keshavarzian_2000

IEEE TRANSACTIONS ON POWER DELIVERY, VOL. 15, NO. 2, APRIL 2000 777

Sag and Tension Calculations for OverheadTransmission Lines at High Temperatures—

Modified Ruling Span MethodMehran Keshavarzian and Charles H. Priebe

Abstract—The ruling span concept is widely used to calculatesags and tensions for new overhead transmission lines and for up-grading existing lines. It provides satisfactory results for a level linewith relatively uniform spans at any temperature, or for any spanlength of a level line at low temperature. The ruling span conceptmay result in an unacceptable error if it is used to calculate sagsand tensions in a line segment with significantly unequal spans athigh temperature. This paper presents a method to calculate sagsand tensions of multi-span line segments at different temperaturesbased on the rotational stiffness of suspension insulator strings. Asimple equation, based on the parabolic approximation, is derivedto calculate changes in the span lengths and conductor sags andtensions. The method follows the ruling span concept but relaxesthe Fundamental assumption of the ruling span method. The accu-racy of the method is compared with the more complex, nonlinear,finite element method.

Index Terms—overhead line, ruling span, high temperature, sag,tension, insulator swing, conductor.

NOMENCLATURE

D = Sag.S = Span length.L = Length of conductor.L = Length of conductor in a single dead-end span.� = Slack.� = Slack of a single dead-end span.T0 = Conductor stringing temperature.H = Horizontal conductor tension.H = Horizontal conductor tension in a single dead-end

span.h = Length of suspension insulator.w = Unit weight of conductor.W = Conductor weight plus 1/2 weight of insulator

string.A = Cross sectional area of the conductor.K = Span stiffness.k = Suspension insulator stiffness.E = Modulus of Elasticity of the conductor.� = Coefficient of thermal elongation of the conductor.� = Longitudinal horizontal movement of an infinitely

flexible suspension clamp.

Manuscript received January 26, 1999; revised September 30, 1999.The authors are with the Commonwealth Edison Company, P.O. Box 767,

Chicago, IL 60690 USA.Publisher Item Identifier S 0885-8977(00)03506-8.

� = Longitudinal horizontal movement of a suspen-sion clamp with a finite stiffness.

(�i�

�i�1)= Change in thei-th span length.

Subscripts:i = Thei-th span of the line section.0 = At temperatureT0.t = At temperatureT .n = Number of spans.R = Ruling span.

I. INTRODUCTION

T HE ruling span formula is based on the fundamental as-sumption that the attachments of the conductor to suspen-

sion structures between dead-end structures are flexible enoughto allow for longitudinal movement to equalize the tensions inadjacent spans to the ruling span tension [1]. In other words, theconductor at each suspension structure is supported by an ele-ment which is infinitely flexible in the longitudinal direction.

If the temperature of a line segment with unequal spans israised uniformly, conductor in each span elongates in responseto the temperature change. This elongation increases the sag,thereby decreasing the tension. If the suspension insulatorsremained stationary (without any rotation), there would be atension difference in adjacent spans of different lengths. How-ever, the suspension clamps displace longitudinally to provideforce resolution at each suspension clamp. The longitudinalhorizontal movements of suspension clamps move the insula-tors from a vertical position and the horizontal component ofthe tension in the nonvertical insulator string resists the tensiondifferential in the adjacent spans. This tension differentialis also called the unbalanced force. The unbalanced force atthe equilibrium position of the insulator string is a functionof the suspension insulator rotational stiffness as well as thestiffnesses of the adjacent spans of conductor.

The main objective of this paper is to present a simple methodto calculate the required suspension clamp longitudinal move-ments and conductor sags and tensions which resolve the un-balanced force at each conductor support due to a temperaturechange. The distinctive feature of this method is the introductionof the horizontal force balance at each conductor support. Themethod, itself, is based on a simple iterative procedure withoutany finite element nonlinear analysis.

