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4 Percent Power (or Copp Er) Loss

4 Percent Power (or Copp Er) Loss

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124

495

1,110

1,980

3,100

4,460

6,070

7,920

10,000

12,400

15,700

19,300

23,400



82

329

740

1,320

2,060

2,960

4,030

5,260

6,660

8,220

10,400

12,800

15,500

18,500

21,700



55

218

491

873

1,360

1,960

2,670

3,490

4,420

5,460

6,910

8,530

10,300

12,300

14,400

16,700

19,200

21,800



2 ACSR



4

Copper



37

149

335

596

932

1,340

1,830

2,390

3,020

3,730

4,720

5,820

7,050

8,390

9,840

11,400

13,100

14,900

18,900

23,300



1/0

ACSR



2

Copper



62

248

557

990

1,550

2,230

3,030

3,960

5,010

6,190

7,830

9,670

11,700

13,900

16,300

18,900

21,800

24,800



8

Copper

41

164

370

658

1,030

1,480

2,010

2,630

3,330

4,110

5,200

6,420

7,770

9,250

10,900

12,600

14,400

16,400

20,800

25,700



4 ACSR



6

Copper



27

109

246

437

682

982

1,340

1,750

2,210

2,730

3,450

4,260

5,160

6,140

7,210

8,360

9,590

10,900

13,800

17,100

20,600

24,600



2 ACSR



4

Copper



19

75

168

298

466

671

913

1,190

1,510

1,860

2,360

2,910

3,520

4,190

4,920

5,710

6,550

7,450

9,430

11,600

14,100

16,800



1/0

ACSR



2

Copper



“V” Phase



25

99

224

398

621

895

1,220

1,590

2,010

2,490

3,150

3,880

4,700

5,590

6,560

7,610

8,740

9,940

12,600

15,500

18,800

22,400



4 ACSR



6

Copper



16

63

141

250

391

563

766

1,000

1,270

1,560

1,980

2,440

2,960

3,520

4,130

4,790

5,500

6,260

7,920

9,780

11,800

14,100



2 ACSR



4

Copper



10

39

88

157

245

353

481

628

795

982

1240

1530

1860

2210

2590

3010

3450

3930

4970

6140

7420

8840



1/0

ACSR



2

Copper



31

70

125

195

280

382

498

631

779

986

1220

1470

1750

2060

2380

2740

3120

3940

4870

5890

7010



2/0

ACSR



1

Copper



Three Phase



56

99

154

222

302

395

500

617

780

964

1170

1390

1630

1890

2170

2470

3120

3850

4660

5550



3/0

ACSR



1/0

Copper



Source: Rural Electrification Administration, U.S. Department of Agriculture: Economic Design of Primary Lines for Rural Distribution Systems, REA Bulletin, 60, 1960.

Note: This table is calculated for a PF of 90%. To adjust for a different PF, multiply these values by the factor of k = (90)2/(PF)2.

For 7.62/13.2 kV, multiply these values by 0.893; for 14.4/24.9 kV, multiply by 0.25.



20

40

60

80

100

120

140

160

180

200

225

250

275

300

325

350

375

400

450

500

550

600



4 ACSR



Load (kW)



8 Copper



6

Copper



Annual

Peak



Single Phase



Table 7.5

Conductor I2R Losses, kWh/(mi year), at 7.2/12.5 kV and a Load Factor of 0.6



78

122

176

240

313

396

489

619

764

925

1100

1280

1500

1720

1960

2480

3060

3700

4400



4/0 ACSR



2/0

Copper



Voltage-Drop and Power-Loss Calculations

409



410



Electric Power Distribution Engineering



7.5  Method to Analyze Distribution Costs

To make any meaningful feeder-size selection, the distribution engineer should make a cost study

associated with feeders in addition to the voltage-drop and power-loss considerations. The cost

analysis for each feeder size should include (1) investment cost of the installed feeder, (2) cost

of energy lost due to I2 R losses in the feeder conductors, and (3) cost of demand lost, that is, the

cost of useful system capacity lost (including generation, transmission, and distribution systems),

in order to maintain adequate system capacity to supply the I2 R losses in the distribution feeder

conductors. Therefore, the total annual feeder cost of a given size feeder can be expressed as





