IEEE TRANSACTIONS ON POWER DELIVERY, VOL. 15, NO. 2, APRIL 2000
777
Sag and Tension Calculations for Overhead
Transmission Lines at High Temperatures—
Modified Ruling Span Method
Mehran Keshavarzian and Charles H. Priebe
Abstract—The ruling span concept is widely used to calculate
sags and tensions for new overhead transmission lines and for upgrading existing lines. It provides satisfactory results for a level line
with relatively uniform spans at any temperature, or for any span
length of a level line at low temperature. The ruling span concept
may result in an unacceptable error if it is used to calculate sags
and tensions in a line segment with significantly unequal spans at
high temperature. This paper presents a method to calculate sags
and tensions of multi-span line segments at different temperatures
based on the rotational stiffness of suspension insulator strings. A
simple equation, based on the parabolic approximation, is derived
to calculate changes in the span lengths and conductor sags and
tensions. The method follows the ruling span concept but relaxes
the Fundamental assumption of the ruling span method. The accuracy 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.
D
S
L
L
T0
H
H
h
w
W
A
K
k
E
NOMENCLATURE
= Sag.
= Span length.
= Length of conductor.
= Length of conductor in a single dead-end span.
= Slack.
= Slack of a single dead-end span.
= Conductor stringing temperature.
= Horizontal conductor tension.
= Horizontal conductor tension in a single dead-end
span.
= Length of suspension insulator.
= Unit weight of conductor.
= Conductor weight plus 1/2 weight of insulator
string.
= Cross sectional area of the conductor.
= Span stiffness.
= Suspension insulator stiffness.
= 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.
1
(1 0
1 01)
i
= Longitudinal horizontal movement of a suspension clamp with a finite stiffness.
= Change in the i-th span length.
i
Subscripts:
= The i-th span of the line section.
0
= At temperature T0 .
t
= At temperature T .
n
= Number of spans.
R
= Ruling span.
i
I. INTRODUCTION
T
HE ruling span formula is based on the fundamental assumption that the attachments of the conductor to suspension structures between dead-end structures are flexible enough
to allow for longitudinal movement to equalize the tensions in
adjacent spans to the ruling span tension [1]. In other words, the
conductor at each suspension structure is supported by an element which is infinitely flexible in the longitudinal direction.
If the temperature of a line segment with unequal spans is
raised uniformly, conductor in each span elongates in response
to the temperature change. This elongation increases the sag,
thereby decreasing the tension. If the suspension insulators
remained stationary (without any rotation), there would be a
tension difference in adjacent spans of different lengths. However, the suspension clamps displace longitudinally to provide
force resolution at each suspension clamp. The longitudinal
horizontal movements of suspension clamps move the insulators from a vertical position and the horizontal component of
the tension in the nonvertical insulator string resists the tension
differential in the adjacent spans. This tension differential
is also called the unbalanced force. The unbalanced force at
the equilibrium position of the insulator string is a function
of the suspension insulator rotational stiffness as well as the
stiffnesses of the adjacent spans of conductor.
The main objective of this paper is to present a simple method
to calculate the required suspension clamp longitudinal movements and conductor sags and tensions which resolve the unbalanced force at each conductor support due to a temperature
change. The distinctive feature of this method is the introduction
of the horizontal force balance at each conductor support. The
method, itself, is based on a simple iterative procedure without
any 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
= 8 3 (D2 0 D2 )=(3 3 S 2 )
= 8 3 (D2 0 D2 ) 3 S 2 =(3 3 S 4 )
= Change [ =S ] 0! 3 (S =S )2
Change [i; t=Si ]T 0!T
A. Ruling Span Method
For level spans, sag and slack in each suspension span at temperature T call be calculated from the following parabolic equations, [2], [3]:
= D = (w 3 S 2 )=(8 3 H )
= D 3 (S =S )2
Sag
i; t
i
= = L 0 S = (8 3 D2 )=(3 3 S )
= (S 3 3 w2)=(24 3 H 2 )
Slack
i; t
i; t
i; t
i
n
n
For most practical situations, the above parabolic equations
are accurate since sags are usually less than 5% of the span
lengths, [4].
