cost lectures
TRANSCRIPT
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Dr. GP Rangaiah
CN4121 Design Project:
Cost Estimation &
Economic Evaluation
Prof. G.P. RangaiahDepar tm ent o f Chemical & Biom olecularEng ineering @ NUS
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Introduction
What, When, Where, Why and How
Contents
Learning Outcomes
Text-Book
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Introduction
What, When, Where, Why and How
Consider investing in a new plantand/or modifying an exis t ing plant.
Greenfield or Developed Site
Plant Operation for Many Years
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Introduction
Life
Cycle
of a
Process
Plant
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1
Feasibility Analysis Numerous Choices Few Months
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Process & Equipment Design Many Choices 4-12 Months
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Engineering & Construction Process Refinements 1-3 Years
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Operation
Enhancements 10-20 Years
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End of Plant Life Salvage Value When will it be?
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Introduction
Why does a company invest?
Is some risk involved in the investment?
When is cost estimation and economicevaluation required?
Many Times during the Plant Life
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What is cost estimation?What is economic evaluation?How do you estimate the cost and evaluateprofitability?
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Introduction
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Cost estimation and economicevaluation are required forSustainability Assessment
Environmental- unpolluted environment
- availability of resources- ecosystem health
Social- inclusion
- equity- human health
Economic- wealth creation- material goods
- employment
Sustainable natural& built environment
Sustainable economicdevelopment
Equitable socialenvironment
Sustainable
Planet
People Profits
Sustainable
Bearable Viable
Equitable
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Introduction
Contents
Introduction
Cost Estimation Estimation o f Capi tal Cost Est im at ion o f Manufactur ing Cost
Economic Evaluation Engineer ing Econo m ic An alys is Profi tabi l i ty An alysis
Summary
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Introduction
Learning Outcomes
The student should be able to:
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Estimation of Capital and OperatingCosts1. Describe
Capital and Operating Costs ofEquipment and Plants2. Estimate
Time Value of Money, Cash FlowDiagram, Depreciation andProfitability Criteria3. Describe
Economic Evaluation of Equipmentand Plants4. Perform
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Introduction
Text-BookTurton R., Bailey R.C., White W.B.,Shaeiwitz J.A., and Bhattacharyya D.,An alysis , Synth esis and Design o fChem ical Proc esses,
4 th Edition, Prentice Hall (2013)
TP155.7 Ana 20139; Chapters 7 to 10
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I hear and I forget.I see and I remember.I do and I understand.
- Confucius
Study/SolveExercises andProblems in the
Text-Book
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Introduction
What, When, Where, Why and How
Contents
Learning Outcomes
Text-Book
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Cost Estimation & Economic Evaluation
Contents
Introduction
Cost Estimation Estimation o f Capi tal Cost Est im at ion o f Manufactur ing Cost
Economic Evaluation Engineer ing Econo m ic An alys is Profi tabi l i ty An alysis
Summary
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Cost Estimation Capital Cost
Types of Capital Cost EstimatesPurchased Equipment CostCost IndexCapital Cost of a Plant
Six-Tenths RuleLang Factor TechniqueModule Costing Technique
Summary and Software
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Capital Cost Estimation Types of Estimates
Approximate to Accurate Estimates
Minimal to Extensive Data/Effort
5 Types of Cost Estimates
Error in the Most Accurate Estimate:
+6 to -4%
Effort in the Least Accurate Estimate:0.015 to 0.3% of plant cost
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Capital Cost Estimation Types of Estimates
* Relative to the accuracy of the detailed estimate# Relative to the cost of the order-of-magnitude estimate
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Type Purpose Accuracy Cost
Order-of-Magnitudeor Ratio
Screening,Feasibility
4 to 20* 0.015 to 0.3%of plant cost
Study or Factored orMajor Equipment
Concept Study,Feasibility
3 to 12* 2 to 4 #
Preliminary Designor Scope
Budget, Approval,Control
2 to 6* 3 to 10 #
Definitive or ProjectControl
Control,Bid/Tender
1 to 3* 5 to 20 #
Detailed or Firm orContractor's
Check Estimate orBid/Tender
Error:+6% to - 4%
10 to 100#
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Capital Cost Estimation Types of Estimates
Example 7.1Capital cost of a plant is $2 million by the StudyEstimate method.
Lowest Expected Cost Range: $1.8 to $2.4 millionHighest Expected Cost Range: $1.0 to $3.4 million
Example 7.2Cost/effort of cost estimation for a plant (costing$5 million) varies from $750 (low accuracy) to $1.5million (high accuracy)
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Cost Estimation Purchased Equipment Cost
Estimation of capital cost of a plant oftenrequires pu rchase co s t of major equ ipm ent in the plant.
Order-of-magnitude and study estimates(the f irs t tw o types ) are based on histor icalco s t da ta .
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Cost Estimation Purchased Equipment Cost
Effect of Size (or Capacity)Purchase cost of equipment, C is correlated with itssize, S via the exponential equation :
Subscript r refers to the reference/base case.Exponent n is in the range 0.4 to 0.8
The above equation can be re-written as:
What is K in terms of C r and S r ?