0885–8977/00$10.00 © 2000 IEEE

778 IEEE TRANSACTIONS ON POWER DELIVERY, VOL. 15, NO. 2, APRIL 2000

II. FORMULAS

A. Ruling Span Method

For level spans, sag and slack in each suspension span at tem-peratureT call be calculated from the following parabolic equa-tions, [2], [3]:

Sag = Di; t = (w � S2

i )=(8 �HR; t)

= DR; t � (Si=SR)2 (1)

Slack = �i; t = Li; t � Si; t = (8 �D2

i; t)=(3 � Si)= (S3

i �w2)=(24 �H2

R; t) (2)

Rate of Slack= �i; t=Si = (8 �D2

i; t)=(3 � S2i )= (S2i �w2)=(24 �H2

R; t) (3)

Ruling Span= SR =p(S3

1+ S3

2+ . . . + S3n)

=(S1 + S2 + . . . + Sn) (4)

For most practical situations, the above parabolic equationsare accurate since sags are usually less than 5% of the spanlengths, [4].

The rate of slack at temperatureT can be calculated from thefollowing equations:

(Rate of Slack)i; t = �i; t=Si = (Li; t � Si; t)=Si

Where

Li; t = Li[1 + �(T � T0) + (HR; t �HR)=E �A]= Li � LR; t=LR (5)

Si; t = Si + (�i � �i�1);t

Substituting forLi; t andSi; t in the rate of slack equation:

�i; t=Si = (Li; t � Si; t)=Si

= [Li � LR; t=LR � Si � (�i � �i�1);t]=Si

= (Li � LR; t=(Si � LR) � 1� (�i � �i�1);t=Si (6)

�R; t=SR = (LR; t � SR)=SR = (LR; t=SR � 1)

The rate of slack can also be calculated from Eq. 3:

�i; t=Si = (Li; t � Si; t)=Si = (8 �D2

i; t)=(3 � S2i )= 8 �D2

R; t � S2i =(3 � S4

R) (7)

Finally the change in the rate of slack and span length due toa change in temperature fromT0 to T can be calculated fromequations 8 to 11:

Change[�i; t=Si]T0!T = Change[�R; t=SR]T0!T

� (�i � �i�1)=Si (8)

Change[�i; t=Si]T0!T = 8 � (D2

i; t �D2

i )=(3 � S2i )= 8 � (D2

R; t �D2

R) � S2i =(3 � S4R)= Change[�R; t=SR]T0!T � (Si=SR)2 (9)

(�i � �i�1);t = Change[�R; t=SR]T0!T � Si�[1� (Si=SR)

2]

= 8 � (D2

R; t �D2

R) � Si � [1� (Si=SR)2]=(3 � S2R)

(10)

Si; t = Si + (�i � �i�1);t (11)

An equation for the change in the span length can also bederived from the expressions for the slacks of the two extremeconductor support conditions – - - infinitely rigid and infinitelyflexible. The result is Eq. 12:

(�i � �i�1);t = (�i; t � �i; t) + (HR; t �Hi; t) � Si=(E �A)= (S3i �w2)f1=(24 �Hi; t

2) � 1=(24 �H2

R; t)g+ (HR; t �Hi; t) � Si(E �A): (12)

When the conductor temperature increases fromT0 toT , sagsand changes in the rates of slack of all spans within the ridingspan can be calculated from 1 and 9, respectively. For the linesegment consisting of “n” spans, the movement of the conductorattachment at the first and last dead-end structure is equal tozero. In other words, both�0 and�n are equal to zero. The move-ment of the suspension clamps at supports 1 to “n � 1” (�1 to�n�1) can be evaluated from Eq. 10 or Eq. 12.

It is important to note that the sum of changes in the spanlengths of all spans within the ruling span should be equal tozero since the total span length (S1+ S2+ . . . : :+ Sn) will notchange with changes in temperature.

