532 3532

3:

P 1 z

3.956 3.956 3.956

41 41

0 E

:48

5

4.666 4.666

54

4.666

61

m

5 z r

Table 33.-Right-of-way

kV’

Conductor

69

No. 4/O AWG ACSR 6/l

115

266.8 kcmil ACSR 2617

138

477 kcmil ACSR 2417

161

477 kcmil ACSR 2417

230

954 kcmil ACSR 4517

345

954 kcmil ACSR 4517 duplex

Ruling span, ft 700 1000 700 1000 700 1000 700 1000 1000 1200 1400 1000 1200 1400

Maximum conductor tension,2 pounds per conductor

Conductor sag at 60 OF, ft

a3500 b3900 a4300 b4800 a6000 b6400 a6000 b6400 b8400 b8300 b8200 b8400 b8300 b8200

’ 69 through 161 kV are H-frame wood-pole construction; ’ Maximum conductor tensions are limited by:

values-NESCmedium

13.16 24.63 12.37 22.75 13.49 25.15 13.49 25.15 29.38 42.07 57.38 29.38 42.07 57.38

Insulator string length, ft 2.5 2.5 4.0 4.0 4.5 4.5 5.5 5.5 7.5 7.5 7.5 12.0 12.0 12.0

loading (U.S. customary) Conductor swing 9-lb/ft2 wind l/3 low point Degrees ft 6S019’ 65O19’ 63OO2’ 63OO2’ 57009’ 57009’ 57009’ 57OO9’ 50°38’ 50°38’ 50°38’ 50’38’ 50°38’ 50°38’

230 and 345 kV are steel tower construction.

a 25 percent ultimate strength at -20 OF final, no load. b 18 percent ultimate strength at 60 OF final, no load. ’ At 3300-ft elevation, and at 60 OF with a 9-lb/ft’ wind 4 At 3300-ft elevation, and rounded off to next highest 5’f&.

14.23 24.65 14.59 23.84 15.11 24.91 15.95 25.75 28.51 38.33 50.16 31.99 41.81 53.64

Outside phase to structure centerline, ft

Minimum horizontal clearance to buildings,3 ft

10 10 12 12 14 14 17 17 25 25 25 30 30 30

10.00 10.00 10.66 10.66 11.12 11.12 11.59 11.59 12.98 12.98 12.98 15.31 15.31 15.31

Right-of-way,4 ft

70 90 ;: 85 100 1?8 135 155 180 155 175 200

II TJ -F ZJ a =I 5 z g 2

Table 34.-Right-of-way

values-NESC heavy loading (metric)

84 mm2 ACSR 6/l

213

Maximum conductor tension,? newtons per conductor a18 200

115

135 mm2 ACSR 26/l

305 213

a182OO b24400

12530 4452

869 1219

65O19' 63OO2'

12.1751 5.0547

3.048 3.658

3.048 3.249

138

242

305

9524 4665 8 101 4 665 8 101 8 954 12 844

1219 1372

2286 2286

63OO2' 57009' 57009' 57009' 57009' 50°38' 50°38'

9.5755 5.0716 1.9582 5.3270 8.2186 8.6905 11.6982

3.658 4.267 4.267 5.182 5.182 7.620 7.620

3.249 3.389 3.389 3.532 3.532 3.956 3.956

34 41 41

Rulins span,

Conductor

kV’ 69

m

230

345

Insulator string length, mm

Conductor swing 0.43kPa wind l/3 low point Degrees m

5194

869

65O19'

Outside phase to structure centerline, m

Minimum horizontal clearance to buildIngs,3 m

6.0544

3.048

3.048

25

is

31 24

f

;z

$

239”

r z

Rightof-way,4 m i

242 mm2 ACSR 24/l

213

483 mm2 ACSR 4517

305 305 366

a24900 b33300 a382OO b33 300 a38200 c51 100 c50700

421

c50 300

11515

2286

50°38'

15.3097

7.620

3.956

54

305 366 421

c51100 c50700 c50 300

128954 844 17515

3658 3658

50'38' 50°38' 50°38'

12.7590 9.7513 16.3705

9.144 9.144

4.666 4.666

54 48 61

mm2ACSR

24/l

213

305 161

Conductor sag at 15.5 oc, mm

483 mm2 ACSR 45/l duplex

: 69 through Maximum

161 kV are H-frame wood-pole conductor

construction;

1372 1676

1676

230 and 345 kV are steel tower construction.

tensions are limited by:

a 50 percent ultimate strength at -18 OC initial, full load. b 33-l/3 percent ultimate strength at -40 OC initial, RO load. c 18 percent ultimate strength at 15.5 OC final, no load. 3 At 1006-m elevation, and at 15.5 OC with a OA3kPa wind. 4 At 1006-m elevation, and rounded off to next highest meter.

m

0 FJ s 5 f

r

Table

Conductor

kV’

35 .-Right-of-way

values-NESC heavy loading (U.S. customary)

Ruhng spa% ft

Maximum conductor tension,2 pounds per conductor

Conductor sag at 60 OF, ft

Insulator string length, ft

a4 100 a4 100

18.95 41.00

2.5 2.5

65O19’ 65O19’

19.49 39.53

10 10

10.00 10.00

80 120

12 12 14 14 17

10.66 10.66 11.12 11.12 11.59 11.59 12.98

80 110 85 105 95 115 135

12.98 12.98

155 180

: =i 6 z

Conductor swing 9-lb/ft’ wind l/3 low point Degrees ft

Outside phase to structure centerline, ft

Minimum horizontal clearance to buildings,’ ft

Rightof-way,4 ft

69

No. 4/O AWG ACSR 6/l

700 1000

115

266.8 kcmil ACSR 26/7

700 1000 700 1000 700 1000 1000

b5 a5 b7 a8 b7 a8 cl1

500 600 500 600 500 600 500

14.57 31.23 15.47 26.54 15.47 26.54 29.35

4.0 4.0 4.5 4.5 5.5 5.5 7.5

63OO2’ 63OO2’ 57009’ 57009’ 57OO9’ 57009’ 50°38’

