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Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
Project management techniques
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Project management techniques

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  • 1. Toolbox 7: Economic Feasibility Assessment Methods Dr. John C. Wright MIT - PSFC 05 OCT 2010
  • 2. Introduction We have a working definition of sustainability We need a consistent way to calculate energy costs This helps to make fair comparisons Good news: most energy costs are quantifiable Bad news: lots of uncertainties in the input data Interest rates over the next 40 years Cost of natural gas over the next 40 years Will there be a carbon tax? Today’’s main focus is on economics Goal: Show how to calculate the cost of energy in cents/kWhr for any given option Discuss briefly the importance of energy gain 3
  • 3. Basic Economic Concepts Use a simplified analysis Discuss return on investment and inflation Discuss net present value Discuss levelized cost 4
  • 4. The Value of Money The value of money changes with time 40 years ago a car cost $2,500 Today a similar car may cost $25,000 A key question –– How much is a dollar n years from now worth to you today? To answer this we need to take into account Potential from investment income while waiting Inflation while waiting 5
  • 5. Present Value Should we invest in a power plant? What is total outflow of cash during the plant lifetime? What is the total revenue income during the plant lifetime Take into account inflation Take into account rate of return Convert these into today’’s dollars Calculate the ““present value”” of cash outflow Calculate the ““present value”” of revenue 6
  • 6. Net Present Value Present value of cash outflow: PVcost Present value of revenue: PVrev Net present value is the difference NPV = PVrev –– PVcost For an investment to make sense NPV > 0 7
  • 7. Present Value of Cash Flow $100 today is worth $100 today –– obvious How much is $100 in 1 year worth to you today? Say you start off today with $Pi Invest it at a yearly rate of iR%=10% One year from now you have $(1+ iR)Pi=$1.1Pi Set this equal to $100 Then $100 $100 = = $90.91 Pi = 1 + iR 1.1 This is the present value of $100 a year from now 8
  • 8. Generalize to n years $P n years from now has a present value to you today of P PV (P ) = n 1 + iR ) ( This is true if you are spending $P n years from now This is true for revenue $P you receive n years from now Caution: Take taxes into account iR=(1-iTax)iTot 9
  • 9. The Effects of Inflation Assume you buy equipment n years from now that costs $Pn Its present value is PV (Pn ) = Pn n (1 + iR ) However, because of inflation the future cost of the equipment is higher than today’’s price If iI is the inflation rate then n Pn = (1 + iI ) Pi 10
  • 10. The Bottom Line Include return on investment and inflation $Pi n years from now has a present value to you today of 1 + iI PV = 1 + iR n Pi Clearly for an investment to make sense iR > iI 11
  • 11. Costing a New Nuclear Power Plant Use NPV to cost a new nuclear power plant Goal: Determine the price of electricity that Sets the NPV = 0 Gives investors a good return The answer will have the units cents/kWhr 12
  • 12. Cost Components The cost is divided into 3 main parts Total = Capital + O&M + Fuel Capital: Calculated in terms of hypothetical ““overnight cost”” O&M: Operation and maintenance Fuel: Uranium delivered to your door ““Busbar”” costs: Costs at the plant No transmission and distribution costs 13
  • 13. Key Input Parameters Plant produces Pe = 1 GWe Takes TC = 5 years to build Operates for TP = 40 years Inflation rate iI = 3% Desired return on investment iR = 12% 14
  • 14. Capital Cost Start of project: Now = 2000 year n = 0 Overnight cost: Pover = $2500M No revenue during construction Money invested at iR = 12% Optimistic but simple Cost inflates by iI = 3% per year 15
  • 15. Construction Cost Table Year Construction Dollars Present Value 2000 500 M 500 M 2001 515 M 460 M 2002 530 M 423 M 2003 546 M 389 M 2004 563 M 358 M 16
  • 16. Mathematical Formula Table results can be written as PVCAP Pover = TC Tc -1 n=0 1 + iI 1 + iR n Sum the series PVCAP TC 1 Pover = = $2129M 1 TC 1 + iI = = 0.9196 1 + iR 17
  • 17. Operations and Maintenance O&M covers many ongoing expenses Salaries of workers Insurance costs Replacement of equipment Repair of equipment Does not include fuel costs 18
  • 18. Operating and Maintenance Costs O&M costs are calculated similar to capital cost One wrinkle: Costs do not occur until operation starts in 2005 Nuclear plant data shows that O&M costs in 2000 are about POM = $95M / yr O&M work the same every year 19
  • 19. Formula for O&M Costs During any given year the PV of the O&M costs are n 1 + iI (n ) PVOM = POM 1 + iR The PV of the total O&M costs are PVOM = TC +TP -1 (n PVOM) n=TC = POM TC +TP 1 n =TC = POM TC 1 + iI 1 + iR 1 1 TP n = $750M 20
  • 20. Fuel Costs Cost of reactor ready fuel in 2000 K F = $2000 / kg Plant capacity factor fc = 0.85 Thermal conversion efficiency = 0.33 Thermal energy per year Wth = fcPeT = (0.85)(106 kWe ) (8760hr ) (0.33) = 2.26 × 1010 kWhr 6 Fuel burn rate B = 1.08 × 10 kWhr /kg Yearly mass consumption Wth MF = = 2.09 × 104 kg B 21
