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Lecture NotesTKP4165 Process

Design

Magne Hillestad

About the course

Process Design

Projects

Optimization inDesign

Flow Diagrams

Mass and EnergyBalances

Flowsheeting

FlowsheetDevelopment

UniSim/HysysSimulation

Heat Integration

Economy

Sizing ofEquipment

Example - Estimating cost of major equipment

1. Estimate the cost, as of 2011, of a kettle type heatexchanger for a distillation column boiler where bothshell and tubes are made of carbon steel. The requiredheat transfer area is 200 m2 and the design pressure is45 bar

2. Estimate the cost, as of 2011, of a centrifugal pumpwith a delivery of 300 gal/min (gpm) at a dischargepressure of 150 psi. The inlet pressure is 20 psi. Thepump has to be made of stainless steel. You know thata pump of similar design in carbon steel and delivering800 gpm at a pressure difference of 50 psi had a cost ofUSD 8000 in 1999. Assume a capacity exponent ofn = 0.65 based on the capacity measured as (delivery xpressure difference). The material factor is 1.3

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Example - Estimating cost of major equipmentpossible solution

1. Kettle type heat exchanger from different sources:I In the previous edition of the textbook we read from Figure 6.3b graph 1 (carbon steel)

USD 55,000 where we have multiply with 1.3 to get a kettle type exchanger andpressure factor 1.3 (30-50 bar). And including cost index from mid 2004 to 2011 the

cost will be 55000 · 1.3 · 1.3 · 585.7444.2

= 122600I In the textbook Table 6.6, under heat exchangers, we can read data for a U-tube kettle

reboiler (as of Jan 2007). The validity is from 10-500 m2:

C = 25000 + 340(200)0.9 = 65032. There is no mentioning of pressure, so weassume this is low pressure (1-10 bar) and we multiply with 1.3 as pressure factor.

Projection of cost from 2007 to 2011 is 585.7509.7

and the cost is USD 97,150I By logging on to the web-site ”http://matche.com/EquipCost/” for a kettle reboiler of

2153 ft2 and internal pressure of 600 psi made of carbon steel, the cost is USD 109,200as of 2007. Projected to 2011 the cost is 585.7

509.7109200 = 125500.

2. The cost of a centrifugal pump. The cost model is C2 = C1(S2S1

)n. The scaling variable is

S = q∆p (volume flow by pressure difference). S2 = 300(150− 20) and S1 = 800 · 50,both are on the same units. The 1999 cost of the stainless steel pump is

C2 = 8000(300(150−20)

800·50 )0.651.3 = 10230. Projected to 2011 the cost is estimated to be585.7390.6

10230 = 15340

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Sizing ofEquipment

Example - Estimating cost of major equipment

I Calculate the cost, as of 2011, of the a shell and tubeheat exchanger , floating head, made of carbon steelshell and stainless steal tubes and with an area of 100m2. The design pressure is 45 bar.

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Sizing ofEquipment

Summary Invesment

I The basis for sizing equipment is the flowsheetcalculation, mass and energy balances.

I Size the equipment and select material of construction

I Estimate the purchase cost of major equipment

I Calculate the total fixed capital cost using factors,direct and indirect cost

I Estimate the working capital 10-20 % of ISBL

I Total investment (Fixed + Working capital)

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Heat Integration

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Revenues

I Revenues are the income generated from sales of mainproduct and byproduct.

I The problem statement usually tells us the mainproduct, but determining byproduct value can be tricky.

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Revenues

Every compound leaving the process is either a product or awaste stream

XYZ Co.

Raw

Materials

Byproduct or

Waste?

Main Product

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Estimating Waste Disposal Costs

I Streams that can be burned have fuel value based onheat of combustion (see later slide for fuel prices inUSD/MMbtu)

I Inert solids can be costed as landfill (USD 50/ton) orused to make roads, etc.

