steam demand. One of the generators is controlled
automatically and the other manually. The output for the day was 5,255,320 pounds of steam
from and at 212 degrees Fahrenheit, and the input
1,675,200 kwhr; this corresponds to a conversion
efficiency of 89.3 per cent. The chart reproduced
in figure 9 shows the steam demand of the mill on
the same day.
In tables I, II, and III are listed the electric steam
generators of various types installed in Canada to
Transmission Line
ary
aate|;n
enar
a cu
ions
Calculations
a
A simple and accurate methods is describedd
For
fo
roverh
for
compiling stringing charts for
overhead
transmission lines, For either horizontal or
inclined
spans. All calculations can be
ncarined
out wi A 20 inchlide rul, an
carried out with a 20 inch slide rule, and
no tables of functions are required other
than those included in this paper. The
procedure is the same for horizontal and
, ,los
Moreover no
no
loss oF
for obliqueobliqu
spans.spans.Moreover,
as
the
inclination
is
suffered
of
accuracy
the span is increased. The final result is
obtained by means of a graphical solution
compleiang accuringing chrtn
which, in principle, is based
Thomas chart.
By
D. 0. EHRENBURG
upon
the
date and table IV gives a summary of pertinent data
on these installations. Complete information regarding installations in the United States was not
available to the author; however, the best information available indicates that there are no installations of the baffle type, and that the total capacity
of the 2-compartment and single-compartment types
is in the neighborhood of 100,000 kw each. Table V
gives a partial list of installations in the United
States.
rim of the canyon. In this case it was found that
the procedures commonly used in sag and tension
calculations lacked either simplicity or accuracy, or
both.
In the proposed method, the length, sag, and
tension are calculated in terms of an arbitrary
parameter z. All formulas are first rigorously
derived from catenary relations, and then simplified
by neglecting small quantities. The accuracy of the
calculations is made consistent with good designing
practice, on the one hand, and with the limitations of
field measurements on the other.
Although originally intended only for steep spans,
the method is applicable to spans of any inclination
whatever. The only limitation is that the parameter
should be less than 0.5 if accuracy and simplicity
practical problems ofthistransmissionvalue.
design z is always well
line
within
In order limiting
to use the method intelligently, it is not
to
necessary follow all derivations in this paper. It
will be sufficient to read from "Graphical Solution"
to "Example 2," inclusive.
z
are to be maintained. However, in
NOTATION AND UNITS
T
H
T,
u. S. Bureau of Reclamdtion,
Denver Colo.
T2
Te
= tension at any point in cable
= horizontal component of tension
= tension at upper support
= tension at lower support
w
h
v
W
1*HE
method for carrying out sag
and tension calculations for overhead transmission
lines described in this paper .was developed when the
U. S. Bureau of Reclamation was confronted with
the problem of designing a number of steep spans
between the transformers on the roof of the Boulder
Canyon power plant and the switch yard on the
__________________________________________________________
A paper recommended for publication by the A.I.E.E. committee on power
transmission and distribution. Manuscript submitted June 4, i934; released
for publication Sept. 5, 1934.
The investigation described in this paper was carried out under the supervision
of R. E. Glover, engineer. All design work and design studies for the Bureau
are under the direction of J. L. Savage, chief designing engineer. All engineering
and construction work is under the direction of R. F. Walter, chief engineer,
Denver, Colo. All activities of the Bureau are under the general charge of
Dr. Elwood Mead, commissioner, Washington, D. C.
JULY 1935
+4s
So
cd
D
x, y
=
effective (or average) tension of cable
pounds
pounds
pounds
pounds
pounds
= weight of cable per unit length
pounds per foot
= wind load per unit length of cable
pounds per foot
= ice load per unit length of cable
pounds per foot
= VI(W + V)2 + h2, resultant force per unit length
pounds per foot
= angle between w and w'
= actual length of cable
feet
= unstressed length of cable
== straight-line
distance between supports
sag of cable
of cable
-defection
= co-ordinates of any point on cable
xl, yi = co-ordinates of upper support
of lower support
x2, yY2' == co-ordinates
x'~ ~ ~ ~ ~ ~ x, co-ordinates of any point on line of supports
a
b
= horizontal spacing of supports feet
= vertical spacing of supports feet
b'
vector w'
feet
= spacing of supports in plane of w', in the direction of
vector w'
feet
a'
rao rs
A
E
feet
feet
feet
feet
feet
feet
feet
feet
= spacing of supports in plane of w', at right angles to
= rao rs-section of cable
= modulus of elasticity of cable
AEpud
~pud
719
= temperature of cable
= coefficient of linear expansion of cable
0
a
feet per foot per degree fahrenheit
pure number
)-1pure number
= aw/2H, a parameter
sinhZ 2 _
F(z) = (
z
1 / Z22
(z)
+
degrees fahrenheit
Z4\
pure
number
LENGTH, TENSION, AND SAG
Length of Cable. By virtue of equation 2 one has
at once (see figure 1):
x2
xl
-(sinh Hw
- snh/)
Hl
1
Transform and make use of the first equation 4.
Then:
s
=
arc PP,
S
=
2H sih aw cosh xl + X2
W
2H
2H/w
= arc
OP- arc O
=
Use equations 5, 6, and 7, and then simplify.
The length, tension, and sag will be written as
functions of the parameter z. It will be assumed for This gives:
the time being that there is no wind load or ice load.
(8
s
2I (sinlh Z)2
a
Y
p
I-1v
P
--------_---X2
-------------a---------.,- --------- a----- <
------------------
-
Fig. 1. Repre-
8
+
z
By putting
of
the cable arc
sentation
(z)F(z ( mz)s )
6~~~~~~~~~~~~
Yi
one has:
2
=
a2(
F)+b2
=
V2+aF
where c is the straight-line distance between supports.
Standard Formulas. Let the cable be represented
by the arc PIP2 of any catenary drawn through the
points P1 and P2. The horizontal and vertical
spacings are a and b, respectively (see figure 1).
Expand F(z) into a power series:
25
23
2
and:
4!
6!
8!