The rate of slack at temperature T can be calculated from the
following equations:
= =S = (L 0 S )=S
i; t
i
i; t
i; t
i
Where
L = L [1 + (T 0 T0 ) + (H
= L 3 L =L
i; t
i
i
R; t
R; t
0 H )=E 3 A]
R
(5)
R
S = S + ( 0 01)
i; t
i
i
i
;t
=Si = (L 0 S )=Si
= [Li 3 L =L 0 S 0 ( 0 01 ) ]=Si
= (L 3 L =(S 3 L ) 0 1 0 ( 0 01) =S
i; t
i
R; t
R
R; t
i
=S = (L
R
R; t
i
i
R
i
i
i
;t
i
R
R; t
i; t
R; t
i; t
i
i; t
(6)
0 ( 0 01 )=S
i
i
i
;t
i
i; t
i
R
i
i
i
R
(11)
;t
i; t
R; t
i; t
i; t
R; t
i; t
R; t
i
6( 0 01 ) = Change [ =S ] 0! 3 6
fS [1 0 (S =S )2 ]g
= Change [ =S ] 0! 3 f(S1 + S2 + . . . : + S )
0 (S13 + S23 + . . .. . .. . . : + S 3 )=S 2 g = 0:0
i
i
;t
i
R; t
R T
T
R
R T
T
n
R
or
i
i
;t
n
i
(7)
Change [R; t=SR ]T 0!T
(8)
i
When the conductor temperature increases from T0 to T , sags
and changes in the rates of slack of all spans within the riding
span can be calculated from 1 and 9, respectively. For the line
segment consisting of “n” spans, the movement of the conductor
attachment at the first and last dead-end structure is equal to
zero. In other words, both 0 and n are equal to zero. The movement of the suspension clamps at supports 1 to “n 0 1” (1 to
n01 ) can be evaluated from Eq. 10 or Eq. 12.
It is important to note that the sum of changes in the span
lengths of all spans within the ruling span should be equal to
zero since the total span length (S1 + S2 + . . . : : + Sn ) will not
change with changes in temperature.
i
R
=
T
6( 0 01) = (1 0 0 ) + (2 0 1 )
+ . . . : + ( 0 01) = 0:0
6( 0 01) = ( 0 0 ) = 0:0
Finally the change in the rate of slack and span length due to
a change in temperature from T0 to T can be calculated from
equations 8 to 11:
Change [i; t =Si]T 0!T
R T
i
R
n
R
=S = (L 0 S )=Si = (8 3 D2 )=(3 3 S 2 )
= 8 3 D2 3 S 2 =(3 3 S 4 )
i
R
R; t
The rate of slack can also be calculated from Eq. 3:
i; t
i
i
0 S )=S = (L =S 0 1)
R
i
R; t
i
;t
i
;t
( 0 01 ) = ( 0 ) + (H 0 H ) 3 S =(E 3 A)
= (S 3 3 w2 )f1=(24 3 H 2) 0 1=(24 3 H 2 )g
(12)
+ (H 0 H ) 3 S (E 3 A):
i; t
R; t
(9)
R
An equation for the change in the span length can also be
derived from the expressions for the slacks of the two extreme
conductor support conditions – - - infinitely rigid and infinitely
flexible. The result is Eq. 12:
Substituting for Li; t and Si; t in the rate of slack equation:
i; t
i
i; t
i
R
(Rate of Slack)i; t
i
S = S + ( 0 01 )
(4)
i; t
R; t
Ruling Span
T
(10)
= S = p(S13 + S23 + . . . + S 3 )
=(S1 + S2 + . . . + S )
i
R
(2)
(3)
i
i
R; t
= =S = (8 3 D2 )=(3 3 S 2 )
= (S 2 3 w2 )=(24 3 H 2 )
i; t
i
i
i; t
R; t
Rate of Slack
R T
i
( 0 01) = Change [ =S ] 0! 3 S 3
[1 0 (S =S )2 ]
= 8 3 (D2 0 D2 ) 3 S 3 [1 0 (S =S )2]=(3 3 S 2 )
(1)
R
i
R
R; t
R; t
i
R; t
R; t
i; t
i
;t
n
n
(13)
The following observations can be made from a review of the
ruling span equations 8 through 13:
1) The change in the span length of any span is independent
of the location of the span in the line segment. However,
the longitudinal horizontal movement of each suspension
clamp depends on the location of the suspension insulator
string within the line segment, Eq. 10.
