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CONFIDENTIAL EH/JAN 2013/CHE604

UNIVERSITI TEKNOLOGI MARAFINAL EXAMINATION

COURSECOURSE CODEEXAMINATIONTIME

PLANT DESIGN AND ECONOMICSCHE604JANUARY 20133 HOURS

INSTRUCTIONS TO CA NDIDATES1 . This question paper consists of FIVE (5) questions.2. Answer ALL the questions only in the Answer Booklet. Start each answer on a new page.3. Do not bring any ma terial into the exam ination room unless perm ission is given by theinvigilator.4. Please check to make sure that this exa mination pack consists of:

i) the Question Paperii) a one - page Append ix 1iii) a one - page Append ix 2iv) a one - page Append ix 3v) a one - page Append ix 4vi) a three - page Appendix 5vii) a one - page Append ix 6viii) a one - page Appendix 7ix) a one - page Appendix 8x) a one - page Appendix 9x i) a o ne -p ag e Appendix 10xii) an An sw er Book let - provided by the Faculty

DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SOThis examination paper consists of 5 printed pages Hak Cipta Universiti Teknologi MARA C O N F ID E N T IA L

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CONFIDENTIAL 2 EH/JAN 2013/CHE604

QUESTION 1( P 0 1 . P 0 2 , C 0 1 . C 1 . C 4 )a) The process flow diagram (PFD) contains the bulk of the chemical engineering datanecessary for the design of a chemical process. For all of these diagrams, there is nouniversally accepted standard even though they convey very similar information.Describe the information that should be included in a typical comm ercial PFD

(8 marks)b) A pipeline conne cting a separator and a tank contain six standard radius elbow , a globevalve that is half open and a gate valve that is fully open. The line is made of commercialsteel pipe with 30mm internal diameter and 130 m length. The viscosity of the fluid is0.62 mNM"2 and density is 874 kg/m 3. The steady state flow rate of the fluid in the

pipeline is 4000 kg/h.i) Calculate the miscellaneou s losses of the pipeline (9 marks)ii) Using the equivalent pipe diame ters me thod, calculate the total pressure drop due tofriction during the steady state condition (3 marks)

QUESTION 2( P 0 6 , C 0 1 . C 0 2 , C 3 .C 4 )

A distil lation column is to be designed to separate the chemical mixture, whose molarcomposition and relative volatilities are given in Table 1 below. The feed is at its boilingpoint. Propane, C3 and iso-pentane, i-C5 are identified as the light and heavy keycomponents in the mixture.Table 1 Molar composition and relative volatilities of the chemical mixture

PropaneIso-butaneN-butaneIso-pentaneN-pentane

Feed (moles)155202535

Distillate (moles)1551240

Bottoms (m oles)0019135

Relativevolatility2.65120.85

a) Determine the minimum number of stages required for the separation using Fenskeequation. (3 marks)b) Evaluate minim um reflux ratio required using Underwood equation. (10 m arks) Hak Cipta Universiti Teknologi MARA C O N FI D E N T IA L

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CONFIDENTIAL EH/JAN 2013/CHE604

c) If the reflux ratio used is 4 , then determ ine the num ber of stages by using E rbar-Maddoxcorrelation. (3 marks)d) Using Kirkbride equ ation, determine the feed point location. (4 marks)

QUESTION 3(P05, P06, C03, C5)

Perform a Problem-Table analysis to estimate pinch temperature and synthesize a heatexchanger network (HEN) for the Reactor-Separator system shown in Figure 1 such thatthe resulting heat exchanger network requires minimum hot and cold util it ies. Assume theminimum approach tem perature is 20C. The units for the CP values are in SI .

20C CP = 5 0Reactorfeed

- H H W 160CReactor

H)nI"* % CP=40

l S C C ^ - ^ Overhead

120CReactoreffluent

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QUESTION 4( P0 1 . P0 5 , P0 6 , C0 1 . C4 )a) Determ ine the bare module cost for a long tube evapo rator with the following operatingconditions for the year 201 2. Maximum operating pressure (tube side) = 40 barg Ma ximum operating pressure (shell side) = 5 barg Ma terial of construc tion = stainless steel Heat exchange area = 200 m 2

(15 marks)b) Explain the difference betwe en direct costs, fixed costs and general expenses. G ive oneexample of each.