The equation of the catenary, referred to the point S2 = C2 + a2F(z) = £2 +T, a2Z2 + 62, 2z4 + a2z6 +
(
0 where it has a horizontal tangent, is
H (
x
Assume that
V\Y =H+a2F = c +mz2
+m4z42+mz +
(10)
The length of arc from (0, 0) to (x, y) is given by
By squaring both sides of equation 10 and com= H sinh x
(2) paring with equation 9, one evaluates the coefficients
w
m. Remembering that C2 = a2 + b2, one finally
Hlw
The tension at the point (x, y) is
obtains:
T_
a2H
c
o
s
h
-
(3)
5
=
^v/c2
+a2F
c
+
Z2
F +
Z4
( / 2
+
H/w
Locating the End Points. Before one can actually
Since this series converges at least as rapidly as
make use of equations 1, 2, and 3, the end points the right-hand side of equation 9, one can neglect
xi, y, and x2, Y2 must be located with respect to the
co-ordinate axes (see figure 1).
From equation 1
YW H/Yw= cosh
cosh X2h
-
H/wi
H/lw
H/lw H/lw
By transforming and making the substitutions
X1-X2
= a,
y,-y=b
Q
(4)
---
P2Y =
\
\
\
one obtains:
b
Hl
H/w
=
2 sinh aw
2H
2x1-a
-~H sinh2iw
2H/w
N
Upon solving this for xl and putting\X
~~~
Z = 2H'
one has
H/w = arbih
zsn
mai
H-/=ar
+z
z(6
6
and
H/2-/
720
=
H/wa
az
sn
/ = arc Snh asn
-z
(7)
Fig. 2.
~~~~~~(5)
Components of forces on cabole
4) = the angle MP1M '
b = MP1
b's = M'Ip1
Draw P2M' perpendicular
to P2N
M 'Q parallel to MP2
Draw
P1M' is perpendicular to P2M' by construction
P1M' is perpendicular to M'Q, since M'Q is parallel to MP2
Therefore P1M' is perpendicular to the plane of MM'P,
Therefore P1M' is perpendicular to MM'
Therefore the
MM'P1 = 90 degrees
Therefore b' =angle
b cos4)
ELECTRICAL ENGINEERING
all terms beginning with the fourth. One then has,
upon rearranging:
a2 I
f (z)
a2z
4
S-cz2+201 +-~z4()
C
6c
72c'
The quantity S
the length of the
Byputting
is the slack of the cable, or
cable in excess of the span length c.
- c
~
By~ putting
f(z) =
Table I-Functions oF z
~
~
~
~
~
~
~
0.010... .0.00001667'... 100.003 .... .100.000
50.000
0.020..: .0.0000666 .:. 50.007
....
0:040:. :00026
~~~~~~~~~~~~~~~~~~0.050...
.0.0004670
has:
c 72c3
VCaluso f(Z)
..0.003740
~~~~~~~~~~0.150.
~~~~~~~~~~~~~~~~~0.160...
.0.0042729
2b2
and
2
z(ae1ien i3tbe
.
I
most casesit
will be found that the term involving
can be neglected.
Uppuersupo rtz
Bye
gviten of
3
Z
.
~~~~~~~~~~~~~0.170....0.0048240
0.1850... .0.003540
0 ..0.006680
.0.007019
~~~~~~~~~~~~~0.2705...
Z4
...
ZS
.0.00000001.... .0.0000010
.0.00000016.::...0.0000080
....
O::000000256:... 0.0000640
201687::1667 :0:000016296....:00002150
25:013
....
25.0000
...20000..
:::
~ ~ 000...00107... 12.527 .... 12.5000
(12) ~~~~~~~~~~~~~~~~0.070....0.0001351 ... 11.41.309
14.2857
.0.00106687
10.0333.... 10.50000
~~~~~~~~~~~~~~~0.1080...
~
1 Z4
0.110... .0.002018
one
l/z
coth z
..9.1275 .... 9.0909
...
...
...
...
6.37166....
6.73032....
5.93895....
5671664....
5.30665....
5..9462 ....
4.83174.:.
.0.00736609.
...
....
...
....
.0.00004096 .....0.0002120
0.00006561... .0.0003290
.0.000104006 .... 0. 0001200
.....0.0001464
... .0.001331
6.6667 ...0.0005063 .... 0.0033758
769250 .... .0.0006554 ... 0.004096
5.8824 .... .0.00038352 .... 0.002491
6.6667 ::: 0.0010506
0.003375
6.5000
...0.0016004 .... :0:008000
4.87805... .0.0017662 .. .0.008615
4.756190..: .0.001945 ::...0.009826
...
'O.i180::
0.90...0.007722
4.72263
4.265116... .0.0021303 ...0.006838
most asesit
wllb foud tht theterm 'Mvo'vi'g ~~~~~~~~~~~~~~~~~0.220...
.0.006808 5..046865.... 4.54545.... 0.002343 .... 0.010680
4
z can be neglected.
~~~~~~~0.22'5...
.0.008459
4.5942 .... 4.44444 .... 0.002563 ... .0.011395
0.230... .0.0088466 . 4.4242.... 4.37839... .0.0027985 .... 0.0192617
UpperSup ot esin y itu feqaio
4.3334.... 4.25532.... 0.002305
0.235... .0.00972302
.... 0.012938
...
4.26486....
4.16667 0.0023431 '....0.013682
XI
~~~~~~~~~~~~~0.2240..0.009062
By maig s oHquton6cadthnsipifig
0:2245:::.0.01003459
4.51630.... 4.08163
0.003563 .... .0.01413
one has:
4.0830. .... 4.00000
.0.0037906 .... 0.01563
Hlw0.250... .0.01084490
4.33362.... 3.921573... .0.004228 ... ..O.012958
0.25... .0.0108730
...
0.260...
.0.019305
3.92464.... 3.77135...
3.846157... .0.004320
.0.004370 ....
0.017582
usofeuto
hn
By main
6ad
= - ~~~.tIa~~
(Sinh
+ b2 cothMP
zf
+
... 0.01861
91 0.26.5... .0.0117345 ... 3.1615.....
Tendsion.
tbequatIon
..
...
...
,
....
....
....
2n
has
/0.270... .0.0121494
0.275... .0.012652
0.280... .0.013118
or,
byequationh8,
W a2+ bl coth z + b
2 T, z
2
= - (S cothb)z+ = 1/2wS) coth z + 1/2wb
Lower-Support
Tension.
