Design of Equipments

Assume feed is liquid at its bubble point =231.67 deg.So q=1Slope =(q\q-1) = ∞Composition of feedNaphthalene = 0.60 mole fraction (Z f )Toluene =0.0012 mole fractionDimethyl Naphthalene =0.3988 mole fractionFeed flow rate =1253.166 Kmole\hrComposition of DistillateX n = naphthalene mole fraction in distillate =0.9911X t = toluene mole fraction in distillate = 0.0027848X m =Dimethylnaphthalene mole fraction in distillate = 6.5x10- 3Flow rate of distillate = 532.31 Kmole \ hrComposition of bottom streamX w =0.3040X m1 = 0.6959Using material balanceF =D +WFZ f =DX d +WX w1235.166 x 0.6 =532.31 x0.9911 +Wx0.3040W = 703.046 Kmole \ hrSince the amount of toluene is very less , for design the tower club toluene withnaphthalene and design the tower as binary distillation column for naphthalene anddimethylnaphthalene

From graph: minimum ( X d \R m +1) = 0.53From graph :minimum reflux = 0.87Operating reflux ratio (1.5 R m ) = 1.305L = DR= 532.31 x1.035= 694.667 Kmole \ hrG =L +d= 694.66 + 532.31= 1226.97 Kmole \hrL = L +qF= 1767.428 Kmole \hrG = G +(q-1)F= 1226.97 Kmole \hrMolecular Wt. Of feed =139.124 Kg\ KmoleENRICHING SECTIONTOPBOTTOM694.67 Liq. (Kmole \hr ) 694.671226.98 Vap. (Kmole \hr ) 1226.98128.14 (M) avg liq 139.124128.14 (M) avg. vap. 139.1240.9911 X 0.600.9911 Y 0.80217.78 T liq in deg. 231.11218.33 T vap in deg. 231.1188929.16 Liq flow rate (Kg \hr) 96645.265157225.21 Vap. Flow rate (Kg\hr) 170702.36849.30 ρ liq. (kg \m 3 ) 833.703.17 ρ vap.( kg \m 3 ) 3.22

STRIPPING SECTIONTOPBOTTOM1767.42 Liq. (Kmole \hr ) 1767.421226.98 Vap. (Kmole \hr ) 1226.98139.124 (M) avg liq 147.47133.6 (M) avg. vap. 147.40.60 X 0.30100.80 Y 0.52231.67 T liq in deg. 245.56231.11 T vap in deg. 246.11245891.65 Liq flow rate (Kg \hr) 260642.63169342.53 Vap. Flow rate (Kg\hr) 180942.74833.70 ρ liq. (kg \m 3 ) 819.693.22 ρ vap.( kg \m 3 ) 3.45⎡ L ⎤⎡ρ⎢ ⎥⎢⎣G⎦⎣ρGL⎤⎥⎦0.5= 0.03455(at top)0.5⎡ L ⎤⎡ρG⎤⎢ ⎥⎢⎥ = 0.03518 (at bottom)⎣G⎦⎣ρL ⎦Maximum at bottom of enriching section. Hence all calculations are based on propertiesat bottom of enriching section

TRAY TOWER DESIGNAssume :Plate spacing , t s = 610 mmHole diameter , d h =5 mmHole pitch , l p =15 mmTray thickness , t T =3 mmHole areaPerforated areaA=Ah =p0.10Assume equilateral triangular pitchColumn diameter, Dc:Based on entrainment flooding.ll relations from Perry’s handbook, 6 th edition.⎡ 20 ⎤Fig. 18-10 , C sb = U nf⎢⎣ σ ⎥⎦C sb∴ σU nf= 0.0883 m\s0.2= 17.29 dyn\cm= 1.3990 m\s⎡ ρg⎢⎢⎣ρl− ρg⎤⎥⎥⎦0.5ft/s.Assume U n= 0.8 U nf= 1.119 m/s.Net area for flow , A n = A c - A d

170702Vapour flow rate = = 14. 56 m 3 /s.3600 x 3.22A n = 14.56 / 1.119 = 13.159m 2 .L w weir lengthLAssume w = 0. 75D⎛ θSin ⎜⎝cL⎞⎟ =⎠ Dwc2 c22θ c = 97.2 0πD 2 cA c = = 0.78522A d = 0.0878 D c2D cA n = 0.785 D c 2 – 0.0878 D c2D c =4.34 mA c = 14.78 mA d =1.626 m 2Active area, A a = A c – 2A d =11.52 m 2 .L w = 4.43 x.75 = 3.255 mLets take 3.3 mPerforated area,A pcorrected θ c = 99.o 0 Cα = 180 - θ c = 81 0 CArea of calming + distribution zone, A czA cz =2(L w * t) , t=thicknessA cz =2(3.3 * 120* 10 -3 ) =0.792 m 2 (that is 5.3% of A c )A wz = 2πD42cα π− (D360 42c− 0.052α)360

A wz = 0.0.348m 2 . (2.3% of A c )A p = A c –2 A d – A cz - A wz= 14.56-3.552-0.198-0.348 = 10.682 m 2A h = 0.1 A p = 1.0682 m 2No. of holes, n h =1.0682π4−3( 5x10) 2= 54500holes.Weir height, h w =50 mm (atmospheric pressure)Weeping check:TOP of enriching sectionVapor flow rate = 13.77 m 3 /s ( at top)Vapor flow rate = 14.72 m 3 /s ( at bottom)A h =1.0682 m 2U n = (13.77/1.0682) = 12.89 m/s.hd= K1+ KAssume sieve plates2K 1 = 0 , K 2 = 50.8/C vA h / A a =1.0682 /11.52 =0.09t T /d h = 0.6C v = 0.75K 2 = 90.33∴ h d =90.33 x (3.17 x12.89 x12.89) \849.14∴ h d =56.0 mm2ρgρUl2h⎡ q∴ h ow = F w x 664 x ⎢⎣ L W⎤⎥⎦2388929.16q 1 ( liquid load ) = = 0. 0293600x849L w = 3.255 mm 3 /s

