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TRANSMISSION LINE PROBLEM.

THESIS FOR DEGREE OF E. E.
FRANCIS E. ANDREWS

1918
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A GRAPHICAL SOLUTION OF THE SHORT
TRANSMISSION LINE PROBLEM,

Thesis for the Degree of Electrical Ingineer.
Michigan Agricultural College.

By
Francis E. Andrews.

1918.
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A GRAPHICAL SOLUTION OF THE SHORT TRANSMISSION
LINE PROBLEM.

The writer has felt for some time that there is need
of a comprehensive treatment of the short transmission line
problem (by short distance lines are meant those in which the
capacity effect is negligible) in such a way ‘that the average
individual who has to deal with it , may, without the use of in-
volved calculations and formulae, be able to ascertain the per=
formance of lines of different characteristics, and under differ-
ent operating conditions such as power-factor, and to readily
make comparison of these things. To this end, a set of curves
showing voltage drop for the various sizes of wire at a stand-
ard spacing arbitrarily chosen as 24 inches, for various power-
factors, has been plotted. This constitutes the groundwork for
the method and is chart A. A means is given for applying a
correction for other frequencies and spacings. In order to
show the limitations of the method, and that they may not be
exceeded in its anplication, curves are given showing the error
under various conditions, and the effect of-capacity at no
load, the latter giving an idea of the limits beyond which it
can no longer be neglected.

THE LINE DROP CHART.

The line drop chart is drawn from points located by
the formula, D=IRcos6 +IXsinf§, I=line currrent, R=resistance
in ohms, X>=reactamce in ohms, =the angle of lag or power=
factor angle at the receiver end of the circuit, and D=line

drop in volts. These quantities are of course per single con-

93'783
2

ductor or to neuftal. This formula was published in an article
in the General Electric Review of June, 1913, entitled, "A
Graphical Mehtod for the Calculation of Short Transmission
Lines", by Fredrik Waern. It will be readily understood by ref=
erence to Figure 1, which is really nothing more than the vector
diagram of the voltages in an alternating current circuit cone

sisting of resistance and inductance.

 

 

Let E=receiver volts.
Let I=line current.
Let R=total resistance of circuit.
Let X=total reactance @¢ circuit.
Let coaf= power-factor of receiver.
Then f=angle of lag, or power-factor angle.
It is then evident from an inspection of Figure l,
that Eo is the generator voltage, this being the vector sum of
the receiver voltage, the resistance drop, and the reactance
drop.
Now let the E vector be produced and BG erected
perpendicular to it from G, |
Also let HF be erected perpendicular to E from H.
Since H¥ and BG are perpendicular to E, the angles FHS
and JGB are each equal to fp.

Angle FAH also equals #.

Therefore, IRcos#+JHsing+JGsin6S= AB,
or LReog$ + (SH+TG) sing = AB,
or tReos#+ IXsing =a. (1)

If an arc, GC, be drawn with O as a center,

Eg-E=AC,
or, taken numerically, AC equals the voltage drop.

It will be noticed that AB very nearly equals AC, and

we may accordingly write with negligible error in most cases,
AB=voltage drop (numerically).

and from (1), D=IRcos6+IXsing, (2)

Which is the formula used in this method for the calculation of

line drop.

The error in this formula is represented by BC, the
difference between Ac and AB. This error is fally discussed
in a subsequent section.

In the construction of chart A, the Line Drop Chart,
formula (2) was used to compute the voltage drop for the vare
ious sizes of wire from No. 10 solid to 300,000 c.m stranded,
for power-factorsfrom unity to .6, lagging. Curves were then
drawn, one for each size of wire, with volts drop as abscissae
and powerfactor as ordinates. This was all done an the basis
of drop per single conductor, per 1000 fect, per ampere, at 60
cycles, with twenty-four inch spacing. Twenty-four inch space
ing was arbitrarily chosen as being a spacing (equivalent)
commonly used in low and medium voltage distribution.
An example of one calculation is given for illustrat-
ive purposes, using No.4 solid wire at 85% power-factor.
R=Resistance per 1,000' No. 4wire=.2530 ohms.
xX =—Reactance ‘per 1,000' No. 4 wire, 24" spacing=.1309 ohms.
(at 60 cycles)
Cons =. 85.
Sin(cos-+.85) =. 527.
Roosé =, 2150.
Xsinf =. 0690.
D=Reos6+Xsing=. 2840, which value is the
one plotted.