(14)
Proceeding
as
T, =w (Scot z/2(W)
+b) cth z+ '2wb
on finds:
r)oh ee-di/
P/(
rt / Tesin
Lower-iSupofinds:
2
2
.2
above,
g
one
14)
(15)ove
Effective Tension. Divide the unstressed length
of athescabl into aelemvents n table IVhe the abl
ousendedt the lenguth of
the
weight
Aenso.
eac
...
...
...
...
..
0.325....0.015769
0.330... .0.0182949
~~~~~~~~~~~~~~~~~0.335...0.0188209
~~~~~~~~~~~~~0.340...
.0.019785
0.345...
0.355..: .0.018809
~~~~~~~01365:... 10102352
...
...
...
0.400... .0.0268852
...
...
.
.
..
As=
I'S.
1j'So,~~~~~~~~~~ ('S.
~~~~0.4395... .0.031856
(17
0.440... .0.0325790
S.S+OA EJo TdSWAE0.445...
.0.033331
0.450... .0.0340922
Byee tra sforming,a one cobtis seto
ofeatct.
1moduu- TdS.
S
_
So
ThS
AE
SS
jEo
T
fetvoenin T
.0.0348651
n
ste 0.455...
0.460... .0.039564
:0.465... .0.036427
.0.037122
~~~~~~~~~~~~~~~~~~~~0.470...
(18)
do(7
TS =S
ofthetensio T whc
e
sdfne
sta
atsisHok' aw i
Tiv etensioniof thenscable is dTind
2', = ~~~~~ f' TdS. or, very nearly
s
au
0.475... .0.0318028
0.480... .0.038842
0.4850... .0.0349665
0:4650:::00304247
)t 0.450... .0.038218
taHvaue
3.13985.... 3.03030 .....0.011865
3.09590....
29570.012984
379...
3.20536.... 2.911800.... 0.01336
3.01265.... 2.879855.....0.014176
2.19729....
2.85714...
3.09593.... 2.98169 ..0.::
.0.01518
12~59
2739738 ... .0.01775
0.01968
.0.020580
.0.021958
...
...
.... .0.02496
.0.02156
..0.02567
0020
.....0.028375
.0.02497
.0.03126256
:.:: 0.032770
....
.0.02343
..
.0.035949
...0.037260
.... 0.03930
...0.034106
0.04288
....0.043760
....
...
.0104863
..
...
...
...
=
3.5444.... 3.44828 .....0.007031
3.4876 .... 3.638936... .0.0075739
3.64327.... 3.33333.....0.00810047
3.63797.... 3.207869.....0.0086548
3.3285.::: 3.242581.....0.0092735
.487
3..17460.... 0.075
3.42309.::: 3.12500.:...0.001049
3.078692... .0.011165
3.1845...
3..0258603...
...
...
S
3. 8615714....0.0064570
3.86033.... 3.7750877.... 0.0065982
...
....
2.70270.1:. 0.11 ...:: 010
3.026
0.370....00195
0.1..0.020424 .. 297570:... 2.637158.....0.02085 ... 10.054287
0.4705... .0.0227563
Dvd h ntesdlnt
0.410... .0.028625
i
0.415.::.0.024240
'ithe aealofecros seto n anden thecbe
0.425.1.10:0303876
moduluspenof elatic
th ty. To en: le etbeo e
0.4390. ...0.031102
E \fctv
wfher Aae i
403.733..
34.703000.... O.0.90051
4.03.727..~. .3.63636... .0.005729
3..9324.....
.
0.29...0.01567
g:
0.015068:..
to 0.385.: 1.0.024887
elemestnt,beomes
/ ~~~~~~~~~J~~~~~~\ ~~~~0.395... .0.0262407
(16)ecale
(
+oa
0.295... .0.0146567
.0.015068
~~~~~~~~~~~~~~~~~~~~~0.300...
0.305... .0.0135576
~0.310.... 0.016094
.0.017995
0.350... .0.012042
are givenS int table I.b
rdc
S sacntn,eult
Noeta2h
the total weightof the cable.
Values ofcoth z
is
.0.013759
~~~~~~0.285...
.0.0121407
~~~~~~~~~~~~~~~0.290...
...
...
...
...
2.7245 .:::
2.8662....
2.86031....
2.60279....
2.7742....
2.51970.:...0.02197
2.53165 ... .0.02434
2.50000... .0.02756
2.46914 ....
.0.02690
2.43902. .. .0.02982
2.75464.::: 2.406418.::.:002085
2174929....
2.3529740.... 0.032637
2.46728.... 2.355810.....0.023419
2.4421 ....
2.43175....
2.3936 ....
2.3702 ....
2.3474 ....
2.29885... .0.02358
2.27273 ... .0.03748
2.247919.....0.036921
2.22222 .....0.04101
2.109780.... 0.042866
..
2.2821 ....
2.42613....
2.24095....
2.2210 ....
2.13891.... 0.04477
2.115054.... 0.042675
2.12766 ... .0.034880
2.108526.... 0.035091
2.082333.....0.053048
2.06186... .0.05533
...
2120161111.
2.105408.... 0.046765
...
...
...
..
...
...
2.35193....
2.303329....
2.2130.... 2.100000.... 0.06501
037
0.046163
.... 0.04640
:
...
...
....
..
0.05664
..0.06827
:.... 0.071487
0057077
....
... .0.07951
.... 0.082632
... 0.08518
....
.0.08812
0.06911
....
....
.0.091420
.0.09734.9
....
...
....
....
0.10057
.0.10381
.0.10723
....
....
.0.510
0.0.1141
.....0.1176
....
001125
..017
c..00449
2d01
(20)048..0.56
Now,95--0 01RR
inegat equation.. 2.0:00.. -60
..( 2
.40
Substitute from equations 5, 6, and 7, and simplify:
wS (S + b2 othz + a2)
(21)
Sa
2
Te= 4
Expand z coth z into a power series:
lating the weight of the cable, use the zero-tension
room-temperature value of S, that is, the unstressed
length So at 60 to 70 degrees fahrenheit.