16.55 31.40 16.78 26.08 17.62 26.92 28.49

1200 1400

c11 400 cl1 300

42.13 57.51

7.5 7.5

50°38’ 50°38’

38.37 50.26

1000 1200

c11 400 500 cl1

29.35 42.13

12.0

50°38’

31.97 41.85

ii 30

15.31

155 175

57.51 12.0 50°38’ 1400 c11 300 t 69 through 161 kV are H-frame wood-pole construction; 230 and 345 kV are steel tower construction. 2 Maximum conductor tensions are limited by:

53.74

30

15.31

200

138’

477 kcmil ACSR 2417

161

477 kcmfl ACSR 2417

230

954 kcmil ACSR 4517

345

954duplex kcmil ACSR 4517

a 50 percent ultimate strength at 0 OF initial, full load. b 33-l/3 percent ultimate strength at -40 OF initial, no load. ’ 18 percent ultimate strength at 60 OF final, no load. ’ At 3300-ft elevation, and at 60 OF with a g-lb/f? wind. 4 At 3300-ft elevation, and rounded off to next highest 5 feet.

:: 25

? D : Y 5

.

: 2

282

TRANSMISSION

28.

Armor

Rods

of vibrations may

and

produced

well

result

by very

turbulence

steady

on the leeward

1 to possibly

millimeters aeolian

the

hertz,

reinforced

induces

vibrations

of an inch

the

aluminum

on a clear, conductor

cold

the conductor support, which and

The

The

a few

light

Therefore, dampers,

effect on vibration value is through

range

and

to 200

amplitudes nodes,

of

tension

force per unit length. On short spans, by the humming sound produced-like and

is usually

this type or both.

strung

to fairly

of conductor

requires

and reduce the amplitude from the reinforcing of the conductor

some protection to the conductor due to flashovers. Armor rods

for

high special

10 to 20 percent; at the point of

against vibration, the armor aluminum conductors are

to a stranded cable of larger diameter-thereby region of maximum bending stress.

A set of 7 to 13 rods, depending length of the rods vary with

are eddy

frequencies

between

and consist of a spiral layer of short, round rods surrounding conductor to its support is made in the middle of the armored

equivalent is in the

which varying

millimeters

frequencies distance

and

morning.

is comparatively

rods have some damping their greatest protective

frequencies

from

types

conductor

periodically

the excitation.

length,

and other

in the

natural

It is the

range

or more. span

to aeolian

stresses

of the

and the conductor and is evident only

Armor however,

to offering from burns

those

produces

velocity,

so it is quite susceptible to vibration. by the use of armor rods, vibration

maximum stress. In addition rods protect the conductor

are subject bending

normally

inches),

wind

of the conductor, small amplitude lines

that

to several

of the

MANUAL

(1 to 30 mi/h).

amplitudes

tensions, protection

made of aluminum attachment of the

are

of 1 to 48 km/h and

are functions

singing of telephone Steel

repeated

Aeolian

winds

in the conductor, diameter the vibration is of extremely the

conductors

which

side of the conductor

100

(a fraction vibrations

Dampers.-All

wind,

in its failure.

stimulated from

Vibration

by

LINE DESIGN

the conductor. length. This

strengthening

The makes

it at the

on conductor size, is required to armor a conductor. The size the size of the conductor. Generally, because of the ease of

application, and for both

and removal if necessary, preformed armor rods are used for all sizes of ACSR conductors steel and Alumoweld overhead ground wires. Formed rods are manufactured with a spiral

shape

the

to fit

diameter

of the

conductor

on which

they

are to be used.

The

ends

of each

rod

are

discharge of armor

or parrot-billed to reduce the chance of abraiding the conductor and the tendency for corona at these points. Clips or clamps are not required on this type of armor rod. Older types rods, now seldom used by the Bureau, include the straight rod and the tapered-rod types.

Straight

armor

rounded

rods,

having

a constant

diameter

sizes of 15 to 62 mm2 (No. 6 AWG to No. with long tapered ends and are used for straight at the

and tapered types time of installation

for their

full

length,

of rods are furnished using special armor

straight and the spiral rod wrenches. These

on the conductor by the installation of armor rod clips or clamps been formed. Normally, armor rod clamps are used on transmission and higher, and armor mostly to the possibility Through experience, effective device against

are used

l/O AWG), inclusive. Tapered 79 mm2 (No. 2/O AWG) and

has found and, when

damper will greatly reduce vibration. We use both armor rods and vibration

dampers

the Stockbridge-type properly installed,

for ACSR

conductor

rods are straight rods conductors. Both the

is formed around the conductor types of rods are held in place at each end after the spiral has lines for voltages of 115 kilovolts

rod clips are used for voltages of 69 kilovolts of corona loss off the sharper edges of the the Bureau vibration

armor larger

the

on our transmission

suspension points may be eliminated if sized clamps are used for the be an almost perfect fit, with extremely small tolerance, to provide strand breakage at this stress point.

and clips. vibration latest lines.

lower.

