chapter 6 mechanical design

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6-1

CHAPTER 6

MECHANICAL DESIGN

6.1 Mechanical Design Of Esterification Reactor 1 (R-101)

6.1.1 Determination Of Orientation Of Reactor

The esterification reactor is designed as a vertical vessel. Since we are dealing with

boiling process, this vertical orientation will provide bigger surface area for

vaporization of the liquid reactants. Bigger surface will hence increases the rate of

product’s formation per unit time.

6.1.2 Determination Of Material Of Construction

The material of construction selected for the reactor system is stainless steel (316).

Acetic acid and para-toluene sulphonic acid utilized in this process give a very

significant corrosion effect to the reactor. Hence, stainless steel is the best material

to be used since it can resist such effect very well. The components in the stainless

steel (316) has provide following functions.

a) Nickel increases toughness and improves low temperature properties and

corrosion resistance.

b) Chromium improves hardness, abrasion resistance, corrosion resistance and

resistance to oxidation.c) Molybdenum provides strength at elevated temperature. Greater strength

can permits thinner walls in process equipments.

6.1.3 Selection Of Impeller Arrangement

The turbine with flat vertical blades extending to the shaft is suited to the vast

majority of mixing duties up to 100,000 cP or so at high pumping capacity. Because

of that the turbine with flat vertical blades is selected [Sinnot 1999].



6-2

Size of impeller depends on the type of impeller and operating conditions.

For turbine impeller, the ratio of diameter of impeller and vessel is in the range of

d/D = 0.3 to 0.6.

Take the ratio of diameter of impeller and vessel as 0.4, hence

The diameter of impeller,

435.24.0 d (6.1.1)

m974.0d

The width of impeller,

8d (6.1.2)

12m.08974.0

The offset of baffle

2d (6.1.3)

m0.5m487.02/974.0

The baffle width,

12 D (6.1.4)

m203.012435.2

The space between impeller and vessel bottom

= H/6 (6.1.5)

= 7.305/6 = 1.2175 m

The baffle height,

6 H H (6.1.6)

mmm 0875.62175.1305.7



6-3

The space between impeller

3 H (6.1.7)

2.435m3305.7

6.1.3.1 Shaft Design

Before shaft design can begin, the motor horsepower, shaft speed and impeller

must be selected first for a turbine agitator application. Shaft design requires two

steps:

1) Establishing the minimum shaft diameter to withstand the forces

acting upon the shaft

2) Determining the natural frequency of vibration for the shaft andturbines

The shaft and impeller must not rotate close to their natural frequency, N c. As stated

in Journal of Liquid Agitation on August 1976 by Wayne D. Ramsey and Gerald C.

Zoller, the operating speed,N of the shaft must be sufficiently far from the systems

natural frequency to prevent the deflections that exceed the yield stress. Then,

N N c .

6.1.3.2 Speed Impeller

51

3 )(394 N nS

H D

g

p

(6.1.8)

where D = blade diameter

Hp = horsepower

n = number of blade

S ggg === specific gravity of 2-EHA

N = speed impeller

For typical power consumption for blending of low viscosity liquids. It is about 0.2

kW/m 3 (Coulson & Richardson, Chemical Engineering, Volume 1, page 293)



6-4

Then, power for the agitator,P is;

V x p P (6.1.9)

where p = power consumption per volume (kW/m 3)

V = volume of reactor in m 3

33 342.0 m x

mkW

P

P = 6.8 kW @ 9.1189 hp

The speed of impeller can be calculated as below

51

3 ))887.0(41189.9

(394974.0 N

314 5702.2

102324.9 N

54.303073 N

Speed impeller, N 158.63 rpm

The torque transmitted by the shaft will have the maximum value above the

uppermost turbine. Since the power drawn by sealing devices is in significant, the

maximum torque,T Q becomes:

N

H T p

Q 63025(max) (6.1.10)

Where H p= Agitator power

N = speed impeller

Then, the torque value can be calculated as below;

63.1589.1189

63025(max)

QT

014.3623(max)

QT

The maximum bending moment, M max , is the sum of product of the hydraulic forces

and the distance from individual impellers to the first bearing

ND L H M p

'

max 19000 (6.1.11)



6-5

Where H p= Agitator power

N = speed impeller

L’ = Length of vessel

D = blade diameter

Then, the bending moment value is

974.063.1587.3059.1189

19000max M

6614.8191max M

The minimum shaft diameter values can be calculated due to relationship to meet

the shear stress and tensile stress, respectively:

Shear stress diameter;

31

2max

2(max)

]16

[ s

Q

s

M T d

(6.1.12)

Tensile stress diameter;

31

2max

2(max)max

])(16

[t

Q

t

M T M d

(6.1.13)

Then, the shear stress and tensile stress can be calculated as below:

31

22

]6000

6614.8191014.362316

[ sd

incd s 9664.1

31

22

]10000

)6614.8191014.36236614.8191(16[

t d

cmincd t [email protected]

Assume diameter = 12 cm



6-7

Then;

5.30305.74763.11305.7

)112.0(10388.0 26 x N c

1583.152119017

c N

rpm N c 1947.782

Since the natural frequency of 782.1947 rpm for the shaft and impeller is more than

the operating speed of 158.63 rpm, then the 12 cm shaft is acceptable.

6.1.4 Cooling Elements Of The Reactor

6.1.4.1 Selection of Cooling Element and Utility

For cooling of the reactants, an internal helical coil is used. The coil is the simplest

and cheapest form of heat transfer surface and it is installed inside the reactor

vessel. The utility supplied to the coil is cooling water.

6.1.4 .2 Determination of Coil’s Dimensi on and Heat Transfer

Now, the length of the cooling coil is determined using following formula [Incropera,

2002];

r imom P it s AQT T mC L D A ,,,

Rearraging, it r imom P D AQT T mC L ,,,

Where, A s is the heat transfer area of the coil, A r is the heat transfer area of the

reactor and D t,i is the inside diameter of the coil (which is assumed to be 0.3 m)

= −3.495 × 10 6 /

Q absorbed by the process is 3.495x10 6 kJ/hr per unit area of the vessel.

By using equation [Incorpera, 2002].

= ∆= ( − )



6-8

By assuming the inlet cooling water at 22°C, the outlet cooling water at 80°C with

Cp= 4.18 kJ/kg.K.

=

∆ (80 −22)=

3.495 × 10 6 /

4.181 . × 80 −22 ° . °

= 14,412.49 /

Assume no heat loss from the coil’s wall, Q transferred by the coil is equal to the Qabsorbed by the process in the vessel.

= = × 2.435 × 7.305= 55.8616 2 (6.1.18)

= 55.8616 × 4.181 58 × 14,412.493.495 × 10 6 × × 0.3

= 59.3 ≈60

The number spiral formed by the cooling coil around the reactor can be calculated

by dividing the length of the coil by the reactor circumference,

=59.29× 2.435

= 7.75 ≈8

6.1.5 Determination Of The Thickness Of Wall Vessel

There is a minimum thickness requires to ensure that any vessel is sufficient rigid to

withstand its own weight and any accidental loads. For a cylindrical vessel, the

minimum thickness required to resist internal pressure can be determined from the

following equation:

= 2 − (6.1.19)

The design conditions for esterification process between acrylic acid and

2EHOL is set to be: 6 bar as the design pressure and the design temperature is at

10% above operating temperature which is 132°C.

The design pressure is taken as 6 bar because during the shutdown

process, the caustic wash of the equipment will be conducted in normal atmospheric



6-9

pressure and it is also due to avoid counterfeit operation during minor process upset

and safety condition.

Therefore, the design stress for stainless steel (316) is taken as 140.4 N/mm 2 at

132°C.