In the "Summary of Formulas," which follows,
wS has been replaced by wSo, where w denotes the
(22) weight per foot of the cable at zero tension and room
-5Z4
z4 + .,.(22)tmeaue
..
1 + 33=1+
Z2x2_- 45
z coth z = zcothz
temperature.
and Sag. Deflection is defined as the
Deflectiondeviation
all. terms
than one-half,
Since z is always less.
>
.
max-imum
the cable from the llne of
s
..
<
beginning with the third can be neglected. Sub- asuxmorts measured atof nht
anbles to the line of
the first 2 terms into
one has:
supports,
equation 21,
stituting
Sag is defined as the maximum deviation of the
cable from the line of supports, measured in the
direction of the resultant force acting on the cable.
From equation 9, by long division, one obtains
case there is no wind, the sag is vertical.)
a21
c2 1
mathematical
relation between the deflection D
1
+
=
-i = 1-_ s 3
Z2
+
+
C2z32+,,. 'l~33 ( C2')Zl ...
and the sag d is given in equation 28.
The equation of the line of supports is
Substitute this for the second term in braces, but
replace 52 by c2 in the last term in braces (which is
small). Then:
y= Y + ab (x - xi)
(24)
(23)
From equations 6 and 5, one obtains
T. = 1/2(S). 1 + (w6S)b s
Tc =
Ws
4zj
+
1
b2 Z2;
+ 3 Z2 + 3S2
Y2
,fb',\(In
-
Values of l/z are given in table I. In most cases xi = a arc sinh bz + a
a sinhz 2
2z
it will be found that the term involving z can be
neglected.
As mentioned before, the product wS is equal to
By equations 1 and 3,
the total weight of the cable, and therefore remains
T_ H
Yw
T=8500 TE T T8650
0
2z(7
4
6500te\ ,/O
~~1~~
z 6000
'b0
O-o
°5000
- e4500
4000
35000 00
0-~~~0
-
_
-0
0
_
0
w-
(26)
w
and, therefore, one has
T~ ~ ~ ~ ~ ~_H
~7, T7,a
8000wiw w w
The deflection is given by
87500 ot < 5/ ]y
07000 T
(25)
s
+\'-,
.-
i
t°= °
- I-
0 DP =
max.L[
a
a2
r,?
I_Y
Y_y)]
= amax. [y_yf
= ad
(8
(28)
V here y and y' are defined by equations 1 and 24,
respectively, and d is the sag. The difference
__
-y
is<maximum
< z |yTs -~when
z~ -~~
0
MAX. DFLECTIN
-
(27)
zarcinh
-
(29)
0.30 0.40 0.50 0LAC60 0.70
IN 0.80 0.901.001.1Substitute equations 25, 27, and 29 into equations
SLACK IN FEET
1 and 24, subtract equation 1 from 24 and simplify:
Fig. 3. Tension-temperature chart
512,000 circular mil, 1.4 inch diameter, type HH hollow
copper conductor
Horizontal spacing = 475 feet
= 435 feet
Vertical spacing
= 8,650 pounds
Maximum tension
Minimum temperature = 10 degrees fahrenheit
Maximum load = 12 pounds per square foot-wind load
EXPLANATION
T', = Upper support tension, 12 pounds per square foot
wind
1_j
Te= Effective tension, 12 pounds per square foot wind
load
Tl= Upper support tension, light
Te = Effective tension, ight
Z
Szcothz - c + barc sinh
a
si
)
Substitute for S from equation 11, for z coth z from equation
and for the ratio z/sinh z as follows:
z
22,
_ /6 Z2 +
The sinh being small, replace the arl sinh bv the sinh.
Drop all terms containing powers of z higher than the third.
One then has:
3-b
d =max.[y'-_y] =c z + 3a2-b
constant (for a given load). To establish the value and:
of wS, one must know the values of w and S at the
same temperature and same tension. Since w is D = a d = L5. + 3a2-42b' a z3
generally measured in the laboratory, at zero tenValues of Z3 are given in table I.
sion and room temperature, one should, in calcu722
z
(30)
Now simplify equation 30 in the following manner:
max. [y' - YA =
(31)
(32)
ELECTRICAL ENGINEERING
For short spans the second term in equations 31
and 32 may be dropped (see "Example 1," which
follows). Then:_
Sag = d =
1/4 cz,
Deflection
=D =1/4 a
z
- - l
T=34,000o LB - - - - -
32AND-
30,000 - -
For long spans with a relatively small difference
in elevation between supports, b2 is negligible com-
2 28,000
therefore:
a2 26000A
pared to a2 (see "Example 2," which follows) and
d = 1/4c'z + 3/144C'Z3,
D = '/4a z + 3/4a .z3
(33)
CORRECTION FOR WIND LOAD AND ICE LOAD
If the cable is covered with ice, the force of gravity
per unit length of the cable is w + v, where w is the
weight of the conductor (in pounds per foot) and v
is the weight of ice (in pounds per foot of cable).
If the cable is acted upon by a horizontal wind
exerting a force h per unit length of the cable (at
right angles to the no-wind plane of the cable), the
cable will hang in an oblique plane determined by
the line of supports and the direction of the resultant
w' of the gravity force w + v and the wind force h.
The formulas developed above are still applicable
in case of the wind load and ice load, provided one
replaces:
by w', where w' = V\(W + V)2 + h2
a and b by a' and b', respectively, where:
b' is the spacing between supports in the oblique plane, measured in
w
dieto
of th frc w';
'
th*iecino
a'the~~~~~~
is the spacingth.oc
between supports in the oblique plane, measured
at right angles to the direction of w'.
It is easily seen from figure 2 that:
b cos4t
b'
a' =
=
b w +V
w'
(34)
GRAPHICAL SOLUTION
By assigning arbitrary values to the parameter z,
mutually corresponding values of S - c, Te, and.
D can be calculated. The effective tension T, can
then be plotted against the slack S - c. Such a
length-tension chart is shown in figure 3. However,
before one can make use of the chart, the elastic
properties of the cable must be taken into consideration.