I Concentrated liquids (& some solids) are costed ashazardous waste

I Dilute aqueous streams are sent to wastewatertreatment (with some exceptions) and costed as extrawastewater capacity (USD 6 per 1000 gal)

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Sources of Price DataI ICIS Chemical Business Americas

I Lists prices for 80 chemicals and 44 fuels online atwww.icispricing.com, updated weekly

I Chemicals, minerals and natural raw materialsI Multiple product grades, spot/contract, markets for

someI Chemical Week (Access Intelligence)

I 22 commodity chemicals, contract & spot, U.S. andEurope

I Oil & Gas Journal (Pennwell)I Lists prices for crude oils, natural gas, fuel productsI Updated weekly

I On-line ResourcesI Commodities (crude oil, gasoline, copper, soy beans,

etc.) & their futures contracts are traded and listed inreal time on major market sites (e.g. Barron’s atwww.online.WSJ.com )

I Chemical Market Associates www.cmaiglobal.com has15-year archive for major petrochemicals, polymers andsome commodities

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Operating Costs

1. Variable operating costsI Raw materialsI Handling of byproducts (Avfallsbehandling)I Miscellaneous material (Driftshjelpemidler)I Utilities (Energi; el. damp, vann)I Packing and shippingI Consumable (Solvents, additives, etc)

2. Fixed operating costsI Operating labour (Driftsoperatører, ++)I Maintenance labour and material (Vedlikehold)I Laboratory costs (Laboratoriekostnader)I Plant overheads (Felleskostnader)I Sales & marketing (Salgskostnader)I Property taxes (eiendomsskatt ++)I AssurancesI Capital wear and tear

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Variable Operating CostsRaw material

I This is where a process model can be useful!I We must account for all raw materials that are

consumed:I In forming the main productI In forming byproducts and waste streamsI As unreacted raw material species in any stream leaving

the plantI In any stream generated by the process that is used as

fuel

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Variable Operating CostsUtility Consumption

I The utility consumption of a process is difficult toestimate without completing the design (after heatintegration)

I Heater duties are set by heat integration, equipmentand control issues

I Steam consumption & production are determined byheat recovery design

I Cooling requirements are also set by heat recovery

I Electric power consumption is mainly determined byhydraulics and solids handling

I There may be opportunities for utility export and powerrecovery that off-set utility costs

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Simplified Steam System

I We can expand from HP to MP to LP in steam turbinesinstead of throttling

I Hence recover shaftwork that can be used in the processor converted to electricity

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Sizing ofEquipment

Fixed operating costs

I LaborI Operators, supervisors, etc.

I Maintenance - workshopI Plant overhead

I Finance, Sales, IT, R&D, etc.

I Property taxes & insurance

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Heat Integration

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Sizing ofEquipment

Fixed operating costsOperators

I For a plants running 24-7

I 5-6 shifts

I Number of operators at each shift can be correlated tothe number of major processing units.

Noperators = (6.29 + 0.23Nunits)0.5

I The processing units include - compressors,towers/columns, reactors, boilers, heat exchangers etc.

I The total labor is about the double of the number ofoperators

I The cost of labor includes direct (salary, sick leave, )and indirect cost (social sec)

I Cost data can be found on SSB (Norway), depends ontype of industry, oil/gas, chemical processing

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Fixed operating costsOperators

I On a USGC basis, a shift position salary of USD 40000/yr is typical

I Supervision is typically taken as 25% of operating labor

I Maintenance - 3-5% of ISBL investment

I Plant overhead - 65% of (labor + maintenance)

I Property taxes & insurance - 2% of fixed capital

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Fixed operating costsSimilar cost

I Basis operatørkostnad (a)

I Driftsledere - 15-25% av a

I Vedlikehold (drift) - 2-20% av a

I Laboratoriedrift - 10-20% av aI Felleskostnader - 20-40% av a

I HelsevernI SikkerhetI KantinedriftI VelferdstiltakI Generelle ingeniørtjenesterI PortvaktI Transporttjenester mm

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Heat Integration

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Sizing ofEquipment

Economic Evaluation of Projects

I How do we decide whether a project is worth pursuing?

I How do we decide if a project is better than thealternatives?