�(�i � �i�1);t = Change[�R; t=SR]T0!T ��fSi[1� (Si=SR)

2]g= Change[�R; t=SR]T0!T � f(S1 + S2 + . . . :+ Sn)

� (S31+ S3

2+ . . . . . . . . . :+ S3n)=S

2

Rg = 0:0

or

�(�i � �i�1);t = (�1 � �0) + (�2 � �1)

+ . . . :+ (�n � �n�1) = 0:0

�(�i � �i�1);t = (�n � �0) = 0:0 (13)

The following observations can be made from a review of theruling span equations 8 through 13:

1) The change in the span length of any span is independentof the location of the span in the line segment. However,the longitudinal horizontal movement of each suspensionclamp depends on the location of the suspension insulatorstring within the line segment, Eq. 10.

2) The change in the span length of any span due to a tem-perature increase will be negative if the span length is

KESHAVARZIAN AND PRIEBE: SAG AND TENSION CALCULATIONS FOR OVERHEAD TRANSMISSION LINES 779

larger than the ruling span and it will be positive if thespan length is smaller than the ruling span. The sum ofchanges in span lengths of all spans larger than the rulingspan (negative values) should be equal in magnitude tothe sum of changes in span lengths of all spans smallerthan the ruling span (positive values), Eq. 10 and Eq. 13.

3) The maximum increase in the span length due to a tem-perature increase will occur at spanSc = SR=

p3. This

maximum increase is equal to:

Max. (�i � �i�1);t = 16 � (D2R; t �D2

R)=(9p3 � SR) (14)

This span length,Sc, is defined as the critical span. Whena temperature increase occurs, any span shorter than theruling span will be subjected to a smaller increase in spanlength than the critical span. In addition, for any spanshorter than the critical span, there is a unique span,S�i ,that, if substituted forSi, will not change the ruling spanlength and will be subjected to the same increase in spanlength for a given increase in temperature. The length ofthis span,S�i , can be calculated from Eq. 15:

S�i = [�Si +p(4 � S2R � 3 � S2i )]=2 (15)

or

(S�i + Si)2 � S�i � Si = S2R

where:

SR > S�i � Sc = SR=p3 � Si

4) The longitudinal suspension clamp movement is thelargest when all spans larger than the ruling span arelocated at one end of the line segment. The suspensioninsulator string between the span larger and the spanshorter than the ruling span will be subjected to thelargest longitudinal movement. In general, for lines withspans larger than the critical span, if the spans are locatedin consecutive order by span length from one end to theother end of the line segment the sags calculated by theruling span method will have the maximum errors ascompare with any other arrangement of the spans.

B. Modified Ruling Span Method

For standard suspension insulator lengths, the longitudinalmovements of the suspension clamps rotate the insulators froma vertical position and cause some tension differential in the ad-jacent spans. The longitudinal horizontal movements of suspen-sion clamps due to temperature changes can be calculated fromthe equilibrium condition, Fig. 1:

�i; t = [(Hi+1; t �Hi; t) +Ki; t ��i�1; t +Ki+1; t ��i+1; t]

=(Ki; t +Ki+1; t + ki) (16)

Ki; t = (HR; t �Hi; t)=(�i � �i�1);t (17)

ki =Wi=h (18)

Fig. 1. Force balance at the suspension clamp.

Hi; t = Hi; t +Ki; t � (�i ��i�1);t (19)

Si; t = Si + (�i ��i�1);t (20)

Di; t = (w � S2i; t)=(8 �Hi; t): (21)

It should be realized that the effects of temperature and sup-port stiffness on the conductor tension are considered separately.First, the effects of temperature on the conductor tension are cal-culated from the tension-temperature relationship based on in-finitely rigid insulator supports. Then the differential conductortension at each support is redistributed based on the insulatorstiffness and conductor span stiffness at each side of the sup-port. Therefore, the conductor span stiffness,Ki; t, which relateschange in the horizontal conductor tension to change in the hori-zontal span length should be evaluated at the condition at whichthe force balance is required, i.e. at temperatureT . In addition,the conductor span stiffness is independent of the location ofthe span within the ruling span section and the stiffness of thesuspension insulators. Therefore, changing the stiffness of sus-pension insulators should not have an effect on the span stiff-nesses and the span stiffnesses can be directly evaluated fromthe equilibrium condition of the two extreme conductor supportconditions, 17. Substituting the expression for span change, Eq.(�i � �i�1);t, from Eq. 12 into Eq. 17 yields the following ex-pressions:

1=Ki; t = f(S3i �w2)[1=(24 �Hi; t2)� 1=(24 �H2

R; t)]

+ (HR; t �Hi; t) � Si=(E �A)g=(HR;t �Hi; t)

1=Ki; t =S3i �w2 � (HR; t +Hi; t)=(24 �Hi; t

2 �H2R; t)

+ Si=(E �A) (22)


1=Ki; t = (�i; t=HR; t) + (�i; t=Hi; t) + Si=(E �A) (23)

1=KR; t = (2 � �R; t=HR; t) + SR=(E �A) (24)

Where:

�i; t = (S3

i �w2)=(24 �H2

R; t)

For infinitely flexible suspension supports between dead-endstructures, the unbalance forces are equal to zero, the con-ductor horizontal tensions (HR; t) are calculated from thetension-temperature relationship, and the changes in the spanlengths(�i � �i�1);t can be evaluated from Eq. 10 or Eq. 12.For infinitely rigid conductor supports, the changes in the spanlengths are equal to zero; the conductor horizontal tensions(Hi; t) and sags can be evaluated as if the line were a series of“n” single dead-end spans. The conductor span stiffnesses attemperatureT are evaluated from Eq. 17 or Eq. 23.

The actual longitudinal movements of the suspension clampsat supports 1 to “n � 1” (�1 to �n�1) can be evaluated fromEq. 16. The value of�i; t depends on the value of�i+1; t whichis not known at the time of calculation of�i; t. Therefore, aniterative procedure was adopted to calculate the longitudinalmovements of the suspension clamps.�i; t. For the first iter-ation, the initial values of�i; t are assumed to be equal to�i; t.The imbalance forces resulting from initial approximations of�i; t are iteratively eliminated. The iterative process convergesvery rapidly from the initial values of�i; t. It should be men-tioned that the sum of changes in the horizontal span lengths,�(�i ��i�1);t of all spans within the line segment must alsobe equal to zero.

The proposed method is based on elastic conductor behaviorwith a constant modulus of elasticity,E, and a constant linearthermal elongation,�, at all temperature/loading conditions. Itcan be generalized to determine conductor sags and tensionsfor other conductor loading conditions such as extreme wind,heavy ice, or combined ice and wind loading. It is a simpleand straightforward numerical method which can easily be pro-grammed, especially by a spreadsheet approach.

Equations 10 and 16 are the focal point of this paper. The(n � 1) equilibrium equations in terms of longitudinal move-ment of the conductor suspension supports, Eq. 16, are solvedby using an iterative procedure rather than the standard numer-ical matrix method. Their usefulness and accuracy in calculatingconductor sags and tensions will be demonstrated and discussedin the following two examples.

III. N UMERICAL EXAMPLES

Example-1: Consider the example line used in the IEEETask Force paper [1] which consists of 10 unequal spans of1590 Kcmil, ACSR Lapwing conductor with a 1000-foot rulingspan. The first and the last structures are dead-end structuresand the nine structures in between are suspension structureswith 60-inch long I-string assemblies. At normal everydaycondition (50�F, no ice, no wind) the horizontal tension inthe conductor in all tell spans is about 8410 lbs and the

suspension I-strings are in a vertical position. The ruling spansag at this temperature is equal to 26.63 feet. The conductorcoefficient of thermal elongation and modulus of elasticity are11:8� 10�61=�F and9:5� 106 psi, respectively.