Hooke's Law. When the temperature is constant,
the effective tension and the length of the cable are
related by Hooke's law. In differential notation,
one has
aTe
AEAE
as.
bS
S0~;
sO
(0
=
constant)
(36)
The unstressed length So is, strictly speaking, a
function of temperature. However, since the quantity
AE is only approximately constant, the temperature
variation of So may be neglected entirely. The
symbol So may be taken to denote unstressed length
at room temperature.
Change of L>oad. Assume that one is given a point
(S -c, T,') on the wind load curve in figure 3,
and that subsequently the wind load is removed,
Here one has a change at constant temperature, and
therefore equation 36 applies. Draw a straight line
JULY 1935
z
- -
- -
-
- - - - - - - __
- _ -
T-
?PXw°° °
T
24,000
-
T
I
- -
Z2,000
|
|
-
20,000
is,oocI
e
LIGHT D
-__
- - - - - - - - Te-
6,ococ - -
235
3O8F, IE A w230
o
0
Z
WIND
N
0
OIT L 225 SA
OLIGH LOAD
31
32
33
SLACK IN FEET
34
-I
35
Fig. 4. Tension-temperature chart
Stranded aluminum conductor with steel core
Horizontal spacing
Vertical spacing
=
=
4,279 feet
feet
185.5
Minimum temprtr = zero degrees fahrenheit
Minimum temperature = zerodrsf
Maximum load
= 12 pounds per square foot wind
and 1/2 inch ice
anATin
EXPLANATION
T1 = Upper support tension
(35)
b
|-
T=33P00 LB
-
Te = Effective tension
Unprimed = light load
Primed = 12 pounds
per square foot wind and 1/2 inch ice
load
Double primed = 1/2 inch ice load
through (S - c, T7,') with slope AE/So. The intersection of this line with the light load curve locates
the new point.
Change of Temperature. Now, assume that one is
given a point (S - c, Te) on the light load curve in
figure 3 and that the temperature rises by AO degrees
fahrenheit. Since the length S, tension T,, and
temperature 0 all vary simultaneously, one can write:
dS = as do +
.s dT6
e
(37)
Obviously, one may analyze dS into 2 components
which can be accounted for one at a time. First,
let
S vary with 0 only. Then, one is followlng a
constant-tension path, that is, a horizontal line.
This shifts the point to the position
[(5 + &I AO) -CJ Te],
where ar is the thermal coefficient of expansion.
Now, let 0 remain constant and let S vary with T,
only. This time one is followving a constant-temperature line, to its intersection with the light load
curve (since the point must be on the light load curve).
Construction of Chart. In practice, a number of
constant-temperature lines are drawn, as in figure 3
or figure 4. Changes in length and tension due' to
varying temperature are found by following the
723
proper constant-load curve to the intersection with
For short spans (see "Example 1"), equation 38
the proper constant-temperature line. The chart can be replaced by:
must be so arranged that the minimum temperature
c
l _
corresponds to the maximum permissible tenslon at z = - max. [T,'] - 1/2 w'b'
(40)
maximum load.
Unstressed Length. It will be noted that formulas
SUMMARY OF FORMULAS
B, C, D, and G involve the quantity So, which is
the unstressed length at room temperature.
If the tension of the cable is low, the maximumBasic Formulas. Slack (length of cable minus
load minimum-temperature value of S, or that value
span length):
of S which corresponds to the initial z, can be used
a2
OFb Z4
s - c = -af(z) +
in
(A)
place of SO (see "Example 1").
c72cc
In case of high tension, one can easily estimate the
value of So by applying a correction to the maximumTension at upper support:
load minimum-temperature value of S, as illustrated
(
TI = (1/2 WSO) coth z + 1/2wb
in "Example
2."
Tension at lower support:
Tables. The equations in "Summary of Formulas"
772 = (1/2 oS0) coth z - I/2wb
(C) involve the following functions of z:
Effective or average tension:
(z) = 66 Z2 + 20,'
1
1?~~b
coth z
Z
=
(D)
T7 (l/2 wSo) -+ (1/2 wSO) T-2 *
z~~~~~~~~~~~~~~~~~~~~
-z4, and zs
Deflection (maximum deviation from line of sup- z ,
ports, at right angles):
These functions are given in table I, for values of
0 to 0.5.
z
from
3a2 -. 2b2
D = 1/4 a z + 1 bc az3
(E)
Illustrative Examples. The details of the calculations are illustrated in 2 practical examples (see
Sag (maximum deviation from line of supports, in below).
direction of force):
1. Example 1 deals with a proposed span at the Boulder Canyon
3a2 -2b
d = 1/4 C z + 144c s3
(F)
Slope of constant-temperature lines:
plant, and illustrates the calculations for a short span of considerable inclination.
2. Example 2 is based upon published data of an actually existing
power
river crossing, and illustrates the calculations for a long span with a
(G)relatively small angle of inclination.
2)Te 5T AE
_ A
(G)
as SO
To Correct for Wind and Ice. Replace a, b, and
w by a', b', and w', where:
w'= V(w + V)
w=
bW + v
+h
w'
Table II-Sample Calculations, Example I
Proposed Span at Boulder Canyon Power Plant
(H)
UpperTension
Slack
Support
S-c
(I)
CALCULATIONS
a
N
I
The practical application of the equations given
under "Summary of Formulas" will now be exInitial Value of Parameter.
The initial value of
z
is that value for which T1' is equal or nearly equal
to the maximum permissible tension ("working ten-
sion").
To establish the initial value of z, use the relation
coh
max.
6(lc)'21
[77,'] - '/2w'b'
(38)
~/2Wc
724
S
I
"
1
I
1j
0.08 .
0.512 .
8,520 .
0.12 .
1.153.
5,690 . 6,030. 5,670 . 16.7
.0
8,860 .
8,500 .
0.648. 7,580 . 7,920 . 7,560
11.1
125
Light Load
a
N
N
()
For nearly horizontal spans, the ratio a'/c may
be taken equal to unity (see "Example 2," whlich
follows).
Deflection
Wind Load
a' = c2-b(J)
plained.
Effective
Tension
N
9ZO
Hl
00\
0.07.0.286..7,270 .7,610.7,260 .
