7 General Case: Substation Service Area with n Primary Feeders
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Design of Subtransmission Lines and Distribution Substations
207
In Figure 4.28, the following relationship exists:
tan θ =
y
x + dx
(4.12)
or
y = ( x + dx ) tan θ
≅ x × tan θ
(4.13)
The total service area of the feeder can be calculated as
ln
∫ dA
An =
x =0
= ln2 × tan θ
(4.14)
The total kilovolt-ampere load served by one of n feeders can be calculated as
ln
Sn =
∫ dS
x =0
= D × ln2 × tan θ
(4.15)
This total load is located, as a lump-sum load, at a point on the main feeder at a distance of
(2/3) × l4 from the feed point a. Hence, the summation of the percent voltage contributions of all
such areas is
%VD n =
2
× ln × K × S n
3
(4.16)
or, substituting Equation 4.15 into Equation 4.16,
%VD n =
2
× K × D × ln3 × tan θ
3
(4.17)
or, since
n(2θ ) = 360°
(4.18)
2
360°
× K × D × ln3 × tan
3
2n
(4.19)
Equation 4.17 can also be expressed as
%VD n =
Equations 4.18 and 4.19 are only applicable when n ≥ 3. Table 4.2 gives the results of the application
of Equation 4.17 to square and hexagonal areas.
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Electric Power Distribution Engineering
Table 4.2
Application Results of Equation 4.17
n
θ
tan θ
4
45°
1.0
6
30°
1
3
%VDn
2
× K × D × l43
3
2
× K × D × l63
3 3
For n = 1, the percent voltage drop in the feeder main is
%VD1 =
1
× K × D × l13
2
(4.20)
%VD2 =
1
× K × D × l23
2
(4.21)
and for n = 2 it is
To compute the percent voltage drop in uniformly loaded lateral, lump and locate its total load at
a point halfway along its length and multiply the kilovolt-ampere-mile product for that line length
and loading by the appropriate K constant [5].
4.8 Comparison of the Four- and Six-Feeder Patterns
For a square-shaped distribution substation area served by four primary feeders, that is, n = 4, the
area served by one of the four feeders is
A4 = l42 mi 2
(4.22)
The total area served by all four feeders is
TA4 = 4 A4
= 4l42 mi 2
(4.23)
The kilovolt-ampere load served by one of the feeders is
S4 = D × l42 kVA
(4.24)
Thus, the total kilovolt-ampere load served by all four feeders is
TS4 = 4 D × l42 kVA
(4.25)
The percent voltage drop in the main feeder is
%VD 4,main =
2
× K × D × l43
3
(4.26)
Design of Subtransmission Lines and Distribution Substations
209
The load current in the main feeder at the feed point a is
I4 =
S4
3 × VL − L
(4.27)
I4 =
D × l42
3 × VL − L
(4.28)
or
The ampacity, that is, the current-carrying capacity, of a conductor selected for the main feeder
should be larger than the current values that can be obtained from Equations 4.27 and 4.28.
On the other hand, for a hexagonally shaped distribution substation area served by six primary
feeders, that is, n = 6, the area served by one of the six feeders is
A6 =
1
× l62 mi 2
3
(4.29)
TA6 =
6
× l62 mi 2
3
(4.30)
The total area served by all six feeders is
The kilovolt-ampere load served by one of the feeders is
S6 =
1
D × l62 kVA
3
(4.31)
Therefore, the total kilovolt-ampere load served by all six feeders is
TS6 =
6
× D × l62 kVA
3
(4.32)
The percent voltage drop in the main feeder is
%VD6,main =
2
3 3
× K × D × l63
(4.33)
The load current in the main feeder at the feed point a is
I6 =
S6
3 × VL − L
(4.34)
or
I6 =
D × l62
3 × VL − L
(4.35)
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Electric Power Distribution Engineering
The relationship between the service areas of the four- and six-feeder patterns can be found
under two assumptions: (1) feeder circuits are thermally limited (TL) and (2) feeder circuits are
voltage-drop-limited (VDL).
1. For TL feeder circuits: For a given conductor size and neglecting voltage drop,
I 4 = I6
(4.36)
Substituting Equations 4.28 and 4.35 into Equation 4.36,
D × l42
D × l62
=
3 × VL − L 3 × VL − L
(4.37)
from Equation 4.37,
2
l6
l = 3
4
(4.38)
Also, by dividing Equation 4.30 by Equation 4.23,
TA6 6 / 3l62
=
TA4
4l42
2
=
3 l6
2 l4
(4.39)
Substituting Equation 4.38 into Equation 4.39,
TA6 3
=
TA4 2
(4.40)
TA6 = 1.50 TA4
(4.41)
or
Therefore, the six feeders can carry 1.50 times as much load as the four feeders if they are
thermally loaded.