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Heat Integration

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Sizing ofEquipment

Economic Evaluation of ProjectsImportant concepts

I Project Cash Flows

I Simple Measures of Economic Return

I Taxes

I Depreciation

I Time Value of Money

I Discounted Cash Flow Methods

I Annualized Costs

I Sensitivity Analysis

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Cumulative Cash Flow

cum

ula

tive

cash

flow

time

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Cumulative Cash Flow

I Initially, money is spent on engineering to design the plant

I This may take 6 to 24 months

I The rate of spending then increases significantly during procurement, construction and start-up

I This typically takes 1 to 2 years

I When the plant is ready to start up, the working capital is committed

I This represents the maximum investment

I The plant typically runs at reduced capacity in its first year

I The plant then runs steadily, bringing in a positive cash flow

I The time taken to recover the total capital invested is called the pay-back time

I Towards the end of plant life, the profitability starts to fall off as operating costs increase

I It may also be harder to sell product if competitors have developed lower cost routes oralternative products

I Finally, at the end of the plant life, the working capital is recovered

I The scrap cost of the plant may also be recovered, but this is usually not considered, as thereare often site remediation costs

I Cash flow diagrams are conceptually simple and give an understanding of what goes on duringthe life of a project

I They are not used in industry as they all look the same and they don’t convey quantitativeinformation efficiently

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Sizing ofEquipment

Quick Measures of Economic Return onInvestment

I The simplest measures of economic return are ratiosthat can be estimated quickly without worrying abouttaxes, depreciation or schedule of cash flows (time valueof money)

I Simple pay-back time = Total investment / Averageannual cash flow

I Return on investment = Average annual cash flow /Total investment

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Example

I

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Depreciation

I Depreciation is a non-cash expense that reduces incomeand therefore less tax

I The depreciation charge is not a real cash flow (nomoney is paid to anyone)

I Depreciation can be thought of as an allowance for the”wear and tear, deterioration or obsolescence of theproperty” as a result of its use.

I Different countries have different rules, so thedepreciation method that applies may be different forinternational projects

I Only capital assets (plant, equipment, buildings,software, intellectual property) can be depreciated (notworking capital - not land)

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Depreciation

I Taxable income I = P −DP Gross profitD Depreciation

I Taxes T = I · trtr Tax rate

I Depreciation is added back to income after taxes to givecash flow after tax

CF = P · (1− tr) +D · tr

I So the effect is to increase after-tax cash flow, andhence make investments more attractive

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Net Cash Flow

I The fixed capital investment

I Working capital

Annual revenues Year 1 Year 2 ..− Annual operating costs Year 1 Year 2 ..

= Gross profit (P ) Year 1 Year 2 ..− Depreciation (D) Year 1 Year 2 ..

= Result before tax Year 1 Year 2 ..− Tax (tr = 0.28) Year 1 Year 2 ..

= Result after tax Year 1 Year 2 ..+ Depreciation Year 1 Year 2 ..

= Net cash flow (CF ) Year 1 Year 2 ..

I The net cash flow is

CF = P (1− tr) +Dtr

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Depreciation

I Depreciation can be calculated in many different ways

I In all methods, the book value is the initial cost of theproperty minus the accumulated depreciation charges

Bm = C −m∑i=1

Di

Bm book value in year mC initial cost of depreciable assetDi depreciation charge in year i

I The book value is not the same thing as the re-sale ormarket value of the asset

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Straight Line Depreciation

I In straight line depreciation the depreciable value, C, iswritten off over the asset life (n years) at a constantrate

Di =C

n

I The book value is Bm = C − mn C

I When m = n, the book value is equal to zero or thesalvage value of the asset, and no further depreciationcharge can be taken

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Declining Balance Depreciation

I In declining balance depreciation, the depreciationcharge is a fixed fraction, a, of the book value

D1 = C · aB1 = C −D1 = C · (1− a)

D2 = B1a = C · (1− a)a

B2 = B1 −D2 = C(1− a)2

hence

Dm = C(1− a)m−1a

Bm = C(1− a)m

I This method gives higher depreciation charges in theearlier years of an asset life, which helps investmentspay out sooner

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Time Value of Money

I Which is worth moreI 500 kNOK now orI 500 kNOK in two years time ?