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Cost Estimation Purchased Equipment Cost
Example Data
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Equipment Size Rangewith Units n
C r in US$(S r in brackets)
Reciprocating Compressor(Motor Drive)
0.75 to 1490kW
0.84 133,000(224 kW)
Heat Exchanger (Shell andTube, Carbon Steel)
1.9 to 1860m2
0.59 21,700(9.3 m 2)
Centrifugal Blower(Excluding Motor)
0.24 to 71std m 3/s
0.60 67,000(4.72 m 3/s)
Jacketed Kettle (GlassLined)
0.2 to 3.8m3
0.48 53,000(0.38 m 3)
C i i h d i C
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Cost Estimation Purchased Equipment Cost
Cost Correlation:
can be rearranged to:
What is the effect of increasing size (S) onthe equ ipm ent co s t per un i t capaci ty ?
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C E i i P h d E i C
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Cost Estimation Purchased Equipment Cost
Cost Plots: Example Heat Exchanger PurchaseCost from Figure A.5 in Turton et al. (2013)
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Uses:1. Confirm CostEstimate usingthe Correlation
2. Know HeatExchangerTypes and theirRelative Costs
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3
4
6
5
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C t E ti ti P h d E i t C t
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Cost Estimation Purchased Equipment Cost
Example 7.3Application of Exponential Equation: Cost of anequipment increases by 52% when its size doubles
Example 7.4Effect of Exponent ( n ) in the Exponential Equationon the Cost Estimate
Example 7.5Application of the Exponential Equation
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C t E ti ti C t I d
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Cost Estimation Cost Index
Equipment cost varies with time due toinflation and technological developments.
Historical cost may have to be used.
Equipment cost in the next few months toyears, has to be estimated.
Suitable cost index (similar to ConsumerPrice Index, CPI ) is required to updatehistorical equipment cost.
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Cost Estimation Cost Inde
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Cost Estimation Cost Index
Cost, C in certain year when the cost index
= I is given by
Here, C b is the known cost in the year when theindex was I b .
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Cost Estimation Cost Index
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Cost Estimation Cost Index
What are the cost indices used by the
chemical industry?
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Cost Index, I YearWhen
I = 100
Valuein Year
2000 Chemical Engineering Plant CostIndex, CEPCI
1957 394
Engineering News-Record ConstructionIndex
1967 579
Marshall & Swift Equipment Cost Index 1926 1103 Nelson-Farrar Refinery ConstructionIndex
1946 1543
Cost Estimation Cost Index
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Cost Estimation Cost Index
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No. Component/Item Weight1 Fabricated Equipment 0.22572 Process Machinery 0.08543 Pipes, Valves and Fittings 0.1224 Process Instruments and Controls 0.04275 Pumps and Compressors 0.04276 Electrical Equipment and Materials 0.03057 Structural Supports, Insulation & Paint 0.0618 Erection and Installation 0.229 Buildings, Materials and Labor 0.07
10 Engineering and Supervision 0.1
CEPCI is a Com po si te Ind ex consisting of:
Cost Estimation Cost Index
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Cost Estimation Cost Index
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Process Machinery:
bought off-the-shelf
examples are centrifuges, filters,agitators, dryers, conveyors, vacuum/
refrigeration systems, crushers/grinders,
thickeners/ settlers, fans/blowers etc.
Cost Estimation Cost Index
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Cost Estimation Cost Index
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CEPCI available on
the last page ofChemicalEngineering
technical magazine
(www.che.com)
Cost Estimation Cost Index
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Cost Estimation Cost Index
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Variation of CEPCI
and CPI during1995-2013
Do you need CEPCI in the future?If required, extrapolate with caution.
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Cost Estimation Cost Index
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Cost Estimation Cost Index
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Which cost index should be used?
Example 7.6Effect of Using Two Different Cost Indices on
the Cost Estimate: Difference is ~7%
Cost Estimation Capital Cost of a Plant
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Cost Estimation Capital Cost of a Plant
Six-Tenths Rule
Exponential equation with n = 0.6 (six-tenths) forestimating capital cost of a plant (or an equipment)app roxim ately and q uickly .
= .
ExampleCapital cost of a plant for producing 100 thousand
tons/year of styrene from benzene and ethylene in1994 was $40 million. Estimate the capital cost forsetting up a plant to produce 200 thousand tons/yearof styrene, next y ear .
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Cost Estimation Capital Cost of a Plant
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Cost Estimation Capital Cost of a Plant
Step 1: Use six-tenths rule to find the capital cost (CC)
of a plant for producing 200 thousand tons/year ofstyrene, in 1994 itself.
CC of the bigger plant = $40.
= $60.6 million
Step 2: Use cost index to estimate CC of the biggerplant next year.
CEPCI in 1994 = 368; CEPCI next y ear = 650 (estimated)
CC for the new plant = $60.6 = $107 million
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Cost Estimation Capital Cost of a Plant
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Cost Estimation Capital Cost of a Plant
Lang Factor TechniqueA simple method to estimate CC of a plant from thepurc hase cos t of m ajor equipm ent , in the plant.
CTM = F Lang ,
Lang factor, F Lang = 4.74, 3.63 and 3.10 for a f luid, so l id- f lu id and so l id processing plant, respectively.Summation covers major equipment (pumps,compressors, vessels, towers etc.) in the process f low
diagram for the plant.