The wire data used was taken from the Standard Hand-
Book for Electrical Engineers, Fourth Edition. Resistance
values were taken at 25 degrees, Centigrade.

THE FREQUENCY AND SPACING CORRECTION CHART.

The principal drawback to the line drop chart is that
it provides for only one frequency and spacing. To overcome this
Objection, the Frequency and Spacing Correction Chart, Chart fF,
haga been drawn.

This chart is practically the same as that published
as a supplement to the article above refared to in the General
Electric Review of June, 1913, but is applied in a different
way. By means of it, the correction to be applied to the values
of Chart A, for other frequencies and spacings can readily be
determined,

Chart F consists essentially of set of diagrams by
means of which the last member of formula (2), Xsinf, can be
determined.
9

A resistance scale was laid off along the Y axis, and
the various sizes of wire up to No. 4 inclusive, were plotted on
this scale according to their respective resistances. The scale
was shortened for numbers 6, 8, and 10, to bring them within the
limits of the chart.

A scale of ohms reactance was laid off along the
negative X axis, the divisions being given the same value as
used for the resistance scale. A set of curves of reactance per
100 cycles was then plotted, one for each spacing. Then the proe
gection of the intersection of any one of these curves with a
size of wire line, upon the X axis, gives the reactance per 100
cycles for that size of wire and spacing.

Radial lines were then drawn from the origin, in the
left hand quadrant, at such an angle for each frequency, tit
the ordinate of any point is in the same proportion to the diste
ance along the radial line from the origin, as the frequency is
to 100 cycles. Arcs of circles with the origin as the center
were also drawn between the 100 cycle line, or X axis, and the
upper radial frequency line, to facilitate the transferring of
the reactance per 100 cycles to th other frequency lines. The
values of reactance for any frequency and spacing can then be
read from the Y axis, direct. |

Radial lines were next drawn from the origin, in the
right hand quadrant, one for each power-factor from .95 to .60,
changing by .05, in such a way that the vertical projection of
any point on one of these lines, upon a horizontal scale of ohms,

equal to the scale used for resistance and reactance, will equal
the horizontal projection of that point upon the Y axis, multie«
plied by sing. Values for Xsing can thus be read from the chart.

For instance, let it be desired to determine X sing
for No. 2 solid wire , at 53 ft. spacing, 25 cycles, and 80%
powere-factor. Project vertically from the intersection of the
No. 2 size line with the 60 inch spacing line, to the X axis,
which will give a value of .244 ohms reactance at 100 cycles.
Follow the circle from this point to the 25cycle line and pro-
ject horizontally to the 80% powerefactor line in the right hand
Quadrant. This will be at.063 on the Y axis. Then project vert-
ically to the X sin§ scale, which will give the value, X sing
equal to .038, the value for which we are solving.

It will be noticed that the spacing curves break where
th change is made from solid to stranded wire, all values
larger than No. 00 being stranded.

It will be observed in the use of the formula, D=Ir
cosé+IXsing, that when the frequency and spacing are changed,
only the tern, 1Xsing changes. Therefore to correct the curves
of the line drop chart for any different frequency and spacing,
it is only necessary to determine the difference in the tern,
tXsinf, due to the change., and apply that difference to the vale
ues of chart A by addition or subtraction as the cape may be.

