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90
10EE53 TRANSMISSION
SUBJECT CODE: 10EE53
No. OF LECTURE HRS'/ WEEK: 04
TOTAL No. OF LECTURE HRS. : 52
AND DISTRIBUTION
IAMARKS: 25
EXAM HOURS: 03
EXAM MARKS: 100
PART-A
UNIT -1
TYPICAL TRANSMISSION & DISTRIBUTION SYSTEMS SCHEME- GEN~RAL
LAYOUTOF POWER SYSTEM, STANDARD VOLTAGES FOR TRANSMISSION, ADVANTAGES OF
HIGH VOLTAGE TRANSMISSION. TRANSMISSION LINE EFFICIENCY AND LINE DROP. FEEDERS,
DISTRIBUTORS & SERVICE MAINS.
. .5 HOURS
UNIT-2
OVERHEAD TRANSMISSION
LINES-TYPES OF SUPPORTING STRUCTURE'SAND LINE
CONDUCTORS USED. SAG CALCULATION-SUPPORTS AT SAME LEVEL AND AT DIFFERENT
LEVELS. EFFECT OF WIND AND ICE, SAG AT ERECTION, STRINGING CHART AND SAG
TEMPLATES. LINE VIBRATORS.
5 HOURS
UNIT-3
INSULATORS-INTRODUCTION,
MATERIALS USED, TYPES, POTENTIAL DISTRIBUTION OVER
A STRING OF SUSPENSION INSULATORS. STRING EFFICIENCY & METHODS OF INCREASING
STRINGS EFFICIENCY, GRADING RINGS AND ARCING HORNS. TESTING OF INSULATORS.
6 HOURS
UNIT-4
(A)CORONA-PHENOMENA,
DISRUPTIVE AND VISUAL CRITICAL VOLTAGES, CORONA
POWER LOSS. ADV ANTAGES AND DISADVANTAGES OF CORONA.
4 HOURS
(B) UNDERGROUND
CABLES-TYPES, MATERIAL USED, INSULATION RESISTANCE,
THERMAL RATING OF CABLES, CHARGING CURRENT, GRADING OF CABLES, CAPACITANCE
GRADING & INTER SHEATH GRADING, TESTING OF CABLES.
6 HOURS
PART-B
UNIT -5 AND 6
LINE PARAMETERS: CALCULATION OF INDUCTANCE OF SINGLE PHASE LINE, 3PHASE LINES
WITH EQUILATERAL SPACING, UNSYMMETRICAL SPACING, DOUBLE CIRCUIT AND
TRANSPOSED LINES. INDUCTANCE OF COMPOSITE CONDUCTOR LINES.
CAPACITANCE-OF SINGLE-PHASE LINE, 3PHASE LINES WITH EQUILATERAL SPACING,
UNSYMMETRICAL SPACING, DOUBLE CIRCUIT AND TRANSPOSED LINES. CAPACITANCE OF
COMPOSITE CONDUCTOR LINES.
12 HOURS
Dept ofEEE, SJBIT
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UNIT-7
PERFORMANCE OF POWER TRANSMISSION LINES-SHORT TRANSMISSION LINES, MEDIUM
TRANSMISSION LINES-NOMINAL T, END CONDENSER AND P MODELS, LONG TRANSMISSION
LINES, ABeD CONSTANTS OF TRANSMISSION LINES, FERRANTI EFFECT, LINE REGULATION.
8 HOURS
•
UNIT-8
DISTRIBUTION-REQUIREMENTS OF POWER DISTRIBUTION, RADIAL & RING MAIN SYSTEMS,
AC AND DC DISTRIBUTION:
CALCULATION FOR CONCENTRATED LOADS AND UNIFORM LOADING. 6 HOURS
TEXTBOOKS:
1. A COURSE IN ELECTRICAL POWER-SON! GUPTA & BHATNAAGAR, DHANP AT RAI &
SONS.
2. ELECTRICAL POWER SYSTEMS-Co
EDITION ,2009.
L. W ADHWA, NEW AGE INTERNATIONAL, 5TH
REFERENCE BOOKS:
1. ELEMENTS OF POWER SYSTEM ANALYSIS- ~.D. STEVENSON, TMH,4TH EDITION
2. ELECTRIC POWER GENERATION TRANSMISSlON & DISTRIBUTION-S. M. SINGH, PHI,2ND
EDITION, 2009.
.
3. ELECTRICAL POWER-DR. S. L. UPPAL, KHANNA PUBLICATIONS
..
Dept ofEEE, SJBIT
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SL.NO
CONTENTS
PAGE
1
UNIT-1 TYPICAL TRANSMISSION &
DISTRIBUTION
SYSTEMS SCHEMETYPICAL TRANSMISSION
&
DISTRIBUTION
SYSTEMS SCHEMETRANSMISSION LINE
GENERAL LAYOUT OF POWER
SYSTEM,
STANDARD VOLTAGES FOR TRANSMISSION,
6-16
No
ADVANTAGES OF HIGH VOLTAGE
TRANSMISSION.
EFFICIENCY AND LINE DROP.
FEEDERS,
DISTRIBUTORS & SERVICE MAINS.
UNIT2-
2
OVERHEAD TRANSMISSION
LINESTYPES OF SUPPORTING STRUCTURES AND
LINE CONDUCTORS USED.
SAG
CALCULATION-SUPPORTS AT SAME LEVEL
AND AT DIFFERENT LEVELS.
EFFECT OF WIND AND ICE,
17-38
SAG AT ERECTION,
STRINGING CHART AND SAG TEMPLATES.
LINE VIBRATORS.
UNIT 3-
3
INSULATORS
INTRODUCTION
39-47
MATERIALS USED, TYPES,
POTENTIAL DISTRIBUTION OVER A STRING
OF SUSPENSION
INSULATORS.
STRING EFFICIENCY & METHODS OF
INCREASING STRINGS EFFICIENCY,
GRADING RINGS AND ARCING HORNS.
TESTING OF INSULATORS.
4
UNIT-(A)CORONA
CABLES
PHENOMENA.
(B)UNDERGROUND
48-53
DISRUPTNE AND VISUAL CRITICAL
Dept of EEE,
SJBIT
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and Distribution
VOLTAGES.
CORONA POWER LOSS.
AovANTAGES AND DISADVANTAGES OF
CORONA.
TYPES, MATERIAL USED.
INSULATION RESISTANCE.
THERMAL RATING OF CABLES.
CHARGING CURRENT.
GRADING OF CABLES.
CAPACITANCE GRADING & INTER SHEATH
GRADING.
TESTING OF CABLES.
5
UNIT5AND6
54-81
..
LINE PARAMETERS: CALCULATION OF
..
INDUCTANCE OF SINGLE PHASE LINE.
3PHASE LINES WITH EQUILATERAL SPACING.
UNSYMMETRICAL SPACING.
DOUBLE CIRCUIT AND TRANSPOSED LINES.
INDUCTANCE OF COMPOSITE CONDUCTOR
LINES.
CAPACITANCE-OF SINGLE-PHASE LINE,
3PHASE LINES WITH EQUILATERAL SPACING.
UNSYMMETRICAL SPACING.
DOUBLE CIRCUIT AND TRANSPOSED LINES.
6
f--
CAP ACITANCE OF COMPOSITE CONDUCTOR
LINES.
UNIT 7
•
81-120
PERFORMANCE OF POWER TRANSMISSION
LINES-SHORT TRANSMISSION LINES.
MEDIUM TRANSMISSION LINES-NOMINAL T,
END CONDENSER AND P MODELS.
LONG TRANSMISSION LINES.
ABCD CONSTANTS OF TRANSMISSION
LINES.
FERRANTI EFFECT.
LINE REGULATION.
-
7
IEEE, SJBIT
UNIT
8
120-155
4
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DISTRIBUTION -REQUIREMENTS OF POWER
DISTRIBUTION.
RADIAL & RING MAIN SYSTEMS
-
AC AND DC DISTRIBUTION:
I
I
I
CALCULATION FOR CONCENTRATED LOADS
AND UNIFORM LOADING.
•
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UNIT -1
TYPICAL TRANSMISSION & DISTRIBUTION SYSTEMS SCHEME- STANDARD
VOLTAGES FOR TRANSMISSION. ADVANTAGE OF HIGH VOLTAGE
TRANSMISSION. FEEDERS, DISTRIBUTORS & SERVICE MAINS.
5 HOURS
•
INTRODUCTION
ELECTRICAL
ENERGY IS GENERA TED BY CONVERSION OF SOURCES:
WATER
FUEL
NUCLEAR
SUN
WIND
BIOMASS
OCEAN
TURBINES-MECHANICAL
TO ELECTRICAL
POWER PLANTS
NUCLEAR PLANT
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WIND POWER TOWERS
PLANT
SOLAR THERMAL
T&DININDIA
•
•
•
•
•
•
GENERATION HAS INCREASED BY 60 TIMES
THE FIRST 100KV LINE COMMISSIONED IN 1911
132 KV LINE IN 1931
220 KV LINE IN 1962
400 KV LINE IN 1978
765 KV LINE IN 2007-IN UP, PANJAB AND MP
•
COUNTRY IS DIVIDED INTO 5 REGIONS
NORTHERN
SOUTHERN
EACH WITH REGIONAL
M
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•
•
•
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EASTERN
ELECTRICITY BOARD
WESTERN
NORTH-EAST
CENTRAL GOVT. ALSO CONTROLS MANY GEN. PLANTS, TRN. LINES
AND SUBSTATIONS THROUGH
NTPC, NHPC, PGCIL ETC.
ADVANTAGES OF HIGH VOLTAGE TRANSMISSION
1.TRANSMISSION EFFICIENCY INCREAASES AS THE TRANSMISSION
VOL TAGE INCREASES
2.
FOR A GIVEN POWER,
P.U RESISTANCE DROP DECREASES
VOLUME OF CONDUCTOR MATERIAL REQUIRED REDUCES
SINCE CURRENT DECREASES WITH INCREASE IN VOLTAGE,
3. POWER TRANSMITTING CAPACITY OF THE TRAMN. LINE IS
PROPORTIONAL TO SQUARE OF THE OPERATING VOLTAGES.
THEREFORE, OVERALL CAPITAL COST DECREASES.
4, COST OF TRANSMISSION LINE PER KM DECREASES WITH INCREASE IN
VOLTAGE LEVEL.
5. WITH INCREASE IN VOLTAGE, SIL(SURGE IMPEDANCE LOADING) OF THE
LINE INCREASES - SO, POWER TRANSFER INCREASES.
HOW DOES POWER REACH US?
.'
•
•
•
•
ELECTRIC POWER IS NORMALLY GENERATED AT llKV IN A POWER STATION.
To TRANSMIT OVER LONG DISTANCES, IT IS THEN STEPPED-UP TO 400KV, 220KV
OR 132KV AS NECESSARY.
POWER IS CARRIED THROUGH A TRANSMISSION NETWORK OF HIGH VOLTAGE
LINES.
USUALLY, THESE LINES RUN INTO HUNDREDS OF KILOMETRES AND DELIVER
THE POWER INTO A COMMON POWER POOL CALLED THE GRID.
THE GRID IS CONNECTED TO LOAD CENTRES (CITIES) THROUGH A SUBTRANSMISSION NETWORK OF NORMALLY 33KV (OR SOMETIMES 66KV) LINES.
Dept of EEE, SJBIT
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THESE LINESTERMINATEINTO A 33KV (OR 66KV) SUBSTATION,WHERE THE
VOLTAGEIS STEPPED-DOWNTO llKVFORPOWERDlSTRIBUTION TO LOAD
POINTS THROUGHA DISTRIBUTIONNETWORKOF LINESAT llKV ANDLOWER.
TRANSMISSION
•
•
A TRANSMISSION GRID IS A NETWORK OF POWER STATIONS,
TRANSMISSION CIRCUITS, AND SUBSTATIONS.
ENERGY IS USUALLY TRANSMITTED WITHIN THE GRID WITH
THREE-PHASE AC.
TRANSMISSION
SYSTEM
BOOkV,400kV lines (Extra High Voltage)
132kV lines (High Voltage)
llkV feeders
(Distribution
network)
11kV/41SV
Distribut:ion
TranSformer on pole
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WHAT IS POWER TRANSMISSION?
•
•
A PROCESS IN THE DELIVERY OF ELECTRICITY TO CONSUMERS,
IS THE BULK TRANSFER OF ELECTRICAL POWER.
POWER TRANSMISSION IS BETWEEN THE POWER PLANT AND A
SUBSTATION NEAR A POPULATED AREA.
ELECTRIC POWER TRANSMISSION ALLOWS DISTANT ENERGY
SOURCES (SUCH AS HYDROELECTRIC POWER PLANTS) TO BE
CONNECTED TO CONSUMERS IN POPULATION CENTERS.
!){STRIBUTION
WHA T IS DISTRIBUTION?
ELECTRICITY DISTRIBUTION IS THE DELIVERY FROM THE
SUBSTATION TO THE CONSUMERS.
•
ELECTRICITY IS USUALLY TRANSMITTED OVER LONG DISTANCE
THROUGH OVERHEAD POWER TRANSMISSION LINES.
UNDERGROUND POWER TRANSMISSION IS USED ONLY IN DENSELY
POPULATED AREAS DUE TO ITS HIGH COST OF INSTALLATION AND
MAINTENANCE,
A POWER TRANSMISSION SYSTEM IS REFERRED TO AS A "GRID";
A TRANSMISSION GRID IS A NETWORK OF POWER STATIONS,
TRANSMISSION CIRCUITS, AND SUBSTATIONS. ENERGY IS USUALLY
TRANSMITTED WITHIN THE GRID WITH THREE-PHASE AC.
A SUBSTATION IS A HIGH-VOLTAGE ELECTRIC SYSTEM FACILITY.
IT IS USED TO SWITCH GENERATORS, EQUIPMENT, AND CIRCUITS
OR LINES IN AND OUT OF A SYSTEM.
;:
•
IT ALSO IS USED TO CHANGE AC VOLTAGES FROM ONE LEVEL TO
ANOTHER, AND/OR CHANGE ALTERNATING CURRENT TO DIRECT
,;'UPP.ENT OR DIRECT CURRENT TO ALTERNATING CURRENT.
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SUBSTATION TYPES
•
AL THOUGH, THERE ARE GENERALLY FOUR TYPES OF
SUBSTATIONS THERE ARE SUBSTATIONS THAT ARE A
COMBINATION OF TWO OR MORE TYPES.
- STEP-UP TRANSMISSION SUBSTATION
STEP-DOWN TRANSMISSION SUBSTATION
- DISTRIBUTION SUBSTATION
_ UNDERGROUND DISTRIBUTION SUBSTATION
- SUBSTATION FUNCTIONS
- SUBSTATION EQUIPMENT
STEP-UP TRANSMISSION SUBSTATION
•
A STEP-UP TRANSMISSION SUBSTATION RECEIVES ELECTRIC
POWER FROM A NEARBY GENERATING FACILITY AND USES A
LARGE POWER TRANSFORMER TO INCREASE THE VOLTAGE FOR
TRANSMISSION TO DISTANT LOCATIONS.
Dept ofEEE, SJBIT
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•
•
•
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A TRANSMISSION BUS IS USED TO DISTRIBUTE ELECTRIC POWER
TO ONE OR MORE TRANSMISSION LINES. THERE CAN ALSO BE A
TAP ON THE INCOMING POWER FEED FROM THE GENERATION
PLANT TO PROVIDE ELECTRIC POWER TO OPERATE EQUIPMENT
IN THE GENERATION PLANT.
A SUBSTATION CAN HAVE CIRCUIT BREAKERS THAT ARE USED TO
SWITCH GENERATION AND TRANSMISSION CIRCUITS IN AND OUT
OF SERVICE AS NEEDED OR FOR EMERGENCIES REQUIRING SHUTDOWN OF POWER TO A CIRCUIT OR REDIRECTION OF POWER.
THE SPECIFIC VOLTAGES LEAVING A STEP-UP TRANSMISSION
SUBSTATION ARE DETERMINED BY THE CUSTOMER NEEDS OF THE
UTILITY SUPPLYING POWER AND TO THE REQUIREMENTS OF ANY
CONNECTIONS TO REGIONAL GRIDS.
TYPICAL VOLTAGES ARE:
HIGH VOLTAGE (HV) AC:69 KV, 115 KV, 138 KV, 161 KV, 230 KV
EXTRA-HIGH VOLTAGE (EHV) AC:345 KV, 500 KV, 765 KV
ULTRA-HIGH VOLTAGE (UHV) AC:1100 KV, 1500 KV
DIRECT-CURRENT HIGH VOLTAGE (DC HV): ±250 KV, ±400 KV, ±500
KV
DIRECT CURRENT VOLT AGE IS EITHER POSITIVE OR NEGATIVE
POLARITY. A DC LINE HAS TWO CONDUCTORS, SO ONE WOULD BE
POSITIVE AND THE OTHER NEGATIVE.
HVDC TRANSMISSION
•
•
;;-
•
•
HIGH VOLTAGE DIRECT CURRENT (HVDC) IS USED TO TRANSMIT
LARGE AMOUNTS OF POWER OVER LONG DISTANCES OR FOR
INTERCONNECTIONS BETWEEN ASYNCHRONOUS GRIDS.
WHEN ELECTRICAL ENERGY IS REQUIRED TO BE TRANSMITTED
OVER VERY LONG DISTANCES, IT CAN BE MORE ECONOMICAL TO
TRANSMIT USING DIRECT CURRENT INSTEAD OF ALTERNATING
CURRENT.
FOR A LONG TRANSMISSION LINE, THE VALUE OF THE SMALLER
LOSSES, AND REDUCED CONSTRUCTION COST OF A DC LINE, CAN
OFFSET THE ADDITIONAL COST OF CONVERTER STATIONS AT
EACH END OF THE LINE.
ALSO, AT HIGH AC VOLTAGES SIGNIFICANT (ALTHOUGH
ECONOMICALL Y ACCEPT ABLE) AMOUNTS OF ENERGY ARE LOST
DUE TO CORONA DISCHARGE, THE CAPACITANCE BETWEEN
PHASES OR, IN THE CASE OF BURIED CABLES, BETWEEN PHASES
AND THE SOIL OR WATER IN WHICH THE CABLE IS BURIED.
Dept ofEEE, SJBIT
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A SUBSTATION
ADVANTAGES OF USING HIGH VOLTAGE FOR TRANSMISSION
•
•
•
TRANSMISSION EFFICIENCY IS IMPROVED BY INCREASING THE
VOLTAGE USING A STEP-UP TRANSFORMER, WHICH REDUCES THE
CURRENT IN THE CONDUCTORS, WHILE KEEPING THE POWER
TRANSMITTED NEARLY EQUAL TO THE POWER INPUT.
THE REDUCED CURRENT FLOWING THROUGH THE CONDUCTOR
REDUCES THE LOSSES IN THE CONDUCTOR AND SINCE,
ACCORDING TO JOULE'S LAW, THE LOSSES ARE PROPORTIONAL
TO THE SQUARE OF THE CURRENT, HALVING THE CURRENT
MAKES THE TRANSMISSION LOSS ONE QUARTER THE ORIGINAL
VALUE.
DC SYSTEMS REQUIRE RELATIVELY COSTLY CONVERSION
EQUIPMENT WHICH MAY BE ECONOMICALLY JUSTIFIED FOR
PARTICULAR PROJECTS.
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SINGLE PHASE AC IS USED ONLY FOR DISTRIBUTION TO END USERS
SINCE IT IS NOT USABLE FOR LARGE POLYPHASE INDUCTION
MOTORS
TRANSMISSION TOWERS
;:
FEEDERS, DISTRIBUTORS AND SERVICE MAINS
•
•
A DISTRIBUTION SYSTEM CAN BE SUBDIVIDED INTO:
FEEDERSCONDUCTORS OF LARGE CURRENT CARRYING CAPACITY
CARRY CURRENT TO THE FEEDING POINTS IN BULK
• DISTRIBUTORSCONDUCTORS FROM WHICH CURRENT IS TAPPED OFF FOR SUPPLYING
THE CONSUMER
• SERVICE MAINSSMALL CABLES BETWEEN THE DISTRIBUTORS AND THE
CONSUMERS PREMISES
PI CAL DISTRIBUTION SYSTEM
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5 RVICE AINS
GENERATING STATION
OR SUBSTATION
TYPICAL DISTRIBUTION SYSTEM
•
THE POWER NETWORK, WHICH GENERALLY CONCERNS THE
COMMON MAN, IS THE DISTRIBUTION NETWORK OF llKV LINES OR
FEEDERS DOWNSTREAM OF THE 33KV SUBSTATION.
EACH llKV FEEDER WHICH EMANATES FROM THE 33KV
SUBSTATION BRANCHES FURTHER INTO SEVERAL SUBSIDIARY
llKV FEEDERS TO CARRY POWER CLOSE TO THE LOAD POINTS
(IJOCALITIES, INDUSTRIAL AREAS, VILLAGES, ETC.,).
•
AT THESE LOAD POINTS, A TRANSFORMER FURTHER REDUCES THE
VOLTAGE FROM llKVT0415VTOPROVIDE
THE LAST-MILE
CONNECTION THROUGH 415V FEEDERS (ALSO CALLED AS LOW
TENSION (LT) FEEDERS) TO INDIVIDUAL CUSTOMERS, EITHER AT
!)V(AS SINGLE-PHASE SUPPLY) OR AT 415V (AS THREE-PHASE
.:..;
L)1)PLY).
Dept of EEE, SJBIT
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A FEEDER COULD BE EITHER AN OVERHEAD LINE OR AN
UNDERGROUND CABLE. IN URBAN AREAS, OWING TO THE DENSITY
OF CUSTOMERS, THE LENGTH OF AN llKV FEEDER IS GENERALLY
UPT03KM.
ON THE OTHER HAND, IN RURAL AREAS, THE FEEDER LENGTH IS
MUCH LARGER (UP TO 20 KM). A 41SV FEEDER SHOULD NORMALLY
BE RESTRICTED TO ABOUT 0.5-1.0 KM. UNDULY LONG FEEDERS
LEAD TO LOW VOLTAGE AT THE CONSUMER END.
i
ADVANTAGES OF HIGH VOLTAGE TRANSMISSION
1. TRANSMISSION EFFICIENCY INCREAASES AS THE TRANSMISSION
VOLTAGE INCREASES
2.
FOR A GIVEN POWER,
P.U RESISTANCE DROP INCREASES
- VOLUME OF CONDUCTOR MATERIAL REQUIRED REDUCES
SINCE CURRENT DECREASES WITH INCREASE IN VOLT AGE,
3. POWER TRANSMITTING CAPACITY OF THE TRAMN. LINE IS
PROPORTIONAL TO SQUARE OF THE OPERATING VOLTAGES.
THEREFORE, OVERALL CAPITAL COST DECREASES.
4. COST OF TRANSMISSION LINE PER KM DECREASES WITH INCREASE IN
VOL TAGE LEVEL.
5. \VITH INCREASE IN VOLTAGE, SIL(SURGE IMPEDANCE LOADING) OF
THE LINE INCREASES - SO, POWER TRANSFER INCREASES.
COZ, PSIL = V2/Z0
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UNIT-2
OVERHEAD TRANSMISSION LINES- SAG CALCULATION IN CONDUCTORS A)
SUSPENDED ON LEVEL SUPPORTS B) SUPPORT AT DIFFERENT LEVELS. EFFECT OF WIND & ICE
TENSION.
5 HOURS
INTRODUCTION
Conductor sag and tension analysis is an important consideration in overhead
distribution line design as well as in overhead transmission line design. The quality
and continuity of electric service supplied over a line (regardless of whether it is a
distribution, a sub-transmission, or a transmission line) depend largely on whether,
the conductors have been' properly installed. Thus, the designing engineer must
determine in advance the amount of sag and tension to be given the wires or cables
of a particular line at a given temperature. In order to specify the tension to be used
in stringing the line conductors, the values of sag and tension for winter and
summer conditions must be' known .. Tension in the conductors contributes to the
mechanical load on structures at angles in the line and at dead ends. Excessive
tension may cause mechanical failure of the conductor itself.
The factors affecting the sag of a conductor strung between supports arc:
1. Conductor load per unit length.
2. Span, that is, distance between supports.
3. Temperature.
4. Conductor tension.
in order to determine the conductor load properly, the factors that need to be taken
into account are:
1. Weight of conductor itself.
2. Weight of ice or snow clinging to wire.
3. Wind blowing against wire.
The maximum effective weight of the conductor is the vector sum of the vertical weight and
the horizontal wind pressure. It is very important to include the most adverse condition. The
wind is considered to be blowing at right angles to the line and to act against the projected
area of the conductor, including the projected area of ice or snow that may be clinging to it.
Economic design dictates that conductor sag should be minimum to refrain from extra pole
height, S8 provide sufficient clearance above ground level, and to avoid providing excessive
horizontal spacing between conductors to prevent them swinging together in mid-span.
Conductor tension pulls the conductor up and decreases its sag. At the same time, tension
elongates the conductor, from elastic stretching, which tends to relieve tension and increase
sag. The elastic property of metallic wire is measured by its modulus of elasticity. The
Dept ofEEE, SJBIT
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modulus of elasticity of a material equals the stress per unit of area divided by the
deformation per unit of length. That is, since
T
u= A
pSI
where u a=stress per unit area in pounds per square inches
T = conductor tension in pounds
A = actual metal cross section of conductor in square inches
The resultant elongation e of the conductor due to the tension is
stress
e ~ modulus or elasticity
Of course, if the modulus of elasticity is low, the elongation is high, and vice versa. Thus, a
small change in conductor length has a comparatively large effect 'on conductor sag and
tension. Sags and stresses in conductors are dependent on the initial tension put on them
when they are clamped in place and are due to the weight of the conductors themselves, to
ice or sleet clinging to them, and to wind pressure.
The stress in the conductor is the parameter on which everything else is based. But the
stress itself is determined by the sag in the conductor as it hangs between adjacent poles or
towers. Since the stress depends on sag, any span can be used provided the poles or towers
are high enough and strong enough. The matter is merely one of extending the catenary in
both loading. Thus, the problem becomes the balancing of a larger number of lighter and
shorter poles or towers against a smaller number of heavier and taller ones.
LINE SAG AND TENSION CALCULATIONS
A conductor suspended freely from two supports, which are at the same level and
spaced L unit length apart, as shown in Figure 2 takes the form of a catenary curve
providing the conductor is perfectly flexible and its weight is uniformly distributed
along its length. if the conductor is tightly stretched (i.e., when sag d is very small in
comparison to span L), the resultant curve can be considered a parabola. If the
conductor's sag is less than 6 percent of its span length, the error in sag computed
by the parabolic equations is less than 0.5 percent. If the conductor's sag is less than
10 percent of the span, the error is about 2 percent.
In distribution systems, determining accurate values of sag is not so important as it
is in transmission systems. Nevertheless, even in the distribution lines, if the
conductor is strung with too low tension, the resultant sag will be excessive, with
the likelihood of wires swinging together and short-circuited. The usual tendency,
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however, is to pull the conductor too tight, which causes the conductor to be
overstressed and stretched when the heaviest loading takes place and the normal sag
after this loading becomes excessive. Then the excessive sag needs to be pulled out
of the conductor, a process that also causes the conductor to be overstressed on
heaviest loading. This process of overstressing and pulling up may cause the
conductors, especially the smaller ones, to he broken. This can be eliminated by
measuring the line tension more accurately.
•
Supports at Same Level
catenary Method
Figure 1 shows a span of conductor with two supports at the same level and
separated by a horizontal distance L.
_
Let '0' be the lowest point on the catenary curve and '1' be the length of the
conductor- between two supports.
Let 'w' be the weight of the conductor per unit length, 'T' be the tension of the
conductor at any point 'P' in the direction of the curve, and Hbe the tension at
origin O.
Further, let's' be the length of the curve between points 0 and P, so that the weight
of the portion s is ws.
Tension T can be resolved into two components, Tthe horizontal component and
T" the vertical component. Then, for equilibrium,
T1 ~ Hand
TY ~ WS
Thus, the portion OP of the conductor is in equilibrium under the tension Tat P. the
weight Ws acting vertically downward, and the horizontal tension H.
In the triangle shown in Figure 2, ds represents a very short portion of the
conductor, in the region of point P When s is increased by ds the corresponding x
and yare increased by dx and dy, respectively.
Hence,
Dept ofEEE, SJBIT
20
--------_.
-~--,---------------------
-
Transmission and Distribution
--
10EES3
I
I
v
~~~_j
Figure 1. Conductor
. , Xl
SJBIT
suspended between supports at same elevation.
21
10EE53
Transmission and Distribution
dy
ws
tan () = -. =-.
dx
H
smce
.tben
((is)2
dx
= 1+
(WS.)2
If
Therefore,
dx ==
..
tis
VI + (wsIH)'1
Figure 2
Integrating both sides gives
. 1
- j Vi + (wsIH)2
r=
.
ds
Therefore,
Dent c f EEE, SJBIT
22
.msmission and Distribution
d
--;".
10EE53
.' _'·;.z;·;....... iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
.
x=
Whc: " ,\
i;; the
H
°
w Sinh
-1
w
v' "ij, -
).
2'
+
K
constant of integration. When x = 0, $ = 0, and K = 0,
s ......H
'1.1 t: .,"
Y' lJ', •.h
ws
H
°
SIn
h -wx
H
L•
I
.. S
= :2 =
H.
W
wL
sinh 21]
.' J ='-$10'
. 2H inh -wL
,
'w
2H
, ....roximately
'-
'I..
1)
w2L
t::::L ( 1+
2.
24H
. , ., .:8IT
23
Transmission
1DEES3
and Distribution
dy ~ ws = sinh ~
dx
H·····
H
or
dy~sm". h wx
H dx.,
Jntegrating both sides,
y ==
J sinh j;
dx
H
wx
cos
-'_'
+ KI
w
H
y== -
I f the lowest pain t of the curve is taken as the origin,when x=O, y = 0,
then KI = Hlw,
=
since, by the series, coshO=1. Therefore,
H (
wx
)
.y = -w cosh--H - 1
is the equation of the curve that is called a catenary .