2) The change in the span length of any span due to a temperature 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 the
span length is smaller than the ruling span. The sum of
changes in span lengths of all spans larger than the ruling
span (negative values) should be equal in magnitude to
the sum of changes in span lengths of all spans smaller
than the ruling span (positive values), Eq. 10 and Eq. 13.
3) The maximum increase in the span length due to a temperature increase will occur at span Sc = SR = . This
maximum increase is equal to:
p3
(
Max. i 0 i01
) = 16 3 (D2 0 D2 )=(9p3 3 S )
;t
R; t
(14)
R
R
This span length, Sc , is defined as the critical span. When
a temperature increase occurs, any span shorter than the
ruling span will be subjected to a smaller increase in span
length than the critical span. In addition, for any span
shorter than the critical span, there is a unique span, Si3 ,
that, if substituted for Si , will not change the ruling span
length and will be subjected to the same increase in span
length for a given increase in temperature. The length of
this span, Si3 , can be calculated from Eq. 15:
Si3
= [0S + p(4 3 S2 0 3 3 S2 )]=2
i
R
(15)
i
Fig. 1.
Force balance at the suspension clamp.
or
(S3 + S )2 0 S3 3 S = S2
i
i
where:
SR > Si3
i
i
Hi; t
R
S = S =p3 S
c
R
For standard suspension insulator lengths, the longitudinal
movements of the suspension clamps rotate the insulators from
a vertical position and cause some tension differential in the adjacent spans. The longitudinal horizontal movements of suspension clamps due to temperature changes can be calculated from
the equilibrium condition, Fig. 1:
i; t
i; t
i
i; t
;t
i
;t
i
;t
i
;t
i
= (H 0 H )=( 0 01)
R; t
ki
i; t
= W =h
i
i
i
;t
i
i
i
;t
;t
= (w 3 S2 )=(8 3 H ):
(21)
i; t
i; t
1=K = f(S 3 3 w2)[1=(24 3 H 2) 0 1=(24 3 H 2 )]
+ (H 0 H ) 3 S =(E 3 A)g=(H 0 H )
i; t
i
i; t
i
R; t
i
i
R; t
R; t
1=K = S3 3 w2 3 (H + H )=(24 3 H
+ S =(E 3 A)
i; t
(18)
i
i
)
R; t
(17)
i; t
It should be realized that the effects of temperature and support stiffness on the conductor tension are considered separately.
First, the effects of temperature on the conductor tension are calculated from the tension-temperature relationship based on infinitely rigid insulator supports. Then the differential conductor
tension at each support is redistributed based on the insulator
stiffness and conductor span stiffness at each side of the support. Therefore, the conductor span stiffness, Ki; t , which relates
change in the horizontal conductor tension to change in the horizontal span length should be evaluated at the condition at which
the force balance is required, i.e. at temperature T . In addition,
the conductor span stiffness is independent of the location of
the span within the ruling span section and the stiffness of the
suspension insulators. Therefore, changing the stiffness of suspension insulators should not have an effect on the span stiffnesses and the span stiffnesses can be directly evaluated from
the equilibrium condition of the two extreme conductor support
conditions, 17. Substituting the expression for span change, Eq.
i 0 i01 ;t , from Eq. 12 into Eq. 17 yields the following expressions:
i; t
Ki; t
(20)
Di; t
1 = [(H +1 0 H ) + K 3 1 01 + K +1 3 1 +1 ] (
(16)
=(K + K +1 + k )
;t
= S + (1 0 1 01)
Si; t
B. Modified Ruling Span Method
i
(19)
i; t
i
4) The longitudinal suspension clamp movement is the
largest when all spans larger than the ruling span are
located at one end of the line segment. The suspension
insulator string between the span larger and the span
shorter than the ruling span will be subjected to the
largest longitudinal movement. In general, for lines with
spans larger than the critical span, if the spans are located
in consecutive order by span length from one end to the
other end of the line segment the sags calculated by the
ruling span method will have the maximum errors as
compare with any other arrangement of the spans.