When the conductor temperature increases to 212�F, the sagand tension of the ruling span can be calculated by the ten-sion-temperature relationship. The changes in the relative slacksand span lengths can be calculated from ruling span equations8 & 11. Table I summaries these calculations. The suspensionclamps at the two sides of the 1500-foot span must move towardthe 1500-foot span by 2.17 feet and 1.05 feet to equalize con-ductor tensions to the ruling span tension and to decrease the1500-foot span length by approximately 3.22 feet. There areonly two spans with lengths larger than the ruling span. Thesum of changes in the span lengths of these two spans due totemperature change is equal to�3.86 feet which is also equalin magnitude to the sum of changes in span lengths of all othereight spans,+3.86 feet. This also indicates that if one rearrangesthese 10 unequal span lengths and locates the 2 spans with spanlengths larger than the ruling span at one end of the line, the sus-pension clamp between the span larger and the span shorter thanthe ruling span would be subjected to the largest longitudinalmovement and this movement would be approximately equal to3.86 feet.

At 212�F conductor temperature, the sags, tensions, and lon-gitudinal horizontal movements of the conductor suspensionclamps can also be calculated from the modified ruling spanforce balance Eqs. 16 through 23. Table I summarizes these cal-culations. The adequacy of the proposed method can be testedin the following two ways:

Method 1: If the longitudinal horizontal movements of theconductor suspension clamps (�i; t) provides the equi-librium force balance at each suspension clamp at tem-perature 212�, the line can be treated, for the purposeof calculations, as 10 single dead-end spans with spanlengths equal to the span lengths at 212�F tempera-ture, calculated by equation 20. Using the new spanlengths and knowing the initial conductor tension of8410 pounds at 50�F and increase in the length of theconductor due to temperature changes, the sag and ten-sion for each span at 212�F temperature can be calcu-lated from the tension-temperature relationship of theconductor. These calculated sags and tensions shouldbe equal to the sags and tensions as calculated by Eq. 19and Eq. 21.

Method 2: The other method which call be used to verify theaccuracy of Equations 16 through 23, is the finite el-ement method. In this case, the IEEE sample line ismodeled using the finite element computer programSAGSEC [5].

The results of these two methods as well as the valueof sags reported in the IEEE paper [1] are also summa-rized in Table. Comparison of sags and tensions cal-culated by the proposed method, by Method-1 and bythe finite element method (Method-2) are in very goodagreement. The discrepancy in computed sags betweenthe proposed method and the finite element method areless than 2% These discrepancies are believed to be


TABLE ISUMMARY OF EXAMPLE-1 CALCULATIONS

due to the parabolic and numerical approximations in-cluded in the modified ruling span method equations.

Example-2: Consider an eight span line consisting of a long1500-foot span and seven equal 821-foot spans of 795 Kcmil,ACSR Drake conductor with a 1000-foot ruling span. This lineis also studied by Seppa [6]. At 60�F unloaded condition, thehorizontal tension in the conductor in all eight spans is about5000 lbs and the suspension I-strings are in a vertical position.The ruling span sag at this temperature is equal to 27.33 feet.The conductor coefficient of thermal elongation and modulusof elasticity are10:4554 � 10

�61=�F and 10:5 � 10

6 psi,respectively. When the line temperature increases to 212�F,the calculated sags from the three methods—the proposedmethod, Method-1, and Method-2, as well as the values of thesags reported in the Seppa paper [6], are summarized in TableII. Again, the discrepancies between the calculated sags fromthese methods are less than 2%.

In this example, the line section consists of equal spans withthe exception of the first span, which is almost twice as longas the other spans. The long span pulls all suspension insula-tors, creating a large longitudinal horizontal movement of thesuspension clamp adjacent to the 1500-foot span. The actualhorizontal movement of the suspension clamp adjacent to the

1500-foot span is approximately equal to 1.1 feet as comparedto 2.7 feet as calculated from the ruling span method. This large2.7-foot horizontal movement would create a very large unbal-anced force, violating the assumption of equalization of tensionsin all spans. Therefore, the ruling span method is not accurateand should not be used in a line of this type at high conductortemperature.