This

choice

is due

damper to be a very models of this type of Armor

conductor. the desired

rods

at conductor

These clamps must protection against

Each

construction

application,

contractor

and

transmission field office

location

in the

to furnish

DATA

the

dampers

vibration

middle

of the

of possible

A damper

the

problem loop

centerline

could

formed

frequencies

effective,

however, absolutely

from

of the

should

be handled

in the is almost

are

be located

conductor

to be furnished

must

be located

in the

middle

suspension

regardless

of size, span

simply.

and

the

midpoint point

A vibration

problem

third

another

of a loop

would becomes

of a loop for

and

are transmitted dampers. The

so the problem

at the

the midpoint could be a node no effect (see fig. 112).

recommendations

conductor

quite

unlimited

283

manufacturer’s

that

or compression dead end. vibrated at the same frequency

the

to be most wind; have

is required

of the‘ vibration

distance

of a strain clamp If all conductors

number

V-ADDITIONAL

line. The data are checked and, if found satisfactory, as the criteria to use for installation of the vibration

at a prescribed

velocity,

CHAPTER

created frequency,

to be effective.

for size,

installed

on the

to the appropriate dampers are installed

clamp

or from

length,

tension,

damper

could

he solved. more

the

and Studies

and

wind

be placed

However,

complex.

in the

mouth

conductor the

the

A damper, by the

damper should

would be made

so that a damper installed at the chosen location will be effective on as many probable frequencies as possible. Numerous laboratory studies have been made by manufacturers of dampers over the years. The new, more sophisticated dampers have been developed through these laboratory studies and should be applied as recommended by the manufacturer. Formulas for computing the frequency and loop length and the basic theory of vibration can be found in most physics books. Two such formulas are:

For frequency :

Metric

U.S. Customary

Hz 51.4534 km/h

Hz 3.26 mi/h

mm

in

f&l d

where f = frequency k= a constant (for air) V= velocity of wind d= outside diameter of conductor For loop length:

where L = loop length f= frequency T= tension in conductor g = acceleration due to gravity W= force of conductor

A standing but

of opposite Reduction

wave,

such

direction of span

as the vibration

loop,

mm Hz N 9.8066 m/s* N/m

is the result

of two

traveling

of motion. length

and

tension

reduces

the

severity

of vibration.

in Hz lb 32.2 ft/s* lb/ft

waves

equal

in magnitude

TRANSMISSION

284

LINE DESIGN

MANUAL

Midpoint of loop f

(Vibration waves are exaggerated vertically for illustmtion) Figure

Galloping

or dancing

112.-Schematic

conductors

by strong gusty winds blowing of eliminating this phenomenon melt

it off

as quickly

value.

Corona

are large-amplitude,

after

it forms

loss on a transmission of conductors when the

occurs

when

waves in a conductor.

low-frequency

vibrations.

Galloping

the

potential

and

before

damage

occurs

(see sec.

line is the result of the ionization electric stress (or voltage gradient)

of a conductor

in air

is raised

conductors

will

result.

There

is always

a power

When and where will corona occur on a given be? What can be done to reduce or eliminate investigators have studied over the years. Three Rockwell Peterson

[ 161, and Peterson formulas have been

to such

used for calculating of obtaining good

the expected data is to take

This

true

of the

Recent available

corona the data

loss for these from the line

extra-high-voltage

lines,

so care

study

based

In fair conductor.

up to a voltage near voltage is an indicator

surface For

the

of a given same

a smooth will

increase

size of the

conductor

diameter,

conductor.

approaches a stranded

Any

corona-and conductors

distortion their

the

spacings

line

cylinder,

is good

to the surface

the higher and

a smooth

conductor

voltage,

also

the

corona.

country. region,

The below

Carroll-Rockwell 3.1 kilowatt

higher voltages. Actually, the being studied after it has been

a published

is small disruptive

that

and per phase

work has been directed toward corona loss in the information for this range should be explored and

and the method of calculation from to that which you propose using. weather, corona The calculated

a value

tufts or streamers the odor of ozone. enough, corrosion

transmission line ? How much power loss will there it? These are some of the questions that many methods of calculation by Peek [15], Carroll and

[ 171 are in general use in this the most accurate in the low-loss

kilometer (5 kilowatt per phase mile). extra-high-voltage range, and the latest

is especially

loss with

14).

process which takes exceeds a certain

dielectric strength of the surrounding air is exceeded. Corona is visible as bluish around the conductor; the visible discharge is accompanied by a hissing sound and In the presence of moisture, nitrous acid is produced and, if the corona is heavy of the

is caused

across irregularly ice-covered conductors. The only known methods are to either prevent the ice from forming on the conductor, or to

as possible

29. Corona.-Corona place on the surface

of vibration

have

the

be exercised

to select line

the disruptive voltage of corona-performance. the

for

of the

must

on transmission

higher

about

considerable

the critical

80 to 85 percent

conductor more

(raised

critical effect

these

best method constructed.

strands,

data

test

very

data

similar

for a particular The closer the disruptive

voltage.

of the

voltage

burrs,

scratches)

of

rough spots become. The

on corona

loss. Fair

weather,

CHAPTER rain,

snow,

hoarfrost,

corona loss. rain produces the

same

loss is observed line

to know

In earlier When

rates

of rainfall

behavior.

Corona

can and

and

dimension

instead

between

text

reduce

overvoltage surge. for

figures

of the

were

illustrations,

from

we have

are

present,

open-circuited

loss to be expected an entire

lines,

transmission

because become corona and

in a

be necessary line.

of energy

more can

affect

it will

loss.

important. system

attenuate

both

corona

reference

chosen

along

studying

The presence of voltage at which

do so, it would

has probably

expected

SI metric

To

when

factor. of the

was avoided-strictly

of corona

the

be considered

of corona

simultaneously

on long

taken

value

switching)

calculating

preferred

peak

corona

aspect

(lightning,

switching

related and

transmission,

must

to determine.

exist

285

than any other single as low as 65 percent

The

could

DATA

temperature

impossible,

influence

voltages

is a procedure

procedure

weather.

that

radio

high

voltage

Following

the

and

loss more at voltages

if not

of high-voltage

years,

abnormally

lightning

fair

difficult,

years

recent

during

is very

all of the

more

pressure,

Rain probably affects corona corona loss on a conductor

transmission

In

atmospheric

V-ADDITIONAL

loss on a transmission line. is reference used centimeters Th

[18].

dimension

of millimeters.