TAC = AIC + AEC + ADC $/mi



(7.61)



where

TAC is the total annual equivalent cost of feeder, $/mi

AIC is the annual equivalent of investment cost of installed feeder, $/mi

AEC is the annual equivalent of energy cost due to I2 R losses in feeder conductors, $/mi

ADC is the annual equivalent of demand cost incurred to maintain adequate system capacity to

supply I2 R losses in feeder conductors, $/mi



7.5.1  Annual Equivalent of Investment Cost

The annual equivalent of investment cost of a given size feeder can be expressed as





AIC = IC F × iF $/mi



(7.62)



where

AIC is the annual equivalent of investment cost of a given size feeder, $/mi

ICF is the cost of installed feeder, $/mi

iF is the annual fixed charge rate applicable to feeder

The general utility practice is to include cost of capital, depreciation, taxes, insurance, and operation and maintenance (O&M) expenses in the annual fixed charge rate or so-called carrying charge

rate. It is given as a decimal.



7.5.2  Annual Equivalent of Energy Cost

The annual equivalent of energy cost due to I2 R losses in feeder conductors can be expressed as





AEC = 3I 2 R × EC × FLL × FLSA × 8760 $/mi



(7.63)



where

AEC is the annual equivalent of energy cost due to I2 R losses in feeder conductors, $/mi

EC is the cost of energy, $/kWh

FLL is the load-location factor

FLS is the loss factor

FLSA is the loss-allowance factor

The load-location factor of a feeder with uniformly distributed load can be defined as







FLL =



s







(7.64)



Voltage-Drop and Power-Loss Calculations



411



where

FLL is the load-location factor in decimal

s is the distance of point on feeder where total feeder load can be assumed to be concentrated for

the purpose of calculating I2 R losses

ℓ is the total feeder length, mi

The loss factor can be defined as the ratio of the average annual power loss to the peak annual

power loss and can be found approximately for urban areas from

2

FLS = 0.3FLD + 0.7 FLD









(7.65)



and for rural areas [6],

2

FLS = 0.16 FLD + 0.84 FLD







The loss-allowance factor is an allocation factor that allows for the additional losses incurred

in the total power system due to the transmission of power from the generating plant to the distribution substation.



7.5.3  Annual Equivalent of Demand Cost

The annual equivalent of demand cost incurred to maintain adequate system capacity to supply the

I2 R losses in the feeder conductors can be expressed as

ADC = 3I 2 R × FLL × FPR × FR





× FLSA [(CG × iG ) + (CT × iT ) + (CS × iS )] $ /mi



(7.66)



where

ADC is the annual equivalent of demand cost incurred to maintain adequate system capacity to

supply 12R losses in feeder conductors, $/mi

FLL is the load-location factor

FPR is the peak-responsibility factor

FR is the reserve factor

FLSA is the loss-allowance factor

CG is the cost of (peaking) generation system $/kVA

CT is the cost of transmission system, $/kVA

CS is the cost of distribution substation, $/kVA

iG is the annual fixed charge rate applicable to generation system

iT is the annual fixed charge rate applicable to transmission system

iS is the annual fixed charge rate applicable to distribution substation

The reserve factor is the ratio of total generation capability to the total load and losses to be supplied. The peak-responsibility factor is a per unit value of the peak feeder losses that are coincident

with the system peak demand.



7.5.4  Levelized Annual Cost

In general, the costs of energy and demand and even O&M expenses vary from year to year during

a given time, as shown in Figure 7.16a; therefore, it becomes necessary to levelize these costs over

the expected economic life of the feeder, as shown in Figure 7.16b.



412



Electric Power Distribution Engineering

1



2



F1



3



4



n



5



F3



F2



F4



F5



Fn–1

Fn



(a)

1



(b)



2



A



3



4



A



A



5



n



A



A



A



A



Figure 7.16  Illustration of the levelized annual cost concept: (a) unlevelized annual cost flow diagram and

(b) levelized cost flow diagram.