Lq2.5WL= 4.92 W, = 0. 75DcwF w = 1.03h ow = 25.326 mmHead loss due to bubble formation ,⎡ σhσ = 409 ⎢⎣ ρld n⎤⎥⎦⎡ 17.29 ⎤h σ= 409⎢⎣833x5⎥⎦∴ h σ = 1.69 mm liq.Now, h d + h σ = 56+1.69 = 57.69mm liq.h w + h ow =50 + 25.32 =75.32 mm(A h / A a ) = 0.09Minimum value to avoid weeping , h d + h σ = 18mm ( from fig. 18-11)Since actual >minimum there is no weepingDowncomer flooding :h dc =h t + h w + h ow + h da + h hgHydraulic gradient :h d > 2.5 h hg mmLet h hg = 0.5 mmh ds = h w + h ow +h hg /2h ow =0.9 * 664 * (0.0325\3.255) 2\3h t = h d +h 1 lh 1 L = βh ds= 27.28mmh ds = 50 +27.28 +0.5\2 =77.53 mmU a = 1.27 m/s = 4.19 ft/sρ a = 3.22 kg/m 3 = 0.2008 lb/ft 3F a =U ρa0.5g= 1.8877

β = 0.60φ = 0.2h L ’= 0.60 x77.53 = 46.518 mh f = h 1 1 / φ t = 232.59 mm.D f =(L w +D c )\2 = (4.34 +3.255) \2 =3.7625m.r h =u f =fhfD2 h +1000D1000qh 1D fff=0.2069 m.1000x0.032246.51x3.76D f= 0.184 m\sReynolds modulesN reh. =r hu fρµ10.2069x0.184x8330.0003From pg. 18-19 : f=0.031000fuh hg =grh12fLL f =D c Cos(θ c /2) = 1.396m.h hg = 0.75mm.f=105706.58(b) Loss under downcomer, h dah da = 165.2 ( q/A da ) 2Assume clearance C = 1” =25.4 mm.h ap = 77.53 – C = 52. 13 mmA da = L w h ap = 3.225 x0.05213 = 0.1681 m 2h da = 165.2 ( 0.0322/0.1681) 2= 31.64 mmh dc = h l + h w + h ow + h da + h hg

=121.82+ 50 + 27.28 + 31.64 + 0.75= 231.49 mmhh dc ’ = dc231.49= = 462. 98 mm liq.φ 0.5dct s = 610 mmAs h dc ’ < t s there is no floodingSummary of tray calculationsD c = 4.34mL w = 3.255mh w = 50 mmt s = 500 mmd h = 5 mml p = 15 mmt t = 3 mmn h = 54500

COLUMN EFFICIENCYEnriching SectionStripping SectionLiquid flow rate kmol/hr.694.67 1767.428Liq. flow rate : kg/hr. 92787.2145 253267.12Vap. flow rate : kmol/hr1226.98 1226.98Vap. flow rate : kg/hr. 163963.789 175133.63ρL(kg/m 3 ) 841.5 826.0ρV(kg/m 3 ) 3.195 3.34Tliq( o C ) 224.72 138.61Tvap( o C ) 225.12 238.66µliq, cP 0.300 0.1µvap , cP 8.48x 10- 3 .0095D L , cm 2 /s 8.8752 x 10 -5 9.204 x 10 -5D G , cm 2 /s 0.04326 0.0449σ 17.29 17.5x 0.7955 0.4505y 0.8955 0.66

Enriching sectionColumn Efficiency:(a)Point Efficiency, E OG0.776 + 0.2285hW− 0.238UNg=0. 5Nscgaρ0.5g+ 105WW = liq. flow rate = (.0.03006 /3.7625)= 8.13* 10 -3 m 3 / m.sU a = 1.273m/sh w = 50 mm , N Scg = (µ g \ρ g D g ) = 0.824N g = 1.13N L = K L aθ LK L a = ( 3.875 x 10 8 D L ) 0.5 (0.4 U a ρ 0.5 g + 0.17 )= 1.681 * 1.053= 1.771m/sθ L = (h L A a ) / (1000q)h L = ϕ e (h w + 15330 C (q \ϕ e ) 2\3C =0.327 + 0.0286exp((-0.1378 h w )C = 0.327K s = U a (ρ g \ ρ l -ρ g )K s = 0.07ϕ e =0.3303h L =332.92 mmN L = 22.14G= MLmλ M TOP = 0.5125mGLmm157225.21= =1.7688929.21λ t = 0.902M BOTTOM = 0.67λ b = 1.183

Nogλ = 1.04=1Ng1λ+NL1=0.993 + 0.046∴ N og = 0.974E OG = 1 – e -NOG = 0.6181(b) Murphree Plate Efficiency, E mvN Pe =DZE2LθLZ L = D c cos (θ L /2) = 2.792 m.D E = 6.675 * 10 -3 (U a ) 1.44 + 0.922 * 10 -4 h L – 0.00562= 6.48 x 10 -3 m 2 /sN Pe = 34.41λ E OG = 0.636EmvFrom fig. 18.29a = 1. 03E∴ E mv = 0.66OG(c) Overall column efficiency, E ocE oc =EEaMVNN=TAlog=1 + EMV[ 1+E ( λ −1)]1⎡⎢⎣1alog λψ− ψ⎤⎥⎦LG⎡ ρ⎢⎣ ρGL⎤⎥⎦0.5= 0.0348Considering 80% flooding,From fig, Ψ = 0.08

∴ E a = 0.614E oc = 0.587E oc = N t \ N AN A = 8.605 ≈ 9 traysN A = 9 traysTower height, = t s * N A= 610 * 10 -3 * 9 = 5.5m∴ H = 5.5m