To

to present

the

in centimeters:

procedure

ensure

This as a

compatibility

Nomenclature: Pk =

corona

loss,

kW/km

P,

=

corona

loss,

kW/mi

E

=

average

surface

critical

visual

Eo= =

line

to ground

>

=

line

frequency,

6

=

air

n

=

number

r

=

conductor

=

spacing

g

=

mean

g,,

=

surface

;=

density

at 50 Hz at 60 Hz

voltage corona

(per

phase)

(per

3-phase)

gradient gradient

voltage,

kV

Hz factor

of conductors radius,

in bundle

cm

of conductors

equivalent

phase

between

spacing, average

voltage

m = conductor

in bundle, cm and

gradient

surface

cm

factor

maximum

surface

gradient,

at which

corona

starts,

(assumed

0.88,

average

kV/cm

kV/cm weathered

conductor)

Assume: 345-kV

transmission

483-mm2 457-mm 10.06-m The

basic

line

(954-kcmil) (18~in) (33-ft)

formula

for

at 1829-m

ACSR,45/7

spacing on conductor flat phase spacing reading

the

corona

(6000-ft) conductor

elevation (duplex)

e =

199.2

r

1.48

=

cm

s = 45.72 cm

bundle

D = loss from

kV

the

curves

shown

on figures

1005.84 113

cm and

114

is:

286

TRANSMISSION

LINE DESIGN

MANUAL

pk

g is analogous to E so, y22 g,

= F A. 0go

0

For a duplex conductor, l+$ (

g=

e )

(2r) log, A+For a single conductor,

g=

e r log, f

g=

(1 +~)WW

(2) (1.48) log,

Calculateg, Results

from

for air density the two-thirds

a high-altitude

test

205.65 == 14.45 kV/cm 14.23

(1 ;;;;k8p72, . fromg,

project

= 21.1 m 6%

at Leadville,

Colo.

0.301 1+ fi > ( [19]

8 varies as the one-half power in lieu of the first power power as indicated by Peterson’s investigations [17].

concluded

that

as suggested

the

= 20.72 kV/cm Calculate

g/go

and

read

corresponding

value

for

14.45 g -z-z go 20.72 From

figure

113,

-&

curve

Pk /n

2 r2 at 50 Hz

from

figure

o 7. ’

A :

= 0.04 Pk = 0.04(2)2 (1.48)2 = 0.2368 kW/km at 50 Hz (per phase)

correction

by Peek

113:

[15]

or

CHAPTER

V-ADDITIONAL

287

DATA

‘k n2r2

0.4

0.6

Figure 113.-Corona

As read test.

from

Because

should multiplying

1.0

1.2

0.3

loss curves for (A) fair weather,

figures

113

the corona

be multiplied

by the

in kilowatts

per

for

~7

QS

(B) rainfall,(C)

for

a 60-Hz

factors, three

phases

The

0.3 OA Q5 Q6 Q7 0.9 09 1.0 I.1 1.2 13

and(D)

value

phases,

should

60-Hz

snow. 104-D-1116.

From [18].

for each phase from a 50-Hz the value read from the chart

kilometer

for all three

for

---

per kilometer to frequency,

system.

the figure

1.1

hoarfrost,

Pk is in kilowatts is in direct proportion

if the loss is desired

three

mile

~5

114,

60/50

and

Combining

and

loss factor

by 1.6093,

by three. value

0.9

may the

be changed

answer

be multiplied

should

by 5.79

to

mile

by

be multiplied to obtain

a loss

systems:

Pk = 0.2368 kW/km at 50 Hz (per phase) PC = 5.79 (0.2368) = 1.371 kW/mi at 60 Hz (per 3-phase)

When curve the

rainfall

two,

three,

100 percent. give

is to be considered,

B. Similarly,

the

for hoarfrost or four

Taking

expected

(whichever the assigned

corona

loss for

the

or snow,

corona

is applicable) percentages the

line

loss due

losses are obtained values times ‘in question.

for

to rain

must

from corona

the corresponding

be read

curves

Cor

loss must

from

figure

ZI, respectively. be apportioned

losses and

summing

113 using Then, to make these

will

288

TRANSMISSION

LINE DESIGN

MANUAL

1.0 0.8

0.6

0.1

0.08 0.06

0.01

0.5

0.6

0.7

0.8

0.9

1.0

I.1

Figure 114.-Average values of corona loss under fair weather with different conductor bundles. (1) single conductor (2) two-conductor bundle (3) three-conductor bundle (4) four-conductor bundle (5) average curve. 104-D-1117. From [18].

CHAPTER

V-ADDITIONAL

DATA

289

Example: Assume

that

is fair,

the

line

5 percent

previously

of the

used

time

is located

it rains,

and

such

that

10 percent

85 percent

of the

time

of the

time

it snows-all

the

during

weather a period

of a year.

pk

-

= 0.04 for fair weather (curve A, fig. 113)

n2 Y2

Pk = 0.04(2)2 (1 .4Q2 = 0.2368 kW/km at 50 Hz (per phase) PC = 5.79(0.2368) ‘k

-

= 1.37 1 kW/mi at 60 Hz (per 3-phase)

= 0.90 for rainfall (curve B, fig. 113)

n2 r2

Pk = 0.90(2j2(

= 7.885 kW/km at 50 Hz (per phase)

1 .48)2

PC = 5.79(7.885) = 45.654 kW/mi at 60 Hz (per 3-phase) pk n2 r2

for snow (curve D, fig. 113)

= 0.15

Pk = 0.1 5(2)2 (1 .48)2 = 1.3 14 kW/km at 50 Hz (per phase) PC = 5.79(1.314) = 7.608 kW/mi at 60 Hz (Per 3-phase)

Summation of losses times percentages: (0.85)(1.371)

+ (0.05)(45.654)

+ 0.10 (7.608) = 4.21 kW/mi at 60 Hz (per 3-phase)

This is the average corona loss for the year. Although justified

this

method

for practical

due to weather conductor. decrease Factors

conditions.