, = 6 = 5.9215 = 6 = 0.6 2

, =0.6 (2435)

2 140.4 −(0.5)

= 5.2123 ≈5.3

The corrosion allowance is the additional thickness of metal added to allow

for material loss by corrosion and erosion. For esterification process, there will be a

severe condition of corrosion arise from the acetic acid and p-toluene sulfonic acidused. According to Sinnot (1999), when this condition occurs, the allowance for

corrosion should be increased from 2 mm to 4 mm.

= 5.3 + 4 = 9.3 ≈10

According to Sinnot, 1999 also, for a vessel diameter of 2m to 2.5m, the

minimum wall thickness required should not be less than 9 mm. Therefore, the wall

thickness is acceptable.

The reactor is insulated to avoid loss of heat from the reactor (to conserve

energy) and to keep process conditions from fluctuating with ambient conditions.

Type of insulator used is rockwool. Rockwool is a ceramic material conceived with

fibres of molten stone. Its main features are its thermal insulation (low thermal

conductivity), non-combustible, fire-resistance and environmental friendly material.

Thickness of insulation is depends on process temperature as shown in the

following table 1.6.

Table 6.1 Thickness of insulation as a function of process temperature

T(°C) 93 205 316

t (mm) 12.7 25.4 31.75

tinsulation for the process with T = 200°C can be estimated by interpolation:



6-10

mmmm

t insulation

258330.24

7.12)7.124.25(9320593200

Therefore, the density of insulator at 25mm thickness is found as 800 kg/m 3.

Total thickness, t (including insulation) = 25 mm + 10 mm = 35 mm.

6.1.6 Head and Closure

The end of a cylindrical vessel is closed by heads of various shapes. The common

types used are:

a) Flat heads

b) Hemispherical heads

c) Ellipsoidal heads

d) Torispherical heads

The heads used for the vessel may be flat if they are suitably buttressed, but

preferably they are some curved shape as the hemispherical, ellipsoidal or

torispherical heads. However, the hemispherical heads are commonly used for high

pressures hence they are not suitable to be used for this esterification reactor. To

calculate the thickness of the head, the following equations can be used:

I. Ellipsoidal heads

=2 −0.2

(6.1.20)

By taking the value of joint factor, J as 1 which implies that the joint is equally

as strong as the virgin plate which achieve by radiographing the complete weld

length and cutting out and remaking any defects. The use of lower joint factors in

design, though saving cost on radiography, but it will result in thicker, heavier &

increase the cost of materials.

Assume J=1,

=0.6 (2435)

2 1 140.4 −0.2 (0.5)= 5.075 ≈5.1

By considering corrosion allowance of 4 mm,

Total thickness = 5.1 + 4 mm = 9.1 mm

T (°C) t (mm)

93 12.7

200 t insulation

205 25.4



6-11

II. Torrispherical heads

=2 −−0.2

(6.1.21)

=

1

4 3 + (6.1.22)

Since the crown radius, R c should not be greater than the diameter of the

cylindrical section, R c is taken as equal to D i. The ratio of the knuckle to crown

radius should not be less than 0.06 to avoid buckling. Hence, R k is taken as 6% of

R c.

R c = D i =2.435m.

Rk = (0.06)(2.435) = 0.1461 m.

=14

3 + 2.4350.1461

= 1.7706

=0.5 2435 1.7706

2 120 1 −0.5 1.7706 −0.2= 9.0116 ≈9.5

Total thickness = 9.5 mm + 4 mm = 13.5mm

III. Flat heads

= (6.1.23)

By assuming plate welded to the end of the shell with a fillet weld is used, C p is

taken as 0.55 and D e = D i.

= 0.55 2435 0.5120

= 86.4482 mm ≈ 86.5 mm

Total thickness = 86.5 + 4 = 90.5 mm.

The largest value of the thickness obtained by using flat head shows the

inefficiency of a flat cover. Hence, for the head of the vessel, ellipsoidal head is

much preferred since it is the most economical whereby the thickness of the head is

the same as the thickness of the vessel.



6-12

6.1.7 Determination of Pipe (Nozzle) Size

Stainless steel pipe is used for the inlet and outlet pipe of the reactor. Optimum

diameter for the pipe can be calculated using the following equation:

= 260 0.52 −0.37 (6.1.24)

I. Inlet pipe (At 110°C)

= =22920.8736

24.79833 = 924.2921 3

= 260 22920.87363600

0.52

924.2931 −0.37

= 260 2.6184 0.0799 = 54.3946 ≈55

Therefore, the pipe used is 55mm pipe.

II. Outlet pipe vent (At 110°C)

= =5052.2543

5.15843 = 979.4227 3

= 2605052.2543

3600

0.52

979.4227 −0.37

= 260 1.1927 0.0782 = 24.25 ≈25


III. Outlet pipe liquid (At 110°C)

= =17868.6194

19.75473 = 904.5250 3

= 26017868.6194

3600

0.52

904.5250 −0.37

= 260 2.3 0.08056 = 48.1749 ≈50




6-13

6.1.8 Determination Of Bolt & Flange Joint

Flanged joint are used for connecting pipes and instrument to vessel, for manholes

cover and for removable vessel head when ease of access is required. Flanged also

used on the vessel body, when it is necessary to divide the vessel into sections for

transport or maintenance. Flanged joints are also used to connect pipe to the

equipments such as pumps and valves.

Flanges dimension must be able to withstand the hydrostatic ends loads and

the bolt loads necessary to ensure tight joint in service. For the design of this heat

exchanger, welding-neck flange are used. It is because welding-neck flanges have a

long tapered hub between the flange ring and the welded joint. This gradual

transition of the section reduces the discontinuity stresses between the flange and

branch. It is also can increase the strength of the flange assembly.

Welding-neck flanges are suitable for extreme service conditions, where

flange are likely to be subjected to temperature, shear and vibration loads. They will

normally be specified for the connections and nozzles on process vessels and

process equipment. The dimensions of welding-neck flanges is chosen base on the

nominal pipe size of the nozzle pipe. From the interpolation made from table in

Appendix H1 in R.K. Sinnot, 2009, by using D nominal of 55mm for the inlet pipe,

25mm for the vent and 50 nm for the outlet pipe. The following values obtained for

bolt and flange at the reactor.

Table 6.2 Values for bolt and flange of the inlet pipe

Nominal

sized1

FlangeRaised

Face BoltingDrilling Boss

D B H d 4 f No. d 2 k d 3 55 65.6 146.7 14 38. 96.7 3 M12 4 14 116.7 78.7

Table 6.3 Values for bolt and flange of the vent pipe

Nominal

sized1

FlangeRaised


D B H d 4 f No. d 2 k d 3

25 33.7 100 14 35 60 2 M12 4 11 75 42



6-14

Table 6.4 Values for bolt and flange of the outlet pipe

Nominal

sized1

FlangeRaised


D B H d 4 f No d 2 k D 3

50 140 320 14 28 90 3 M12 4 14 110 80

Figure 6.1 Flanged Joint Standard (R.K.Sinnot, 1999)



6-15

6.1.9 Design Of Vessel Subject To Combined Loading

Instead of the pressure, the vessel is also subject to other loads. Hence, it must be

designed to withstand these loads without failure. The main sources of load to be

considered are:

a) Dead weight of vessel and contents

b) Wind

c) Earthquake (seismic)

d) External loads imposed by piping and attached equipment

From the previous page, the minimum thickness required for pressure loading is

9 mm. A much thicker wall will be needed at the column base to withstand the wind

and dead weight loads. As a first trial, divide column into five sections, with the

thickness increasing by 2mm per section. Try 11.1, 13.1, 15.1, 17.1, 19.1 mm.