(5 marks)QUESTION 5(P 01 .P0 2 , C04 , C4, C6 )

a) The an nual variable production costs of a vinyl chloride plant is RM 980,000. This plantcurrently operates at 85% of full capacity. On top of that, the sum of annual fixedcharges, overhead costs and general expenses is RM700,000 and considered not tochange with the production rate. Total annual sales of the plant are RM1.96 million andthe product sell for RM 1 4,000 per tonne.i) Calculate the break-ev en point (8 marks)ii) Determ ine the gross annua l profit with depreciation (2 marks)iii) Determ ine the net annua l profit at 1 00% capacity if the incom e tax rate is 25% of grossprofit. (2 marks)b) A consultant company just finished with few alternative designs of processes to produceformaldehyde from methanol. Cost data for these alternatives are shown in Table 2.

For all parts of this problem, assume that the required before-tax return on investment is1 5% per annum and the equipment life is considered to be 12 years.i) As the person-in-charge of doing the evaluation, do you recommend the construction ofthis plant using the base case design? (4 marks)ii) Th ere are two (2) alternative processes that require more expensive equipments, asindicated in Table 2. Recommend whether these two (2) alternatives profitable or not byusing incremental analysis. (4 marks) Hak Cipta Universiti Teknologi MARA CONFIDENTIAL

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Table 2 Details on projects considered

Capital investmentAdditionalinvestmentAnnual productrevenueAnnual raw m aterialcostsAll other annu aloperating costs

Base Case D esign(USD 106/year)11 .9

-11 . 13.16

0.299

Project B(USD 106/year)-

1.0211 .13.160.206

Project C(USD 106/year)-

0.511 .13.160.258

END OF QUESTION PAPER

Hak Cipta Universiti Teknologi MARA CONFIDENTIAL

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CONFIDENTIAL APPENDIX 1 EH/JAN 2013/CHE604

TABLE 1.0 UNIT CONVERSION TABLE

Quantity' MassLength

Volume

ForcePressure

Energy

Power

Equivalent Valuest k g =* lOOOg = 0.QQ1 metr ic ton 22046215^= 35;27392ozllbfa = 16 pz .*=. 5 X l&f4. ton = 453j593ig = 0.453593 kgl m = 100 cm: 1000mm'== 1 06microns (fun) #= 10 1angstroms= 393 7 in. = 3.2808 ft - 1.0936 yd = O.0O06214mile1 ft = l 2 l m = l/3yd = 0304801 = 3048 011l m 3 = 1000 L = M ^cm 3 = 1 0 6 m L= 35.3145 ft3 = 220.83 imperial gaflon s= 264.17 gal= 1056.68 qt1 ft3 - 1728 in .3 = 7.4805 gal = 0.028317 m 3 = 28317 L= 28,317cm 3I N = 1 k g-m /s2 = 10s dynes - 10 5g-cm/s2 = 0.22481 lbf1 lb f = 32.174 Ibm-ft/s2 - 4A482.K = 4.4482 X W dynes1 arm = 1.01325 X 10 5 N/m 2 (Pa) = 101325 kPa = 1.01325 bar= 1M325X :10s dynes/cm2- 760m m H g at 0C (torr) == 10333 m H a O a t 4C= 146 96 Ibf/in^ (m ) = 33.9 ft H 2 0 a t 4C= 29.92 1in, Hg at 0C1J = lN-m = 107ergs = lQ 7dyne'Cm 2.778 X 10 ~ 7kW -h = 023901 cal= 0.7376 ft4bf = 9.486 X 1 0 ^ 4 B tu1 W = 1 J/s = 0.23901 cal/s = 0.7376 ft-lbf/a = 9.486 X i d * 4 Btu/s= 1.341 X 10~ 3 hp

(A)

' . -

*= i . c * * _ . _ / 2 .2 0 4 6 2 1 b m \Example: Th e factor to convert grams to lb m is _ ^ 1\ lOUQ g y


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CONF IDENTIAL APPEN DIX 2 EH/JAN 2013/CHE604

Table 1.1 Pipe roughnessMaterial Absolute roughness, mmDrawn tubingCommercial steel pipeCast iron pipeConcrete pipe