i; t
= H + K 3 (1 0 1 01)
i; t
i; t
i; t
2 3 H2
R; t
)
(22)
780
IEEE TRANSACTIONS ON POWER DELIVERY, VOL. 15, NO. 2, APRIL 2000
1=K = (
i; t
i; t
=H
R; t
) + (
i; t
1=K = (2 3 R; t
R; t
=H
=H
R; t
i; t
) + S =(E 3 A)
i
) + S =(E 3 A)
R
(23)
(24)
Where:
i; t
= (S 3 3 w2)=(24 3 H 2 )
i
R; t
For infinitely flexible suspension supports between dead-end
structures, the unbalance forces are equal to zero, the conductor horizontal tensions (HR; t ) are calculated from the
tension-temperature relationship, and the changes in the span
lengths i 0 i01 ;t can be evaluated from Eq. 10 or Eq. 12.
For infinitely rigid conductor supports, the changes in the span
lengths are equal to zero; the conductor horizontal tensions
(H i; t ) and sags can be evaluated as if the line were a series of
“n” single dead-end spans. The conductor span stiffnesses at
temperature T are evaluated from Eq. 17 or Eq. 23.
The actual longitudinal movements of the suspension clamps
at supports 1 to “n 0 ” ( 1 to n01) can be evaluated from
Eq. 16. The value of i; t depends on the value of i+1; t which
is not known at the time of calculation of i; t . Therefore, an
iterative procedure was adopted to calculate the longitudinal
movements of the suspension clamps. i; t . For the first iteration, 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 converges
very rapidly from the initial values of i; t . It should be mentioned that the sum of changes in the horizontal span lengths,
i 0
i01 ;t of all spans within the line segment must also
be equal to zero.
The proposed method is based on elastic conductor behavior
with a constant modulus of elasticity, E , and a constant linear
thermal elongation, , at all temperature/loading conditions. It
can be generalized to determine conductor sags and tensions
for other conductor loading conditions such as extreme wind,
heavy ice, or combined ice and wind loading. It is a simple
and straightforward numerical method which can easily be programmed, especially by a spreadsheet approach.
Equations 10 and 16 are the focal point of this paper. The
(n 0 ) equilibrium equations in terms of longitudinal movement of the conductor suspension supports, Eq. 16, are solved
by using an iterative procedure rather than the standard numerical matrix method. Their usefulness and accuracy in calculating
conductor sags and tensions will be demonstrated and discussed
in the following two examples.
(
)
1 1
1
1
1
1
1
1
1
1
6(1 1 )
1
III. NUMERICAL EXAMPLES
Example-1: Consider the example line used in the IEEE
Task Force paper [1] which consists of 10 unequal spans of
1590 Kcmil, ACSR Lapwing conductor with a 1000-foot ruling
span. The first and the last structures are dead-end structures
and the nine structures in between are suspension structures
with 60-inch long I-string assemblies. At normal everyday
condition (50 F, no ice, no wind) the horizontal tension in
the conductor in all tell spans is about 8410 lbs and the
suspension I-strings are in a vertical position. The ruling span
sag at this temperature is equal to 26.63 feet. The conductor
coefficient of thermal elongation and modulus of elasticity are
: 2 06 = F and : 2 6 psi, respectively.
When the conductor temperature increases to 212 F, the sag
and tension of the ruling span can be calculated by the tension-temperature relationship. The changes in the relative slacks
and span lengths can be calculated from ruling span equations
8 & 11. Table I summaries these calculations. The suspension
clamps at the two sides of the 1500-foot span must move toward
the 1500-foot span by 2.17 feet and 1.05 feet to equalize conductor tensions to the ruling span tension and to decrease the
1500-foot span length by approximately 3.22 feet. There are
only two spans with lengths larger than the ruling span. The
sum of changes in the span lengths of these two spans due to
temperature change is equal to 03.86 feet which is also equal
in magnitude to the sum of changes in span lengths of all other
eight spans, 3.86 feet. This also indicates that if one rearranges
these 10 unequal span lengths and locates the 2 spans with span
lengths larger than the ruling span at one end of the line, the suspension clamp between the span larger and the span shorter than
the ruling span would be subjected to the largest longitudinal
movement and this movement would be approximately equal to
3.86 feet.
