PROCESS DESIGN:

Let ,

Mole fraction of MEG XF= 0.9889 XD = 0.999 XW = 0.4918 9.2.1 Nos of theoretically stages

Use McCabe Theile method for determining the theoretical stages.

McCabe Theile Method: [17,21]

61 So, top = P^satMEG / P^satDEG

= 1.334 / 0.1333 = 10.077

Now at bottom P^satMEG = 1.4757 KPa P^satDEG = 0.1673 KPa

So, _bottom = P^sat_MEG / P^sat_DEG

= 1.4757 / 0.1673

= 8.8174

Thus, average = { top x bottom }^0.5 = {10.007 x 8.8174}^0.5

average = 9.3933 Now we have the equation

Y =  X / {1+ (-1)}

From the above eqⁿ we can generate the vapor- liquid equilibrium data given as follow.

Table-9.2 Vapor-Equilibrium data

X 1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Y 0 0.51069 0.70134 0.80102 0.8623 0.9037 0.9337 0.9563 0.9740 1

Now plotting x-y diagram & Using McCabe Theile method for determining the theoretical stages.

no. of theoretical stages = 18

9.2.2 Packed Column Diameter And Column Height:

The necessary data required is as follows : Reflux Ratio = 0.71

Feed = saturated liquid (assume) Kmol of Feed F = 188.306 kmol

Kmol of Distillate D = 184.54 kmol Kmol of Bottom W = 184.54 kmol Liquid density L = 954 Kg/m³

62 Vapor Density G = 2.099Kg/m³

Liquid Viscosity  = 2.2 x 10^-3 Kg/m s

No. of theoretical stages = 18 ( from vapor liquid Equilibrium diagram) Owing to its low pressure drop per theoretical stage metal Pall Ring is often preffered to other packings for vacuum distillation.

Here, L/D =0.71

63 Owing to its low-pressure drop, the metal pall ring is used as a packing for vacuum distillation,

Taking vapor mass velocity as 60% if flooding velocity.

Hence, Area of tower =

64

2

4 D A_{t }

= 18.19

Diameter of tower = D = 4.81  5 m Total height of packed bed,

Z = No.of theoretical stages x H.E.T.P.

HETP = height equivalent to one theoretical plate For pall ring metal (1.5”) H.E.T.P. = 0.5425

Z = No .of theoretical stages x H.E.T.P.

Z = 18 x 0.5425 Z = 9.765 m

= 10 m

In packed tower, height of each bed is approximately ≥ its diameter.

Hence, 2 beds of 5 m heights are provided.

Spacing between each bed is = 2 m Hence, total spacing = 0.5 x 3 = 1.5 m

Taking both disengaging space and distribution space = 2 m Hence total height of tower = 10 + 2 + 2

= 14 m MECHANICAL DESIGN [23, 24, 26]

9.2.3 Shell Design:

Shell Thickness (based on external pressure):

Diameter of tower = Di = 5m Height of column = 14 m

Pressure inside column is vacuum.

Outside pressure is 1 atm = 0.1 MN /m² Hence design pressure = Pd = 0.1 x 1.05

= 0.105 MN /m² Shell is I.S. 2825-1969

Allowable stress = f = 98.1 MN /m² Welding joint efficiency factor = J = 0.85 Thickness of shell,

The inside depth of the end can be calculated from the following correlation, hi = Ri -



RiDi/2



x(RiDi/22ri

65

Effective length of tower W/O stiffner = (Tangent to Tangent length + 2/3 x hi) = 12 + 2/3x 0.9688

= 12.6459m Do / L = 5/ 12.6459

= 0.3953

K value corresponding to this = 0.246 & m =2.43 Modulus of elasticity E = 1x 10⁵ MN/m₂

Corroded shell thickness,

P = KxE x (t/Do)^m

0.105 = 0.246 x 1 x 10⁵ (t/4)^2.43 t = 0.02467 m

Take standard thickness = 26 m t = 26 m

66

= 454.68 cm Temperature = Td = T + 50 ^oC

= 95 + 50

= 145 ^oC Table-9.3 Head thickness

t(cm) (assumed)