Parabolic Method
In the case of short spans with small sags, the curve can be considered as a parabola.
When the horizontal tension is the same, the radius of curvature at the lowest point
of the conductor is the same for both the parabola and the catenary. but the outlines
of the two curves are different at all points between the lowest point of the curve
:'~.id the point of support. As the span length and sag are increased, the difference in
outline of the two curve becomes more significant. Changes in the loading will
produce different changes in the length of the two curves to make the sags different.
Since the sag by the parabola solution is smaller than the sag by the catenary
solution for the same horizontal tension, the angle ()will be smaller. Thus, the
vertical component of tension is smaller for the parabola solution than for the
Dept of ;':EE, SJBIT
24
10EE53
Transmission and Distribution
catenary. This difference increases along the curve toward the support, becoming
maximum at the supports. However, for the sake of simplicity, the following
assumptions are made:
1. The tension is considered uniform throughout the span, the slight excess of
tension at the supports over that in the middle being neglected.
2. The change in length of the conductor due to elastic stretch or temperature
expansion is taken as equal to the change of length of a conductor equal in
length to the horizontal distance between the points of supports .
.In Figure 3, let P be any point on the parabolic curve such that arc OP is equal to x.
The portion OP of the conductor is in equilibrium under the action ofT, H. and wx.
As previously done, the tension T can be resolved into two components, T, and Ty.
Theq, for equilibrium,
T', == Hand
T.,. = wx
Figure 3. Parameters of parabola.
TAKING MOMENTS ABOUT P.
moments clockwise= moments anticlockwise
or
Hy
x
= WX?
...
2
y:;; _WX.•
2H
Distribution lines usually have comparatively short spans with small sag. The difference
between the maximum tension T and the horizontal tension H is relatively small because of
Dept of EEE, SJBIT
25
Transmission and Distribution
10EES3
short spans and small sags. Under such conditions, a slight error will result if T is
substituted for H in the equation for sag. Therefore,
wx2
2T
y=-. When,
x=!L
,
And y is equal to sag or deflection d; therefore, the sag is
2
d= wL
8T
1=
L(l + ~1:)
EXAMPLE 10.1
A sub-transmission line conductor has been suspended freely from two towers and has
taken the form of a catenary that has c = 1600 ft. The span between the two towers is 500 It,
and the weight of the conductor is 4122 Ib/mi.
Calculate the following:
(a) Length of conductor
(b) Sag.
(c) Maximum and minimum values of conductor tension using catenary method.
(d) Approximate value of tension by using parabolic method.
Dept of EEE, SJBIT
26
Tra: smission and Distribution
10EE53
iiiiiiiiiiiiiiiiiiiiii"""",,~' Oiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii~
Solutioi.
(a) C,i:',g equation (10.6),
lH, inh _,
wt.
I =~Sln
w
2H
or
I=
2C( sinh ~)
H
c=~
w
e-r-
j;
1 nererore,
I :;.2 x 1600 . h.
sm
500
2 X 1600
= 3200 sinh 0.15625
= 502.032
'i
ft
equation (10.8),
1= L(1 + ~;;:)
1= L(l +
= 500(1
2~;2)
+
24
= 502.0345 ft
5002
X
)
16002
U·) thing equation (lO.16L
H ( cosh --wt: - 1)
d::=: _,
2H
w
d= C(COSh L -1)
2c
c=
if
-w
27
10EES3
Transmission and Distribution
Therefore,
d = 1600(cosh 2
:i~oo-
1)
= 1600(cosh 0.15625 - 1)
=- 19.6 ft
(c) Using equation (10.33),
Tmax ;;;;
w( c + d)
4)22
= 52sO (1600
+ 19.6)
-== 1264.4 lb
Using equation (10.35).
= 4122 x 1600
5280
::' 1249.llb
(d) From equation (10.46),
WL2
T=·_·· - .
3d
= (4122/5280) x 5002
8 x 19.6
--1244.7Ib
Supports at Different Levels: Unsymmetrical
Spans
Consider a span
L between two supports, as shown in Figure 10.5, whose elevations
differ by a distance h. Let the horizontal distance from the lowest point of the curve to
the lower and the higher supports be xl and x2, respectively.
By using equation (I 0.46), that is,
Dept of EEE, SJBIT
28
10EES3
Transmission and Distribution
wx2
y=_._
2T
d I and d, sags can be found as
and
B
h
L
Figure 4. Supports at different levels.
29
10EE53
Transmission and Distribution
Therefore,
or
or
since
L = XI
+ X2
Therefore,
2Th
wL
=X2
~
Xl
By adding the above two equations,
2Th
2x2= L + -.-
wL
= -L2
X2
Dept of EEE, SJBIT
Th
wL
+~-
30
,,
33
; ii, SJBIT
1DEES3
Transmission and Distribution
B
Figure 5. Case of negative x,
On Subtracting the same two equations,
2x
=i
L _ 2Th
wL
J
L
Th
x=---1
In
2
wl:
then
Xt
uation,
;:L-:'
is positive
then ..l'l is zero
then x 1 is negative
iL'gative, the lowest point (the point 0) of the imaginary curve lies outside
:-<,~_~f
span, as shown in Figure 5
., i
r
j
'l .. SJBIT
31
Dept ofEEE, SJBIT
Transmission and Distribution
34
10EE53
10.5 EFFECTS OF ICE AND WIND LOADING
The span design consists in determining the sag at which the line is constructed so that
heavy winds, accumulations of ice or snow, and excessive temperature changes will not
stress the conductor beyond its elastic limit, cause a serious permanent stretch, or result in
fatigue failures from continued vibrations, in other words, the lines will be erected under
warmer and nearly still-air conditions and yet must comply with the worst conditions.
10.5.1 Effect of Ice
In mountainous geographic areas, the thickness of ice formed on the conductor becomes
very significant. Depending on the circumstances, it might be as much as several times the
diameter of the conductor. Ice accumulations on the conductor affect the design of the line
(1) by increasing the dead weight per foot of the line and' .
(2) by increasing the projected surface of the line subject to wind pressure.
Mostly used for distribution lines.
Figure 6. Probable configuration of ice covered conductor cross-sectional area.
Even though the more likely configuration of a conductor with a coating of ice is as shown
in Figure 6, for the sake of simplicity, it can be assumed that the ice coating, of thickness t,
inches, is uniform over the surface ofa conductor, as shown in Figure 7.
Then the cross-sectional area of the ice is
Dept ofEEE, SJBIT
32
Transmission and Distribution
iDEES3
for a I-f: : »gth of conductor,
Swi = de ~ 2tr I
ft2/ft
Therefore, the horizontal force exerted on the line as a result of the wind pressure with no
ice (Figure 8) is
•
for a I Jt L.T;,thof conductor,
where
r
\:,rizontal wind force (i.e., load) exerted on line in pounds per feet
ind pressure in pounds per square feet
whereas with ice (Figure 9), it is
Figure 9. Force of wind on conductor covered with ice.
FIGURE
•
10
P == SWiP lb/unit length
i.>
\
!::
:-JGTH OF CONDUCTOR,
<:jBIT
35
10EE53
Transmission and Distribution
p_ de +21,
-
12
P
Ib/ft
THEREFORE, THE EFFECTNE LOAD ACTING ON THE CONDUCTOR IS
W..
=:
V pZ + (w + W,)Z
Ib/ft
ACTING AT AN ANGLE e TO THE VERTICAL, AS SHOWN IN FIGURE
10.
By REPLACING W BY WE IN THE PREVIOUSLY DERIVED EQUATIONS FOR TENSION AND SAG
OF THE LINE IN STILLAIR, THESE EQUATIONS CAN HE APPLIED:rO A WIND- AND ICE-LOADED
LINE. FOR EXAMPLE, THE SAG EQUATION BECOMES
d=
W~L2
8T
ft
EXAMPl.F.lO.2
A STRESS-CROSSING OVERHEAD SUB-TRANSMISSION qNE HAS A SPAN OF 500 FT OVER THE
STREAM. THE LINE IS LOCATED IN A HEAVY-LOADING DISTRICT IN WHICH THE HORIZONTAL
WIND PRESSURE IS 4 LB/FT2 AND THE RADIAL THICKNESS OF THE ICE IS 0.50 IN. USE AN
ACSR CONDUCTOR OF 195 KCMIL HAVING AN OUTSIDE DIAMETER OF 1.093 IN., A WEIGHT
OF 5399 LB/MI. AND AN ULTIMATE STRENGTH OF 28,500 LB. ALSO USE A SAFETY FACTOR
OF 2 AND 57 LB/FT3 FOR THE WEIGHT OF ICE. USING THE PARABOLIC METHOD, CALCULATE
THE FOLLOWING:
(A) WEIGHT OF ICE IN POUNDS PER FEET.
(H) TOTAL VERTICAL LOAD ON CONDUCTOR IN POUNDS PER FEET.
(C) HORIZONTAL WIND FORCE EXERTED ON LINE IN POUNDS PER FEET.
(D) EFFECTIVE LOAD ACTING ON CONDUCTOR IN POUNDS PER FEET.
(E) SAG IN FEET
Dept of F,EE, SJBIT
36
Transmission and Distribution
10EE53
Solution
(a) Using equation (10.63),
w, = l.25t,(dc + t
l)
= 1.25 x 0.50(1.093 + 0.50)
;=: 0.9956
..
(hi
lb
sing equation (10.64)
•
WT=,W+
Wi
The weight of the conductor is
W
= 5399 1b/mi
1
~
w = 52,805399 = l.02251blft
Therefore.
Wr = 1.0225 + 0.9956
"'"2.01811b/ft
(c) From equation (10.72),
d; + 2t .
•
37
8£
lJ Zp'v
£~£['Z,
.
[810'(: x 89
(} soo
=
r=
p
= :iJp,s rU:>fl.l~A (J)
~ L·O~·~un6t.:l
r Hlq( csst '2:
I
I
U1ql £t.69 0
OS;Z'PI x 8
~OO~x £~£l' z
-
LS~ ._
."
- t:
z7
.1>1·
,
'(teOI) uonenba
WOld
qI o~ttI = . Z ... = J.. (a)
OOS"8Z
ZrOI
~ln1Jf.1U! UA\oqs su
lJ/qf £~£I'Z =
{STOOl + ''lLL69°0 j\
l;
if'M
=
+ M) + zd A = ..M
~)JnjH pUR (£L"Or) uonenbo gU!Sn(p)
'tror
lI/q( LL69·0 =
t X ZI.
O~·O
£S3301
X
Z + £60· I =
Transmission and Distribution
iDEES3
UNIT-3
CORONA- PHENOMENA, EXPRESSION FOR DISPUTATIVE & VISUAL CRITICAL VOLTAGES &
CORONA POWER LOSS
4 HOURS
.,
"
Introduction
Corona on transmission lines causes power loss, radio and television interference, and
audible noise (in terms of buzzing, hissing, or frying sounds) in the vicinity of the line. At
extra high-voltage levels (i.e., at 345 kV and higher), the conductor itself is the major
source of audible noise, radio interference, television interference, and corona' loss. The
audible noise is a relatively new environmental concern and is becoming more important
with increasing voltage level. For example, for transmission lines up to 800 kV, audible
noise and electric field effects have become major design factors and have received
considerable testing and study. It had been observed that the audible noise from the corona
process mainly takes place in foul weather. In (Icy conditions, the conductors normally
operate below the corona detection level, and therefore, very few corona sources exist. In
wet conditions, however, water drops on the conductors cause large number of corona
discharges and a resulting burst of noise. At ultrahigh-voltage levels (1000 kV and higher),
such audible noise is the limiting environmental design factor.
Phenomena
.,
Succinctly put, corona is a luminous partial discharge due to ionization of the air
surrounding a conductor caused by electrical overstress. Many tests show that dry air at
normal atmospheric pressure and temperature (25°C and 76 em barometric pressure)
breaks down at 29.8 kV/cm (maximum, or peak, value) or 21.1 kV/cm (rms, or effective,
value). There are always a few free electrons in the air due to ultraviolet radiation from the
sun, cosmic rays from outer space, radioactivity of the earth, etc. As the conductor becomes
energized on each half cycle of the ac voltage wave, the electrons in the air near its surface
are accelerated toward the conductor on its positive half cycle and away from the conductor
on its negative half cycle. The velocity attained by a free electron is dependent on the
Intensity of the electric field. If the intensity of the electric field exceeds a certain criti cal
value, any free electron in this field will acquire a sufficient velocity and energy to knock
one of the outer orbit electrons clear out of one of the two atoms of the air molecule. This
process is called ionization, and the molecule with the missing electron is called a positive
on. The initial electron, which lost most of its velocity in the collision, and the electron
According to Nasser, "coronas have various industrial applications, such as in high-speed
printout devices, in air purification devices by electronic precipitators, in dry-ore separation
systems, as chemical catalysts, in radiation detectors and counters, and in discharging
undesirable electric charges from airplanes and plastics. Coronas are used as efficient means
1\:pt ofEEE, SJBIT
39
Transmission and Distribution
10EE53
of discharging other statically electrified surfaces of wool and paper in the manufacturing
industry.
Knocked out of the air molecule, which also has a low velocity, are both accelerated by the
electric field, and therefore, each electron is capable of ionizing an air molecule at the next
collision. Of course, after the second collision, there are now four electrons to repeat the
process, and so on, the number of electrons doubling after each collision. All this time, the
electrons arc advancing toward the positive electrode, and after many collisions, their
number has grown enormously. Therefore, this process is called the avalanche process.
Note that each so-called 'electron avalanche is initiated by a single free electron that
finds itself in an intense electrostatic field. Also note that the intensity of the electrostatic
field around a conductor is nonuniform. Therefore, it has its maximum strength at the
surface of the conductor and its intensity diminishes inversely as the distance increases
from the center of the conductor. Thus, as the voltage level in the conductor is increased,
the critical field strength is approached, and the initial discharges take place only at or near
the conductor surface. For the positive half cycle, the electron avalanches move toward the
conductor and continue to grow until they hit the surface. For the negative half cycle, the
electron avalanches move way from the conductor surface toward a weaker field and ceases
to advancewhen the field becomes too weak to accelerate
•
. ~,['T!.r;E,SJBIT
40
Transmission
and Distribution
10EE53
40r-------~------.-----~_r------~------~
30
•
~
~
.
~
~
~
20
"0
.!?
0..
0.
«
10
o O~--------~--
--~----------L--N~O~iO:n:iz:at:lO~n~(:da:r:k~re~g~jo:n:)J
1
2
Gap spectng, d,
3
4
5
em
Figure 1
THE ELECTRONS TO IONIZING VELOCITY. THE CORONA CONSISTED OF A SEQUENCE OF LOWAMPLITUDE CURRENT PULSES WHOSE REPETITION RATE DEPENDED ON THE SHARPNESS OF
THE POINT. FIGURE 1 ILLUSTRATES THE FACT THAT WHEN THE VOLTAGE ACROSS A POINT-
•
TO-PLANE GAP IS GRADUALLY INCREASED, A CURRENT (IN THE ORDER OF ioA) IS
MEASURED. HERE THERE IS NO IONIZATION THAT TAKES PLACE, AND THIS CURRENT IS
KNOWN AS THE SATURATION CURRENT.
S,2.3 Factors Affecting Corona
As a rule of thumb, if the ratio of spacing between conductors w the radius conductors
before corona phenomenon occurs. Since for overhead lines this ratio is much greater than
15, the flashover can be considered as impossible under normal circumstances. At a given
voltage level, the factors affecting corona include line configuration, conductor type.
Dept of EEE, SJBIT
41
Transmission and Distribution
10EE53
condition of conductor surface, and weather. In a horizontal configuration, the field near the
middle conductor is larger than the field near the outer conductors. Therefore, the disruptive
critical voltage is lower for the middle conductor, causing larger corona loss than the ones
for the two other conductors. If the conductors are not spaced cquilaterally, the surface
gradients of the conductors and therefore the corona losses ate not equal. Also, the
conductor height affects the corona loss, that is, the greater the height, the smaller the
corona loss. The corona loss is proportional to the frequency of the voltage. Therefore, the
higher the frequency, the higher the corona losses. Thus, the corona loss at 60 Hz is greater
than the one at 50 Hz. Of course, the corona loss at zero frequency, that is, direct current, is
far less than the one for ac current.
The irregularity of the conductor surface in terms of scratches, raised strands, die burrs, die
grease, and the particles of dust and dirt that clog the conductor can significantly increase
the corona loss.
For the same diameter, a stranded conductor is usually satisfactory for about 80-85
percent of the voltage of a.smooth conductor. As said before, the size of the conductors and
their spacings also have considerable effect on corona loss. The larger the diameter, the less
. . likelihood of corona. Therefore, the use of conductors with large diameters or the use of
hollow conductors or the use of bundled conductors increases the effective diameter by
reducing the electric stress at the conductor surfaces.
The breakdown strength of air varies with atmospheric conditions. Therefore, the
breakdown strength of air is directly proportional to the density of the air. The air-density
factor is defined as
0= 3.9211p
273 + t
where p = barometric pressure in centimeters of mercury
t = ambient temperature in degrees Celsius
Rain affects corona loss usually more than any other factor.
For example. it may cause the corona loss to be produced on a conductor at voltages as low
as 65 percent of the voltage at which the same loss takes place during fair weather. Heavy
winds have no effect on the disruptive critical voltage or on the loss, but presence of smoke
lowers the critical voltage and increases the loss. Corona in fair weather may be negligible
up to a voltage close to the disruptive critical voltage for a particular conductor. Above this
voltage, the impacts of corona increase very quickly.
A transmission line should he designed to operate just below the disruptive critical voltage
in fair wether so that corona only takes place during adverse atmospheric conditions.
Therefore, the calculated disruptive critical voltage is an indicator of corona performance of
the line. However, a high value of the disruptive critical voltage is not the only criterion of
satisfactory corona performance. The sensitivity of the conductor to foul weather should
also be considered (e.g., corona increases more slowly on stranded conductors than on
Dept ofEEE, SJBIT
42
•
Transmission and Distribution
10EE53
smooth conductors). Due to the numerous factors involved, the precise calculation of the
peak value of corona loss is extremely difficult, if not impossible. The minimum voltage at
which the ionization occurs in fair weather is called the disruptive critical voltage and
can be determined from
E =
o
V.
0
r In(Dlr)
•
where Eo = value of electric stress (or critical gradient) at which disruption starts in
kilovolts per centimeters
Vo = disruptive critical voltage to neutral in kilovolts (rms)
r = radius of conductor in centimeters
D = spacing between two conductors in centimeters
Since, in fair weather, the EO of air is 21.1 kV/cm rrns,
D
Vo == 21.1r In ~
r
kV
which is correct for normal atmospheric pressure and temperature (76cm Hg at 25°C). For
other atmospheric pressures and temperatures,
D
Vo = 21.10r In t
•
kV
where 0 is the air density factor. Further, after making allowance for the surface condition
of the conductor by using the irregularity factor, the disruptive critical voltage can be
expressed as
where m = irregularity factor (0< mt) I)
..: for smooth, polished, solids, cylindrical conductors
(;.93-0.98 for weathered, solid, cylindrical conductors
t.: 'SJBIT
43
Transmission and Distribution
10EE53
=0.87-0.90 for weathered conductors with more than seven strands
=0.80--0.87 for weathered conductors with up to seven strands
Note that at the disruptive critical voltage v" there is no visible corona. In the event that the
potential difference (Or critical gradient) is further increased, a second point is reached at
which a weak luminous glow of violet color can be seen to surround each conductor. The
voltage value at this point is called the visual critical voltage and is given by
where Vv -visual critical voltage in kilovolts (rms)
mv=irregularity factor for visible corona (0< mI, 1)
=1 for smooth, polished. solid, cylindrical conductors
=0.93-0.98 for local and general visu corona on weathered, solid, cylindrical
conductors
= O.70-0. 75 for local visual corona on weathered stranded conductors
=0.80-0.85 for general visual corona on weathered stranded conductors
Note that the voltage equations given in this section are for fair weather. For wet weather
voltage values, multiply the resulting fair weather voltage values, multiply the resulting fair
weather voltage values by 0.80. For a three-phase horizontal conductor configuration, the
calculated disruptive critical voltage should be multiplied by 0.96 and 1.06 for the middle
conductor and for the two outer conductors, respectively.
8.1
Assume that a three-phase overhead transmission line is made up of three equilaterally
spaced conductors, each with overall diameter of 3 em. The equilateral spacing between
conductors is 5.5 m. The atmosphere pressure is 74 em Hg and the temperature is lOoC. If
the irregularity factor of the conductors is 0.90 in each case, determine the following:
(a) Disruptive critical rms line voltage.
(b) Visual critical rrns line voltage.
EXAMPLE
Solution
Dept ofEEE, SJBIT
44
•
iDEES3
nsmlssion and Distribution
'1'
""", ,,,,,,,· ••ioiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
~
(a)
,
3.9211p
I
:n3 + t
C
=
3.9211 x 74
273 + 10
=.
025'3
L
D
.18mo' In ~
r
"')'11 x 1.0253 x 0.90 x 1.51n ~~~
•
72.4 kV/phase
l;.
the rrns line voltage is
Vo = V3 x 172.4 = 298.7 kV
(t·
\'isual critical rrns line voltage is
..
:t1.18mr.o(1
+ .J:r)ln ;
2].1 x 1.0253 x 0.90 x 1.5 x
(1 + 'vi1.0253
0.3,
)In 5152
x 1.5
m
.J
'...:,.U4.2 kV/phase
.ire, the rms line voltage is
VI) = V3 x 214.2
= 370.9 kV
na Loss
to Peek,the fair weather corona loss per phase
:tor can be calculated from
j
•
241
r, = T
(f
' ( r .)lf2
+ 25) D
(V - VO)2
X
10-5 kW!km
.cncy in hertz
(,: IT
45
10EE53
Transmission and Distribution
V -Iine-to-neutral operating voltage in kilovolts
Vo = disruptive critical voltage in kilovolts
The wet weather corona can be calculated from the above equations by multiplying Vo by
0.80. Peek's equation gives a correct result if
(1) the frequency is between 25 and 120 Hz,
(2) the conductor radius is greater than 0.25 em. and
(3) the ratio of V to va is greater than 1.8.
r,
(X
(~rI2
The power LOSS IS proportional to the square root of the size of the conductor.
Therefore, the larger the radius of the conductor, the larger the power loss. Also, the larger
the spacing between conductors, the smaller the power loss. Similarly,
r, tx (V -
VO)2
that is, for a given voltage level, the larger the conductor site, the larger the disruptive
critical voltage and therefore the smaller the power loss.
According to Peterson (11], the fair weather corona loss per phase or conductor can be
calculated from
r - 1.11066 X
c -
r, =
10~4
[In(2Dld)Y
1.78738 X 10-4
[In(2Dld)f.
tv 2F
kW/km
.
:!
tv F
kW/mi
where d = conductor diameter
D = spacing between conductors
f
frequency in hertz
V= line-to-neutral operating voltage in kilovolts
F= corona factor determined by test and is a function of ratio of V to Va
=
..
In ;~eneral,the corona losses due to fair weather conditions arc not significantly large at
ext:ra-high-voltage range. Therefore, their effects are not significant from technical andlor
economic points of view. Whereas the corona losses due to foul weather conditions are
very significant.
Dept ofEEE, SJBIT
46
Transmission and Distribution
iiiiiiiiiiii~",,,",;;q;;;,"
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
TP',RW
where
TP:',RW
~
TP<.FW
10EE53
___
+ [ ~ jr'ln(l + KR)] ~ E;'
= total three-phase corona losses due to rainy weather in
kilowatts per kilometer
fT', ,FW;:;; total three-phase corona losses due to fair weather in
kilowatts per kilometer
V = line-to-line operating voltage in kilovolts
r = conductor radius in centimeters
n =: total number of conductors (number of conductors per
bundle times 3)
E, = voltage gradient on underside of conductor i in kilovolts
(peak) per centimeter
m= "an exponent (~ 5)
J = Joss current constant (--4.37x 10-10 at 400 kV and 3.32 x
10-10 at 500 kV and 700 kV)*
R = rain rate in millimeters per hour or inches per hour
K = wetting coefficient (10 if R is in millimeter per hour or 254
if R is in inches per hour)
.
•
0)1'.
TH
1 .,,! XAMPLE 8.1 AND ASSUME THAT THE LINE OPERATES AT 345 KV AT 60 Hz AND
[',). ! ,'JGTH IS 50 ML. DETERMINE THE TOTAL FAIR WEATHER CORONA LOSS FOR THE
•
. : U . S.;LHT
47
10EES3
Transmission and Distribution
LINE BY USING PEEK'S FORMULA.
Solution
According to Peek, the fair weather corona loss per phase is
390
1/2
r, = T (/+ 25) rD
390
=
1.0253 (60 + 25)
(V - VO)2
X
10-J~
1 SU2
;50 (199.2 - 172.4)2 X 10-5
= 12.1146 kW/mi/phase
or J for the total line length,
P, = 12.1146 x 50:::;605.7 kW/phase
Therefore, the total corona loss of the line is
r, = 3 x 605.7 =-1817.2 kW
UNIT-4
INSULATORS- TYPES, PIN INSULATORS, SUSPENSION INSULATORS, STRAIN INSULATORS,
SHACKLE INSULATORS, CAUSES OF INSULATION FAILURE. TESTING OF INSULATORS.
4 HOURS
INSULATORS
•
rr L~fi iL ;vU:CHANICAL SUPPORT FOR CONDUCTORS AND ALSO PREVENT CURRENT
TO
FLOW TO GROUND VIA SUPPORTS.
1. PROVIDES NECESSARY INSULAnON BETWEEN LINE CONDUCTORS AND SUPPORT
(GROUND).
Dept ofEEE, SJBIT
48
Transmission and Distribution
10EE53
PROPERTIES:
1) HIGH MECHANICAL STRENGTH IN ORDER TO WITHSTAND CONDUCTOR LOAD,
WINDLOAD.
2) HIGH ELECTRICAL RESISTANCE OF INSULATOR MATERIAL IN ORDER THAT
DIELECTRIC
STRENGTH IS HIGH.
•
1) MATERIAL SHOULD BE NON-POROUS, FREE FROM IMPURITIES AND CRACK.
2) HIGH RATIO OF PUNCTURE STRENGTH TO FLASH OVER •
MATERIALS
PORCELAIN,
PORCELAIN
LEAKAGE.
COMMONLY
GLASS, STEATITE OR SOME OTHER COMPOSITE
LESS EFFECTED
MATERIALS.
BY TEMP. CHANGE AND GIVES LESS TROUBLE FROM
USED INSULATORS
1PIN INSULATORS
2 SUSPENSION INSULATORS
3 STRAIN
INSULATORS
4 SHACKLE INSULATORS
12Smm.
HC-PN11OS
•
Dept ofEEE, SJBIT
49
Transmission and Distribution
10EE53
CAUSES OF INSULATION FAILURE
- INSULATORS REQUIRED TO WITHSTAND BOTH
MECHANICAL AND ELECTRICAL STRESS.
- HIGH ELECTRICAL STRESS MAY CAUSE
BREAKDOWN OF INSULATORS.
-- FLASHOVER**
(DUE TO ARC FORMATION IN THE
SURFACE BETWEEN CONDUCTOR AND PIN)
-- PUNCTURE**
(DISCHARGE OCCURS FROM
CONDUCTOR TO PIN VIA BODY OF INSULATOR)
FLASH OVER VOLTAGE
SAFETY FACTOR OF INSULATOR = PUNCTURE STRENGTH
Dept ofEEE, SJBIT
50
Transmission and Distribution
10EES3
..
•
SUSPENSION INSULATORS:
- USED FOR mGHVOLTAGE (>33KV)
- CONSISTS OF NUMBER OF PORCELAIN DISCS
CONNECTED IN SERIES BY METAL LINKS IN THE FORM
OF STRING.
- CONDUCTOR IS SUSPENDED AT THE BOTTOM END OF
THE STRING AND OTHER END IS FIXED IN THE CROSS
ARM OF THE TOWER.
- EACH DISC CAN INSULATE UPTO 11 KV. So NUMBER
OF DISC IN A STRING DEPENDS UPON THE VOLTAGE
LEVEL OF TRANSMISSION. (FOR 66 KV, 6 DISC)
"I'
:" ,:.".,
....
I I.;;
j~.
l"'l~'r.' SJBIT
-, -_.
I. _• ..,
51
Transmission and Distribution
10EE53
Anv ANTAGES OF SUSPENSION INSULATORS:
- CHEAPER THAN PIN TYPE BEYOND 33 KV
- EACH DISC IS DESIGNED FOR 11 KV. HENCE DEPENDING UPON THE
WORKING VOLTAGE, DESIRED NUMBER OF DISC CAN BE CONNECTED IN
SERIES.
- ANY DAMAGE DISC CAN BE REPLACED. D WHOLE DISC WILL NOT BE
USELESS IF ONE DISC DAMAGES.
- ARRANGEMENT PROVIDES GREATER FLEXIBILITY TO THE LINE.
- INSULATOR IS FREE TO SWING IN ANY DIRECTION AND CAN TAKE UP THE
POSITION OF MINIMUM MECHANICAL STRESS.
- GENERALLY USED WITH STEEL TOWERS. SINCE CONDUCTOR RUNS BELOW
EARTHED CROSS ARM, IT CAN PROVIDE PROTECTION FROM LIGHTNING.
29
STRAIN INSULATORS:
- USED IN DEAD END OF THE LINE WHERE
THE LINE IS SUBJECTED TO GREATER
TENSION.
- FOR HV X-MISSION LINE, IT CONSISTS OF
ASSEMBLY OF SUSPENSION INSULATORS AS
SHOWN.
- FOR HIGH TENSION IN THE LINE TWO OR
MORE STRINGS OF STRAIN INSULATORS CAN
BE USED IN PARALLEL.