New

rapidly

Fair

weather lines

with

various

for

As indicated

transmission

rather for

of calculation

purposes.

corona

loss is only

in the example, corona

tend

there

loss depends

to have

higher

an approximation,

mostly

losses;

on the

however,

changes surface

these

conductors

weather

conditions

conductor

in fog,

mist,

Weathered

conductor

in fair

weather

Corona

loss curves and

computed

by the

for

ACSR

higher

of the

values

will

are:

in rain

Weathered

elevations

in the losses

condition

time.

Range All

it is apparently

are substantial

different conductor

Carroll-Rockwell

and

voltages sizes-from method

snow

are shown which for

fair

0.47

to 0.60

0.54

0.60

to 0.80

0.70

0.80

to 0.95

0.88

on figure may

A verage value

115.

Curves

be determined

the

weather

are shown estimated

at 25 o C (77 o F).

for corona

different loss as

TRANSMISSION

290

a 0

LINE DESIGN MANUAL

CHAPTER

V-ADDITIONAL

DATA

291

TRANSMISSION

292 30.

Stringing

Sag

furnished

for stringing

sag data the

sag and

studies,

tension

data

determination

electrical For

field

results

installation

for

spans

than

sag table ruling

conductor

for

to cover length

a range

increments,

and

from

are not

the

without

high

become

On

free

suspension from

the

tensions,

and

if the stringing

sags

and

are usually

loading

a temperature

form

lengths

in 5-m

elevation,

range

Spans.-When

conductors

spans.

If the

if the

terrain

however,

sag and

loading

is

use and

F in 10’

are

increments,

above

-18

if

conductor

in field

from

tension

conditions

if the

to 50 percent

should

at a lower

the

ruling

to 49 ‘C

in 5’

increments.

Offset Data for Inclined low

below

tables

unloaded

0 to 120’

span.

Stringing

for the preparation

conditions

from

span.

generally

for convenience

range the

but

on initial

loading

in table

sag tables

on the ruling

required

are the

are based

on final

listed

data

to furnish of each

lengths,

basic spans

values

50 percent

strings

span

and

of sag in that sheaves

conductor

hanging

terrain

is not

is quite

conductor

supports

in stringing

sheaves

very

this

steep,

to hang put span

to obtain

dead

intermediate

ends

to the

conductor. the

steep,

the

proper

dead

dead

should

tend

problem sagging

on adjacent to run

downhill

can be handled of the

one

to isolate

be such as to minimize sag and offset dead end (the last structure clipped

the

steep

conductor

see figure

is clipped

a way

that

the amount

the

in, slack

conductors

of slack

in a given

span

the sag while

has been

clipped

dead

ends,

must

be sagged the

For

either

sagged

structure from

in one

conductor

purposes

tension

permanent

ahead, during the

the

the suspension between

the

the

clip-in

comparatively

the conductor

the

the

distance

it is necessary temporary clamp

dead

sections

dead

will

end

operation. level

for

since

Where

in one operation,

where

of

is changed,

in. Calculations

operation.

the

be taken

or temporary,

of calculation,

For

components

to change

conductor

116. must

horizontal

it is necessary

of line being

to maintain

be equal;

the

sagging ends.

be at least

is snubbed, ends,

ends

T2 must

in such

Whenever

between

to permit

the

span

sag after dead

in the section

There

conductor

temporary

great

temporary

be the last structure

where

between

is too

after

upper

correct

of spans

of conductor

establish

the

is also changed-so

the

in a series

between

vertically

into

Tl and

tensions

HI and Hz are equal.

tension

length

these

suspension

These

units,

insulator

are made

shall

level

about

sheaves,

the amount

entire

sags and

based

length

span

The

from

running

the conductor

offsets

on exact line.

metric

concern;

on

line

and

lengths

the individual

For

the

the

directly

including

it is necessary

of span

lengths

into

final

right,

If

based

complex.

lower

is in the

upon

are also based

conductor.

wires,

range

of span

at the same

spans

much

may

ground

based

as to cover

Sag and Insulator

(b)

for

important.

are not

design

he exactly

increments.

span

extremely

wires

calculations will

the entire

is unstressed,

an extent

in lo-ft

structures

the

sag values

to such

span

cover

approximately for

the

strengths,

separately

to be installed

and

complete

calculations

and overhead

which

to a substation

span

Stringing

expanded

line,

used on the rest of the transmission

of a stringing

prestressed.

ground

he disastrous.

spans

values the

of the

are

overhead

of the

sizes and

None

sag data

and

design

of conductors

spans

that

at the

the

are computed

for approach

Tables-Stringing

conductors

for

may

suspension

Dead-ended tension

used

is wrong.

far off-the

prepared

Sag the

of structure

clearances

are too

data

Data.-(u)

LINE DESIGN MANUAL

be clipped

and

The

to end

the point

selection

of line,

calculations. The insulator string ofthe in) must be held in a vertical position

last previous temporary while the next section

The tension in the conductor while after the suspension clamp is clipped

in the stringing to the conductor,

or lower than the tension insulator string clipped

in may

the

line

is brought

swing

to the

towards

proper

or away

of

should of

sag.

from

new

section

sheaves may be higher so the last suspension of line

being

brought

to sag if the

insulator

is

CHAPTER

V-ADDITdONAL

DATA

293

Figure 116.-Conductor tensions running stringing sheaves.

not properly held sagging and clipping to sheave a reference string clamp for

sag by mark

in a vertical position. in of the conductor

is equal

to the

and

sag correction

at any point

in a conductor

length

of the

ordinate

data

is given

of uniform of the

curve

in the

error in the is brought

sag plus correction) of several spans, under the point where each insulator

following

cross section at the

using free

vertical, a serious After the conductor

Clipping in is then started at any structure by placing offset distance and direction from the reference mark.