STRIPPING SECTION :⎡ L ⎤⎡ρ⎢ ⎥⎢⎣G⎦⎣ρGL⎤⎥⎦0.5= 0.0902(at top)0.5⎡ L ⎤⎡ρG⎤⎢ ⎥⎢⎥ = 0.0933 (at bottom)⎣G⎦⎣ρL ⎦Maximum at bottom of stripping section. Hence all calculations are based on properties atbottom of stripping sectionAssume :Plate spacing , t s = 610 mmHole diameter , d h =5 mmHole pitch , l p =15 mmTray thickness , t T =3 mmHole areaPerforated areaA=Ah =p0.10Assume equilateral triangular pitchColumn diameter D c :Based on entrainment flooding.All relations from Perry’s handbook, 6 th edition.⎡ 20 ⎤Fig. 18-10 , C sb = U nf⎢⎣ σ ⎥⎦C sb =0.0.43 ft\s∴ σ =17.5 dyn\cmU nf = 1.232 m\sAssuming 80% floodingU n = 0.8 U nf=0.9858 m/s.0.2⎡ ρg⎢⎢⎣ρl− ρg⎤⎥⎥⎦0.5ft/s.Net area for flow , A n = A c - A d180942.74Vapour flow rate = = 14. 56m 3 /s.3600 x 3.45

A n = 14.56 / 0.98858 = 14.77m 2 .L w weir lengthLAssume w = 0. 75D⎛ θSin ⎜⎝cL⎞⎟ =⎠ Dwc2 c22θ c = 97.2 0πD 2 cA c = = 0.78522A d = 0.0878 D c2DcA n = 0.785 D c 2 – 0.0878 D c2D c =4.66 mA c = 16.61 mA d =1.85 m 2Active area, A a = A c – 2A d =12.89 m 2 .L w = 4.6 x.75 = 3.45mPerforated area,Aα = 180 – 97.2 = 82.82 0 CArea of calming + distribution zone, A czA cz =2(L w * t) , t=thicknessA cz =2(3.45 * 120* 10 -3 ) =0.828 m 2 (that is 5.05% of A c )A wz = 0.3928m 2 . (2.3% of A c )A p = A c –2 A d – A cz - A wz= 16.61-2x1.85 –0.828 –0.3928= 11.67 m 2A h = 0.1 A p = 1.167 m 2No. of holes, n h =1.167π4−3( 5x10) 2= 59468holes.Weir height , h w =50 mm (atmospheric pressure)

Weeping check :TOPof enriching sectionU n = 12.5 m/s.hd= K1+ K2ρgρUAssume sieve plates2K 1 = 0 , K 2 = 50.8/C vA h / A a =1.0682 /11.52 =0.09t T /d h = 0.6C v = 0.75K 2 = 90.33∴ h d =90.33 x (3.22 x12.5 x12.5) \833.7h d =54.5mml2h⎡ qh ow = F w x 664 x ⎢⎣ L W⎤⎥⎦24445891.65q 1 ( liquid load ) = = 0. 08193600x833.70L w = 3.255 mF w = 1.04h ow = 56.36 mmHead loss due to bubble formation ,23m 3 /s⎡ σhσ = 409 ⎢⎣ ρld n⎤⎥⎦⎡ 17.5 ⎤h σ= 409⎢⎣833.70x5⎥⎦∴h σ = 1.717mm liq.Now, h d + h σ = 54.5+1.715 = 56.25mm liq.h w + h ow =50 +56.34 =106.21 mm(A h / A a ) = 0.09Minimum value to avoid weeping , h d + h σ = 16mm ( from fig. 18-11)

Since actual >minimum there is no weepingDowncomer Floodingh dc =h t + h w + h ow + h da + h hgHydraulic gradient :h d > 2.5 h hg mmLet h hg = 0.5 mmh ds = h w + h ow +h hg /2h ow =1.03 * 664 * (0.088\3.45) 2\3= 58.68mmh ds = 50 +58.68 +0.5\2 =108.93 mmh t = h d +h 1 lh 1 L = βh dsU a = 1.13 m/s = 3.70 ft/sρ a = 3.45kg/m 3 = 0.2150lb/ft 3F a =U ρa0.5g= 1.7177β = 0.61φ = 0.22h L ’= 0.61 x1108.93= 66.26 mh f = h 1 1 / φ t = 301.18mm.D f =(L w +D c )\2 = (4.6 +3.45) \2 =4.025m.r h =u f =fhfD2 h +1000D1000qh 1D fff=0.2619m.1000x0.08866.26x4.025= 0.3311 m\sReynolds modulesN reh. =r hu fρ1µ1

0.2619x0.3311x819.69=710986.050.0001From pg. 18-19 : f=0.031000fuh hg =grh2fLfL f =D c Cos(θ c /2) = 0.3.042m.h hg = 4.02mm.(Loss under downcomer, h dah da = 165.2 ( q/A da ) 2Assume clearance C = 1” =25.4 mm.h ap = 108.93 – C = 83.23 mmA da = L w h ap = 3.45 x0.08323= 0.2945 m 2h da = 165.2 ( 0.088/0.2942) 2= 14.81 mmh dc = h l + h w + h ow + h da + h hg=109.80+ 50 + 58.68+ 14.81 + 4= 237.29 mmh dc237.29h dc ’ = = = 474. 58 mm liq.φ 0.5dct s = 610 mmAs h dc ’ < t s there is no floodingSummary of tray calculationsD c = 4.6mL w = 4.01mh w = 50 mmt s = 610 mmd h = 5 mml p = 15 mmt t = 3 mmn h = 59468holes