- FOR LOW VOLTAGE « 11 KV), SHACKLE
INSULATOR USED AS STRAIN INSULATOR
Dept of EEE, SJBIT
52
•
Transmission and Distribution
iDEES3
SHACKLE INSULATORS:
- CAN ALSO BE USED AS STRAIN INSULATOR.
- FREQUENTLY USED FOR LOW VOLTAGE
(400 V) DISTRIBUTION LINE.
- CAN BE USED BOTH HORIZONTAL AND
VERTICAL POSITION.
- DIRECTLY FIXED TO THE POLE WITH BOLT
OR CAN BE FIXED IN THE X-ARM.
- USED WHERE WE NEED TO RUN THE LINE
WITH CERTAIN BENDING WITH SOME
ANGLE. (WHERE TURNING OF THE ROAD
ETC. OCCURS)
..-
UNIT - 5
UNDERGROUND
CABLES-
TYPES,
MATERIAL USED,
INSULATION RESISTANCE,
THERMAL RATING OF CABLES, CHARGING CURRENT, GRADING OF CABLES, CAPACITANCE
GRADING & INTER SHEATH GRADING, TESTING OF CABLES.
of EEE, SJBIT
6 HOURS
53
Transmission and Distribution
10EE53
UNDERGROUND CABLES
INTRODUCTION
Underground cables may have one or more conductors within a protective sheath.
The protective sheath is an impervious covering over insulation, and it usually is
lead. The conductors are separated from each other and from the sheath by
insulating materials. The insulation materials used are
1. rubber and rubberlike compounds.
2. varnished cambric, and
3. oil-impregnated paper.
Rubber is used in cables rated 600 V-35. kV, whereas polyethylene (FE),
propylene (PP) and polyvinyl chloride (PVC) arc used in cables rated 600 V138kV. The high-moisture resistance of rubber makes it ideal for submarine cables..
Varnished cambric is used in cables rated 600 V-28 kV. Oil-impregnated paper is
used in solid-type cables up to 69 kV and in pressurized cables up to 345 kV. In.the
solid-type cables, the pressure within the oil-impregnated cable is not raised above
atmospheric pressure. In the pressurized cables, the pressure is kept above
atmospheric pressure either by gas in gas pressure cables or by oil in oil-filled
cables. Impregnated paper is used for higher voltages because of its low dielectric
losses and lower cost.
Cables used for 59 kV and below are either (1) low pressure, not over 15 psi, or (2)
medium pressure, not over 45 psi. High-pressure cables, up to 200 psi, installed
pipes arc not economical for voltages of 69 kV and below. Voids or cavities can
appear as the result of faulty product or during the operation of the cable under varying
load. Bending the cable in handling and on installation, and also the different thermal
expansion coefficient of the insulating paper, the impregnating material and the lead sheath
result in voids in the insulation of cable not under pressure. The presence of higher
electrical field strength ionization that appears in the voids in the dielectric leads to
destruction of the insulation. The presence of ionization can he detected by means of the
power factor change as a test voltage is applied. The formation of voids is avoided in the
case of the oil-filled cable. With the gas-filled cable, the pressure in the insulation is
increased to such a value that existing voids or cavities are ionization free. Ionization
increases with temperature and decreases with increasing pressure.
CONDUCTORS
The conductors used in underground cables can be copper or aluminum. Aluminum dictates
larger conductor sizes to carry the same current as copper. The need for mechanical
flexibility requires stranded conductors to be used. The equivalent aluminum cable is lighter
in weight and larger in diameter in comparison to copper cable. Stranded conductors can be
in various configurations, for example, concentric, compressed, compact, and rope.
UNDERGROUND
CABLE TYPES
Cables are classified in numerous ways. For example, they can be classified as (1)
underground, (2) submarine, and (3) aerial, depending on location. They can be classified
DeptofEEE,SJBIT
54
Transmission
and Distribution
10EE53
according to the type of insulation, such as (1) rubber and rubberlike compounds, (2)
varnished cambric, and (3) oil-impregnated paper. They can be classified as (1) single
conductor, (2) two conductor duplex, three conductor, etc., depending on the number of
conductors in a given cable. They can be classified as shielded or nonshielded (belted),
depending on the presence or absence of metallic shields over the insulation. Shielded
cables can be solid, oil filled, or gas filled. They can he classified by their protective finish
such as (1) metallic (e.g., a lead sheath) or (2) nonmetallic (e.g., plastic).
insulation shields help to (1) confine the electric field within the cable; (2) protect cable
better from induced potentials; (3) limit electromagnetic or electrostatic interference; (4)
equalize voltage stress within the insulation, minimizing surface discharges; and (5) reduce
shock hazard (when properly grounded) [1].
In general, shielding should be considered for nonmetallic covered cables operating at a
circuit voltage over 2 kV and where any of the following conditions exist [2]:
1. Transition from conducting to nonconducting conduit. _
2. Transition from moist to dry earth.
3. In dry soil, such as in the desert.
4. In damp conduits.
5. connections to aerial lines.
6. Where conducting pulling compounds are used.
7. Where surface of cable collects conducting materials, such as soot, salt, or
cement deposits.
8. Where electrostatic discharges are of low enough intensity not to damage cable
but are sufficient in magnitude to interface with radio or television reception.
In general, cables are pulled into underground ducts. However, if they have to he buried
directly in the ground, the lead sheath (i.e, the covering over insulation) has to be protected
mechanically by armor. The armor is made of two steel tapes wound overlapping each other
or heavy steel wires.
Where heavy loads are to be handled, the usage of single-conductor cables are
advantageous since they can be made in conductor sizes up to 3.5 kcrnil or larger. They are
also used where phase isolation is required or where balanced single-phase transformer
loads are supplied. They are often used to terminate three-conductor cables in singleconductor potheads, such as at pole risers, to provide training in small manholes. They can
be suppLied triplexed or wound three in parallel on a reel, permitting installation of three
single-conductor cables in a single duct. Figure 4.1shows a single- conductor, paperinsulated power cable.
1he belted cable construction is generally used for three-phase low- voltage operation, up to
5 kV, or with the addition of conductor and beLt shielding, in the 10-15 kV voltage range.
It receives its name from the fact that a portion of the total insulation is applied over
partially insulated conductors in the form of an insulating belt, which provides a smooth
"cushion" for the lead sheath.
Figure 1. Single-conductor, paper insulated power cable. (Courtesy ofOkonite Company.)
Dept of EEE, SJBIT
55
Transmission
and Distribution
10EE53
Even though this design is generally more economical than the shielded (or Type H)
construction, the electrical field produced by three-phase ac voltage is assymmetrical, and
the fillers are also under electric stress. These disadvantages restrict the usage of this cable
to voltages below 15 kV. Figure 2, shows a three-conductor, belted, compact-sector, paperinsulated cable. They can have concentric round, compact round, or compact-sector
conductors.
The three-conductor shielded, or type H, construction with compact sector conductors is the
design most commonly and universally used for three-phase applications at the 5-46-kV
voltage range. Three-conductor cables in sizes up to lkcmil are standard, hut for larger
sizes, if overall size and weights are important factors, single-conductor cables should be
preferred.
It confines the electric stress to the primary insulation, which causes the voltage rating
(radial stress) to be increased and the dielectric losses to be reduced. The shielded paperoil dielectric has the greatest economy for power cables at high voltages where reliability ..
and performance are of prime importance. Figure 3 shows a three-conductor, shielded (type
H), compact-sector, paper-insulated cable.
..
Figure 4, presents various protective outer coverings for solid-type cables, depending on
installation requirements. Figure 5, shows the recommended voltage ranges for various
types of paper-insulated power cables.
Most cable insulations are susceptible to deterioration by moisture to varying degrees.
Paper and oil, which have had all the moisture completely extracted in the manufacture of a
paper cable, will reabsorb moisture when exposed to the atmosphere, and prolonged
exposure will degrade the exceptionally high electrical qualities. Because of this, it is
mandatory in all paper cable splices and terminations to reduce exposure of the insulation to
moisture and to construct and seal the accessories to ensure the complete exclusion of
moisture for long and satisfactory service life.
Therefore, it is important that all cable ends are tested for moisture before splicing
or potheading. The most reliable procedure is to remove rings of insulating paper from the
section cut for the connector at the sheath, at the midpoint, and nearest the conductor and
immerse the tape "loops" in clean oil or flushing compound heated to 280-300°F. If any
traces of moisture are present, minute bubbles will exclude from the tape and form "froth"
in the oil.
The shields and metallic sheaths of power cables must he grounded for safety and
reliable operation. Without such grounding, shields would operate at a potential
considerably above the ground potential. Therefore, they would be hazardous to touch and
would incur rapid degradation of the jacket or other material that is between shield and
ground. The grounding conductor and its attachment to the shield or metallic sheath,
norman" at a termination or splice, needs to have an ampacity no lower than that of the
Dr : of EEE, SJBIT
56
T r;J ~<mission and Distribution
iDEES3
shield: . ";case of a lead sheath, the ampacity must be large enough to carry the available
fault ,:' "i.' : 'llld duration without overheating. Usually,
l' SJBIT
57
lJ8fS 'gmT.1 0 l\)~:.j
..
£53301
uounqt.ustrj
pUE UOJSSJlliSUE.l.L
Transmission and Distribution
10EE53
Figure 4 (Continued).
Each cable type can _De.but usually
is not, used beyond indicated VOltages
High oil pressure"oilo!'ltatic"
low oil pressure oil filled
0f'iSh gas
....pre~StJre
,
I
I
UIlltMdlum gas pre~ure
IllI Low gas pressure
Solid
Belted solid
I
1
I
I
I
I
2,5 7.S
15
l
I I
I
23 35 46 69
Voltage, kV
If I
I
115 161 230
138
Figure 5. Recommended voltage ranges for various of paper-insulated paper
cables. (Courtesy of Okonite Company.)
the cable shield lengths are grounded at both ends such that the fault current would divide
and flow to both ends, reducing the duty on the shield and therefore the chance of damage.
The capacitive charging current of the cable insulation, which is on the order of 1 mAlft of
conductor length, normally flows, at power frequency, between the conductor and the earth
electrode of the cable, normally the shield. Of course, the shield, or metallic sheath,
provides the fault return path in the event of insulation failure, permitting rapid operation of
the protection devices [1].
CABLE INST ALLA TION TECHNIQUES
There are a number of ways to install the underground cables; for example:
1. Direct burial in the soil, as shown in Figure 6. The cable is laid in a trench that is
:,1l:,1]. I,j by machine.
2. in ducts or pipes with concrete sheath, as shown in Figure 7. For secondary network
-ystems, duct lines may have 6-12 ducts.
'1. Wherever possible, in tunnels built for other purposes, for example, sewer lines, water
IT'qins. gas pipes, and duct lines for telephone and telegraph cables.
59
I
345
I
5~
10EE53
Transmission and Distribution
Protective
barrier
(ll)
(0)
Figure 6. Direct burial: (a) for single-conductor cables; (b) for
triplexed cables.
In general, manholes are built at every junction point and comer. The spacing of
manholes is affected by the types of circuits installed, allowable cable-pulling
tensions, and utility company's standards and practice. Manholes give easily .
accessible and protected space in which cables and associated apparatus can he
operated properly. For example, they should provide enough space for required
switching equipment, transformers, and splices and terminations. Figure 4.8 shows a.
straight-type manhole. Figure 4.9 shows a typical street cable manhole, which "is
usually used to route cables at street intersections or other locations where cable
terminations are required.
TABLE 1 Typical Values of Various Dielectric Materials
Dept of EEl':, SJBIT
Dielectric Material
K
Air
1
Impregnated paper
3.3
Polyvinyl chloride (PVC)
3.5-8.0
Ethylene propylene insulation
2.8-3.5
Polyethylene insulation
2.3
60
Transmission and Distribution
iiiiiiiiiiioo.;;-...;.... ;;:,,~-.
10EE53
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
1 Cross-Linked polyethylene
12.3-6.0
1
Dielectric Constant of Cable Insulation
The dielectric constant of any material is defined as the ratio of the capacitance of a
condenser with the material as a dielectric to the capacitance ofa similar condenser with air
as the dic"Jc'ctric,it is also called the relative permittivity of specific inductive capacity.
It is \ ,:icl';> denoted by K. (It is also representedby e, or SIC.) Table I gives the typical
valu .. : Ci' the dielectric constants for various dielectric materials -.
.__'
C
O.0073K ..
F/IOOO it
106 JoglO(Dld)
(4.31
where C == capacitance in farads per 1000 it
K = dielectric constant 'Qf. cable insulation
D = diameter over insulation in unit length
d= diameter over conductor shield in unit length
Charging Current
By definition of susceptance.
b=
we
S
(4.32
or
b=21TfC
S
(4.33
Then the admittance Y corresponding to C is
V=jb
Of·
i
,
61
10EES3
Transmission and Distribution
or
Y = j2'1TfC
(4.3
S
Therefore, the charging current is
(4.3
Of,
.
. .
.
ignormg j,
(4.3
.. For example, substituting equation (4.31) into equation (4.36), the chargii
current of a single conductor cable is found as
O.OO73KVo...N)
l061oglO(Dld)
21(/
1(. =
X
(4.3'
or
O.0459fKV(L_N)
It
where
= 103 JoglO(Dld)
(4.3,
A/IOOOft
f = frequency in hertz
D == diameter over insulation in unit length
d == diameter over conductor shield in unit length
K = dielectric constant of cable insulation
V ::= line-to-neutral voltage in kilovolts
At 60 Hz frequency,
2.752KV(L.N)
I, == l031oglO(DJd)
A/lOOOft
The charging current and the capacitance are relatively greater for insulated cables than in
overhead circuits because of closer spacing and the higher dielectric constant of the
insulation of the cable. In general, the charging current is negligible for overhead circuits at
distribution voltages, contrary to high-voltage transmission circuits.
Dept ofEEE, SJBIT
62
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Transmission and Distribution
Determination of Insulation Resistance of
Single-Conductor Cable
Assume that the cable shown in Figure 8 has a length of 1 m.
Then the incremental insulation resistance of the cylindrical element in the radial direction
IS
Figure 8. Cross section of single-conductor cable.
(4.40)
Therefore, the total insulation resistance between the conductor and the
lead sheath is
P
R, == 2Trlln
Vi
R
r
(4.41 )
Iii = total insulation resistance in ohms
p == insulation (dielectric) resistivity in ohm meters
I = total length of cable in meters
.R = outside radius of insulation or inside radius of lead sheath in
meters
r = radius of conductor in meters
(illife
practical version of equation (4.41) is given by the Okonite
." .: ".yt as
\.... -_,_tl
(4.42)
R, = total insulation resistance in megohms per 1000 ft for particular
;.....lJle construction
Tsi = specific insulation resistance in megohms per 1000 ft at 60 of
D == inside diameter of sheath
d '= outside diameter of conductor
bIT
63
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Transmission and Distribution
TABLE 4.2 GNES TYPICAL R VALUES OF VARIOUS INSULATION MATERIALS. EQUATION
(4.19) INDICATES THAT THE INSULATION RESISTANCE IS INVERSELY PROPORTIONAL TO THE
LENGTH OF THE INSULATED CABLE. AN INCREASE IN INSULATION THICKNESS INCREASES
THE DISRUPTIVE CRITICAL VOLTAGE OF THE INSULATION BUT DOES NOT GIVE A
PROPORTIONAL DECREASE IN VOLTAGE GRADIENT AT THE CONDUCTOR SURFACE.
THEREFORE, IT DOES NOT PERMIT A PROPORTIONAL INCREASE IN VOLTAGE RATING.
EXAMPLE 4.3
A 250-KCMIL, SINGLE-CONDUCTOR, SYNTHETIC RUBBER, BELTED CABLE HAS A
CONDUCTOR DIAMETER OF 0.575 IN. AND AN INSIDE DIAMETER OF SHEATH OF 1.235 IN. THE
. CABLE HAS A LENGTH OF 6000 FT AND IS GOING TO BE USED AT 60Hz AND 115KV.
CALCULATE THE FOLLOWING:
(A) TOTAL INSULATION RESISTANCE IN MEGOHMS AT 60 of.
(B) POWER LOSS DUE TO LEAKAGE CURRENT FLOWING THROUGH INSULATION RESISTANCE.
Solution
(a) By using equation (4.42),
D
R, =: r$,log d
From Table 4.2, specific insulation resistance r $1 is 2000 f\t1fi I 1000 ft.
Therefore, the total insulation resistance is
.
·1.235
RI
= 6 x 2000 log 0.575
=3.984MO
(b) The power loss due to leakage current is
V~
115,0002
R, == 3984 X 106
::::3.3195 W
TABLE 4.2 TYPICAL VALUESOF R
Insulation Material
Dept of EEE, SJBIT
'ai
(Mfl I1000it)
Synthetic rubber
2,000
Ethylene propylene insulation
20,000
Polyethylene
50,000
64
1DEES3
Transmission and Distribution
PVC
2,000
Cross-linked polyethylene
20,000
For a sector-type three-conductor cable,
D3S
where D
= overall
= D - O.35d
(4.67)
diameter of cable with circular cross-sectional conductors
== overall diameter of cable with sector-type three conductors
-d;:; diameter-of conductor
D3s
S = lead sheath thickness of cable
t = belt insulation thickness of cable
T = thickness of conductor insulation in inches
Geometric Factors
The geometric factor is defined as the relation in space between the cylinders formed by
sheath internal surface and conductor external surface in a single-conductor betted cable.
For a three-conductor belted cable, this relation (i.e., geometric factor) is sector shaped, and
by relative thicknesses of conductor insulation T and belt insulation 1. For a -single-conductor cable, the geometric factor G is given by
D
G = 2.303 loglo d
(4.68)
where D = inside diameter of sheath
d == outside diameter of conductor
Table 4.3 presents geometric factors for single-conductor and three- conductor belted
cables. In this table, G indicates the geometric factor for a single-conductor cable, Go
indicates the zero-sequence geometric factor, and G1 indicates the positive-sequence
geometric factor for three-conductor belted cables. Also, Figures 9 and 10 give geometric
factors for single-conductor and three-conductor belted cables. In Figure 11, Go indicates
the zero-sequence geometric factor and G1 indicates the positive- sequence geometric
factor.
in Table 4.3 and Figures 9 and 10,
Dept of EEE, SJBIT
65
10EE53
Transmission and Distribution
T= thickness of conductor insulation in inches
t = thickness of belt insulation in inches
d=outside diameter of conductor in inches
For single-conductor
cables,
1=0
EXAMPLE
4.8
A single-conductor, paper-insulated, belted cable will be used as an underground feeder of
3 ml. The cable has a 2000-MCM (2000-kemil) copper conductor.
(a) Calculate the total de resistance of the conductor at 25 "C.
(b) Using Table 4.5, determine the effective resistance and the skin effect on the effective
resistance in percent if the conductor is used at 60-Hz alternating current.
. (c) Calculate the percentage of reduction in cable ampacity in part (b).
Solution
{a) From Table 4.5, the de resistance of the cable is
R<l<; ::::
0.00539 n/lQOO ft
or the total dc resistance is
Rdc
= O~OO539x 5280 x 3' = 0.0854 n
(b) From Table 4.5, the skin effect coefficient is 1.233; therefore, the
effective resistance at 60 Hz is
Rf:ff = (skin effect coefficient) x
Rdc
= (1.233) x 0.0854
=0.10530.
or it is 23.3 percent greater than for direct current.
(c) The reduction in the cable ampaeity is also 23.3 percent.
SHEATH CURRENTS IN CABLES
The flow of ac current in the conductors of single-conductor cables induces ac voltages in
the cable sheaths. When the cable sheaths are bonded together at their ends, the voltages
induced give rise to sheath (eddy) currents, and therefore, additional12R losses occur in the
sheath. These losses are taken into account by increasing the resistance of the relevant
conductor. For a single-conductor cable with bonded sheaths operating in three phase and
arranged in equilateral triangular formation, the increase in conductor resistance is
Dept of EEE, SJBIT
66
Transmission and Distribution
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(4.77)
•
i
,JBIT
67
10EE53
Transmission and Distribution
where Xm
= mutual
reactance between conductors and sheath per phase I
ohms per mile
ra = sheath resistance per phase in ohms per mile
The mutual reactance between conductors and sheath can be calculate
from
}{m = 0.2794
f
60 loglo
2S
70
+ r,
(4.78
and the sheath resistance of a lead sheath cable can be determined fron
0.2
r = -::---~:---~
J
(TO + r,)(ro - rJ
where
(4.79
f = frequency
S
=
in hertz
spacing between conductor
centers in inches
r 0 ::=; outer radius of lead sheath in inches
T f == inner radius of Iead sheath in inches
In equation (4..78).
(4.80
and
Dept ofEEE, SJBIT
68
1'1'... 1 <mission and Distribution
_
10EES3
.....•. ;..." ....,.,., ..:..... iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
GMD=DIII
=s
Therefore, for other conductor arrangements, that is, other than equilaters
triangular formation,
f JoglO o;
Xm = 0.2794 60
D
..
n' !mi/phase
(4.81
Dm
D
(4.82
s
and if the frequency used is 60 Hz,
Xm
= O.27941og
10
k
or
· 1 .Drn
X m= O.1213
n lJ,'
(4.83
s
Hence. in single-conductor cables, the total resistance to positive- 0
negative-sequence current flow, including the effect of sheath current, is
(4.84
1 ),':
\
!.
69
Transmission and Distribution
1DEES3
where ta = total positive- or negative-sequence resistance, including shea
current effects
rc = ac resistance of conductor, including skin effect
The sheath loss due to sheath currents is
(4.8t
p.J = [2 Ar
or
(4.8
or
(4.8'
where r.J = sheath resistance per phase in ohms per mile
1= current in one conductor in amperes
Xm =:; mutual reactance between conductors and sheath per phase
ohms per mile
De~t of i:EE, SJBIT
70
: .:nsmission and Distribution
10EE53
For a three-conductor cable with round conductors, the increase
conductor resistance due to sheath currents is
Ar
= 0.04416
( ')2
r, '0 +
'I
fi/mi I phase
{4.S
where
•
_ d + 2T
S - v'3
(4.8
and rJ = sheath resistance, from equation (4..79)
r 0 = outer radius of lead sheath in inches
r1 = inner radius of .lead sheath in inches
S = distance between conductor center and sheath center for thre
conductor cable made of round conductors
d = conductor diameter in inches
T = conductor insulation thickness in inches
For sector-shaped conductors, use equations (4.88) and (4.82) but cc
ductor diameter is
d = 82-86% of.diameter of round conductor having same
cross-sectional area
gle-conductor cables may be operated short-circuited or open-circuited. If the
hort-circuited, they are usually bonded and grounded at every manhole. This
sheath voltages to zero but allows the flow of sheath currents. There are
! iques of operating with the sheaths open-circulated:
, :ound wire is used, one terminal of each sheath section is bound to the ground
ner terminal is left open so that no current can flow in the sheath.
" .. c i ';
"
'\
"
,
1,
. ':"
', ii;i .. \~.;.)
bonding,
':
; c
.
"c
shown in Figure 12, so that only the sheaths are transposed electrically. The
bonded together and grounded at the end of each complete transposition. Thus,
sheath voltages induced by the positive-sequence currents becomes zero.
-<'dancebonding, impedances are added in each cable sheath to limit sheath
termined values without eliminating any sequence Currents.
,'.;»g transformers [5].
i;
c'
;,,:1'
i\
at each section, connections are made between the sheaths of cables
c
'c
c.'
'
, IT
71
Transmission and Distribution
r-----
10EE53
------1
-.z-
'"':"
Figure 12. Cross bonding of single-conductor cables.
EXAMPLE 4.9
Assume that three 35-kV, 350-kcmil, single-conductor belted cables are located in touching
equilateral formation with respect to each other and the sheaths are bounded to ground a
several points. The cables are operated at 34.5kV and 60Hz. The cable has a conductor
diameter of 0.681 in., insulation thickness of 345 emil, lead sheath thickness of 105 emil,
and a length of lOmi.Conductor ac resistance is O.l9Q/Wmi per phase at 50°C. Calculate
the following:
(a) Mutual reactance between conductors and sheath.
(b) Sheath resistance of cable.
(c) increase in conductor resistance due to sheath currents.
(d) Total resistance of conductor including sheath loss.
(e) Ratio of sheath loss to conductor loss.
T
of FEE, SJBIT
72
Transmission and Distribution
10EES3
(f) Total sheath losses of feeder in watts if current in conductor is 400 A.
Solution
(a) By using equation (4.78).
Xm
= 0.279410glQ
2S
'0 + r,
O/mi/phase
where
r,» O.~l
+ 0.345 == 0.686 in.
'0 = r, + 0.105 == 0.791 in.
S = 1.582 in.
Therefore,
2 x 1.582
Xm = 0.2794Iog1() 0.791 + 0.686
= 0.09244 nl mi
or
Xm ;;;;:
0.9244 0.1phase
(b)
using equation (4.79).,
0.2
r = ----:-:----
(ro + 'j)('o -
S
rJ
0.2
----~~~~---::--::-::-:::(0.791 + 0.686)(0.791 - 0.686)
= 1.2896!l/mi
(1-;
's = 12.896!l/phase
iV")
'sing equation (4.77),
0.092442
+ 0.092442
= 1.2896 1.2896l
= 0.00659 Ofmi
;-.
'--".;
!. L:.
:"JBIT
73
vL
Lt£trO
=
06I'O
6~90(rO
~seqd/U 6S96'l = OJ
JO
!lU/U6~6rO =
6S'90(fO+ 061'0
=
'(MfV) uonsnba gU!s[l (p)
<)Suqd/U 6S90'O =.LV
10
£53301
, .. J'
mission and Distribution
..
1DEES3
1~;
; i
Tlut i
Sheath loss = 3.47% x conductor
]2R
loss
,; U .ing equation (4.87),
tor three-phase loss
1
,: 3 X 4002 X 0.00659
= 3163.2 W/mi
or, for total feeder length,
r, =31,632W
,
;
: i
J
,
'1'
75
10EE53
Transmission and Distribution
Therefore,
-/1.3326
1.9+ j4.64
-j2.1734
1.9+ j4.64
-}0.4918
-;1.3326
-j0.4918
-j1.8244
-j2.4441
-jO.4918
-/1-8'244
-jO.4918
-jO.4918
1.9+ j4.64
-j].3326
-1'2.1734
-jl.3326
-jl.8244
-j0.4918
-j0,4918
-j1.8244
-j1.8244
-jO.4918
-j0.4918
1.9+ /4.64
-j1.3326
-jl.3326
1.9 + j4 ..64
-/2.1734
-J1.3326
-;2.4441
-j1.8244
~j0.4918
-;2.1734
..
- j1.3326
1.9 + j4.64
By Kron reduction.
[Zncwl = [ZJ] - HZ';;} - [Zj]}[Zkrl {[Zmr - [Zs]}
or
4.521 + j2.497
[Zobel = 1.594 - J:Z.604
[ 1.495 - )3.090
1.594 - j2.604
2.897 + J?802
0.930 - Jl.929
1.495 - j3.090]
0.930 -. jl,929 •
2 ~97 + j3.802 .
(b) By doing the similarity transformation,
[Z')121 == [Ar1[Zlloc][A]
n
LOCATION OF FAULTS IN UNDERGROUND CABLES
There arc various methods for Locating faults in underground cables. The method used for
locating any particular fault depends on the nature of the fault and the extent of the
experience of the testing engineer. Cable faults can be categorized as
1. Conductor failures or
2. Insulation failures.
In general, conductor failures are located by comparing the capacity of the insulated
conductors. On the other hand, insulation failures are located by fault tests that compare the
resistance of the conductors.
In short cables, the fault is usually located by inspection, that is, looking for smoking
manholes or listening for cracking sound when the kenetron is applied to the faulty cable.
T . j ccation of ground faults on cables of known length can be determined by means of the
balanced-bridge principle.
Kenetron is a two-electrode high-vacuum tube. They arc used as power rectifiers tor
applications requiring low currents at high de voltages, such as for electronic dust precipitation
and high-voltage test equipment.
;)'
Y EEE, SJBIT
76
Transmission and Distribution
10EE53
Fault Location by Using Murray Loop Test
It is the simplest of the bridge methods for locating cable failures between conductors and
..
ground in any cable where there is a second conductor of the same size as the one with the
fault. It is one of the best methods of locating high-resistance faults in low-conductorresistance circuits. Figure 13 shows a Murray loop.
The faulty conductor is looped to an un faulted conductor of the same cross-sectional area,
and a slide-wire resistance box with two sets of coils is connected across the open ends of
the loop. Obviously, the Murray loop cannot be established if the faulty conductor is broken
at any point. Therefore, the continuity of the loop should he tested before applying the
bridge principle. In order to avoid the effects of earth currents, the galvanometer is
connected as shown in the figure. A battery energizes the bridge between the sliding contact
or resistance box center and the point at which the faulty line is grounded. Balance is
obtained by adjustment of the sliding contact or resistance. If the non-grounded (un-faulted)
line and the grounded (faulted) line have the same resistance per unit length and if the slide
wire is of uniform cross-sectional area,
A
2L-X
-=--_
B
X
or
X ~
2L
A
units of length
( 4.212)
units of length
(4.213)
1+ B
or
x=
2 LB
A+B
where X= distance from measuring end to fault point
L= length of each looped conductor
A=resistance of top left-hand side bridge arm in balance
B=resistance of bottom left-hand side bridge arm in balance
Therefore, the distance X from the measuring end to the fault can be found directly in terms
of the units used to measure the distance L.
of EEE, SJBIT
77
Transmission
and Distribution
10EES3
Fault Location by Using Varley Loop Test
It can be used for faults to ground where there is a second conductor of the same size as the
one with the fault. ft is particularly applicable in Locating faults in relatively high-
resistance circuits. Figure 4.35 shows a Varley Loop.
The resistance per unit length of the un-faulted conductor and the faulted conductor must be
known. Therefore, if the conductors have equal resistances per unit length (e.g., rc), the
resistance (2L-X)rc constitutes one arm of the bridge and the resistanc
1
Be pcsltien
varl<y loop positioo
Figure 14. Varley loop.