I. Weight Loads

The approximate weight of a cylindrical vessel with domed head ends and

uniform thickness steel vessel can be estimated from the following equation:

= 240 + 0.8 (6.1.25)

By taking:

C v = 1.08 for vessel with only few fitting (internal coil)

Dm = [D i + (t x10-3)] m = 2.435 m + 0.015 m = 2.45m

Hv = 7.305 m , t = 9.1 mm

= 240 1.08 2.45 7.305 + 0.8 2.45 9.1

= 53,541.1750 = 53.54

II. Weight Of Insulation

Rock wool density = 800 kg/m 3.

Approximate volume of insulation

= = 2.435 7.305 25 10−3 = 1.3970 3 (6.1.26)



6-16

Weight Of Insulation,

= = 1.3970 3 800 3 9.81 2 = 10,963.656

= 10.9637 (1.25)

Double insulation for fittings = 2 10.9637 = 21.9274

III. Weight of external fittings

External fitting used at the reactor is a plain steel ladder. From Nelson

(1963), weight of the ladder is estimated to be 150 N/m length.

Hence,

= 150 7.305 = 1095.75 = 1.0958

IV. Weight of internal coil

= (6.1.28)

= 8000 3 , = 9.81 2 , 2 ,

Approximate volume of internal coil

= = 0.085 34 0.002 (6.1.29)

= 18.1584 10−3 3

= 8000 18.1584 10−3 9.81 = 1425.0712

Double this value to allow for f ittings,

= 2 1425.0712 = 2850.1424 ≈2.85

V. Weight Of Ellipsoidal Head

, = 9 ,

,

=13

112

3 −3 (6.1.30)

=1

36× × 2.45 3

−2.435 3 = 0.02343 3



6-17

= . (6.1.31)

∴ = 8000 0.02343 9.81 = 1838.7864

Double this value to allow fittings;

∴ = 2 1838.7864 = 3677.5728

∴ ,= + + + + (6.1.32)

= 53,541.1750 + 21,927.312 + 1,095.75 + 2,850.1424 + 3,677.5728

= 83,091.9522 ≈83.092

VI. Wind Loads

A vessel installed in the open must be designed to withstand the weight

bending stress caused by wind loading. The wind loading is a function of the wind

velocity, air density and the shape of structure. A column must be designed to

withstand the highest wind speed that is likely to be encountered at the site during

the life of the plant. For our plant in which located at Gebeng Industrial Site 2,

Kuantan, the worse-case wind speed that has occurred is 50 km/hr. (Malaysian

Meteorological department, 2011). However, this wind load does not give effect to

the equipment. Therefore, the value for preliminary design is taken as reference to

R.K.Sinnot, 2009 which is 160 km/hr.

,= 0.05 2 , / 2 (6.1.33)

= , /

= 0.05 × (160

)2 = 1280 2

Mean diameter including insulation,

= + 2 + 10−3 (6.1.34)

= 2.435 + 2 15.1 + 25 10−3 = 2.5152 ≈2.52

Wind loading (per linear metre), Fw = (6.1.35)

= 1280 2.5152 = 3219.456



6-18

Bending moment at bottom tangent line

=2

2 (6.1.36)

=3219.456 7.305 2

2 = 85,899.956 ≈85.90

VII. Analysis of stresses

At bottom tangent line

Pressure stresses:

= 4 =

0.5 2 2435

4 19.1 = 15.9359 2 (6.1.37)

=2

=0.5 2 2435

2 19.1= 31.8717 2 (6.1.38)

Dead weight stresses:

=+

=83,091.9522

2435 + 19.1 19.1 2

= 0.5643 2 (6.1.39)

Bending stress:

= + 2 = 2435 + 2 19.1

= 2473.2 (6.1.40)

=64

4 −4 =64

2473.2 4 −2435 4 = 1.1087 10 11 4 (6.1.41)

= ± 2

+ (6.1.42)

= ±85,899.956

1.1087 10 11

10001

4

24352

+ 19.1 = ± 0.9581 2

The resultant longitudinal stress is

= + ± (6.1.43)

,

= 15.9359

−0.5643 + 0.9581 = 16.3297 2

= 15.9359 − 0.5643 −0.9581 = 14.4135 2



6-19

As there is no torsional shear stress, the principal stresses will be .

The radical stress is negligible

≈2= 0.25

2

The greatest difference between the principal stresses will be on the down-wind side

whereby, [ 31.8717 – 14.4135] = 17.4582 2 which is below maximum allowable

design stress.

VIII. Check elastic stability (buckling)

Critical buckling stress:

= 2 10 4 (6.1.44)

= 2 10 4 19.12473.2

= 154.4558 2

Maximum compressive stress will occur when the vessel is not under pressure

= + = 0.5643 + 0.9581 = 1.5224,

Therefore, the vessel will be able to withstand in the case of increase in

external pressure, triggering collapse, or buckling of the tank. Thin-walled cylindrical

tanks are prone to buckling (or inward collapse) due to accidentally induced internal

vacuum. In industrial application, during the sterilisation process, steam cancondense, causing a reduction in volume.

16.3297

31.8717 30.2632

14.4135

Up-wind Down-wind



6-20

6.1.10 Reactor Support

The method used to support a vessel depends on the size, shape and weight of the

vessel; the design temperature and pressure; the vessel location and arrangement;

and the internal and external fittings and attachments. Since the design reactor is a

vertical vessel, a skirt support is recommended as it does not impose concentrated

loads on the vessel shell. Supports will impose localized loads on the vessel wall,

and the design must be checked to ensure that the resulting stress concentrations

are below the maximum allowable design stress.

I. Determination of total weight of vessel

= 90°

, = 135 2

, = 200,000 2

=4

2 (6.1.45)

= 4 2.4352

7.3051000

3 9.81 2 = 333,716.0082 ≈333.72

Total weight of vessel, W T = 83,091.9522

∴Overall total weight = W T + approximate weight (6.1.46)

= 83,091.9522 + 333,716.0082 = 416,807.9604 ≈416.81

II. Determination of dead weight

Wind loading, F w = 3219.456 Skirt height, H s = 1 m

Bending moment at base of skirt, M s

= + 2

2 (6.1.47)

= 3219.456 7.305 + 1 2

2= 111,027.8096 ≈111.03

1 st trial, skirt thickness = bottom section of vessel, 19.1 mm.



6-21

, =4+

(6.1.48)

=4 111,027.8096 1000

12435 + 19.1 2435 19.1

= .

,

=+

(6.1.49)

=333,716.0082

2435 + 19.1 × 19.1= .

,

=+

(6.1.50)

= 83,091.9522

2435 + 19.1 × 19.1= .

III. The resulting stress in the skirt

Maximum σ s (compressive) = + (6.1.51)

= 1.2386 + 2.2662 = .

= − (6.1.52)

= 1.2386 −0.5643 = .

∴Take joint factor, J as 0.85 (double-welded butt/equivalent type of joint with spot

degree of radiography).

IV. Criteria for design

≯ (6.1.53)

0.6743 ≯135 2 0.85 90°

0.6743

≯

114.75



6-23

I. ,

=1 4 − (6.1.55)

=1

16 125 2

4 111,027.80962.5 − 83,091.9522N = 47.2763 2

II. Bolt root diameter

= 4 47.2763= 7.7585

III. Total compressive load on the base ring per unit length

=4

2 + (6.1.56)

=4 111,027.8096

2.435 2 + 83,091.9522

2.435= 23,842.07004 + 10,862.0082

= 34,704.0782 = 34.70

∴Take f c, bearing pressure as 5 N/mm 2 is one of the concrete foundation pad.

IV. Base ring width, L b

= 1

10 3 (6.1.57)

= 34,704.07825 2 10 3

= 6.9408 ≈7

∴Use M24 bolts (BS 4190:1967) where the root area is 353 mm 2. This

is the minimum width required; actual width depends on the chair

design.

