ELECTRIC ENERGY ECONOMICS

Transcription

1 ELECTRIC ENERGY ECONOMICS The cost to the consuer for electric energy can be roughly apportioned according to Generation 60% Transission 0% Distribution 20% Adinistrative/Profit 0% Econoics drives the selection of an appropriate power generation schee for the given situation. The need ay be one only during high electricity deand hours (peak load) or the new power ay be needed 24 hours a day (base load). Base load is that load below which the deand never falls, that is, the base load ust be supplied 00% of the tie. The peaking load occurs less than about 5% of the tie; the interediate load transpires between 5% to 00% of the tie. To iprove industrial plant cost effectiveness, soe copanies are ipleenting cogeneration, which is the production of electricity in-house along with industrial process stea. Cogeneration was essentially encouraged by the passage of the 978 Public Utility Regulatory Policies Act (PURPA). Since costs are expressed on a per kwe hr basis, a high capacity factor is desired so the capital cost is spread out. The capital and operating (including fuel) costs generally dictate how a plant is used on the grid (hydroelectric units are an exception to the following): Loading Capital Costs O&M and uel Costs Exaple Plants Base High Low Coal, Nuclear Peak Low High Oil, Old natural gas units The base load plants can be further differentiated: nuclear plants have high initial fixed (capital) charges and low fuel costs, whereas coal-fired plants have lower fixed charges and higher fuel costs. The plant selection for various load deands can be suarized as Loading Power Plants Econoics Capacity actor* Base Large coal and nuclear units Large capital outlay is acceptable, but ust have reasonable operating and aintenance costs C > 57% Interediate Peak Old coal, oil or gas units with low theral efficiency and cobined cycle plants Cobustion turbines and diesel engines; and hydroelectric Have id range operating and aintenance costs Need to iniize capital costs, but fuel costs can be high 23% < C < 57% C < 23% * These capacity factors are fro A. W. Culp, Principles of Energy Conversion, 2 nd ed., 99, McGraw-Hill, p. 3; however, these C values should be considered approxiate. The capacity factor (C) is the energy produced during soe tie interval ratioed to the energy that could have been produced at net rated power during the sae tie interval. Another plant perforance terinology is the availability factor (A), which is the tie period that the plant is operable divided by the total period considered (A C). The availability factor does not directly appear in the econoics equations below; however, a peaking plant ust aintain a high availability factor copared to its capacity factor. ElectricEnergyEconoics K.E. Holbert, Sept. 20 Page of 7

2 The load following plants include both the interediate and the peak load units. The spinning reserve is excess capacity that is running and synchronized with the syste, and is equal to the largest unit in the power network in case the unit should trip. Another consideration is the reserve capacity (historically aintained at about 20%), which is the difference between the total rated capacity of all the units in the grid and the expected peak deand on the syste. The cobined value of all the capital and operating assets of a utility for its rate base that is used by state regulators in deterining a fair return on the utility investent. A point of contention in the rate base is whether construction work in progress (CWIP) is allowed. or soe electricity generation schees (e.g., solar and wind), it ay be necessary to include soe type of energy storage device or echanis in order to supply power at all ties it is required or for peaking power needs. The cost of such energy storage facilities should be included in the overall costs. Alternatively, base-load electric ay be eployed during off-hours (idnight to early a..) to "charge" the energy storage device. Possible energy storage approaches include puped water storage (hydro), copressed air for gas turbine use, theral energy storage (solar theral facility), battery storage, flywheels, and agnetic energy storage (using superconducting aterials). Electric Power Generation Costs In calculating the cost of electricity production ( /kwe hr), the energy costs are divided into three categories:. capital (e C ): including land, equipent, construction, interest; 2. operating and aintenance (e OM ): wages, aintenance, soe taxes and insurance; and 3. fuel costs (e ). Historically, electricity costs were often expressed in ills per kilowatt-hour where,000 ills equal $. The overall electricity generation cost is deterined fro Capital +O & M + uel e e kwe hr Electric Energy Produced Let us consider calculating the electricity cost for a single year. The electric energy (E) produced in a given year can be deterined using the noinal (net) plant power rating (P e,rate ) and the capacity factor (C) for that year via C e OM e E P, C (365 days/) (2) e rate O&M cost Operating and aintenance (O&M) costs ay be divided into () fixed O&M costs ($/kw ) that vary with the plant size, and (2) variable O&M costs ( /kw hr) which are proportional to the unit energy production. The fixed costs coprise wages and overhead for peranent eployees, routine aintenance and other fees; whereas variable costs consist of equipent outage aintenance, utilities, and consuables such as cheicals. If the total annual O&M costs are known, then. O & M [$/] eom P, C (8760 hrs/) (3) e rate () ElectricEnergyEconoics K.E. Holbert, Sept. 20 Page 2 of 7