Dept of EEE, SJBIT
78
Tra: .mission and Distribution
RJ
R2
=
iDEES3
R3 + XTc
(2L - X)rc
or
(4.214)
"
Dr
X= R·~2R
J
.
2
(2L
~l
2
R'j)
rc
units of length
(4.215)
where X = distance from measuring end to fault point
L = length of each looped conductor
R I = resistance of bottom left-hand side bridge arm in balance
t. - = resistance of top left-hand bridge arm in balance
/' == adjustable resistance of known magnitude
conductor resistance in ohms per unit length.
T
Tf
11,
ti,
C~
',)
i,'tor resistance is not known, it can easily be found by changing the switch to
;':hi'.!Cn and measuring the resistance of the conductor 2L by using the Wheatstone
j.
c
Cable Checks
cables should be subjected to a nondestructive test at higher than normal
\ ~,!:;c; ', egger testing is a common practice The word
trade name of a line of ohmmeters manufactured by the James G. Diddle
'y '<crtain important information regarding the quality
1\
in",',
Figure 15. Lightweight battery-operated
:i " .unes O.Biddle Company.)
. 'HiT
cable route tracer. (Courtesy
79
Transmission and Distribution
10EES3
Figure 16. Automatic digital radar cable test set. (Courtesy of James G
BiddleCompany. )
and condition of insulation can be determined from regular Megger readings that is a form
of preventive maintenance.
For example, Figure 13 shows a portable high-resistance bridge for cable-fault-locating
work. Faults can be between two conductors or between a conductor and its conducting
sheath, concentric neutral, or ground. Figure 14 shows a heavy-duty cable test and faultlocating system, which can be used for either grounded or ungrounded neutral 15-kV
cables. The full 100 rnA output current allows rapid reduction of high-resistance faults on
cables rated 35 kV ac or higher to the level of 25 kV or lower for fault- locating purposes.
Figure 15 shows a lightweight battery-operated cable route trace that can be used to locate,
trace, and measure the depth of buried energized power cables'. Figure 16 shows an
automatic digital radar cable test set that requires no distance calculations, insulation
calibrations, or zero pulse alignments,
Dept ofEEE, SJBIT
80
..
Transmission and Distribution
10EES3
UNIT-6
•
LINE PARAMETERS: CALCULATION OF INDUCTANCE OF SINGLE PHASE, 3PHASE LINES WITH
•
EQUILATERAL & UNSYMMETRICAL SPACING. INDUCTANCE OF COMPOSITE CONDUCTOR
LINES. CAPACITANCE-CALCULATION FOR TWO WIRES & THREE PHASE LINES, CAPACITANCE
CALCULATION FOR TWO WIRE THREE-PHASE LINE WITH EQUILATERAL
SPACING.
& UNSYMMETRICAL
10 HOURS
TRANSMISSION LINE PARAMETERS
THE POWER TRANSMISSION
LINE IS ONE OF THE MAJOR COMPONENTS
OF
AN ELECTRIC POWER SYSTEM. ITS MAJOR FUNCTION
IS TO TRANSPORT
ELECTRIC
ENERGY, WITH MINIMAL
LOSSES, FROM THE POWER SOURCES
TO THE LOAD CENTERS, USUALLY SEPARATED BY LONG DISTANCES.
THE
DESIGN OF A TRANSMISSION
LINE DEPENDS ON FOUR ELECTRICAL
PARAMETERS:
1. SERIES RESISTANCE
2. SERIES INDUCTANCE
3. SHUNT CAPACITANCE
4.
SHUNT CONDUCTANCE
THE SERIES RESISTANCE RELIES BASICALLY ON THE PHYSICAL COMPOSITION OF THE
CONDUCTOR
AT A GIVEN TEMPERATURE.
THE SERIES INDUCTANCE
AND SHUNT
CAPACITANCE ARE PRODUCED BY THE PRESENCE OF MAGNETIC AND ELECTRIC FIELDS
AROUND THE CONDUCTORS, AND DEPEND ON THEIR GEOMETRICAL ARRANGEMENT.
THE
SHUNT CONDUCTANCE
1NSULATORS AND AIR.
TO NOMINAL
CONDUCTANCE
IS DUE TO LEAKAGE CURRENTS
FLOWING ACROSS
As LEAKAGE CURRENT IS CONSIDERABLY SMALL COMPARED
CURRENT,
IT IS USUALLY NEGLECTED,
IS NORMALLY NOT CONSIDERED
AND THEREFORE,
SHUNT
FOR THE TRANSMISSION
LINE
MODELING.
EOUIY ALENT CIRCUIT
Dept ofEEE, SJBIT
81
10EE53
Transmission and Distribution
ONCE
EVALUATED, THE
TRANSMISSION
LINE AND
ARRANGEMENT
OF THE
LINE PARAMETERS ARE USED
TO
PERFORM
PARAMETERS
TO
MODEL
DESIGN CALCULATIONS.
(EQUIVALENT
CIRCUIT
THE
THE
MODEL)
REPRESENTING THE LINE DEPENDS UPON THE LENGTH OF THE LINE.
•
I.
Dent Cl;' EEE, SJBIT
82
"
r:
I ,',
LOAD
LOAD
\'
I ,j, I
EQUIVALENTCIRCUIT OF A SHORT-
FIGURE 13,2 EQUIVALENT CIRCUIT OF A
MEDIUM- LENGTH TRANSMISSION LINE,
" ISMISSIONLINE.
..
..
83
A TRANSMISSION LINE IS DEFINED AS A SHORT-LENGTH LINE IF ITS LENGTH IS LESS THAN 80 KM (50
MILES). IN THIS CASE, THE SHUT CAPACITANCE
EFFECT IS NEGLIGIBLE AND ONLY THE RESISTANCE
AND INDUCTIVE REACTANCE ARE CONSIDERED. ASSUMING BALANCED CONDITIONS, THE LINE CAN BE
REPRESENTED BY THE EQUIVALENT CIRCUIT OF A SINGLE PHASE WITH RESISTANCE
REACTANCE XL IN SERIES (SERIES IMPEDANCE), AS SHOWN IN FIG.13.l.
A LENGTH BETWEEN
80 KM (50
MILES) AND
LENGTH LINE AND ITS SINGLE-PHASE
CIRCUIT CONFIGURATION
[1].
240
KM
(150
R,
MILES), THE LINE IS CONSIDERED A MEDIUM-
EQUIVALENT CIRCUIT CAN BE REPRESENTED IN A NOMINAL P
THE SHUNT CAPACITANCE OF THE LINE IS DIVIDED INTO TWO EQUAL
PARTS, EACH PLACED AT THE SENDING AND RECEIVING ENDS OF THE LINE. FIGURE
EQUIVALENT CIRCUIT FOR A MEDIUM-LENGTH
BOTH
SHORT-
PARAMETER
AND
AND INDUCTIVE
IF THE TRANSMISSION LINE HAS
MEDIUM-LENGTH
13.2
SHOWS THE
LINE.
TRANSMISSION
LINES
MODELS. HOWEVER, IF THE LINE IS LARGER THAN
240
USE
APPROXIMATED
LUMPED-
KM, THE MODEL MUST CONSIDER
PARAMETERS UNIFORMLY DISTRIBUTED ALONG THE LINE. THE APPROPRIATE
SERIES IMPEDANCE AND
SHUNT CAPACITANCE ARE FOUND BY SOLVING THE CORRESPONDING DIFFERENTIAL EQUATIONS, WHERE
VOLTAGES AND CURRENTS ARE DESCRIBED AS A FUNCTION OF DISTANCE AND TIME. FIGURE
13.3 SHOWS
THE EQUIVALENT CIRCUIT FOR A LONG LINE.
THE CALCULATION OF THE THREE BASIC TRANSMISSION LINE PARAMETERS IS PRESENTED IN THE
FOLLOWING SECTIONS
[17].
13.2 RESISTANCE
THE AC RESISTANCE OF A CONDUCTOR IN A TRANSMISSION LINE IS BASED ON THE CALCULATION OF ITS
DC RESISTANCE. IF DC CURRENT IS FLOWING ALONG A ROUND CYLINDRICAL CONDUCTOR, TH~ CURRENT
IS UNIFORMLY DISTRIBUTED OVER ITS CROSS-SECTION AREA AND ITS DC RESISTANCE IS EVALUATED BY
Roc = pI (fl)·
A.
(I 3. 1)
where p = conductor resistivity at a given temperature (H-m)
1 = conductor length (m)
A = conductor cross-section area (rn")
Z
sin hr/
rf
Load
y tan h (rJ!2)
2
rl/2
FIGURE ll.3 Equivalentcircuit ofa long-length transmission line. Z = zl = equivalenttotal seriesimpedance (0),
y= yl=equivalent total shunt admittance (:S), z=sedes impedance per unit length (O/m), y=shunt admittance
per unit length (5/m), y =,;Z}' = propagation constant.
IF AC CURRENT IS FLOWING, RATHER THAN DC CURRENT, THE CONDUCTOR EFFECTIVE RESISTANCE IS
HIGHER DUE TO FREQUENCY OR SKIN EFFECT.
Dept of EEE, SJBIT
84
..
FREQUENCY
THE
EFFECT
FREQUENCY
OF
THE
AC
VOLTAGE
RESISTANCE DUE TO THE NONUNIFORM
KNOWN
AS SKIN EFFECT. As
PRODUCES
SECOND
OF THE
EFFECT
ON
CURRENT.
THE
CONDUCTOR
THIS PHENOMENON
IS
FREQUENCY INCREASES, THE CURRENT TENDS TO GO TOWARD THE
SURFACE OF THE CONDUCTOR AND THE CURRENT
REDUCES THE EFFECTIVE CROSS-SECTION
DENSITY DECREASES AT THE CENTER. SKIN EFFECT
AREA USED BY THE CURRENT,
AND THUS, THE EFFECTIVE
RESISTANCE INCREASES. ALSO, ALTHOUGH
IN SMALL AMOUNT,
OCCURS WHEN OTHER CURRENT-CARRYING
CONDUCTORS ARE PRESENT IN THE IMMEDIATE VICINITY. A
SKIN CORRECTION FACTOR K, OBTAINED
•
A
DISTRIBUTION
A FURTHER
BY DIFFERENTIAL EQUATIONS
CONSIDERED TO REEVALUATE THE AC RESISTANCE. FOR
RESISTANCE INCREASE
AND BESSEL FUNCTIONS,
60Hz, KISESTIMATED
AROUND
IS
1.02
(13.2)
OTHER VARIATIONS IN RESISTANCE ARE
CAUSED BY
'TEMPERATURE .
SPIRALING OF STRANDED
CONDUCTORS
BUNDLE CONDUCTORS
ARRANGEMENT
13.2.2
TEMPERATURE
EFFECT
THE RESISTIVITY OF ANY CONDUCTIVE MATERIAL VARIES LINEARLY OVER AN OPERATING TEMPERATURE,
AND THEREFORE, THE
TEMPERATURE
RESISTANCE
RISES, THE
OF
ANY
CONDUCTOR
SUFFERS THE
SAME
VARIATIONS.
As
CONDUCTOR RESISTANCE INCREASES LINEARLY, OVER NORMAL OPERATING
TEMPERATURES, ACCORDING TO THE FOLLOWING EQUATION:
(13.3)
I
where R2 = resistance at second temperature t2
R] = resistance at initial temperature tj
T = temperature coefficient for the particular material eC)
Resistivity (p) and temperature coefficient (1) constants depend upon the particular conductor
r.t:
Table 13.1lists resistivity and temperature coefficients of some typical conductor materials [31.
13.2.3
SPIRALING
AND
BUNDLE
CONDUCTOR
EFFECT
THERE
.:
ARE TWO TYPES OF TRANSMISSION
: :'HEAD CONDUCTORS, MADE
DepL of EEE, SJBIT
OF
NAKED
LINE CONDUCTORS:
METAL
AND
OVERHEAD
SUSPENDED
ON
AND
UNDERGROUND.
INSULATORS,
85
ARE
iDEES3
Transmission and Distribution
PREFERRED
OVER
UNDERGROUND
CONDUCTORS
BECAUSE
OF
THE
LOWER
COST
AND
EASY
MAINTENANCE. ALSO, OVERHEAD TRANSMISSION LINES USE ALUMINUM CONDUCTORS, BECAUSE OF THE
LOWER COST AND LIGHTER WEIGHT COMPARED TO COPPER CONDUCTORS,
SECTION AREA IS NEEDED TO CONDUCT
TYPES OF
COMMERCIALLY
REINFORCED
CONDUCTOR
(ACSR),
(AAC),
AVAILABLE
THE SAME AMOUNT
ALUMINUM
OF CURRENT.
CONDUCTORS:
THERE ARE DIFFERENT
ALUMINUM-CONDUCTOR-STEEL-
ALUMINUM-CONDUCTOR-ALLOY-REINFORCED
AND ALL-ALUMINUM-
ALTHOUGH MORE CROSS-
(ACAR),
ALL-ALUMINUM-
ALLOY-CONDUCTOR (AAAC).
..
TABLE 13.1 Resistivity and Temperature Coefficient of Some Conductors
Material
Resistivity at 2WC (H.m)
Temperature Coefficient (0C)
1.59X 10-8
1.72 X 10-8
1.77 X 10-8
2.83 X 10-8
243.0
234.5
241.5
Silver
Annealed copper
Hard-drawn copper
Aluminum
228.1
ALUMINUM STRANDS
2 LAYERS,
30 CONDUCTORS
STEEL STRANDS
7
FIGURE
ACSR
13.4
STRANDED ALUMINUM CONDUCTOR WlTH STRANDED STEELCORE (ACSR).
IS ONE
ALTERNATE
---~
CONDUCTORS
OF THE
MOST
USED
CONDUCTORS
LAYERS OF STRANDED CONDUCTORS,
IN TRANSMISSION
LINES. IT
CONSISTS
OF
SPIRALED IN OPPOSITE DIRECTIONS TO HOLD THE
STRANDS TOGETHER, SURROUNDING A CORE OF STEEL STRANDS. FIGURE
13.4
SHOWS AN EXAMPLE OF
ALUMINUM AND STEEL STRANDS COMBINATION.
THE PURPOSE OF INTRODUCING A STEEL CORE INSIDE THE STRANDED ALUMINUM CONDUCTORS IS TO
OBTAIN A HIGH STRENGTH-TO-WEIGHT
EASIER TO MANUFACTURE
RATIO. A STRANDED CONDUCTOR OFFERS MORE FLEXIBILITY AND
THAN A SOLID LARGE CONDUCTOR.
HOWEVER, THE TOTAL RESISTANCE IS
INCREASED BECAUSE THE OUTSIDE STRANDS ARE LARGER THAN THE INSIDE STRANDS ON ACCOUNT OF
THE SPIRALING
[8].
THE RESISTANCE OF EACH WOUND CONDUCTOR AT ANY LAYER, PER UNIT LENGTH, IS
BASED ON ITS TOTAL LENGTH AS FOLLOWS:
Dept of EEE, SJBIT
86
,
Transmission and Distribution
Rcooo = ~
VII + (1)2
'ifp
1DEES3
(13.4)
(Dim)
,,,hereRcand = resistanceof wound conductor (D)
J
I
.
+ (1T~)'
Prond
~
length of wound conductor (m)
limn}..= re anve prtc h
= -.-.-
0
f woun d con d uctor
2rJaycr
•
lwtn = length of one turn of the spiral (m)
.21'111}'t!r~diameter
ofthe layer (m)
THE PARALLEL COMB INAnON
OF N CONDUCTORS, WITH SAME DIAMETER PER LAYER, GIVES THE
RESISTANCE PER LA YERAS FOLLOWS:
Rlayer
.
I
= -»-. (n1m)
I:_!_'
..
(13.5)
. IRI
l""
,
Similarly, the total resistance of the stranded conductor is evaluated by the parallel combination of
resistancesper layer.
In high-voltage transmission lines, there may be more than one conductor per phase (bundle cOnfig·
uration) to increase the current capability and to reduce corona effect discharge. Corona effect occurs
when the surface potential gradient of a conductor exceedsthe dielectric strength of the surrounding air
(30 kVfern during fairweather), producing ionization in the area dose to the conductor, with consequent
corona losses,audible noise, and radio interference. As corona effect isa function of conductor diameter,
fine configuration, and conductor surface condition) then meteorological conditions playa kej' role in
~s evaluation. Corona lossesunder rain or snow, for instance, are much higher than in dry weather.
Corona, however, can be reduced by increasing the total conductor surface. Although corona losses
rely on meteorological conditions) their evaluation takes into account the conductance between conductors and between conductors and ground. By increasing the number of conductors per phase} the
\nal cross-section area increases, the current capacity increases, and the total AC resistance decreases
proportionally to the number of conductors per bundle. Conductor bundles may be applied to any
Depl ofEEE, SJBIT
87
10EE53
Transmission and Distribution
1
1'1(
,
'..,
d
1
I
I,.
d
~
d
,.,
... ...
a
•
•
(a)
(c)
(b)
FIGURE 13.5 Stranded conductors arranged in bundles per phase of (a) two, (b) three) and (c) four.
voltage but are 'alwaysused at 345 kVand above to limit corona. To maintain the distance between
bundle conductors along the line,spacers made of steel or aluminum bars are used. Figure 13.5shows
some typical arrangement of stranded bundle:configurations.
CURRENT-CARRYING
(AMPACITY)
IN OVERHEAD TRANSMISSION
CAPACITY
LINES, THE CURRENT-CARRYING
CAPACITY IS DETERMINED
THE CONDUCTOR RESISTANCE AND THE HEAT DISSIPATED FROM ITS SURFACE
[8].
MOSTLY BY
THE HEAT GENERATED IN
A CONDUCTOR (JOULE'S EFFECT) IS DISSIPATED FROM ITS SURFACE AREA BY CONVECTION AND RADIATION
GIVEN BY
Dept DfEEE, SJBIT
88
Transmission and Distribution
10EES3
(13.6)
•
where R = conductor resistance (0)
I = conductor current-carrying (A)
S = conductor surface area (sq. in.)
W<; = convection heat loss ('IN/sq. in.)
Wr= radiation heat loss (W/sq. in.)
Heat dissipation by convection is defined as
i' ( )
_ 0..0128#
we where
p
= atmospheric pressure (atm)
v
= wind velocity (ft/s)
~.!23~
air
( 13.7)
ut W
dcond
= conductor diameter (in.)
Talc = air temperature (kelvin)
~d
.:l t .=:= Tc - Tair = temperature rise of the conductor (0C)
Heat dissipation by radiation is obtained from Stefan-Boltzmann law and is defined as
Wr =
where
36.8£
[(
t;
T
) 4 ( 1000)4] •
air
1000'-
(W/sq. in.]
(13.8)
We = radiation heat loss ('tVI sq. in.)
E = emissivity constant (1 for the absolute black body and 0.5 for oxidized copper)
T. = conductor temperature (OC)
[ "ccc ambient temperature
(OC)
Substituting Eqs. 03.i) and (13.8) in Eq. (13.6) we can obtain the conductor ampacity at given
ternperarures
1=
V/s(wc + w
r}
(A)
R
1=
R
'T'I\
1 ~.l.<
al! \i
AND
"'." !'IL),SJBIT
.(T:lOO()4
- ~r)~(A)
S (At(O.OJ28v'PV)
'r
'3 ItC:i + 36.8E
-
nd
(13.9)
(13.10)
INDUCTIVE
89
10EES3
Transmission and Distribution
A
CURRENT-CARRYING
CONDUCTOR.
CONDUCTOR
IF THE CURRENT
VOLTAGE IS INDUCED.
PRODUCES CONCENTRIC
VARIES WITH
THEREFORE,
THE
TIME,
AN INDUCTANCE
MAGNETIC FLUX LINES AROUND
THE
THE
MAGNETIC FLUX CHANGES AND
A
IS PRESENT, DEFINED AS THE RATIO OF THE
MAGNETIC FLUX LINKAGE AND THE CURRENT. THE MAGNETIC FLUX PRODUCED
BY THE CURRENT
IN
TRANSMISSION LINE CONDUCTORS PRODUCES A TOTAL INDUCTANCE WHOSE MAGNITUDE DEPENDS ON
THE LINE CONFIGURATION.
To
DETERMINE
THE INDUCTANCE
OF THE LINE, IT IS NECESSARY TO
CALCULATE,AS IN ANY MAGNETIC CIRCUIT WITH PERMEABILITY M, THE FOLLOWING FACTORS:
1.
MAGNETIC FIELD INTENSITY
2.
3.
MAGNETIC FIELD DENSITY
•
H
B
•
FLUX LINKAGE L
l3.4.1
INDUCTANCE
CONDUCTOR
OF A SOLID,
ROUND, INFINITELY
LONG
CONSIDER AN INFINITELY LONG, SOLID CYLINDRICAL CONDUCTOR WITH RADIUS R, CARRYING CURRENT
I
AS SHOWN IN FIG.
CURRENT
13.6.
IF THE CONDUCTOR
IS MADE OF A NONMAGNETIC
MATERIAL,
AND THE
IS ASSUMED UNIFORMLY DISTRIBUTED (NO SKIN EFFECT), THEN THE GENERATED INTERNAL
AND EXTERNAL MAGNETIC FIELD LINES ARE CONCENTRIC CIRCLES AROUND THE CONDUCTOR
WITH
DIRECTION DEFINED BY THE RIGHT-HAND RULE.
l3.4.2 INTERNAL
MAGNETIC FLUX
INDUCTANCE
DUE
TO
INTERNAL
To OBTAIN THE INTERNAL' INDUCTANCE, A MAGNETIC FIELD WITH RADIUS X INSIDE THE CONDUCTOR
OF LENGTH L IS CHOSEN, AS'SHOWN IN FIG.
13.7.
THE FRACTION OF THE CURRENT Ix ENCLOSED IN THE AREA OF THE CIRCLE CHOSEN IS DETERMINED BY
7rX2
Ix = J-, (A)
(13.11 )
1Tr
Extarna! Field
FIGURE 13.6 External and internal concentric magnetic flux lines around the conductor.
Dept of EEE, SJBIT
90
iDEES3
Transmission and Distribution
_---
... ;0""..0;. ...
-
---..
.'
_-
...
,
I
I
!
Jx --------\
,
.>::
FIGURE 13.7 Internal magnetic flux.
AMPERE'S
LAW DETERMINES· THE MAGNETIC FIELD INTENSITY
Hx,
CONSTANT AT ANY POINT ALONG
THE CIRCLE CONTOUR AS
Hx =
I
I'
-== -x (AIm)
21Tr2 .
(13.12)
21TX
The magnetic flux density Bxis obtained by
(13J3)
where J.l = fLo = 4-rr x 10-7 HIm fur a nonmagnetic material.
The differential flux: d<,6enclosed in a ring of thickness dx lOr a I-m length of conductor and the
differential flux linkage dA in the respective area are
d<,6= Bxdx
= ~ (~)
(Ix!)r
1Tx2 d<,6= -fLo .d.\ = --rrr2
2-rr
(13.14)
dx (Wb/m)
4
dx (Whim)
(13.15)
The internal flux: linkage is obtained by integrating the differential flax linkage from x = 0 to
"-in!
=
f'
.0
X = r
dA = fLo I (Wb/m)
81T
(13.16)
.
Therefore, the conductor inductance due to internal flux linkage, per unit length, becomes
- "-in!
1.; rtt---I
Dept ')1' LFE, SJBIT
-
fLo (H I m ).
81T
(13.17)
r
91
Transmission
10EES3
and Distribution
EXTERNAL INDUCTANCE
THE
EXTERNAL INDUCTANCE IS EVALUATED ASSUMING THAT THE TOTAL CURRENT
AT THE CONDUCTOR
SURFACE (MAXIMUM
FIELD CIRCLE OF RADIUS Y (FIG.
DENSITY
By,
I
IS CONCENTRATED
SKIN EFFECT). AT ANY POINT ON AN EXTERNAL MAGNETIC
13.8), THE
MAGNETIC FIELD INTENSITY
H,
AND THE MAGNETIC FIELD
PER UNIT LENGTH, ARE
Hy
By
= .L. (Aim)
(13.18)
2-rry
= /LHy =
/La
!_ (T)
(13.19)
21T Y
The differential flux d</>enclosed in a ring of thickness
dy, from point DI to point D2, for a l-rn length of
conductor is
(13.20)
As the total current I flows in the surface conductor,
then the differential flux linkage dA has the same
magnitude as the differential flux d</>.
D2
ciA = d</>= J.Lo ~ dy (Wb/m)
270 y
(B.21)
The total external flux linkage enclosed by the ring is
obtained by integrating from DI to D2
FIGURE 13.8 External magnetic field.
Dl.
'\1-2
=
f
(D)
J.L 1JD1 -dy = _Q_
J.L I In
ciA = _Q_
270
. D,
D,
Y
270
_I
~
(Wb/m)
(13.22)
In general, the total external flux linkage from the surface of the conductor to any point D, per unit
length, is
Aext =
J
f)
r
J1. lin
ciA = _.0
270
(D)'r'
[Wb/rn]
(13.23)
.
The summation of the internal and external flux linkage at any point D permits evaluation of the total
inductance of the conductor 401> per unit length, as follows:
Dept of EEE, SJBIT
92
and Distribution
'frd 1)("mission
Amtl+A
at
=~
I [l+ln(~)]
=
iDEES3
:I InC_~4r)
J.Lo I.(__!!_)
tot = Amt+I Aut = 21f
n (ill,1R
L
\\'h cte
W :
','
(HI
m
(13.24)
(Vvbjm)
).
(13.25)
H. (geometric mean radius) = e-1/4, = 0.7788r
c considered as the radius of a fictitious conductor assumed to have no internal flux but
t> inductance as the actual conductor with radius r.
I'['.;L)LC L\.;'~CE OF A TWO-WIRE SINGLE-PHASE LINE
c '" nsider a two-wire single-phase line with solid cylindrical conductors A and B \\1th the same
.. arne length 1,and separated by a distance D, where D > f, and conducting the same current I, as
- : ~':g. 13.9. The current flows from the source to the load in conductor A and returns in
B (IA=-In).
l;'.lgnetic flux generated by one conductor links the other conductor. The total flux linking
,;c, ; -\, for instance, has two components: (a) the flux generated by conductor A and (b) the flux
'~,';
conductor B which links conductor A.
.. in Fig. 1110, the total-flux linkage from conductors A and B at point Pis
(13.26)
(13.27)
EIT
93
10EE53
Transmission and Distribution
•
D
.',
A
Q
FIGURE 13.9
where
EXternal magnetic flux around conductors in a two-wire single-phase line.
flux linkage from magnetic field of conductor A on conductor
flux linkage from magnetic field of conductor B on conductor
ASBP= flux linkage from magnetic field of conductor B on conductor
AsAP = flux linkage from magnetic field of conductor A on conductor
The expressions of the flux linkages above, per unit length, are
AAAP=
AABP=
A",AP
J.Lo
=21T
P
P
P
P
Itn. ( --DAP ) (\\'bjm)
GMRA
DaP
J.Lo
A.>lJlp
=
BspdP= --Itn
J
A at point
A at point
B at point
B at point
_
2'lT
D
AsAP = fDJo.P BApdP = _ J.Lo I
2'lT
•D
(Dsp)
--
(Wb/m)
(13.29)
(DAP)
D
(Wb/m)
(13.30)
D
In
(13.28)
Assp = J.l.-o I In( _DnP __)_ (Wb/m)
21T
GMRB
.
(13.31)
The total flux linkage of the system at point P is the algebraic summation of AAP and Asp
(13.32)
D
J.l.o
_ [. ( -_-AP ) _( - D ) ( -DBP
Ap =-11n
2'lT
GMRA
Dept ofEEE, SJBIT
DAP
GMRs
)
(
D)]
--
DsP
J.Lo _ (
rY
=-11n
2'lT
GMRAGMRB
) (Wb/m)
(13.33)
94
•
•.
Transmission and Distribution
10EE53
If the conductors have the same radius.
TA= TS = T, and the point P is shifted to
infinity, then the total flux linkage of the
system becomes
A = JLo I1n(__E_)
GMR
p
,.
r h',
(vVb/m)
(13.34)
'IT
and the total inductance per unit length
becomes
~'.L )).1 0
Rux linkage of (a) conductor A at point Pand
or B on conductor A at point P. Single-phase system.
(13.35)
g Eqs, (13.25) and (13.35), it can be seen that the inductance of the single-phase system is
t', (' i nd uctance of a single co nductor,
r ,I line with stranded conductors, the inductance is determined using a new GMR value
',,11:1ed GM~trnndo:h evaluated according to the number of conductors. If conductors A and B in the
;,~-phasesystem, are formed by nand m solid cylindrical identical subcondnctors in parallel, respectt~hen
" h ','
..
GMRA-!trllnd~= n2
n
II II
Dij
(13.36)
Dij
(13.37)
i=l j=1
m
GMRB-!tI'il~ = m1
m
II II
i=1 ]=1
GMR.tmnded for a particular cable can be found in conductor tables given by the
-r,
[f
,'j
.;
'_<_ ('
!,L
c":
. _. \,!!':,
,e conductor is composed of bundle conductors, the inductance is reevaluated taking
t the number of bundle conductors and the separation among them. The GM~und1c is
to determine the final inductance value. Assuming the same separation among bundle
the equation for GMRbundl£,up to three conductors per bundle, is defined as
GMR"
i
.~
;~:~-.lIilT
bundle conductors
=
Vi d-1 GMR.t~
(13.38)
95
10EE53
Transmission and Distribution
GMRII bundle
conductors
= yt dn-J GMRsmoocd
(13.38)
'where n = number of conductors per bundle
GMR.trand.,d = GMR of the stranded conductor
d = distance between bundle conductors
For four conductors per bundle with the same separation between consecutive conductors, the
GIvlRlxmdl., is evaluated as
GMRt
bundle conductors
= 1.09 {/ d3GMR$lranded
(13.39)
•
INDUCTANCE OF A THREE-PHASE LINE
The derivations for the inductance in a single-phase system can be extended to obtain the inductance per
phase in a three-phase system. Consider a three-phase, three-conductor system with solid cylindrical
conductors with identical radius rA, rB, and ro placed horizontally with separation P'B' Doc, and Dc~
(where D> r) among them. Corresponding currents lA' In, and lc flow along each conductor as shown
in Fig. 13.11.