Lang fac tor d oes no t take in to accou nt m ater ial ofcon st ruct ion and op erat ing pressure .
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Capital Cost of a Plant is Significantly More
than Sum of Purchase Cost of All Equipment
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Factor ( See boo k for d etails) Cost Symbol/Equation (a) Equipment Purchase (b) Materials for Installation
(c) Labor for Installation
Direct Project Expense
CP CM = M CP
CL = L (CP + C M)
CDE = C P + C M + C L
(a) Freight, Insurance & Taxes (b) Construction Overhead (c) Contractor Eng. Expenses
Indirect Project Expense
CFIT = FIT (CP + C M) CO = O CL CE = E (CP + C M)
CIDE = C FIT + C O + C E
Bare Module Cost CBM = C DE + C IDE
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Som e More Contr ibut ing Factors
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Factor ( See boo k fo r d etails) Cost Symbol/Equation Bare Module Cost CBM = C DE + C IDE
(a) Contingency (b) Contractor Fee
CCont = Cont CBM CFee = Fee CBM
Total Module Cost CTM = C BM+CCont +CFee Total Module Cost for Developed Sites & Additions to Existing Plants
Auxiliary Facilities (a) Site/Land and its Development
(b) Auxiliary Building (c) Off-sites and Utilities
CSite = Site CBM
C Aux = Aux CBM COff-sites = Off-sites CBM
Grass Roots Cost CGR = C TM+CSite +C Aux+COff-sites
Grass Roots Cost for Green Field Sites and New Plants
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Values of s depend on Equipment
Example 7.9 for a Heat ExchangerCarbon Steel, Ambient Pressure
Purchase Cost = $10,000Costs for Materials, Labor, Freight, Overhead,Engineering as % of Purchase CostBare Module Cost = $32,910
Bare Module Cost Factor, F BM = 3.291
FBM Values available in Appendix A
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Module Costing Technique
Introduced by Guthr ie around 1970
Many Techniques using this Approach
Bare Module Cost, C BM CP0
(B1 + B 2 FP FM)CP0: Purchase Cost of Equipment at BaseConditions (Carbon Steel Material and AmbientPressure Operation) B1 and B 2 are Coefficients, whose values dependon the EquipmentFP and F M are Pressure and Material Factors
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Module Costing Technique
Correlations/Plots for C P0 of Different Equipment
in Appendix A of Turton et al. (2013)
CP 0 = ( )
Here, S is the Capacity/Size of the Equipment, and
are Coefficients.
Use the Correlation/Plot w ithin the range stated. A void extrapolat ion.
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Pressure Factor, F P
Equipment cost increases with increasingpressure (vacuum) due to wall thickness
Recall equation(s) for calculating wall thicknessfrom the mechanical design lectures
For equipment operating below 0.5 bar, F P = 1.25.
Pressure factor is correlated with P by:
Log 10 FP = C 1 + C 2 log 10(P) + C 3 [log 10(P)] 2
Values of coefficients, C n are in Table A.2.
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Pressure Factor, F P Equipment cost increases with increasingpressure (vacuum) due to wall thickness
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Example 7.11:
Effect of Pressureon Cost of Shell-
and-Tube HeatExchanger and
Discussion 1
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Material Factor, F M
Different materials of construction are needed tomeet bu dg et , co rros ion, temp erature and p ressur e requirements of operation.
FM for different materials and equipment are inFigure A.18 and Table A.3.
Example 7.12 demonstrates the effect of pressureand material of construction on bare module costof a heat exchanger.
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Materials of Construction Used in Process Industry
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Material Cost Comments CarbonSteel
BaseCase
Less than 1.5 wt % carbon, most economical andcommon
Low-alloySteel
Low-Moderate
With chromium (for resistance to mildly acidic & oxidizingconditions) and molybdenum (for strength at high T)
StainlessSteel Moderate
More than 12 wt % chromium, high resistance tochemicals & rusting
Aluminum &its Alloys
Moderate High strength-to-weight ratio, good corrosion resistance
Copper & its Alloys
Moderate High thermal conductivity, resistance to seawater
Titanium &its Alloys
High Good strength-to-weight ratio and resistant to oxidizingagents
Nickel & its Alloys
High Nickel with copper ( Monel ), chromium ( Inconel ) andmolybdenum ( Hastelloy ), excellent chemical resistanceat high temperature
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Example 7.13 presents cost estimation of a tower
with sieve t rays .Cost of Tower + Cost of Trays
Summary
Bare Modu le Cos t , CBM = CP 0 [B 1 + B 2 FP FM] Values o f B 1 and B 2 are in Tables A .4 to A .6 & Fig ur e A.19.
B 1 and B
2 are in th e range 0.96 to 2.25 (Table A .4).