3 Exaple: Consider a hypothetical 600 MWe power plant, whose operating costs are purely due to the salaries of the dozen workers eployed there. If the annual wages and overhead average $00,000 per eployee, and the capacity factor is 35%, what is the O&M coponent of the electricity cost? (2 eployees)($00,000/)(000 ills/$) e OM 0.65 ills/kwe hr 3 (600 0 kw)(0.35)(8760 hrs/) Clearly, this exaple lacks consideration of all the various O&M costs. uel cost The fuel cost is probably the next easiest coponent to deterine. The annual theral heat required to produce a given aount of electricity can be deterined fro the plant theral efficiency. The theral efficiency (η th ) is the net electricity generated (P e ) divided by the heat input ( Q th ) required to produce that electricity, i.e., th Pe Q th. The theral heat produced is the product of the fuel input rate ( fuel ) and the fuel heat content (e.g., heating value, HV), that is, Q th fuel HV. The annual electric energy produced, with soe representative units for the various quantities, is thus E fuel HV th kwehr ton/gal (4) Btu/kJ kwehr The annual fuel cost is deterined fro the aount (e.g., ass or volue) of fuel used and its cost per unit ass or volue ( C ) uel fuel C (5) ton/gal ton/gal $ $ or a fossil-fired unit, C is coonly expressed in ters of dollars per ton or gallon, and the heating value is given in Btu/lb or Btu/gal. or a nuclear power plant, C is typically quoted in dollars per kilogra of uraniu, and the burnup, B in MWth day per etric ton of uraniu (MTU), is the terinology eployed for expressing fuel heat content. By cobining the above two expressions, we arrive at the fuel cost of electricity uel C e (6) E HV th As should be expected, the fuel cost is independent of the plant capacity factor. Exaple: An 800 MWe coal-fired unit has a theral efficiency of 38% and a capacity factor of 82%. The utility is in negotiations for a long-ter contract for coal having a heating value of 5,700 Btu/lb. To hold the fuel cost below 0.9 /kwe hr, what is the axiu price, in dollars per short ton, that the utility should be willing to pay? e HV C th ton/gal Btu/kJ Btu kw hr lb $ 0.9 5,700 (0.38) 2000 $3.47 / ton kw hr lb The given capacity factor was extraneous inforation. 342 Btu Capital cost Coputation of the capital cost is the ost difficult due to the large expenditure and tie period involved. As a substantial siplification, the capital cost can be likened to a hoe ortgage payent, which ostly consists of principal and interest. Significant differences between the hoe ortgage and power plant cases do exist however. or exaple, power plants take years (vs. onths) to build; hence, utilities ust borrow oney during the plant construction period (granted soe individuals obtain hoe construction ton 00 ElectricEnergyEconoics K.E. Holbert, Sept. 20 Page 3 of 7