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Time Value of Money

I Money that is available now is more valuable to acompany than the same amount in the future, becausethe money could be invested and earn interest

I Interest rate p (e.g. 12% p = 0.12I After one year: C0 · (1 + p) = C0 · 1.12I After five years: C0 · (1 + p)5 = C0 · 1.76I After n years: C0 · (1 + p)n

I Conversely, C0 · 1.12 in one year from now is worth thesame as C0 today

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Time Value of Money

I To allow for time value of money, future cash flows arediscounted to convert them into a present value

I Present value = future value × discount factor

I The discount factor is just the inverse of the compoundinterest rate (1 + p)n

I Present value = future worth in year n / (1 + p)n

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Time Value of Money

I Discounting future cash flows allows for the fact thatmoney available earlier in the project life could earnmore interest than money available later in the project

I The time value of money is not the same as inflation(although the math is similar)

I Inflation is a rise in the general level of prices, measuredagainst a standard level of purchasing power

I Chemical companies usually do not allow for inflation ineconomic analysis, as it is assumed that inflation affectsall prices equally

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Net Present Value

I We can discount future cash flows to find their presentvalue

I The sum of the present values is then the Net PresentValue (NPV) of the project

I The net present value

NPV =n∑i=0

CFi(1 + p)i

Time (yrs)

Cu

mu

lati

ve

Cas

h F

low

Time (yrs)

Cu

mu

lati

ve

Cas

h F

low

CF

PV of CF

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Net Present Valueis one of the most widely used economic measures

I It captures time value of money

I It captures the value of investment incentives such asaccelerated depreciation schedules

I It captures variations in construction schedule andoperating rate

I It allows for price forecast models that include cyclicbehavior

I It can help understand the best timing for investmentsas well as investment scope

I It is easily coded into spreadsheets

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Net Present Value

I The only drawback of NPV is that bigger projects withlarger capital investment have bigger NPV

I It can’t be used for optimization unless an upper boundis set on plant size (as bigger is always better)

I Sometimes companies get around this by usingNPV/capital investment as a measure to compareprojects

I The DCFROR (also known as internal rate of return,IRR) is the interest rate at which the NPV is equal tozero at the end of the project life

I A more profitable project will have a higher DCFROR

I DCFROR is more useful than NPV for comparingprojects of different size

I DCFROR is directly analogous to an interest rate, andso allows projects to be compared with otherinvestments

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ProfitabilityExample 1

I Et prosessanlegg produserer 100 kton/ar (produkt)

I Utbyttet er 0.8, d.v.s. 1 kg ravare gir 0.8 kg produkt

I Du har utviklet en ny reaktor som gir et utbytte pa 0.9

I Alle kostnader knyttet til installasjon av ny reaktor erestimert til a være 45 MNOK

I Ravarepris er 1.0 krone pr. kg mens produktprisen er2.0 kroner pr. kg.

I Dersom forbruket av ravarer holder konstant, hva ertilbakebetalingstiden av denne produksjonsøkningen?

I Hva er arlig avkastning (ROI)?

I Hva er NPV av investeringen (bruk 10 ars horisont ogrente pa 15%)

I Hva er internrenten av prosjektet?

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Project Invest NCF (MNOK)(MNOK) 1 2 3 4 5

A 100 30 30 30 30 30B 100 10 20 30 40 50

Calculate

I Return on investment

I Payback time

I Net present value

I Internal interest rate

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Project Capital FuelCost savingsmill MMBtu/h

A 1.5 15B 0.6 9C 1.8 16D 2.2 17E 0.3 8

I A process heat-recoverystudy identifies 5potential modificationprojects

I All projects areindependent and can berealized

I Fuel costs is USD6/MMBtu and the plantoperates 350 days/year

I Which project has thesimple pay-back timelower than 1 year?

I If all projects are realizedimmediately, what is theNVP with 15% interestand 10 years horizon.

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Annualized Costs

I An alternative way of comparing a current capitalexpense with a future cash flow is to convert the capitalexpense into a recurring annual capital charge

I This method is useful when comparing the costs ofassets with different life

Example:

I A stainless steel reactor costs USD 500,000 and lasts 6years. A carbon steel reactor costs USD 300,000 andlasts 3 years. Which is better?