Facto r, F B M = [B 1 + B 2 F P F M ] is in the range 1 to 16
(Figu re A .19)
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Summary
Total Modu le Cost , CTM = = . ,
Grass Roots Plant Cost,
CGR = C TM + C Site + C Aux + C Off-sites
CGR = C TM + . ,
Update the Estimated Cost to the Present/FutureTime using Cost Index
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Example 7.14
Estimate Cost of a Column with a Cooler
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Equipment Capacity/Size Material, PressureE-101 Condenser 170 m 2, Shell & Tube
(Floating Head)Tube: CS, 5 bargShell: CS, 5 barg
E-102 Reboiler 205 m 2, Shell & Tube(Floating Head)
Tube: SS, 19 bargShell: CS, 6 barg
E-103 Product Cooler 10 m 2, Double Pipe CS, 5 barg (all)P-101A & B Reflux
Pumps
Shaft Power = 5 kW
Centrifugal
CS,
Discharge P = 5 bargT-101 Column 2.1 m diameter, 23 mheight, 32 sieve trays
CS, 5 bargSS Trays
V-101 Reflux Drum 18 m diameter6 m length, horizontal
CS, 5 barg
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Estimate Cost of Each Equipment usingData/Correlations in Appendix A
Consider E-102 Reboiler Purc hase Cos t, C P 0 = $36,900 usi ng th e Cor relation :
log 10(CP0) = K1 + K2 log 10(A) + K 3 [log 10(A)]2
= 4.8306 - 0.8509 log 10(205) + 0.3187 [log 10(205)] 2
F M = 1.8 from Figu re A.18
F P = 1.024 us ing the Co rrelation :
log 10 FP = C 1 + C 2 log 10(P) + C 3 [log 10(P)] 2
= - 0.00164 - 0.00627 log 10(19) + 0.0123 [log 10(19)] 2 45
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B are Mod ule Cost, C B M = C P 0 [B 1 + B 2 F P F M ]
= $36,900 (1.63 + 1.66 1.024 1.8)
(w ith B 1 & B 2 fro m Table A.4)
= $36,900 4.7 = $173,500
Estimate C BM of Each and Every Equipment
Total Bare Module Cost, , = $797,000
Total Module Cost, C TM . , = $940,000
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Capital Cost Estimation
Straightforward ProceduresRequires Lot of Data/CorrelationsSize, Material of Construction & OperatingPressure of Each Equipment
Programs for Capital Cost Estimation
Study Section 7.3.8 on CAPCOST programTry this program for estimating cost of a columnwith a cooler (Example 7.14) discussed above
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Other Capital Cost Estimation Programs
CCEP based Capital Cost Estimation correlations inSeider W.D., Seader J.D., Lewin D.R. and WidagdoS., Product and Process Design Principles:Synthesis, Analysis, and Evaluation, 3 rd edition,
John Wiley (2010)
DFP based on Detailed Factorial method and costdata in Sinnott R.K. and Towler G., ChemicalEngineering Design 5 th edition, Butterworth andHeinemann (2009)
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Other Capital Cost Estimation Programs
DFP and CCEP programs are available in IVLE-Workbin.
Which Reference/Program should be used? Read the Cover Story in Chemical Eng ineering , p. 22-29,August (2011), available in IVLE-Workbin
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Fixed Capital Investment (FCI)
Total Module Cost OR Grass Roots Cost Procedures and Terminology are notStandard
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Turton et al. Seider et al.Bare Module Cost SameTotal Module Cost ???Grass Roots Cost Total Depreciable
CapitalTotal PermanentInvestment
Advice: Follow One Book
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Contents
Introduction
Cost Estimation
Estimation o f Capi tal Cost Est im at ion o f Manufactur ing Cost
Economic Evaluation
Engineer ing Econo m ic An alys is Profi tabi l i ty An alysis
Summary
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Cost Estimation Manufacturing Cost
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Cost of Manufacture:
COM = Direct Manufacturing Costs (DMC)+ Fixed Manufacturing Costs (FMC)+ General Manufacturing Expenses (GE)
DMC includes Many Factors: see the tableon the next slide
Also, DMC depends on Production Rate
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Cost Estimation DMC
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Factor Symbol/Equation
Remarks
Raw Materials C RM Require flow rates & pricesWaste Treatment C WT Require quantity & treatment costUtilities C UT Fuel, electricity, steam, water,
refrigeration, instrument air, N 2 Operating Labor C OL Cost of plant operatorsDirect Supervisory &Clerical Labor (DSCL)
(0.1 - 0.25) C OL Cost of admin, engineering &support staff
Maintenance &Repairs (M&R)
(0.02 - 0.1) FCI Cost of labor & materials
Operating Supplies (0.1 to 0.2) ofM&R
Lubricants, chemicals, filters, safetygear & uniforms for operators
Laboratory Charges (0.1 to 0.2) C OL Tests for product quality &troubleshooting
Patents & Royalties (0 to 0.06) COM Cost of using licensed technology
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Using mid-value of the range,
DMC = C RM + C WT + C UT + 1.33 C OL
+ 0.069 FCI + 0.03 COM
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Fixed Manufacturing Costs (FMC)
FMC = 0.708 C OL + 0.168 FCI
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Factor Symbol/Equation Remarks Depreciation 0.1 FCI Approximate, depends on local
tax laws Taxes and
Insurance
(0.014 to 0.05) FCI Depends on local taxes
PlantOverheadCosts
(0.5 to 0.7) ofCOL, DSCL and M&R
Accounting, fire, safety &medical services; canteen &recreation; general engineering
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General Manufacturing Expenses (GE)
GE = 0.177 C OL + 0.009 FCI + 0.16 COM
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Factor Symbol/Equation Remarks Administration Costs 0.15 of C OL, DSCL
and M&R Costs for administration,buildings etc.