V. Actual width required = L r + t s + 50mm (6.1.58)

= 76 + 21 + 50

= 147 mm



6-24

VI. Actual bearing pressure on concrete foundation,

f ′c =Fb

Aw (6.1.59)

= 34.70 × 10 3

147 × 10 3 = 0.2361 N/mm 2

VII. Minimum thickness,

t b = L r 3 f′c

f r (6.1.60)

Where,

Lr = distance from the edge of the skirt to the outer edge of the

ring (mm)

tb = base ring thickness (mm)

f’c = actual bearing pressure on base (N/mm 2)

f r = allowable design stress in the ring material, typically 140

N/mm 2

t b = 76 3 (0.2361)140

= 5.4058 mm ≈ 5.5 mm

Chair dimension as tabulated in the tables A12 for bolt size M24.

6.1.12 Gasket

Gaskets are used to make a leak-tight joint between two surfaces. It is impractical to

machine flanges to the degree of surface finish that would be required to make a

satisfactory seal under pressure without a gasket. Gaskets are made from “semi -

plastic” materials; which will deform and flow under load to fill the surface

irregularities between the flange faces, yet retain sufficient elasticity to take up the

changes in the flange alignment that occur under load. The following factors must be

considered when selecting a gasket material:



6-25

1. The process conditions: pressure, temperature, corrosive nature of the

process fluid.

2. Whether repeated assembly and disassembly of the joint is required.

3. The type of flange and flange face

In the spigot and socket, and tongue and grooved faces, the gasket is confined

in a groove, which prevents failure by “blow-out”. Matched pairs of flanges arerequired, which increases the cost, but this type is suitable for high pressure and

high vacuum service.

Figure 6.2 Spigot and socket flange (R.K.Sinnot,1999)

Figure 6.3 Spiral wound gasket (Bikudo.com)

By referring to the operating condition of the process, the gasket material

chosen is the spiral-wound metal, asbestos filled. Even though the cost to buy this

gasket is relatively expensive, the price of the product is expensive. Due to its long

lasting lifetime, it can decrease the capital cost of the plant. Followings are the

advantages of spiral wound gasket (Donit Tesnit, 2011):

1. Sealing under heavy operating conditions

2. Strong stress compensation, stable and reliable sealing performance even

under frequent pressure fluctuation condition



6-26

3. Solid construction provides stability and seallability even when the sealing

surfaces are slightly corroded or bent

4. Easy installation

Figure 6.4 The cross-sectional area of esterification reactor

6.1.13 Summary Of The Reactor Design

Table 6.5 Summary Of Mechanical Design

Thickness of reactor (mm) 9Type of head Ellipsoidal headMaterial of construction Stainless steel (316)

Thickness of head (mm) 9Inlet pipe diameter (mm) 55Outlet vent pipe diameter (mm) 25Outlet pipe diameter (mm) 50Skirt thickness (mm) 19.1



6-27

Table 6.6 Reactor specification sheet (R-101)

SPECIFICATION DATA

MECHANICAL DESIGN

IdentificationItem no R-101

Design orientation VerticalOperating Condition

Operating temperature, oC 110Operating pressure, bar 0.29Design temperature, oC 132

Design pressure, bar 6

Material of construction Austenitic Stainless Steel (18Cr/8Ni,316)Design stress, kN/m 2 140.4

Wall thickness, mm 10Thickness insulation, mm 25

Wind loading, N/m 3222Head and Closure Design

Type Ellipsoidal HeadThickness, mm 10

Internal Cooling Coil

Length, m 60Number of spiral 8

Impeller ArrangementType of impeller Turbine with flat vertical blades

Material of construction Austenitic Stainless Steel (18Cr/8Ni,

316) Diameter of impeller, m 0.974

Width of impeller, m 0.12Offset of baffle, m 0.50

Baffle width, m 0.203

Baffle height, m 6.0875Shaft Design


316) Speed impeller, rpm 158.63

Maximum torque 3623.014Maximum bending moment 8191.6614

Diameter of shaft, cm 12Weight of Load

Vessel, kN 58.865

Insulation, kN 21.9274External fittings, kN 1.0958



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Internal Coil, kN 5.584Ellipsoidal head, kN 3.924

Total weight, kN 91.396Analysis of stress

Bending moment, kNm 85.961Critical buckling stress, N/mm 2 162

Maximum compressive stress, N/mm 2 1.5078Support design

Type Skirt support


316) Skirt thickness, mm 20

Skirt height, m 1Base ring and Anchor bolts

Nu. Of bolts 12 Area of bolt, mm 2 73.74

Bolt root diameter, mm 9.6899

6.2 MECHANICAL DESIGN OF ESTERIFICATION REACTOR 2 (R-102)

6.2.1 Material of construction

The material used for esterification reactor R-102 is stainless steel (18Cr/8Ni, Mo 2½%, 316). The stainless steels are the most frequently used corrosion resistant

materials in the chemical industry. The important corrosion resistance the chromium

content must be above 12%, and the higher the chromium content, the more

resistant is the alloy to corrosion in oxidizing conditions. Nickel is added to improve

the corrosion resistance in non-oxidizing environment.

Grade 316 is the standard molybdenum-bearing grade, second in

importance to 304 amongst the austenitic stainless steels. The molybdenum gives

316 better overall corrosion resistant properties than Grade 304, particularly higher

resistance to pitting and crevice corrosion in chloride environments. It has excellent

forming and welding characteristics. It is readily brake or roll formed into a variety of

parts for applications in the industrial, architectural, and transportation fields. Grade

316 also has outstanding welding characteristics. Post-weld annealing is not

required when welding thin sections.



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6.2.2 Thickness of the vessel

=2 − (6.2.1)

Design conditions for esterification process between Acrylic Acid and 2-Ethylhexanol

is set to be:

Pressure, P i = 6 bar absolute (5.9216 atm)

= 4.9216 atm (gage pressure)

Temperature = 200 0C

Design stress, f = 120 N/mm 2

Material joint efficiency, J = 1

Inner diameter, D i = 2632.2 mm

Corrosion allowance = 4 mm

Therefore design pressure, P i 4.9216 atm = 0.4987 2


for material loss by corrosion and erosion. For esterification process, there will be a

severe condition of corrosion arise from the acrylic acid and acid catalyst used. According to Sinnot (1999), when this condition occurs, the allowance for corrosion

should be increased from 2 mm to 4 mm.

Calculation of thickness;

=0.4987 × 2632.2

2 1 120 −0.4987= 5.4809 ≈6

= 6 + 4 =

The total thickness obtained is appropriate referring to table below, for a

vessel diameter of 2.5m to 3.0m, the minimum wall thickness required should not be

less than 10 mm. So, the thickness of 10mm esterification reactor is appropriate.



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Table 6.7: Minimum thickness required corresponds to vessel diameter

Vessel Diameter (m) Minimum thickness (mm)1.0 5

1.0 to 2.0 72.0 to 2.5 92.5 to 3.0 103.0 to 3.5 12

6.2.3 Insulation of the Vessel

The material used for insulator on this esterification reactor is mineral wool. Mineral

wool is made from molten glass, stone or slag that is spun into a fiber-like structure.

The reactor is insulated to avoid heat loss from the reactor (to conserve energy) and

to keep the process conditions from fluctuating with the ambient conditions.

Table 6.8: Thickness of insulation as a function of process temperature

T ( 0C) t (mm)

93 12.7

205 25.4

316 31.75

insulatiot for the process with T = 120 0C can be estimated through interpolation:

T ( 0C) t (mm)

93 12.7

200 insulatiot

205 25.4

mm25mm83.24

7.12)7.124.25(9320593200t insulation



6-31

Hence, the thickness of insulation for esterification reactor R-102 is 25 mm.

