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1、Copyright The Aluminum Association Inc. Provided by IHS under license with AA Licensee=IHS Employees/1111111001, User=Wing, Bernie Not for Resale, 04/18/2007 11:35:19 MDTNo reproduction or networking permitted without license from IHS -,-,- STD-AA 54-ENGL a978 E Ob04500 0014327 145 5 THE ALUMINUM AS
2、SOCIATION The member companies of The Aluminurn Association, Inc., represent approximately 85 percent of domestic production of primary ingot and shipments of U.S. aluminum mill products. Mill products include sheet and plate; foil; extrusions; electrical conductor and wire, rod and bar. In addition
3、 to producers of primary ingot and mill products, the associations membership also includes secondary smelters, foundries and producers of master alloy and additives. The association is a primary source of statistics, standards, and economic and technical information on aluminum and the aluminum ind
4、ustry in the United States. NOTICE Disclaimer The use of any information contained herein by any member or non-member of The Aluminum Associa- tion is entirely voluntary. The Aluminum Association has used its best etforts in compiling the information contained in this book. While the Association bel
5、ieves that its compilation procedures are reliable, it does not warrant, either expressly or impliedly, the accuracy or completeness of this information. The Aluminum Association assumes no responsibility or liability for the use of the information herein. All Aluminum Association published standard
6、s, data, specifications and other technical materials are reviewed at least every five years and revised, reaf- firmed or withdrawn. Users are advised to contact Th Aluminum Association to ascertain whether the info4 mation in this publication has been superseded in the interim between publication a
7、nd proposed use. Copyright The Aluminum Association Inc. Provided by IHS under license with AA Licensee=IHS Employees/1111111001, User=Wing, Bernie Not for Resale, 04/18/2007 11:35:19 MDTNo reproduction or networking permitted without license from IHS -,-,- STD-AA 54-ENGL 1998 M Ob04500 0034328 081
8、I . The Evaluation of Losses in Conductors Contents Section1 Introduction . 1 Section II General Method and Assumed Parameters 1 Section I I Transmission Conductor Examples 3 Section IV Distribution Conductor Examples . 5 SectionV Appendix . 8 Second Edition. 1998 Copyright The Aluminum Association
9、Inc. Provided by IHS under license with AA Licensee=IHS Employees/1111111001, User=Wing, Bernie Not for Resale, 04/18/2007 11:35:19 MDTNo reproduction or networking permitted without license from IHS -,-,- STD-AA 54-ENGL L99B I Ob04500 0034329 TL8 111 Section I Introduction The high costs of energy
10、and generation facilities have made it extremely important to evaluate power losses when selecting conductors. Construction costs and energy costs have increased dramatically during the past decade and this trend may well continue. This publication provides a simple method for evaluating the power l
11、osses in transmission and distribu- tion conductors. It is intended to serve as a guide to aid in the selection of conductors which will be the most economical over their design life. of total system design. Factors such as structure sitting, structure loading, ground clearance, etc. are not taken i
12、nto account in this evaluations. choice on every line which is to be built or reconductored. Often the selection of a conductor with lower resistance will prove to be the most economical choice even though its first cost may be higher. The lower resistance may, of course, be achieved in several ways