Distribution & Selling
Costs
(0.02 to 0.2) COM Marketing expenses
Research &Development
0.05 COM For enhancing theprocess and products
Cost Estimation Manufacturing Cost
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Add DMC, FMC and GE, and re-arrange to solvefor COM
COM = 0.28 FCI + 2.73 C OL + 1.23 (C RM + C WT + C UT)
COM without Depreciation
COM = 0.18 FCI + 2.73 C OL + 1.23 (C RM + C WT + C UT)
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Cost of Operating Labor
Number of operators per sh i f t isNOL = [6.29 + 31.7 P 2 + 0.23 N np ]0.5
Here, P is the number of processing steps involvingpart iculate so l ids and N np is the number of equipmenthandling fluids (such as compressors, towers,reactors, heaters & exchangers).
P and N np are counted based on the process flowdiagram. P is zero for fluid-processing plants, and N np does no t include pumps, vessels and tanks.
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Cost of Operating Labor
Each operator works for 5 eight-hour shifts per weekfor 49 weeks per year. Hence, the total number ofoperators required per y ear is 4.5 N OL.
Cost of one operator per hour or year?
Example 8.2: Computation of number ofoperators required per shift and annual labor costfor a process with compressors, exchangers,furnaces, pumps, reactors, towers and vessels.
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Cost Estimation Manufacturing Cost
Utility Cost (US $)
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Utility Cost (US $) Air supply (3.3 barg) $0.35/100 std m 3 Steam (5 barg, 160 0C) $27.70/1000 kg
Steam (10 barg, 184 0C) $28.31/1000 kgSteam (41 barg, 254 0C) $29.97/1000 kg
Cooling Water at 30 0C (returned at 40-45 0C) $0.0148/1000 kg Process Water $0.067/1000 kg
Chilled Water at 5 0C (returned at 15 0C) $0.185/1000 kg Potable/Drinking Water $0.26/1000 kg
Boiler Feed Water $2.45/1000 kg Electric Power $0.06/kWh
Fuel Oil $549/m 3 Natural Gas $0.42/std m 3
Waste Water Treatment: Primary (Filtration)Secondary (Filtration & Activated Sludge)
Tertiary (Secondary & Chemical Processing)
$41/1000 m 3 $43/1000 m 3 $56/1000 m 3
Solid & Liquid Waste Disposal: Non-hazardousHazardous
$36/ton $200-2000/ton
Utility
Costs :Typicaldatataken
fromTable 8.3in Turtonet al.(2013)
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Utility Systems : Extensive and Complex
Example8.6 forSteamCosts
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Raw Material Costs
See Table 8.4 for Costs of Some Chemicals
Yearly Costs and Stream Factor
Require Number of Days of Operation
Stream Factor = Number of Days of Operation
per Year / 365
Stream Factor: 0.90 ~ 0.96
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Example 8.9
Estimate quantities and yearly cost of utilities forseveral equipment in a plant
Example 8.10Estimate cost of manufacture withoutdepreciation (COM d) of benzene from toluene.
Raw Materials: Toluene = $53.9 million/year;Hydrogen = $6.6 million/year
Total = $60.5 million/year
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Cost Estimation Manufacturing Cost
E l 8 10 ( i d)
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Example 8.10 (continued)
Utilities: Steam = $3.4 million/year;Cooling Water = $0.2 million/yearFuel Gas = $2.8 million/year;
Electricity = $0.04 million/yearTotal = $6.4 million/year
No Waste Streams and Waste Treatment Cost!
Labor: $0.7 million/year
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Example 8.10 (continued)
Fixed Capital Investment, FCI = $11.7 million
COMd = 0.18 FCI + 2.73 C OL + 1.23 (C RM + C WT + C UT)= $86.5 million/year
Benzene Production = 8,210 kg/hour
= 8,210 24 365 0.95 = 68.3 million kg/year
Cost Price of Benzene = $1.27/kg(Compare with that in Table 8.4)
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Contents
Introduction
Cost Estimation
Estimation o f Capi tal Cost Est im at ion o f Manufactur ing Cost
Economic Evaluation Engineer ing Econo m ic An alys is Profi tabi l i ty An alysis
Summary
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Several Concepts and Relations for
Profitability Analysis
Investment, Interest and Time Value of Money
DepreciationRevenue, Tax, Profit and Cash Flow
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Investment, Interest and Time Value of
Money
Investment to earn Interest (or Return)
By Individuals and Companies
Money, w hen Inv ested , m akes Mo ney
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Invest P $ today for n Years
Interest/Return of i $/($.year) (say, 0.1 or 10% per
year)
Value/Amount after n Years
( ) assuming Simple Interest
assuming Annual Compound Interest
Which interest simple or annual compound,is better?
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Amount at Present Time Amount at Future Time(say, 10 Years from Now)
$10,000 $10,000 (1+0.1) 10 = $25,937
$3,855
[= $10,000/(1+0.1)10
]
$10,000
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Is $10,000 at the Present Time better than
the same amount at the Future Time?
Money tod ay is w or th m ore than th at in the fu ture .
