Sieve plate Distillation Column

Process Equipment Design

Problem: A process stream of solution of methanol, acetic acid having zF = 0.5 has to be concentrated by Distillation. The distillate product stream has to contain methanol of xD = 0.95. The bottom product stream contain methanol of xB = 0.05. The operating pressure for the operation is 94.13 kPa. Design the distillation column using data supplied.

Group VI

Himanshu Meena (12CH30015)

Kamble Manish Vijay (12CH30016)

K.S.S.Viswanat (12CH30017)

Monish Kumar (12CH30018)

Vapor-Liquid Equilibrium Data

Vapour liquid Equilibrium Data of Methanol and Acetic Acid is presented below:

Pressure-94.13 kPa (Source: Vapor-Liquid Equilibria of Binary Mixtures Methanol-Acetic Acid, Ethanol-

Acetic Acid, n-Propanol-Acetic, n-Butanol Acetic Acid. Chem.Eng.Sci.Journal 10 (1959) 105-111)

T [K] x1 [mol/mol] y1 [mol/mol]

388.95 0 0

385.15 0.037 0.109

383.05 0.058 0.165

378.15 0.107 0.303

375.95 0.137 0.353

370.55 0.208 0.491

367.95 0.25 0.556

365.85 0.28 0.603

364.85 0.306 0.63

359.95 0.387 0.735

357.85 0.427 0.776

354.85 0.492 0.831

353.05 0.517 0.853

349.15 0.601 0.911

346.85 0.657 0.934

344.75 0.713 0.951

342.25 0.785 0.971

340.45 0.839 0.983

339.05 0.882 0.989

336.95 0.959 0.996

336.15 0.995 0.999

336.05 1 1

First, the equilibrium data was plotted and the 45° feed line was drawn. Then xD point was plotted on

the 45° line and the operating line for enriching section was drawn .From the intercept of this line on y

axis the min reflux ratio was determined on the y-axis. The intercept on y axis was (xD/Rm+1) and

minimum reflux ratio(Rm)= 0.55.

Reflux ratio

n.o.stages(with reboiler)

Tray Efficiency Real n.o stages Column Diameter Total height

0.5 inf

0.55 9.8 0.6 15 1.0598 16.2

0.6 8.9 0.6 14 1.0768 15.3

0.65 8 0.6 12 1.0935 13.5

0.7 7.7 0.6 12 1.1099 13.5

0.75 7.4 0.6 11 1.1261 12.6

0.8 7 0.6 10 1.1421 11.7

0.85 6.9 0.6 10 1.1578 11.7

0.9 6.8 0.6 10 1.1734 11.7

0.95 6.7 0.6 10 1.1887 11.7

1 6.6 0.6 10 1.2039 11.7

1.1 6.3 0.6 9 1.2336 10.8

1.2 6 0.6 9 1.2626 10.8

1.3 5.9 0.6 9 1.291 10.8

1.5 5.7 0.6 8 1.346 9.9

But from the table, it is seen that as the reflux ratio is increased; The no of stages hence the tray

installation costs would come down. At the same time, the reboiler and condenser load were found to

be increasing. In order to achieve a balance between the two, we draw a graph between reflux ratio and

the total costs. A sample cost calculation for R=0.75 is shown below:

Sample Cost Calculation

F=10 000kg/h;

M.W of CH3OH=32(distillate);M.W. Of CH3COOH= 60

Mol.Wt of feed=0.5*32+0.5*60

=46

Feed=10000/46=217.39 kmol/h

F=D+B

Fzf=DXd+BXb and xb=0.05;xd=0.95;zf=0.5

Solving D=108.69 kmol/h=B=108.69

Vapour Rate in distillate= V = D(1+0.75)=190.21 kmol/h = 6352.93 kg/h = 1.764 kg/s

(V’-V)/F + 1 =q = 0.85

Solving

Vapour flow rate below feed=V’=4852.93 kg/hr=1.348 kg/s

Column Diameter Dc

a) At top

ΡL = 791.8 kg/m3

ρv = PM/RT = 1.122 Kg/m3 P = 94.13 KPa , T = 337 K

maximum allowable vapour velocity = Uoperational =(-0.171l2 + 0.27l-0.047)*((pl - ρv)/ρv)0.5

l,Tray Spacing =0.9 gives Uoperational = 1.582 m/s

Column Diameter =√(4Vw/πρvUoperational)