13、. The use of a large size conductor, a higher conduc- tivity aluminum or, for ACSR constructions, an aluminum-clad steel core all contribute to lower resistance and result in lower power losses. In determining the economics in conductor selection, there are three primary factors to be considered. Th
14、ey are: conductor investment cost; energy cost; and demand or capacity costs. For convenience, these are usually calculated on an annual basis. It is necessary to take into account, therefore, not only the annual cost of providing the conductor, but also the annual cost of energy to generate the I2R
15、 losses in the conductor and the additional system capacity to provide those losses. This publication has four additional sections: Section II describes the method for evaluating and compar- ing the total annual conductor costs and it lists several assumed factors. Section III presents an example of
16、 evaluating alternative transmission conductors. Section IV presents similar examples but for distribution conductors. Section V is an appendix which provides background on the development of equations in Section II. No attempt is made in the method described here to develop the optimum conductor fr
17、om the standpoint Very significant savings can be realized in both transmission and distribution by evaluating the conductor Section II General Method and Assumed Parameters For convenience, Table 1 lists the values of the parameters which will be used in the examples for evalu- ating both ac transm
18、ission and distribution conductors. Note that for small changes in conductor size (resistance), the cost comparison between two conductors will be approximately the same regardless of operating temperatures. However, for large differences in con- ductor size (resistance) where one conductor may oper
19、ate at a substantially lower temperature than the other, the actual conductor resistance at the operating temperature under peak load can be used to more accurately predict cost savings. Techniques for calculating resistance at various temperatures can be found in the Aluminum Electrical Conductor H
20、andbook published by the Aluminum Association. systems by using dc resistances. The evaluation method demonstrated is based on a conductor one mile long and may also be used for dc 1 Copyright The Aluminum Association Inc. Provided by IHS under license with AA Licensee=IHS Employees/1111111001, User
21、=Wing, Bernie Not for Resale, 04/18/2007 11:35:19 MDTNo reproduction or networking permitted without license from IHS -,-,- STD.AA 54-ENGL 3998 11111 Ob04500 0034330 73T Table 1 Conductor Evaluation Factors Used in Examples Transmissions - 19% Fixed charge rate on conductors (Fd) Fixed charge rate o
22、n conductors (Ft) Distri butions 20% - 800 i 7% Cost of installed generating capacity (Cg) - $/kW Fixed charge rate on generating capacity (Fg) Cost of transmission lines (Ct) - $/kW Cost of distribution substations (Cs) - $/kW Fixed charge rate on substations (Fs) 800 60 20 20% 17% 1.2 0.95 1 1 50%
23、 1 Reserve factor (RF) Peak responsibility factor (PRF) Load distribution factor (DF) Loss allowance factor (LAF) Load factor (LF) Growth factor (GF) 1.2 0.65 0.764 1 .O3 40% 1.65 0.020 0.0313 30 4% Inflation rate - r 8% Interest rate - i Present cost of energy - $/kW Equivalent annual cost of energ
24、y (EAC) - $/kW Service life of conductor - years 0.025 0.0329 15 4% 8% 1. Annual Demand Costs The power loss (PL) is calculated: I peak)2 R 1 O00 PL (in kW) = This value is used to calculate annual demand cost (Cd) from adjusted power loss (APL): APL=PL X RF X PRF X DF X GF X LAF Transmission: Cd =