.8.3
0.08.0.373.6,360.6,700.6,350.9.5
0.09.0.473.5,660.6,000.5,640.10.7
0.12.0.841.4,250.4,590.4,230.14.3
0o.14..1.144..3,650..3,990..3,630..16.7
ELECTRICAL ENGINEERING
EXAMPLE 1- PROPOSED SPAN AT
BOULDER CANYON POWER PLANT
the tension at the upper support at light load T,
and at wind load T1' and the effective tension at
light load Te and at wind load T6' are plotted
Given the following:
against the slack S-c.
The 10 degree fahrenheit line is drawn through
512,000 circular mil, 1.4 inch diameter, type HH hollow copper conthat point on the Te curve which is directly under
ductor
E = 16,000,000 pounds per square inch
the 8,650 pounds point on the T1' curve.
a = 0.0000096 foot per foot per degree fahrenheit
The deflection D is also plotted against the slack.
Maximum tension = 8,650 pounds
As in most cases, the deflection curves for light load
ww = 145775
1.577 pounds per foot
pounds per footand wind load are found to coincide (see figure 3).
a = 475 feet
6 = 435 feet
Stringing Chart. The points marked by circles on
Maximum wind load = 12 pounds per square foot
the T1 curve in figure 3 comprise the data for the
Minimum temperature 10 degrees fahrenheit, maximum 150 degrees stringing chart.
fahrenheit
In this case it is assumed that the cable will be
strung at no wind (or, during a relapse of the
Tension to be measured at upper support with and that the dynamometer will be used at the wind),
upper
dynamometer.
support. Therefore, the stringing chart should be
made up as follows:
Wind Load Curve,
Horizontally, plot temperature in degrees fahrenh = 12 pounds per square foot times the projected area
heit.
=12 X1X 1=1.40poundsperfoot
Vertically, plot the upper-support tension T1
W-= w+h = 2.11 pounds per foot
(light load) in pounds, and, if desired, also the light
_v 0.748
load deflection D or sag d = a
b=
c2 feet
that the horizontal scale is the temperature
c = 6441
644. 1 feet
b' = 325.3 feet (by equation I)
of the cable, not air temperature.)
=
=
414900(Note
=
b'12= 105,800
a2 - 309,100 (by equation J)
a'
556 feet
EXAMPLE 2-MISSISSIPPI RIVER CROSSING
To find the initial value of z:
'/2 w'c = 680
t/2 w'b' = 340,
8,650680- 340 =0.082
Z =
Let z be 0.080, then 0.090, etc. (see table II).
To calculate '/2w'So: from table II, when z =
0.080,
S-c = 0.5
S = 644.1 + 0.5 = 644.6 feet
be sufficlently
to let
case, it
sufficie'ntly accurate
InInthis
this case,
lt will
wlll be
accurate to
let
So
=
644.6 feet. Then,
'1/2w'So
=
=
AE = 11,860,000 pounds
= 0.00006
po o
508 pounds
d
e
Maximum tension = 33,000 pounds
n
w = 1.684 pounds per foot
a = 4,279 feet
b = 185.5 feet
Maximum load = 12 pounds per square foot wind and 1/2 inch ice.
Minimum temperature 0 degrees fahrenheit, maximum 120 degrees
fahrenheit.
680 pounds
Remaining calculations are in table II.
Light-Load Curve. To calculate '1/2WSO:
1/awSo = 1/2 X 1.577 X 644.6
(Compare with "Mississippi River Crossing of
Crystal City Transmission Line," by H. W. Eales
atnd E. Ettlinger, A.I.E.E. TRANSACTIONS, volume
44, 1925, pages 378-97.)
Maximum Load Curve.
(Compare with
1.")
w +v
w'
w±+v
=
=
"Example
2.623 pounds per foot
c = 4,283 feet
3.322 pounds per foot
b' = 146.5 feet
2.623 =086a' =4,280 feet
07896
Remaining calculations are in table II.
W
.262
3
Constant-Temperature Curves. To calculate hori. . .
zontal interval of the constant-temperature lines for
To find initial value of z:
z0= 10 degrees fahrenheit:
1/2w'b' = 240,
172W'c = 7,114
S
a=- AO = Sa X 10
AS = 644.6 X 0.0000096 X 10 = 0.062 foot,
or, very nearly, AS =
foot
As
=
SX
To calculate slope of constant-temperature lines:
A = 512,000 circular mils = 0.402 square inch
AE = 0.402 X 16,000,000 = 6,430,000 pounds
Slope = 61', - ASE = 6,4630
9,980 pounds per foot,
0060
A
=33,000 -24046
A=212
Al
4.61,
7,114
coth z =
1 1
1 +
=
21.25
457
From table I, the initial value of z = 0.220.
To calculate '/,w'So: from table III, when z =
0.220,
S-4C8
= 34.6
431 fe
or. very nearly,
slope = 1,000.opfoutnds
This length corresponds to about 0 degree fahrenheit
foo
and 33,000 pounds tension. If the tension
is reConstruction of Chart. From the data in table II, leased, the cable will shrink by some 12 feet (since
JULY 1935
725
Calculations. As a first approximation for b, one can take
AE = 11,860,000 pounds). If the temperature is
raised to 60 degrees fahrenheit, it will gain some 2
(42)
b,= 22
feet (see below). One may take:
So = 4,310 feet,
which is sufficiently accurate to use in equation 41. As a second
approximation, one has, by virtue of equation 9,
'/2W'So = 7,160 pounds
-
Remaining calculations are in table III.
Light Load Curve. To calculate '/2wSo:
1/2WSo
=
1/2 X 1.684 X 4,310
=
3,630 pounds
_ Z2
(43)
The slack can be calculated from the relation
a2
Remaining calculations are in table III.
Constant-Temperature Curves. To calculate hori- s- = 6c
zontal interval of the constant-temperature lines for where
average values of a and c may be used.
z<O=
10 degreesfahrenheit:
10 degTeesfahrenheit:
of g, one has
~~~~~By definition
AS = Z-S AO = Sa X 10
AS = 4,318 X 0.00000662 X 10
285
foot
or, very nearly, AS = -100
=
g a641oxima)+(35
or, approximately,
6+475 435-
0.286 foot,
To calculate slope of constant-temperature lines:
Slope
pounds
AE
so - 11,860,000
4,310 feet
11,000 pounds
Slope = 2,750 pounds per foot =
4 feet
Tas
By assigning arbitrary values to a, the following table of values
is obtained:
a
11. 50 ... .11.70 ... .12.00 .1 .12.30 ... .12.60
8.83 .... 8.56 .... 8.14 .... 7.68 ....
bi
z.