0,00150.0460.260.3 to 3.0

Table 1.2 Pressure losses in pipe fittings and valves (for turbulent flow)Fitting or valve K, number of number of equivalentvelocity heads pipe diameters45 standard elbow45 long radius elbow90 standard radius elbow90* standard long elbow90" square elbowTee-entry from legTee-entry into legUnion and couplingSharp reduction (tank outlet)Sudden expansion (tank inlet)Gate valve fully open1/4 open1/2 open3/4 openGlobe valve, bevel seat-fully open1/2 openPlug valve - open

0.350.20.6-0.80.451.51.21.80.040,51.00.15164168.50.4

151030-4023756090225507.580020040

300450IS


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9TJTJmzoXw

mi>zOWOXma>o

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CONFIDENTIAL APPEND IX 4 EH/JAN 2013/CHE604

too


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CONFIDENTIAL APPENDIX 5(1) EH/JAN 2013/CHE604

USEFUL FORMULAE

P = pMRT

is m4 if

A P , = 8 / pu

Re Divp

logN mm lOgOfLK

E f V -+ I* - ' a,- - 6

log AT, 0.2061og

Class 1 estimate: +6% to -4 % accurate10. C = KA"

11 . U vC V ^ W12. (T \

K* XJ


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USEFUL FORMULAE

12 .( A \C2 C j

0.6

\A b J( T \v A y

13 . log10 FP=C,+ C2 log10 P + C3 (log10 P)2

14. Cm=CP[B l+B2FPF u]15. log10 CP = Kx + K2 log10 A + K3(logw Af

16. COMd = 0.180FC7 + 2.73COL + 1 . 2 3 ^ +CWT+CPM )17. D M 7 = 0 ^ + ^ + 0 ^ . + 1.33COL + 0.069FCI + 0.03COMd18. FMC = 0.708COi + 0.068FC7 + depreciation

19. .EL40C = (Capital Inv estm ent) x ( , /, ne I + YOC

20 . G = 0.1 llCn, + 0.009FCI + 0.16COM,'0Z,21. ( i + 0 n - i

22 . F F = - F C 7 + C F x , i,

23 . p . ^ ( 1 + Q - - l.^ J /! + /"

24 . 8j~ sf coj25 . lS i = sjT_coi-dj26 . i ^=%a-eo27 4f = % + d /


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USEFUL FORMULAE

28 .

29 .

30 .

N PV = -FCIttcm + Net Revemte(PfAri*n)

I1VP = Additional Investment+ [Annual Op erating Cost Base Armwd Operating Cost Jfl( / ^ Un)

YearCEPCI

2011585.7

2010550.8

2009521.9

2008575.4

2007525.4

2006499.6

2005468.2

2004444.2

2003402

2002395.6

2001394.3


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Table A. l Equipment Cost Data to Be Used w ith Equation A. l {Continued)Equipment TypeDust Collectors

Evaporators

Fans

Fitters

> , ' * "

o

Furnaces

Equipment DescriptionBaghouseCyclone scrubbersElectrostatic precipitatorVenhiri scrubberForced circulation (pump ed)Falling filmAgitated film (scraped wall)Short tubeLong tubeCentrifugal radia lBackward curveAxial vaneAxial tubeBentCartridgeDisc and drumGravityLeafFa nPlate and frameTableTubeReformer furnacePyrolysis furnaceNonreactive fired heater

K*4.50073.6298&629S3.62985.02383.91195.00005,23664642035391334713.17613X11145,1055331074.81234,27563.8187481234,27565.10555.10553.06802385973488

*z-0.5818-04991-0.4991-0.4991

0.34750.86270.1490

-0.6S7203698

- 0 3 5 3 3-0.0734-0.1373- 0 3 3 7 5-05001-0.2403-0.7142-016480-0.3765-0.7142-O.64S0-03001-0.5001

0.65970.9721

- L 1666

K t0.08130.04110.04W0.04110.0703

-0.0088-0.0134

035000.00250,447703090034140.4722000010OO270.04200,07140.01760.04200.07140.00010.00010O194

-0O2O60.2028

Capacity, UnitsVolum e, nv*Volume, m 3Volume, m 3Volume, m 3Area, m 2Area,m 2Area ,m 2Area,m JArea, in 1Gas flowrate, m J / sGas flowrate, m*/sGas flowrate, m 3 / sGas f lowrate, m VsArea rm 2Area, m 2Area,m*Area, m 2Arearm 2Area ,m 2Area ,m 2Area,m JArea ,m 1Duty,kWDuty,kWDuty,kW