COLUMN EFFICIENCY 0F STRIPING SECTIONStriping section(a)Point Efficiency, E OG0.776 + 0.2285hW− 0.238UNg=0. 5Nscgaρ0.5g+ 105WW = liq. flow rate = 0.021 m 3 / m.sU a = 1.163m/sh w = 50 mm , N Scg = (µ g \ρ g D g ) = 0.834N g = 1.154N l = K L aθ LK L a = ( 3.875 x 10 8 D L ) 0.5 (0.4 U a ρ 0.5 g + 0.17 )= 1.8m/sθ L = (h L A a ) / (1000q)h L = ϕ e (h w + 15330 C (q \ϕ e ) 2\3C =0.327 + 0.0286exp((-0.1378 h w )C = 0.327K s = U a (ρ g \ ρ l -ρ g )K s = 0.0742ϕ e =0.3081h L =80.86 mmN L =21.36G= MLmλ M TOP = 0.8mGLmm157225.21= =0.988688929.21λ t =0.8164M BOTTOM = 1.6λ b = 0.9582λ = 0.88

Nog=1Ng1+λNL1=0.86695 + 0.044E OG = 1 – e -NOG = 0.66Murphree Plate Efficiency, E mvN Pe =DZE2LθLZ L = D c cos (θ L /2) = 3.04 m.D E = 6.675 * 10 -3 (U a ) 1.44 + 0.922 * 10 -4 h L – 0.00562= 6.48 x 10 -3 m 2 /sN Pe = 75.41E OG =0.64EmvFrom fig.18.29 = 1. 04EOG∴ E mv = 0.672Overall column efficiency, E ocE oc =EEaMVNN=TAlog=1 + EMV[ 1+E ( λ −1)]1⎡⎢⎣1alog λψ− ψ⎤⎥⎦0.5L ⎡ ρG⎤⎢ ⎥ = 0.038G ⎣ ρL ⎦Considering 93% flooding,From fig,Ψ = 0.093∴ E a = 0.632E oc = 0.625

E oc = N t \ N AN A = 4.8 ≈ 5 traysN A = 9 traysTower height, = t s * N A= 610 * 10 -3 * 5= 3.05m∴ H = 3.05m

Mechanical design of Distillation columnDiameter =4.6mOperating pressure =1 atm =1.032 Kg\ cm 2Design pressure = 1.1 x1.032 = 1.132 Kg\ cm 2Operating temperatur =245.5 0 CDesign temperatur = 250 0 CShell material - carban steelShell – double weled but jointSkirt height =2mTray spacing = 610mmTop disengaingn space =0.3mBottom separator space = 0.4mAllowable stress = 1187 Kg\ cm 2Insulation material – asbestosDensity of insulation =575 kg\ m 3Head torsphericalMaterial –carbon steelSkirt supportHeight = 2mMaterial =carbon steelNozzlesNo. Of nozzles =5Tray sieve typeNo. Of tray =8Hole diameter =5mmThickness =3mmWeir height =50 mmMaterial for tray – stainless steel

Calculations of shell thickness :Considering the vessel as an internal pressure vessel.t s = ( PD I \2fj-P)+ Ct s = Thickness of shell (mm)P = Design pressure (kg/cm 2) =1.1362 kg/cm 2D i = Diameter of the shell (mm) = 4600mmf = Allowable /permissible tensile stress (kg/cm 2 ) = 1187C = Corrosion allowance (mm) =2mmJ = Joint Efficiency.Considering double welded bolt joints with backing stripJ = 85% = 0.85t s = 1.1362 x 4600 + 2 = 4.95 mm2(950 x 0.85)-1.1362Taking the thickness of the shell as t s = 6mmHead shallow dished & torispherical head.t h = PR c W2fJR c = Crown radius = outer diameter of the shell = 4600+ 2(6) =4612mmR L = knuckle radius = 0.06 R cW= Stress intensification factorW =0.25 ( 3+ (Rc\Rk) 0.5t h = 1.1362x4612x1.77 = 4.59mm2x1187x0.85Thickness of head is t h =6mm =0.236 inchesWeight of head:

Diameter = OD + OD + 2Sf + 2 icr3 3OD = outside diameter of shell = 4612mm = 181.5(inches)icr = inside cover radius = 0.75 inchesSf = straight flange length = 1.5 inchesDiameter = 181.5+ 181.5 + 2 (1.5) + 2 ( 0.75)3 3Diameter (d) = 192.56 inchesWeight of head = π ( 4.89) 2 (6 x 10 -3 )x77004= 867.22 kgweight of head ∼ 2675kg(3) Calculation of stresses:(i) Axial tensile stress due to pressuref ap = P di = 1.1362x4600 = 653.315 Kg/cm 24(t s -c) 4 (6-2)This is same throughout the column height(ii) Circumferential stress :2 f ap = 2x 653.315 = 326.66 Kg/cm 2(iii) compressive stress due to dead loads:Compressive stress due to weight of shell up to a distance x metre.f ds = weight of shellCross-section area of shellf ds = (π/4) ( D 2 o – Di 2 )f s x(π/4) ( Do 2 – Di 2 )

Di&Do - Internal & external diameters of shellS s- density of shell.Also,f ds = weight of shell per unit height x XπD m (t s -c)Dm = Mean diameter of the shell (cm)t sC= thickness of the shell (cm)= Corrosion allowance (cm)f ds = S s (x)S S = 7700kg/cm 2= 0.0077kg/cm 2f ds = 0.77x kg/cm 2Compressive stress due to weight of insulation at height (x) mf d(ins) = π Dins tins Sins (x)π D m (t s –c)Dins = Diameter of insulationtins = Thickness of insulationSins = Density of insulationD m = Mean diameter of shell= [ Dc + ( Dc+2 t s )]2Assume : asbestos in the insulation material.Sins = 575 kg/m 3 = 0.000575 kg/cm 3tins = 75mm = 7.5cmDins = Dc+2 t s +2tinsDins = 4600 + 2(6) + 2(75) = 4762mm = 476.2cmD m = 4600+ (4612) =4606mm = 460.6cm2f d (ins) = π (476.2) 7.5 x 0.000575π (460.2) (0.6 – 0.2)= 2.23 x kg/cm 2

