CO2 Absorber

CO2 Absorber The Carbon dioxide gas produced resulting due to the reforming reactions, which is used as a raw material for Urea. It is separated from the gas mixture by the absorption in the solvent MDEA (Methyl Di ethanolamine). The gas mixture contains 16.95 mole % of Carbon dioxide which is separated by absorption in MDEA. The overhead contains 5.91% of carbon dioxide.

Material Balance Material balance of CO2 La*Xa + V*y = L*X + Va*Ya X = { (0.130871*50265.29)- (0.000233*80.42) }/6588.79 = 0.9984 y1 = 0.1308 y2 = 0.000233 x1 = 0.9984 x2 = 0 (if we consider the pure solvent) Y1 = 0.1308/(1-0.1308) = 0.15 Y2 = 0.000233/(1-0.000233) = 0.00023305 X1 = 0.9984/(1+0.9984) = 0.499 X2 = 0 L/V = (0.15-0.000233)/(0.499-0) =0.3 Now, The height of the column, Z= (HOG)*(NOG) X

Y

0

0

0.601

0.064

Fig. 9.1 Equilibrium Curve of CO2-MDEA system

From that, Y1* = 0.14X1 & Y2* = 0.14X2 NOG = (Y1-Y2)/{(Y1-Y1*)-(Y2-Y2*)}/ln((Y1-Y1*)/(Y2-Y2*)) = (0.15-0.000233)/{(0.0802)-(0.000233)}/ln((0.0802)/(0.000233)) = 3.7 ≈ 4 HTU, HOG= G /{Kya * (1-Y)*LM} (1-Y)*LM = {(Y1-Y1*)-(Y2-Y2*)}/ln((Y1-Y1*)/(Y2-Y2*)) = (13473.2)/(3000*0.4667) = 5.34 ≈ 6 So, Z= (HOG)*(NOG) = (4)*(6) = 24 unit (m)

Mechanical Design: If we consider L/D = 8 So, D = 3 m L = 24 m Design Pressure = 10.5 kg = 1.029N/mm2 Temperature = 70° C Packing Height = 2.4 m

Packing type: Pall Rings Thickness of insulation = 80 mm (PUF) MOC: Carbon Steel (SA-516 grC) Fall = 150.48 N/mm2 E = 2 x 105 MN/m2 Corrosion Allowance = 2mm Joint efficiency Factor = 0.85 Top disengaging space = 1 m Weight of attachments (pipes, ladders & platform) = 750 N/m Wind velocity Vw = 1.24 m/s

Weight of fluid and Packing = 900 N/m2 Sp. Gravity of MOC = 7.85 gm. /cm3 Column Shell Thickness: ts = PD / 2fJ – P ts = 1.029*3000 / 2*105.48*0.85 – 1.029 ts = 17.3 mm ~ 18 mm td = 18 + 2 = 20 mm Axial Stress due to pressure: fap = PD/4(t - c) = 1.029 * 3000 / 4(18) = 42 N/mm2 Now, Do = Di + 2t Do = 3040 mm ρshell = 7850 kg/m3 ρshell = 7850*9.087/106 = 0.077N/mm2 1) Stresses due to load: 

Due to weight of shell: Fdx = weight of shell / cross section of shell = Π/4(Do2 – D 2) * ρ* H Π/4(Do2 – Dii2) =ρH = 0.077H N/mm2



Due to weight of insulation:

Fdins

=



Dins Hinstins Dm t s  C 

Dm = (Do+ Di)/2 = 3020 mm Dins = Do + 2*tins = 3040 + 2(80) = 3200 mm Fdins = 3200 * 80 * 5640 * 10-6*H / 3020*10 = 0.0478H N/mm2  Due to weight of liquid & tray: Weight in terms of height = (H/ packing spacing)* (Π/4*Di2)*(Wt of liq. + tray) ρliq = 1040 kg/m3

Weight of each tray = 0.8 kN/m2 = 0.8 * 103 / 9.81 = 81.55 kg/m3 Weight = (H/4.13)* (Π/4*32)* 81.55 = 139.5 H kg Fliq = ( 139.5 H) / ( Dm ts  C  ) = 139.5 H*9.807/ Π*3020*18