Min Size0-080.060.060.06

550as10100

1111

0.915

0.90,50,6OS0 50.90.9

300030001000

Max Size35 0200200200

1000500

5100

10,000100100100100115200300

8023 5300

8011 5115

100,000100,000100,000

(continued) I

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oozTable A.Z Pressure Factors for Process Equipment (Correlated from

Equipment TypeCompressorsDrives

Evaporators

Fans*

Fu rnaces

Heat exchangers

Equipment DescriptionCentrifugal, axial, rotary, and reciprocatingGas turbineIntern, comb, en gineSteam turbineElectricexplosion-proofElectrictotally enclosedElectricopen/drip-procrfForced circulation (pump ed), falling film,agitated film (scraped wall), short tube,and long tubeCentrifugal radial, a nd centrifugal backward curveAxial vane and axial tubeReformer furnacePyrolysis furnaceNonreactivefiredheaterScraped wail

Teflon tube

Data in Guthrie [ 1 , 2 } ,Q

000000000.1578

0G0000.140500.101700.134700.6072

13.14670

Q0000o.000-0.299200.2089900.208990

- 026980

-0.1957-P- 02368

o-0.9120 *-12.6574

0

and Ulrich8000000000.1413

0-0.03280

-0.032800.129300.094030040210033273.07050

[ 3 ] )Pressure Range (barg)

-------

P

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Table A.5 Equations for Bare Module Costai tdA.4

for Equipment Not Covered by Tables A.3

Equipment Type Equation for Bare Module CostComp ressors and blowers without drivesDrives for comp ressors and blowersEvaporators and vaporizersFans with electric d rivesFired heaters and furnaces

Power recovery equip meitlSieve trays, valve trays, anddemister pad s

T ower packing

FTfeQm saperheatisee&in jfcdorfarsteamlaereCF r - 1 far f c h e a J e a ^and is given byFT - 1 + 0.0Q1MAT- 0.00000335(aT)1

where ATk the amount of supedwat in "C

Where W is Ihe number of toys and F , is aquantity factor far (raysoaky given fey ->-logu F U 0.4771 + 0.08516Iog3() N -0.l4730ogwN)Jf6rN

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Table A.6 Iden tificatio n of Materia l Factors for Equipme nt Listed in Table A.5 to Be Used with FigureIdentification Number

1234567891011121314IS161718192021222324 ' *"

Equipment TypeCompressors/blowers

Drives for compressors and blowers-

Evaporators arid vaporizers

*v"*-

Equipment DescriptionCentrifugal comp ressor or blowerCentrifugal comp ressor or blowerCentrifugal com pressor or blowerAxial compressor or blow erAxial compressor or blowerAxial compressor or blow erRotary compressor or b lowerRotary compressor or blower Rotary compressor or b lowerReciprocating compresso r or blowerReciprocating com pressor o r blowerReciprocating compressor or blowerEiexmo"-*explosioxipjorElectrictotally enclos edElectricopen/dripproofGas turbineSteam turbineInternal combustion engineEvaporatorforced circ> short or longtubeEvaporatorforced ar c, short or long tubeEvaporator-^faroed circ, short or long tubeEvaporatorforced circ, short or long tubeEvaporatorforced cir c short or long tube -Evaporatorfalling film, scraped-wall

A.19Material of ConstructionCSSS *Ni alloyCSSSNi alloyCSSSNi alloyCSSSNi alloyzOWOIm

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Table 9.1 Commonly Used Factors for Cash Flow Diagram C alculationsConversion Symbol Common Name Eq. No. Formula

P to F (P[P, i, n) Single Payment Compo und Amo untFactorF to P (PIP, i, n) Single Payment Present Worth FactorA to F (F/A, i, n) Uniform Series Compound AmountFactor, Future Worth of AnnuityFtoA (A/F,i,n) Sinking Fund FactorP to A (AJP, i, n) Capital Recovery FactorA to P (P/A, i, n) Uniform Series Present W orth Factor,Present W orth of Annuity

(9.5)(9.6)(9.11)(9.12)(9.13)(9.14)

(1 + 0"1

(l + 0"(1 + if - 1

ii

(1 + 0" - 1i(i + 0"

. ( i + 0" - 1(l + 0" - 1

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