The total magnetic flux enclosing conductor A at a point Paway from the conductors is the sum of the
flux produced by conductors A, B, and C as follows:
cPAP = cPAAP + cPABP + cPACP
(13.40)
where cPAAP= flux produced by current IA on conductor A at point P
cPABP= flLLX produced by current In on conductor A at point P
cPAep=flux produced by current lc on conductor A at point P
Considering l-rn length for each conductor, the expressions for the fluxes above are
Dept ufEEE, SJBIT
96
Transmission and Distribution
10EE53
..
,
FIGURE B.ll
Magnetic flux produced by each conductor in a three-phase system.
cPAAP =
JLo In ( GMR
DAPA
2"ITIA
)
(13.41)
(Whim)
(13.42)
(13.43)
, SIBIT
97
Transmission and Distribution
1DEES3
The corresponding flux:linkage of conductor A at point P (Fig. 13.12) is evaluated as
(13.44)
having
(13.45)
(a)
p
(b)
p
FIGURE 13.12 Flux linkage of (a) conductor A at point P, (b) conductor B on conductor A at point 1>, and (c)
conductor C on conductor A at point P. Three-phasesystem.
Dept of EEE, SJBIT
98
;\nsmission and Distribution
L>!IP
=
J
AACP =
J
AASP
JLo.
BBP dP = -
10EE53
IB
in
(~p)
-D
DAli
2Tl'
Lb
JLo Ic In -D
BcpdP= -.
DA(;
(Dcp)
2Tl'
\\'1<\e AAP=total flax linkage of conductor A at point
(Wb/m)
(13.46)
(WhIm)
(13.47)
AD
AC
P
'- .r= flax linkage from magnetic field of conductor A on conductor A at point P
A !BP= flux linkage from magnetic field of conductor B on conductor A at point P
<"
", = flux linkage from magnetic field of conductor C on conductor A at point P
"'uting Eqs. (13.45) through (13.47) in Eq. (13.44) and rearranging, according to natural
,i,hin.> law, we have
DAP )
AM = -fLo [ IA In ( --R-
GM
21T
AAP
=;; [IA
A
+ In In(Lnp)
-D
+ lc
~B
In(G1~:RJ
+18 In(D:B)
(U:p)]
In -
DAC
+ Ietn(D:J
+ 21T
fLo [IAIn(DAP) + 18 In(Dsp) + Ic
(Wb/m)
(13.48)
]
In(Dcp)] (\"lb/m)
,(13.49)
':hc arrangement ofEq. (13.48) into Eq, (13.49) is algebraically correct according to natural logarithms
" .. r-wever, as the calculation of any natural logarithm must be dimensionless, the numerator in the
,;,i'.I'I$ In(l!GMRA), tn(l/DAB),
and In(l/D.",c) must have the same dimension as the denominator.
;, "'.':1:" applies for the denominator in the expressions In(DAP)' In(DBP)' and In(Dcp).
, a balanced three-phase system, where h + Is + Ic = 0, and shifting the point P to infinity in
'Lat DAP= DBP = DCi'> then the second part of Eq. (13.49) is zero, and the flux linkage of
k< A becomes
AA = J10
21T
[1...
In (G IR .)
1M
A
+ Is in (DlAU.) + Ie In (D1AC.)].
(13.50)
(\Vb/ m)
•
99
10EE53
Transmission and Distribution
Similarly. the flux linkage expressions for conductors Hand C are
An
=~
[IA In(~)
Ac
=~
[h In(~A) + In(~B) + In(G~~Rc)]
+ In In(~~Ra)
Ia
+ Ie In(~e)]
Ie
(13.51)
(\~!m)
(13.52)
(Wb/m)
The
linkage of each phase conductor depends on the three currents, and therefore, the inductance
per phase is not only one as in the' single-phase system. Instead, three different inductances (self and
. mutual conductor inductances) exist. Calculating theinductance values from the equations above and
.arranging the equations in a matrix form we can obtain the set of inductances in the system
(13.53)
where AA, As. Ae = total flux linkages of "conductors A, H, and C
LAA:t LSII, Lee = self-inductances of conductors A, B, and C field of conductor A at point P
LAB. Lac, l..cA, LIlA> LeB, LAC = mutual inductances among conductors
V{Ith nine different inductances in a simple three-phase system the analysis could be a little
more complicated. However, a single inductance per phase can be obtained if the three conductors
are arranged with the same separation among them (symmetrical arrangement), where
D:..= DAB = Doc = L\:.A. For a balanced three-phase system (fA + Ia + Ie = O. or IA = -Is - Ie), the flux
linkage of each conductor, per unit length, will be the same. From Eq, (13.50) we have
(13.54)
If GMR value is the same for all conductors (either single or bundle GMR), the total flux linkage
expression is the same for all phases. Therefore, the equivalent inductance per phase is
Lph...., = Po
2'lT
.iNDUCTANCE
I"INES
OF TRANSPOSED
Dept of} ':EE, SJBIT
In(
D
) (Him)
GMRphas<:
THREE-PHASE
(13.55)
TRANSMISSION
100
10EE53
Transmission and Distribution
IN
ACTUAL
TRANSMISSION
LINES,
THE
PHASE
CONDUCTORS
CANNOT
MAINTAIN
SYMMETRICAL
ARRANGEMENT ALONG THE WHOLE LENGTH BECAUSE OF CONSTRUCTION CONSIDERATIONS, EVEN WHEN
BUNDLE CONDUCTOR
DIFFERENT
FOR
SPACERS ARE USED. WITH ASYMMETRICAL SPACING, THE INDUCTANCE WILL BE
EACH
PHASE,
WITH
A CORRESPONDING
UNBALANCED
VOLTAGE DROP ON EACH
CONDUCTOR. THEREFORE, THE SINGLE-PHASE EQUIVALENT CIRCUIT TO REPRESENT THE POWER SYSTEM
CANNOT BE USED.
HOWEVER, IT IS POSSIBLE TO ASSUME SYMMETRICAL ARRANGEMENT IN THE TRANSMISSION LINE BY
TRANSPOSING THE PHASE CONDUCTORS. IN A TRANSPOSED SYSTEM, EACH PHASE CONDUCTOR OCCUPIES
THE LOCATION OF THE OTHER TWO PHASES FOR ONE-THIRD OF THE TOTAL LINE LENGTH AS SHOWN IN
FIG.
13.13. IN
DISTANCE
D,
(GMD)
THIS CASE, THE AVERAGE DISTANCE GEOMETRICAL MEAN DISTANCE
AND
THE
CALCULATION
OF
PHASE
INDUCTANCE
DERIVED
FOR'
SUBSTITUTES
SYMMETRICAL
ARRANGEMENT IS STILL VALID.
THE INDUCTANCE PER PHASE PER UNIT LENGTH IN A TRANSMISSION LINE BECOMES
J.Lo ( GMD ) . .'
Lp"""" = 21T In G~fRp"""" (Him)
(13.56)
Once the inductance per phase is obtained. the inductive reactance per unit length is
(13.57)
A
C
A
B
B
G
><=
c
><=
A
B
1---------------------+-------------------------+---------------------.,
//3
//3
1/3
FIGURE 13.13
Arrangement of conductors ill a transposed line.
CAPACITANCE
REACTANCE
CAPACITANCE
EXISTS
CAPACITIVE
AND
AMONG
TRANSMISSION
LINE
CONDUCTORS
DIFFERENCE. To EVALUATE THE CAPACITANCE BETWEEN CONDUCTORS
WITH PERMITTIVITY
«,
DUE
TO
THEIR
POTENTIAL
IN A SURROUNDING
MEDIUM
IT IS NECESSARY TO DETERMINE THE YOLTAGE BETWEEN THE CONDUCTORS, AND
THE ELECTRIC FIELD STRENGTH OF THE SURROUNDING.
CAPACITANCE
CONDUCTOR
CONSIDER
OF
A SOLID, CYLINDRICAL,
PERMITTIVITY
«0,
A
SINGLE-SOLID
LONG CONDUCTOR
AND WITH A CHARGE OF Q"
WITH
RADIUS
COULOMBS PER METER, UNIFORMLY DISTRIBUTED ON
THE SURFACE. THERE IS A CONSTANT ELECTRIC FIELD STRENGTH
Dept of EEE, SlBIT
R, IN A FREE SPACE WITH
ON THE SURFACE OF CYLINDER
101
1DEES3
Transmission and Distribution
(FIG.
13.14).
THE RESISTIVITY OF THE CONDUCTOR IS
ASSUMED TO BE ZERO (PERFECT CONDUCTOR),
WHICH RESULTS IN ZERO INTERNAL ELECTRIC FIELD DUE
TO THE CHARGE ON THE CONDUCTOR.
THE CHARGE QI> PRODUCES AN ELECTRIC FIELD RADIAL TO THE CONDUCTOR WITH EQUIPOTENTIAL
SURFACES CONCENTRIC
TO THE CONDUCTOR. ACCORDING TO GAUSS'S LAW, THE TOTAL ELECTRIC FLUX LEAVING A CLOSED
SURFACE IS EQUAL TO THE TOTAL
THEREFORE, AT AN OUTSIDE POINT
CHARGE
P
INSIDE THE
VOLUME
ENCLOSED
BY THE
SURFACE.
SEPARATED X METERS FROM THE CENTER OF THE CONDUCTOR,
THE ELECTRIC FIELD FLUX DENSITY AND THE ELECTRIC FIELD INTENSITY ARE
Density],
=!1 = _1_ (C)
A
(13.59)
21TX
Bectrie Field Lines
FIGURE 13.14 Electricfieldproduced from a single conductor.
Dept of EEE, SJBIT
102
-----
-
---------------------------------------
Trausunssion and Distribution
10EES3
~.~A!§.:::::;.~~·i<iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
Ep = Densityp = _q_ (Vim)
S
(13.60)
21TSOX
where Den"ityp= electric flux density at point P
electric field intensity at point P
A = surface of a concentric cylinder with l-m length and radius x (nr')
E::::' t'7)
= ~:
= permittivity of free space assumed for the conductor (F/m)
The potential difference or voltage difference between two outside points PI and Pz 'with correspondipi! jl':- r., ,", Xl and Xl from the conductor center is defined by integrating the electric field intensity
fro n:
1 .' ;.'~
Vi-2 =
X:
J
::r,
dx
Ep-- =
X
JX! -'-"q ---dx =.--.q x,
2»so x
21TSo
[Xl]
ns.sn
In __ (V)
Xl
Then, the capacitance between points PI and Pz is evaluated as
q
C'-2
2»&0
=V
= -[
1-2
I
X2
n-
.
.
]
(13,62)
(F/m).
XI
If point Pj is located at the conductor surface (XI = r); and 'point P2 is located at ground surface below
the conductor (Xl = h), then the voltage of the conductor and the capacitance between the conductor
and. :C"'.'.,' . are
q
Vcond = --.
2'lT&0
In [It]
- (V)
q
4ond-groWld
=
Vo:md
(13.63)
r
-rhl
2m>"o....
=
In -
(13.64)
(F/m)
r
,\ P.\( IT ".>l'CE OF
A SINGLE-PHASE
LINE
WITH
Two
, TWO-WIRE SINGLE-PHASE LINE WITH CONDUCTORS A AND B WITH THE SAME RADIUS R,
<:FT'..\p.:.
'·:liCL
'; T~V A DISTANCE
r:
'.ONDUCTOR
D > RA
AHASA
AND RB. THE CONDUCTORS ARE ENERGIZED BY A VOLTAGE SOURCE
CHARGE QP AND CONDUCTOR BA CHARGE Q
AS SHOWN IN FIG. 13.15.
THE CHARGE ON EACH CONDUCTOR GENERATES INDEPENDENT ELECTRIC FIELDS. CHARGE QP ON
COND\JCTC~
(
r.; ~: l~_,\
•>
.. '
A
/'. VOLTAGEVAH_A
BETWEEN BOTH CONDUCTORS. SIMILARLY,CHARGE
Q
ON CONDUCTOR
:, , ;~SA VOLT AGE V AB-B BETWEEN CONDUCTORS .
'JBrr
103
10EE53
Transmission and Distribution
D
D
A
FIGURE 13.15 Electricfield produced from a two-wire single-phasesystem.
V:-O\B-A is
calculated by integrating the electric field intensity, due to the charge on conductor A, on
conductor B from r; to D
(13.65)
VAB-S
is calculated by integrating the electric field intensity due to the charge on conductor B from D to
VAB-U
J'" Eudx =--In-q
=
D
2~EO
[f'- B]
YB
(13.66)
D
The total voltage is the sum of the generated voltages VAH-A and VAB-B
Vw =
VAll-A
+ VAll-B
= --
[D]
q
2~EO
In -
rA
- --
q
2~EO
In [rB].
- = -- q
.
D
2~EO
.[..D2
In-. -
]
If the conductors have the same radius, fA = fS= 1; then the voltage between conductors
capacitance between conductors CAB' for a l-m line length are
~~B
=
.s. In [D]r
1TEO
(V)
~EO
CAB
=
[D]
In -
(F/m)
(13.67)
.rAlll
VAS>
and the
(13.68)
(13.69)
r
The voltage between each conductor and ground (G) (Fig. 13.16) is one-half of the voltage between the two
G~;;dl1cl.(':-~.
Therefore, the capacitance from either line to ground is twice the capacitance between lines
Dept ofEEE, SJBIT
104
Transmlssion and Distribution
VAB
VBG = 2 (V)
VAG =
CAG = -
10EE53
q
VAG
21TSo
= -_
(F/m)
(13.70)
(13.71)
In [~]
..
q+
A ~~--~~I--~
B
. VAG
') Capacitance between line to ground in a two-wire single-phaseline.
{·.\P \{!! .'.NCE OF A THREE-PHASE
LINE
105
10EE53
Transmission and Distribution
Consider a three-phase line with the same voltage magnitude between phases, and assuming a balanced
system with abc (positive) sequence such that llA+ qa + qc =0. The conductors have radii rA' ra. and rc.
and the space between conductors are DAB>DBC> and DAC (where DAB. Dac. and DAc > rA. ra. and rd·
Also, the effect of earth and neutral conductors is neglected.
The expression for voltages between two conductors in a single-phase system can be extended to
obtain the voltages between conductors in a three-phase system. The expressions for VAB and VAC are
(13.72)
(13.73)
If the three-phase system has triangular arrangement with equidistant conductors such
that DAB = DBC = DAC = D, with the same radii for the conductors such that rA=rs = rc = r (where
D > r), the expressions for \fAB and v...c are
VAS ~ 2:"
~ 2~"
VAC ~
2:eo [qAIO [~] + ~m
~2~e.
Dept;' r' j:.' ~E, SJBIT
[q.m [~] +q.ln
[q.m [~] + q,m
[qAm [~]
[~l [~ll
[~ll
+qcln
(V)
(13.74)
[~l
+ qcln [;]]
+90k[~]] (V)
(13.75)
106
Transmission and Distribution
1DEES3
Balanced line-to-line voltages with sequence abc, expressed in terms of the line-to-neutral voltage are
where VAN is the line-to-neutral voltage. Therefore, VAN can be expressed in terms of VAD and ~c as
VAN
= VAn + VAC
(13.76)
3
and thus, substituting.
~<\B
and
VAN ~ ~e,
VAC
from Bqs, (13.67) and (13.68) we have
[[qAln [~]
.. ~ ~so [2q.ln
+ qsln
+ [qAill
[;]]
[~l(qs +qc) ill [;]]
+
[~] + ~ln
[;]]]
(V)
(13.77)
Under balanced conditions qA + tin + lJc= 0, or -q ....= (qB + qc ) then, the final expression for the lineto-neutral voltage is
VAN
= -2 1 qA In
'ITSo
[D]r
(V)
(13.78)
The positive sequence capacitance per unit length between phase A and neutral can now be obtained.
The same result is obtained for capacitance between phases Band C to neutral
qA
CAN
=
v-
2'ITSo
= .
AN
CAPACIT ANCE
OF
STRANDED
[D]
In-
(13.79)
CP/m)
r
BUNDLE
CONDUCTORS
THE CALCULATION OF THE CAPACITANCE IN THE EQUATION ABOVE IS
BASED ON
1.
SOLID CONDUCTORS WITH ZERO RESISTIVITY (ZERO INTERNAL ELECTRIC FIELD)
2. CHARGE UNIFORMLY DISTRIBUTED
3.
EQUILATERAL SPACING OF PHASE CONDUCTORS
IN ACTUAL TRANSMISSION LINES, THE RESISTIVITY OF THE CONDUCTORS PRODUCES A SMALL INTERNAL
ELECTRIC FIELD AND THEREFORE,
THE ELECTRIC FIELD AT
THE
CONDUCTOR
SURFACE IS SMALLER
THAN THE ESTIMATED. HOWEVER, THE DIFFERENCE IS NEGLIGIBLE FOR PRACTICAL PURPOSES.
BECAUSE OF THE PRESENCE OF OTHER CHARGED CONDUCTORS,
NONUNIFORM,
is
THE CHARGE DISTRIBUTION
IS
AND THEREFORE THE ESTIMATED CAPACITANCE IS DIFFERENT. HOWEVER, THIS EFFECT
NEGLIGIBLE FOR MOST
PRACTICAL CALCULATIONS. IN A LINE WITH STRANDED CONDUCTORS, THE
CAPACITANCE IS EVALUATED ASSUMING A SOLID CONDUCTOR WITH THE SAME RADIUS AS THE OUTSIDE
RADIUS OF THE STRANDED CONDUCTOR.
MOST TRANSMISSION
THIS PRODUCES A NEGLIGIBLE DIFFERENCE.
LINES DO NOT HAVE EQUILATERAL SPACING OF PHASE CONDUCTORS.
THIS
CiiLSLS U;FtERENCES BETWEEN THE L1NE-TO-NEUTRAL CAPACITANCES OF THE THREE PHASES. HOWEVER,
f)cni (' 1'1 FE.. SJBIT
107
10EES3
Transmission and Distribution
TRANSPOSING THE PHASE CONDUCTORS BALANCES THE SYSTEM RESULTING IN EQUAL LINE-TO-NEUTRAL
CAP ACITANCE FOR EACH PHASE AND IS DEVELOPED IN THE FOLLOWING MANNER
Consider a transposed three-phase line with conductors having the same radius r, and with space
between conductors DAB, ~,and
DAC , where DAB,Doc, and DAc > r.
Assuming abc positive sequence, the expressions for VABon the first, second, and third section of the
transposed line are
VABflut = 2~eo [qA In [D;B]
+ ~ In [v'AB] + qcln[~:]]
= 2:'80
[qAIn
(13.80)
[~~]J
(V)
(13.81)
[D;c] + qs In [O:c) + qc In [~]] (V)
(13.82)
v.~second = 2~eo [qAln[r;c] + qnln[~]
v.~third
(V)
+ qC1n
Similarly, the expressions for VAC on the first. second, and third section of the transposed line are
VAC
firn
V.~C5«ond
= 2~eo
[~ln[ D;c] + qsIn [::]
= 2~eo [qAln[D;B] + qnln[~~J
VACthird= 2~80
[~dn[n;c]
+_--qB
In [~:]
+ ~ In [;~]]
+ qC1n[;AB]]
+ ~ln[;oc]]
(13.83)
(l3.84)
(13.85)
Taking the average value of the three sections, we have the final expressions of VAS and VAC in the
transposed line
Dept of EEE, SJBIT
108
Transmission and Distribution
10EE53
(13.86)
(13.87)
For a balanced system where =q« = (qB + qc), the phase-to-neutral voltage
VA.>;!
(phase voltage) is
(13.88)
where GMD = ~/DABDBCDO\ = geometrical mean distance for a three-phase line.
For bundle conductors, an equivalent radius r., replaces the radius r of a single conductor and is
determined by the number of conductors per bundle and the spacing of conductors. The'expression of rc
is similar to GMRbundle used in the calculation of the inductance per phase, except that tbe actual outside
tJdhlo ,;f the conductor is used instead of the G~IRpb"sr. Therefore, the ex-pressionfor YAN is
¥:\N transp
= -2-1
1rEo
[GMD]
qA. In-re
(V)
(13.89)
where r" = (d"-lr)l/n=equivalent
radius for up to three conductors per bundle (m)
fr
1.09 (d3t)I/4 = equivalent radius for four conductors per bundle (m)
(I distance between bundle conductors (rn)
i, -:number of conductors per bundle
i ,;:'" , l~iC capacitance and capacitive reactance, per unit length. from phase to neutral can be
CAN tramp =
XAN trsnsp
I I·J. SJBIT
=
qA
21rso
(')
= --:r;-::G:-:-"-1D=-=-]
FI'm
AN tramp
In •..~i;e .
1T
I'
1
21r/C r\Ntransp
= _.._1_. In [GMDJ. (Him)
41rfEo
f.,.
(13.90)
(13.91)
109
Transmission and Distribution
1DEES3
CAPACITANCE DUE TO EARTH'S SURFACE
CONSIDERING
A SINGLE-OVERHEAD
SEPARATED A DISTANCE
H FROM
CONDUCTOR
WITH A RETURN
PATH
THROUGH
TH.E EARTH,
EARTH'S SURFACE, THE CHARGE OF THE EARTH WOULD BE EQUAL IN
MAGNITUDE TO THAT ON THE CONDUCTOR BUT OF OPPOSITE SIGN. IF THE EARTH IS ASSUMED AS A
PERFECTLY CONDUCTIVE HORIZONTAL PLANE WITH INFINITE LENGTH, THEN THE ELECTRIC FIELD LINES
WILL GO FROM THE CONDUCTOR TO THE EARTH, PERPENDICULAR
TO THE EARTH'S
SURFACE (FIG.
13.17).
H
Earth's Surface
FIGURE 13.17 Distribution of electric field lines from an overhead conductor to earth's surface.
To calculate the capacitance, the negative charge of the earth can be replaced by an equivalent charge
of an image conductor with the same radius as the overhead conductor, lying just below the overhead
conductor (Fig. 13.18).
The same principle can be extended to calculate the capacitance per phase of a three-phase system.
Figure 13.19 shows an equilateral arrangement of identical single conductors for phases A, B, and C
carrying the charges qA, qa, and qc and their respective image conductors A', B', and C'.
DA, DB, and Dc are perpendicular distances from phases A, B,and C to earth's surface. DAA" DBll', and
Dcc' are the perpendicular distances from phases A, B, and C to the image conductors A', B', and C',
\hltage VAncan be obtained as
V:"8= _1_
qA
In[~AB]
+... q8 In [D'IlJ + QC.I0. [~k] -..j
:\
AS
21iEo [ -iJAln [D'_'-.\B
1]
D.~A!.
Dept ofEEE, SJBIT
AC
[Dsc]
- qn In [DaB
- QC1n 1]
DAB!
(V)
(13.92)
D,K'
110
•
Transmission and Distribution
iDEES3
•
,
Equivalent Earth Charge
FlC,L'RE 13.18
Equivalent image.conductor representing the charge of the earth.
III
10EES3
Transmission and Distribution
Beqs
Overhead Conductors
crOcI
:
,,
Dc
I
,
Image Conductors
FIGURE 13.19 Arrangement of image conductors in a three-phase transmission line.
As overhead conductors are identical, then r =
arrangem~nt, D=DAO = Doc = DCA
rs: =
ro= rc. Also, as the conductors have equilateral
Similarly, expressions for Voc and \j.c are
Dept of EEE, SJBIT
112
Transmission and Distribution
~2~; [-qAln
VBC
VAC ~ 'uo
The
[q+
.~\ oltage
VAN
+q.H~]
-In [~])
[~]
[0,] -In [~])
-
qsm
iDEES3
+qc( ln [ti] - In[;;])]
(V) (13.94)
[~:l qc [~l- 1
+
(In
In[~:])
(V)
(13.95)
becomes, through algebraic reduction,
•
(13.96)
Therefore, the phase capacitance CAN' per unit length, is
C:<\N
=-
IJA
v'\N
2'lT80
.3
=
In
(~J ln
=
(F/m)
[t~::.~:~:]
(13.97)
Equations (13.79) and (13.97) have similar expressions, except for the term In «DAlY Dsc DCA,)113/(DAA'
Dln,
'1/3) included in Eq. (13.97), That term represents the effect of the earth on phase
increasing its total value. However, the capacitance increment is really small, and is usually
aI)""';"'6
-rause distances from overhead' conductors to ground are always greater than distances
;·..·,;..luctors.
(' Characteristics
), ~
1 ,"')
,
of Overhead Conductors
, and 13.2b present typical values of resistance, inductive reactance and capacitance react.t length, of ACSR conductors. The size of the conductors (cross-section area) is specified in
, , ,'<11eters and kcmil, where a emil is the cross-section area of a circular conductor 'with a
; 1/ 1000 in. The tables include also the approximate current-carrying capacity of the
j assuming 60 Hz, wind speed of 1.4 mijh,
and conductor and air temperatures of 75"C
, .spectively Tables 13.3a and 13.3b present the corresponding characteristics of AACs.
". ,3IT
113
10EE53
Transmission and Distribution
UNIT-7
PERFORMANCE
TR.-LINES,
NOMINAL
T
OF POWER TRANSMISSION LINES- SHORT TR.-LlNES, MEDIUM
METHOD, END CONDENSER METHOD, IT METHOD AND LONG
TRANSMISSION LINES, ABeD CONSTANTS OF TRANSMISSION LINES, POWER FLOW
THROUGH
LINES,
P-V
&
Q-V
COUPLING.
10
HOURS
SHORT TRANSMISSION UNES (UP TO 50 rnl, OR 80 km)
In the case of a short transmission line, the capacitance and leakage
resistance to the earth are usually neglected, as sbown in Figure 2.23.
Therefore, the transmission line can be treated as a simple, lumped, and
constant impedance, that is.
..
Z==R+jXL
=zl
;; rl+ jxl
n
(2.163)
where Z = total series impedance per phase in ohms
z = series impedance of one conductor in ohms per unit length
XL = total inductive reactance of one conductor ill ohms
x ==inductive reactance of one conductor in ohms per unit length
I;;;; length of line
The current entering the line at the sending end of the line is equal to the
current leaving at the receiving end. Figures 2.24 and 2.25 show vector (or
phasor) diagrams for a short transmission line connected to- an inductive
load and a capacitive load, respectively, It can be observed from the figures
that
V.l =Vn + IRZ
(2.164)
Is= III = I
(2.165)
Vfl==VS-JRZ
(2.166)
where Vs "'"sending-end phase (line-to-neutral) voltage
VR = receiving-end phase (line-to-neutral) voltage
Is = sending-end phase current
Dept ofEEE, SJBIT
114
..
IDEES3
Transmission and Distribution
IS~
IR_""
.....
y
Sending
end
+
Vs
y
a'
y
+
Vn
(50urce)
end
(source)
NO--------------------------------------ON'
. Figure 2.23. Equi..'alent circuit of short transmissionline.
.. ' '~':E,SJBIT
Receiving
115
Transmission and Distribution
lOEES3
Agure 2.24. Phasor diagram of short transmission line to inductive load.
IR = receiving-end phase current
Z = total series impedance per phase
Therefore, using VB.as the reference. equation (2.164) can be written as
(2.167)
where the plus or minus sign is determined by 4> If' the power factor angle 'of
the receiving end or load. If the power factor is lagging. the minus sign is
employed. On the other hand, if it is leading. the plus sign is used.
However. if equation (2.166) is used, it is convenient to use Vs as the
reference. Therefore.
VR =Vs - (Is cos <Ps± Jls sin <lls)(R
(2.168)
+ jX)
where €J>s is the sending-end power factor angle, that determines. as before.
whether the plus or minus sign will be used. Also, from Figure 2.24, using
VIt as the reference vector
7
Vs = V(VR + IR
Dept of EEE, SJBIT
COS<l>R
+ IX sin (I>}l)2 + (IA'''COS4>R
± IR sin 4>R)4
(2.169)
116
.msmission and Distribution
""~">">,~,,.,,,,,,~, iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii~iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii~~~
iDEES3
Ys
VB
Figure 2..25. Phasor diagram of short transmission line connected to capacitive load.
and load angle
(2.170)
or
~
(I
= tan
-1
IX cos <l>R ± lR sin <})R
VR + lR cos4tR + IX sin <PH
>
(2.171)
>
The generalized constants, or ABeD parameters, can be determined by
inspection of Figure 2.23. Since
[::]=[~ ~][~;]
(2.172)
and AD - BC = 1, where
A=l
B=Z
C=o
(2.173)
0= 1
then
(2.174)
and
Lic
'I
;!
r:
S.JBIT
117
10EE53
Transmission and Distribution
The transmission efficiency of the short line can be expressed as
.,.,=
output
input
_ V3VRI cos 4>R
- V3Vsl cos <l>s
_ VR cos <l>R
- Vs costl>s
(2.175)
Equation (~.175) is applicable whether the line is single phase or three
phase.
The transmission efficiency can also be expressed as
_
Y1 -
output.
output + losses
For a single-phase line,
(2.176)
For a three- phase line,
(2.177)
2.17.1 Steady-State Power Limit
Assume that the impedance of a short transmission line is given as Z = Zj_g_.
Therefore, the real power delivered, at steady state, to the receiving end of
the transmission line can be expressed as
.
PA =
Vs X VR
Z
.
cos(8 - 0) -
2
It
V
z
cos 0
(2.178)
and similarly, the reactive power delivered can be expressed as
(2.179)
of EEE, SJBIT
118
lOEES3
T!"-mslnission and Di~;trlbution
If Vs and VR are the line-to-neutral voltages, equations (2.178) and
(2.179) give PR and QR values per phase. On the other hand, if the obtained
PRand Q8 values are multiplied by 3 or the line-to-line values of Vs and VIl
are employed, the equations give the three-phase real and reactive power
delivered to a balanced load at the receiving end of the line.