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Annualized Costs

I Costs are annualized by using an amortization function(similar to the method used for paying off a homemortgage with interest)

I This function is derived by calculating the constantamount of money that you would have to pay into afund each year such that at the end of a fixed term theprinciple and interest would all be paid off

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Annualized Costs

I If we invest an amount A each year, also at interestrate p, then it matures to a sum S after n years.

S = A(1 + p)n−1 +A(1 + p)n−2 + · · ·+A(1 + p) +A

I By multiplying with (1 + p)

S(1+p) = A(1+p)n+A(1+p)n−1+· · ·+A(1+p)2+A(1+p)

I By subtracting the two:

Sp = A(1 + p)n −A

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Annualized Costs

I If the annual payments A have accumulated to thesame value as investing I at interest rate p, then:

S = I(1 + p)n =A

p[(1 + p)n − 1]

I The annualized cost A of an investment I is

A =I[p(1 + p)n]

[(1 + p)n − 1]

I So we can define an annual capital charge ratio, ACCRas:

ACCR =A

I=

[p(1 + p)n]

[(1 + p)n − 1]

I This is the fraction of the principal (investment) I thatmust be paid each year to recover the investment at thetarget interest rate

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Annualized Cost

Interest rate p ∗ 100 ACCR (n = 10) ACCR (n = 20)

10 0.163 0.11712 0.177 0.13415 0.199 0.16020 0.239 0.20525 0.280 0.25330 0.323 0.302

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Annualized CostExample

I A stainless steel reactor costs USD 500,000 and lasts 6years. A carbon steel reactor costs USD 300,000 andlasts 3 years. Which is better?

I If the interest rate is 15%, then

I For stainless steel reactor (6-yr life)

ACCR =[0.15(1 + 0.15)6]

[(1 + 0.15)6 − 1]= 0.264

I Annual cost is ACCR × initial cost, being USD 132,100

I For carbon steel reactor (3-year life)

ACCR =[0.15(1 + 0.15)3]

[(1 + 0.15)3 − 1]= 0.438

I Annual cost is ACCR × initial cost, being USD 131,400

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ACCR rule of thumb

I Remember from previous lectures that OSBLinvestments, engineering costs, contingency, workingcapital and several fixed costs are often estimated asratios of ISBL investment

I If we have 15% interest and 10-year life, ACCR = 0.199

I If OSBL is 0.4 ISBL, Engineering is 0.1 ISBL, andworking capital is 0.15 ISBL, then total capital is 1.65ISBL.

I Annual capital charge is 0.199 × 1.65 ISBL = 0.328ISBL

I This is the basis of the widely used rule of thumb”annualize capital costs is ISBL cost divided by three”

I Note: No contingency and no fixed costs related tocapital

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ACCR rule of thumb

I If OSBL = 0.4 ISBL, Engineering = 0.1 (ISBL+OSBL), Contingency = 0.15 ISBL and working capital= 0.15 ISBL, then: Total capital is 1.84 ISBL

I Annual capital charge = 0.199 × 1.84 ISBL = 0.366ISBL

I If we assume maintenance cost is 3% of fixed capital(ISBL + OSBL), property tax and insurance 2% andplant overhead 65%, then the fixed costs are: (0.03 +0.02) × 1.65 × (1 + 0.4) ISBL = 0.116 ISBL

I Hence the total annualized cost due to ISBL capital ismore like 0.482 ISBL

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Annualized Capital Charges

I Be very careful when using ACCR rules of thumb, andmake sure you use the right factor depending onwhether you are working from ISBL, fixed capital(=ISBL + OSBL) or total capital investment

I Also remember to use the appropriate discount rate foryour company, p = r + i+ u, (interest rate + inflation+ uncertainty)

I Annualized capital charges can also be added to CCOPto give a total cost of production. In this case the totalinvestment should be multiplied by the ACCR to givethe annual capital charge.

