12 App Lication Of The A, B, C, D General Ci Rcuit Constants To Radial Feeders
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307
Design Considerations of Primary Systems
Vs
δ
Vr
θr
0°
Ir
Figure 5.30 Phasor diagram.
I r = I r ∠ − θr
(5.48)
where
–
Vr = receiving-end voltage phasor
–
Vs = sending-end voltage phasor
–
Ir = receiving-end current phasor
The sending-end voltage in terms of the general circuit constants can be expressed as
Vs = A × Vr + B × I r
(5.49)
A = A1 + jA2
(5.50)
B = B1 + jB2
(5.51)
I r = I r (cos θr − j sin θr )
(5.52)
Vr = Vr ∠0° = Vr
(5.53)
Vs = Vs (cos δ + j sin δ )
(5.54)
where
Therefore, Equation 5.49 can be written as
Vs cos δ + jVs sin δ = ( A1 + jA2 )Vr + ( B1 + jB2 )( I r cos θr − jIr sin θr )
from which
Vs cos δ = A1Vr + B1Ir cos θr + B2 I r sin θr
(5.55)
Vs sin δ = A2Vr + B2 I r cos θr − B1I r sin θr
(5.56)
and
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Electric Power Distribution Engineering
By taking squares of Equations 5.55 and 5.56, and adding them side by side,
Vs2 = ( A1Vr + B1I r cos θr + B2 I r sin θr )2 + ( A2Vr + B2 I r cos θr − B1I r sin θr )2
(5.57)
or
(
)
(
Vs2 = Vr2 A12 + A22 + 2Vr I r cos θr ( A1B1 + A2 B2 ) + B12 Vr2 cos2 θr + I r2 sin 2 θr
(
)
+ B22 I r2 sin 2 θr + I r2 cos2 θr + 2Vr I r sin θr ( A1B2 − B1 A2 )
)
(5.58)
Since
Pr = Vr I r cos θr
(5.59)
Qr = Vr I r sin θr
(5.60)
Qr = Pr tan θ
(5.61)
and
Equation 5.58 can be rewritten as
(
) (
)(
Vr2 A12 + A22 + B12 + B22 1 + tan 2 θr
2
r
2
r
) VP
= Vs2 − 2 Pr [( A1B1 + A2 B2 ) + ( A1B2 − B1 A2 ) tan θr ] (5.62)
Let
K = Vs2 − 2 Pr [( A1B1 + A2 B2 ) + ( A1B2 − B1 A2 ) tan θr ]
(5.63)
Then Equation 8.62 becomes
(
) (
)(
Vr2 A12 + A22 + B12 + B22 1 + tan 2 θr
2
r
2
r
) VP
−K =0
(5.64)
or
(
) (
)(
Vr2 A12 + A22 + B12 + B22 sec 2 θr
2
r
2
r
) VP
−K =0
(5.65)
Therefore, from Equation 5.65, the receiving-end voltage can be found as
(
1/ 2
1/ 2
2
2
2
2
2
2
2
K ± K − 4 A1 + A2 B1 + B2 Pr sec θr
Vr =
2 A12 + A22
(
)(
)
)
(5.66)
309
Design Considerations of Primary Systems
Also, from Equations 5.55 and 5.56,
Vs sin δ = A2Vr + B2 I r cos θr − B1I r sin θr
and
Vs cos δ = A1Vr + B1I r cos θr − B2 I r sin θr
where
Ir =
Pr
Vr cos θr
(5.67)
Therefore,
Vs sin δ = A2Vr +
B2 Pr B1Pr
−
tan θr
Vr
Vr
(5.68)
Vs cos δ = A1Vr +
B1Pr B2 Pr
+
tan θr
Vr
Vr
(5.69)
and
By dividing Equation 5.68 by Equation 5.69,
tan δ =
A2Vr2 + B2 Pr − B1Pr tan θr
A1Vr2 + B1Pr + B2 Pr tan θr
(5.70)
tan δ =
A2Vr2 + Pr ( B2 − B1 tan θr )
A1Vr2 + Pr ( B1 + B2 tan θr )
(5.71)
or
Equations 5.66 and 5.71 are found for a general transmission system. They could be adapted to
the simpler transmission consisting of a short primary-voltage feeder where the feeder capacitance
is usually negligible, as shown in Figure 5.31.
To achieve the adaptation, Equations 5.63, 5.66, and 5.71 can be written in terms of R and X.
Therefore, for the feeder shown in Figure 5.31,
[ I ] = [Y ][V ]
(5.72)
Vr
Vs
Is
Figure 5.31 A radial feeder.