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Amount in Year n ( ) can be discounted
(brought) to the Present Time
Rearrange to
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Time value of money refers todifferent values of the investment
at different times due to theearning capability of money.
Time value of moneydoes not include
inf lat ion o r p urchas ingpow er of m oney.
1
2
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Depreciation
Accounts for Decrease in Value of Equipment/Plant with Time
Strategy to Recover the Investment in the Plant
and to Decrease Tax Payable on ProfitDepends on Local Tax Laws
Does Land Depreciate?
FCI excluding Land Cost can be Depreciated
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Cost Estimation Engineering Economic Analysis
Working Capital (WC)
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Working Capital (WC)
Start up and production will take a few monthsNo products and revenue in this periodMoney is needed for salaries, raw materials etc.
WC is the investment required for start up andfor financing first few months of operation
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WC = 15 to 20% of FCI
Total Capital Investment =FCI + WC
1
WC can be Recoveredand so it can not be
Depreciated.
2
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Types of Depreciation
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Types of Depreciation
1. Straight Line (SL) Depreciation: d k = D/n2. Sum of the Years (SOY) Depreciation3. Double Declining Balance (DDB) Depreciation
See the Text Book for Details.
Depreciation Calculations require:Fixed capital investment exclud ing land (FCI L)Life of the equipment/plant (n years)Salvage value, S (often assumed to be zero)Total capital for depreciation, D = FCI L - S
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Example 9 21
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WhichMethodis Better?
Example 9.21Fixed capital investment (excluding land cost) in a newproject is $150 million and the salvage value of the plant is$10 million at the end of 7 years of equipment life. What isthe depreciation according to different methods?
76
0
5
10
15
20
25
30
35
40
45
0 1 2 3 4 5 6 7 8
Year
Dep
reciation ($ MM) SL
SOY
DDB
1
23
Cost Estimation Engineering Economic Analysis
Revenue, Tax, Profit and Cash Flow
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Revenue, Tax, Profit and Cash Flow
Revenue = R ( How d o yo u f ind revenue? )Expenses = COM = COM d + d
COMd = Manufacturing Costs Excluding Depreciation
d = DepreciationTax rate = t
Income Tax = (R - COM d - d) t
After Tax (Net) Profit = (R - COM d - d) (1 - t)
After Tax Cash Flow = (R - COM d - d) (1 - t) + d
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Cost Estimation Engineering Economic Analysis
Example 9.23
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Example 9.23Calculation of profit and cash flowAfter-Tax Profit: Figure E9.23 in Turton et al.(2013)
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After-TaxCash Flow
1
2
Cost Estimation & Economic Evaluation
Contents
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Contents
Introduction
Cost Estimation Estimation o f Capi tal Cost Est im at ion o f Manufactur ing Cost
Economic Evaluation Engineer ing Econo m ic An alys is Profi tabi l i ty An alysis
Summary
79
Cost Estimation Profitability Analysis
Cash Flow Diagrams
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Cash Flow Diagrams
Criteria for Profitability AnalysisPayback Period (Time, Years)Cumulate Cash Position (Cash, $$$)Rate of Return (Interest Rate, Fraction)
Criteria Without Time Value of Money
Criteria Considering Time Value of Money(Discounted Profitability Criteria)
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Cost Estimation Profitability Analysis
Cash Flow Diagrams (CFDs)
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Cash Flow Diagrams (CFDs)
Discrete and Cumulative CFDs
Show Investment and Profit (Outward and Inward
Cash Flows) in Each Year of the Project
Examples 10.1 and 10.2
Investment/Costs/Profit for a New Chemical Plant
Cost of Land, L = $10 Million (at time = 0 Years)
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Cost Estimation Profitability Analysis
FCIL (excluding Land) = $150 Million
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L ( g ) Year 1 of Co ns truc tion Phase = $90 Mill ion Year 2 of Co ns truc tion Phase = $60 Mill ion
Ok to require m ore investment in year 1?
Plant Start-up at the End of Year 2? Wo rkin g Capital = 20% of FCI L
COM excluding Depreciation (COM d)= $30 Million/year (after Start-up)
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Cost Estimation Profitability Analysis
Sales Revenue = $75 Million/Year
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Ok to assum e the sam e revenue throu gho ut the plant l i fe?
Taxation Rate, t = 45%
Salvage Value of the Plant = $10 Million A fter Plant Life of 10 Years
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Cost Estimation Profitability Analysis
Depreciation according to Mod ified Ac celerated
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gCost Recov ery Sys tem (Table 9.2, Turton et al.)