For convenience, a table of values of Xsinf, for 24
inch spacingat 60 cycles, is given on chart F, so that to mke
a correction, it is only necessary to take the value for the
different frequency and spacing from the chart, using the tabu-

lar value for the other term of the correction equation.
THE LINE LOSS CHART,

 

The line loss chart, chart B, has been drawn to show
the line loss for the various sizes of wire at different power=
factors. From it can be read directly, the watts loss per single
conductor, per ampere at unity power-factor, at any desired powers

factor above .60. The expression, per ampere at unity powere
factor, needs some explanation. The values of curves at the point
of power-factor equal to 1 were calculated for a current of one
ampere per conductor. The values at the other powere-factors were
computed on the basis of the power remaining constant at the
value due to the one ampere current at unity power-factor, as
the power-factor decreases. The current in the line will theree
by increase, with consequent increase of loss, the power com-
ponent of current however remaining constant. This current will
vary inversely as the powerefactor, and since the loss varies
as the square of the current, it will also vary inversely as the
square of the power-factor.

The line loss chart then shows the variation of loss
under constant kilowatts load, as the power-factor varies. in
its application, if the real amperes~resultant of power and
wattless components- are known, the loss is read from the inter-
section of the size of wire line with the unity power-factor
line. If the power component of current is known, the loss is
read from the intersection of the size line with the line cor=
responding to the power-factor at which the calculation is being
made. To determine the loss for other loads than for that of one

ampere at unity power-factor, it is of course necessary to mule
tiply by the number of amperes flowing in the line.

Chart B then gives a ready comparison of the losses
for the various sizes of wire at different power-factors, and

a quick means of determining the loss for any condition.
THE ERROR IN THE LINE DROP CHART,

In the discussion of the line drop chart, the state-
ment was made that in most cases, the error involved by the use
of formula (2) is negligible. That this is really so is shown
by Chart C, which consists of curves of error and size of wire
for power=factors from 1.0 to .60, with D equal to 10% of E,.
The sizes of wire are laid off along the X axis so that the
error can be found for any one. The values of error were deter=
mined by means of the formula, :

% Error=100 | a ° 1/2, \/z8 - tan®(tan71x/R ~ A) 2 (3)

Referring again. to Figure 1], and remembering that

 

AB=>D=Drop,
Eo= generator voltage,
EB=receiver voltage,
it is readily seen that error=E, + (E+D). (4)
(R+D)2=2% - pet—x? = tan®(angle GaB)D? (5)
Angle GAB=angle GEH - angle JAH.
Angle GAB=tan"1x/R - f,
Substituting in (5), (E+D)4=—x* ~ tan2(tan71x/R = 6)D%,
E+D=\r4 - tan4(tan"1x/R - 6) D2,
Substituting in (4), error=Eo = a8 tan*(tan"1x/R - A )D&.
g error=error/E, (100).
& error=Eo/B,(100) = 100/B, \/z8 - tan®(tan~1x/R - 6)D* .
% error=100 [2 - 1/8 ,\/ 38 - tan*(tan~1x/R - ) D2 |.

 

 

 

 
Solutions of this formula were made for the various
sizes of wire and power-factors used in chart C, and the results
were plotted as shown by the curves. All calculations were made
for the standard conditions of spacing, frequency, etc. used in
chart A, and for a drop equal to 10% of the generator voltage:
To show the application of the formula, an example of the cal-
culation of error for No. 6 wire at a powerefactor of .70, with
generator voltage equal to 100 and drop equal to 10 volts, is
given.

% error=100 [1-2/200 \f008-tan?(tan-2. 13632/. 403-cos~1. 7)104 |.
=100 [1 - 1/100\/10,000 = tan4(tan-1, 338 = cos~4.7)100 |.
=100[1 « 1/100\10, 000 - tan*(18.68° -« 45. 58°)100 |.
=100 (1- 1/100V/10,000 - tan*( - 26. 90°) 100 .
=100 [2 - 1/100\f10, 000 - 5073(200) | .
=100[1 - 1/100\f0,000 = 25.75 | , ,
=100[1 + .1/100\f974. 25 | .
=100[1 - 1/100(99.875)] ,
=100(1 - .99875).