(b) Compressive stress due to liquid & tray in the column up to height (x) m.Liquid & tray weight fox height (x)F liq = (x-1) + 1 πDi 2 x S liquid(0.4572) 4F liq = (x – top disengaging space) +1 πDi 2 x S liquid (Ref: 3, pg :294)Tray spacing 4= x –0.3 + 1 π (4.6) 2 x 8500.61 4f d (liq) == (x +0.31)3145.9 kg.F liqπ D m (t s –c)= (x +0.31) 3145.9 = (40.40x – 12.4) kg/cm 2π (460.6) (0.6-0.2)(d) Tensile stress due to wind loads in self supporting vesself wx = M wzM w = bending moment due to wind load= wind load x distance2= 0.7 P w D m X 22z=modulus for the area of shell = π D 2 m (t s - c)4f wx = 0.7 P w D m x 2 = 1.4 Pw x 2

22 π D m (t s -c) π D m (t s -c)4P w = wind pressureP w = 25 lb/ft 2= 121.9 kg/m 2Mw = (0.7 x 121.9 x 4.606) x 2 = 392.518 x 22Z = π ( 4.606) 2 (0.006 – 0.002) = 0.06644392.518f wx = (392.518)\ 0.0664 = 5902.4 x 2 kg/m 2 . = 0.5902 x 2 kg/m 2Stresses due to seismic load are neglected.Calculations of resultant longitudinal stress ( upwind side )Tensile:ft,max = f wx + f ap – f dsf wx = stress due to wind load.f ap = Axial tensile stress due to pressuref ds = Stress due to dead loads.ft,max = 0.5902 x 2 + 653.315 – 0.77xft,max =fJf = allowable stress = 1187kg/cm 2J= Joint factor = 0.85ft,max = 1187 (0.85 ) = 1008.95 kg/cm 20.5902 x 2 - 0.77x +653.315 = 1008.950.5902 x 2 – 0.77x – 355.63 =0a =0.5902, b = -0.77, c = - 355.63x = - b 1 ± ¥E 2 – 4 ac = 0.77 ± ¥ 2a 2(0.5902)

x = 25.07mCalculation of resultant longitudinal stress (downwind side) compressive :ft,max = fwx - fap + fdsfc,max = 0.125 E tDoE= Elastic modulus = 2x10 5 MN/m 2 =2x10 6 kg/cm 2t = Shell thickness = 6mm.D o = 4612 mmfc,max = 0.125 x 2x10 6 6 = 325.23kg/cm 24612Consider, 325.23= 1.0604 x 2 – 163.33 + 0.77 x1.0604 x 2 + 0.77x – 812.12 = 0a= 0.5902, b = 0.77, c= - 978.41x= - 0.77 ± ¥ 2 + 4 (0.5902) (978.41)2(0.5902)x= 40.06m∴ The calculated height is greater than the actual tower height. So we conclude that thedesign is safe and thus design calculations are acceptable.∴ A thickness of 6mm is sufficient throughout the length of the shell.Design of skirt support :Total height of column including skirt height (H) = 8.2+2+0.3+0.4= 10.9mMinimum weight of vessel (W min ) ={ π(D i +t s )t s (H-skirt height )s+2 (2670) }D i = diameter of shell = 4.6mt s = 0.006mS s = Density of material

W min = π (4.6+ 0.006) 0.006 (10.9-2) 7700+2(2670)= 11286.83kg.Maximum weight of column (W max ) = W w + W i + W e + W aW s = weight of shell during test = 5452.13 kgs.W i = weight of insulation = π (d 2 ins - d 2 o) H S ins4= π{ 4.891 2 – 4.612 2 } 10.9 (575)4= 13037.96 kgsW e = weight of water during test = π Di 2 (H-2) Swater4= π (4.6) (10.9-2) 10004= 156139.64kgsWa = weight of attachments = 7100 kgsW max = 11286.83+13037.96+5452.13+7100 = 63195.55kgsPeriod of vibration at minimum dead weightT min = 6.35 x 10 -5 H 3/2 W min ½D= 0.312 s∴ K 2 = a coefficient to determine wind load =2period of vibration at maximum dead weightt s3/2 ½(Ref: 5, pg:147)T max = 6.35 x 10 -5 H W maxDt s3/2 ½= 6.35 x 10 -5 10.9 36876.924.60 0.006= 0.952sk 2 =2

Total load due to wind acting on the bottom & upper part of vesselP W = k 1 k 2 P w HDK 1 = coefficient depending upon safe factor= 0.70 (for cylindrical surface )P W = wind loadP W = wind pressure = 1000 N/M 2 = 100 kg/m 2For minimum weight condition D = D i = 4.6 mFor maximum weight condition D =D ms = 4.89 m∴( P W ) min = 0.7x2x100x4.6x10.9= 7019.6kg(P W)max = 0.7x2x100x4.89x10.9= 7462.14kgMinimum & maximum wind moments(M W ) min = (P W ) min x H =38256.82 kg2(M W ) max = (P W ) min x H = 40668.66 kg.m2As the thickness of the skirt is expected to be small, assumeD i ∼ D o = 1.7mσ zwm (min) = 4 M W(min)πD 2 t= 4x 38256.82π(4.6) 2 t= 2303.15/t Kg/m 2fzwm(max) = 4×40668.663π(4.6) 2 t= 2448.35 Kg/m 2t