Z = R + jX
Ir
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Electric Power Distribution Engineering
or
I s Y11
=
I r Y21
Y12 Vs
Y22 Vr
(5.73)
where
Y11 =
1
Z
Y21 = Y12 −
1
Z
(5.74)
(5.75)
1
Z
(5.76)
Y22
=1
Y21
(5.77)
Y22 =
Therefore,
A1 = −
or
A1 + jA2 = 1
(5.78)
A1 = 1
(5.79)
A2 = 0
(5.80)
where
and
Similarly,
B1 = −
1
=Z
Y21
(5.81)
or
B1 + jB2 = R + jX
(5.82)
B1 = R
(5.83)
B2 = X
(5.84)
where
and
311
Design Considerations of Primary Systems
Substituting Equations 5.79, 5.80, 5.83, and 5.84 into Equation 5.66,
1/ 2
1/ 2
2
2
2
2
2
K ± K − 4( R + X )Pr sec θr
Vr =
2
(5.85)
or
K
Vr =
2
1/ 2
1/ 2
4( R 2 + X 2 )Pr2
1
1
±
−
2
K cos2 θr
(5.86)
or
K
Vr =
2
2 ZPr
1 ± 1 −
K cos θr
1/ 2
1/ 2
(5.87)
where
K = Vs2 − 2 × Pr ( R + X × tan θr )
(5.88)
Pr ( X − R × tan θr )
Vr2 + Pr ( R + X × tan θr )
(5.89)
Also, from Equation 5.71,
tan δ =
Example 5.1
Assume that the radial express feeder, shown in Figure 5.31, is used on rural distribution and is
connected to a lumped-sum (or concentrated) load at the receiving end. Assume that the feeder
impedance is 0.10 + j0.10 pu, the sending-end voltage is 1.0 pu, Pr is 1.0 pu constant power load,
and the power factor at the receiving end is 0.80 lagging. Use the given data and the exact equations for K, Pr, and tan δ given previously and determine the following:
a. Compute Vr and δ by using the exact equations and find also the corresponding values of
the Ir and Is currents.
b. Verify the numerical results found in part (a) by using those results in
V s = Vr + (R + jX )Ir
Solution
a. From Equation 5.88,
K = Vs2 − 2 × Pr (R + X × tanθ r )
= 1.02 − 2 × 1[0.10 + 0.1× tan(cos−10.80)]
= 0.65 pu
(5.90)
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Electric Power Distribution Engineering
From Equation 5.87,
K
Vr =
2
1/ 2
1/ 2
2ZPr
1± 1−
K cos θ r
1/ 2
1/ 2
0.65
2 × 0.141× 1.0
=
1± 1−
2 0.65 × 0.8
= 0.7731 pu
From Equation 5.89,
tan δ =
=
Pr (X − R × tanθ r )
Vr2 + Pr (R + X × tanθ r )
1.0 [0.10 − 0.10 × tan(cos−1 0.80)]
0.77312 + 1.0 [0.10 + 0.10 × tan(cos−1 0.80)]
= 0.0323
therefore
δ ≅ 1.85°
Ir = Is =
=
1 .0
∠ − 36.8°
0.7731× 0.80
= 1.617∠ − 36.8° pu
Pr
∠ − θr
Vr cos θ r
b. From the given equation,
Vr = Vs − (R + jX ) Ir
= 1.0∠1.85° − (0.10 + j 0.10)(1.617∠ − 36.8°)
≅ 0.7731∠0° pu
5.13 Design of Radial Primary Distribution Systems
The radial primary distribution systems are designed in several different ways: (1) overhead primaries with overhead laterals or (2) URD, for example, with mixed distribution of overhead primaries
and underground laterals.
5.13.1 Overhead Primaries
For the sake of illustration, Figure 5.32 shows an arrangement for overhead distribution, which
includes a main feeder and 10 laterals connected to the main with sectionalizing fuses. Assume
that the distribution substation, shown in the figure, is arbitrarily located; it may also serve a second
313
Design Considerations of Primary Systems
144 services
(518 kVA)
Laterals
144 services
(518 kVA)
a΄
b
b΄
c
c΄
d
d΄
e
e΄
10 blocks
(3300 ft)
a
Main
6 blocks
(5760 ft)
Circuit breaker
6 blocks
(5760 ft)
Figure 5.32 An overhead radial distribution system.
area, which is not shown in the figure, that is equal to the area being considered and, for example,
located “below” the shown substation site.
Here, the feeder mains are three phase and of 10 short block length or less. The laterals, on the
other hand, are all of six long block length and are protected with sectionalizing fuses. In general,
the laterals may be either single phase, open wye grounded, or three phase.