Assume that the costs occur discretely at the end of each year, as shown in Figure 7.16a. The

levelized annual cost* of equal amounts can be calculated as

 P

P

P

A =  F1   + F2   + F3   +

 F 2

 F 3

  F 1

i







i



i



i

i

 P   A 

+ Fn     

 F n   P n



(7.67)



or









A=





n





j =1



i

i

 P   A 

Fi     

 F  j   P n



(7.68)



where

A is the levelized annual cost, $/year

Fi is the unequal (or actual or unlevelized) annual cost, $/year

n is the economic life, year

i is the interest rate

(P/F )in is the present worth (or present equivalent) of a future sum factor (with i interest rate and

n years of economic life), also known as single-payment discount factor

(A/P)in is the uniform series worth of a present sum factor, also known as capital-recovery factor

The single-payment discount factor and the capital-recovery factor can be found from the compounded-interest tables or from the following equations, respectively,

i







1

P

 F  = (1 + i )n

 n



* Also called the annual equivalent or annual worth.



(7.69)



413



Voltage-Drop and Power-Loss Calculations



and

i



i(1 + i )n

 A

 P  = (1 + i )n − 1

 n







(7.70)



Example 7.15

Assume that the following data have been gathered for the system of the NL&NP Company.

Feeder length = 1 mi

Cost of energy = 20 mills/kWh (or $0.02/kWh)

Cost of generation system = $200/kW

Cost of transmission system = $65/kW

Cost of distribution substation = $20/kW

Annual fixed charge rate for generation = 0.21

Annual fixed charge rate for transmission = 0.18

Annual fixed charge rate for substation = 0.18

Annual fixed charge rate for feeders = 0.25

Interest rate = 12%

Load factor = 0.4

Loss-allowance factor = 1.03

Reserve factor = 1.15

Peak-responsibility factor = 0.82

Table 7.6 gives cost data for typical ACSR conductors used in rural areas at 12.5 and 24.9 kV.

Table 7.7 gives cost data for typical ACSR conductors used in urban areas at 12.5 and 34.5 kV.

Using the given data, develop nomographs that can be readily used to calculate the total annual

equivalent cost of the feeder in dollars per mile.



Table 7.6

Typical ACSR Conductors Used in Rural Areas

Installation



Conductor

Ground Wire

Size



Conductor Ground Wire



Total



Cost and Hardware Installed



Size



Weight (lb)



Weight (lb)



$/lb



Cost ($)



Feeder Cost ($)



#4

#2

1/0

1/0

1/0

1/0



356

769

1223

1542

1802

3642



356

566

769

769

769

769



0.6

0.6

0.6

0.6

0.6

0.6



6,945.6

7,176.2

7,737.2

8,563

9,985

10,967



7,800

8,900

10,400

11,800

13,690

17,660



#4

#2

1/0

1/0

1/0

1/0



356

769

1223

1542

1802

3462



356

566

769

769

769

769



0.6

0.6

0.6

0.6

0.6

0.6



7,605.6

7,856.2

8,217.2

8,293

9,615

11,547



8,460

9,580

10,880

11,530

13,320

18,240



At 12.5 kV

#4

1/0

3/0

4/0

266.8 kcmil

477 kcmil

At 24.9 kV

#4

1/0

3/0

4/0

266.8 kcmil

477 kcmil



414



Electric Power Distribution Engineering



Table 7.7

Typical ACSR Conductors Used in Urban Areas

Conductor Ground Wire



Conductor Ground Wire



Cost



Installation and

Hardware Installed



Total



Size



Size



Weight (lb)



Weight (lb)



$/lb



Cost ($)



Feeder Cost ($)



#4

#4

#4

1/0



356

769

1223

3462



356

356

356

769



0.6

0.6

0.6

0.6



21,145.6

22,402.2

24,585

28,307



22,000

24,000

27,000

35,000



#4

#4

#4

1/0



356

769

1223

3462



356

356

356

769



0.6

0.6

0.6

0.6



21,375.6

22,632.2

24,815

28,537



22,230

24,230

27,230

35,230



At 12.5 kV

#4

1/0

3/0

477 kcmil

At 34.5 kV



12.00



12.00



10.00



10.00



8.00



8.00

kV



.9



4.00



4.00



2.00



2.00



0.00

0.00

(a)



2.00 4.00



0.00

0.00



6.00 8.00 10.00 12.00 14.00



Demand in MVA (A. W. G. I/0)



12.5



6.00



24



6.00



kV



14.00



kV



14.00



12.5



Total annual cost (× $ 1000)



#4

1/0

3/0

477 kcmil



(b)



9



.