4 loans too). In addition, the tie value of oney should be considered, that is, the value of one dollar ten years fro now will likely be different than today, and depreciation should also be accounted for. The reainder of this docuent concerns onetary calculations related to capital costs. To keep the capital calculations reasonable, and perhaps to assist soon-to-be graduates in thinking about future financial decisions, several exaples dealing with personal finances are presented below. Effective Interest Rate In any cases, interest is copounded on a onthly or daily basis. Soeties it is necessary to convert an interest rate fro one copounding period to another basis. The stated interest rate (i), which is copounded n ties a year, can be adjusted to an ties a year copounding basis with an effective interest rate of n i j (7) n Exaple: Let the stated annual ortgage interest rate be i=8% which is copounded onthly (n=2). Therefore, the effective annual (=) interest rate is j=8.3%: 0.08 j uture Value Single Payent/Investent If a single deposit (P) of oney is ade, and we wish to deterine the aount of oney () in the bank at soe future tie, then the future value of that initial investent at the end of t years with an annual interest rate of i copounded n ties a year is where P is also known as the present value. P nt i P (8) n Exaple: Consider a deposit of $0,000 into a CD earning 8% interest that is copounded onthly. The value after 5 years can be coputed directly fro the stated interest rate of 8% via Eq. (8), or fro the effective interest rate of 8.3% as shown below: 2(5) P 0.08 $0,000 2 $0,000( 0.083) Question: Can you use an Excel function to find this sae answer? 5 $4, $4, uture Value Periodic Payent/Investents If repeated equal deposits of A are ade on a regular periodic basis of ties a year with the payents ade at the end of each interval, then the future worth after t years is A j A (9) j ElectricEnergyEconoics K.E. Holbert, Sept. 20 Page 4 of 7

5 Exaple: An IRA is started and a onthly deposit of $00 is ade by the individual into the account, which earns 8% copounded onthly. At the end of 40 years of eployent, the IRA has ($00)(2) 2(40) 0.08 A $349, Question for thought: Why didn't we use 8.3%? uture Value Periodic Payent/Investents with Escalation To copute the future value of periodic payents when the payents are increased at a constant annual rate of e (for exaple, due to inflation), we eploy Ae A j e A 0 j j A e j j e j e where A is the aount of the first payent. The aount of the -th payent can be found fro: e 0 A (0) e A A () Exaple: If you are anticipating raises over the course of your eployent period, then the onthly deposits of $00 into the earlier IRA exaple can be increased (escalated) by soe annual rate, e=3%, and the future value at the end of 40 years is then Ae ($00)(2) (40) (40) $502,998 Breakeven To deterine the aount of tie to breakeven on the purchase/installation of soe syste (versus just investing the oney), we equate the future value of an initial investent to the su of the future value of the savings or periodic payents received (for exaple, as sales) and the syste salvage value (SV), if any. P ( A or Ae) SV (2) The nuber of years (t) required to reach breakeven is found fro substituting Equations (8) and (0) (or (9) for no escalation) into (2) nt i A j P n j e e SV Exaple: You are building a custo hoe and the contractor suggests an upgraded air conditioning (A/C) unit that is expected to save $400 in electricity costs in the first year. If the upgraded A/C unit costs an additional $5,000 cash, how any years before it pays for itself? Assue that the interest rate for a $5,000 CD is 5% copounded daily and electricity costs are expected to clib by 2% per year; while bank interest rates are 3% copounded daily for any savings. Take the salvage value to be a noinal $,000. Because the electricity cost escalation rate and interest rates are on different bases, we adjust the to an annual basis. (3) ElectricEnergyEconoics K.E. Holbert, Sept. 20 Page 5 of 7