Here, in the event of a permanent fault on a lateral line, only a relatively small fraction of the
total area is outaged. Ordinarily, permanent faults on the overhead line can be found and repaired
quickly.
5.13.2 Underground Residential Distribution
Even though a URD costs somewhere between 1.25 and 10 times more than a comparable overhead
system, due to its certain advantages, it is used commonly [4,5]. Among the advantages of the
underground system are the following:
1. The lack of outages caused by the abnormal weather conditions such as ice, sleet, snow,
severe rain and storms, and lightning
2.The lack of outages caused by accidents, fires, and foreign objects
3.The lack of tree trimming and other preventative maintenance tasks
4.The aesthetic improvement
For the sake of illustration, Figure 5.33 shows a URD for a typical overhead and underground
primary distribution system of the two-way feed type. The two arbitrarily located substations are
assumed to be supplied from the same subtransmission line, which is not shown in the figure, so
that the low-voltage buses of the two substations are nominally in phase. In the figure, the two
overhead primary-feeder mains carry the total load of the area being considered, that is, the area of
314
Overhead main 2
10 blocks
(3300 ft)
10 blocks
(3300 ft)
1036 kVA per two laterals
Overhead main 4
120 blocks per area
288 services per two laterals
Overhead main 3
Substation
Overhead main 1
Electric Power Distribution Engineering
120 blocks per area
288 services per two laterals
1036 kVA per two laterals
One-phase URD lateral
One-phase URD lateral
Normally open
12 blocks
(11,520 ft)
Figure 5.33 A two-way feed-type underground residential distribution system.
the 12 block by 10 block. The other two overhead feeder mains carry the other equally large area.
Therefore, in this example, each area has 120 blocks.
The laterals, in residential areas, typically are single phase and consist of directly buried (rather
than located in ducts) concentric neutral-type cross-linked polyethylene (XLPE)-insulated cable.
Such cables usually insulated for 15 kV line-to-line solidly grounded neutral service and the commonly used single-phase line-to-neutral operating voltages are nominally 7200 or 7620 V.
The installation of long lengths of cable capable of being plowed directly into the ground or
placed in narrow and shallow trenches, without the need for ducts and manholes, naturally reduces
installation and maintenance costs. The heavy three-phase feeders are overhead along the periphery
of a residential development, and the laterals to the pad-mount transformers are buried about 40 in.
deep. The secondary service lines then run to the individual dwellings at a depth of about 24 in. and
come up into the dwelling meter through a conduit. The service conductors run along easements and
do not cross adjacent property lines.
315
Design Considerations of Primary Systems
The distribution transformers now often used are of the pad-mounted or submersible type. The
pad-mounted distribution transformers are completely enclosed in strong, locked sheet metal enclosures and mounted on grade on a concrete slab. The submersible-type distribution transformers are
placed in a cylindrical excavation that is lined with a concrete bituminized fiber or corrugated sheet
metal tube. The tubular liner is secured after near-grade level with a locked cover.
Ordinarily, each lateral line is operated NO at or near the center as Figure 5.33 suggests. An
excessive amount of time may be required to locate and repair a fault in a directly buried URD
cable. Therefore, it is desirable to provide switching so that any one run of primary cable can be
de-energized for cable repair or replacement while still maintaining service to all (or nearly all)
distribution transformers.
Figure 5.34 shows apparatus, suggested by Lokay [2], which is or has been used to accomplish
the desired switching or sectionalizing. The figure shows a single-line diagram of loop-type primary-feeder circuit for a low-cost underground distribution system in residential areas. Figure 5.34a
shows it with a disconnect switch at each transformer, whereas Figure 5.34b shows the similar setup
without a disconnect switch at each transformer. In Figure 5.34a, if the cable “above” C is faulted,
the switch at C and the switch or cutout “above” C are opened, and, at the same time, the sectionalizing switch at B is closed. Therefore, the faulted cable above C and the distribution transformer at
C are then out of service.
Figure 5.35 shows a distribution transformer with internal high-voltage fuse and with stickoperated plug-in type of high-voltage load-break connectors. Some of the commonly used plug-in
types of load-break connector ratings include 8.66 kV line-to-neutral, 200 A continuous 200 A load
break, and 10,000 A symmetrical fault close-in rating.
Overhead primary feeder
Overhead primary feeder
A
A
Lightning
arresters
and fuse
at cable
termination
Underground primary
feeders
Fused
lateral
D
Underground primary
feeders
C
Normally
closed
sectionalizing
switch
B
B
(a)
Normally open
sectionalizing switch
(b)
Normally open
sectionalizing switch
Figure 5.34 Single-line diagram of loop-type primary-feeder circuits: (a) with a disconnect switch at
each transformer and (b) without a disconnect switch at each transformer. (From Westinghouse Electric
Corporation, Electric Utility Engineering Reference Book-Distribution Systems, Vol. 3, East Pittsburgh,
Pittsburgh, PA, 1965.)
