Cost Lectures

8/12/2019 Cost Lectures

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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


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

4

1

Feasibility Analysis Numerous Choices Few Months

2

Process & Equipment Design Many Choices 4-12 Months

3

Engineering & Construction Process Refinements 1-3 Years

4

Operation

Enhancements 10-20 Years

5

End of Plant Life Salvage Value When will it be?



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

5

What is cost estimation?What is economic evaluation?How do you estimate the cost and evaluateprofitability?



Introduction

6

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



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

7



Introduction

Learning Outcomes

The student should be able to:

8

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



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

9

I hear and I forget.I see and I remember.I do and I understand.

- Confucius

Study/SolveExercises andProblems in the

Text-Book



Introduction


Contents

Learning Outcomes

Text-Book

10



Cost Estimation & Economic Evaluation

Contents

Introduction



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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* 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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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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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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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 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 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

1

3

4

6

5

2

C t E ti ti P h d E i t C t


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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, 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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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



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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:



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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.



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CEPCI available on

the last page ofChemicalEngineering

technical magazine

(www.che.com)



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Variation of CEPCI

and CPI during1995-2013

Do you need CEPCI in the future?If required, extrapolate with caution.

1

2

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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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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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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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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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p

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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p

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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p

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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p

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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p

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



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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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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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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


Summary

67

Cost Estimation Engineering Economic Analysis


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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


D i ti


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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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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


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?

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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


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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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


Contents


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Contents

Introduction



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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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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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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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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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

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Year 1 2 3 4 5 6

Percent 20 32 19.2 11.52 11.52 5.76


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

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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?

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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.

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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)


End ofY k

DiscreteC h Fl

Discounted CashFl


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+




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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

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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




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Net Present Value,NPV

Cum ulat ive Discou ntedCash Flow at the End of

th e Proj ect = $17.12

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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


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




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Can NPV be negative?

Is large or small NPV preferred?

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Present Value Ratio

..

.

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End of Year, k

DiscreteCash Flow

Discounted CashFlow


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




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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

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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



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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?

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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%.

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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 ( )

Evaluating Equipment Alternatives



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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:

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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.


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?


Contents


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Introduction



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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Cost Lectures

Documents

Transcript of Cost Lectures