=(4*1.764/π*1.12*1.582)0.5

=1.1261 m

b) At bottom

ΡL = 1049 kg/m3

ρv = PM/RT = 1.705 Kg/m3 P = 94.13 KPa , T = 389 K

maximum allowable vapour velocity = Uoperational =(-0.171l2 + 0.27l-0.047)*((pl - ρv)/ρv)0.5

l =0.9 gives Uoperational = 1.425 m/s

Column diameter =√(4Vw/πρvUoperational)

=(4*1.348/π*1.705*1.425)0.5

=0.84 m

Column Height

Column has 11 trays i.e., 10 tray spacings i.e., 10*0.9=9m

Bottom Space:Provide 0.8m liquid depth at bottom 1.6m clear space above liquid

Top Space:Provide 1.2m space above

Total height=12.6m

Purchase Cost

Density of steel=490 lb/ft3

Thickness=5/8 in

Total Weight of shell=3.71*(5/8)*(1/12)*3.14*41.34*490*1.12

=13772 lb

Purchased cost of steel=43000$

Cost of 10 stainless steel sieve trays = 11*900*1.45 (1.45 – quantity factor) Timmerhaus

=14355$.

Cost of 12 18 in manholes = 12*116*18 = 25056$

Cost of 2 10inch vapour line nozzle = 1380$

Cost of 3 4inch nozzle( for feed) = 3*69*4=828$

Cost of 2 4inch nozzle( 1for reflux,1 for bottoms prod )=414$

Total purchase cost = 76090$

Annual Cost of Condenser

Rate of heat transfer/hour =moles of vapour condensed/hr * molal latent heat of vap

Methanol =1104 KJ/kg , Acetic acid = 402 KJ/kg

= 6352.93 x 1104=6790647 KJ/hr = 6436632.23 Btu/hr

Cold water available at 90oF, Methanol boiling point = 148.47 F

Heat transfer area = q/h(T-Tcoolant) = 6436632.23/[100*(148.47-90)] = 1100.84 ft2

Assuming h = 100 btu/ hr ft2 oF for condenser

Cost = 20.68*1100 = 22748 $

Cost of cooling water = 0.238$/10000 kg

6790647 Kj/hr= m(kg/hr) *1*27.8, m=244267 kg/hr ,

Assuming 8 hrs of operation in 1 day and 250 days in a year , cost = 11627 $.

Annual cost of reboiler

Heat in = heat out

Heat capacity of acetic acid = 123.1 J/mol-K, Heat capacity of methanol= 79.5 j/ mol k

Heat capacity of mixture = 2.20 kj/kg-K, taking 148.47 F =337.7 K as reference

Q + 10000*2.2*(354.85-337.7) = Qcondenser (=6790647 Kj/hr )+ 108.69*79.5*(391-337.7)