offset

Procedure The tension

is not held could occur.

checking the corrected sag (stringing chart should be placed on the conductor directly

is supported. at the proper

calculating

If the insulator in the new section

when

given

the center of the suspension An explanation of a method paragraphs.

suspended point

in the form times

the

unit

of a catenary force

conductor. At

support

A on figure

117:

Directrix 2

+”

1

Figure 117.-Dimensions required for calculating stringing operations. 104-D-1 119.

insulator

offset

and sag correction

data during

of the

294

TRANSMISSION

LINE DESIGN MANUAL

T, = WYA in span 1

+Y, - Y,)inspan2

T, =W(YA

T, - T, = W(Y, - Y,)

where : W = force of conductor in newtons per meter (pounds per foot) T, and T, = conductor tensions in newtons (pounds)

= difference in elevation between the directrices of the two cantenaries, which is also the difference in elevation between the low points of sag in the two spans, in meters (feet)

y2 - Yl

A table with the following column headings should be made: Column 1: Station number. This shows the survey station 2: Span length L, in meters (feet) 3: Yz- Yt in meters (feet). This value

Column Column

the low points at the stringing be essentially sheets. Column

4:

of sags in spans adjacent temperature; however, the

same

at any

W(

yZ- Y,),

given

( W)

(col.

shows

the

where

each

difference

structure

in elevation

is located. Yz- Yl between

to each structure. These sags should be the initial sags because the difference between sags in the two spans will Ys- Yr may be measured on the plan-profile temperature, 3),

in newtons

(pounds).

This

value

shows

the

Tz-Tl, on the two sides of the structure. between the conductor tensions, Column 5: Assumed tension Hin newtons (pounds). This value shows an assumed component Hof the tension (called horizontal tension for convenience) in the conductor in the stringing at the stringing

sheaves. For temperature

this assumption, use the initial horizontal as shown on the sag-tension calculation

difference horizontal as it hangs

tension of the conductor form. Assume this tension

to be in a certain span (generally, it is best to use one of the longer spans) and compute the tensions in other spans by adding or subtracting increments from column 4. (pounds). This value shows the difference between Column 6: H,H, (& -col. S), in newtons the horizontal

H

in each Column

tension span with 7: Offset

in the the Kin

conductor

conductor millimeters

at the

ruling

K= lOOOW2 L3 mm/N 12H,,3 This value in tension. slack per Column in slack in this

shows the change The sum of the

column

span

corresponding

is the overall

or K=-

change

and

w2 L3

in slack in a span corresponding values in this column gives the

pound) change in the tension 8: Trial offset, (col. 6) (col. for each

Ho

span

the assumed

horizontal

tension

hanging in the stringing sheaves. per newton (inches per pound).

for the complete 7), in millimeters

to the of slack

unbalanced

Hl13

in/lb

to a one-newton (one-pound) change total change in slack per newton (or section of line (inches). This

tensions.

for the complete

The

section

being value

considered. shows the

algebraic of line,

based

sum of the

change values

on the assumed

CHAPTER tensions. sum the

This

is a positive line.

The

sum

must

he zero

value,

If the

sum

should

sum

of column

be applied

that

to the

Ho-H

9:

Column

10:

Column

11: Modulus

by

Corrected

offset,

of column

(col.

295

has been

he subtracted

sum

he added

to the

7 is the

assumed

in each

from

complete

the

complete

total

span.

section

section

correction

If the

of the

in tension

of line.

which

of line.

column

correction

must must

section

corrected,

DATA

tension

amount

the

complete

correct

of slack

that

8 divided

Column

if the

amount

is negative,

V-ADDITIONAL

6 - (2 col. 7)

(col.

8/X

col.

7), in newtons

9), in millimeters

in millimeters

(pounds).

(inches).

(inches)

1OOOL (col. 9) mm or AE

12L (col. 9) in AE

where : A = area of conductor in square millimeters (square inches) E = modulus of conductor in gigapascals (pounds per square inch) Column

12: Final

correction

in millimeters

(inches).

(Z col. 10 + x col. 11) (col. 7) z col. 7 Columns (col.

9, 10, 11, and

11) is the

in columns corrections

in length

13: Final

amount Column

the amount

of the

11 should

proportional

Column the

change

10 and

12 are used equal

of offset

(col.

required

14: Sum

of offsets,

necessary

to offset

span.

(running

sum

insulator

change

be made

in column

13),

in millimeters

string

from

the vertical.

in millimeters

modulus

The

sum

correction of the

values

in either

of these

12 to offset

these

remainders.

(inches).

This

value

shows

(inches).

This

value

shows

12), in millimeters

of col.

spans. while in sheaves

The

in tension.

is a remainder

11 + col.

in each each

of the offsets in the individual Column 15: Sag correction

must

col.

in the offsets.

with

If there

length

10 +

corrections

conductor zero.

to the span offset,

to make

This

offset

columns,

is the summation

(feet)

(3H, )(col. 10) (3H,,)(col. 10) (2W)(col. 2) mm Or (2W)(col. 2) (1/12) ft

This

column

conductors

shows

are in the

the

amount

stringing

that sheaves

will

be necessary

to obtain

to correct

the

correct

the sag in each

sag after

the

span

while

conductor

the

is clipped

in. The long rough and

offset

as the terrain accurate

and

sag correction

individual where

offset the offset

computer

data

as computed

for one span

is not

in any one span

program

should

in such in excess

in a section

be used

instead

a table

should

of 381 mm of line

of

may this

(15 in).

be sufficiently For

accurate

installations

exceed

381 mm,

simplified

method.