24



kV



2.00 4.00 6.00 8.00 10.00 12.00 14.00



Demand in MVA (A. W. G. 4,7 strands)



Figure 7.17  Total annual equivalent cost of ACSR feeders for rural areas in thousands of dollars per mile:

(a) 477 cmil, 26 strands, (b) 266.8 cmil, 6 strands, AWG 4/0, and AWG 3/0.

Solution

Using the given and additional data and appropriate equations from Section 7.5, the following

nomographs have been developed. Figures 7.17 and 7.18 give nomographs to calculate the total

annual equivalent cost of ACSR feeders of various sizes for rural and urban areas, respectively, in

thousands of dollars per mile.



Example 7.16

The NL&NP power and light company is required to serve a newly developed residential area.

There are two possible routes for the construction of the necessary power line. Route A is 18 miles

long and goes around a lake. It has been estimated that the required overhead power line will cost

$8000 per mile to build and $350 per mile per year to maintain. Its salvage value will be $1500

per mile at the end of its useful life of 20 years.



415



18.00



18.00



15.00



15.00



12.00



12.00



kV



0.00

0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00

Demand in MVA (477 cmil 26 strands)



0.00

0.00



(b)



14.00



12.00



12.00



10.00



10.00



8.00



8.00



.9



kV



14.00



24



4.00



2.00



2.00



0.00

0.00

(c)



0.00

0.00



3.00 6.00 9.00 12.00 15.00 18.00 21.00

Demand in MVA (A. W. G. 4/0)



(d)



kV



3.00 6.00 9.00 12.00 15.00 18.00 21.00

Demand in MVA (266.8 cmil 6 strands)



6.00



4.00



9



.

24



V



3.00



5k



3.00



12.



6.00



6.00



12.



.9



9.00



6.00



kV



Total annual cost (× $ 1000)



(a)



24



9.00



5k

V



21.00



12.

5k

V



21.00



12.5



Total annual cost (× $ 1000)



Voltage-Drop and Power-Loss Calculations



2.00 4.00



9



.

24



kV



6.00 8.00 10.00 12.00 14.00



Demand in MVA (A. W. G. 3/0)



Figure 7.18  Total annual equivalent cost of ACSR feeders for urban areas in thousands of dollars per mile:

(a) 477 cmil, 26 strands, (b) AWG 3/0, (c) AWG 1/0, and (d) AWG 4, 7 strands.

On the other hand, route B is 6 miles long and is an underwater line that goes across the lake.

It has been estimated that the required underwater line using submarine power cables will cost

$21,000 per mile to build and $1,200 per mile per year to maintain. Its salvage value will be $6000

per mile at the end of 20 years. Assume that the fixed charge rate is 10% and that the annual ad

valorem (property) taxes are 3% of the first cots of each power line. Use any engineering economy

interest tables and determine the economically preferable alternative.

Solution

Route A: The first cost of the overhead power line is





P = ($8, 000 /mile)(18 miles) = $144, 000



and its estimated salvage value is





F = ($1, 500 /mile)(18 miles) = $27, 000



416



Electric Power Distribution Engineering



The annual equivalent cost of capital invested in the line is

10%



10%



A

A

A1 = $144, 000   − $27, 000  

P

 20

 F 20





= $144, 000(0.11746) − $27, 000(0.01746) = $16, 443



The annual equivalent cost of the tax and maintenance is





A 2 = ($144, 000)(3%) + ($350 /mile)(18miles) = $10,620



Route B: The first cost of the submarine power line is





P = ($21,000/mile)(6miles) = $126,000



and its estimated salvage value is





F = ($6,000/mile)(6miles) = $36,000



Its annual equivalent cost of capital invested is

10%



10%



A

A

A1 = $126, 000   − $36, 000  

 P 20

 F 20





= $14,171



The annual equivalent cost of the tax and maintenance is





A 2 = ($126, 000)(3%) + ($1, 200 /mile)(6miles) = $10, 980



The total annual equivalent cost of the submarine power line is

A = A1 + A2





= $14,171+ $10, 980 = $25,151



Hence, the economically preferable alternative is route B. Of course, if the present worth of the

costs are calculated, the conclusion would still be the same. For example, the present worth of

costs for A and B are

10%







P 

PWA = $27, 063  

= $230, 414

 A 20



and

10%







P 

PWB = $25,151 

= $214,136

 A 20



Thus, route B is still the preferred route.