 

 

 

 

 

 

=100(. 00125).
% error=. 145.

Inspection of the formula shows that the per=-cent
error depends upon the size of the angle GAB, the angle between
voltage drop vector and the receiver voltage vector, and upon
the relation of D to BE When the ratio YE, is small, the error
is small. When the angle GAB is Zero, that is, when the imped-
ence or voltage drop vector is in line with the receiver voltage

vector, the error is zero. the curves show this latter point -
10

Where they are tangent to the line of Zero error.

The curves of chart C show that for most cases, the
error is less than 1%. At unity power-factor, for sizes larger
than No. 00, and at a power-factor of .95 for sizes larger than
350,000 circular mils, the error exceeds this value, and judge=-
ment will accordingly have to be used in the application of the
method by taking this Limitation into consideration.

In order to give some idea of the variation in error
for other drops than 10%, calculations have been made by the
same method for drops from 0 to 15%, for No. 10 wire at .7
power=factor, for No. 00 wire at the power-factor of 1.0, and
for 300,000 circular mil cable at .9 power-factor. These sizes
and power-factors were chosen as representing a range of error
at 10% drop up to about 1%. These results were put into curves
as shown by chart D. It will be noticed that for the first few
per-cent drop below 10, the error decreases quite rapidly.

THE EFFECT OF CAPACITY.

Another thing necessary to consider in transmission
line problems is the effect of line capacity. In most cases,
where the line is short, this effect can be neglected. The cap-
acity effect causes the voltage to rise at the receiver end of
the line, especialy at no load, where there is no lagging com-
ponent of voltage due to inductive drop to counteract the lead-
ing component due to the line capacity. This rise in voltage at
the end of the line at no load may be taken as a measure of the
capacity effect. The per=cent increase in voltage depends upon

the length of the line, the frequency, and the spacing.
Ll

The per-cent rise in voltage has been computed for
numbers 4 and O wire, and for 250,000 and 500,000 circular mil
cable, at 25 and 60 cycles, and for five foot equivalent spac@
ing, and for distances up to slightly above 300 miles. Curves
have been plotted showing these results in chart E,

The formula used in these calculations was published
by FP. W. Peek in the General Electric Review of June, 1913, in
the article entitled, Practical Calculations of Long Distance
Transmission Line Characteristics?’ The formula is,

EgeE (1+ Z¥/2)+ 23 -(1+-27/6) .

The nomenclature is as follows.

¥g—eenerator volts to neutral.

¥,-=receiver volts to neutral.

$,-=amperes per conductor at receiver end,

Z=impedence of one conductor=r = jx.

r=resistance of one conductor in ohms.

x=inductive reactance of one conductor to neutral, ohms.

CO=capacity in farads per single conductor to neutral.

=frequency in cycles per econd.

Y= - on ECs.

The dot beneath a symbol denotes a vector value. j is
the operator used to denote vector addition, being chosen to
represent lag when positive and lead when negative.

Since in this case, application is made of the formue
1a at no load only, Ip=0, and the last term of the second men-
ber=0.

The formula then becomes,
12

E,=E,(1+2Y/2),

or E fFp=1+2y/2,
Or B/ Fr - 1= Zy/2,
or Ep/Sr > Br/Br= (Bg = By)/Fr= 27/2,

or (B, - B,)/¥,(100) = 1002y/2=% increase in receiver volts,
or % increase in receiver volts=100ZY/2=100/2( r+ jx) (-2mfC).
An example of the calculation is given for No. O wire.
r per mile=.528. ohms.
x per mile=.746 ohms, with 5° spacing, at 60 cycles.
C per mile=(.01511) (1075) farads, with 5" spacing.
% increase =100/2(.528-+. 746) (-29760j.01511)1079.
=(.528-+. 746) (235. 71)10°4,
=(-3.015j - 4. 26)1074,
=107* \{-3.015)2+(=4. 26)2 .

=1074 \27.25 .

= (5. 22)1074,

 

—. 000522, for one mile.