0.0637...0. .0789.... 0.1017 .... 0.1244.... 0.1474
7.17
b2
b
2+b2,
=
Using equlations B and C, one has:
wSo coth
z - wb
zeSo
coth z-wb
wS coth zwS+coth
wb z=+wb=
In this particular case, it is permissible to use the approximate
1'
-oh=
z
a[z-w)b +w(b +b)]
726
0.230
8.55
....
8.11
14.39
....
14.31
0.353
....
....
0.586
....
7.65
....
....
7.03
.14.09
....14.21
0.879
....
1.230
6,300
4,890
4,000 .... 3,370
Graphical Solution. To establish a better comparison between
....
....
g I14.40 9 P
e 14.30 0.205 F7E
--14.20
---.-
14.10
- PF =
TeI650C
-
_ _2
_
2
6000
5500
z
z
50_
_
4500
Z
4000
Lu
F 3500 _
0.30 0.40
0.50
0.057
FEET
0.60
0.70
0.205 FEET0.80 0.90
SLACK IN FEET
-
1.00
1.10
Fig. 5. Study of insulator efect
1.4 inch
HH conductor
512,000 circular
Horizontal
spacingmil,= 475 feetdiameter, type
Vertical spacing
=
435 feet
Weight of insulator = 754 pounds
the performances with and without the insulator, it will be assumed
that the effective tension at no-wind and 10 degrees fahrenheit is
Te = 6,450 pounds (the same as in figure 3).
~~~Since c in this case is a variable, one has
ebSSd9 +-dTe -dc
b
d(S -c) = dS -dc =
T
By definition of g,
1
c =644.1 -g,
aw z
dc = -dg,
and therefore:
Solving for z one then has:
_w[a(So ± S) - (a + a)S]
....
7,800
No attention has been paid so far to the effect of, insulators.
However,-in short dead-ended spans of high voltage lines the influence of the insulators on the mechanical performance of the cable
may be appreciable. To gain a general idea of the effect of insulators, a correction will be worked out for the span considered in
"Example 1." For the sake of simplicity, only the insulator at the
lower support will be taken into account.
Notation. The following notation will be used in this appendix:
a, b = projections of cable span
a, = projections of insulator span
o
S =length of insulator
So = unstressed length of cable
w = weight of cable per unit length
w = weight of insulator per unit length
T, = tension at cable end of insulator
T2 = tension at insulator end of cable
c = Vaa2 + P, z = aw/2H
z = aw/2H
c = Va
gA/(a
(a + a)2 + (bS + b)2 _ Va2
g ==
a2 + b2
=
Data. Let w 52 pounds per foot, S = 14.5 feet.
From "Example 1" it follows that:
a +a = 475feet
644.1 feet
c+ g
b + b = 435 feet
So + S = 644.6 feet, very nearly
Condition for Equilibrium. Obviously, for equilibrium T2
T2.
z
....
S-c....
Appendix I-EKfect of Insulators
_ aw
8.82
14.43
..
g..
coh
(45)
..
Construction of Chart. See figure 4.
relations-a
(44)
(41
(4)
d(as)= O+
d(-)= Oda+O
d~
g(6
dTe +dg(6
...ELECTRICAL ENGINEERING
Table II-Sample Calculations, Example 2
of maximum vertical deviation is also the point of maximum normal
deviation from the line of supports.
In a horizontal span, the point of maximum deviation from the
line of supports is exactly in the middle of the span. In an inclined
span, the point of maximum deviation is offset from the middlepoint toward the upper support. By equations 6, 7, and 29, the
Mississippi River Crossing
UpperSupport Effective
Slack
S -c
Tension
Wind Load and Ice Load
K
horizontal offset is:
Sag
Tension
_
aa *h
+
,
,
Offset =-arc sinh
X W2z
uzs o
X
>
»,,,1O+
2azOffset
gN^
@
N
+'
-
a
9
o-z~
0.205... 30.0... 17,950...
0.210... 31.5... .17,540...
0.215 ... .33.0.. . 17,140..,.
0.220.. .34.6.. 16,770..
X
|
~~~~~~~~~N>N
c
-
,W
Xdm
>
>
;Z
i
*,
s
18,110... 17,710... 219.6... 0.8... .220.4
17,700 .... 17,290... 224.9... 0.8... 225.7
17,300 ... 16,880 ... .230.3... 0.9 ... 231.2
16,930.. .16,500.. .235.6.. .0.9.. .236.5
Ice Load (w" = w + v)
N
N0
a
kd
0
02202.... 34.6
ffi
0.220. ...34.6.25,690
.26.....
25,690 .....
se4
.
a
Distance from point of maximum deviation =- arc sinh
2z
a
b
(48)
-
Angle of Inclination. The cosine of'the angle which the cable
makes with the horizontal at the upper support is
cos (angle of inclination) = H = awl1
2T-
J
SAMPLE
U _ ICALCULATIONS
N
>
t3
(47)
=
Point of Horizontal Tangency. If the cable has a horizontal
tangent, the point of horizontal tangency is the lowest point of the
cable.
In a horizontal span, the horizontal tangent occurs exactly in the
middle of span. In an inclined span, the point of horizontal tangency moves toward the lower support.
By means of equation 47, the point of maximum deviation can be
4> > o with considerable accuracy. By equation 29, the distance
located
<:
s the latter point to the point of horizontal tangency can be
UXfrom
found at once:
Light Load
u
+
NN
N"
-:
afl eJ
+
oo
o + INI t
d;
23g
o
a2 sinh z
nearly:
or, very
^
0.220 .. 34.6.. .33,070 ... .33,310 ... 32,540
0.225... .36.2 .. 32,360 ... 32,600 . .31,820
,
bz
- - ar sinh
as
nhh
a
By transforming and making use of equation 8, one has:
4N
N .
U)i
b
arc sin
2z
0
236.5
.2
23
oRequired: clearance to water, Mississippi River crossing in
°
"Example 2," at 32 degrees fahrenheit, ice load, no wind.