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Electric Power Distribution Engineering
Plug-in type
HV load break connectors
HV fuse
Figure 5.35 A distribution transformer with internal high-voltage fuse and load-break connectors.
HV load
break switches
HV fuse
Figure 5.36 A distribution transformer with internal high-voltage fuses and load-break switches.
Figure 5.36 shows a distribution transformer with internal high-voltage fuse and with stickoperated high-voltage load-break switches that can be used in Figure 5.34a to allow four modes of
operation, namely, the following:
1. The transformer is energized and the loop is closed
2.The transformer is energized and the loop is open to the right
3.The transformer is energized and the loop is open to the left
4.The transformer is de-energized and the loop is open
In Figure 5.33, note that, in case of trouble, the open may be located near one of the underground
feed points. Therefore, at least in this illustrative design, the single-phase underground cables should
be at least ampacity sized for the load of 12 blocks, not merely six blocks.
In Figure 5.33, note further the difficulty in providing abundant overvoltage protection to cable
and distribution transformers by placing lightning arresters at the open cable ends. The location of
the open moves because of switching, whether for repair purposes or for load balancing.
Example 5.2
Consider the layout of the area and the annual peak demands shown in Figure 5.32. Note that the
peak demand per lateral is found as
144 customers × 3.6 kVA/customer ≅ 518 kVA
Assume a lagging-load power factor of 0.90 at all locations in all primary circuits at the time of the
annual peak load. For purposes of computing voltage drop in mains and in three-phase laterals,
assume that the single-phase load is perfectly balanced among the three phases. Idealize the
voltage-drop calculations further by assuming uniformly distributed load along all laterals. Assume
nominal operating voltage when computing current from the kilovoltampere load.
317
Design Considerations of Primary Systems
For the open-wire overhead copper lines, compute the percent voltage drops, using the
recalculated percent voltage drop per kilovoltampere–mile curves given in Chapter 4. Note that
p
Dm = 37 in. is assumed.
The joint EEI-NEMA report [6] defines favorable voltages at the point of utilization, inside the
buildings, to be from 110 to 125 V. Here, for illustrative purposes, the lower limit is arbitrarily
raised to 116 V at the meter, that is, at the end of the service-drop cable. This allowance may
compensate for additional voltage drops, not calculated, due to the following:
1. Unbalanced loading in three-wire single-phase secondaries
2. Unbalanced loading in four-wire three-phase primaries
3. Load growth
4. Voltage drops in building wiring
Therefore, the voltage criteria that are to be used in this problem are
Vmax = 125 V = 1.0417 pu
and
Vmin = 116 V = 0.9667 pu
at the meter. The maximum voltage drop, from the low-voltage bus of the distribution substation
to the most remote meter, is 7.50%. It is assumed that a 3.5% maximum steady-state voltage drop
in the secondary distribution system is reasonably achievable. Therefore, the maximum allowable
primary voltage drop for this problem is limited to 4.0%.
Assume open-wire overhead primaries with three-phase four-wire laterals, and that the nominal voltage is used as the base voltage and is equal to 2400/4160 V for the three-phase fourwire grounded-wye primary system with copper conductors and Dm = 37 in. Consider only the
“longest” primary circuit, consisting of a 3300 ft main and the two most remote laterals,* like the
laterals a and a′ of Figure 5.32. Use ampacity-sized conductors but in no case smaller than AWG
#6 for reasons of mechanical strength. Determine the following:
a. The percent voltage drops at the ends of the laterals and the main.
b. If the 4% maximum voltage-drop criterion is exceeded, find a reasonable combination of
larger conductors for main and for lateral that will meet the voltage-drop criterion.
Solution
a. Figure 5.37 shows the “longest” primary circuit, consisting of the 3300 ft main and the most
remote laterals a and a′. In Figure 5.37, the signs //// indicate that there are three phase and
one neutral conductors in that portion of the one-line diagram. The current in the lateral is
Ilateral =
=
Sl
3 × V L −L
518
≅ 72 A
3 × 4.16
(5.91)
Thus, from Table A.1, AWG #6 copper conductor with 130-A ampacity is selected for the
laterals. The current in the main is
Imain =
=
Sm
3 × V L −L
1036
≅ 144 A
3 × 4.16
* Note that the whole area is not considered here, but only the last two laterals, for practice.
(5.92)