Minimum and maximum dead load atresses:fzw (min) = W min = 11286 .83 = 781.41 Kg/m 2πDt π(4.6)t tfzw (max) = W max = 36826.92 = 2553.09 Kg/m 2πDt π(4.6)t tMaximum tensile stress without any eccentric load is computed as follows.(tensile) fz = σ zwm (min) - fzw (min)fz = fJ118.7×10 5 ×0.85 = 2303.15 – 781.41 Kg/m 2tTherefore t = 1.5 x10 -4 m= 0.15 mmMaximum compressive load:Compressive: fz = fzwm(max) + fzw (max)fz = 0.125 E(t/Do)= 0.125×2×10 5 ×10 6 ×( t/4.612)2448.35+2553.09 = 5.42 ×10 9 ttTherefore t = 7.14×10 -4 mAs per IS 2825-1969,minimum corroded skirt thickness is 7 mm. Providing 1 mmcorrosion allowance, a standard 8 mm thick plate can be used for skirt.Design of skirt bearing plate:Maximum compressive stress between plate and foundation:fc = Wmax + Ww(max)A ZA = π(4.6 - l) ll - 0uter radius of bearing plate - outer radius of skirt supportZ = πR 2 m lR m = Do - l

2A = π(4.6 - l )lR m = (4.6 - l )2Z = π (4.6 - l ) 2 l4fc = 36876.92 + 40668.663π(4.6- l )l π (1.82 - l ) 2 l4Allowable compressive strength of concrete foundation values from 5.5 - 9.5 MN/m 2Assume :5.5 -9.5 MN/m 25.5 × 10 5 = 36876.92 + 40668.663π(4.6- l )l π (1.82 - l ) 2 l4l = 0.0191 + 0.0302(4.6- l ) (4.6 - l ) 2By hit and trial method. l = 0.02 mTherefore 20 mm is the width of the bearing plateThickness of bearing plate ,t bp = l(3fc/t) 1/2fc- maximum compressive load at l = 0.02 m = 0.23 × 10 6 Kg/m 2t bp = 20[(3×0.23 × 10 6 )/(118.7 × 10 5 )] 1/2= 4.23 mm∼ 6 mmMinimum stress between the bearing the plate and the concrete foundation:fmin = Wmin - Mw(min)A Z= 156139.64 - 40668.63[π(1.82-0.02)0.02] [π(1.82-0.02) 2 0.02]

= 7548.63 Kg/m 2J = WminD(Mw)max= (7513.23×4.6)13054.05= 0.8498Therefore this is less then 1.5,the vessel will not be steady by its own weightTherefore anchor bolts has to be used.P bolt (n) = fmin A= 7548.63×3.14×(4.6 - 0.02)0.02= 853.73 Kg

MINOR EQUIPMENT DESIGNCondenserProcess DesignDesign of a overhead condenser for the vapor mixture of the given in mole fraction,the temp. of the mixture is 218.33deg. and water comes at 25deg. And out at 40.Hot fluid enters at 218.33deg. and leaves at 218.33deg.Cold fluid enters at 25deg. and leaves at 40deg..PROCESS DESIGN :(Vapor)F 10.91 Kg/sF(liq.)CoolingCondensation occurs at isothermal condition, correction factor F t =1.Assume feed is saturated at 218.33deg.Properties are evaluated at 218.33deg.µ g = Viscosity = 9.47x10 -3 cp.ρ g = Density = 3.177 kg/m 3 .c pg = Heat capacity =1.73 KJ/kg deg.λ =Latent Heat =320.24KJ/KgVapor flow rate = 43.67 Kg\sec.Here flow rate is high so we use four condensers in parallel.So flow rate = 43.67\4 =10.91kg\secHeat load =mλ=10.91x320.24=3483.14x10 3 Kg\sec

h i = 3386.56W/m 2 KFILM COEFFICIENTShell side – DistillateTemp. of wall=124.165deg.Film temp. =171.24deg.Property at this temp.ρ = 855.12 kg/m 3k = 0.112 W/mKµ = 0.28cp.c p = 1.71 KJ/kg degBaffle spacing B = 0.406 m.G s = 10.91kg\sReynold’s number ( N Re ) =4×Mass flow rate of condensateµ × N=1178.15323t ×Length of tube1.151−1h s = [] 3 [ ] 31NReh s =678.313 W/m 2 Kthe dirt factor is 5.283x10 -4 W/m 2 Kover all heat transfer cofficient1\U =(D o \D i )x1\h i +1\ h s +5.283x10 -4U = 426.92 W/m 2 KCalculated U = 426.92W/m 2 K assuming U= 300 W/m 2 K∴ Design is okay.PRESSURE DROP CALCULATIONTUBE SIDEN Re = 3386.54

F = 0.079 ( N Re ) -1/4 = .0.01305H=(4 x f x v 2 x L )\(2gD)=0.332mP= ρgH=3.161kN\ m 2∆P =(2.5ρ v 2 )\2=1.378 kN\ m 2∆P Total = 2(1.378+3.161)=9.078 kN\ m 2which is very less than permissible, therefore design is okaySHELL SIDET Vapour =218.33deg.δ m = (p 1 - D)l s D s \ p 1here ,p 1 = pitch =1”D s = shell diameter = 635 mmδm =(( 31.75 –25.4) x635x635) \(2x31.75)δm =.040 m 2PRESSURE DROP CALCULATIONEnd zones ∆p e ,two end zones.Cross flow zones ∆p c ,( N b –1) crossWindow zones ∆p w , N b zones∆p c =b f k w 2 N c (µ w \ µ b )b =2x10- 3w = 10.91 Kg\sδm =0.040 m 2N c = D s (1 –2(l c \ D s ) )\ P pP p == (1.732\2) p 1 mm