Figure 6.5: Reactor Cross Sectional Area of R-102

6.2.4 Cooling element inside the vessel

Internal coil

For cooling purposes of reactants inside the reactor, an internal helical coil is used.

The coil is the simplest and cheapest form of heat transfer surface and it is installed

inside the reactor vessel. The utility supplied to the coil is cooling water. The

assumptions as listed below:-

1. No heat loss from the coil’s wall 2. Heat transferred by process is equal to heat absorbed by coils

The diameter of the coil is subscribed as

=30

(6.2.2)

The pitch is taken as twice the diameter coil. Hence;

=2.6322

30= 0.0877

Pitch is times two of coil diameter;

= 2 × 0.0877 = 0.1755



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Energy, Q produced by the process is 9.7616×10 5 kJ/h per unit area of the vessel.

By assuming no heat loss from the coil ’s wall, Q transferred by the process is equalto Q absorbed by the coils.

=

(6.2.3)

= = 2.6322 7.8966 = 65.3000 2

= 9.7616 × 10 5 Ĥ = 7167.9

= Ĥ =

9.7616 × 10 5

7167.9= 136.1849

Tout of the coil is assumed to be as ambient temperature, 27 0C

= ∆

(6.2.4)

= 7.0122 /. o C

= 137 × 7.0122 × 120.27 = 8.9342 × 10 4

=. =

8.9342 × 10 4 (65.3)9.7616 × 10 5 = 5.9765 2

Hence, area of the coil is 5.9765 m 2.

Equation for area of the coil as below;

=

(6.2.5)

So, Length of the coil;

= =5.9765(0.0877)

= 21.69

Hence, length of coil in esterification reactor R-102 is 21.69 m.



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3. Flat heads

= (6.2.9)

By assuming plate welded to the end of the shell with a fillet weld is used, C p is

taken as 0.55 and D e = D i.

= 0.55 × 2632.2 0.4987120

= 93.33 ≈94

Adding corrosion allowance of 4mm

= 94 + 4 = 98

From the three values of thickness, the thickness of flat heads is the the

greatest value. This value is not an effective idea for angle of economical. Hence,

ellipsoidal head is decided to be the vessel head since it would probably be the most

economical. The thickness of the head is as the same as the vessel thickness. This

is more preferable.

6.2.6 Pipe Size Selection

Material types used for inlet and outlet pipe also stainless steel. The formula to

determine the pipe size as shown below;

= 260 0.52 −0.37

(6.2.10)

=

(6.2.11)

Inlet Pipe S6

= 1.677 × 10 4

= 20.793

=

1.677 × 10 4

20.79 = 806.6378 3



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= 2601.677 × 10 4

3600

0.52

(806.6378) −0.37 = 48.64

Therefore, the pipe size used is 50mm pipe.

Outlet Pipe S7

= 874.8

= 15063

=874.81506

= 0.5809 3

= 260874.8

3600

0.52

(0.5809)

−0.37 = 152

Therefore, the pipe size used is 200mm pipe.

Outlet Pipe S8

= 1.59 × 10 4

= 20.063

=1.59 × 10 4

20.06= 792.622

3

= 2601.59 × 10 4

3600

0.52

(792.622) −0.37 = 47.62

Therefore, the pipe size used is 50 mm pipe.

6.2.7 Bolt Flanged Joint

Flanged joints are used for connecting pipes and instruments to vessels, for

manhole covers, and for removable wessel heads when ease of access is required.

Flanges may also be used on the vessel body, when it is necessary to divide the

vessel into sections for transport or maintenance. Flanged joints are also used to

connect pipes to other equipment, such as pumps and valves. Flanges range in size

from a few millimeters diameter for small pipes, to several metres diameter for those

used as body or head flanges on vessels. Standards flanges are available in a

range of types, sizes and materials; and are used extensively for pipes, nozzles andother attachments to pressure vessel



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Figure 6.6: Typical standard flange design

Refer to Coulson & Richardson'sChemical Engineering Design Volume 6 (Bolted

flanged joints), we got the values of bolt and flange for each pipe. The tables below

are showing the values of each pipe.

S6: D nominal = 50 mm

Table 6.9: Values for bolt and flange of the inlet pipe (S4)

d 1 Flange Raised Face

BoltingDrilling Boss

D b h d 4 f No. d 2 k d 3

60.3 140 14 28 90 3 M12 4 14 110 80


Table 6.10: Values for bolt and flange of the outlet pipe (S6)




219.1 320 20 44 258 3 M16 8 18 280 240


Table 6.11: Values for bolt and flange of the outlet pipe (S8)




60.3 140 14 28 90 3 M12 4 14 110 80



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6.2.8 Design of Vessel Subject To Combine Loading

Pressure vessels are subjected to other loads in addition to pressure must be

designed to withstand the worst combination of loading without failure. The main

sources of load to consider are:

a) Dead weight of vessel and contents

b) Wind

c) Earthquake (seismic)

d) External loads imposed by piping and attached equipment.

However, in this country, the impact of earthquake is not that important to be

considered according to the geographical location.

6.2.8.1 Dead Weight of Vessel

1. Weight of vessel

The approximate weight of a cylindrical vessel with domed head ends and

uniform thickness can be estimated from the following equation:

= 240 + 0.8

(6.2.12)

= 1.08 , = 10

= + × 10 −3 = 2.6322 + 10 × 10 −3 = 2.6422

= 7.8966

= 240 1.08 2.6422 7.8966 + 0.8 × 2.6422 10 = 68556.77

= 68.557

2. Weight of Ellipsoidal Head

= (6.2.13)

= 8000 3 , =9.81

, = 10

=13

×12

2.6522 3 −2.6322 3 = 0.0366 3

= 8000 × 0.0366 × 9.81 = 2872.37 Double this value to allow fittings; hence = 5744.74 = 5.74



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3. Weight of insulation

In this esterification reactor, the material of insulator used is mineral wool.

The density of mineral wool is 130 kg/m 3.

=

6.2.14)

= 2.6322 + 0.010 + 0.010 7.8966 25 × 10 −3 = 1.6449 3

= = 130 × 1.6449 × 9.81 = 2097.74

Double this value to allow fittings; hence = 4195.48 = 4.20

4. Weight of internal coil

=

(6.2.15)

= 8000 3 , =9.81

, = 2

= = 0.0877 21.69 0.002 = 0.0120 3

= 8000 × 0.0120 × 9.81 = 941.76

Double this value to allow fittings; hence = 1883.52 = 1.88

5. Weight of external fittings

Plain steel ladder is used for external fitting at the reactor. From Nelson

(1963), weight of the ladder is estimated to be 150 N/m length.

= 150 × 7.8966 = 1184.49 = 1.18

Total weight of the vessel;

= + + + +

= 68556.77 + 4195.48 + 1184.49 + 1883.52 + 5744.74 = 81565



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6.2.8.2 Wind Loading

Dynamic wind pressure is 1280 N/m 2

= 2.6322 + 2 10+25 × 10 −3 = 2.7022

So, loading per linear meter of vessel, is;

=1280

2 × 2.7022 = 3458.82 /

Bending moment at bottom tangent line;

=2 2

(6.2.16)

= ( )

= 3458.82 × 7.8966 = 107839.60

6.2.9 Analysis of Stress

From bottom tangent line;

1. Longitudinal pressure stress

=2

(6.2.17)

=0.4987 × 2702.2

2(10)= 67.38 / 2

2. Circumferential pressure stress

=4

(6.2.18)

=0.4987 × 2702.2

4(10)= 33.69

3. Dead weight stress

=+

(6.2.19)

=81565

2632.2 + 10 (10)= 0.98 / 2



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4. Bending stress

= 2632.2 + 2 × 1 0 = 2652.2

=64

4 −4

(6.2.20)

=64

2652.2 4 −2632.2 4 = 7.2437 × 10 10 4

= ±2

+ == ±107839.60 × 10 3

7.2437 × 10 10

2632.22

+ 10 = ±1.97 / 2

6.2.9.1 The resultant longitudinal stress, is;

= + + = 33.69 −0.98 + 1.97 = 34.68 / 2

= + −= 33.69 −0.98 −1.97 = 30.74 / 2

From the above resultant longitudinal stress, the highest difference between the

principle stresses will be on the down-wind side which is . / .