Ro / t Ro / 100t Factor B

P = xRo t factB

/ 22 . 14

1.2 378.9 3.78 4400 0.7402

1.3 349.75 3.49 4500 0.8201

1.4 324.77 3.24 4800 0.9421

1.5 303.2 3.03 5000 1.051› Pdesign

So, we take thickness of head t = 1.5 cm Consider 2mm corrosion allowance,

t_h = 1.5 +0.2 t_h = 17 mm Standard thickness available t_h = 18 mm

9.2.5 Check For Stresses In MEG Column [Design Of Vertical Column]:

DATA:

Vessel Type = Class 1

Pdesign =1.05 MN/m2 Total Column height = 14 m

Column Diameter = 5m Skirt height = 4 m

Packing Type = 38mm Metal Pall ring Column MOC = SS304

Wind Pressure = 1000 N/m²

Weight of attachments (ladder + pipes + liquid distributor) = 1373.4N ρ_liquid = 1113.2 Kg/m³ =10.917 x 10³ N/m³

Head = Torispherial Head thickness =18mm T_design = 145^o C

Permissible Stress =91.8 MN/m²

67 ρs = 78480 N/m³

Insulation = Asbestos ρ_in = 5.64*10³ N/m³ t_in =50 *10^-3 m

Shell thickness without corrosion allowance = 26 = 0.026m Do = 5.052m

Vessel is Located in non seismic zone

 Determination Of Longitudinal Stresses:

 Axial Stress due to pressure _p = P x Do / 4 x t_s

= 1.05 x 5.052 / 4 x 0.026 = 51.005 MN/ m²

Axial Stress Due to dead Load up to X m (a) Stress due to Shell weight s

f_s = Ws / Π x D x t_s

Ws = weight of shell = Π x D x t x ρ_s x X f_s = Π x D x t x ρ_s x X / Π x D x t_s

= ρs x X

= 0.7848 X MN / m² ( b) Stress due to insulation in

in = tin x ρin x X/ ts

= 0.05 x ( 5.64*10³ ) x X / 0.026 in = 0.01084 X MN / m²

(c ) Stress due to liquid and packing load in column Volume of bed = Π x D²x H /4

= Π x 5² x 10 = 196.3495 m³ Consider void fraction for packing є = 0.5 Liquid hold up = Vb x є

= 196.3495 x 0.5 = 98.1747 m³

68 Weight of liquid = Vl x ρl

= 137.44 x 954 = 93658 kg = 918.511 x 10³ N

Total weight of packing = packed bed volume x density of packing = Π x D²x H /4 x ρp

= Πx5²x 10 /4 x 430

= 83.43 x 10³ kg = 824.08 x 10³ N l = Weight of (liquid + packing) /  x Dx ts

Weight of ( liquid + packing) =Weight of liquid + total weight of packing = ( 918.511 x 10³ + 824.08 x 10³ )

= 1.7465MN _l = 1.7465 x X /  x ( 5.052) x( 0.026 ) = 4.243 x X MN/m²

( d ) Stress due to attachments

Weight of head = /4 x D² x thead x ρs

= 0.785 x ( 5.052 )² x ( 0.018 ) x ( 784480 )

W_head =0.02831 MN

Weight of ladder + liq distributor = 1373.4 x X N a = Whead + (Wladder & liquid distributor) /  x Di x t = 0.02831 + 0.0013734 x X /  x ( 5 ) x (0.026) a = 0.06931 + 0.003362 x X MN/m²

Total dead load stress dead = s + in + l + a

= 0.7848 X + 0.01084 X + 4.243 X + 0.06931 + 0.003362 X dead =0.06931 +4.3246 xX N/m²

 Longitudinal Stress due to dynamic loads

The axial Stress due to wind load in self supporting tall vessel,

69

The total load due to wind acting on the bottom and upper part of thr vessel are determined by