Voltage-Drop and Power-Loss Calculations



417



Example 7.17

Use the data given in Example 6.6 and assume that the fixed charge rate is 0.15, and zero salvage

values are expected at the end of useful lives of 30 years for each alternative. But the salvage value

for 9 MVA capacity line is $2000 at the end of the 10th year. Use a study period of 30 years and

determine the following:









a. The annual equivalent cost of 9 MVA capacity line.

b. The annual equivalent cost of 15 MVA capacity line.

c. The annual equivalent cost of the upgrade option if the upgrade will take place at the end of 10

years. Use an average value of $5000 at the end of 20 years for the new 15 MVA upgrade line.

Solution







a. The annual equivalent cost of 9 MVA capacity line is

15%



A

A1 = $120, 000  

er year

= $120, 000(0.15230) = $18, 276 per mile pe

 P 30







15%



A

A2 = $150, 000  

= $150, 000(0.15230) = $22, 845

 P 30







b.







c. The annual equivalent cost of 15 MVA capacity line is

15%

15%

15%

15%



P 

P   A 

P 

A 2 = $120, 000 − $2, 000   + $200, 000   − $5, 000     

 F 10

 F 30   P 30

 F 10





= [$120, 000 − $2, 000(0.2472) + $200, 000(0.2472) − $5, 000(0.0151)](0.15230)





= $25, 718.92



As it can be seen, the upgrade option is still the bad option. Furthermore, if one considers the 9 MVA

vs. 15 MVA capacities, building the 15 MVA capacity line from the start is still the best option.



7.6  Economic Analysis of Equipment Losses

Today, the substantially escalating plant, equipment, energy, and capital costs make it increasingly

more important to evaluate losses of electric equipment (e.g., power or distribution transformers)

before making any final decision for purchasing new equipment and/or replacing (or retiring) existing ones. For example, nowadays it is not uncommon to find out that a transformer with lower losses

but higher initial price tag is less expensive than the one with higher losses but lower initial price

when total cost over the life of the transformer is considered.

However, in the replacement or retirement decisions, the associated cost savings in O&M costs

in a given life cycle analysis* or life cycle cost study must be greater than the total purchase price

of the more efficient replacement transformer. Based on the “sunk cost” concept of engineering

economy, the carrying charges of the existing equipment do not affect the retirement decision,

regardless of the age of the existing unit. In other words, the fixed, or carrying, charges of an existing equipment must be amortized (written off) whether the unit is retired or not.

* These phrases are used by some governmental agencies and other organizations to specifically require that bid evaluations or purchase decisions be based not just on first cost but on all factors (such as future operating costs) that influence

the alternative that is more economical.



418



Electric Power Distribution Engineering



The transformer cost study should include the following factors:













1. Annual cost of copper losses

2.Annual cost of core losses

3.Annual cost of exciting current

4.Annual cost of regulation

5.Annual cost of fixed charges on the first cost of the installed equipment



These annual costs may be different from year to year during the economical lifetime of the equipment. Therefore, it may be required to levelize them, as explained in Section 7.5.4. Read Section 6.7

for further information on the cost study of the distribution transformers. For the economic replacement study of the power transformers, the following simplified technique may be sufficient. Dodds

[10] summarizes the economic evaluation of the cost of losses in an old and a new transformer step

by step as given in the following text:

















1. Determine the power ratings for the transformers as well as the peak and average system loads.

2.Obtain the load and no-load losses for the transformers under rated conditions.

3.Determine the original cost of the old transformer and the purchase price of the new one.