Q = 6873904 kj /hr= 6515549 btu/hr

Assuming heat transfer coeff of reboiler = 80 btu/ hr ft2 oF

A = Q/h*(Tsteam – T), steam is at 60 psi , so steam temp = 292.7 F

A = 6515549/[80*(292.7-244.4)]= 1686 ft2

Cost = 25.61*A = 43178$

Cost of steam =3.31 $/ 1000 kg

6873904 kj/hr =( 2.13*1000)*m , m= 3227 kg/hr

Cost of steam = 21363 $ for 1 year operation

Total cost = 184370 $

The data for other values of Reflux ratio is tabulated as below:-

Reflux Ratio

Shell trays Manhole Nozzles condensor

cooling

water

Reboiler

steam Total cost

0.55 49000 16875 33408 2622 20163 10298 38308 18951 189625

0.6 48000 16380 31320 2622 20814 10630 39527 19554 188847

0.65 46000 15320 27144 2622 21764 11098 40746 20257 184951

0.7 44000 15410 27144 2622 22015 11194 41965 20360 184710

0.75 43000 14753 25056 2622 22765 11627 43184 21363 184370

0.8 41000 13988 22968 2622 23415 11959 44402 21966 182320

0.85 41700 14123 22968 2622 24066 12291 45621 22569 185960

0.9 42000 14248 22968 2622 24716 12623 46840 23172 189189

0.95 42500 14300 22968 2622 25367 12955 48059 23775 192546

1 42800 14500 22968 2622 26017 13288 49278 24378 195851

1.1 41000 14850 20880 2622 27318 13952 51716 25585 197923

1.2 42000 15221 20880 2622 28619 14616 54154 26791 204903

1.3 42500 15592 20880 2622 29920 15281 56591 27997 211383

1.5 40500 15120 18792 2622 32522 16610 61467 30409 218042

And a graph between Reflux ratio and total costs is plotted as shown:

Optimum reflux ratio is found out to be 0.8.

Feed, Distillate and Bottom product Calculations for Optimum Reflux Ratio:

F=10 000kg/h;

M.W of CH3OH=32(distillate); M.W. Of CH3COOH= 60

Mol.Wt of feed=0.5*32+0.5*60

=46

Feed=10000/46=217.39 kmol/h

F=D+B

Fzf=DXd+BXb and xb=0.05;xd=0.95;zf=0.5

Solving D=108.69 kmol/h=B=108.69

Vapour Rate in distillate= V = D(1+0.8)=195.64 kmol/h = 6534.44 kg/h = 1.81 kg/s

As Vapor flow rate above the feed tray = Vapor flow rate below the feed tray + Amount of Vapor in feed,

We have V’ = V-(1-q)*F

Vapor flow rate below feed=V’= 5034.44kg/hr= 1.4 kg/s

As L = R*D = 0.8*108.69 = 86.95 kmol/hr = 2904.13 kg/hr

L’-L = F*q, Liquid flow rate below feed=L’=11404.13 kg/hr

Column Diameter Dc

At top

PL = 752 kg/m3

ρv = PM/RT = 1.122 Kg/m3 P = 94.13 KPa , T = 337 K

Flv = L/V * (ρv/ ρL)0.5 = 0.0172

K1 = 0.12*(0.0195/0.02)0.2 = 0.1193

Uf = 3.08 m/s

Design for 85 percent flooding rate

Uf = 2.618 m/s

Volumetric Flow rate = Vapour flow rate above feed/Density = (1.81 Kg/s)/(1.122 kg/m3) = 1.613 m3/s

Area = Volumetric flow rate / Uf

= 0.617 m2

So Diameter = 0.92 m

At bottom

PL = 987 kg/m3

ρv = PM/RT = 1.705 Kg/m3 P = 94.13 KPa , T = 389 K

Flv = L’/V’ * (ρv/ ρL)0.5 = 0.094

K1 = 0.12*(0.0196/0.02)0.2 = 0.1193

Uf = 2.87 m/s

Design for 85 percent flooding rate

Uf = 2.44 m/s

Volumetric Flow rate = Vapour flow rate below feed/Density = (1.4 Kg/s)/(1.705 kg/m3) = 0.821 m3/s

Area = Volumetric flow rate / Uf

= 0.336 m2

So Diameter = 0.65 m

Liquid flow pattern:-

Max. volumetric liquid rate=L’/3600 * 987

=3.21 x 10-3 m3/s

From the graph, pg 568, Coulson & Richardson it is clear that a single pass tray can be used.

Provisional Plate Design:-

Column Diameter Dc =0.92 m

Column Area Ac =0.6648 m2

Downcomer area Ad = 0.12 x 0.6648 = 0.0797 m2, at 12 per cent

Net area An = Ac - Ad = 0.6648 – 0.0797 = 0.5850 m2

Active area Aa = Ac - 2Ad = 0.6648 – 2*0.0797 = 0.5054 m2

Hole area Ah take 5 per cent Aa as first trial = 0.0257 m2

Weir length (from Figure 11.31)

Ad/Ac =12%, lw =0.74*0.92=0.681 m

Take weir height hw - 50 mm

Hole diameter dh - 5 mm

Plate thickness - 5 mm

a)Check weeping

Max. liquid rate Lw = 3.167 kg/s

Minimum liquid rate, at 70 per cent turn-down = 0.7 x 3.167

= 2.217 kg/s

The height of the liquid crest over the weir how = 750[Lw /ρL lw]2/3

Maximum how = 20.7 mm

Minimum how = 16.3 mm

At minimum rate how + hw = 66.3 mm

From fig 11.30 pg 571,

K2 = 30.4

The vapour velocity at the weep point is the minimum value for stable operation.