TCOP = CCOP + ACCR · Investment

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Break Even DiagramWhen the plant need to reduce production

I May be due to production overcapacityI Fixed operating costs that cannot be avoided

(grunnkostnad)

Production / Capacity

5025 75 1000

Fixed operating costs:Labor, overhead, ++

Sales revenues

Total operating costsBreak-even point

Shut-down point

Fixed operating costs:Insurance, property tax, capital wear

Mill

NO

K / y

ear

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Operating Costs

1. Variable operating costsI Raw materialsI Handling of byproducts (Avfallsbehandling)I Miscellaneous material (Driftshjelpemidler)I Utilities (Energi; el. damp, vann)I Packing and shippingI Consumable (Solvents, additives, etc)

2. Fixed operating costsI Operating labour (Driftsoperatører, ++)I Maintenance labour and material (Vedlikehold)I Laboratory costs (Laboratoriekostnader)I Plant overheads (Felleskostnader)I Sales & marketing (Salgskostnader)I (G) Property taxes (eiendomsskatt ++)I (G) AssurancesI (G) Capital wear and tear

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Simple Sensitivity Analysis

Base 110% 120% 130% 140% 60% 70% 80% 90%

0

10

20

30

40

NPV (MM$) Gross margin

Fixed costs

Capital cost

Utility costs

I Gives a qualitative indication of the relative sensitivity ofNPV or DCFROR to major cost or process parameters

I Usually plotted as base ± a percentage

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Sizing of equipment

I Here, only selected equipment will be shown

I You should be able to lookup design and read designmethods

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Shell and Tube Heat Exchangers

I

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Shell and Tube Heat Exchangers

I

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Heat Exchanger Design

I Heat exchange design must:I Provide required areaI Contain process pressureI Prevent leaks from shell to tubes or tubes to shellI Allow for thermal expansionI Allow for cleaning if fouling occursI Allow for phase change (some cases)I Have reasonable pressure drop

I S&T heat exchangers are built to standards set by theThermal Exchanger Manufacturers Association (TEMA)

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Temperature Cross

I When Th,out < Tc,out we have a temperature cross.

I Temperature cross causes problems if exchanger is notcounter-current and gives low Ft factors

I If Tc,out − Th,out > 0.05∆TLM, then Ft < 0.8 and it isusually best to split the exchanger into multiple shells inseries

I Number of shells can be estimated by stepping off onT-H diagram

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Temperature Cross in Simulation

I Most simulatorsshow an error ifthere is a low Ft

factor

I For example, inUniSim Design theexchanger shows upyellow

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Selection of Sides

Process Fluid Side Selection Reason

Fouling fluid Tube Easier to cleanViscous fluid Shell Lower ∆pSuspended solids Tube No dead spots for settlingHighest Temp Tube Cheaper, mechanically strongerHighest pressure Tube Cheaper, mechanically strongerCooling water Tube Easier to cleanCorrosive fluid Tube Cheaper, easier to replace tubeMuch larger flow Shell Lower ∆pCondensing fluid Shell Drains better

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Heat Exchanger Design

1. Determine duty, check for temp cross,Q = Wcp∆T + δW∆Hvap

2. Estimate U and hence calculate area,A = Q/(UFt∆TLM)

3. Determine exchanger type and tube layout

4. Pick d, L and calculate number of tubes, hence shelldiameter

5. Calculate hi, ho and confirm U . Return to 2 if needed.

6. Calculate ∆p. Return to 3 if needed

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Standard Dimensions of Steel Tubes

I As a guide 19 mm (3/4 inch) is a good trial diameterwith which to start design calculations

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Plug Flow Reactor

V

W,ω0,k

V + dV

W,ωk

I Let Rk be the formation rate of component k on massbasis [kg/(sm3)] and W be the total mass flow rate[kg/s].

I Mass balance for component k:

0 = (Wωk)V − (Wωk)V+dV + RkdV

I Simplified since total mass flow rate is constant (massconservation)

WdωkdV

= Rk

I Given initial condition as feed composition ωk(0) = ω0,k

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Plug Flow Reactor

I The energy balance can be simplified:

WcpdT

dV=∑j

(−∆H)jrj − Ua(T − Tw)

I Where cp =∑

k ωkcp,k

I Component reaction rate [kmol/(m3-s)]; Ri =∑

j νi,jrj

I And component reaction rate [kg/(m3-s)]; Ri = MiRi

I The heat transfer area a = A/V ; for multiple tubesa = 4/d

I The overall heat transfer coefficient U .