At 212 F conductor temperature, the sags, tensions, and longitudinal horizontal movements of the conductor suspension
clamps can also be calculated from the modified ruling span
force balance Eqs. 16 through 23. Table I summarizes these calculations. The adequacy of the proposed method can be tested
in the following two ways:
Method 1: If the longitudinal horizontal movements of the
conductor suspension clamps ( i; t ) provides the equilibrium force balance at each suspension clamp at temperature 212 , the line can be treated, for the purpose
of calculations, as 10 single dead-end spans with span
lengths equal to the span lengths at 212 F temperature, calculated by equation 20. Using the new span
lengths and knowing the initial conductor tension of
8410 pounds at 50 F and increase in the length of the
conductor due to temperature changes, the sag and tension for each span at 212 F temperature can be calculated from the tension-temperature relationship of the
conductor. These calculated sags and tensions should
be equal to the sags and tensions as calculated by Eq. 19
and Eq. 21.
Method 2: The other method which call be used to verify the
accuracy of Equations 16 through 23, is the finite element method. In this case, the IEEE sample line is
modeled using the finite element computer program
SAGSEC [5].
The results of these two methods as well as the value
of sags reported in the IEEE paper [1] are also summarized in Table. Comparison of sags and tensions calculated by the proposed method, by Method-1 and by
the finite element method (Method-2) are in very good
agreement. The discrepancy in computed sags between
the proposed method and the finite element method are
less than 2% These discrepancies are believed to be
11 8 10 1
9 5 10
+
1
KESHAVARZIAN AND PRIEBE: SAG AND TENSION CALCULATIONS FOR OVERHEAD TRANSMISSION LINES
SUMMARY
OF
781
TABLE I
EXAMPLE-1 CALCULATIONS
due to the parabolic and numerical approximations included in the modified ruling span method equations.
Example-2: Consider an eight span line consisting of a long
1500-foot span and seven equal 821-foot spans of 795 Kcmil,
ACSR Drake conductor with a 1000-foot ruling span. This line
is also studied by Seppa [6]. At 60 F unloaded condition, the
horizontal tension in the conductor in all eight spans is about
5000 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 modulus
of elasticity are 10:4554 2 1006 1= F and 10:5 2 106 psi,
respectively. When the line temperature increases to 212 F,
the calculated sags from the three methods—the proposed
method, Method-1, and Method-2, as well as the values of the
sags reported in the Seppa paper [6], are summarized in Table
II. Again, the discrepancies between the calculated sags from
these methods are less than 2%.
In this example, the line section consists of equal spans with
the exception of the first span, which is almost twice as long
as the other spans. The long span pulls all suspension insulators, creating a large longitudinal horizontal movement of the
suspension clamp adjacent to the 1500-foot span. The actual
horizontal movement of the suspension clamp adjacent to the
1500-foot span is approximately equal to 1.1 feet as compared
to 2.7 feet as calculated from the ruling span method. This large
2.7-foot horizontal movement would create a very large unbalanced force, violating the assumption of equalization of tensions
in all spans. Therefore, the ruling span method is not accurate
and should not be used in a line of this type at high conductor
temperature.
The effects of change in the temperature from 60 F to 250 F
on the span stiffnesses of the 1500-foot span, the 1000 foot
ruling span, and the 821-foot span are shown in Fig. 2. At 60 F
temperature, in order to decrease the span length of the 1500 and
821 foot spans by one foot, the conductor horizontal tensions
should decrease by 346 lbs and 1825 lbs, respectively. However,
at 212 F temperature, the required reduction in the conductor
tensions are only 191 and 806 lbs. This illustrates that changes
in temperature have more dramatic effects in the span stiffnesses
of shorter spans than in those of longer spans.