The minimum design vapour velocity ũh = [K2-0.90(25.4- dh)]/ρv0.5

= 9.234 m/s

actual minimum vapour velocity = min vapour rate/ Ah

= 0.7 x Actual vapour rate/ Ah = 0.7 x 0.82/0.0257

= 22.33 m/s

>9.23 m/s

So minimum operating rate will be well above weep point.

b)Plate pressure drop

Dry plate drop

Maximum vapour velocity through holes ûh = vapour rate/ Ah

= 31.9 m/s

From Graph 11.34, for plate thickness/hole dia. = 1, and Ah/Ap ~ Ah/Aa = 0.05, C0 = 0.81

The pressure drop through the dry plate, hd = 51 [uh/C0]2 [ρV/ρL]

= 136 mm liquid

Correction factor, Residual head hr = (12.5 x 103)/ ρL

= 16.49 mm liquid

total plate pressure drop= pressure drop through dry plate and of liquid crest

= 138 + 16.49 + (50 + 20.3)

= 228 mm liquid

c)Downcomer liquid back-up

Downcomer pressure loss

Take hap = hw — 10 = 40 mm.

Area under apron, Aap = hap x lw = 0.681 x 40 x 10-3 = 0.0272 m2.

The head loss in the downcomer hdc = 166 [Lwd/ρLAm]2

Where Am = either the downcomer area Ad or the clearance area under the downcomer

Aap; whichever is the smaller, m2.

= 0.0272 m2

hdc = 3.97 mm say 4 mm

Back-up in downcomer hb = (50 + 20.7) + 4 + 228

= 303 mm or 0.303 m

< 0.5(plate spacing + weir height)

so tray spacing is acceptable.

d)Residence Time

Check residence time tr = AdhbcρL/Lwd = 0.06*.303*987/3.167

= 5.611 sec

>3 sec satisfactory

d)Entrainment

Also un = 1.61 m3/s ÷ An

= 1.61/0.5850= 2.75 m/s

e)Flooding

Percent flooding = un actual velocity/uf

= 2.75/3.08 = 81%

Less than Design value of 85%

Tray Geometry

Lw/D =0.74 12% downcomer area , angle subtended at centre = 98

Consider 5 cm unperforated tip

Angle subtended at plate edge by unperforated strip == 180 — 98 = 82

Mean length, unperforated edge strips = (0.92 – 0.05)*3.141 * 82/180 = 1.245 m

Area of unperforated edge strips =1.245 * 0.050=0.0622 m2

Mean length of calming zone = (0.92 – 0.05) sin (99/2) = 0.661 m

Area of calming zone = 2(0.661 x 0.05) = 0.0661 m2

Total area for perforations, Ap =0.5054 -0.0622 - 0.0661 = 0.3771 m2

Ah/Ap = 0.0257/0.3771 = 0.068

From 11.33 lp/dh=3.7, satisfactory

Ah = 0.0257

Area of 1 hole = 1.9634 x 10-5 m2

Number of holes = 1309

Mechanical Design of Distillation Column:

a) Shell:

Diameter of the tower =Di =0.920 m

Working/Operating Pressure = 94.13 KPa =0.959 kg/cm2

Design pressure = 1.1*Operating Pressure = 1.0549 kg/cm2

Working temperature = 389K

Design temperature = Operating temp. + 50⁰F(10 K) = 399K

Shell material - IS: 2002-1962 Carbon steel (specific gravity 7.7)

Permissible tensile stress (ft) = 70 MN/m2 = 713.8 kg/cm2

Elastic Modulus (E) = 1.98x105 MN/m2

Insulation material - asbestos

Insulation thickness = 2”= 50.8 mm

Density of insulation = 2700 kg/m3

b) Head - tori spherical dished head:

Material - IS: 2002-1962 Carbon steel

Allowable tensile stress = 70 MN/m2 = 713.8 kg/cm2

c) Support skirt:

Height of support = 1000 mm = 1.0 m

Material - Carbon Steel

d) Trays-sieve type:

Number of trays = 10

Tray spacing: 900 mm

e) Support for tray:

Purlins - Channels and Angles

Material - Carbon Steel

Permissible Stress = 95 MN/cm2 = 970 kg/cm2

1. Shell minimum thickness:

Considering the vessel as an internal pressure vessel.

ts = ((P*Di)/ ((2*ft*J)- P)) + C

where ts = thickness of shell, mm

P = design pressure, kg/cm2

Di = diameter of shell, mm

ft = permissible/allowable tensile stress, kg/cm2

C = Corrosion allowance, mm

J = Joint factor

Considering double welded butt joint with backing strip

J= 85% = 0.85

Thus, ts = ((1.0549*920)/ ((2*713.8*0.85)- 1.0549)) + 3 = 3.8 mm

Taking the thickness of the shell = 6 mm (standard)

2. Head Design- Shallow dished and Tori spherical head:

Thickness of head = th = (P*Rc*W)/ (2*f*J)

P =internal design pressure, kg/cm2

Rc = crown radius = diameter of shell, mm

W= stress intensification factor or stress concentration factor for torispherical

head,

W= ¼ * (3 + (Rc/Rk)0.5)

Rk = knuckle radius, which is at least 6% of crown radius, mm

Now, Rc = 920 mm

Rk = 6% of Rc = 0.06*920 = 55.2 mm

W= ¼ * (3 + (Rc/Rk)0.5) = ¼ * (3 + (16.67)0.5) = 2.833 mm

So thickness = 2.27 mm

Including corrosion allowance of 3mm, the thickness of head = 5.27 mm

Standard thickness = 6 mm

Pressure at which elastic deformation occurs

P (elastic) = 0.366 E (t/ Rc)2 = 7.86 MN/ m2

The pressure required for elastic deformation, P(elastic)>3(Design Pressure) Hence, the thickness

is satisfactory.

Weight of Head:

Diameter = O.D + (O.D/24) + (2*sf) + (2*icr/3)

Where O.D. = Outer diameter of the shell, inch

icr = inside cover radius, inch

sf = straight flange length, inch

From Table , sf =1.5”

icr = 2.31”

Also, O.D.= 940 mm = 36.22”

Diameter = 36.22+ (36.22/24) + (2*1.5)+(2/3*2.31)

d = 41.01” = 1041.85 mm.

Weight of head = (π*D2*t/4)*ρ

= 39.39 Kg where ρ for Carbon Steel = 7700 kg/m3

3. Shell thickness:

a) Axial Tensile Stress due to Pressure:

σap = P*Di /[4(ts -c)] = 1.0549*920/[4*(6-3)] = 80.875 kgf/cm2

This is the same throughout the column height.

b) Axial stress due Dead Loads:

I. Compressive stress due to Weight of shell up to a distance ‘X’ meter from top

σds = weight of shell/cross-section of shell

= (π*D*t*X*ρs)/(π*t*D)

= X*ρs

ρs = density of shell material = 7700 kg/m3

So σds = 7700*X kg/m2 = 0.77X kg/cm2

The vessel contains manholes, nozzles etc., additional weight may be estimated 20%

of the weight of the shell.

So σds = 1.2*7700*X kg/m2 = 0.924*X kg/cm2

II. Compressive stress due to weight of insulation at a height X meter:

σds = weight of shell/cross-section of shell

= (π*Dins*tins*X*ρins)/(π*(ts-c)*D)

asbestos is to be used as insulation material.

density = 2700 kg/m3

tins = 2” = 5.08 cm.

Dins =Dc+2ts+2tins =1033.6 mm

So σzi = 51484.69 X kg/m2

= 5.148469 X kg/cm2

III. Stress due to the weight of the liquid and tray in the column up to a height X

σd, liq. = .weight of liquid and tray per unit height X / π*Dm* (ts - c)

Tray spacing is 900 mm.