6.2.10 Check Elastic Stability (Buckling)

= 2 × 10 4

(6.2.21)

= 2 × 10 4 102644.2

= 75.64 / 2

The maximum compressive stress will occur when the vessel is not under

pressure which exceeds the total value of dead weight and bending stress, 2.95

N/mm2

, well below the critical buckling stress. Hence, we can say the design of R-102 is satisfactory.

6.2.11 Vessel Support Design

Skirt supports are used for tall and vertical columns. The support must be designed

to carry the weight of the vessel and contents and any superimposed load, such as

wind loads. In 2-EHA plant, the reactor used is cylindrical and vertical vessel. So,

the type of skirt used is straight cylindrical support.



6-41

Type of support : Straight cylindrical skirtθ s : 90 0

Design stress, f s : 135 N/mm 2 at ambient temperature, 27 0C

Skirt height : 1.0 m

Young modulus : 200,000 N/mm 2

=4

× 2.6322 2 × 7.8966 1000 × 9.81 = 160146.67

= 160.15

= 81565 + 160146.67 = 241711.67 = 241.71

1. Bending moment at skirt base;

= + 2

2

(6.2.22)

= 3458.28.8966 2

2= 136881.92 = 136.88

2. Bending stress in skirt;

Taking skirt thickness as the same as the thickness of the bottom section of

the vessel;

= 10 ,

= 4

+

(6.2.23)

=4(136.88 × 10 6 )

[ 2652.2 + 10 10 × 2652.2]= 2.47 / 2

Where;

Ms = maximum bending moment at the base of the skirt

ts = skirt thickness

Dv = outside diameter of the vessel, 2.6522 m



6-42

3. Dead weight in skirt;

=2

[ ( + ]

(6.2.24)

=2(160146.67)

[ ( 2652.2 + 10 10]= 3.83 / 2

= 2

[ ( + ]

(6.2.25)

=2(81565)

[ ( 2652.2 + 10 10]= 1.95 / 2

4. Resulting stress in skirt;

Maximum = + = 2.47 + 3.83 = 6.3 / 2

Maximum = − = 2.47 −1.95 = 0.52 / 2

5. General consideration for skirt design;

Taking the joint factor, J as 1;

i) <

0.422 < 135 1 90

0.422 < 135 / 2

ii) < 0.125

6.3 2 < 0.125 × 200,00010

2652.290

6.3 2 < 94.26 / 2



6-43

6.2.12 Base Rings and Anchor Bolts

Assume pitch circle diameter =3m

Circumference of bolt circle =3000π

Bolt design stress, f b =125 N/ mm 2

Recommended space between bolts =600 mm

Minimum number of bolt required, N b =3000π/600=15.71

Closest multiple of 4 = 16

Bending moment at the base skirt, M s =136881.92 N

Total weight (operating value) =81565 N

1. Area of bolt;

= 1

4 − (6.2.26)

=1

16(125)4(136881.92)

3 −81565 = 50.47 3

2. Bolt root diameter;

= 4(50.47)= 8.02 ≈9

3. Total compressive load on the base ring per unit length;

=4

2 +

(6.2.27)

=4(136881.92)

(2.6522) 2 +81565(2.6522)

= 34565.92 /

Assume a pressure of 4 N/mm 2 is one of the concrete foundation pad, f c.

Minimum width of the base ring,

= ×1

10 3 =34565.924 × 1 0 3 = .



6-44

6.3 MECHANICAL DESIGN OF DISTILLATION COLUMN (T-102)

6.3.1 Introduction

Several factors need to be considered in the mechanical design of distillation

column such as:

1. Design pressure

2. Design temperature

3. Material of construction

4. Design stress

5. Wall thickness

6. Welded joint efficiency

7. Analysis of stresses

a. Dead weight load

b. Wind load

c. Pressure stress

d. Bending stress

8. Vessel support

9. Insulation

6.3.1.1 Design Pressure

Generally, design pressure is taken as 5 to 10% above the operating pressure at the

bottom of column to avoid spurious operating during minor process upsets.

6.3.1.2 Design Temperature

The design temperature at which the design stress is evaluated should be taken asthe maximum working temperature of the material, with due to allowance for any

uncertainty involved in predicting vessel wall temperature.



6-45

6.3.1.3 Material of Construction

Selection of suitable material must be taking into account the suitability of material

for fabrication (particularly welding) as well as the compatibility of the material with

the process environment. The chosen material of construction must meet the

several purposes, such as:

1. Readily available

2. Low cost

3. Subjected to welding

4. Corrosion resistant to feed and product

5. Easily fabricated

6. High strength

6.3.1.4 Design Stress

It is necessary to determine the maximum allowable stress that the material can

withstand without failure under operating condition.

6.3.1.5 Wall Thickness

Design of wall thickness, e, is determined by using this formula :

e =Pi Di

2 Jf − 0.2 Pi ⋯⋯(6.3.1)

Where,

e = minimum thickness of the plate required

P i = internal pressure

Di = internal diameter

F = design stress

J = joint factor (J = 1 for ellipsoidal head)



6-46

6.3.1.6 Welded Joint Efficiency

The strength of welded joint will depend on the types of joint and the quality of the

welding. Take welding joint as 1.0 implies that the point is equally as strong as the

virgin plate; this is achieved by radio graphing the complete weld length, and cutting

out and remarking any.

6.3.1.7 Analysis of Stresses

The column also subjected to other loads such as vessel shell, plate fittings and

weight of liquid to fill into the vessel. Total weight of column can be calculated by

using formula:

WV = 240 × C V × D m × HV + 0.8 D m t × 10 −3 kN ⋯⋯(6.3.2)

Where,

W v = total weight of shell, excluding internal fitting such as plate

C v = a factor to account for the weight of nozzle, manways and internal

support

Dm = mean diameter of vessel (D c + t x 10 -3)

Hv = height or length between tangent lines, m

t = wall thickness

a. Wind loading

Wind loading will only be important on tall column installed in the open. Columns are

usually free standing, mounted on skirt support, and not attached to structural steel

work.

b. Pressure stress

The longitudinal and circumferential stresses due to pressure can be calculated

using equation:

σL =P D4 t

and σh =P D2 t

⋯⋯(6.3.3)

Where,

P = operating pressure

D = column diameter



6-47

t = thickness

c. Bending stresses

Bending moments will be caused by the following bending condition:

1. Wind load on tall self supported vessels

2. Seismic loads on tall column

3. Dead weight and wind loading on piping and equipment.

Bending stress can be calculated using this formula:

σb = ±MIv

Dc

2 + t ⋯⋯(6.3.4)

Where,

Iv = π64

Do 4 – Di4

Do = (D i + 2t)

6.3.2 Calculation on Mechanical Engineering Design of Distillation Column

6.3.2.1 Column Design Specification

Total column height = 8.9 m Allow 2 m for clearance height = (8.9 + 2) m

= 10.9 m ≈ 11 m

Internal diameter, Dc = 2.04 m

Operating pressure, Top column = 0.04 bar

Bottom column = 0.11 bar

Material of column = Stainless steel

Tray type = Sieve tray (15 trays equally spaced)

Material of tray = Stainless steel (316)Operating temperature = 120 0C



6-48

(i) Design Pressure

Distillation column must be designed to withstand the maximum pressure to which it

is likely to be subjected in operation. Usually, the design pressure is taken above the

normal working operation. The purpose is to avoid counterfeit operation during

minor process upset and safety condition, therefore the design pressure is taken as

6 bars as it is used in the real industry.