P_w = K₁ x K₂ x P_windpress x H x Do ( insulated ) Where,

Pw = wind pressure = 1000 n/m2

K₁ = coefficient depending upon the shape factor = 0.7 for cylindrical surface K₂ = coefficient depending upon the period of vibration of the vessel

= 1 if period of vibration is 0.5 sec or less =2 if period of vibration is exceed 0.5 Do( insulated ) = Di + 2 tin

Period of Vibration is given by

T = 6.35 x 10^-5 x ( H / D )^1.5 x ( W / t ) ^0.5 H = L + Skirt height

W = total weight Now,

Weight of shell

Ws =  x Dits x L x ρs

= (3.14) x (5) x (0.026) x ( 14 ) x (78480) Ws = 0.4487 M N

Weight of insulation

Win =  x D x tin x L x ρin

= ( 3.14 ) x ( 5 ) ( 0.05 ) x ( 14 ) x ( 5.64*10³ ) Win = 0.06266 MN

Weight of ( liquid + packing) = 1.7465 MN

Weight of attachment = Weight of two heads + Weight of ladder, pipes, liq distributor = 2 (0.02831) + 1.3734 x 10^-3

Wa = 0.058 N

Total Weight = W_S + W_in + W (liquid + packing) + W_a = 0.4487 + 0.06266 + 1.7465 + 0.058 W = 2315.86 KN

H = L + Skirt height = 14 + 4

H = 18 m

a) Period of Vibration

T = 6.35 x 10^-5 x ( H / D )^1.5 x ( W / ts ) ^0.5

70

T = 6.35 x 10^-5 x ( 18 / 5.052 )^1.5 x ( 2315.86 / 0.026) ^0.5 T = 0.127 sec < 0.5 sec



K2 =1

K1 = 0.7 for cylindrical vessel b) Wind load

Pw = K1 x K2 x Pwindpress x H x Do ( insulated )

Do( insulated ) = Di + 2 tin

= 5 + 2 (0.05)

P_w = ( 0.7 ) x ( 1 ) x ( 1000 ) x ( X ) x ( 5.1 ) Do = 5.1 m Pw = 3.571 x X KN / m

 Bending moment

i.e Wind moment at the base of the vessel due to wind load is given by Mwind = { Pwmd x H }/2

= 3.571 x X x X /2 M_wind = 1.785 x X² KN m

 Resulting bending stress

i.e Wind Stress in axial direction is given by

wind = 4 x Mwind /  x ts x D²

= 4 x (1.785 x X²) / 3.14 x ( 0.02 6) x ( 5.052 )²

wind = 0.00349 x X² MN/m²

 Calculation of Resultant longitudinal stress : Tensile stress on upwind side at X m from top

tensile = -p - dead + wind

f (max) = f x J

= 98.1 x 0.85 = 83.385 MN/m²

Substituting f _max for f _tensile

83.385 = 0.00349 X² - 4.3246X - 0.0693 – 51.005 Solving this equation

0.00349X² – 4.3246 X -134.46 = 0 ax² + bx + c =0

By Solving equation

71 X = 1496.59 >>> 18m

So This design is Ok

Compressive stress on downwind site at X m from top Down wind side ,

f compression =p + dead +wind

compressive = 0.125 x E x ( t / Do )

= 0.125 x ( 2 x 10⁵) x ( 0.026 / 5.052 ) = 128.661 MN/m²

Equating maximum value of f comp.

128.661 = 0.00349X² – 4.3246 X+ 0.0693 – 51.005 Solving this Equation

 0.00349X² – 4.3246 X -77.5867 = 0 ax² + bx + c = 0

X = 18.683 >>> 18 m So, This design is ok.