=27.4 mmP n =0.5 p 1 = 15.875 mml c = 25% of shelldia.N c =12∆p c =3.41 KN\ m 2∆p w =( b x ( W) 2 (2 +0.6 x N cw ) )\ (S w x δm xρ )N cw = 0.8 l c \ P p=(0.8 x0.25 x635) \ 27.4= 4.63S w = S wg - S wtS wg =0.1806 m 2 ) ( from perry hand book)S wt =( (N t ) ( 1 - F c )ΠD 2 0 )\8S wt = 0.026∆p w =14.265 KN\ m 2∆p e = ∆p c ( 1+ N cw \ N c )= 3.41(1+4.63\12)= 4.725 KN\ m 2Total pressure drop:0 +14.265 +2 x 4.72523.716 KN\m 2But actual pressure is 40% of this = 9.486 KN\m 2MECHANICAL DESIGN OF CONDENSER:

SHELL SIDE:Material: Carbon steelNo. Of shells: 1No. Of passes: 2Fluid: naphthalene vapourInternal diameter : 635mWorking pressure: 0.1 N/mm 2Design pressure: 0.0.106 N/mm 2Inlet temperature: 218 .33 0 COutlet temperature: 218 .33 0 CAllowable stress: 950 Kg/Cm 2TUBE SIED:Material: Stainless steel (IS grade 10)No. Of tubes: 278Outside diameter: 25.4mmLength: 3.048mFluid: waterPitch: 31.75mm (triangular)Allowable stress: 10.06 Kg/mWorking pressure: 1atmDesign pressure: 1.06Kg/cm 2Inlet temperature: 25ºCOutlet temperature: 40ºCSHELL SIDE

SHELL THICKNESS:t s = (P d × D S )/(2fJ-P d )= (0106×635)/((2×95×0.85)-0.106)= 0.41 mmBut minimum thickness of shell is 6 mmTherefore with corrosion allowance of 2mmThickness of shell = 8 mmNOZZLE DIAMETERM = Mass velocity/sec= 10.91Kg/secDensity ρ =850 kg/m 3Assume velocity to be 10 m/sec(π ×d 2 n ×ρ × v)/4 = Md 2 n = (10.91×4)/(10×3.14×850)d n =0.0404mNOZZLE THICKNESSt n = (P d ×d n )/(2fJ-P)= (0.106×40)/(2×95×0.85-0.106)= 0.026 mmNozzle thickness with corrosion allowance = 5 mmHEAD THICKNESS:t h = (P d ×R c ×W)/(2fJ)W = (1/4)(3+(R c /R K ) 1/2 )R c - crown radius = 635mmR K - Knuckle radius 6% of shell I.D.= 38.1mmW = 1.77t h = 0.737mmUsing same thickness as that of the shell = 8 mmBAFFLE ARRANGEMENT:Transverse bafflesNumber of baffles =1

Baffle Spacing = D s = 635 mmThickness of baffles = 6 mmHeight of baffle = 0.75 × D s= 477mmTIE RODS AND SPACERS:For shell diameter D s = 635 mmNo. Of tie rods = 6No. of spacers =Diameter of rods = 13 mmFLANGE CALCULATION:Flange material = IS: 2004-1962 class 2Bolting steel =5% Cr M o steelGasket material =asbestos compositionShell inside diameter =635 mmShell thickness t s = 9mmShell outside diameter = (2×t s )+635= (2×8) +635= 653 mmAllowable stress of flange material = 100 MN/m 2Allowable stress of bolting material = 138 MN/m 2GASKET WIDTH:G o /G i =[(y-P d m)/(y-p d (m+1))] 1/2m-Gasket factor =2.75Y-Minimum design seating stress =25.5 MN/m 2G o /G i =[25.5-(0.106×2.75)/25.5-(0.106×3.75)] 1/2= 1.003Minimum gasket width N =10 mmBasic gasket seating width b o = N/2=5 mm< 6.3 mmLet, inside diameter of gasket =inside diameter of shell =635 mmG i = 656

Mean gasket width =G i +N=657mmtherefore G =657mmEstimation of bolt load:Load due to design pressure H =(πG 2 P d )/4=(π×0.657 2 ×0.106)/4= 0.035 MNEffective gasket sitting width b =b o =6.27 mm since bW o , controlling load =0.328MNMinimum bolting area:Total cross sectional area of bolt under operating condition A m1 =W o /S bS b -nominal bolt stress at design temperature of 218.33ºC=138 MN/m 2Therefore A m2 =0..328/138=2.37 x10 -3 m 2Total cross sectional area of bolt required for gasket seating A m2 =W b /S aS a -nominal bolt stress at ambient temperature (30ºC)=138 MN/m 2Therefore A m1 =0..042/138=3.04 x10 -4 m 2Since A m2 > A m1 ,A m = A m2 = 2.37 x10 -3 m 2Calculation of optimum bolt size:C =2(R+g 1 )+BChoosing bolt M-18 × 2

Total number of bolts = G/(18 × 2)= 664/36= 18so we take No. Of bolt = 20Actual number of bolts = 220R-radial clearance from bolt to point of connection of hub and back of flange = 27mmB-inside diameter of flange =outside diameter of shell =0.654 mg 1 =g o /0.707,let g o = 8 mmg 1 =1.415 g oC =2(0.027 + 1.415 x 0.008) + 0.654= 0.7306mTherefore bolt circle diameter = 0.7307 mFlange outside diameter:A = C + bolt diameter + 0.02= 0. 7307 + 0.018 + 0.02= 0.795 mCheck of gasket width:A b -root area of bolt (m 2 )-1.54 ×10 -4 m 2S g -allowable stress for bolting material at atmospheric temperature =138 MN/ m 2Therefore, A b S g /πGN = 20.69Since 20.69 > 2yCondition is satisfied.Flange moment computations:W o = W 1 + W 2 + W 3 (under operating condition)W 1 = (πB 2 P d )/4= (3.14×0.654 2 ×0.106)/4= 0.035MNW 2 = H – W 1H =(πG 2 P d )/4=(3.14×0.657 2 ×0.106)/4=0.0359 MN