2. For VDL feeder circuits: For a given conductor size and assuming equal percent voltage drop,
%VD 4 = %VD6
(4.42)
Substituting Equations 4.26 and 4.33 into Equation 4.42 and simplifying the result,
I A = 0.833 × I 6
(4.43)
From Equation 4.30, the total area served by all six feeders is
TA6 =
6
× l62
3
(4.44)
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Design of Subtransmission Lines and Distribution Substations
Substituting Equation 4.43 into Equation 4.23, the total area served by all four feeders is
TA4 = 2.78 × l62
(4.45)
Dividing Equation 4.44 by Equation 4.45,
TA6 5
=
TA4 4
(4.46)
TA6 = 1.25 TA4
(4.47)
or
Therefore, the six feeders can carry only 1.25 times as much load as the four feeders if they
are VDL.
4.9 Derivation of the K Constant
Consider the primary-feeder main shown in Figure 4.29. Here, the effective impedance Z of the
three-phase main depends upon the nature of the load. For example, when a lumped-sum load is
connected at the end of the main, as shown in the figure, the effective impedance is
Z = z × l Ω/phase
(4.48)
where
z is the impedance of three-phase main line, Ω/(mi phase)
l is the length of the feeder main, mi
When the load is uniformly distributed, the effective impedance is
Z=
1
× z × l Ω/phase
2
(4.49)
When the load has an increasing load density, the effective impedance is
Z=
2
× z × l Ω/phase
3
(4.50)
Taking the receiving-end voltage as the reference phasor,
Vr = Vr ∠0°
(4.51)
Vs
Vr
I
Z = R + jX
P
Q
I
Figure 4.29 An illustration of a primary-feeder main.
Load
Pr + jQr
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Electric Power Distribution Engineering
Vs
Iz
JLX
δ
0°
Vr
θI
IR
Vs – Vr
I
Figure 4.30 Phasor diagram.
and from the phasor diagram given in Figure 4.30, the sending-end voltage is
Vs = Vs ∠δ
(4.52)
I = I∠ − θ
(4.53a)
The current is
and the power-factor angle is
θ = θVr − θ I
= 0° − θ I = −θ I
(4.53b)
and the power factor is a lagging one. When the real power P and the reactive power Q flow in
opposite directions, the power factor is a leading one.
Here, the per unit voltage regulation is defined as
VR pu
Vs − Vr
Vr
(4.54)
Vs − Vr
× 100
Vr
(4.55)
and the percent voltage regulation is
VR pu =
or
%VR = VR pu ×100
(4.56)
whereas the per unit voltage drop is defined as
where VB is normally selected to be Vr.
VR pu
Vs − Vr
VB
(4.57)
Design of Subtransmission Lines and Distribution Substations
213
Hence, the percent voltage drop is
Vs − Vr
× 100
VB
(4.58)
%VD = VD pu ×100
(4.59)
%VD =
or
where VB is the arbitrary base voltage. The base secondary voltage is usually selected as 120 V. The
base primary voltage is usually selected with respect to the potential transformation (PT) ratio used.
Common PT Ratios
VB (V)
20
60
100
2,400
7,200
12,000
From Figures 4.29 and 4.30, the sending-end voltage is
Vs = Vr + IZ
(4.60)
Vs = (cos δ + j sin δ ) = Vr ∠0° + I (cosθ − j sin θ )( R + jX )
(4.61)
or
The quantities in Equation 4.61 can be either all in per units or in the mks (or SI) system. Use
line-to-neutral voltages for single-phase three-wire or three-phase three- or four-wire systems.
In typical distribution circuits,
R≅X
and the voltage angle δ is closer to zero or typically
0° ≤ δ ≤ 4°
whereas in typical transmission circuits,
δ ≅ 0°
since X is much larger than R.
Therefore, for a typical distribution circuit, the sin δ can be neglected in Equation 4.61. Hence,
Vs ≅ Vs cos δ
and Equation 4.61 becomes
Vs ≅ Vr + IR cos θ + IX sin θ
(4.62)
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Electric Power Distribution Engineering
Therefore, the per unit voltage drop, for a lagging power factor, is
VD pu =
IR cos θ + IX sin θ
VB
(4.63)
and it is a positive quantity. The VDpu is negative when there is a leading power factor due to shunt
capacitors or when there is a negative reactance X due to series capacitors installed in the circuits.