9.2.6 Skirt Support Design For MEG Column:

 The minimum weight of vessel with two head and the shell is given by W_min = W_s + 2 W_head

= 0.4487 + ( 2x 0.03831) Wmin =0.50532 MN

 The maximum weight of vessel with two head and the shell is given by W_max = W_min + W_insulation + W_attachment + W( liquid + packing )

= 0.50532 + 0.062665 + 0.058 + 1.74657 Wmax = 2.37248 MN



Period of Vibration is given by At minimum dead weight

Tmin = 6.35 * 10^-5 x ( H / D)^1.5 x ( Wmin / ts )^0.5

= 6.35 * 10^-5 x ( 18 / 5.052 )^1.5 x ( 505.32 / 0.026 )^0.5 Tmin = 0.059 sec < 0.5 sec

 K₂ =1

K1 = 0.7 for cylindrical vessel At maximum dead weight

Tmax = 6.35 x 10^-5 x ( H/ D )^1.5 x ( Wmax / ts )^0.5

= 6.35 x 10^-5 x ( 18 / 5.052 )^1.5 x ( 2372.48 / 0.026 )^0.5 T_max = 0.13 sec < 0.5 sec

 K₂ = 1

72 K1 = 0.7 for cylindrical vessel

 Minimum and maximum wind moments are computed by Maximum and minimum Wind load

P_w = K₁ x K₂ x ( Pw ) x H x Do (based on insulation thickness) Do = D_i + 2 x t_in

= 5+ 2 x ( 0.05 ) = 5.1 m

For minimum Weight condition Do = 5.052 m

For maximum Weight condition Do = 5.1 m ( insulated ) Pwmin = 0.7 x ( 1 ) x ( 1000 ) x ( 18 ) x ( 5.052 )

= 63.665KN

Pwmax = ( 0.7 ) x ( 1 ) x ( 1000 ) x ( 18 ) x ( 5.1 ) = 64.26 K N

Wind moment

Mwind = Pw x H / 2

Mwindmi = 63.665 x 18 / 2 = 572.89 KN m

Mwinsmax = 64.26 x 18 /2 = 578.34 KN

Wind stress

wind = 4 x Mwind /  x Di2

x tsk

_{wind min} = 4 x 572.89 / 3.14 x ( 5 )² * t_sk = 30.277 / tsk KN / m²

wind max = 4 x 578.34 / 3.14 x ( 5 )² tsk

= 29.4821 / tsk KN / m² dead,min = Wmin /  x Di x tsk

= 0.50532 / 3.14 x 5 x t_sk = 0.0328 / tsk M N / m²

dead max = Wmax /  x Di x tsk

= 2.37248 / ( 3.14 ) x ( 5) x tsk

= 0.151 / t_sk MN / m²

Maximum tensile stress without any eccentric load is given by

tensile = windmin - deadmin

Take J = 0.7 for double weld joint

 tensile = permissible x J = 98.1 x 0.85 = 83.385 MN/m²

73 tensile = windmin - deadmin

83.385 = windmin - deadmin

83.385 = 0.0328 / tsk + 0.030277 / tsk t_sk = 3.11 x10 ⁵ m

Maximum tensile stress without any eccentric load is given by

compressivein skirt wall = 0.125 E tsk / Do

= 0.125 (2 x10⁵⁾ t_sk / 5.052 = 4948.53 t_sk

compressivein skirt wall = windmax + deadmax

4948.53 tsk = 0.02948 / t_sk + 0.151 / t_sk

tsk = 6.054 mm

But minimum corroded skirt thickness = 8mm (as per IS : 2825-1969)  Skirt thickness tsk = 8

Table-9.4 Specification for Distillation Column (Packed column) SR.

NO

.

^PARAMETER DESCRIPTION

1 Tower MOC SS304

2 Tower ID 5 m

3 Tower OD 5.052 m

4 Shell thickness 26 mm

5 Shell head thickness 18mm

6 Height of tower(Without support) 14m

7 Height of packed Bed 10 m

8 Type of Head Torrispherical

9 No. of Beds of packing 2 , each of height 5 m

10 P/Z Selected 50 N/M²/m

11 Tower Support Skirt Support

12 Skirt MOC SS304

13 Skirt Height 4 m

14 Skirt thickness 8 mm

15 Type Pall Ring

16 MOC Metal (S S)

17 Bulk Density 430 kg/m3

18 HETP 0.5425m

74

CHAPTER X

In document Ethylene Glycol (Page 60-74)