= 0.008014 H N/mm2 

Due to attachments:

Fdatt = Wt. of attachments * H / ( Dm t s  C  

= 750*H / (3.14*3020*18) = 0.00439 H N/mm2 Now, Total Stress = (0.077 + 0.0478 + 0.008014 + 0.00439)H Fdh = 0.1372 H N/mm2 Stresses due to wind load: Fwind = 1.4 Pw H2 / П*Do(t-c) Pw = 0.05(3.6*1.242) = 0.2767 N/m2 Fwind = 1.4*0.2767*H2 / Π*3040*18 = 0.06915 H2 N/mm2 Now, AH2 - BH + C = D A = 0.06915 B = 0.1372 C = 40 D = 105.48 On solving the equation, 0.06915H2 – 0.1372H + 40 = 105.48 H = 27.2 M The maximum height that can be withstood 20mm of thickness of this vessel is 27.2m so we can go further with the thickness of 20mm with 24m of height calculated earlier in the process design of absorption column.

Skirt Support : Data : Dia. of vessel = 3000 mm Height of the vessel = 24 m Wt. of vessel + attachments = 351153.27 N Dia. of skirt = 3000 mm Height of skirt = 4 m Wind press. = 1285 N/mm2 288

Calculation: Stress due to dead wt. Fd = ΢W / (Пdsk tsk) = 351153.27 / (П*3000*tsk) = 37.27 / tsk N/ mm2 Stress due to wind load Fwb = Mw/Z = 4Mw / П*dok2*tsk Where Mw = Plw(h1/2) + Puw(h1 +(h2/2)) Fwb = 41.2/tsk N/mm2 Sesmic load Fsb = (2/3)*CWH/(П*Rok*tsk) = 0.0181 N/mm2 Min. tensile stress Ftmax = (41.2 – 37.27)/tsk Ftmax = 3.93/tsk N/mm2 Permissible tensile stress = 1400 N/mm2 tsk = 130/1400 = 0.929 mm Min. compressive stress Fcmax = (41.2 + 37.27)/tsk = 78.47 /tsk N/mm2 Fcpermissible < (1/3) yield point < 200/3 = 66.6 N/mm2 tsk = 78.47/66.6 = 11.78 mm Use thickness of 12 mm

Skirt bearing plate Assume bolt circle diagram = skirt dia. + 32.5 cm = 332.5 cm Compressive stress between bearing plate and concrete foundation. fc = (΢W/A) + (MW/z) = (351153.27/A) + (0.7*1285*39.82/2*z) = 1.22 N/mm2 Mmax = fc*bl2/2 = 1.22*162.52/2 f = 6Mmax/btb2w tb = 24.81 mm > 20mm So, a bolting chair can be used. Anchor bolts: Min. wt. of the vessel = 350403.27 N Fo = (Wmin/A) – (Mw/z) = -0.91 N/mm2 Since it is negative, the vessel skirt must be anchored to concrete foundation by an anchor bolt. 24 bolts should be used (Since 3000mm dia. is max. for 24 bolts) Ab = 1/Nb*fb[(4Ms/Db)-W] = 1/(24*1.22)[(4*1254/332.5) – 351153.27] = 182.52 mm2 Bolt Diameter = Db = (4 Ab/π) ½ = (4*182.52/3.14) ½ = 15.24 mm

Nozzle Design

For Nozzle Nozzle diameter (m) pressure (N/m^2) all. Stress f (N/m^2) thickenss of nozzle (m) design thickness (m) thickness (mm) shell dia (m) thickness of shell (m) head thickness As (m^2) A' (m^2) Ho (m) Hi (m) Ao (m^2) Ai (m^2) Ar (m^2) Tr (mm) Ring pad req.

