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Unformatted text preview: Toolbox 7: Economic Feasibility Assessment Methods
Dr. John C. Wright MIT  PSFC 05 OCT 2010 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 Basic Economic Concepts
Use a simplified analysis Discuss return on investment and inflation Discuss net present value Discuss levelized cost
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 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 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 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 Pi = $100 $100 = = $90.91 1 + iR 1.1 This is the present value of $100 a year from now
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=(1iTax)iTot
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 Pn = (1 + iI ) Pi
10 n 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 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 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 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 Capital Cost
Start of project: Now = 2000 Overnight cost: Pover = $2500M No revenue during construction Money invested at iR = 12% Optimistic but simple Cost inflates by iI = 3% per year year n = 0 15 Construction Cost Table
Year Construction Dollars Present Value 2000 2001 2002 2003 2004 500 M 515 M 530 M 546 M 563 M 500 M 460 M 423 M 389 M 358 M
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 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 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 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=TC (n PVOM) = POM = POM TC +TP 1 n =TC TC 1 + iI 1 + iR 1 1
TP n = $750M
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 MF = Wth = 2.09 104 kg B
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 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 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 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 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 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 Byproduct credits heat Different fc base load or peak load?
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 Why?
Consider a spherical tank Cost Power Material Surface area: C 4 R2 Volume: P (4 / 3) R 3 COE scaling: Conclusion: C /P 1/R
1 1 / P 1/ 3 C cap P = e C ref Pref This leads to plants with large output power
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 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
p pref
1 C 1 = C ref = C ref N ref N 1 31 Example cont.
Cost of the nth unit C n = C 1n = C ref
1 Total capital cost: sum over separate units
C cap =
N n =1 1 C ref N ref = N 1 N ref N 1 n C n = C ref N ref N N N n
n =1 C ref N ref N 1 1 N n dn > If we want a few large units It's a close call need a much more accurate calculation
32 Dealing With Uncertainty
Accurate input data Uncertain data Risk Quantify risk accurate COE estimate error bars on COE calculate COE standard deviation size of error bars Several ways to calculate , the standard deviatiation
Analytic method Monte Carlo method Fault tree method We focus on analytic method
33 The Basic Goal
Assume uncertainties in multiple pieces of data Goal: Calculate all uncertainties Plan: Calculate Calculate for a single uncertainty for multiple uncertainties for the overall COE including 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 The Average Value
The average value of the cost is just C = 0 CP (C )dC 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 ) = 1
(2 )
1/ 2 C exp (C C ) 2 2( C ) 2 36 Uncertainties Multiple Uncertainties
Assume we know C and 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 j ) Tot = j Cj
37 SE T6 Economic Assessment 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, c = 0.1 CO&M = 1.27, OM = 0.15 Cfuel = 0.56, f = 0.4 38 SE T6 Economic Assessment Uncertainties Nuclear Power Example Continued Continued The total standard deviation is then given by (c C cap )2 + (OM C O&M )2 + (f C fuel )2 = C cap + C O&M + C fuel 0.130 + 0.0363 + 0.0502 = = 0.086 5.4 Large f has a relatively small effect. Why is the total uncertainty less than the individual ones? (Regression to the mean) 39 SE T6 Economic Assessment 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
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This note was uploaded on 02/24/2012 for the course MECHANICAL 2.650J taught by Professor Johnc.wright during the Fall '10 term at MIT.
 Fall '10
 JohnC.Wright

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