4.Obtain the carrying charge rate, system capital cost rate, and energy cost rate for your

particular utility.

5.Calculate the transformer carrying charge and the cost of losses for each transformer. The

cost of losses is equal to the system carrying charge plus the energy charge.

6.Compare the total cost per year for each transformer. The total cost is equal to the sum of

the transformer carrying charge and the cost of losses.

7. Compare the total cost per year of the old and new transformers. If the total cost per year of

the new transformer is less, replacement of the old transformer can be economically justified.



Problems

7.1 Consider Figure P7.1 and repeat Example 7.5.

7.2 Repeat Example 7.7, using a transformer with 75 kVA capacity.

7.3 Repeat Example 7.7, assuming four services per transformer. Here, omit the UG SL. Assume

that there are six transformers per block, that is, one transformer at each pole location.

7.4 Repeat Problem 7.3, using a 75 kVA transformer.

7.5 Repeat Example 7.8, using a 100 kVA transformer and #3/0 AWG and #2 AWG cables for the

SLs and SDs, respectively.

7.6 Repeat Example 7.10. Use the nominal primary voltage of 19,920/34,500 V and assume that

the remaining data are the same.

7.7 Assume that a three-conductor dc overhead line with equal conductor sizes (see Figure P7.7) is

considered to be employed to transmit three-phase three-conductor ac energy at 0.92 power factor.



0.004 + j0.012 Ω/



Distribution

transformer



0.12 + j0.04 Ω/



0.12 + j0.04 Ω/

A



30 A

cos ΘA = 0.7

lagging



Figure P7.1  One-line diagram for Problem 7.1.



B



40 A

cos ΘB = 1.0



C



50 A

cos ΘC = 0.8

lagging



419



Voltage-Drop and Power-Loss Calculations

IL (dc)

V = Vm

2V

V = Vm



Figure P7.7  Illustration for Example 7.7.



7.8







7.9







7.10











7.11







If voltages to ground and transmission line efficiencies are the same for both direct and alternating

currents, and the load is balanced, determine the change in the power transmitted in percent.

Assume that a single-phase feeder circuit has a total impedance of 1 + j3 Ω for lines and/or

transformers. The receiving-end voltage and load current are 2400∠0° V and 50 ∠−30° A,

respectively. Determine the following:

a. The power factor of the load.

b. The load power factor for which the voltage drop is maximum, using Equation 7.51.

c. Repeat part (b), using Equation 7.52.

An unbalanced three-phase wye-connected and grounded load is connected to a balanced threephase four-wire source. The load impedance Za, Zb, and Zc are given as 70 ∠30°, 85 ∠−40°, and

50 ∠35° Ω/phase, respectively, and the phase a line voltage has an effective value of 13.8 kV:

Use the line-to-neutral voltage of phase a as the reference and determine the following:

a. The line and neutral currents

b. The total power delivered to the loads

Consider Figure P7.l and assume that the impedances of the three line segments from left to right

are 0.1 + j0.3, 0.1 + j0.1, and 0.08 + j0.12 Ω/phase, respectively. Also assume that this three-phase

three-wire 480-V secondary system supplies balanced loads at A, B, and C. The loads at A, B, and

C are represented by 50 A with a lagging power factor of 0.85, 30 A with a lagging power factor

of 0.90, and 50 A with a lagging power factor of 0.95, respectively. Determine the following:

a. The total voltage drop in one phase of the lateral using the approximate method

b. The real power per phase for each load

c. The reactive power per phase for each load

d. The kilovoltampere output and load power factor of the distribution transformer

Assume that bulk power substation 1 supplies substations 2 and 3, as shown in Figure P7.11,

through three-phase lines. Substations 2 and 3 are connected to each other over a tie line,

as shown. Assume that the line-to-line voltage is 69 kV and determine the following:

a. The voltage difference between substations 2 and 3 when tie line 23 is open-circuited

b. The line currents when all three lines are connected as shown in the figure



j1



1.8



1+



Ω/



1



09



+j



Ω/



2



1 + j 2 Ω/



125 A

cos Θ = 0.90 lag



Figure P7.11  Distribution system for Problem 7.11.



3



195 A

cos Θ = 0.85 lag



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