a more It

as

on very detailed is usually

TRANSMISSION

296 unnecessary in one

to consider

operation,

the

the

offset

and

corrections

LINE DESIGN MANUAL

sag correction

calculated

data

are within

if, in a section all three

of line

of the

following

Line conductor mm w (1) Maximum summation of offsets at any structure (2) Maximum difference between summations of offsets at adjacent structures (3) Maximum sag correction in any span

The

same

conductors

procedure should

A sample

as described

he used

problem

for

to calculate

has been

worked

out

in both

kcmil),

ACSR,

sagged

limits:

Overhead ground wire mm (id

(6)

76

(3)

76

(3)

51

(2)

305

(12)

305

(12)

the

data

is being

152

calculating

similar

that

for

offset

the

metric

and

overhead

and

U.S.

sag correction ground

customary

data

for

line

wires. units

to illustrate

this

procedure:

Example Conductor: Full

load

644 mm2 conditions:

Maximum Initial

tension tension

(1272 13-mm

under

at 15.5

(l/2-in)

full ‘C

load

(60

45/7

radial conditions

“F)

(Bittern)

ice with =

is 26 040

a 0.19-kPa

53 378

N (5854

(4-lh/ft2)

wind

at -18

’ C (0 ’ F)

N (12 OOd lb)

lb) widh

a korresponding

sag of 12 485

mm

(40.96 ft). Area

of conductor

A =

Initial

modulus

of conductor

689 mm2

AE = 32 162 542 N H,, = T- Ws = 26 040 -

Initial

E

(1.068 =

(7 230

46.678

Figures shows

in the

118 and 119 the stationing,

procedure

is shown

GPa

(6.77~10~

lb/in2)

=

N

360 lb)

(20.9277)(12.485)

= 5854 - (1.434)(40.96)

120

in2)

25 778

= 5795 lb

show the sag and tension calculations elevations, and span lengths for the in tables

36 and

37.

for the given conductor, and figure sample problem. The table described

CHAPTER

OCm-678

V-ADDITIONAL

DATA

297

(3-78)

:":nT:"L' SAGCALCULATIONS

Ml

2 CONDUCTORst'jmm R/7%7"

//3Sf

.4547

LOADING&

Code Nams

Linear Force Fnclor:

@3

Rated Breaking Strength/q Diameter .A

N

Dead Load Force (W’) i?t&d!

Permanenr set 0.00 Qa. CreepO.OOQ

*s

Tenston Llmlt*tions: Initial.-

%.Ai-K-

Final.!.

;;;“y+f$-

N 25

8g9

N

o.ow

Exp.: jc

A=

Initial

+!ZZt// Final AE

OzL&perOC

Dare ~

Initial */977

T~cp-/UNSTRESSEG LENGTN

LOADING Ice

HEX.

./

N/l? Modulus. (E) Final &.c&&-

Temp. Coeff. of Linear

%-

Computed by

yd,

sqj

Total O.CQd6?

Area (A)amd

-N

Final .- 15.5% .-

,“;

(W”‘)

Resultant:

% -N

Loaded.~oC.5oW

/dn?m

N/m

mm

jb.

GPa

/,,79

44I//9

GPa

009

AE .&

,”

x=d3782

SAG, mfn

1 SAGFACTOR /

3 '; I SPAN LENGTH(S

1

SW,N

1

TENSION, N

NO Ice. No Wind (W’)

Figure

DC-578

118.Sag

and tension

calculation

form for example

problem

on insulator

offset

and sag correction

(metric).

(3.73)

!;,p CONDUCTOR&J

7

Code Nams

Riftcrn

Rated Breaking Diameter Tenston

LOADING k'eavv

14, Load

J!!.?+

Weight Factors: ,?

,

/ofi

Dead Weight

lb + ‘/Lin.

inch

Llmltatlons: o

lb

Final *OF*%-

Compuled by -

LOADING

IbItt 3,gy

0

Wind Resultant:

lb OFa%-

(W’) ./,

Ice (W”)

&lb

Initial ,OF&s. Loaded, Final. *OF-

SAGCALCULATIONS

jb

Area (A) /,& in2 Temo. Coeff. of Linear Exp.:

lb

Total

0.000 0 a

jc

TEMP.. oF UNSTRESSED LFJGTH

i!

per “F

1

Modulus. (E) Final 51.36 Initial mx

f=o.3/

x 103 lb/i&? 105 lb/in2

NE% lzl

k=O.&7 SAGFACTOR 1

Y

0.00~~.

:iIi:

?R/977

% -lb Date

7

Creep o.ooni2f-

78/T

3.‘0073

(W”‘)

Permanent Set O.OOti?

Ib/ft

SAG.fl

A”,’ +z?z?-

1

SW,Ib

:,”

1 TENSION,Ib

SPAN LENGTH(S) -FEET

No Ice. No Wind (W) 120

Figure 119.~Sag customary).

I/,nn3

and tension

922 IO. d/l/3 a & 7 lo. QQ&&@

calculation

form

/ I / 1

for example

I

problem

I

on insulator

i/hi/9. JL49.

offset

/

I

/

I

and sag correction

(U.S.

298

TRANSMISSION

LINE DESIGN MANUAL

365.76 (1200)

457.2

-I

335.28

(1500)

426.72

(1400)

396.24

(1300)

(1100)

g + E 03

\

fcr; .l”.J.

j 7c I ”

fImn\ \trvv,

-. -.38 (1225)

304.8 (IqOO)

*.

381 (1250)

\x L

335.28

(D g

I

(1100)

\ylooo~

A 2 243.84 ZOO)

I

Y-1

213.36 (700)

182.88 (600:

1 All values shown are in meters (feet) 152.4 (500) Figure

120.-Profile

of spans for example

problem

on insulator

offset and sag correction.

104-D-1120.

Table 36.-Data from example problem on insulator offset and sag correction (metric) 1

Station

2

SP k@h L.

3

4

y, - y, . In

m

WY2 - Yt) W(3), N

5

Asrunltd H, N

6

7

Ho-H. H,

- (5).