Different % of FCI L in 6 years after plant start-up
84
Year 1 2 3 4 5 6
Percent 20 32 19.2 11.52 11.52 5.76
Cost Estimation Profitability Analysis
Discrete and Cumulative Cash Flows ($ Million) in the Project
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End of Year, k Invest-ment d k After Tax Cash Flow:(R-COM d-d k) (1-t) + d k DiscreteCash Flow CumulativeCash Flow
0 (10) - - (10) (10)1 (90) - - (90) (100)2 (60+30) - - (90) (190)3 - 30 38.25 38.25 (151.75) 4 - 48 46.35 46.35 (105.40) 5 - 28.8 37.71 37.71 (67.69) 6 - 17.28 32.53 32.53 (35.16) 7 - 17.28 32.53 32.53 (2.64) 8 - 8.64 28.64 28.64 26.00
9 - - 24.75 24.75 50.7510 - - 24.75 24.75 75.5011 - - 24.75 24.75 100.2512 10+30 - 30.25 * 70.25 170.50* Revenue in Year 12 includes salvage value ($10 Million) of the plant
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Discrete CashFlow Plot
CumulativeCash Flow Plot
-250
-200
-150
-100
-50
0
50
100
150
200
0 1 2 3 4 5 6 7 8 9 10 11 12
Year
$MM
1
2
Profitability Criteria
Cost Estimation Profitability AnalysisSome Data for the Example
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ywithout Time Value
of Money
Payback Period, PBP (orPayout/Payoff or CashRecovery Period/Time):
Time requ ired, after start-upfor recov ering f ixed capi talinvestm ent excluding land
co st (FCI L ) FCI L does not inc lud ewo rking capi tal
87
End of
Year, k
Discrete
Cash Flow 0 (10) 1 (90) 2 (90) 3 38.25
4 46.35 5 37.71 6 32.53 7 32.53 8 28.64 9 24.75
10 24.75 11 24.75 12 70.25
FCIL =90+60 =
150
38.25 +46.35 +37.71 +32.53 =153.84
PBP = ?
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Profitability Criteria
Cost Estimation Profitability AnalysisSome Data for the Example
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without Time Valueof Money
Rate of Return
Rate of Return o nInv estm ent = Av erage NetProfit/FCI L
ROROI = (170.5/10)/150 =0.114 (11.4%)
An y Other Defini t ions?
89
End of Year, k
DiscreteCash Flow
CumulativeCash Flow
0 (10) (10) 1 (90) (100) 2 (90) (190) 3 38.25 (151.75) 4 46.35 (105.40) 5 37.71 (67.69) 6 32.53 (35.16) 7 32.53 (2.64)
8 28.64 26.00 9 24.75 50.75 10 24.75 75.50 11 24.75 100.25 12 70.25 170.50
Discounted Profitability Criteria
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Assu m e adiscou nt rate(i.e., exp ect ed
return oninvestments) . Discou nted cashflow s for Examp le8.1 (w ith a disco un trate o f 0.1) are asfol lows.
90
End of Year, k
DiscreteCash Flow
DiscountedCash Flow
CumulativeDiscounted CF
0 (10) (10) (10) 1 (90) (90)/1.1 =
(81.82)
(91.82)
2 (90) (90)/1.1 2 =(74.38)
(166.20)
3 38.25 38.25/1.1 3 =28.74
(137.46)
4 46.35 46.35/1.1 4 =31.66
(105.80)
Cost Estimation Profitability Analysis
End ofY k
DiscreteC h Fl
Discounted CashFl
CumulativeDiscounted CF
DiscountedC h Fl
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Year, k Cash Flow Flow Discounted CF
0 (10) (10) (10) 1 (90) (90)/1.1 = (81.82) (91.82)
2 (90) (90)/1.1 2 = (74.38) (166.20) 3 38.25 38.25/1.1 3 = 28.74 (137.46) 4 46.35 46.35/1.1 4 = 31.66 (105.80) 5 37.71 37.71/1.1 5 = 23.41 (82.39) 6 32.53 32.53/1.1 6 = 18.36 (64.03) 7 32.53 32.53/1.1 7 = 16.69 (47.34) 8 28.64 28.64/1.1 8 = 13.36 (33.98)
9 24.75 24.75/1.1 9 = 10.50 (23.48) 10 24.75 24.75/1.1 10 = 9.54 (13.94)
11 24.75 24.75/1.1 11 = 8.67 (5.26) 12 70.25 70.25/1.1 12 = 22.38 17.12
Cash Flows
forExample
10.1
(Discountrate, i = 0.1)
For annualcompound
interest,
or+
Discounted Profitability Criteria
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Discounted PaybackPeriod (DPBP):
Time requ ired, after start- up fo r recovering FCI L w i th
al l cash f lows d iscoun tedback to t ime zero
Discou nted FCI L = 90/1.1 +
60/1.1 2
= 131.4
92
Sum of Discounted Cash Flows in Years 3 to 8= 28.74+31.66+23.41+18.36+16.69+13.36 = 132.2
DPBP~ 6 years
Discounted Profitability Criteria
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Net Present Value,NPV
Cum ulat ive Discou ntedCash Flow at the End of
th e Proj ect = $17.12
93
NPV is th e extra cash f low(express ed as p resent v alue)
generated by th e pro jectafter recov ering th e entire
investm ent along w ith areturn at the discou nt rate.
End of Year, k
DiscreteCash Flow
Discounted CashFlow
CumulativeDiscounted CF
0 (10) (10) (10) 1 (90) (90)/1.1 = (81.82) (91.82)
2 (90) (90)/1.1 2 = (74.38) (166.20)
3 38.25 38.25/1.1 3 = 28.74 (137.46)
4 46.35 46.35/1.1 4 = 31.66 (105.80) 5 37.71 37.71/1.1 5 = 23.41 (82.39)
6 32.53 32.53/1.1 6 = 18.36 (64.03)
7 32.53 32.53/1.1 7 = 16.69 (47.34)
8 28.64 28.64/1.1 8 = 13.36 (33.98)
9 24.75 24.75/1.1 9 = 10.50 (23.48) 10 24.75 24.75/1.1 10 = 9.54 (13.94)
11 24.75 24.75/1.1 11 = 8.67 (5.26)
12 70.25 70.25/1.1 12 = 22.38 17.12
Discounted Profitability Criteria
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Can NPV be negative?