Gas Inlet

Gas Outlet

1 0.45 1961331 98000000

2 0.3 1961331 98000000

0.00570012 0.00770012 7.7 3

0.00380008 0.00580008 5.8 0.3

0.02 0.02 0.03 0.03 0.0417556 0.0042256 0.0018614 0.0012464 0.050870424 0.033988233 0.040985363 0.023392306 -1.218E-08 -5.42526E-09 0.000303292 8.42123E-05 -0.04019748 -0.003063407 -92 -12 no No

Flange Design ffa = permissible tensile stress in flange under atmospheric conditions ffo = permissible tensile stress in flange under operating conditions. fba = permissible tensile stress in bolt under atmospheric conditions fb0 = permissible tensile stress in bolt under operating conditions. Nmin = minimum gasket width NozzleDesign pressure nozzle Diameter j Welded shape factor thickness of nozzle(mm) ya (min gasket seating stress)

= = = = =

1.029 450 0.8 5 124.11

(N/mm2) (mm) mm N/mm2

m(min. gasket factor) shell diameter (mm) gasket inside dia Gi (mm)

= = =

ffa ffo fba fbo

= = = =

129 113.76 103.42 86.87

Nmin = minimum gasket width Nmin Nmin(mm)

= =

(Go-GI)/2 15.5

N N(mm)

= =

Nmin + 0.005 cm 15.505

Go/Gi Go Gi gasket outside dia Go1(mm)

= = = =

(Y-pm/Y-(m+1)*p)0.5 ((Y-pm/Y-(m+1)*p)^0.5)*Gi 300 mm 451.9674419 mm

gasket outside dia Go2(mm) gasket outside dia Go2(mm) Go(mm)

= = =

2*Nmin+Gi 481 mm 481 mm

b0 = basic gasket seating width Bo bo(mm)

= =

N/2 7.7525

bo>6.3mm then b=(bo/2)^0.5 b(mm)

=

1.392165579 mm

G G(mm) Hp= Hp Hp(N) H= H H(N) BOLT LOAD CALCULATIONS

= = = =

5.5 3000 Mm 450 Mm N/mm2 N/mm2 N/mm2 N/mm2

(Gi+Go)/2 465.5 mm force required to keep gasket from leaking ∏(2b)Gmp 23032.88213 N total hydrostatic force

= =

∏/4*G2*p 175034.7998 N

Wm1 Wm1(N)

Wm2 Wm2(N)

Exists under the operating condition Wm1 = =

H+Hp 198067.6819 N

Because of tightening up of the bolts Wm2 = 3.14*G*b*Y = 252549.7836 N calculation of Minimum Bolting Area Required (Am)

Am1 Am1(mm2)

= =

Wm1/fbo 2280.046989 mm2

Am2 Am2(mm2)

= =

Wm2/fba 2441.98205 mm2

Am Am(mm2)

= maximum of Am1&Am2 = 2441.98205 mm2

Bolt calculation (Ab, db)

Ab Ab(mm2)

Actual cross sectional area of bolt should not exceed = 2*3.14*Y*G*N/fba = 225.8 mm2

number of bolts. N N

= =

Am/Ab 10.81480093

n1

=

12

Bolt Diameter(db)

Am Db db(mm)

= = =

3.14/4*db2 (Am*4/3.14)0.5 16.10073797 mm

g1 g1(mm)

= =

1.5go 7.5 mm

R R(mm)

= =

(G-Do)/2+10+db/2 20.5 mm

Bolt circle diameter (C) C1 C1(mm)

= =

ID+2(g1+R) 506 mm

Bolt Spacing (Bs) Bs Bs(mm)

= =

3.14*C1/n1 132.4033333 mm

thickness from baffle spacing tf1 tf1(mm)

= =

(Bs-2*db)*(m+0.5)/6 100.2018574 mm

E = edge clearance, mm E(mm)

=

20.4 mm

Outside Diameter of Flange(DFo) DFo = Dfo(mm)

= =

C + 2E 546.8 mm

H*G (mm^2)

=

Hg hG(mm)

81478699.3 mm2

hG = Radial distance from gasket load reaction to bolt circle = C-G/2 = 20.25 mm

Wm = total bolt load Wm Wm(N)

= =

K K

= =

Flange thickness Tf tf(mm)

= =

max(wm1,wm2) 252549.7836 N

1/(0.3+(1.5*Wm*hg/HG)) 2.537106732

G*(P/K*f)^0.5+C 27.79477816 mm