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7 General Case: Substation Service Area with n Primary Feeders

# 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

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

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.

Pr + jQr

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Electric Power Distribution Engineering

Vs

Iz

JLX

δ

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

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)

217

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

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