W 2 = .0359 –0.035= 9.175 X 10 -4 MNW 3 = W o – H= 0.042 – 0.0359= 0.0061MNTotal flange moment,M o = W 1 a 1 + W 2 a 2 + W 3 a 3a 1 = (C – B)/2= 0.065 ma 3 = (C-G)/2= 0.064 ma 2 = (a 1 +a 3 )/2= 0.0645 mM o =0.065×0.035 + 0.0645×0.00091 + 0.0064×0.0061= 2.724 × 10 -3 MNmBolting up condition:Total flange moment M g = Wa 3W =(A m + A b ) Sg/2=0.376 MNa 3 =0.064mTherefore M g = 0.0240 MNmSince M g > M o for moment under operating condition,M g is controlling.Therefore M = M g = 24×10 -3 MNmFlange thickness:t 2 = (MC F Y)/BS TK =A/B= Outer diameter of flange/ inner diameter of flange= 0.795/0.654= 1.21Y =14Assume C F =1

Therefore thickness 't' = 0.0719mActual bolt spacing B S =πC/n= (3.14×0.785)/20= 0.123 mBolt pitch correction factor C F = [B S /(2d+t)] 1/2=[0.123/(2×0.018+0.0719)] 1/2=1.06Therefore C 1/2 F = 1.03Actual flange thickness = C 1/2 F t= 1.03×0.0719= 0.074 mTUBE SIDE:TUBE THICKNESS:t t =Pd o /2fJ+PJ =1 foe seamless tubeTherefore t t =(1..06×25.4)/(2×1006+1.06)= 0.0136 mmNo corrosion allowance since the tube is made of stainless steelThickness of tube = 1mmTUBE SHEET:t S = FG [0.25P/f] 1/2F-The value of F varies according to type of heat exchanger, for most cases it is taken asG =657 mmTherefore t S =657[(0.25×1.55)/1006] 1/2= 11.25mmCHANNEL AND CHANNEL COVER:t =G[KP/f] 1/2K =0.3 for ring type gasketMaterial of construction is carbon steelSo allowable stress f =950 Kg/cm 2

Therefore t =657[(0.3×1.06)/950] 1/2= 12.68 mmWith corrosion allowance t =15 mmNOZZLE THICKNESS:Assume inlet and outlet diameter = 75 mmThickness of nozzle t h =Pd/2fJ-P=(1.06×75)/(2×0.85×950-1.06)=0.0426 mmwith corrosion allowance thickness = 6mmSADDLE SUPPORT DESIGN:Material -Low carbon steelVessel diameter=0.654 mmLength of shell =3.048 mTorispherical head:Crown radius =635 mmKnuckle radius =63.5 mmWorking pressure =1AtmShell thickness = 8 mmHead thickness = 8mmCorrosion allowance = 2mmPermissible stress = 950 Kg/cm 2R-Vessel radius = 425.5Distance of saddle center line from shell end A =0.45×R= 147.15mm < 0.2LLongitudinal bending moment:The bending moment at the support is;M 1 =QA[1-{(1-A/L)+(R 2 -H 2 /2AL)/(1+4H/3L)}]A =147.15 mmQ =W/2[L+4H/3]W-Weight of fluid and vessel.

Weight of shell material:W 1 =[π(D 2 o -D 2 i )Lρ shell material ] /4ρ shell material - 7700 Kg/m 3W 1 =[3.14(0.654 2 -0.635 2 )×3.048×7700] /4= 451KgWeight of tubes:W 2 = n[π(D 2 o -D 2 i )Lρ tube material ] /4ρ tube material -7800 Kg/m 3W 2 =278[3.14(0.0254 2 -0.022 2 )×3.048×7800] /4= 869.55KgWeight of tube sheet:W 3 = (2πD 2 tρ)/4= (2×3.14×0.635 2 ×0.0120×7800)/4= 62.85KgLiquid load in the shell:W 4 =(shell volume -tube volume)ρ liquid=[(πD 2 s L)/4 - (nπd 2 o l)/4 ]×1175=[(π×0.654 2 ×3.048)/4 - (278×π×0.0254 2 ×3.08) /4 ]×850= 508.05 KgLiquid load in tubes:W 5 = nπd 2 i lρ liquid /4=(278×3.14×0.022 2 ×3.048×1000)/4= 321.9KgTherefore total weight W T = W 1 + W 2 + W 3 + W 4 + W 5= 450.91 + 869.55+ 62..85 + 508.05 + 321.9= 2213.38 KgHence, Q =2213.38/2[3.048+(4×0.257)/3]= 3752.416KgmM 1 =(3752.416×0.147)[1-{(1-0.147\3.048)+(0.327 2 -0.257 2 /2×0.147×3.048)/(1+4×0.257/3×3.048)}]=57.23 Kgm

The bending moment at the center of the span is given by:M 2 =(QL/4)[{1+2(R 2 -H 2 )/L 2 }/{1+(4H/3L)}- (4A/L)]=(3752.416×3.048/4)[{0.90- (4×0.147/3.048)]= 2021.80KgmStress in shell at the saddle:At the top most fiber of the cross-section,f 1 =M 1 /(K 1 πR 2 t)For an angle of 120º ,K 1 =0.197 mt-thickness of shell = 8 mmf 1 =57.23/(0.197×3.14×0.327 2 ×0.008)=10.81 Kg/cm 2At the bottom most fiber of the cross-sectionf 2 =M 1 /(K 2 πR 2 t)For an angle of 120º ,K 2 =0.192 mf 2 =57.23(0.192×3.14×0.327 2 ×0.008)1=11.09 kg/cm 2Stress in the shell at mid point:f 3 =M 2 /(πR 2 t)=2021.8/(3.14×0.327 2 ×0.008)=75.27Kg/cm 2Thus the values of stresses are within the limited rangeHence the designed support is acceptable.