IV. CONCLUSIONS
A general procedure for the evaluation of conductor sags and
tensions of a level transmission line which includes the force
balance at each suspension clamp is presented. The procedure
782
IEEE TRANSACTIONS ON POWER DELIVERY, VOL. 15, NO. 2, APRIL 2000
TABLE II
SUMMARY OF EXAMPLE-2 CALCULATIONS
APPENDIX
The detailed procedures for obtaining Eqs. 8 through 12 are
discussed in this section. The change in the rate of slack of the
ruling span and i-th span due to a change in temperature from T0
to T can be calculated from Eq. 6 by evaluating at temperature
T and T0 :
Change [R; t =SR ]T 0!T
= [R; t =SR ]@T
0 [
=SR ]@T 0
= ( LR; t 0 SR )=SR 0 (LR 0 SR )=SR = (LR; t 0 LR )=SR :
R; t
(A1)
Change [i; t =Si ]T 0!T
Conductor span stiffness vs. temperature.
i
i
T
0
R; t
i
R
R
i
i
i
R; t
R; t
where:
i
R
i
;t
i
i
i
is developed around the traditional ruling span concept but is applicable for any series of span lengths. The horizontal force balance at each conductor support is achieved by allowing movement of the conductor support. The force imbalance is iteratively eliminated to ensure static equilibrium. The procedure is
simple and it can be easily incorporated into a spreadsheet computer program. Numerical examples are included to demonstrate
agreement between the results of the proposed method and the
more complicated finite element method. A number of qualitative conclusions, based on the derived equations, are also given
in the paper.
i; t
= [(L 3 L
)=(S 3 L ) 0 1 0 (
0 [L =S 0 1]
0 L )=S 0 ( 0 01) =S
= (L 3 S =L 3 S ) 3 (L
= 3 Change [
=S ] 0! 0 ( 0 01 ) =S
(A2)
i
Fig. 2
0 [ =S ]@
0 01 ) =S ]
= [i; t =Si ]@T
T
R
i
i
i
;t
i
;t
i
i
3 S )=(L 3 S )
3 S 4 + 8 3 D 2 3 S 2 )=(3 3 S 4 + 8 3 D2 3 S 2 )
= (Li
= (3
R T
R
R
R
R
i
R
i
R
R
R
The magnitude of i which is evaluated at temperature T0
is very close to one, because the riding span sag at stringing
temperature is usually less than 5% of the ruling span length. In
Example-1 the values of for the 1500-foot and 450-foot spans
are 1.002 and 0.998, respectively. In addition the value of i is
multiplied by the change in the rate of slack of the ruling span,
which is significantly smaller than one (usually on the order of
KESHAVARZIAN AND PRIEBE: SAG AND TENSION CALCULATIONS FOR OVERHEAD TRANSMISSION LINES
0
10 3 ).
Therefore, it is reasonable to substitute i = 1. Eq. 8 is
obtained by replacing i with unity in the above Eq. A2.