The top chamber height is m and it does not contain any liquid or tray.

Average liquid density =869.5 kg/m3

Liquid and tray weight for X meter

No. of Trays = [(X-1)/0.9 + 1]

Weight of Liquid on Trays = (π*D2/4*Weir Height)*ρliq*No of Trays

= (π*D2/4*Weir Height)*ρliq*[(X-1)/0.9 + 1]

σd,liq = Weight of liquid/Cross Section

= (π*D2/4*Weir Height)*ρliq*[(X-1)/0.9 + 1]/ [(π*Dm* (ts - c))]

= 289*[1.11X-0.11]/ [π*1.0336* (0.006 – 0.003)]

= 2.9667*(1.11X-0.11)kg/cm2

= 3.29X-0.33 kg/cm2

IV. Compressive stress due to attachments such as internals, top head, platforms

and ladder up to height X meter.

σd,attachments = weight of attachments/cross-section of shell

Now total weight up to height X meter = weight of top head + pipes +ladder, etc.,

Taking the weight of pipes, ladder and platforms as 25 kg/m = 0.25 kg/cm

Total weight up to height X meter = (Weight of head+25X) kg =(39.37+25X) kg

Total compressive dead weight stress:

σdx = σds + σins +σd (liq) + σd (attch)

= 0.77X + 5.148469 X + 3.29X-0.33 + (39.37+25X)

= 34.208 X + 39.04 kg/cm2

V. Tensile stress due to wind load in self supporting vessels:

σwx = Mw /Z

Where, Mw = bending moment due to wind load = (wind load* distance)/2

= 0.7*Pw*D*X2/2

Z = cross section for the area of shell m

=(π*Dm2/4 * (ts-c))

So σwx = (0.7*Pw*D*X2/2)/ (π*Dm2/4 * (ts-c))

Now Pw = 0.05 Vw2

Where Vw = 140 km/hr upto a height of 20 m

So Pw = 25 lb/ft2 from Brownell Young

= 37.204 kg/m2

Substituting σwx = 11.98 X2 /(π*0.92*(0.006-0.003))

= 1381X2 Kg/m2 or 0.1381X2 kg/cm2

VI. Stresses due to Seismic load:

σsx = Msx / (π*Dm2/4 * (ts-c))

Where, bending moment Msx at a distance X meter is given by

Msx = [C*W*X2/3] * [(3H-X)/H2]

Where, C = seismic coefficient,

W= total weight of column, kg

H = height of column

Total weight of column = W= Cv*π*ρm*Dm*g* (Hv+ (0.8*Dm))*ts*10-3 (eqn. 13.75,

page 743, Coulson and Richardson 6th volume)

Where W = total weight of column, excluding the internal fittings like plates, N

Cv = a factor to account for the weight of nozzles, man ways, internal

supports, etc.

= 1.5 for distillation column with several man ways, and with plate

support rings or equivalent fittings

Hv = height or length between tangent lines (length of cylindrical section)

g = gravitational acceleration = 9.81 m/s2

t = wall thickness

ρm = density of vessel material, kg/m3

Dm = mean diameter of vessel =Di +2t = 0.932 m

So W = 24604.32 N = 2508.08 kg

Weight of plates: ------- (Coulson and Richardson 6th volume)

Plate Area = π*0.922/4 = 0.665 m2

Weight of each plate = 1.2*0.665 = 0.798 KN

Weight of 10 trays = 79.8 KN

= 813.45 Kg

Total weight of column = 2508.08 + 813.45 = 3321.53 Kg

Let, C = seismic coefficient = 0.08

Msx = [C*W*X2/3] * [(3H-X)/H2] = 22.7 X– 0.647 X3

Therefore σsx = 1.1235X-0.032X3 kg/cm2

On the up wind side:

σt,max = (σwx or σsx) + σap –σdx

=0.1381X2 + 161.75 – (34.208 X + 39.04)

= 0.1381X2 +122.71 – 34.208X

But σt,max = fJ = 713.8 * 0.85 Kg/cm2 = 606.3 Kg/cm2

Solving X = 261m >> 11.7 m

On the down wind side:

σt,max = (σwx or σsx) -σap +σdx

=0.1381X2 - 161.75 + (34.208 X + 39.04)

= 0.1381X2 -122.71 + 34.208X

But σc,max = 0.125E*(t/D) = 0.125*1980000*(6/920) Kg/cm2 = 1614.13 Kg/cm2

Solving X = 43.23 >> 11.7 m

So we conclude that the design is safe and thus the design

calculations are acceptable.