Since Z and Y each vary as the distance and since the
per-cent rise in voltage varies as their product, the latter
will vary as the square of the distance.

Therefore the per-cent rise in voltage for 100 miles
in the above example for No. 0 wire is,.00052(100)”=5, 22.

The other values used in the plotting of the curves
were determined in the same way.

The curves of chart E then show that with the larger
sizes of wire and the lower frequencies, the capacity effect is

the least and can be neglected for longer distances. The actual
13

distances for which this may be neglected can be judged from
chart E, depending upon the per=-cent increase in voltage at no
load which is allowable without taking the capacity calculations
into consideration. Also, if the equivalent spacing is greater
than five feet, as used in the calculation of the chart, the
capacity effect will be correspondingly less and may be neglect-
ed for longer distances.

THE APPLICATION TO A PROBLEM,

In order that the application of the method to a prac-
tical proglem may be fully understood, an example is worked out
completely.

Let it be desired to determine the regulation and loss
of a three phase line consisting of three number O solid wires,
horizontally spaced with three and six feet between wires on a
standard ten foot transmission cross-arm. The line is 100 miles
long, operates at 25 cycles, at a voltage of 70,000 at the ree
ceiver end, at 80% power-factor, under a load of 6,000 kiloe
watts.

Equivalent spacing= \/x6x9 =—5.45 ft.

(Timbie & Higbie, A.C. Electricity, Course II, P. 343).

Amp. per conductor =6,000,000/70,000x1. 73x. 80=62,

Refering to chart A, the drop for No. O wire at 60
cycles and 24" spacing is .1522 volts per ampere paer 1,000 ft.
It is necessary to apply a correction to bring this to the con-
ditions of the problem, which is done by means of chart F.

Refering to the table accompanying chart F, Xsinf
14

per 1,000 ft. for No. O wire at 24" spacing, 60 cycles, and .80
power-factor, is .O722. Refering to the diagram, the intersec@-
tion of the No. O solid wire line with the 5.45 ft. spacing line
(taken between the 60" and 72" lines) falls at .24 minus ohms
reactance per 100 cycles. Following the circle from this point
to the 25 cycle radial line and projecting the intersection across
horiZontally to the 80% power-factor line, by projecting verti-
cally to the Xsinf scale, we read for Xsing per 1,000 ft. at
29 cycles, 5.45 ft. spacing and .80 powerefactor, .0364,.

The correction to be applied then is

-0358 - .0722=.0364, —

and the drop per ampere per 1,000 ft.under the conditions of the
problem is 01523 - .0364=.1159 volts.

Applying the factors for length of line and load,

drop=.1159 x 100 x 5.28 x 62=—3790 volts,
and % regulation=3790 x 1.73 x 100 / 70,000=9. 38.

Refering to chart B, the loss per 100 ft. per ampere
for No. O wire is .102 watts. The loss for the conditions of
the problem then will equal per conductor,

100 x 5.28 x 62°x .102=20. 7kw.
or 26.7 x 3 x 100 / 6,000=10,36,
or % loss=10. 36.

Now refering to chart C, it is seen that for No. 0
wire at .8 power-factor, the error is less than 03% and can
therefore be neglected. This error is of course for 10% drop,
which is very nearly that of the problem.

Refering next to chart C, it is seen that for No. 0
15

wire, the capacity effect for 100 miles would cause a rise in
voltage at the receiver end at no load (25 cycles) of about 2%,
which may be considered negligible. :

CONCLUSION.

In conclusion, it may be said that while these curves
and charts may not in every case provide the best solution of
the problem in hand, it is thot that they will offer an easily
applied method for the solhution of a large number of the probe
lens met with in every day practice.

The general survey afforded of the field of appli-
cation is considered a valuable asset of these charts for the
reason that one can often obtain from them by inspection, the
relative characteristics of different lines, and in many cases,
where accurate information is not required, the direct reading
from the chart will suffice even tho the conditions mgy differ

somewhat from those for which the charts were drawn.
 

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