25、APL X FgCg Distribution: Cd = APL X (FgCg + FtCt + FsCs) 2. Annual Energy Loss Costs The annual cost of energy losses (Ce) can be calculated from annual energy losses (Pe): Pe (in kWh per year) = (PL) (Loss Factor X 8760) (DF) (LAF) (GF) Loss Factor (distribution) = 0.15 (LF) -+ 0.85 (LF2) Loss Fact
26、or (transmission) = 0.3 (LF) + 0.7 (LF2) Transmission Ce = Pe (EACtrms) Distribution Ce = Pe (EACdist) See appendix for the derivation of values of Equivalant Annual Cost (EAC) of Energy. 3. Annual Fixed Cost of Conductors The annual fixed cost of conductors (Cc) can be determined from the price of
27、conductor in $/mile: Transmission Cc = Price X (Ft) Distribution Cc = Price X (Fd) 2 Copyright The Aluminum Association Inc. Provided by IHS under license with AA Licensee=IHS Employees/1111111001, User=Wing, Bernie Not for Resale, 04/18/2007 11:35:19 MDTNo reproduction or networking permitted witho
28、ut license from IHS -,-,- STD-AA 54-ENGL 1998 I Ob04500 OOL1133L b7b . I 4. Total Annual Cost of Conductors The total annual costs of conductors (C) can be determined from the sum of the annual component costs: C = Cd + Ce + Cc For convenience Table 2 will be used to summarize calculations. Table 2
29、Economic Comparison Parameters Al ternatives I. I (Peak Current, amperes) 2. Tc (Conductor Temperature, OC) 3. R (AC Resistance at Tc, Ohms/mile) 4. PL (Power Loss, kW/Mile) 5. APL (Adjusted Power Loss, kWIrnile) 6. Cd (Annual Demand Cost, $/mile) 7. Pe (Annual Energy Losses, kWh/mile) 8. Ce (Annual
30、 Energy Loss Cost, $/mile) 9. Price of Conductor ($/mile) 10. Cc (Fixed Annual Cost, $/mile) 11. C (Total Annual Cost, $/mile) 12. S (Annual Savings, $/mile) .13. PV (Present Value of Annual Savings, $/mile) Section 111 Example of Economic Evaluation for Two Transmission Conductor Alternatives. Eval
31、uate the total annual cost of Starling (715.5 kcmil 26/7 ACSR) vs. Drake (795 kcmil 2617 ACSR) for Since there is only a small difference in conductor size, the ac resistance at 50C will be used although a 500 ampere peak load. Starling will operate at a slightly higher temperature than Drake. I2R P
32、owerLoss: PL= - 1 O00 = 35.5 kW/mile 1 O00 Starling PL= Drake 500 o128) = 32.0 kW/mile PL= l0;o Adiusted Power Loss: APL = (PL) (RF) (PRF) (DF) (GF) (LAF) From Table 1 for transmission: RF= 1.2 PRF = 0.95 DF = 1.0 GF = 1.0 LAF = 1.0 Starling APL = 35.5 X 1.2 X 0.95 X 1 X 1 X 1 = 40.47 kW/mile Drake
33、APL = 32.0 X 1.2 X 0.95 X 1 X 1 X 1 = 36.48 kW/mile 3 Copyright The Aluminum Association Inc. Provided by IHS under license with AA Licensee=IHS Employees/1111111001, User=Wing, Bernie Not for Resale, 04/18/2007 11:35:19 MDTNo reproduction or networking permitted without license from IHS -,-,- STD-A
34、A 54-ENGL 1998 I Ob04500 0034332 502 W - Annual Demand Charge: Cd = APL X Fg Cg From Table 1 for generation Fg = 0.17 Cg = $800/kW FgCg = $136/kW Starling Cd = 40.47 X 136 = $5503.92 Drake Cd = 36.48 X 136 = $4961.28 It is convenient to record the results of these and other intermediate calculations
35、 in the Table 2 form; for Annual Energv Losses: Pe = (PL) (Loss Factor X 8760) (DF) (LAF) (GF) From Table 1 for transmission: Load Factor (LF) = 50% and Loss Factor (transmission) = 0.3 (LF) + 0.7 (LF2) = 0.325 See Appendix for source of correlation factor used to convert peak losses to annual losse
36、s. Starling Pe = (35.5) (0.325 x 8760) X 1 X 1 X 1 = 101,069 kWh/mile Drake Pe = (32.0) (0.325 x 8760) X 1 X 1 X 1 = 91,104 kWmile Annual Cost of Enerev Losses: Ce = EAC X Pe From Table 1 for transmission conductor based on a present cost of energy of $0.020 kWh, 4% inflation Starling Ce = 0.03 13 X
37、 101,069 = $3 163.46/mile Drake Ce = 0.0313 X 91,104 = $2851.56/mile Annual Cost of Conductor: Cc = Price X Fixed Charge Rate For illustrative purposes $5200/mile and $5776/mile will be used for the price of Starling and Drake respectively. These prices are arbitrarily based on $1 .O0 per pound for