N

20+72.64 365.760 24+38.40

28642 -50.597

27583 -41.148

-861

-45.110

-944

-1805

.08030

-944

(6) _ z

(8)

(7)(9),

2:0’N

-

13

Fhul mnectlon

lOOOL(9)

[Z(lO) + Z(11#7)

AE’ mm

Z(7)

15

14

FiMl

sum of

ofr33t

ofY3et3

(lo)+ (ll)+ (12). “E). '

mm

.% cofr3ction whikln

SiHwus 3Ho

(10)

2wo’ mm

mm 0

-299

-2445

-255

-28

0

-283

-43.5%

-912

-52.121

-1091

358.140

-13%

-145

-111

-14

0

-1288

-125

-612

.11091

-105

-525

-58

-6

0

-287

-64 -472

25778

365.760 38+93.82

0.104 26

-2864

(6) (7).

Corrected Corrected Moduhu Ho-H offat correction

12

-408 26722

381.000 35+28.06

1000 3Ls 3 128s , mm/N

Trill OffwA

11

-283

373.380 31+47.06

K

10

9

-1059

335.280 27+73.68

8

othet pr Newton

0

,117

84

0

419

49

5

0

238

54 -418

24 866

912

.I0426

95

1331

139

15

0

154

2003

.09788

196

2422

237

27

0

264

702 -264

23775

42+51.%

1223 0

Tot&

0.615

45

-258

l1

-

-1

Numbers in parentheses are column numbers.

Table 37.-Data from example problem on insulator offset and sag correction (U.S. customary) 1

2

3

4

W(Y,-Y,) W(3).

lb

5

A33llmd H, lh

6

7

8

Ho-H.

0ff33t po:d

Ho - (5).

K

lb

W2L3

Trial off3et (6);)~

10

9

cofaected fim

offmt

Ho-H (6)-z,

11 Modulur

12 FillsI

13 FiMl

14 Sumof

whikhl

(7) (9h

3heavel

la

Ho3.

lb

in

3Ho

1200 8o+oo

91+00 103+25

1100 1225 1250

115+75

6439 -lag

2W(2)

ft

-194

-148

-212

-171 1175

-549.6

-10.04

.014 06

-5.725

-311.6

-4.38

-117.6

-2.28

26

-1.09

0 0.0017

-11.13

-0.57

.0013

-4.95

6201

-4%

6007

-212

.01942

-4.113

5795

0

.02064

0

-0.24

.0018

-2.52

55%

205

.01826

5345

450

.01714

94.4

1.95

0.20

.cm19

2.15

3.751

299.4

5.47

0.60

.0017

6.07

7.695

544.4

9.33

1.06

.0016

10.39

-205

1200 127+75

-11.785

0.018

-238

-135

-143

-644

-245

139+50

TOti Numbers in parentheses are column numbers.

-0.107

-78

(10)

-

in

in/lb 68+00

15 SW w ‘II

-10.177

+a05

-0.04

-11.1 -16.1 -18.6 -16.4

-4.2 -2.0 -0.9 0.8 2.3

-10.4 0

4.0

-1 0 12 ,

300

TRANSMISSION

31.

Transmission

Line Equations.-If

and there are no later reroutes; in numerical notation and there start

work

on the

equation

will

Assume crew

the line,

also

that

starts

same result

two

from

4752

2370+66.4

while

second

this point is Sta. 2370+ 66.4Bk belongs to the part of the line that station in the two other

crews

(fig.

121).

is surveyed

from

one end to the other,

he one

of a line ends

meet

or more

after

of a line station

equations

a survey

and which

at a common

toward

is the point

crew

designates

it as Sta.

2374+31.2.

in the

has been

work

on the

line,

the

The

200+364.8 If the station and the length If the

length

ahead line

These

is greater

is shortened

One

length they

will

common

equation

of each

point

as

to identify

= Sta. 2374 + 31.2Ah. The Bk means backand indicates that station behind the common point. Similarly, Ah means ahead and indicates

lengths

may

of the common point. will be 475 200-364.8 be in meters

or feet,

There is a difference of 364.8 = 474 835.2 if there are no depending

= 475 564.8 if there are no other equations (fig. 122). back is greater than the station ahead, then there is an overlap of the line is increased by the amount of overlap (fig. 123).

station of the

An

other.

on the

units

Assume the crew which started at the beginning of the line determines that the meeting Sta. 2374+31:2, and the crew starting at the end of the line says the point is Sta. 2370+66.4. equation will then read Sta. 2374+ 31.2Bk = Sta. 2370+66.4Ah, and the line length will 475

line.

completed.

each

approximate

point, but the values will be different. at the beginning of the line designates

belongs to the section of line ahead designations, and the length of line

equations

will

at an assumed

the two

that common which started the

there

at opposite

other

line

equally spaced stations will increase uniformly in the line. However, if two or more survey crews

of a portion

start

the

+ 00. When

have a station value for Assume that the crew Sta.

and

points,

a reroute

crews

0 +00

a transmission

then all successive, will he no equations

at different

survey

at Sta.

say Sta.

line

LINE DESIGN MANUAL

than by

the

the

station

value

back,

of the

length

then

there of the

of station

is a gap in the gap

(fig.

124).

used. point

is The then be

designations

stationing

and

the

2360+002360+00 -

2370+00 2370+00-

on 2374+31.2Bk

EQUATION on 2370+66.4Ah

-1

:--I

.J

r ---I Station

2370+66.4Bk-

EQUATION St&ion

2374+31.2Ah

1

I 1 I L

---me

I

1--1

2380+00-

2380+00-

TRANSMISSION

302

Figure 123.Station ahead

Figure

124.-Station

LINE DESIGN

designations

designations

when station

when station

MANUAL

back is greater than station

ahead is greater

than station

back.