Is large or small NPV preferred?
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Discounted Profitability Criteria
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Present Value Ratio
..
.
95
End of Year, k
DiscreteCash Flow
Discounted CashFlow
CumulativeDiscounted CF
0 (10) (10) (10) 1 (90) (90)/1.1 = (81.82) (91.82)
2 (90) (90)/1.1 2 = (74.38) (166.20)
3 38.25 38.25/1.1 3 = 28.74 (137.46)
4 46.35 46.35/1.1 4 = 31.66 (105.80)
5 37.71 37.71/1.1 5 = 23.41 (82.39)
6 32.53 32.53/1.1 6 = 18.36 (64.03)
7 32.53 32.53/1.1 7 = 16.69 (47.34)
8 28.64 28.64/1.1 8 = 13.36 (33.98)
9 24.75 24.75/1.1 9 = 10.50 (23.48) 10 24.75 24.75/1.1 10 = 9.54 (13.94)
11 24.75 24.75/1.1 11 = 8.67 (5.26)
12 70.25 70.25/1.1 12 = 22.38 17.12
Discounted Profitability Criteria
Cost Estimation Profitability Analysis
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Disco un ted Cash Flow Rate of Return , DCFRR A lso , kn ow n as Internal Rate of Return (IRR) Specif ic Disco unt Rate for w hich Cum ulat ive Disco unted
Cash Flow (i.e., NPV) is Zero
96
DiscountRate, i
NPV($ Million)
0.00 170.5
0.10 17.12 0.12 0.77 0.13 -6.32 0.15 -18.66
NPV for Several Discount Rates
DCFRR = ?
2
1
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Comparing Several Projects
Cost Estimation Profitability Analysis
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Minim um acceptab le (after-tax) retu rn fo r thecom pany is 10%
Projec t A can be cho sen only onc e.
Which projec t shou ld be se lec ted for inves tm entand w hy?
98
See the incremental analysis including Examples10.4 and 10.5 in Section 10.3 in Turton et al. (2013)
Choose the Project with the Highest NPV
2
1
Evaluating Equipment Alternatives
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h
Exam ple 10.8 : Which of the following pumpsshould be selected for an application? Assume adiscount rate of 8%.
99
Pump Details Capital Cost($)
OperatingCost ($/year)
EquipmentLife (years)
Carbon Steel Pump 8,000 1,800 4
Stainless Steel Pump 16,000 ( ) 1,600 ( ) 7 ( )
Evaluating Equipment Alternatives
Cost Estimation Profitability Analysis
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One method is based on Equivalent AnnualOperating Cost (EAOC) for the alternatives.Refer Tur to n et al . (2013) fo r o th er m etho ds .
EAOC ($/year) = Operating Cost ($/year)+ Amortized Capital Cost ($/year)
Capital Cost ($) is amortized (distributed) over theequipment life considering time value of money.
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Cost Estimation Profitability AnalysisAssume the capital cost of an equipment with life ofn years, is P and the expected return is i.
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The amortized capital cost, A for n years is suchthat NPV of them is equal to P.
( )
What is the significance of the terms?
The above equation can be re-written as:
101
For n = 10 years and i = 15, A/P = 0.2 (i.e.,amortized capital cost is 20% of the capital cost.
Cost Estimation Profitability AnalysisExam p le 10.8 : Which of the following pumps shouldbe selected for an application? Discount rate = 8%.
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Pump Details Capital Cost($)
OperatingCost ($/year)
EquipmentLife (years)
Carbon Steel Pump 8,000 1,800 4
Stainless Steel Pump 16,000 ( ) 1,600 ( ) 7 ( )
For carbon steel pump , . + .+ .
= 0.302 and so
A = $0.302x8000 = $2417.
Cost Estimation Profitability Analysis
For stainless steel pump, . + .+
= 0.192 and
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p p,+ .
so A = $0.192x16000 = $3072.
EAOC for carbon steel pump = $1800 + $2417 =$4217.
EAOC for stainless steel pump = $1600 + $3072 =$4,672.
Which pump do you recommend?
Cost Estimation & Economic Evaluation
Contents
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Introduction
Cost Estimation Estimation o f Capi tal Cost Est im at ion o f Manufactur ing Cost
Economic Evaluation Engineer ing Econo m ic An alys is Profi tabi l i ty An alysis
Summary
104
Cost Estimation Economic Evaluation
Learning Outcomes
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The student should be able to:
105
Estimation of Capital and OperatingCosts1. Describe
Capital and Operating Costs ofEquipment and Plants2. Estimate Time Value of Money, Cash Flow
Diagram, Depreciation andProfitability Criteria
3. Describe
Economic Evaluation of Equipmentand Plants4. Perform
Thank You and All the Best
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