The effects of change in the temperature from 60�F to 250�Fon the span stiffnesses of the 1500-foot span, the 1000 footruling span, and the 821-foot span are shown in Fig. 2. At 60�Ftemperature, in order to decrease the span length of the 1500 and821 foot spans by one foot, the conductor horizontal tensionsshould decrease by 346 lbs and 1825 lbs, respectively. However,at 212�F temperature, the required reduction in the conductortensions are only 191 and 806 lbs. This illustrates that changesin temperature have more dramatic effects in the span stiffnessesof shorter spans than in those of longer spans.

IV. CONCLUSIONS

A general procedure for the evaluation of conductor sags andtensions of a level transmission line which includes the forcebalance at each suspension clamp is presented. The procedure


TABLE IISUMMARY OF EXAMPLE-2 CALCULATIONS

Fig. 2 Conductor span stiffness vs. temperature.

is developed around the traditional ruling span concept but is ap-plicable for any series of span lengths. The horizontal force bal-ance at each conductor support is achieved by allowing move-ment of the conductor support. The force imbalance is itera-tively eliminated to ensure static equilibrium. The procedure issimple and it can be easily incorporated into a spreadsheet com-puter program. Numerical examples are included to demonstrateagreement between the results of the proposed method and themore complicated finite element method. A number of qualita-tive conclusions, based on the derived equations, are also givenin the paper.

APPENDIX

The detailed procedures for obtaining Eqs. 8 through 12 arediscussed in this section. The change in the rate of slack of theruling span andi-th span due to a change in temperature fromT0to T can be calculated from Eq. 6 by evaluating at temperatureT andT0:

Change[�R; t=SR]T0!T = [�R; t=SR]@T � [�R; t=SR]@T0

= (LR; t � SR)=SR � (LR � SR)=SR = (LR; t � LR)=SR:

(A1)

Change[�i; t=Si]T0!T = [�i; t=Si]@T � [�i; t=Si]@T0

= [(Li � LR; t)=(Si � LR)� 1� (�i � �i�1);t=Si]

� [Li=Si � 1]

= (Li � SR=LR � Si) � (LR; t � LR)=SR � (�i � �i�1);t=Si

= �i � Change[�R; t=SR]T0!T � (�i � �i�1);t=Si (A2)

where:

�i =(Li � SR)=(LR � Si)

= (3 � S4R + 8 �D2

R � S2

i )=(3 � S4

R + 8 �D2

R � S2

R)

The magnitude of�i which is evaluated at temperatureT0is very close to one, because the riding span sag at stringingtemperature is usually less than 5% of the ruling span length. InExample-1 the values of� for the 1500-foot and 450-foot spansare 1.002 and 0.998, respectively. In addition the value of�i ismultiplied by the change in the rate of slack of the ruling span,which is significantly smaller than one (usually on the order of


10�3). Therefore, it is reasonable to substitute�i = 1. Eq. 8 isobtained by replacing�i with unity in the above Eq. A2.

The change in the rate of slack of ruling span andi-th spancan also be evaluated in terms of sags from Eq. 7:

Change[�R; t=SR]T0!T = [�R; t=SR]@T � [�R; t=SR]@T0

= [8 �D2

R; t=(3 � S2

R)]� [8 �D2

R=(3 � S2

R)]

= 8 � (D2

R; t �D2

R)=(3 � S2

R) (A3)

Change[�i; t=Si]T0!T = [�i; t=Si]@T � [�i; t=Si]@T0

= [8 �D2

R; t � S2

i =(3 � S4

R)]� [8 �D2

R � S2i =(3 � S4

R)]

= 8 � (D2

R; t �D2

R) � S2

i =(3 � S4

R)

= Change[�R; t=SR]T0!T � (Si=SR)2 (A4)

Substituting the value of change in the rate of slack of thei-thspan from Eq. A4 into Eq. 8 and then solving for(�i � �i�1);tyields the following expressions:

Change[�R; t=SR]T0!T � (Si=SR)2

= Change[�R; t=SR]T0!T � (�I � �i�1);t=Si

(�i � �i�1);t = Change[�R; t=SR]T0!T � Si

� [1� (Si=SR)2] (A5)

Slack of i-th span at temperatureT for infinitely rigid andflexible insulator support can be calculated from the followingtwo equations:

�i; t = Li; t � Si = (S3i �w2)=(24 �Hi; t

2) (A6)

�i; t = Li; t � Si; t = (S3i �w2)=(24 �H2

R; t) (A7)

where:

Li; t = Li; t[1 + (HR; t �Hi; t)=(E �A)]

Si; t = Si + (�i � �i�1);t

Substituting the values ofLi; t andSi; t into Eq. A7 and thensolving for(�i � �i�1);t:

Li; t[1 + (HR; t �Hi; t)=(E �A)]� Si � (�i � �i�1);t

= (S3i �w2)=(24 �H2

R; t)

(Li; t � Si) + Li; t � (HR; t �Hi; t)=(E �A)� (�i � �i�1);t

= (S3i �w2)=(24 �H2

R; t)

(�i � �i�1);t = (S3i �w2)f1=(24 �Hi; t

2) � 1=(24 �H2

R; t)g

+ (HR; t �Hi; t) � Li; t=(E �A) (A8)

In the above equation, the length of the conductor in a singledead-end span can be replaced with the span length for sim-plicity, In addition, Eq. A8 can also be written in terms of slack:

(�i � �i�1);t = (S3i �w2)f1=(24 �Hi; t

2 � 1=(24 �H2

R; t)g

+ (HR; t �Hi; t) � Si=(E �A) (A9)

(�i � �i�1);t = (�i; t � �i; t) + (HR; t �Hi; t) � Si=(E �A)(A10)

ACKNOWLEDGMENT

The authors would like to thank their colleague at ComEd,Mr. Parvez Rashid, Member IEEE, for reviewing this paper.

REFERENCES

[1] Y. Motlis et al., “Limitations of the Ruling Span Method for OverheadLine Conductors at High Operating Temperatures,” inReport ofthe IEEE Task Force “Bare Conductor Sag at High Temperature”,IEEE/PES Winter Meeting, 1998.

[2] E. S. Thayer, “Computing Tensions in Transmission Lines,”ElectricalWorld Magazine, pp. 72–73, 1924.

[3] C. O. Boyse and N. G. Simpson, “The Problem of Conductor Saggingon Overhead Transmission Lines,”Journal AIEE, pt. II, vol. 91, pp.219–231, 1944.

[4] One Southwire Drive30 119Overhead Conductor Manual. Carrollton,Georgia: Southwire Company, 1994.

[5] SAGSEC, , “Computer Program for Sags and Tensions in Multi-SpanSystems, Power Line Systems,”, Madison, WI, 1997.

[6] T. O. Seppa. Sags and Tension Equalization at High Temperatures. pre-sented at 1996 Summer Power Meeting, Symposium on Thermal Rating

Mehran Keshavarzian, Member ASCE, received the B.Sc. degree in Civil En-gineering from the University of Tehran, Iran. He obtained his M.Sc. and Ph.D.in structures from the University of Illinois at Urbana-Champaign in 1981 and1984, respectively. He is a registered Professional Engineer (PE) in the stateof California since 1986 and a registered Structural Engineer (SE) in the stateof Illinois and California since 1990. He is currently senior transmission struc-tural engineer at the Commonwealth Edison Company (ComEd) in [email protected].

Charles H. Priebe, Member ASCE, received his B.Sc. degree in Civil En-gineering from the University of Illinois at Urbana-Champaign in 1973. Hehas been a registered Professional Engineer (PE) in the state of Illinois since1979. He is currently the Technical Lead Engineer in the Transmission LineEngineering Department at the Commonwealth Edison Company (ComEd) inChicago. His professional interests are in the application of computers and newtechnologies to the physical design and analysis of overhead transmission [email protected].

keshavarzian_2000

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