= 6 bar ×0.1 N

mm 2

1 bar

= 0.6 Nmm 2

(ii) Design Temperature

Take design as 10% above the operating temperature,

= 120.2℃ × 1.1

= 132.22℃

(iii) Material of Construction

The material used for construction of this column is Stainless Steel 316. Grade 316is the standard molybdenum-bearing grade and gives better overall corrosion

resistant properties. It has excellent forming and welding characteristics. It is readily

brake or roll formed into a variety of parts for applications in the industrial,

architectural, and transportation fields. Grade 316 also has outstanding welding

characteristics. Post-weld annealing is not required when welding thin sections.

Minimum thickness required for pressure loading, (t),

t = ∆P (D c )2 σdes − ∆P ⋯⋯(6.3.5)

t =0.6 N/mm 2 (2040 m)

2 114.33 N/mm 2 − (0.6 N/mm 2 ) = 5.37 mm


for material lost by corrosion and erosion. Based on table 13.4, Coulson &Richardson, Chemical Engineering, volume 6, page 739, and this minimum



6-50

(Where 1.2 is factor for contacting plates, steel including typical liquid loading in

kN/m 2)

For 15 plates = 15 x 3.92

= 58.8 kN

6.3.2.4 Weight of Insulation

Insulating material: Mineral wool

Characteristics are;

(i) Made from molten glass, stone or slag that is spun into a fiber-like

structure

(ii) Very safe insulation materials

(iii) Provide better acoustical and insulating results than fiberglass

(iv) Outstanding resistance to fire

(v) Higher density (130 kg/m 3)

By taken insulation thickness of 75 mm,

Approximate volume of insulation = π × Dc × H v × (75 × 10 -3)

= π × 2.04 × 11 × (75 × 10-3)

= 5.29 m 3

Weight = 5.29 m 3 x 130 kg/m 3 x 9.81 m/s 2

= 6.75 kN

Double this value to allow for fitting = 13.5 kN

Total weight (W),

Shell = 71.52 kN

Plates = 58.80 kNInsulation = 13.50 kN

Total = 143.82 kN



6-51

6.3.2.5 Wind Loading

Take dynamic wind pressure as 1280 N/m 2, corresponding to 160 kph.

Mean diameter, including insulation = D c + D c (t insulation + t wall)

= 2.04 + 2.04 (0.075 + 0.01)

= 2.21 m

Loading per unit length, F w = 1280 N/m 2 × 2.21 m

= 2828.8 N/m

Bending moment at bottom tangent line,M x:

Where x = H v = 11 m (column height)

Mx = F w x2

2 ⋯⋯(6.3.6)

Mx = 2828.8 11 2

2 = 171 142 N/m

6.3.2.6 Analysis of Stresses

At bottom tangent line,

Pressure stresses:

σL =P D4 t

and σh =P D2 t

Where,

P = operating pressure (0.6 N/mm 2)

D = column diameter (2.04 m)

t = thickness (10 mm)

σL =0.6 (2040)

4 (10) = 30.6 N/mm 2

σh =0.6 (2040)

2 (10) = 61.2 N/mm 2



6-52

(i) Dead Weight Stress

σw =Wv

π Dc + t t ⋯⋯(6.3.7)

σw = 71.52 kNπ 2040 + 10 mm (10mm)

= 1.11 N/mm 2 (compressive stress)

(ii) Bending Stress

σb = ±Mlv

Dc

2 + t

Where,

lv =π64 Do

4 − Di

4

Do = Di + 2t

= 2040 + 2 10 = 2060 mm

lv =π64

2060 4 − 2040 4 = 3.38 × 10 10 mm 4

Therefore,

σb = ±171142 × 10 3 N/mm

3.38 × 1010

mm4

2040

2 + 10 mm

σb = ± 5.22 N/mm 2

The resultant longitudinal stress is:

σz = σL + σW + σb ⋯⋯(6.3.8)

σw is compressive therefore it is negative.

σz upwind = 30.6 − 1.11 + 5.22 = + 34.71 N/mm 2 σz downwind = 30.6 − 1.11 − 5.22 = 24.27 N/mm 2

The greatest difference between the principal stresses will be on the downwind side:

σh - σz (downwind) = (61.2 - 24.27) N/mm 2

= 36.93 N/mm 2

Design stress = 140.33 N/mm 2 (for stainless steel 316)

The value of differences between the principal stresses is well below the maximum

allowable design stress.



6-53

(iii) Elastic Stability (Buckling)

Critical buckling stress,

σc = 2 × 10 4 t

Do

σc = 2 × 10 4 10

2060 = 97.1 N/mm 2

Therefore, a critical buckling stress is 97.1 N/mm 2.

When the vessel is not under pressure (where the maximum stress occur):

Maximum stress = σw + σh

= (1.11 + 61.2) N/mm2

= 62.31 N/mm 2

The maximum stress is well below the critical buckling stress. Therefore, design is

satisfactory.

6.3.2.7 Design of Domed End and Wall Thickness

Ellipsoidal head is the most economical types of head that being used inpetrochemical equipment. Therefore, ellipsoidal head is chosen. Material of

construction for ellipsoidal head is stainless steel.

e =Pi Di

2 Jf − 0.2 Pi

Where,

e = minimum thickness of the plate required

P i = internal pressure, 0.6 N/mm 2

Di = internal diameter, 2.04 mf = design stress, 140.33 N/mm 2

J = joint factor (J = 1 for ellipsoidal head)

Therefore, minimum thickness required:

e =0.6 (2040)

2 1 140.33 − 0.2(0.6) = 4.36 mm

Add 4 mm for corrosion allowance,

e = (4.36 + 4) mm= 8.36 mm



6-54

≈ 9 mm

So, thickness for the domed end with ellipsoidal head is taken as 10mm which is

same as wall thickness.

6.3.2.8 Design for the Skirt Support

Material of construction for skirt support is stainless steel.

Design stress = 140.33 N/mm 2

Young’s modulus = 200000 N/mm 2

The maximum dead weight load on the skirt will occurs.

Aprroximate weight =π4

× Dc

2 × Hv

× ρL

g

⋯⋯(6.3.9)

Aprroximate weight =π4

× 2.04 2 × 11 × (801.7) (9.81)

Aprroximate weight = 282.76 kN

Weight of vessel from previous calculation = 143.82 kN

Total weight = (282.76 + 143.82) kN

= 426.58 kN

Wind loading from previous calculation = 2.828 kN/m

Take skirt support as 3 m height.