From longitudinal stress alone we see that a thickness of 6mm throughout is safe.

4. Design of Support:

The cylindrical shell of the skirt is designed for the combination of stresses due

to vessel dead weight, wind load and seismic load. The thickness of skirt is uniform

and

is designed to withstand maximum values of tensile or compressive stresses.

Data available:

(i) Diameter = 920 mm.

(ii) Height =11.7 m

(iii) Weight of vessel, attachment = 2508.08 kg.

(iv) Diameter of skirt (straight) = 920 mm

(v) Height of skirt = 3.0 m

(vi) Wind pressure = 25 kg/m2

a. Stresses due to dead Weight:

σd=dead weight of attachments and vessel / (π*Dsk* tsk)

= 2508.08/(π*92.0*t)

= 8.67/t

b. Stress due to wind load:

pw = k * p1* h1* Do

p1 = wind pressure for the lower part of vessel,

k = coefficient depending on the shape factor

= 0.7 for cylindrical vessel.

Do = outside diameter of vessel,

The bending moment due to wind at the base of the vessel is given by

Mw = pw * H/2

σwb = Mw/Z

Z- Modulus of section of skirt cross-section

pw = 0.7* 25*3.0*0.92 = 48.3 kg

Mw = pw *H/2 = 48.3×3/2 = 72.45 kg-m

Substituting the values we get,

σwb = 0.2506/tsk kg/cm2

c. Stress due to seismic load:

Load = C*W

C = seismic coefficient = 0.08,

W= total weight of column

Stress at base σsb = (2/3)*(H*W*C)/(π*Dsk2*tsk/4) =0.588/t kg/cm2

Maximum tensile stress:

σt,max= σwb - σd

= 8.419/t

But permissible tensile stress = 713.8 kg/cm2

t=0.1179 mm

Maximum compressive stress:

σc,max= σwb + σd

= 8.9206/t

But σc,max = Yield Point/3 = 1980/3 = 660 kg/cm2

So t=0.135 mm

As per IS 2825-1969, minimum corroded skirt thickness = 7 mm

Thus use a thickness of 7 mm for the skirt.

5. Nozzle Design:

1. Feed Nozzle:

nozzles are required for vapour discharge, reflux inlet, feed inlet, reboiled vapour inlet and

liquid bottoms outlet.

The inner diameter of the nozzle is calculated by the formula:

Di = 0.363 * (ṁ) 0.45 * (ρ) 0.13

Where Di = inner diameter of the nozzle in m

ṁ = mass flow rate in m3/s

ρ = density of the liquid in kg/m3

For feed, ṁ = 0.00376 m3/s and ρ = 738.2 kg/m3

Di = 0.0694 m

and the thickness of the nozzle is calculated by the formula:

t = (P * Di)/ (2 * f * J – P)

Where t = thickness of the nozzle in m

P = design pressure in Pa = 1.1 * maximum pressure

Di = diameter of the nozzle

f = allowable stress

J = joint factor = 0.85

Calculating thickness = 0.006 mm

additional thickness for corrosion be 3 mm.

Total Thickness = 3.006 mm

Standard thickness is 6mm.

2. Nozzle for distillate:

For distillate, ṁ = 0.00128 m3/s and ρ = 752 kg/m3

Following the above procedure,

Di = 0.0428 m

And thickness including corrosion allowance = 3.003 mm

So Standard thickness = 6mm

3. Nozzle for residue: For distillate, ṁ = 0.00184 m3/s and ρ = 987 kg/m3

Di = 0.0523 m

And thickness including corrosion allowance = 3.0045 mm

So Standard thickness = 6mm