(a) Data:
From figure
4, at 32 degrees fahrenheit, ice load, no wind:
= 234 feet
~~~~~~Sag
Te"i
= 25,950 pounds
33.9 feet
(b) To establish value of z:
In this case,
T. - 1'S 1
S
At 10 degrees fahrenheit, the slack and tension of the cable are
given by point 1 in figure 5. Assuming the span length c constant
for the time being and letting dO = 140 degrees, one follows the
path 1-2-3 to the point 3 on the T. curve. As a result, there is a
change in g (dg = -0.205 foot) and consequently one follows the
path 3-4-5 (see figure 5). Again there is a change in g (dg = 0.057
foot), and an infinite process results. However, the final position
of the point on the Te curve is found with sufficient accuracy by
drawing a smooth line through the points 2, 6, and 4.
At 150 degrees fahrenheit, the effective tension is seen to be 4,200
pounds, as compared to 3,930 pounds in "Example 1," a difference
of less than 300 pounds.
Another calculation was carried out in exactly the same manner,
for a much steeper span, 516 feet long. The difference due to the
lower insulator at 150 degrees fahrenheit was found to be 550 pounds.
Couclusions. The following conclusions may be drawn concerning
the effect of insulators:
1. The insulator has a tendency to act as a counterweight, decreas-
and therefore
,
In providing for clearances and in designing towers, it is often
necessary to calculate a profile of the conductor. Several formulas
pertaining to the geometry of the cable will now be derived.
Point of Maximum Deviation. It has been shown that the point
JULY 1935
'
X
0.2178
25,950
Te,
(c) To calculate horizontal distances:
Point of maximum deviation to middle-point:
z =
b(S
=
-
2az
c) _
185.5 X 33.9
=
2 X 4279 X 0.2178
3.4 feet
Point of maximum deviation to point of horizontal tangency:
arc sinh- = b-I1- --,i = 425.7 feet
ing the range of variation of tension.
2. In compiling stringing charts, it is generally permissible to
neglect the effect of insulators.
Appendix Il-Prohile Data
- c =
a
zL
6a
Midepntopitofhrztatngcy
425.7 -3.4 = 422.3 feet
Point of horizontal tangency to lower support:
X2= 2,139.5 - 422.3 = 1,717.2 feet
(d) To calculate elevation of lower support above point of
zontal tangency:
y2
X2 _ lj
=a2z (,cohiE.
a/2z
J
a
2z -2 \a/2z/
hori-
150 fee
-727
state limit of a system consisting of a transmission
line with synchronous machines at both ends accu-
(e) Conclusion: at 32 degrees fahrenheit, ice load, the lowest point
of the cable is 150 feet below the lower support.
According to H. W. Eales and E. Ettinger (page 396 of "Mississippi River Crossing of Crystal City Transmission Line," A.I.E.E.
TRANSACTIONS, volume 44, 1925) this value should be 151 feet.
.ea
U$2 <; 1"
a.e
y
.lnevwithtsnchr machs .atnbot es,acuratel
evalang thee
on power and on voltage.
In steady state problems changes are assumed to
take place slowly enough so that inertia forces can
be neglected. Consequently any relative shift in
phase position of the voltages in different parts of a
system can be neglected; and if the voltage is
known for all currents and power factors at any
point of a system, the stability of any steady load
connected to the system at that point can be calculated without knowing more about the system. The
current diagram (figures 8 and 9) was adopted as
e | $
*, .u ion
or va ura ea tIrcuit s
Methods of steady state power calculation
developed specifically for: (1) systems
having more machines than can be handled
and
covninlyb,by eitigmethods;
existing metnoes; ane
conveniently
(2) systems consisting of a line with magnetically saturated synchronous machines
at both ends are outlined in this paper.
(1) general
that
Two general methods are given: Two
involving the use of equivalent reactance
or impedance, and (2) that involving the
..roiin orOF current alagrams.
diagrams.
superposition
Methods of determining the equivalent
reactance and impedance of any unknown
system from measurements at the terminals
of that system, whether magnetically saturated or not, also are given. When a complete solution is desired, superposition of
current diagrams of saturated circuits is
shown to be better than the method involving the use of either equivalent reactance
or impedance.
By
STERLING BECKWITH*
ASSOCIATE A.I.E.E.
variables. It must be remembered that if the action
of voltage regulators is to be considered, a transient
and not a steady state problem will result. Two
methods of calculating stability from a current
diagram are discussed. One involves the use of
equivalent impedance, and the other involves the
of one diagram upon another.
placing
The solution of the first problem was accomplished
by the use of equivalent impedances calculated from
current diagrams, and the solution of the second
problem was obtained best by the superposing of
current diagrams. Although the method of supercurrent diagrams to take account of saturamtposing
tion is given only small space here because of its
simplicity, it is thought to be the best method for
most
problems, as it is rigorous when test data for
the machines are available, and it enables a comto be obtained with less effort than
pletethesolution
use of equivalent reactances.
by
~~~~~~~~The
Metropolitan
EQUIVALENT REACTANCE
If the terminals of a power system are available,
but nothing is known about the system behind the
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Water District of Southern
CGlifornia, Los Angeles
THE METHODS of steady statee4|1
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Me/\\
A
power calculation here outlined are the outgrowth
of 2 specific problems. One problem was to find
the steady state limit of a system (figure 14) having
more machines than could be handled by existing
methods without detrimental preliminary simplification. The other problem was to find the steady
A paper recommended for publication by the A.I.E.E. committee on power
transmission and distribution, and scheduled for discussion at the A. I.E.E.
l/./
tf
Fig. 1. Vector
diagram with re-
/
l/
/
Fig. 2. Vector diagram with respect to receiving end with Es
spect to sending end constant
hlo
o
oeqa
w.t Es cosatNt htA
even ththAougho e digami
for
the
FigureS 1ostn
Pacific Coast conveution, Seattle, Wash., August 27-30, 1935. Manuscript
submitted January 2, 1935; released for publication March 4, 1935.fiue1vnthghhedgrmsfo
the
* Now with Allis-Chalmers Manufacturing Company, Milwaukee, Wis.
728
\
same
set
of conditions
ELECTRICAL ENGINEERING