The change in the rate of slack of ruling span and i-th span
can also be evaluated in terms of sags from Eq. 7:
Change [R; t =SR ]T 0!T
= [R; t=SR ]@T
0 [
R; t =SR ]@T 0
3 D 2 =(3 3 S 2 )] 0 [8 3 D2 =(3 3 S 2 )]
2 0 D2 )=(3 3 S 2 )
= 8 3 (D
= [8
R; t
R
R; t
R
R
In the above equation, the length of the conductor in a single
dead-end span can be replaced with the span length for simplicity, In addition, Eq. A8 can also be written in terms of slack:
0 01) = (S 3 3 w2 )f1=(24 3 H
+ (H
0 H ) 3 S =(E 3 A)
(i
(A3)
0 [ =S ]@ 0
4
2
2
2
= [8 3 D
3 S =(3 3 S )] 0 [8 3 D 3 S 2 =(3 3 S 4 )]
2 0 D2 ) 3 S 2 =(3 3 S 4 )
= 8 3 (D
=S ] 0! 3 (S =S )2
(A4)
= Change [
= [i; t =Si ]@T
R
i
R; t
R
R; t
i; t
R
i
i
T
i
i
R
R
Substituting the value of change in the rate of slack of the i-th
span from Eq. A4 into Eq. 8 and then solving for (i 0 i01);t
yields the following expressions:
3 (S =S )2
=S ] 0! 0 ( 0 01 ) =S
Change [ =S ] 0! 3 S
Change [R; t =SR ]T 0!T
=
Change [R; t
0 01) =
3 [1 0 (S =S )2]
(i
i
;t
i
R T
i
T
R; t
R
I
i
R T
;t
T
i
i
(A5)
R
Slack of i-th span at temperature T for infinitely rigid and
flexible insulator support can be calculated from the following
two equations:
i; t = Li; t 0 Si
3
= (Si
i; t = Li; t 0 Si; t
3 w2)=(24 3 H
3
= (Si
i
i; t
i; t
2
0 1=(24 3 H 2 )g
R; t
(A9)
i
i; t
2)
3 w2 )=(24 3 H 2
R; t )
0 01 )
i
;t
= ( i; t
0
i; t )
+ (HR; t
0H
i; t )
3 S =(E 3 A)
i
(A10)
T
R
R T
;t
R; t
(i
R; t
i
R
R
Change [i; t =Si ]T 0!T
783
(A6)
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 Overhead
Line Conductors at High Operating Temperatures,” in Report of
the IEEE Task Force “Bare Conductor Sag at High Temperature”,
IEEE/PES Winter Meeting, 1998.
[2] E. S. Thayer, “Computing Tensions in Transmission Lines,” Electrical
World Magazine, pp. 72–73, 1924.
[3] C. O. Boyse and N. G. Simpson, “The Problem of Conductor Sagging
on Overhead Transmission Lines,” Journal AIEE, pt. II, vol. 91, pp.
219–231, 1944.
[4] One Southwire Drive30 119 Overhead Conductor Manual. Carrollton,
Georgia: Southwire Company, 1994.
[5] SAGSEC, , “Computer Program for Sags and Tensions in Multi-Span
Systems, Power Line Systems,”, Madison, WI, 1997.
[6] T. O. Seppa. Sags and Tension Equalization at High Temperatures. presented at 1996 Summer Power Meeting, Symposium on Thermal Rating
(A7)
where:
Li; t = Li; t [1 + (HR; t 0 H i; t)=(E 3 A)]
Si; t = Si + (i 0 i01 );t
Substituting the values of Li; t and Si; t into Eq. A7 and then
solving for (i 0 i01 );t :
Li; t [1 + (HR; t 0 H i; t )=(E 3 A)] 0 Si 0 (i 0 i01 );t
3
2
2
= (Si 3 w )=(24 3 HR; t )
(L i; t 0 Si ) + Li; t 3 (HR; t 0 H i; t)=(E 3 A) 0 (i 0 i01 );t
3
2
2
= (Si 3 w )=(24 3 HR; t )
2
3
2
2
(i 0 i01 );t = (Si 3 w )f1=(24 3 H i; t ) 0 1=(24 3 HR; t)g
(A8)
+ (HR; t 0 H i; t ) 3 Li; t =(E 3 A)
Mehran Keshavarzian, Member ASCE, received the B.Sc. degree in Civil Engineering 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 and
1984, respectively. He is a registered Professional Engineer (PE) in the state
of California since 1986 and a registered Structural Engineer (SE) in the state
of Illinois and California since 1990. He is currently senior transmission structural engineer at the Commonwealth Edison Company (ComEd) in Chicago.
Mehran.Keshavarzian@ucm.com.
Charles H. Priebe, Member ASCE, received his B.Sc. degree in Civil Engineering from the University of Illinois at Urbana-Champaign in 1973. He
has been a registered Professional Engineer (PE) in the state of Illinois since
1979. He is currently the Technical Lead Engineer in the Transmission Line
Engineering Department at the Commonwealth Edison Company (ComEd) in
Chicago. His professional interests are in the application of computers and new
technologies to the physical design and analysis of overhead transmission lines.
Charles.H.Priebe@ucm.com.