If, In equation (2.178), all variables are kept constant with the exception
of I), so that the real power delivered, PR• is a function of ~ only, PI{ is
maximum when I) = e. and the maximum power t obtainable at the receiving
end for 8' given regulation can be expressed as
Z~ (VV;Z - R).
V2
PR,m..,. =
(2.180)
where Vs and VI! are the phase (Iine-to-neutral) voltages whether the system
is single phase or three phase.
The equation can also be expressed as
vi x cos 9
(2.181)
Z
• Also called the steady-state power limit.
SJBIT
119
lOEES3
Transmission and Distribution
If Vs =
vR'
(2.182)
or
(2.183)
aod similarly, the corresponding reactive power delivered to the load is
given by
.
.QJ R =
.v1
- __..!
Z sin (J
(2.184)
As can be observed. both equations (2.183) and (2.184) are independent of
Vs voltage. The negative sign in equation (2.184) points out tha·[ the load is
a sink ofleading vars,' that is, going to the load or a source of Iagging VaTS
(i.e., from the load to the supply). The total three-phase power transmitted
on the three-phase line is three 'times the power calculated by using the
above equations. If the voltages are given in volts, the power is expressed in
watts or vars. Otherwise. if they are given in kilovolts, the power is
expressed in megawatts or megavars.
In a similar manner, the real and reactive powers for the sending end of a
transmission line can be expressed as
v:~Z V
2.
_ Vs
X
Ps-Zcoso-
R
cos(8+o)
(2.185)
sin(8 + 8)
(2.186)
and
QS =
V x
V2
Z sin (J
__l.
-
S
Z
v
R
If, in equation (2.185). as before. all variables are kept constant with the
exception of 8, so that the real power at the sending end, Ps' is a function of
l) only, Ps is a maximum when
(J
+ 0 = 180"
, For many decades, [he electrical utility industry has declined to recognize two different kinds
of reactive power, leading and lagging ~'aT.f.Only magnetizing ~'ar$are recognized, printed on
varmeter seals pJates, bought, and sokl. Therefore, in [he following sections, the leading Of.
lagging van will be referred 10 R!>magnetizing vars.
'(Ii
of EEE, SJBIT
120
........c-
'1r.rnsmisslon and Distribution
".0;<,.......
'·
10EE53
~iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii~~~iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii~~~
Therefore, the maximum power at the sending end. the maximum input
power. can be expressed as
(2.187)
or
(2.188)
However. jf Vs = YR'
(2.189)
and similarly. the corresponding reactive power at the sending end, the
maximum input vars, is given by
v1
Qs = Z$.SIO (J
(2.190)
As can be observed, both equations (2.189) and (2.190) are independent of
VI'fvoltage, and equation (2.190) has a positive sign this time.
2.17.2 Percent Voltage Regulation
The voltage regulation of the line is defined by the rise in voltage when full
oad is removed, that is,
Iv I-IVRI
(2.191)
= . sIVRI' x 100
Percentage of voltage regulation
or
•
Percentage of voltage regulation
t
ee
IV.'tt~ll-IVRFLI
IV'"
x 100
•
II.Ft
(2.192)
where
IVsl = magnitude
of sending-end phase (line-to-neutral) voltage at
no load
IVIfl = magnitude
.
of receiving-end phase (line-to-neutral)
voltage
at full load
• For further information sec: Stevenson
SJBIT
13. p, 97),
121
lOEES3
Transmission and Distribution
IV,q,NL.1
= magnitude of receiving-end voJtage at no load
IVR.FLI=: magnitude of receiving-end voltage at full load with con-
stant IVsl
Therefore, if the load is connected at the receiving end of the line,
and
An approximate expression for percentage of voltage regulation is
,
(R cos¢> :± X sin
Percentage of voltage re~ulatlOn;::;' lit
. ltv
.
EXAMPLE
¢l )
R X
R
100
(2.193)
2.5
A three-phase, 6o..Hz overhead short transmission line has a line-to-line
voltage of 23 leV at the receiving end, a total impedance of 2.48 ± j6.57 01
phase, and a load of 9 MW with a receiving-end I.lgglllg power factor of
0.85.
(a) Calculate line-to-neutral and line-to-line voltages at sending end.
(b) Calculate load angle.
Solution
Method I: Using complex algebra:
(a) The line-to-neutral reference voltage is
VR{L'L)
VR(L-N) =
=
V3
23 X lO} L!!.
v'3 ....."'"13,294.8 is!.. V
The line current is
9Xl~
..'
I ::; V3 )( 23 x l()~x 0.85 x (0.85 - ,0.527)
Dept of EEE, SJBIT
=:
266.1(0.85 - 10.527)
=:;
226.19 - j140.24 A
122
lOEES3
and Di~;l:ribution
Transrnission
Therefore,
IZ
= (226.19 - jI40.24)(2.48 + j6.57)
= 266.1/-31.8° x 7.02/69.32
0
= 1868.95 L37.52°V
Thus, the line-to-neutral voltage at the sending end is
VS(L.N) ... VR(L-N)
+ IZ
= 14,820 /4.4
0
V
The line-to-line voltage at the sending end is
= V3VS(L_N)
VS(L'U
;:: 25,640 V
(b) The load angle is 4.4°.
Method II. Using I as the reference:
(a) VR cos <IlR + IR;;; 13,294.8 x 0.85
+ 266.1 x 2.48 ~ 11.960
VR sin!l>R + IX = 13,294.8 x 0.527 + 266.1 x 6.57;;; 8754
Then
~~jL-N);::
(11,960.52
+ 8754.662)1I2
314,820 V/phase
VS(L-L) =25,640
(
'"
-I
b) tlls='¥R+8=tan
c5
= 4>s -
8754
n,960
V
= 36.20
4>R = 36.2 - 31.8 = 4.40
Method III. Using VR as the reference:
(a)
VS(L-N)
= [(VR + IR cos 4>8 + IX
+ (IX cos ¢lR
IR cos <l>R
=:
-
sin ¢JR)2
IR sin <l>R)2f2
266.1 x 2.48 x 0.85;;;;;;560.9
IR sin <l>R = 266.1 X 2.48 X 0.527 -- 347.8
IX cos <l> R = 266.1
X
6.57 x 0.85 = 1486.0
sin 4tR = 266.1
X
6.57 x 0.527 = 921.0
J){
! 'j.:, SJBIT
123
Transmission
lOEE53
and Distribution
Therefore,
V$(I.._N) = [(13,294.8
+ 560.9 + 921.0)2 + (1486.0 -
347.8)2JI/2
= [14,776.72 + 1138.22)112
s=r 14,820V
V3VS(L_I..)
VS(L-L) ...
S;;25,640V
s: _
-I 1138.2 _
e
(."b)" ....tan 14,776.7 - 4.4
Method IV. Using power relationships:
Power Joss in tbe line is
P10$S
= 3i1.R
"",;
j x 266.12 x 2.48
X
10-6
= 0.521 MW
Total input power to tbe line is
PT= P+ PI"'"
=9
+ 0.527 = 9.52iMW
Var loss in the line is
QI".. =312X
=3
X
266.12
X
6.57
X
10-6:;;; 1.396 Mvar lagging
Total rnegavar input to the line is
Psin 4>R
Q r = cos.p +Q ID~'
R
=
9 x 0.527
0.85
+ 1.396 = 6.976 Mvar Jdgg,mg
Total megavolrampere input to the line is
ST=VP~+Q~
"'"V9.5272
() V
a
+ 6.9762 = 11.81 MVA
~ Sr
S(L.-.L} -
V31
11.81 x 106
= V3 x 266_1 ;;; 25,640
Dert ofEEE, SJBIT
V
124
Transmission and Distribution
10EE53
P
(b) cos e, = S~
9.527
.
= 11.81 :::;0.807 laggmg
Therefore.
= 36.2"
€l>s
5 = 36.2" - 31.8"
= 4.4°
Method V. Treating the three-phase line as a single-phase line and having
Vs and VR represent line-to-line voltages, not line-to-neutral
voltages: .
(a) Power delivered is 4.5 MW
4.5
X
106
IU"e ,:: 23 X 10) x 0.85
-= 230. 18·A
=:? x 2.48 = 4.960
X;oop·= 2 x 6.57 = 13.140
. Rlcop
VR cos <l>R = 23 X 103 x 0.85 = 19,550 V
Vn sin <l>R == 23
X
10l x 0.527:::: 12,121 V
IR =230.18 x 4.96 -= 1141.7 V
IX = 230.18 x 13.14"" 3024.6 V
Therefore.
VS(L-L)
= [(VR
COS<l>R
+ IR)"l + (VR sin <PR + IX)2fl2
= [(19.,550 + 1141.7)2 + (12,121 + 3024.6)2],12
= [20,691.72 + 15,145.61JIIZ
~25.640V
..
Thus,
VS(L.L)
VS(L.N)
-= V3
= 14,820 V
(
..
(b) <ls = tan
-I
a = 36:2" -
"I
f L~EE,SJBIT
15,145.6
20,691.7
36
20
= .
31.8" ""4.4"
125
Transmission
10EES3
and Distribution
2.6
EXAMPLE
Calculate percentage of voltage regulation for the values given in Example
2.5.
(a) Using equation (2.191).
(b) Using equation (2.193).
Solution
(a) Using equation (2.191).
= IVS~;JVklx 100
Percentage of voltage regulation
=
.
14,820 - 13,294.8 x 100
13.294.8
=11.5
(b) Using equation (2.193).
Percentage of voltage regulation
=1
R cos <)JR
+ X sin <)JR
VR
R
= 266 1 2.48 xO.SS
.
xl
00
+ 6.57 x 0.5?7 x 100
13,294.8 .
= 11.2
Representation of Mutual Impedance of Short Lines
Figure 2.26(a) shows a circuit of two lines. x and y, that have selfimpedances of ZH and Zn. and mutual impedance of Z:t)" Its equivalent
circuit is shown in Figure 2.26(b). Sometimes, it may be required to
preserve the electrical identity of the two lines. as shown in Figure 2.27. The
mutual impedance ZXY can be in either line and transferred to the other by
means of a transformer that has a 1: 1 turns ratio. This technique is aJso
applicable for three-phase lines.
ExAMPLE
2.7
Assume that the mutual impedance between two parallel feeders is 0.09 +
jO.3fl/mi per phase. The self-impedances of the feeders are 0.604 /50.4°
and 0.567 /52.9° O/mi per phase, respectively. Represent the mutual impedance between the feeders as shown in Figure 2.26(b).
Dept ofEEE, SJBIT
126
lOEE53
Transmission and Distribution
1 ~1
Zn - Z-'JI
Figure 2.27. Representation of mutual impedance between two circaits by means of
1: 1 transformer.
Solution
Zry == 0.09 + jO.3
n
= 3.85 + jO.465 fl
/52.fJO = 0.342 + jO.452 n
Zu"" 0.604/50.40
Zy, ,."0.567
Therefore,
z...- Zxy
Zy'I -
= 0.295 + jO.l65 fl
Z~y = 0.252
+ jO.152 n
Hence, the resulting equivalent circuit is shown in Figure 2.28.
0,09 + jO,3 0
Fjgure 2.28
,. .E, SJBIT
127
Transmission
IOEES3
and Distribution
2.18 MEDIUM-LENGTH TRANSMISSION LINES (UP TO 150 ml,
OR 240km)
As the line length and VOJt:I~C Increase, the use of the formulas developed
for the short transmission hues grve inaccurate results. Therefore, the effect
of the current leaking through the capacitance must be taken into account
for a better approximation. Thus. the shunt admittance is "lumped" at a few
points along the line and represented by forming either a T or a Il network,
as shown in Figures 2.29 and 2.30.
In the figures,
Z=zl
For the T circuit shown in Figure 2.29,
Vs=IsX
}Z+IRx!Z+V,R
= [Ill + (VR + III x
~Z)YHZ + VR + lR~Z
or
(2.194)
B
A
and
or
Is= .__.
Y X VR + (1
C
Alternatively.
+ 4ZY)IR
(2.195)
D
neglecting conductance so that
and
yields
Ic=VcXY
Vc=VR+IRX
Dept of EEE, SJBIT
!Z
128
6ZI
lJ8fS
puc
(ZOZ'z)
J;)lSUtlJl JO
'x!l~eut
(WZ'z)
J~l;)utl!Jt?d l!n:ll!::J
Ip.Jauaa
S;)WO"~q 'lq..llcm
"lfl <l~nJI!;) .l-ltlU~WOll{! ..I0J
- Zk~ + t =0
(ooz'Z)
X=:)
ZX¥+t=v
(S61'Z)
lZXt +Z=
(661'Z)
a
a
;)
.............
H1(Azi + I) + HA x A =.$"I
(L61'Z)
A(zi x HI + HA) + HI =
X X;)A+ 11'1 =
;;,II+)il "" sl
'081V
v
(961'Z)
..
JO
(Zf)l(ZAf +l)IfI+AIfAJ+Zt
xHI+HA=
z~x sl + JA= SA
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0:
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If[ X
~
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'tEI3 .10
V
Z + HA(ZX¥ + r)::= sA
lO
Z(UI + itA x Ai) + llA= sA
'(SOZ'Z) uouanbe OlU! (LOZ'Z) uouenbo 8U!lnmsqns Ag
(80,'Z)
'OSYV
(LQZ'Z:)
<I
_____..._......
~I(ZA
(90Z'Z)
~
f + I) + lfA(Z,A f + x) =.\'I
JO
iiI + lfAx X ¥ + X HH1Z + NA(ZA
'(SOZ'C:) uopsnba
i + OJ = sl
OlU! (POZ'Z) uonenbo gU!lOlnsqnS Ag
pUB
0:
V
(toz'C;)
JO
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lJ8JS
pu(:
(61Z'Z)
ZzA ~ + A] ""[0
ZAy+t
ZAt
(L.IZ"Z)
(91Z'Z)
V•.
]
J
II
+ 1""0
Zc;At +A =::)
(~IC:"Z)
z=g
ZAt + I =V
(vIZ''l)
o
(£I~'Z)
t11(ZX
J
...----..
t + T) + ~A(ZtA.~ + A) = $1
lO
HIZAY
= sl
+ ~A(zxi + 1)A.f + til + IIAX f
«ZYZ'Z) uonenbe 01O! (LOZ'C:) uO!lunb~ jU!lnlHsqns ~q
..
(ZlZ'Z)
'
HIZ x A. f + HA(Z.\f
(HZ'Z)
'(OIZ'Z) uopanba
+ I)A i= 1.')1
OlU!
(6OZ'l:) uopsnbo
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pue
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10EE53
Transmission and Distribution
As can be proved easily by using a delta-wye transformation, the
nominal-T and nominal-Il circuits are not equivalent to each other. This
result is to be expected since two different approximations are made to the
actual circuit, neither of which is absolutely correct. More accurate results
can be obtained by splitting the line into several segments, each given by its
nominal-T or nominal-Tl circuits and cascading the resulting segments.
Here, the power loss in the line is given as
(2.221)
which varies approximately as the square of the through-line current. The
reactive powers absorbed and supplied by the line are given as
(2.222)
and
(2.223)
respectively. The Q L varies approximately as the square of the through line
current, whereas the Qc varies approximately as the square of the mean line
voltage. The result is that increasing transmission voltages decrease the
reactive power absorbed by the line tor heavy loads and increase the
reactive power supplied by the line for light loads.
The percentage of voltage regulation for the medium-length transmission
line's is given by Stevenson [31 as
.
Percentage of voltage regulation=
IVsl/~AI- iVR,FLI
. IV
I
x
100
. (2.224)
R.FI.
IVsl = magnitude
where
of sending-end
phase
(line-to-neutral)
volt-
of receiving-end phase {line-to-neutral)
age at full load with constant
IAI :.0 magnitude of line constant A
volt-
age
IVR,I't.I"" magnitude
EXAMPLE
IVsl
2.8
A three-phase 138·kV transmission line is connected to a 49-MW load at a
0.85 lagging power factor. The line constants of the 52-mi-Iong line are
Z = 95 /18° nand Y = 0.001/90° S. Using nominal-T circuit representation,
calculate:
(a) The A, B. C, and D constants of the line.
(b) Sending-end voltage.
Dept of EEE, SJBIT
132
Transmission and Distribution
lOEE53
(c) Sending-end current.
(d) Sending-end power factor.
(e) Efficiency of transmission.
Solution
138kV
V}!(L'N)
= v'3
= 79,768.8 V
Using the receiving-end voltage as the reference,
VR(L'N)
= 79,768.8
1St.. V
The receiving-end current is
~x~
..
iR= v'3x138XI03xO,gs:;;241.46A
..
c
or 241.46/-31.8 A
(a) The A, B, C, and D constants for the nominal-T circuit representation
are
A= 1 + ~YZ
= 1+
hO.OOl /9<Y')(95/78<»
= 0.9535 + }0.0099
= 0.9536 (0.6°
B=Z+
}YZ2
:::95 /78°"+ 1(0.001/90")(95/78°)2
:::18.83+ j9O.86
= 92.79 /78.3°
n
C:;; Y = 0.001/9(f S
D"'" 1 + ~YZ=A
= 0.9536/0.6"
>]
..
(b) [VSfL.-N
Is
= [O.9536/0.6Q
0.001/900
92.79/78.3:][?
0.9536 ~
0]
79,768.8iQ:
_41.46 -3!.&_
L
The sending-end voltage is
VS(L.N) = 0.9536/0.6°
x 79,768.8/.!!_
+ 92.79 /78.3
0
x 241.46 / -31.8°
= 91,486 + jI7,0486'" 93,060.9 /10.4" V
Or
'i
SJBIT
133
10EE53
Transmission and Distribution
(c) The sending-end current is
Is = O,OOl/9fr x 79,768,8 ~
= 196.95 -
+ 0,9536/0.6°
j39.5 = 200.88/-11.3°
x 241.46/-31.8°
A
(d) The sending-end power factor is
e, = 10.4°+
11.3°.::::21.7"
cos <lls = 0.929
(e) The efficiency of transmission is
output
II= input
=
=
v'3VRIR cos 4>R
Y3Vsls cos e,
x 100
138 X 103 x 241.46 x 0.85 x 100
160,995.4 x 200.88 x 0,929
=94.27%
EXAMPLE
2.9
Repeat Example 2.8 using nominal-Il circuit representation.
Solution
(a) The A, B, C, and D constants for the nominal-If circuit representation are
A= 1+ !YZ
;::0.9536/0.6°
B ;:;::Z;;:;95/78° {1
C
= Y + 1y2Z
= O.OOl/9QD + 1(0.0011900)2(95
= -4.9379
L78°)
x 10-0 + jl02.375 x lO-s:;;; 0.001L9O.3°S
D= 1 + !YZ=A
= 0.9536/0.6°
(b) [VS(L'N)]
Is .
= [°.9536/0.6°
. 0'(XJl/90.3°
95/7SO][
0.9536/0.6°
79,768.8!.sr
241.46/-31.$°
]
Therefore,
VS(L-N) = 0.9536/0.6° x 79.768.8 ~
+ 95/78
0
x 241.46!~31.8°
= 91,940.2 + j17,352.8 = 93,563.5 L1O.7°V
Dept ofEEE, SJBIT
134
~';nnsmission and Distribution
10EE53
~.,.~*"·"·~~:.,~;"'."""'iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
or
VS(L-L) "'" 161,864.9/4fJ.7"V
(c) The sending-end current is
Is =0.001/90.3" x 79,768.8LQ:+ 0.9536/0.6° x 241.46/-31.8"
= .196.53 -
j39.51 :: 200.46/-11.37"
A
(d) The sending-end power factor is
<1>$ = 10.7" + 11.37" = 22.07"
cos ~s = 0.927
(e) The efficiency of transmission is
71=
=
'V'3VRIR cos ~R
v'3VsIs cos ~s
.. 00
xi
138 X 103 x 241.46 x 0.85 x 100
161,864.9 x 200.46 x 0.927
=94.16%
The discrepancy between these results and the results of Example 2.8 is
due to the fact that the nomlnal-T and nominal-If circuits of a mediumlength line are not equivalent to each other. In fact, neither the nominal-T
nor the nominal-If equivalent circuit exactly represent the actual line.
LONG TRANSMISSION LINES (ABOVE 150 ml,
OR 240km)
A more accurate analysis of the transmission lines require that the parameters of the lines are not lumped, as before, but are distributed uniformly
throughout the length of the line.
Figure 2.31 shows a uniform long line with an incremental section dx at a
distance x from the receiving end, its series impedance is z dx, and its shunt
admittance is y dx, where z and yare the impedance and admittance per
unit length, respectively.
The voltage drop in the section is
dVi = (Vz + dV;o) - V". = dV".
= (I.. + d(z)z dx
or
(2.225)
.'.E, SJBIT
135
lOEE53
Transmission and Distribution
~I
+
Vs
V..
N
x .. l
~~
I~
dx
N'
~I~
x-o
~
FIgure 2.31. One phase and neutral connection of three-phase transmission line.
Similarly, the incremental charging current is
(2.226)
Therefore,
(2.227)
and
dlr
dx =yVx
(2.228)
Differentiating equations (2.227) and (2.228) with respect to .r,
d2V"
dl~
--=zdx'l
dx
(2.229)
and
,.
d2I..
tb?=Y
dV"
(2.230)
dx
Substituting the values of dlxldx and dV;rldx from equations (2.228) and
(2.229) in equations (2227) and (2.230), respectively,
d2V
t =yzV.
(2.231)
dx
Dept ofEEE, SJBIT
136
lOEE53
Transmission and Distribution
and
(2.232)
o. v..
At x =
= V/l and I., = JR- Therefore, the solution of the ordinary
second-order differentia] equations (2.231) and (2.232) gives!
V(x);:; (cosh Yfi,x)VI'l + (
.fy sinh vYiX)IR
(2.233)
A
B
Similarly,
I(x)
= ( Yf- sinh '\IYZx)VR + (cosh v'YZx)IR
(2.234)
D
c
Equations (2.233) and (2.234) can be rewritten as
Vex) = (cosh yx)V/l + (Zc sinh yx)h
(2.235)
= {Y<sinh yx)VR + (cosh YX)IR
(2.236)
and
I(x)
where y = propagation constant per unit length, ;;.;;ffz
Zc <:: characteristic (or surge or natural) impedance of line per unit
length, :;;;;VzlY
Y" == characteristic (or surge or natural) admittance of line per unit
length,
:;.VYTZ
Further,
Y"'" a
where ex
:=
+ jf3
(2.237)
attenuation constant (measuring decrement in voltage and current per unit length in direction of travel) in nepers per unit
length
P '= phase
(or phase change) constant in radians per unit length (i.e.,
change in phase angle between two voltages, or currents, at two
points one per unit length apart on infinite line)
, See Stevenson (J. p. 1031 and Neuenswander 12, p. 35),
: LEE, SJBIT
137
Cst'Z'Z)
(LtZ'Z)
0 ljU!S :>Z =- 'M:J\ tjU!S AlZf\
"" II-..qu~s 'z = 9 .
e qsoo = Yif. qsoo = 11.,qsoa == V
pue
(SVZ'z)
'SlU1l:1SUOO
(I:)gV
JO SUI1:}1 ul
pu~
JO
(Itt·z)
[ sA$'1]. (_ [[1-..4.$.00
IA. qlI!s 'z
puc
1I..qu!S~A]·
=[~I.]
..
I/'" qsoo
sA
(OVZ'Z)
(6£Z'Z)
pue
"I(l/'" 4ulS 'Z) + 11A(/A.qsoo)
= sA
~w003q (9£Z'Z) ptre (~€z·z) suonsnba 'J = x U;}qM.
£53301
'Jr.msmission and Distribution
iiiiiOo,
;,c"';""',,........
lOEE53
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c = Y~sinh ..,1= VYiZsinh VYZ= v, sinh B
(2.249)
D = A = cosh "(I
(2.2S0)
= coshVYZ=cosh"
9=VYZ
(2.251)
sinh ')'1 =
HeY! - e-Y')
(2.252)
cosh "il=
He + e-"{l)
(2.253)
Y{
Equations (2.238)-(2.251) can be used if tables of complex hyperbolic
functions OJ" pocket calculators with complex hyperbolic functions are availble. Alternatively, the following expansions can be used:
sinh "'it "" sinh{al
+jl3t)
= sinh al cos PI + j cosh al sin pi
(2.254)
cosh "(1= cosh(al + fPl) == cosh at cos pi + j sinh at sin pI
(2.255)
Furthermore, by substituting for ..,1 and Zc in terms of Y and Z, that is,
the total line shunt admittance per phase and the total line series impedance
per phase, in equation (2.245) gives
(2.256)
and
(2.257)
or, alternatively,
(2.258)
and
(2.259)
The factors in parentheses in equations (2.256)-(2.259) can readily be
found by using Woodruff's charts, which are oat included here but can be
found in L. F. Woodruff, Electric Power Transmission (Wiley, NY, 1952).
;
,..,'
"
SJBIT
139
iDEES3
Transmission and Distribution
The AB~D parameters in terms of infinite series can be expressed as
YZ
A= 1+ 2
y2Z2
+ 24 +
y3Z3
720
y'Z4
+ 40,320 + ...
(2.260)
(2.261)
(2.262)
where
Z = total line series impedance per phase
=zl
=;
(r
+ ixdl .0.
Y ;;;;:total line shunt admittance
per phase
';;;;;;
yl
== (g+ jb)1
S
In practice, usually not more than three terms are necessary in equations
(2.260)-(2.262). Weedy [7] suggests the following approximate values for
the ABeD constants if the overhead transmission line is less than 500 km in
length:
A=D== 1 + !YZ
(2.263)
+ kVZ)
(2.264)
C=Y(l + lYZ)
(2.265)
B =Z(l
and
However, the error involved may be too large to be ignored for certain
aoolications,
ExAMPLE 2.10
A single-circuit, (>O.Hz, three-phase transmission line is 150 mi long. The
line is connected to a load of 50 MVA at a lagging power factor of 0.85 at
138 kV. The line constants are given as R = 0.1858 fi/mi, L = 2.60 mH/mi,
and C = O.012~F Imi. Calculate the following:
(a) A, B. C, and D constants of line.
(b) Sending-end voltage.
(c) Sending-end current.
Dept ofEEE, SJBIT
140
Transmission
lOEES3
and Distrihution
(d) Sending-end power factor.
(e) Sending-end power.
(f) Power loss in line.
(g) Transmission Hoe efficiency.
(11)Percentage of voltage regulation.
(i) Sending-end charging current at no load.
(j) Value of receiving-end voltage rise at no load if sending-end voltage
is held constant
Solution
+- j21T'X
z =0.1858
60 x 2.6
10-3
X
+ jO 9802::: O.99TI /79.27 fi/mi
=0.1858
D
y = j2w x 60 x 0.012 X to-ii = 4.5239 x 10-6 /9lf'S/mi
The propagation constant of the line is
-r = VYi
= [(4.5239 X
10-6 {9IJ)(O.9977/79.27")]1/2
= [45135 x 1O-6f2/(!9<r
+ 79.27°) = 0.002144/84.63"
The characteristic impedance of the tine is
)1/2
0.9977/79.27"
'Z . (
z, = ~y =
4.5239 x 1O-6~
= (O.9:'~~
6)'11/1<79,27
10
0
_
9O = 469.62/-5.37°n
Q)
The receiving-end line-to-neutral voltage is
138kV
VR(L.N}
= v'3
= 79,674.34 V
Using the receiving-end voltage as the reference,
VR(L?N)
The receiving-end current is
III = v'3
mx~
J
3 x 138 x 10
""
79,674,34!.!t.. V
= 209.18 A
.'
or 209.18/-31.8
0
(a) The A, B, C, and D constants of the line:
A = cosh
"il
== cosh(a + i(3)!
'" Heal elJIl
...
e'-Ill
e-1JJ/)
=
He'll !(JI ...e-al
t:..P.D
141
A
Zvl
£S98'O "" S'<J:I SO;)
060'0£ = o£'91 + 06L'0 = S'¢1
S!lOpCJl~h\od pua-Surpuos ~lf.l (p)
0(,91 -!V80S'9Ll
=
(08·t£.....!81'60Z) x
"
(oLS'O!L.6V6·O)+ (J)7vt'vL9'6L)(o81'06! L900'O) = SI
pU~.gU!PU;)S~U (o
51 lU<ilUO;)
A St· L8t'lL 1
=
("I")S
A
A 06L'n! )O'OLt'66 "'"
(08'1£-!
Sr'60Z)(oVt'6L/ ~·Stt)
+ (:07 pt·PL9·61.)(.JSO! L6V6'O)ee (tVI}S'A
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1'Zt = [CUI_a .;» - lrJ,:iJ /(>3HYz:;:;
1t)3
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+ tote o,a
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IDEES3
ion and Distributlon
(e) The sending-end power is
Ps
'=
V3Vs(L_L)ls cos <1>s
= V3 x
172,287.18 x 176.8084 x 0 8653 = 45,654.46 kW
(f) The receiving-end power is
PR = V3VR(L_L/k cos <P,R
= 'V:J x 138 x 103 x 209.18 x 0.85 = 42,499 kW
Therefore, the power loss in the line is
PL == Ps - PEt "" 3155.46 kw
(g) The transmission line efficiency .is .
Pp
. 42,499
7J ... P $ x 100 = 45,654.46 x 100 == 93.1%
(b) The percentage of voltage regulation is
.
Percentage of voltage regulation
=
99,470.04 -79,674.34
79,-614.34
x 100 = 24.9%
(i) The sending-end charging current at no load is
I"
0)
= ~YVso_.I'I) = (339.2925 x
10-6)(99,470.05) ;;;;;33.75 A
The receiving-end voltage rise at no load is
VR(l...N) ;;;;VS(I..'N) -
ZI"
= 99,470.05 /13.79"
- (149.66/79.27")(33.75/103.79")
== 104,436.74/13.27" V
Therefore. the line-to-line voltage at the receiving end is
VR(L.L)
= V3VR(L.N) "'" 180,889.74 V
2.19.1 Equivalent Circuit of Long Transmission line
Using the values of the ABCD parameters obtained for a transmission line,
it is possible to develop an exact 11 or an exact T, as shown in Figure 2.32.
