Design Analysis for Methanol Production

Assumptions:

1. Fluid Stored = Methanol

2. Flowrate = 27,000 kg/hr and density of methanol = 792 kg/m3
3. Storage Time = 7 days
4. Tank Type = Vertical Cylindrical Storage Tank
5. Operating conditions:
● Pressure = 1 bar(atmospheric)
● Temperature = 25°C
● Design margin = 5%
6. Material = Carbon steel-ASTM A36 steel or equivalent (Metals & Inc, n.d.)
7. Design Code = API 650

Volume Calculation:

The volume of methanol to be stored is:

27,000 kg/hr
Vmethanol =
792kg /m3

= 34.1 kg/m3

For 7 days of storage:

Vtotal = 34.1 × 24 × 7 × 1.05(design margin)

= 6,014.4 m3

6014.4
Volume of a single tank considering 12 tanks is given as V =
12tanks

V = 501.2 m3

V = Vcylinder + Vroof

πD 2 H πD 2
V= +
1/3 × hr
4 4

hr = 0.2 D is the height of conical roof.

The optimum Height to Diameter ratio for a large methanol storage tank ranges from
1 to 1.5 ratio. We have calculated H/D ratio based on minimum cost of the vessel.
Cs = cost of the sides per sq. meter

Ch = cost of the head or top per sq. meter

= 1.5 Cs

Cb = cost of the bottom per sq. meter

= 0.6 Cs

Solution:

π D2
Let, Ct = Cs π D H + (Cb +Ch)
4

π D2H
V=
4

4V
H=
π D2

Substituting the value of H,

4V π D2
Ct = Cs + (Cb +Ch)
D 4

Differentiating with respect to design variable D and equating to zero

dC t = −4 C sV πD =
+ (Cb +Ch) 0
dD D2 2

Cs 8V
D3 =
C b+C h π

Substituting for the volume in terms of D and H gives the following optimum D/H
ratio for minimum cost of the vessel.

D = 2C s
H C b+C h

Substituting for Cb and Ch in terms of Cs gives,

D =
2Cs
H ( 0.6+1.5 ) Cs

H
= 1.05
D
As, H/D ratio is within the range it is acceptable.

πD 2 ×1.05 D πD 2
501.2 = + 1/3 × 0.2 D
4 4

Solving the value of D, we get D = 8.29 m and H = 1.05 × 8.29 m = 8.71 m

Shell Thickness

Using one foot method,

4.9 D( H −0.3 G)
T= +C (Petroleum Institute, n.d.)
Sd

C = Corrosion allowance (mm)

H = Height of the tank (m)

D = Diameter of the tank (m)

Sd = Allowable stress (MPa)

Figure 1: Allowable Stress for material (Petroleum Institute, n.d.)

From the above table, Allowable stress for A-36 carbon steel is 160 MPa.

Corrosion rate for carbon steel is 0.25 mm per year (Tjahjanti et al., 2019)

0.25 mm
C= × 15 years
year

C = 3.75 mm

4.9 ×8.29 (8.71−0.3 × 0.792)

T== +3.75
160

T = 5.35 mm
Tfinal = 6mm for safety and fabrication.

Figure 2: Minimum Shell thickness

As the diameter is 8.29 m < 15 m, the shell thickness of 6 mm is accepted.

Roof Design

Methanol is volatile, flammable liquid with low boiling point of 64.7°C. Pantoon-
type internal floating roof is preferred as it controls vapor emissions, minimizing
evaporation and provides vapour barrier. It is simpler and cheaper providing adequate
vapor suppression for lighter liquid like methanol.

Loads on internal floating roof:

Dead load (Wd):

Wd = Area × Deck plate thickness(aluminium) × Density of material(aluminium) × g

Wd = πr2 × 0.003 × 2700kg/m3 × 9.81

Where, deck plate thickness is taken within the range of 0.2 mm to 0.5 mm for
internal floating room with aluminium (Petroleum Institute, n.d.)

Density of aluminium = 2700 kg/ m3

Wd = 4.28 KN

Live Load (Wl):

API 650 specifies a minimum live load of 0.24 Kpa (Petroleum Institute, n.d.)

