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12 App Lication Of The A, B, C, D General Ci Rcuit Constants To Radial Feeders

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







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)



312



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)





b







c







d







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)



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