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