Wl = 0.24 × A = 0.24 × 53.97 = 12.95 KN

Ftotal = Wd + Wl

Ftotal = 4.28 + 12.95

Ftotal = 17.23 KN
Buoyant force from pontoon

We selected 4 pontoons with the diameter(d) of 0.6m and Length(L) = 3 m

The Volume of one pontoon = π(d/2)2L

= 0.848 m3

Total Submerged volume for 4 Pontoon (Vsub) = 4 × 0.848 = 3.39 m3

B = Vsub × Density of methanol × g

B = 3.39 × 792 × 9.81

B = 26.36 KN

Buoyancy check

The total load acting downward is Ftotal =17.23 KN, and the total buoyant force is B =
26.36 KN.

= B > Ftotal, the buoyancy provided by the pontoon is sufficient to support the total
load with safety margin hence, the diameter and length of the pontoon is accepted.

Base plate

Total force on a base

Wtotal = Wshell + Wmethanol + Wcone + Winternal Floating roof

Wshell = πDH × ts × ρsteel

= π × 8.29 × 8.71 × 0.006 × 7850

= 10,684 kg

Wmethanol = Vmethanol × ρmethanol

= 501.2 × 792

= 396,950 kg

Winternal floating roof = πr2 × tr × ρaluminium

= π (4.145)2 × 0.003 × 2700

= 437 kg
Wcone = πr2 × t × ρsteel

= π (4.145)2 × 0.006 × 7850

= 1271.2 kg

Wtotal = 10,684 kg + 396,950 kg + 437 kg + 1271.2 kg

= 409,342 kg

Ftotal = 409,342kg × 9.81 = 4,015,645 N

Thickness on base plate = Ftotal/allowable stress of material × Area of base plate

4,015,625 N
=
160,000,000 Pa ×53.975 m 2

= 2.5 mm + 3.75 mm (corrosion allowance)

= 6.25 mm.

Drainage slope

A slope of 1:120 is preferred towards the drain to prevent liquid accumulation.

D
Slope Height Difference =
120

= 69 mm, the center of the tank bottom should be 69 mm higher than the edges near
the shell.

Nozzle

Process inlet nozzle:

Flowrate is given by, Q = Volume/Time

Q = 6014.4/168 hours

Q = 0.00996 m3/s

Assume velocity for liquid = 4m/s

Area = Q/Velocity

= 0.00996/4
A = 0.00249 m2

Nozzle Diameter(D) = (4A/π)1/2

D = 56.3 mm

Figure 3:Dimensions for Shell nozzle

From the above table, the reinforcing required for the nozzle is given in column-5 and
calculated by interpolating as;

Interpolation:

( 3068−2763 )
54 + ( ×56.3
( 60−54 )

The reinforcing required for the nozzle of size 56.3 mm is 2951.9 mm.

Figure 4:Dimensions for shell nozzle flange

From the above table, minimum thickness of flange and outside diameter is given by
column-2 and column-3.

Interpolation:

The minimum size of flange is 77.3 mm and outside diameter of flange is 1708 mm.

Manhole
Manhole shall be provided around the roof to provide an effective pattern for access,
lighting, and ventilation of the product storage interior. The manhole shall have a
minimum nominal diameter of 600 mm (Petroleum Institute, n.d.).

The table below represents the size of manhole which is 600 mm along with diameter
of neck as 600 mm, cover plate of 762 mm and bolt circle of 699 mm.

Figure 5:Dimensions for Roof manhole

Wind Load

Wind Pressure:

It is calculated using the formula

(Society of Civil Engineers, n.d.) connected to API-650

P = 0.613CdV2Gh

P = wind pressure (Kpa)

Cd = Drag coefficient for cylindrical tank = 0.7, per API 650.

Gh = Gust factor = 1.1(for exposed location)

V = velocity which is taken 190 km/hr for minimum wind load equivalent to 52.7
m/s (Petroleum Institute, n.d.)

P = 0.613 × 0.7 × 52.72 × 1.1

P = 1.31 Kpa.

Over-turning Moment:

Mw = PDH2/2 (Society of Civil Engineers, n.d.)

= 1.31 × 8.29 × 8.712/2

= 411.9 kNm.

Resisting Moment:

Mr = Weight of the tank × D/2

= 4,015,645 N × 8.29/2

= 16644.8 kNm

Factor of Safety:

Mr
FOS =
Mw

16644.84/411.9

FOS = 40.4 > 1.5

The tank is stable under given wind load.