For the equivalent-If circuit.
Zn = B = Z<sinh ()
(2.266)
""Z, sinh-yl
(2.267)
sinh VYZ
=Z --==--
(2.268)
\1Yi
SJBIT
143
10EES3
Transmission and Distribution
I
I
-I-
+
Vs
-.
Yn
2"
---'-
Zn
1
I
,-"in
T
c.........._
(b)
. . Figure 2.32. Equivalent and T circuits for long transmission line.
and
(2.269)
or
Yn = 2 tanh.[(l 12)"(/)
(2.270)
Ze
or
Yn
Y tanh(1I2)VYZ
T= 2"
(2.271)
(1/2)vYz
For the equivalent-T circuit,
'Lr
A-I
"2 = --c- =
cosh(}- 1
y~sinh ()
(2.272)
or
(2.273)
or
ZT
T=
Z tanh (1J2)vYZ
2" (ll2)vYZ
(2.274)
and
(2.275)
Dept ofEEE, SJBIT
144
H8JS
(LLZ'Z)
JO
(9LZ'Z)
JO
£53301
lOEE53
Transmission and Distribution
Surge Impedance loading
(SIL) of Transmission Line
In power systems, if the line is lossless,' the characteristic impedance Z, of a
line is sometimes called surge impedance. Therefore, for a loss-free line.
R"'O
and
G=O
Thus,
z:c = Vy
rz ~ 'Ieff. n
(2.287)
and is a pure resistor. It is a function of the line inductance and capacitance
as shown and is indepenent of the line length.
The surge impedance loading (SIL) (or the natural loading] of a transmis• When dealing with bigh frequencies or with surges due to lightning, losses are "oneil ignored
(3).
Dept of EEE, SJBIT
146
10EE53
nsrnission and Distribution
sion Hoe is defined as the power delivered by the line to a purely resistive
load equal to its surge impedance. Therefore,
SIL= IkV R(L-Ll 12 MW
Z..
(2.288)
SIL:;' !kV R(L-L) 12 MW
VIAe
(2.289)
SIL = v'3lvRUA.) II]L I W
(2.290)
II I = IV~(L-~)l
(2.291)
e
or
•
or
where
L
V3xVLlC
A
and
SIt. = surge impedance loading in megawatts or watts
IkVR(l.L)1 "" magnitude of line-to-line receiving-end voltage in kilovolts
IVJl(L.ql == magnitude of line-to-line receiving-end voltage in volts
Z. = surge impedance in ohms,
=::YEiC
IL""" Line current at surge impedance loading in amperes
In practice, the allowable loading of a transmission line may be given as a
fraction of its SIL. Thus, SIL is used as a means of comparing the
load-carrying capabilities of Jines.
However, the SIL in itself is not a measure of the maximum power that
can be delivered over a line. For the maximum delivered power, the line
length, the impedance of sending- and receiving-end apparatus, and all of
the other factors affecting stability must be considered.
Since the characteristic impedance of underground cables is very low. the
SIL (or natural load) is far larger than the rated load of the cable.
Therefore, a given cable acts as a source of Jagging vars.
The best way of increasing the SIL of a line is to increase its voltage level,
since, as it can be seen from equation (2.288), the SI.L increases with its
square. However, increasing voltage level is expensive. Therefore, instead,
the surge impedance of the line is reduced. This can be accomplished by
adding capacitors or induction coils. There are four possible ways of
" SJBIT
147
Transmission
10EE53
and Distribution
changing the line capacitance Or inductance, as shown in Figures 2.36 and
2.37.
For a lossless line, the characteristic impedance and the propagation
constant can be expressed as
Z=~
"
(2.292)
C
and
y=VLC
(2.293)
Therefore, the addition of Jumped inductances in series will increase the
line inductance, and thus; the characteristic impedance and the propagation
constant will be increased. which is not desirable.
The addition of lumped inductances in parallel will decrease the line
capacitance. Therefore, the propagation constant will be decreased, but the
characteristic impedance will be increased, which again is not desirable.
The addition of capacitances in parallel will increase the line capacitance.
Hence, the characteristic impedance will be decreased, but the propagation
constant win be increased, which affects negatively the system stability.
However, for the short lines, this method can be used effectively.
Finally. the. addition of capacitances in series will decrease the line
inductance. Therefore, the characteristic impedance and the propagation
constant will be reduced. which is desirable.' Thus, the series capacitor
compensation of transmission Jines is used to improve stability limits and
voltage regulation. to provide a desired load division, and to maximize the
load-carrying capability of the system. However, having the full line current
going through the capacitors connected in series causes harmful overvoltages
L
L
L
)
L
~L
L
)
{oJ
(b)
Figure 2.36. Transmission line compensation by adding lump inductances in: (a)
series; (b) parallel (i.e., shunt).
Dept of EEE, SJBIT
148
..
nsmission and Distribution
lOEE53
L
o~lI~l
O-+----1L
It------t---c
ccc
o--II~'-.....jl...(
--11 ....(
--0
0
0
(aJ
(b)
Figure 2.37. Transmission line compensation by adding capacitances in: (0) parallel
(i.e., shunt); (b) series.
on the capacitors. during short circuits. Therefore, they introduce special
problems for line protective relaying," Under fault conditions, they introduce an impedance discontinuity (negative inductance) and subharmomc
currents, and when the capacitor protective gap operates. they imprc......
high-frequency currents and voltages on the system. All of these factors
result. in incorrect operation of the conventional relaying schemes. The
series capacitance compensation of distribution lines has been attempted
from time to time for many years. However, it is not widely used.
EXAMPLE 2.13
Determine the SIL of the transmission line given in Example 2.10.
Solution
The approximate value of the surge impedance of the line is
z ;; ff
(: 'V C
=' ( ..2.6 X
10-3
0.012 x 10
')'112
= 465..5fl
6
Therefore,
SIL ~.
fkV. R(I."(.) 12 ;;;;:--.11381l
v'LlC
469.62
"'" 40
.
913 MW
which is an approximate value of the SIL of the line. The exact value of the
SIL of the line can be determined as
SIL
SJBIT
. 12 = _._-_._:::
In812 40.552 MW
= IkV .R(l-L)
z,
469.62
149
Transmission
10EE53
and Distribution
GENERAL CIRCUIT CONSTANTS
Figure 2.38 shows a general two-port, four-terminal network consisting of
passive impedances connected in some fashion. From general nework
theory,
Vs=AVR+BIR
(2.294)
Is"" CVR+DIR
(2.295)
VR = DVs - Bls
(2.296)
IR = -CVs + AIs
(2.297)
and
Also,
and
It is always true that the determinant of equations (2.294) and (2.295) or
(2.296) and (2.297) is always unity, that is,
AD-Be=1
Is~
+
Vs
-
(2.298)
IR~
A, 8, C, 0
(pas~ive network)
Figure 2.38. General two-part, four-terminal network.
Dept ofEEE, SJBIT
150
-
-_.
--------,~-----------------------
Transmission and Distribution
10EE53
In the above equations, the A. B, C, and D are constants for a given
network and are caned general circuit constants. Their values depend on the
parameters of the circuit concerned and the particular representation
chosen. In general, they are complex numbers. For a network that has the
symmetry of the uniform transmission line,
A=D
(2.299)
2.20.1 Determina.tionof A. B, C, and 0 Constants
The A, B, C, and D constants can be calculated directly by network
reduction. For example, when IR = 0, from equation (2.294).
Vs
A=-
(2.300)
VB
and from equation (2.295),
(2.301)
Therefore, the A constant is the ratio of the sending- and receiving-end
voltages, whereas the C consrant is the ratio of sending-end current to
receiving-end voltage when the receiving end is open-circuited, When
VB = 0, from equation (2.294),
(2.302)
When Vn = 0, from equation (2.295),
(2.303)
Therefore, the B constant is the ratio of the sending-end voltage to the
receiving-end current when the receiving end is short-circuited. Whereas the
D constant is the ratio of the sending-end and receiving-end currents when
the receiving end is short-circuited.
Alternatively, the A, D, C, and D generalized circuit constants can be
calculated indirectly from a knowledge of the system impedance parameters
as shown in the previous sections. Table 2.4 gives general circuit constants
for different network types. Table 2.5 gives network conversion formulas to
convert a given parameter set into another one.
As can be observed in equations (2.294) and (2.295), the dimensions of
the A and D constants are numeric. The dimension of the B constant is
impedance in ohms, whereas the dimension of the C constant is admittance
in siemens.
EE, SJBIT
151
10EES3
Transmission and Distribution
=- 2400IS!.. + 2.173 /17.7" x 83.33/-300
= 2576.9
- j38.58
= 2.577.2l. -0. 9° V
(c) The input impedance is:
_ Vs _ AVR + BIR
Z,n - Is - CVR + DIn
=
2577.2/-0.9° "" 309·3 /?<9 I" n
83.33/ -30"
..
cW.'
(d) The real and reactive power loss in the line:
SL =SS-SR
where
S5 =V,sls = 2.577.2 1-0.9'" x 83.33 1+30" = 214,758/29.1"
or
SS =1S x z .n·sXP""214758/291"VA
,
..
Therefore.
SL
= 214.758[29.1"- 200,000 130"
= 14,444.5 + j4444.4
that is, the active power loss is 14,444.5 W, and the reactive power
loss is 4444.4 vars,
Power Relations Using A, B, C, and D Line Constants
For a given long transmission line, the complex power at the sending and
receiving ends are
(2.338)
and
(2.339)
Also, the sending- and receiving-end voltages and currents can be expressed
as
Dept ofEEE, SJBIT
Vs =AVn -+- BIR
(2.340)
Is= CVR + DIn
(2.341)
152
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puc
(psrz)
JO
(Ov£'Z) uounnbe
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pun
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(OS£;'Z)
'(Zv£'Z) uouenba wOld
(6t£·Z)
(gp£'Z)
(LP£'Z)
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~HA=lfA
~l.Iso.J=V=(I
~
(~t£·z:)
Mf\ qD!S ~
(9P£'Z)
qU!S
it
;: ~ ;) ;: o
=
ff7 e =fI
D7v
sUI: - sAV"'" lfA
(tti:'Z)
+ sA3- = ~I
SIO
(£t£·Z)
~qso~=
(PV£';::)
=V
pun
£53301
lOEE53
Transmission and Distribution
By substituting equation (2.352) and (2.355) into equations (2.338) and
(2.339), respectively,
(2.356)
and
(2.357)
Therefore, the real and reactive powers at the sending end are
Ps
=
AY;
B
cos(f3 -
VSVR
a) - -----yJ cos({3 + 0)
(2.358)
and
(2.359)
and the real and reactive powers at the receiving end arc
PR =
VyVR.
'B
cos({3 - 8) -
AV~
B
(2.360)
cos(f3 - a)
and
(2.361)
For constant Vs and YR, for a given line, the only variable in equations
(2.358)-(2.361) is a, the power angle. Therefore, treating Ps as a function
of {j only in equation (2.358), Ps is maximum when f3 + 8 = 180°. Therefore,
the maximum power at the sending end, the maximum input power, can be
expressed as
(2.362)
and similarly the corresponding reactive power at the sending end, the
maximum input VaTS, is
(2.363)
On the other hand, PR is maximum when 8 = {3.Therefore, the maximum
Dept ofEEE, SJBIT
154
Transmission and Distribution
lOEE53
power" obtainable at the receiving end can be expressed as
VSVR
PR•mu == ~
-
Avi
iJ
(2.364)
cos(f3 - a)
and similarly, the corresponding reactive power delivered at the receiving
end is
QR.lnu = -
•
AY~ .
B
sm(t3 -
(2.365)
a)
In the above equations, Vs and VR are the phase (Iine-to-neutral) voltages
whether the system is single phase or three phase. Therefore, the total
three-phase power transmitted on the three-phase line is three times the
power calculated by using the above equations. If the voltages are given in
volts, the power is expressed in watts or VaTS. Otherwise, if they are given in
kilovolts, the power is expressed in megawatts or megavars,
For a given value of ,,(, the power loss Pl,. in a long transmission line can
be calculated as the difference between the sending- and the receiving-end
real powers,
(2.366)
and the lagging vars loss is
(2.367)
SJBn
155
10EES3
Transmission and Distribution
TRANSMISSION AND DISTRIBUTION
Unit 1. Basics of Transmission and Distribution
1. Draw the line diagram of a typical power supply scheme indicating the standard
voltages.(5)
2. Write short note on advantages ofHV transmission.(5)
3. Write short note on H V D C transmission.(5)
4. Write short note on feeders, distributors and service mains.(5)
5. Write the comparison betn. Overhead and underground transmission systm.(5)
6. What are the advantages and limitations of high voltage a.c transmission? (6)
Unit 2. Overhead Transmission Lines
1. Show that a transmission
line conductor
suspended
between
level support
assumes the shape of a catenory. Derive the expression for sag.(8)
2. Explain what is sag and why it is inevitable in over head transmission lines? What
are the factors influencing it?(5)
3. With usual notations derive an expression for maximum sag of a tranmn. Line
where the supports are at different levels?(5)
4. Obtain the expression for sag in a freely suspended conductor when the supports
are at equallevels.(8)
5. Explain the effects of sag in overhead trasmn. Line.(5)
6. Obtain the expression for sag in a power conductor when the supports are at equal
levels, taking into the effect of wind and ice loading(8)
7. Write short note on effect of ice load and wind effect on sag oftransmn. Line.(5)
b. From the first principles derive the expression for sag in a freely suspended
conductor when the supports are at unequal levels (6)
Dept of EEE, SJBIT
Page 1
.u.smission and Distribution
; I
Unit 3. Corona
I,
What is corona? Derive expression for the disruptive critical voltage and visual
uitical voltage.(8)
What are the effects of corona? (3)
Write short note on disruptive critical voltage. (6)
'>~cuss the advantages and disadvantages of corona. (6)
.plain the terms with reference to corona.
"
i)disruptive critical voltage
ii) Power loss due to corona (4)
, ': rite short note on corona in transmission lines.(5)
.vrite short note on factors affecting corona and methods of reducing corona effect.
8. Explain the terms with reference to corona.
i)visual critical voltage
ii) Power loss due to corona (7)
Unit 4. Insulators
.xplain the various tests conducted on insulators. (7)
'Vith usual notations, derive the general expression for the metal link of string to
~capacitance, when guard ring is used for the string of insulators.(5)
'/lhat is string efficiency in the context of suspension insulators? Explain the
.nethods of improving the same.(8)
,1
v/rite short note on different types ofOH line insulators.
, efine string efficiency and hence calculate the mathematical expression for it.(8)
'>,plain the methods of improving the string efficiency.(6)
'lhat are the insulators with OH Lines? Discuss the desirable properties of
',)sulators and name the types of insulators.(6)
Unit 5. Under ground Cables
, .av.
the cross sectional view of a single core cable and explain the
, .truction.I S)
,:. SJBlT
Page 2
10EES3
Transmission and Distribution
2. Compare the dielectric stress of a homogeneous cable with that of a capacitance
graded cable. (7)
3. Write short notes on testing ofcables.(5)
4. Derive an expression for insulation resistance of a cable. (5)
5. Explain capacitance grading of cables with appropriate derivation. (10)
6. What is meant by grading of cables? Briefly explain the various methods of
grading. (7)
7. Write short note on thermal rating of cables.( 5)
8. Derive expressions for the maximum and minimum dielectric stress in a single
core cable and ontain the criterion for keeping the diectric stress to a minimum
value. (6)
9. Compare the merits and demerits of underground system overhead system. (6)
10. State five advantages of using underground cables for power distribution. (5)
11. Write short note on laying ofUG Cables. (5)
Unit 6. Transmission Line Parameters
1. Explain the terms self GMD and mutual GMD and prove that the inductance of a
group of parallel wires carrying current can be represented
in terms of their
geometric distances. (10)
2. Derive an expression for the capacitance of a unsymmetrically spaced but regularly
transposed line. (10)
3. Derive the expression for the inductance of a 3 phase transmission line with un
symmetrcal spacing without transposition. Use the Flux linkage concept. (10)
4. Write short notes on transposition oftransmn. Lines. (l0)
5. Derive the expression for the inductance ofa 3 phase unsymmetrically spaced but
transmission line/km. (8)
6. Derive the expression for the capacitance of a 3 phase single ckt. Line with
equilateral spacing (10)
7. Show how the inductanceof3
phase trasmn. Line with equilateral and symmetrical
spacing between conductors can be calculated. (8)
Dept of EEE, SJBIT
Page 3
C;';,;::
.smission and Distribution
lOEE53
-...' ,,,~,.,"'''''''''iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
Derive the expression for the capacitance of a 3 phase line with unsymmetrical
spacing. (8)
5. What is skin effect? Which are the factors influencing skin effect? (4)
"
Derive from first principles, an expression for the inductance per phase per km of
:; 3 phase regularly transposed tm. Line. The conductors are of diamt. d mt. and
; laced at the comer of a triangle of sides a, b, c.(lO)
Ca !culate the inductance of single phase two wire line starting from
! .indamentals.I 10)
Unit 7. Performance of Transmission Line
1.
l>:< vc expressions
for generalized ABeD constants for a long transmission line using
.•:n rous method of analysis. (10)
2, Derive expressions for ABeD constants for a medium transmission line using nominal
T model. Hence prove AD-Be
=
1.
(10)
::, \Vrite short note on classification of transmission lines. (5)
'.'!
;
short note on Ferranti effect.(5)
are ABeD constants of a trasmn. Line? Determine the same for a a medium
r..ission line using nominal TI model. Hence prove AD-Be
=
1.
(10)
, ;::.::, ,: the expressions for sending end voltage and current of a medium trasrnn. Line
(llc'minal T method) interms ofY, Z, Vr and Ir. (8)
\\ ',' short note on surge impedance loading. (5)
-,I,
Pc.'; \ ,: expressions for generalized ABeD constants for equivalent T representation of
,,;
:ransmission line. (10)
i(;,!",.uss the terms voltage regulation and transmission efficiency as applied to
smission line. (4)
l [, \\ rite and explain the classification of overhead trasrnn. Lines. (4)
;, Calculate the effcy. And voltage regulation for medium tm. Line assuming nominal T
1.' !i.,ive an expression for the effcy. And voltage regulation for short tm. Line giving
, " .ctor diagm. (6)
"",~,~-",.,.,_"_"""",,,!,!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
,",'t,
SJBIT
Page 4
,insmission and Distribution
iDEES3
Unit 8. Distribution Systems
1.
Write short note on radial and ring main distributors. (5)
2. What is meant by DC distribution? Explain with diagram different types of DC
distribution and discuss their merits and demerits. (10)
3. Write short note on radial distribution system. (5)
4. Show different types of distribution systems with single line diagrams and state
the merits and demerits of ring main and radial distribn. Systems. (10)
5. Calculate the total voltage drop in uniformly loaded distributor, when it is fed at
one end. (50
6. Write short note onring main distributors. (5)
7. Write short note on feeders, distributors and service mains, (5)
'""'"''''''''''"!'!' !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
D· i of EEE, SJBIT
Page 5
..
lOEE53
T: r:'~:nission and Distribution
TRANSMISSION AND DISTRIBUTION
Unit 1. Basics of Transmission and Distribution
Duw the line diagram of a typical power supply scheme indicating the standard
..
\ .iltages.I.S)
Jul2009/Janll
Write short note on advantages ofHV transmission.tS)
Jan2010/Ju111
'Vrite short note on H V D C transmission.(5)
Jul2009/Janll
Write short note on feeders, distributors and service mains.(5) Ju12010/Augll
Write the comparison betn. Overhead and underground transmission system.(5)
o. What are the advantages and limitations of high voltage a.c transmission? (6)
Jul2009/Janll
Unit 2. Overhead Transmission Lines
'how that a transmission line conductor suspended between level support
..imes
the shape of a catenory. Derive the expression for sag.(8)
Jul2008/Augll
;.xplain what is sag and why it is inevitable in over head transmission lines? What
.re the factors influencing it?(5)
Ju12008/Janll
''''ith usual notations derive an expression for maximum sag of a tranmn. Line
, here the supports are at different levels?(5)
Ju12009 / Janll
Obtain the expression for sag in a freely suspended conductor when the supports
;'i"e at
equallevels.(8)
::: Explain the effects of sag in overhead trasmn. Line.(5)
Jul2010/Janll
Ju12009/Janll
Cutain the expression for sag in a power conductor when the supports are at equal
" , ~k;1ginto the effect of wind and ice loading(8)
Ju12009/Janll
-vrite short note on effect of ice load and wind effect on sag oftransmn. Line.(5)
SJBlT
Page 1
Transmission and Distribution
10EES3
8. From the first principles derive the expression for sag in a freely suspended
conductor when the supports are at unequal levels (6)
Ju1200S/AuglO
Unit 3. Corona
1. What is corona? Derive expression for the disruptive critical voltage and visual
critical voltage.(8)
..
Ju12009/Janll
2. What are the effects of corona?(3)
Jan2010/Ju109
3. Write short note on disruptive critical voltage. (6)
4. Discuss the advantages and disadvantages of corona. (6)
Aug2009/Janll
Ju12009/Janll
5. Explain the terms with reference to corona.
Jan200S!JuIl
i)disruptive critical voltage'
0
ii) Power loss due to corona (4)
6. Write short note on corona in transmission lines.(S)
Ju12009/Janll
7. Write short note on factors affecting corona and methods of reducing corona effect.
6. 8. Explain the terms with reference to corona.
i)visual critical voltage
Ju12009/Janl0
ii) Power loss due to corona (7)
Unit 4. Insulators
1. Explain the various tests conducted on insulators. (7)
Ju12009/Janl
2. With usual notations, derive the general expression for the metal link of string to
3. line capacitance, when guard ring is used for the string of insulators.(S)
Ju12009/Janll
4. What is string efficiency in the context of suspension insulators? Explain the
S. methods of improving the same.(8)
Ju12009/Janll
6. Write short note on different types of 0 H line insulators.
Jul2009/Janll
?
Define string efficiency and hence calculate the mathematical expression for it.(8)
8. Explain the methods of improving the string efficiency.(6)
Ju12010/JanlO
9. What are the insulators with 0 H Lines? Discuss the desirable properties of insulators
and name the types of insulators.( 6)
Dept of EEE, SJB1T
Ju12009/Janll
Page 2
nsrnission and Distribution
10EE53
Unit 5. Under ground Cables
·Taw the cross sectional view of a single core cable and explain the
construction.(5)
Aug2009/Janll
2. Compare the dielectric stress of a homogeneous cable with that of a capacitance
•
,.
graded cable. (7)
Ju12010/Janll
.vrite short notes on testing of cables.(5)
Jul2009/Janll
.ierive an expression for insulation resistance ofa cable. (5) Jul2009/Janlr
.
Explain capacitance grading of cables with appropriate derivation. (10)
Nhat is meant by grading of cables? Briefly explain the various methods of
)·,rading.(7)
Ju12009/Janll
Write short note on thermal rating of cables.(5)
Ju12010/Janll
Derive expressions for the maximum and minimum dielectric stress in a single
core cable and ontain the criterion for keeping the diectric stress·to a 'minimum
value. (6)
Ju12009/Janll
· 'ornpare the merits and demerits of underground system overhead system. (6)
(~Iatefive advantages of using underground cables for power distribution. (5)
/rite short note on laying ofUG Cables. (5)
Ju12009/Janll
Unit 6. Transmission Line Parameters
'<plainthe terms self GMD and mutual GMD and prove that the inductance of a
roup of parallel wires carrying current can be represented in terms of their
!',eometricdistances. (10)
Jul2009/Janll
" .rive an expression for the capacitance of a unsymmetrically spaced but regularly
transposed line. (10)
Ju12009/Janll
·crive the expression for the inductance of a 3 phase transmission line with un
-rnetrcalspacing without transposition. Use the Flux linkage concept. (10)
'Writeshort notes on transposition oftransmn. Lines, (10)
.J.
Ju12009/Janll
,)c:rivethe expression for the inductance of a 3 phase unsymmetrically spaced but
.,:i:.,
SJBIT
Page 3
10EES3
Transmission and Distribution
transmission line/km. (8)
Ju12009/Janll
6. Derive the expression for the capacitance of a 3 phase single ckt. Line with
equilateral spacing (10)
Jul2009/Janll
7. Show how the inductanceof3 phase trasmn. Line with equilateral and symmetrical
spacing between conductors can be calculated. (8)
Ju12009/Janll
8. Derive the expression for the capacitance of a 3 phase line with unsymmetrical
spacing. (8)
Ju12009/Janll
5. What is-skin effect? Which are the factors influencing skin effect? (4)
•
6. Derive from first principles, an expression for the inductance per phase per km of
a 3 phase regularly transposed tm. Line. The conductors are of diamt. d mt. and
placed at the comer ora triangle of sides a, b, c.(10)
Ju12009/Janll
7. Calculate the inductance of single phase two wire line starting from
fundamentals.(10)
Jul2009/Janll
Unit 7. Performance of Transmission Line
1. Derive expressions for generalized ABCD constants for a long transmission line using
rigorous method of analysis. (10)
Jul2009/Janll
2. Derive expressions for ABCD constants for a medium transmission line using nominal
T model. Hence prove AD-BC = 1.
Jul2009/Janll (10)
3. Write short note on classification of transmission lines. (5)
Jul2009/Janll
4. Write short note on Ferranti effect.(5)
Jul2009/Janll
5. What are ABCD constants of a trasmn. Line? Determine the same for a a medium
transmission line using nominal TI model. He.nceprove AD-BC = 1.
(10)
6. Derive the expressions for sending end voltage and current of a medium trasmn. Line
(nominal T method) interrns ofY, Z, Vr and Ir. (8)
8. Write short note on surge impedance loading. (5)
Ju12009/Janll
Jul2009/Janll
9. Derive expressions for generalized ABCD constants for equivalent T representation of
long transmission line. (10)
Ju12009/Janll
10. Discuss the terms voltage regulation and transmission efficiency as applied to
Dept of EEE, SJBIT
Page 4
r:lnsrnission and Distribution
lOEE53
transmission line. (4)
Ju12009/Janll
11.Write and explain the classification of overhead trasmn. Lines. (4) Ju12009/Janll
12. Calculate the effcy. And voltage regulation for medium tm. Line assuming nominal T
method. (8)
Ju12009/Janll
13.Derive an expression for the effcy. And voltage regulation for short tm. Line giving
the vector diagm. (6)
Ju12009/Janll
•
I'
Unit 8. Distribution
Syste'ms
L Write short note on radial and ring main distributors. (5)
Ju12009/Janll
2. What is meant by DC distribution? Explain with diagram different types of DC
distribution and discuss their merits and demerits. (10)
3. Write short note on radial distribution system. '(5) ,
Ju12009/Janll
Ju12009/Janll
4. Show different types of distribution systems with single line diagrams and state
the merits and demerits of ring main and radial distribn. Systems. (10)
.~, Calculate the total voltage drop in uniformly loaded distributor, when it is fed at
r_
one end. (50
Ju12009/Janll
Write short note onring main distributors. (5)
Ju12009/Janll
7. Write short note on feeders, distributors and service mains, (5) Ju12009/Janll
Page 5
06EE53
Transmission and Distribution
Unit 1
1. State the Advantages of High Voltage Transmission
Advantages of high voltage transmission
1. Transmission efficiency increaases as the transmission voltage increasesfor a given
power,
2. P U resistance drop decreases
3. Volume of conductor material required reduces since current decreases with increase in
voltage,
4. Power transmitting capacity of the tramn. line is proportional to square of the operating
voltages. therefore, overall capital cost decreases.
5. Cost of transmission line per km decreases with increase in voltage level.
6. with increase in voltage, sil(surge impedance loading) of the.line increases - so, power
transfer increases.
2. How does power reach us?
Electric power is normally generated at 11kv in a power station.
To transmit over long distances, it is then stepped-up to 400kv,.220kv or 132kv as
necessary.
Power is carried through a transmission network of high voltage lines.
Usually, these lines run into hundreds of kilometres and deliver the power into a common
power pool called the grid.
The grid is connected to load centres (cities) through a sub-transmission network of
normally 33kv (or sometimes 66kv) lines.
These lines terminate into a 33kv (or 66kv) substation, where the voltage is stepped-down
to 11kv for power distribution to load points through a distribution network oflines at
11kv and lower.
3. What are the Substation types
Although, there are generally four types of substations there are substations that are a
combination of two or more types.
- step-up transmission substation
- step-down transmission substation
- distribution substation
- underground distribution substation
- substation functions
substation equipment
Dept of EEE, SJBIT
Page 1
•
Transmission and Distribution
06EE53
Unit 2
1. What are the factors affecting Sag?
The factors affecting the sag of a conductor strung between supports arc:
1. Conductor load per unit length.
2. Span, that is, distance between supports.
3. Temperature.
4. Conductor tension.
•
2. What are the effects of corona? Derive expression for the visual critical voltage (8)
.J:m/Feb-09
Corona on transmission lines causes power loss, radio and television interference, and
audible noise (in terms of buzzing, hissing, or frying sounds) in the vicinity of the line. At
extra high-voltage levels (i.e., at 345 ~V and higher), the conductor itself is the major source
of audible noise, radio interference, television interference, and corona loss. The audible
noise is a relatively new environmental concern and is becoming more important with
increasing voltage level. For example, for transmission lines up to 800 kV, audible noise and
electric field effects have become major design factors and have received considerable
testing and study. It had been observed that the audible noise from the corona process mainly
lakes place in foul weather. In (Icy conditions, the conductors' normally operate below the
corona detection level, and, therefore, very few corona sources exist. In wet conditions,
however, water drops on the' conductors cause large number of corona discharges and a
resulting burst of noise. At ultrahigh-voltage levels (1000 kV and higher), such audible noise
is the limiting environmental design factor.