38、ACSR conductor; actual conductor prices should be obtained from supplier. this example see Figure 1. rate, 8% interest rate and a 30-year service life, the equivalent annual cost of energy, EAC = 0.0313/kWh. ( From Table 1 the fixed charge rate for transmission conductor is 19% Starling Cc = 5200 X
39、0.19 = $988.00/mile Drake Cc = 5776 X 0.19 = $1097.44/mile Total Annual Cost of Conductor: C = Cd + Ce + Cc Starling C = 5503.92 + 3163.46 + 988.00 = 9655.38 Drake C = 4961.28 + 2851.56 + 1097.44 = 8910.28 Differential Annual Savings/Mile, S = $745.10 Present Value of Annual Savinps (1 +i)“ -1 i( 1
40、+ i“ P V = S n = 30 i = 0.08 PV = 745.10 (1 1.26) = $8389.83/mile Conclusions This evaluation indicates that the annual savings from selecting Drake over Starling recovers Drakes ini- tial cost premium in about 9 months for the parameters considered. The present value of this annual savings is almos
41、t 150% of the price of the Drake conductor. 4 I Copyright The Aluminum Association Inc. Provided by IHS under license with AA Licensee=IHS Employees/1111111001, User=Wing, Bernie Not for Resale, 04/18/2007 11:35:19 MDTNo reproduction or networking permitted without license from IHS -,-,- STD-AA 54-E
42、NGL 1998 Ob04500 0034333 449 H Figure 1 Economic Comparison Starling (71 5.5 kcmil) 500 50 0 . 1 42 35.5 40.47 5503.92 1 01,069 3 1 63.46 5200 988 9655.38 - - I Parameters I Alternatives: I Drake (795 kcmil) 500 50 0.128 32.0 36.48 4961.28 91,104 2851.56 5776 1097.44 891 0.28 745.10 8389.83 1. I (Pe
43、ak Current, amperes) 2. Tc (Conductor Temperature, OC) 3 . R (AC Resistance at Tc, Ohms/mile) 4. PL (Power Loss, kW/mile) 5. APL (Adjusted Power Loss, kW/miIe) 6. Cd (Annual Demand Cost, $/mile) 7. Pe (Annual Energy Losses, kWh/mile) 8 . Ce (Annual Energy Loss Cost, $/mile) 9. Price of Conductor ($/
44、mile) 10. Cc (Fixed Annual Cost, $/mile) 11. C (Total Annual Cost, $/mile) 12. S (Annual Savings, $/mile) 13. PV (Present Value of Annual Savings, $/mile) Section IV Examples of economic evaluations for alternative distribution sizes. First, evaluate the total annual cost of Raven (i/O AWG ACSR) vs.
45、 Quail (2/0 AWG ACSR) for a 69 ampere peak load. I2R Power Loss: PL= 1 0 0 0 Raven Quail 692 o866 = 4.12 kW/mile 1000 PL= ( 692 (o686 = 3.27 kW/mile 1 O00 PL= ( Adjusted Power Loss: APL = (PL) (RF) (PRF) (DF) (GF) (LAF) From Table 1 for distribution: RF= 1.2 PRF = 0.65 DF = 0.764 GF = 1.65 LAF = 1.0
46、3 Raven APL = 4.12 X 1.2 X 0.65 X 0.764 X 1.65 X 1.03 = 4.17 kW/mile Quail APL = 3.27 X 1.2 X 0.65 X 0.764 X 1.65 X 1.03 = 3.31 kW/mile Annual Demand Charge: Cd = APL (FgCg + FtCt + FsCs) From Table 1: generation FgCg = 0.17 X $800 = $136/kW transmission FtCt = 0.19 X $ 60 = $ 11.4O/kW substations F
47、sCs = 0.20 X $ 20 = $ 4/kW $151.40/kW Raven Cd = 4.17 X $151.40 = $631.34/mile Quail Cd = 3.31 X $151.40 = $501.13/mile 5 Copyright The Aluminum Association Inc. Provided by IHS under license with AA Licensee=IHS Employees/1111111001, User=Wing, Bernie Not for Resale, 04/18/2007 11:35:19 MDTNo repro
48、duction or networking permitted without license from IHS -,-,- STD-AA 54-ENGL 31998 . I Ob04500 00314334 385 m Annual Enerev Losses: Pe = (PL) (Loss Factor X 8760) (DF) (LAF) (GF) From Table 1 for distribution: Load Factor (LF) = 40% and Loss Factor (distribution) = 0.15 (LF) + 0.85 (LF2) = O. 196 Raven Pe = (4.12) (O. 196 X 8760) (0.764) ( 1.65) (1.03) = 9 185 kWh/mile Quail Pe = (3.27) (0.196 X 8760) (0.764) (1.65) (1.03) = 7290 kWh/mile Annual Cost of Enernv Losses: Ce = EAC X Pe From Table 1 for distribution conducto