  • 21. Fuel Formula Yearly cost of fuel in 2000 PF = K F M F = $41.8M / yr PV of total fuel costs PVF = PF TC 1 1 TP = $330M 22
  • 22. Revenue Revenue also starts when the plant begins operation Assume a return of iR = 12% Denote the cost of electricity in 2000 by COE measured in cents/kWhr Each year a 1GWe plant produces We = Wth = 74.6 × 108 kWhr 23
  • 23. Formula for Revenue The equivalent sales revenue in 2000 is PR = (COE )(We ) (COE ) fcPeT = = ($74.6M ) ×COE 100 100 The PV of the total revenue PVR = PR TC 1 1 TP = $(74.6M ) (COE ) TC 1 1 TP 24
  • 24. Balance the Costs Balance the costs by setting NPV = 0 PVR = PVcons + PVOM + PVF This gives an equation for the required COE 100 Pover 1 1 COE = fcPeT TC TC 1 = 3.61 TC TP + POM + PF + 1.27 + 0.56 = 5.4 cents /kWhr 25
  • 25. Potential Pitfalls and Errors Preceding analysis shows method Preceding analysis highly simplified Some other effects not accounted for Fuel escalation due to scarcity A carbon tax Subsidies (e.g. wind receives 1.5 cents/kWhr) 26
  • 26. More More effects not accounted for Tax implications –– income tax, depreciation Site issues –– transmission and distribution costs Cost uncertainties –– interest, inflation rates O&M uncertainties –– mandated new equipment Decommissioning costs By-product credits –– heat Different fc –– base load or peak load? 27
  • 27. Economy of Scale An important effect not included Can be quantified Basic idea –– ““bigger is better”” Experience has shown that C cap C ref Pref = Pe Pref Pe Typically 1/ 3 28
  • 28. Why? Consider a spherical tank Cost Power Material Surface area: C (4 / 3) R 3 Volume: P COE scaling: C /P 4 R2 1/R 1 / P 1/ 3 Conclusion: C cap Pe = C ref Pref 1 This leads to plants with large output power 29
  • 29. The Learning Curve Another effect not included The idea –– build a large number of identical units Later units will be cheaper than initial units Why? Experience + improved construction Empirical evidence –– cost of nth unit C n = C 1n ln f ln 2 f = improvement factor / unit: f 0.85 = 0.23 30
  • 30. An example –– Size vs. Learning Build a lot of small solar cells (learning curve)? Or fewer larger solar cells (economy of scale)? Produce a total power Pe with N units Power per unit: pe = Pe/N Cost of the first unit with respect to a known reference C 1 = C ref p pref 1 = C ref N ref N 1 31
  • 31. Example –– cont. Cost of the nth unit C n = C 1n N ref N = C ref 1 n Total capital cost: sum over separate units C cap = N C n = C ref n =1 1 C ref N ref = N 1 N ref N 1 N n n =1 C ref N ref N 1 N 1 n dn N > If we want a few large units It’’s a close call –– need a much more accurate calculation 32
  • 32. Dealing With Uncertainty Accurate input data Uncertain data Risk accurate COE estimate error bars on COE size of error bars Quantify risk calculate COE ± standard deviation Several ways to calculate , the standard deviatiation Analytic method Monte Carlo method Fault tree method We focus on analytic method 33
  • 33. The Basic Goal Assume uncertainties in multiple pieces of data Goal: Calculate for the overall COE including all uncertainties Plan: Calculate for a single uncertainty Calculate for multiple uncertainties 34
  • 34. The Probability Distribution Function Assume we estimate the most likely cost for a given COE contribution. E.g. we expect the COE for fuel to cost C = 1 cent/kWhr Assume there is a bell shaped curve around this value The width of the curve measures the uncertainty This curve P(C) is the probability distribution function It is normalized so that its area is equal to unity 0 P (C )dC = 1 The probability is 1 that the fuel will cost something 35
  • 35. The Average Value The average value of the cost is just C = CP (C )dC 0 The normalized standard deviation is defined by 1/ 2 2 1 = (C C ) P (C )dC 0 C A Gaussian distribution is a good model for P(C) P (C ) = 2 1 1/ 2 (2 ) C exp (C C ) 2 2( C ) 36
  • 36. Uncertainties Multiple Uncertainties Assume we know C and σ for each uncertain cost. The values of C are what we used to determine COE. Specifically the total average cost is the sum of the separate costs: C Tot = Cj Pj (Cj )dCj = Cj. j The total standard deviation is the root of quadratic sum of the separate contributions (assuming independence of the Cj ) again normalized to the mean: 2 j (C j σj ) σTot = j Cj 37 SE T-6 Economic Assessment
  • 37. Uncertainties Nuclear Power An Example Example We need weighting - why? Low cost entities with a large standard deviation do not have much effect of the total deviation Consider the following example Ccap = 3.61, σc = 0.1 COM = 1.27, σOM = 0.15 Cfuel = 0.56, σf = 0.4 38 SE T-6 Economic Assessment
  • 38. Uncertainties Nuclear Power Example Continued Continued The total standard deviation is then given by (σc C cap )2 + (σOM C OM )2 + (σf C fuel )2 σ= C cap + C OM + C fuel √ 0.130 + 0.0363 + 0.0502 = = 0.086 5.4 Large σf has a relatively small effect. Why is the total uncertainty less than the individual ones? (Regression to the mean) 39 SE T-6 Economic Assessment
  • 39. MIT OpenCourseWare http://ocw.mit.edu 22.081J / 2.650J / 10.291J / 1.818J / 2.65J / 10.391J / 11.371J / 22.811J / ESD.166J Introduction to Sustainable Energy Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.

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