1\ transmission line should he designed to operate just below the disruptive critical voltage in
lair wether so that corona only takes place during adverse atmospheric conditions. Therefore,
calculated disruptive critical voltage is an indicator of corona performance of the line.
; 1,,w[;VV, a high value of the disruptive critical voltage is not the only criterion of
.uisfactory corona performance. The sensitivity of the conductor to foul weather should also
he considered (e.g., corona increases more slowly on stranded conductors than on smooth
conductors). Due to the numerous factors involved, the precise calculation of the peak value
()r corona loss is extremely difficult, if not impossible. The minimum voltage at which the
ioriization occurs in fair weather is called the disruptive critical voltage and can be
ddcnnined
from
! he
E =
Vo
o
rIn(D/r)
where Eo ~ value of electric stress (or critical gradient) at which disruption starts in kilovolts
centimeters
Va -disruptive critical voltage to neutral in kilovolts (nns)
Page 2
Transmission and Distribution
r
D
radius
06EES3
of
conductor
m
centimeters
= spacing between two conductors in centimeters
Since, in fair weather,
the
EO of air is 21.1 kVlcm rrns,
D
Vo = 21.1r In r
kV
which is correct for normal atmospheric pressure and temperature (76cm Hg at 25°C). For other
atmospheric pressures and temperatures,
D
Vo::; 21.1c5r1n -t
kV
where (5 is the air density factor. Further, after making allowance for the surface condition of the
conductor by using the irregularity factor, the disruptive critical voltage can be expressed as
where m -irregularity factor (0< mO 1)
~1 for smooth, polished, solids, cylindrical conductors
~0.93-0.98
for weathered, solid, cylindrical conductors
~0.87-0.90
for weathered conductors with more than seven strands
~0.8Q-.0.87 for weathered conductors with up to seven strands
Note that at the disruptive critical voltage V" there is no visible corona. In the event that the potential
difference (Or critical gradient) is further increased, a second point is reached at which a weak
luminous glow of violet color can be seen to surround each conductor. The voltage value at this point
is called the visual critical voltage and is given by
where Vv~visual critical voltage in kilovolts (rrns)
mr irregularity factor for visible corona (0< ml , I)
~l for smooth, polished. solid, cylindrical conductors
~0.93-0.98 for local and general visu corona on weathered, solid, cylindrical conductors
~0.70-0.75 for local visual corona on weathered stranded conductors
~0.8Q-.O.85 for general visual corona on weathered stranded conductors
Note that the voltage equations given in this section are for fair weather. For wet weather voltage
values, multiply the resulting fair weather voltage values, multiply the resulting fair weather
voltage values by 0.80. For a three-phase horizontal conductor configuration, the calculated
disruptive critical voltage should be multiplied by 0.96 and 1.06 for the middle conductor and for
the two outer conductors, respectively.
Dept of EEE, SJBIT
Page 3
•
Transmission and Distribution
06EE53
3. What are the Effects of ice and wind loading
The span design consists in determining the sag at which the line is constructed so that
heavy winds, accumulations of ice or snow, and excessive temperature changes will not
stress the conductor beyond its elastic limit, cause a serious permanent stretch, or result in
fatigue failures from continued vibrations, in other words, the lines will be erected under
warmer and nearly still-air conditions and yet must comply with the worst conditions .
•
Effect of Ice
In mountainous geographic areas, the thickness of ice formed on the conductor becomes
very significant Depending on the circumstances, it might be as much as several times
the diameter of the conductor. Ice accumulations on the conductor affectthe design of the
line
(1) by increasing the dead weight per foot of the line and
(2) by increasing" the projected surface of the line subject to wind pressure.
Mostly used for distribution lines .
••••
Ice
Figure 6. Probable configuration of ice covered conductor cross-sectional area.
Even though the more likely configuration of a conductor with a coating of ice is as shown in
Figure 6, for the sake of simplicity, it can be assumed that the ice coating, of thickness t, inches,
. '"form over the surface ofa conductor, as shown in Figure 7.
i :':] the cross-sectional area of the ice is
or
A , == 1Tt(d
, _ c +1,)
A,
=
1!47Tt,(d, + t,)
•
m.2
ft"
d = diameter of conductor In inches
, = radial thickness of ice coating in inches
t
SJBIT
Page 4
Transmission
06EE53
and Distribution
•
t,
t.
Figure 7. Assumed configuration of ice-covered conductor cross-sectional area.
If itie ice load is assumed to be uniform throughout the length of the conductor, the volume of ice
i:
•
V.~rh1Txft,(dc+I,)
fefft
per loot 1S
The weight of the ice is 571b/It3, so that the weight of ice per foot is
or approximately
w, = 1.25I,(c_ic + 'i)
Ib/ft
Therefore, the total vertical load on the conductor per unit length is
wr=
w,'_
w,
where Wr = total vertical load on conductor per unit length
W = weight of conductor per unit length
w, = weight of ice per unit length
Unit 3
l.What is corona? Derive expression for the disruptive critical voltage.(8) Jan/Feb 07
Corona on transmission lines causes power loss, radio and television interference, and audible
noise (in terms of buzzing, hissing, or frying sounds) in the vicinity of the line.
A transmission line should he designed to operate just below the disruptive critical voltage in fair
wether so that corona only takes place during adverse atmospheric conditions. Therefore, the
calculated disruptive critical voltage is an indicator of corona performance of the line. However, a
high value of the disruptive critical voltage is not the only criterion of satisfactory corona
performance. The sensitivity of the conductor to foul weather should also be considered (e.g.,
corona increases more slowly on stranded conductors than on smooth conductors). Due to the
numerous factors involved, the precise calculation of the peak value of corona loss is extremely
Dept of EEE, SJBIT
Page 5
Transmission and Distribution
06EES3
if not impossible. The minimum voltage at which the ionization occurs in fair weather is
,,>d the disruptive critical voltage and can be determined from
E =
o
•
'.
Vo
r In(Dlr)
D
Vo == grln--o
r
where Eo ~value of electric stress (or critical gradient) at which disruption starts in
kilovolts per centimeters
Vo ~ disruptive critical voltage to neutral in kilovolts (rms)
r ~radius of conductor in centimeters
D = spacing between two conductors in centimeters
in fair weather, the EO of air is 21.1 kV/cm rms,
h is correct for normal atmospheric pressure and temperature (76cm Hg at 25°C). For other
«spheric pressures and temperatures,
D
t
Vo:::;;21.10r In -
kV
re 8 is the air density factor. Further, after making allowance for the surface condition of
. conductor by using the irregularity factor, the disruptive critical voltage can be expressed as
where m ~ irregularity factor (0< mO I)
~l for smooth, polished, solids, cylindrical conductors
~0.93-O.98 for weathered, solid, cylindrical conductors
~0.87-0.90 for weathered conductors with more than seven strands
~O.80-0.87 for weather-edconductors with up to seven strands
are the effects of corona? Derive expression for the visual critical voltage (8)
l'if) 09
.ansrnission lines causes power loss, radio and television interference, and audible noise
of buzzing, hissing, or frying sounds) in the vicinity of the line. At extra high-voltage levels
5 kV and higher), the conductor itself is the major source of audible noise, radio
renee, television interference, and corona loss. The audible noise is a relatively new
Page 6
Transmission
06EE53
and Distribution
environmental concern and is becoming more important with increasing voltage level. For example,
for transmission lines up to 800 kV, audible noise and electric field effects have become major design
factors and have received considerable testing and study. It had been observed that the audible noise
from the corona process mainly takes place in foul weather. In (Icy conditions, the conductors
normally operate below the corona detection level, and therefore, very few corona sources exist. In
wet conditions, however, water drops on the conductors cause large number of corona discharges and
a resulting burst of noise. At ultrahigh-voltage levels (1000 kV and higher), such audible noise is the
limiting environmental design factor.
A transmission line should he designed to operate just below the disruptive critical voltage in fair
wether so that corona only takes place during adverse atmospheric conditions. Therefore, the
calculated disruptive critical voltage is an indicator of corona performance of the line. However, a
high value of the disruptive critical voltage is not the only criterion of satisfactory corona
performance. The sensitivity of the conductor to foul weather should also be considered (e.g., corona
increases more slowly on stranded conductors than on smooth conductors). Due to the numerous
factors involved, the precise calculation of the peak value of corona loss is extremely difficult, if not
impossible. The minimum voltage at which the ionization occurs in fair weather is called the
disruptive critical voltage and can be determined from
E =
o
IT.
Yo::::;
v:0
r In(Dlr)
Eorln-D
r
where Eo = value of electric stress (or critical gradient) at which disruption starts in kilovolts
per centimeters
.
Vo = disruptive critical voltage to neutral in kilovolts (rms)
r = radius of conductor in centimeters
D = spacing between two conductors in centimeters
Since, in fair weather, the EO of air is 21.1 kVfcm rms,
D
Vo::::; 21.1r]n r
kV
which is correct for normal atmospheric pressure and temperature (76cm Hg at 25°C). For other
atmospheric pressures and temperatures,
D
Vo = 21.10r In t
kV
where () is the air density factor. Further, after making allowance for the surface condition of the
conductor by using the irregularity factor, the disruptive critical voltage can be expressed as
Page 7
•
Transmission and Distribution
Vo =2Llomorln
•
06EE53
D
-
r
kV
where m = irregularity factor (0< mO I)
=1 for smooth, polished, solids, cylindrical conductors
=0.93-0.98 for weathered, solid, cylindrical conductors
=0.87--0.90 for weathered conductors with more than seven strands
=0.80--0.87 for weathered conductors with up to seven strands
Note that at the disruptive critical voltage v" there is no visible corona. In the event that the potential
difference (Or critical gradient) is further increased, a second point is reached at which a weak
luminous glow of violet color can be seen to surround each conductor. The voltage value at this point
is called the visual critical voltage and is given by
- - ( .ff,.
03) In rD
V" = 21.1om"r 1 +
kV
where Vv -visual critical voltage in kilovolts (rrns)
l11v=irregularityfactor for visible corona (0< m l , I)
=1 for smooth, polished. solid, cylindrical conductors
=0.93-0.98 for local and general visu corona on weathered, solid, cylindrical conductors
= O.70-0. 75 for local visual corona on weathered stranded conductors
=0.80--0.85 for general visual corona on weathered stranded conductors
Note that the voltage equations given in this section are for fair weather. For wet weather voltage
values, multiply the resulting fair weather voltage values, multiply-the resulting fair weather
voltage values by 0.80. For a three-phase horizontal conductor configuration, the calculated
uptive critical voltage should be multiplied by 0.96 and l.06 for the middle conductor and for
:L..;tWO outer conductors, respectively.
3.Write short note on disruptive critical voltage. (6) July/Aug-09
.mission line should he designed to operate just below the disruptive critical voltage in fair
that corona only takes place during adverse atmospheric conditions. Therefore, the
ted disruptive critical voltage is an indicator of corona performance of the line. However, a
:,1 :e of the disruptive critical voltage is not the only criterion of satisfactory corona
The sensitivity of the conductor to foul weather should also be considered (e.g., corona
increases more slowly on stranded conductors than on smooth conductors). Due to the numerous
factors involved, the precise calculation of the peak value of corona loss is extremely difficult, if not
impossible. The minimum voltage at which the ionization occurs in fair weather is called the
disruptive critical voltage and can be determined from
. :;IBIT
Page 8
06EE53
Transmission and Distribution
E =
Vo
o rIn(D/r)
D
v:o = E0rlnr
..
4.Explain the terms with reference to corona.
i)disruptive critical voltage
ii) Power loss due to corona (10)
JullAug-08
The minimum voltage at which the ionization occurs in fair weather is called the disruptive
critical voltage and can be determined from
E
o
=
V.
0
r In(Dlr)
where Eo ~value of electric stress (or critical gradient) at which disruption starts in
kilovolts per centimeters
Vo -disruptive critical voltage to neutral in kilovolts (rms)
r ~radius of conductor in centimeters
D = spacing between two conductors in centimeters
Since, infair weather, the EO of air is 21.1 kV/cm rms,
Yo = 21.1r
D
In r
kV
which is correct for normal atmospheric pressure and temperature (76cm Hg at 25°C). For other
atmospheric pressures and temperatures,
D
Vo == 21.1orln t
•
kV
where 0 is the air density factor. Further, after making allowance for the surface condition of
the conductor by using the irregularity factor, the disruptive critical voltage can be expressed as
Dept of EEE, SJ8IT
Page 9
Transmission and Distribution
06EE53
D
Vo=21.15morln
-
r
kV
where m = irregularity factor (0< rnO 1)
= 1 for smooth, polished, solids, cylindrical conductors
= 0.93-0.98
for weathered, solid, cylindrical conductors
= 0.87-0.90
for weathered conductors with more than seven strands
t.h e
from
J
•
=0.80---0.87 for weathered conductors with up to seven strandsAccording
weather
corona
loss
per
phase
or conductor
~~
,
241
Pc = T (f
+ 25)
(rD.)112 (V -
390
( r )1l2
r. == T
(f + 25) D
(V.,..
Where,f
Vo)2
X
to Peek,
can be calculated
10-5 kW/km
Vo)2 X 1O-s kW/mi
"frequency in hertz
V line-to-neutral operating voltage in kilovolts
=
Vo= disruptive critical voltage in kilovolts
The 'ret ., '.'lther corona can be calculated from the above equations by multiplying
< "i:
: ',:s a correct result if
.
( i \ the frequency is between 25 and 120 Hz,
, I :: .e conductor
radius is greater than 0.25 ern. and
CO ratio of V to' VO is greater than 1.8.. '
Vo by 0.80. Peek's
r,~(~r/l
T1':
P"'" (';'
OSS IS proportional
to the square root of the size of the conductor.
:>; larger the radius of the conductor, the larger the power loss. Also, the larger the spacing
,[uctors, the smaller the power loss. Similarly,
P~ex: (V - VO)2
; "'civenvoltage level, the larger the conductor site, the larger the disruptive critical voltage and
'. smaller the power loss.
Peterson (11], the fair weather corona loss per phase or conductor can be calculated from
P
_ 1.11066 X 10-4
c -
P
[In(2Dld)]2
= 1.78738 X 10-4
c
[In(2Dld)}2
1.
tv
F
kW/km
Z
fV F
kW/mi
nductor diameter
SJBIT
Page 10
Transmission and Distribution
06EE53
D ~spacing between conductors
f = frequency in hertz
V= line-to-neutral operating voltage in kilovolts
F= corona factor determined by test and is a function of ratio of V to Vo
In general, the corona losses due to fair weather conditions arc not significantly large at extra-highvoltage range. Therefore, their effects are not significant from technical andlor economic points of view.
Whereas the corona losses due to foul weather conditions are very significant.
•
where TPc•RW = total three-phase corona losses due to rainy weather in
kilowatts per kilometer
TP, .FW """ total three-phase corona losses due to fair weather in
kilowatts per kilometer
V = line-to-line operating voltage in kilovolts
r "'"conductor radius in centimeters
n = total number of conductors (number of conductors per
bundle times 3)
E, = voltage gradient on underside of conductor i in kilovolts
(peak) per centimeter
m = an exponent (SO; 5)
J = Joss current constant (-4.37 x 10-10 at 400kV and 3.32 x
10-10 at 500kV and 700kV)#
R = rain rate in millimeters per hour or inches per hour
K = wetting coefficient (10 if R is in millimeter per hour or 254
if .R is in inches per hour)
S.Write short note on corona in transmission lines.(S) Ju1lAug-09
Corona on transmission lines causes power loss, radio and television interference, and audible noise
(in terms of buzzing, hissing, or frying sounds) in the vicinity of the line. At extra high-voltage levels
(i.e., at 345 kV and higher), the conductor itself is the major source of audible noise, radio
interference, television interference, and corona loss. The audible noise is a relatively new
environmental concern and is becoming more important with increasing voltage level. For example,
for transmission lines up to 800 kV, audible noise and electric field effects have become major design
factors and have received considerable testing and study. It had been observed that the audible noise
from the corona process mainly takes place in foul weather. In (Icy conditions, the conductors
normally operate below the corona detection level, and therefore, very few corona sources exist. In
wet conditions, however, water drops on the conductors cause large number of corona discharges and
a ~C:C:l:lti'1g
burst of noise. At ultrahigh-voltage levels (1000 kV and higher), such audible noise is the
limiting environmental design factor.
6.Write short note on factors affecting corona and methods of reducing corona effect.
(7) Jan/Feb-lO
Dept of EEE, SJBIT
Page 11
..
T
nsrnission and Distribution
C'
06EE53
transmission lines causes power loss, radio and television interference, and audible noise
of buzzing, hissing, or frying sounds) in the vicinity of the line. At extra high-voltage levels
(i . 'f 345 kV and higher), the conductor itself is the major source of audible noise, radio
if; fcn.:nce,television interference, and corona loss. The audible noise is a relatively new
en' ,,,qnental concern and is becoming more important with increasing voltage level. For example,
f
. nsmission lines up to 800 kV, audible noise and electric field effects have become major design
f" '. or ': and have received considerable testing and study. It had been observed that the audible noise
fi
corona process mainly takes place in foul weather. In (Icy conditions, the conductors
n
operate below the corona detection level, and therefore, very few corona sources exist. In
v
" itions, however, water drops on the conductors cause large number of corona discharges and
burst of noise. At ultrahigh-voltage levels (1000 kV and higher), such audible noise is the
.:,uironmental design factor.
'.A"
(i
j
r.'
..
,
'i' L>'piain the terms with reference to corona.
i)visual critical voltage ii) Power loss due to corona (10)
Q.No-4
•
JullAug-09
ssume that a three-phase overhead transmission line is made up of three equilaterally
conducters, each with overall diameter of 3 em. The equilateral spacing between
:,' 5.5 m. The atmosphere pressure is 74 cm Hg and the temperature is 10°C. If
! Yfactor of the conductors is 0.90 in each case, determine the following:
, uptive critical rms line voltage.
iDlT
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06EE53
Transmission and Distribution
Jan/Feb - 08
(b) Visual critical rms line voltage.
(a)
0= 3.921lp = 3.9211 x 14 = 1.0253
273 + I
273 + 10
D
Vo = 21.1omo' ln = 21.1
r
-
550
x 1.0253 x 0.90 x 1.5 In 1.5
:,~172.4 kV/phase
Thus, the rrns fine voltage is
"
Vo = V3 x 172.4 = 298.ikV
(b) The visual critical rrns line voltage is
( 0.3)
D
V" =21.10111" 1 + V8r In-;
= 21.1
(
0.3) 1.5
x 1.0253 x 0.90 x 1.5 x 1 + v'1.0253
= 214.2
X
550
In 1.5
kVJphase
Therefore, the rms line voltage is
VI' = v'3 x 214.2 = 370.9 kV
•
•
9.Assume that a three-phase overhead transmission line is made up of three equilaterally
spaced conductors, each with overall diameter of 3 cm. The equilateral spacing between
conductors is 5.5 m. The atmosphere pressure is 74 cm Hg and the temperature is
lO°C.Consider Example 8.1 and assume that the line operates at 345 kV at 60 Hz and the
line length is 50 mi. Determine the total fair weather corona loss for the Line by using
Dept of EEE, SJBIT
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Transmission and Distribution
06EE53
Peek's formula.
Solution
laniFeb-lO
According to Peek, the fair weather corona loss per phase is
390
r, = T
(f
,
rl/2
+ 25) D (v-
390
~)2
X 1O-1~
1.5112
=
1.0253 (60 + 25) 550 (199.2 - 172.4)2 x 10-5
=
12.1146 kW/mi/phase
or, for the total line length.
P, = 12.1146 x 50"'" 605.7 kW/phase
Therefore, the total corona loss of the line is
P, = 3 x 605.7
= 1817.2 kW
Unit 3
Overhead Transmission Lines
Show that a transmission line conductor suspended between level support assumes the
shape of a catenory. Derive the expression for sag.(lO)
catenary Method
Figure 1 shows a span of conductor with two supports at the same level and separated by a
hei: :,~,:,:,ddistance L.
Let '0' be the lowest point on the catenary curve and '1' be the length of the conductor between
two supports.
Let \1" be the weight of the conductor per unit length, 'T' be the tension of the conductor at any
point'r'
in the direction of the curve, and H be the tension at origin O.
!,!hu. kt 's ' be the length of the curve between points 0 and P, so that the weight of the
\ WS.
can be resolved into two components, T the horizontal component and
Then, for equilibrium,
T"
the vertical
Tr ~H and Ty ~ WS
"
Thus, the portion OP of the conductor is in equilibrium under the tension Tat P. the weight Ws
acting vertically downward, and the horizontal tension H.
':imgle shown in Figure 2, ds represents a very short portion of the conductor, in the
P When s is increased by ds the corresponding x and yare increased by dx and
. SjBIT
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and Distribution
a ....- ......
--_
noe ..
.. _
. .....,. ..........
_
....
_
.. _,_
-
~v
H
Figure 1. Conductor
suspended between supports at same elevation.
tan 8 = dy = ws
dx
H
since
then
(:r=l+(~;r
Therefore.
dx=
Dept of EEE, SJBIT
ds
Vl + (wslH)i
Page 15
1',.
l, l
.mission and Distribution
:!l::::=_: __....
""----
06EE53
... -_ ....
Figure 2
In:.
>cth sides gives
x=
X
.'
1
. ds
f ,/1 + (WS/H)2
= -Hw
sinh
K
H +
-rL WS
-
l.e constant of integration. When x = 0, $ = 0, and K ~0,
s=
H . h
W Sin
WX
H
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Transmission and Distribution
When x =
!L,
I
H
wL
s = - = - sinh -.
2
w
211
Therefore,
1=
2H.
-;-
1= 2H
w
Of
wL
..
sinh 2H
[.!..I! 2H
wL + .!. ('WL)3
3! 2H
+ ... J'
.
.
approximately
dy
dx
= ws = sinh ~
H
H
or
dY"'"
. h --wx dx
..
H
S10
Integrating both sides,
y "'"
f
sinh ;; dx
H
w
y= -cos-
I f the lowest point
then K,
wx
H
+ KI
of the curve is taken as the origin, when x=O, y ~ 0
=-
f
•
Hlw,
"il,..;e,by the series, coshO=l. Therefore,
y
= -l/ (cosh -.wx - 1)
Dept of EEE, SJBlT
HI
H
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Transmission and Distribution
06EE53
is the equation of the curve that is called a catenary .
Explain what is sag and why it is inevitable in over head transmission lines? What are
the factors influencing it?(7)
,
Conductor sag and tension analysis is an important consideration in overhead distribution
line design as well as in overhead transmission line design. The quality and continuity of
electric service supplied over a line (regardless of whether it is a distribution, a subtransmission, or a transmission line) depend largely on whether the conductors have been
properly installed. Thus, the designing engineer must determine in advance the amount of sag
and tension to be given the wires or cables of a particular line at a given temperature. In order'
to specify the tension to be used in stringing the line conductors, the values of sag and
tension for winter and summer conditions must be known. Tension in the conductors
contributes to the mechanical load on structures at angles in the line and atdead ends.
Excessive tension may cause mechanical failure of the conductor itself.
The factors affecting the sag of a conductor strung between supports arc:
1. Conductor load per unit length.
2. Span, that is, distance between supports.
3. Temperature.
4. Conductor tension.
With usual notations derive an expression for maximum sag of a tranmn. Line where the
supports are at different levels?( 10)
Consider a span L between two supports, as shown in Figure 10.5, whose elevations differ by a
distance h. Let the horizontal distance from the lowest point of the curve to the lower and the higher
supports be xl and x2, respectively.
By using equation (10.46), that is,
...
.,
iJf
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06EE53
'j'c;}t1';mission and Distribution
wx2
y=-
2T
d I and d2 sags can be found as
-
and
B
L
Figure 4. Supports at different levels.
Therefore,
h=d 2 -d
1
or
w
Il=-(X
2
2
2T . 2 -x) I
or
wL
h::: 2T (X2 - Xl)
since
Therefore,
By adding the above two equations,
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06EE53
2Th
2x2=L+ -
wL
Th
-L2 +wi:
x2=
B
t
o
Figure 5. Case of negative
Xl
J;: :)e,L;;""Lng the same two equations,
2x -= L --I
2Th
wl:
L
2
wL
then
XI
x =--I
Th
.on,
"
is positive
then x 1 is zero
then
Xl
is negative
the expression for sag in a freely suspended conductor when the supports are at equal
,.') - refer Q.No-l
"., ., ....,', .,' ''"'",',,,,.''
......... !'II!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
, ,\;IBlT
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Transmission and Distribution
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Explain the effects of sag in overhead trasmn. Line.(5)
Conductor sag and tension analysis is an important consideration in overhead distribution
line design as well as in overhead transmission line design. The quality and continuity of
electric service supplied over a line (regardless of whether it is a distribution, a subtransmission, or a transmission line) depend largely on whether the conductors have been
properly installed. Thus, the designing engineer must determine in advance the amount of sag
and tension to be given the wires or cables of a particular line at a given temperature. In order
to specify the tension to be used in stringing the line conductors, the values of sag and
tension for winter and summer conditions must be known. Tension in the conductors
contributes to the mechanical load on structures at angles in the line and at dead ends.
Excessive tension may cause mechanical failure of the conductor itself.
Obtain the expression for sag in a power conductor when the supports
levels, taking into the effect of wind and ice loading(12)
are at equal
Effect of Ice
In mountainous geographic areas, the thickness of ice formed on the conductor becomes very
significant. Depending on the circumstances, it might be as much as several times the diameter of the
conductor. Ice accumulations on the conductor affect the design of the line
(1) by increasing the dead weight per foot of the line and
(2) by' increasing the projected surface of the line subject to wind pressure.
Mostly used for distribution lines.
•
Ice
Figure 6. Probable configuration of ice covered conductor cross-sectional area.
Even though the more likely configuration of a conductor with a coating of ice is as shown in
Figure 6, for the sake of simplicity, it can be assumed that the ice coating, of thickness t, inches,
is uniform over the surface ofa conductor, as shown in Figure 7.
Then the cross-sectional area of the ice is
Dept of EEE, SJBIT
Page 21
-..
Tran
mission and Distribution
06EE53.
or
. 2
m.
or
.,
,
where d = diameter of conductor in inches
/ = radial thickness of ice coating in inches
I
.
I.
! :i.:~Lre
7. Assumed configuration ofice-covered conductor cross-sectional area.
i " ". '.
1,·•.•
'"ad is assumed to be uniform throughout the length of the conductor, the volume of ice per
V, "'" J~41T
X
{I,
(d, + I.)
ft3/ft
,.1: »fthe ice is 571b/It3, so that the weight of ice per foot is
;'; ; :"
.uutely
w, = 1.25t,{dc + Ii)
(b/ft
.. he total vertical load on the conductor per unit length is
iVT:::
WI'- ':',
total vertical load on conductor per unit length
of conductor per unit length
weight of ice per unit length
w = weight
•• ~.".~ ..",..,...
--~!!!!!!!!!!!!!!!~!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!~
I~,
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Effect of Wind
It is customary to assume that the wind blows uniformly and horizontally across the projected area of the
conductor covered with no ice and ice, respectively.
The projected area per unit length of the conductor with no ice is
SRI
where
= An'!
projected area of conductor covered with no ice in square feet per Unit length
Ani = cross-sectional area of conductor covered with no icc in square feet
l= length of conductor in unit length
for a 1-ft length of conductor with no ice,
Sni=
-
....
Sn. = fidi
whereas with ice, it is
Swi = Aw.i
where SWi= projected area of conductor covered with icc in square feet per unit length
Figure 8. Force of wind on conductor covered with no ice.
AWi ~ cross-sectional
area of conductor covered with ice in square feet
1~length of conductor in unit length
for a 1-ft length of conductor,
de + 2/, I ft2/ft
S
wi
=
12
Therefore, the horizontal force exerted on the line as a result of the wind pressure with no ice (Figure 8) is
P = SniP
Ib/unit length
for a 1-ft length of conductor,
where P= horizontal wind force (i.e., load) exerted on line in pounds per feet
p =wind pressure in pounds per square feet
whereas with ice (Figure 9), it is
Dept of EEE, SJBIT
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Transmission and Distribution
Il!.I---'-"- -""-----"'~--
I
06EE53
......
- ................
--
Figure 9. Force of wind on conductor covered with ice.
w+w ..------
WI(
Figure 10
P == SwiP lb/unit length
fe" :: ; -ft length of conductor,
p_ de +2/,
12 P
; :,.~:',
Ib/n
'c: effective load acting on the conductor is
w", ;:;;\lpZ
+ (w + W,)2
Iblft
acting at an angle 0 to the vertical, as shown in Figure 10.
By replacing w by We in the previously derived equations for tension and sag of the line in still air, these
can he applied to a wind- and ice-loaded line. For example, the sag equation becomes
d=!!',L2
8T
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ft
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Transmission and Distribution
06EES3
Write short note on effect of ice load and wind effect on sag of transmission Line.(5)
The span design consists in determining the sag at which the line is constructed so that heavy winds,
accumulations of ice or snow, and excessive temperature changes will not stress the conductor
beyond its elastic limit, cause a serious permanent stretch, or result in fatigue failures from continued
vibrations, in other words, the lines will be erected under warmer and nearly still-air conditions and
yet must comply with the worst conditions.
In mountainous geographic areas, the thickness of ice formed on the conductor becomes very
significant. Depending on the circumstances, it might be as much as several times the diameter of the
conductor. Ice accumulations on the conductor affect the design of the line
(1) by increasing the dead weight per foot of the line and
(2) by increasing the projected surface of the line subject to wind pressure.
Mostly used for distribution lines.
It is customary to assume that the wind blows uniformly and horizontally across the projected area of
the conductor covered with no ice and ice, respectively.
From the first principles derive the expression for sag in a freely suspended conductor
when the supports are at unequal levels (6)
Refer Q.No-3.
Dept of EEE, SJBlT
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