The complex power at the receiving end is
Pr + jQr = Vr I *
(4.64)
Therefore,
I =
Pr − jQr
Vr
(4.65)
since
Vr = Vr ∠0°
Substituting Equation 4.65 into Equation 4.61, which is the exact equation since the voltage angle δ
is not neglected, the sending-end voltage can be written as
Vs = Vr ∠0° +
RPr + XQr
RQr − XPr
−j
Vr ∠0°
Vr ∠0°
(4.66)
or approximately,
RPr + XQr
Vr
(4.67)
RPr + XQr
VrVB
(4.68)
(Sr /Vr ) R cos θ + (Sr /Vr ) X sin θ
VB
(4.69)
Sr × R cos θ + Sr × X sin θ
VrVB
(4.70)
Pr = Sr cos θ W
(4.71)
Vs ≅ Vr +
Substituting Equation 4.67 into Equation 4.57,
VD pu ≅
or
VD pu ≅
or
VD pu ≅
since
Design of Subtransmission Lines and Distribution Substations
215
and
Qr = Sr sin θ var
(4.72)
Equations 4.69 and 4.70 can also be derived from Equation 7.63, since
Sr = Vr I VA
(4.73)
The quantities in Equations 4.68 and 4.70 can be either all in per units or in the SI system. Use
the line-to-neutral voltage values and per phase values for the Pr, Qr, and Sr.
To determine the K constant, use Equation 4.68,
VD pu ≅
RPr + XQr
VrVB
or
VD pu ≅
(S3φ )(s )(r cos θ + x sin θ )((1/ 3) × 1000)
puV
VrVB
(4.74)
or
VD pu = s × K × S3φ puV
(4.75)
VD pu = s × K × Sn puV
(4.76)
or
where
K≅
(r cos θ + x sin θ )((1/ 3) × 1000)
VrVB
(4.77)
Therefore,
K = f (conductor size, spacing, cos θ, VB)
and it has the unit of
VD pu
Arbitrary no.of kVA ⋅ mi
To get the percent voltage drop, multiply the right side of Equation 4.77 by 100,
so that
K≅
(r cos θ + x sin θ )((1/ 3) × 1000)
× 100
VrVB
(4.78)
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Electric Power Distribution Engineering
which has the unit of
%VD
Arbitrary no.of kVA ⋅ mi
In Equations 4.74 through 4.76, s is the effective length of the feeder main that depends upon the
nature of the load. For example, when the load is connected at the end of the main as lumped sum,
the effective feeder length is
s = 1unit length
when the load is uniformly distributed along the main,
s=
1
× l unit length
2
when the load has an increasing load density,
s=
2
× l unit length
3
Example 4.2
Assume that a three-phase 4.16 kV wye-grounded feeder main has #4 copper conductors with an
equivalent spacing of 37 in. between phase conductors and a lagging-load power factor of 0.9.
a. Determine the K constant of the main by employing Equation 4.77.
b. Determine the K constant of the main by using the precalculated percent voltage drop per
kilovolt-ampere-mile curves and compare it with the one found in part (a).
Solution
a. From Equation 4.77,
K≅
(r cos θ + x sin θ )((1/3) × 1000)
VrVB
where
r = 1.503 Ω/mi from Table A.1 for 50°C and 60 Hz
xL = xa + xd = 0.7456 Ω/mi
xa = 0.609 Ω/mi from Table A.1 for 60 Hz
xd = 0.1366 Ω/mi from Table A.10 for 60 Hz and 37 –in. spacing cos θ = 0.9, lagging
Vr = VB = 2400 V, line-to-neutral voltage
Therefore, the per unit voltage drop per kilovolt-ampere-mile is
K ≅
(1.503 × 0.9 + 0.7456 × 0.4359)((1/3) × 1000)
24002
≅ 0.0001 VDpu /(kVA ⋅ mi)
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Design of Subtransmission Lines and Distribution Substations
or
K ≅ 0.01%VD/(kVA ⋅ mi)
b. From Figure 4.26, the K constant for #4 copper conductors is
K ≅ 0.01%VD/(kVA ⋅ mi)
which is the same as the one found in part (a).
Example 4.3
Assume that the feeder shown in Figure 4.31 has the same characteristics as the one in Example
4.2, and a lumped-sum load of 500 kVA with a lagging-load power factor of 0.9 is connected at
the end of a 1 mi long feeder main. Calculate the percent voltage drop in the main.
Solution
The percent voltage drop in the main is
%VD = s × K × Sn
= 1.0 mi × 0.01%VD/(kVA × mi) × 500 kVA
= 5.0%
Example 4.4
Assume that the feeder shown in Figure 4.32 has the same characteristics as the one in Example
4.3, but the 500 kVA load is uniformly distributed along the feeder main. Calculate the percent
voltage drop in the main.
S = l = 1 mi
# 4 copper
Dm = 37˝
kVLL = 4.16 kV
500 kVA
PF = 0.9 lag
Figure 4.31 The feeder of Example 4.2.
l = 1 mi
1
s = – = 0.5 min
2
Figure 4.32 The feeder of Example 4.4.
a