6 0.05 i We eploy Eq. (3) to deterine the breakeven point ($5,000) () t $5,000 (.0527) t ($400) () $38,277.5 (.03045) 0.03 j () t t t (.02) $, () t ($,000) Nuerical solution of the above equation shows that the syste starts to pay for itself in the twentieth year after purchase. Question: do you think the A/C unit will last 20 years? Present Value Unifor Series or a present value (loan aount) of P A, the periodic payent (A) can be deterined fro nt A n i PA (4) i n Exaple: What is the onthly principal and interest payent for a $00,000 loan borrowed over 30 years at an 8% interest rate? 2 (2)(30) 0.08 $00,000. A A $ / onth This sae value is found in Excel using PMT(8%/2,360,00000,0). Question: what is the total aount paid over the loan period? The capital (or fixed) cost ust be properly distributed throughout the expected operating lifetie of the plant. or financial analyses, plants are generally assued to have a lifetie of 40 years, although they ay operate for either a shorter or a longer period. If the utility spreads the construction costs and interest over 40 years with equal annual (or onthly) payents, then our objective is to find the annual payent aount. To do so, we can think of the payent as being soe percentage of the construction cost ( B ), that is, we can establish a levelized annual fixed-charge rate (I) such that CAP I B B I ec kwe hr E Pe, rate C (8760 hrs/) P (5) e, rate C 8760 where CAP is the annualized aount of the entire capital costs including interest. The ter B / P e, rate is the cost of building a power plant, which is generally expressed in ters of dollars per installed kilowattelectric ($/kwe). As the electric rating is increased, the ratio B / P e, rate generally decreases this is known as econoy of scale. Table I shows representative production costs for electric generating stations in the U.S. Exaple: A 500-MW power plant costs $,200 per kwe. If the expected capacity factor is 87%, deterine the appropriate annual fixed-charge rate (I) and the capital cost of the electricity. irst, deterine the initial cost to build the power plant B ( kwe)($,200 / kwe) $ 600 illion ElectricEnergyEconoics K.E. Holbert, Sept. 20 Page 6 of 7

7 Next, ake a siple deterination of the annual payent for principal and interest using Eq. (4) and assuing a 5% interest rate paid over 40 years: ()(40) 0.05 $ 600 illion A A $ illion/ CAP 0.05 ro Eq. (5), the levelized annual fixed-charge rate is CAP $ illion/ I (00%) 5.83% / B $ 600 illion inally, the capital portion of the electricity cost is B I $, / 00 ec 0.98 /kwe hr, 8760 kwe (0.87)(8760 hr/) $ Pe rate C Table I: Costs of New Central Electric Generating Stations Technology Size Capital Cost Variable O&M ixed O&M Heat Rate (MW) ($/kw) Cost (ills/kw hr) Cost ($/kw) (Btu/kW hr) Scrubbed Coal 600, ,200 Integrated Coal-Gasification Cobined Cycle (IGCC) 550, ,765 IGCC w/carbon Sequestration 380 2, ,78 Conventional Gas/Oil Cobined Cycle ,96 Conventional Cobustion Turbine ,833 uel Cells 0 4, ,930 Advanced Nuclear 350 2, ,400 Bioass 80 2, ,9 Geotheral 50, ,376 Hydropower 500, ,022 Wind 50, ,022 Solar Theral 00 3, ,022 Photovoltaic 5 5, ,022 Source: U.S. DOE EIA, Assuptions to the Annual Energy Outlook 2008, Report DOE/EIA-0554, June 2008, p. 79. Additional generation schees appear in the DOE publication. One ay observe fro the above exaple that the fixed-charge rate can be coputed directly fro I CAP B A P A d d where d is the discount rate, and N is the nuber of years for payent. The discount rate is the su of the actual interest rate and the inflation (escalation) rate. Overall Generation Costs Substituting Equations (3), (6) and (5) into (), yields an overall forula for the electric generation costs e kwe e, rate N I B O & M ec eom e hr P C (8760 hrs/) HV ro the above expression, we can easily deterine which paraeters should be iniized as well as those that should be axiized to achieve reduced electricity costs. C th (6) (7) ElectricEnergyEconoics K.E. Holbert, Sept. 20 Page 7 of 7