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Plug Flow ReactorProgramming in Python or Matlab

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Gas-Liquid SeparatorsChapter 10.9

I Vertical SeparatorsI When V/L is highI Typical knock out of liquid from a gas

I Horizontal SeparatorsI When long liquid hold-up time is requiredI When V/L is low

I Settling velocity (terminal velocity)

ut = 0.07

(ρL − ρV

ρV

)1/2

I If NOT demister us = ut · 0.15 to be on the safe side

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Vertical SeparatorsChapter 10.9.2

I The vessel diameter Dv and thevolumetric gas flow rate VV ⇒ gasvelocity

I When the gas velocity is equal todroplet settling velocity

π

4D2

vus = VV

I The diameter is usually roundedup to the nearest standard vesselsize so that standard vesselclosures can be used; see Section13.5.

I The liquid hold up about 10 min

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Vertical SeparatorsExample 10.7 from textbook

Make a preliminary design for a separator to separate a mixture of steam and water; flow rates: steam2000 kg/h, water 1000 kg/h; operating pressure 4 bar.

I From steam tables, at 4 bar, saturation temperature is 143.6 C, liquid density is 926,4 kg/m3,vapor density is 2.16 kg/m3.

I Droplet settling velocity ut = 0.07( 926.4−2.162.16

)0.5 = 1.45m/s

I Demister is not required; us = 1.45 · 0.15 = 0.218m/s

I Vapor liquid flow rate: 20003600·2.16 = 0.257m3/s

I The diameter of the vessel: D =√

4·0.257π·0.218 = 1.23m

I Round to nearest vessel size (4ft) is 1.25m

I Liquid volumetric flow rate: 10003600·926.14 = 3.0 · 10−4m3/s

I Allow for a liquid residence time of 10 min.

I Required liquid volume: 3.0 · 10−4 · 10 · 60 = 0.18m3

I Liquid height: 0.18π·1.252/4

= 0.15m

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Horizontal SeparatorsChapter 10.9.3

I The most economical lenght/diameter given by pressureOperating pressure, bar Length/diameter ratio

0− 20 320− 35 4> 35 5

I Design criterion: droplets in vapor should have enoughtime to settle into liquid phase

I Vertical droplet residence time equal (or less then)horizontal gas residence time.

I Liquid residence time should at least be 10 min.304 / 306


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Horizontal SeparatorsExample 10.8 from textbook

Design a horizontal separator to separate 10,000 kg/h of liquid (density 962.0 kg/m3), from 12,500

kg/h of vapor (density 23.6 kg/m3). The vessel operating pressure will be 21 bar.

I Droplet settling velocity; ut = 0.07{(962− 23.6)/23.6}0.5 = 0.44m/s

I Separator without demister us = ut0.15 = 0.066m/s

I Take hV = 0.5D and L/D = 4

I Vertical droplet residence time: hV/us = 0.5D/0.066 = 7.58D

I Horizontal gas residence time: L/uV

I Vapor cross section area: πD2

40.5 = 0.393D2

I Volumetric vapor flow: 125003600·23.6 = 0.147m3/s.

I Vapor velocity: uV = 0.1470.393D2 = 0.374D−2

I Horizontal gas residence time: L/uV = 10.70D3

I For satisfactory separation, requires: 7.58D = 10.70D3, gives D = 0.84m, say D = 0.92m(3ft, standard pipe size).

I Check liquid residence time: L/uL = VL/qL

I Liquid volumetric flow rate: qL = 100003600·962 = 0.00289 m3/s.

I Liquid cross section area: π0.922

40.5 = 0.332m2

I Liquid volume: 4D0.332 = 1.23m3

I Liquid residence time: 1.230.00289

= 426s, which is 7 min

I Increase the liquid residence time by increasing D

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Distillation ColumnDiameter

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example - estimating cost of major equipmentfolk.ntnu.no/andersty/4. klasse/tkp4165... ·...

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