Bending moment at base skirt

= 2.828 kN/m ×(column height + skirt support height) 2

2

Bending moment at base skirt,M s = 2.828 kN/m ×(11 + 3) 2

2 = 277.1 kNm

The resultant stresses in the skirt support will be:

σs (tensile) = σbs - σws σs (compressive) = σbs + σws

Where,

σbs = bending stress in the skirt

σws = dead weight stress in the skirt



6-55

σbs =4 M s

π Ds + t s t s Ds and σws test =

Wπ Ds + t s t s

⋯⋯(6.3.10)

Where,

sM = maximum bending moment, evaluated at the base of the skirt

(due to the wind, seismic and eccentric load)

sD = inside diameter of the skirt, at the base.

st = skirt thickness

Therefore,

σbs = 4 277.1 × 103

kNmmπ 2040 + 10 (10) (2040) = 8.44 N/mm 2

σws test =282.76 × 10 3 N

π 2040 + 10 10 = 4.39 N/mm 2

σws operating =143.82 × 10 3 N

π 2040 + 10 10 = 2.23 N/mm 2

Maximum σs (compressive) = σbs + σws

= 8.44 + 4.39

= 12.83 N/mm 2

Maximum σs (tensile) = σbs - σws

= 8.44 - 2.23

= 6.21 N/mm 2

Take joint factor, J =1:

Criteria for design:

σs (tensile) < f s J sin θ

6.21 < 140.33 x 1 x sin 90 0

6.21 < 140.33

σs (compressive) < 0.125 E Y t sDs

sin θ

12.83 < 0.125 (200000)10

2040

sin90 °

12.83 < 122.55



6-56

Both criteria are satisfied and 4 mm is added for corrosion allowance. Therefore for

the design stress thickness (10 + 4) mm is 14 mm.

6.3.2.9 Base Ring and Anchor Bolts

(i) Approximate Pitch Circle Diameter

Approximate pitch circle diameter = 2.2m

Circumference of bolt circle = 2200 π

(ii) Number of Bolts Required at Minimum Recommended Bolt Spacing

2200600

= 11.52

By follow Scheiman rules (Coulson & Richardson’s, 1999) bolt used most bemultiple of 4. Closest multiple of 4 = 12 bolts

Take bolt design stress = 125 N/mm 2

(iii) The Anchor Bolts,

At operating value,

Ms = 277.1 kNm

WTotal vessel = 282.76 kN

The anchor bolts are assumed to share the overturning load equally, and the bolt

area required is given by:

Ab =

1

Nb f b

4 M s

Db − W

⋯⋯(6.11)

Where,

Ab = area of one bolt at the root of the thread (mm 2)

Nb = number of bolts

f b = maximum allowable bolt stress (N/mm 2); typical design value 125

N/mm 2

Ms = bending (overturning) moment at the base (Nm)

W = weight of the vessel (N)

Db = bolt circle diameter (m)



6-57

Ab =1

12 × 1254(277.1 × 10 3 )

2.2 − 282.76 × 10 3 = 147.37 mm 2

Bolt root diameter,

4 Abπ = 4 (147.37)π = 13.7 mm ≈ 14mm

Use M24 bolts (BS 4190:1967) where the root area is 353 mm 2.

(iv) Total Compressive Load on the Base Ring per Unit Length

Fb = 4 M s

π Ds2 +

Wπ Ds

⋯⋯(6.3.12)

Where,

Fb = compressive load on the base ring (N/m)

Ds = skirt diameter (m)

Fb =4 (277.1 × 10 3 )

π (2.04) 2 +

282.76 × 10 3

π (2.04) = 128.9 kN/m

(v) Minimum Width of the Base Ring,

Lb =Fb

f c

110 3 ⋯⋯(6.3.13)

Where,

Lb = base ring width (mm)

f c = the maximum allowable bearing pressure on the concrete foundation



6-58

pad, which will depend on the mix used, and will typically range from

3.5

to 7 N/mm 2

Take the bearing pressure as 5 N/mm 2,

Lb =128.9 × 10 3

5 × 1 0 3 = 25.78 mm

This is the minimum width required; actual width depends on the chair design.

Actual width required = L r + t s + 50mm

= 76 + 14 + 50

= 140 mm

Actual bearing pressure on concrete foundation,

f ′ c =Fb

Aw =

128.9 × 10 3

140 × 10 3 = 0.921 N/mm 2

Minimum thickness,

t b = L r 3 f ′ c

f r

⋯⋯(6.3.14)

Where,

Lr = distance from the edge of the skirt to the outer edge of the ring (mm)

tb = base ring thickness (mm)

f’c = actual bearing pressure on base (N/mm 2)

f r = allowable design stress in the ring material, typically 140 N/mm 2

t b = 76 3 (0.921)140

= 10.67 ≈ 11mm

6.3.2.10 Piping and Flange Design

Optimum diameter of flange can be calculating using equation below.

d, optimum = 260 G 0.52 ρ−0.37

⋯⋯

(6.3.15)



6-59

Where,

G = mass flowrate (kg/s)

ρ = mass density (kg/m 3)

(i) Feed inlet

Flowrate = 1.67 x 10 4 kg/h

Density = 17.56 kg/m 3

Therefore,

d, optimum = 260 (4.64) 0.52 (17.56) −0.37 = 200mm

(ii) Top Column Inlet

Flowrate = 2292 kg/h


Therefore,

d, optimum = 260 (0.64) 0.52 (0.1167) −0.37 = 456mm

As reference to typical standard flange design table, so the standard diameter size

of flange is taken as 500mm.

(iii) Bottom Column Inlet

Flowrate = 1.441 x 10 4 kg/h


Therefore,d, optimum = 260 (4.00) 0.52 (801.7) −0.37 = 45mm

As reference to typical standard flange design table, so the standard diameter size

of flange is taken as 50mm.



6-60

Figure 6.7: Typical Standard Flange Design

(Source: R. K. Sinnott, John Metcalfe Coulson, John Francis Richardson, Chemical

Engineering Design. Volume 6)



6-61

Table 6.12 : Standard Flange Size

Streamd opt

(mm)

Pipe Flange Raised FaceBolting

Drilling Boss

d 1 D b h d 4 f No d 2 k d 3

Feed 200 219.1 320 20 55 258 3 M16 8 18 280 236

Top 500 508 645 24 68 570 4 M20 20 22 600 538

Bottom 50 60.3 140 14 38 90 3 M12 4 14 110 74

Table 6.13 : Summary of Mechanical Design

Operating Condition Dimensions

Design Pressure

Design Temperature

0.6 N/mm 2

132.22 oC

Plate

Material Stainless steel (316)

Design Stress 140.33 N/mm 2

Cylindrical Section 10 mm

Column Weight

Mean Diameter 2.05 mDead Weight 71.52 kN

Weight of Plates 58.8 kN

Weight of Insulations 13.5 kN

Total Weight 143.82 kN

Wind Loading

Mean Diameter 2.21 m

Loading 2828.8 N/m

Vessel Supports

Material Stainless Steel

Design Stress 140.33 N/mm 2


Wind Loading 2.828 kN/m

Skirt Support Height 3 m

Bending Moment 277.1 kNm

Dead Weight (Test)Dead Weight (Operating)

4.39 N/mm2

2.23 N/mm 2



6-62

Thickness 14 mm

Anchor Bolts

Bolts 12

Design Stress 125 N/mm 2

Bending Moment 277.1 kNm

Area 147.37 mm 2

Bolts Root Diameter 13.7 mm

Types M24 bolts (BS 4190:1967)

6.4 MECHANICAL DESIGN OF DISTILLATION COLUMN (T-103)

6.4.1 Introduction

Several factors need to be considered in the mechanical design of distillation

column which are:

1. Design pressure

2. Design temperature

3. Material of construction

4. Design stress

5. Vessel thickness6. Heads and closure

7. Column weight

8. Analysis of stresses

9. Vessel support

10. Piping sizing

6.4.2 Column Design SpecificationTotal column height = tray spacing x no. of stages

= 0.6 x 21

= 12.6 m

Allow, 2 m for clearance height = 12.6 + 2

= 14.6 m

Internal diameter, D c = 1.22 m

Operating pressure,

Top column = 2 kPa

Bottom column = 9 kPa



6-63

Take column operating at = 6.5 x 10 -3 bar (at vacuum condition)

Operating temperature = 390.4K

Material of column = stainless steel

Tray type = sieve tray (20 trays)

Material of type = stainless steel

Corrosion allowance = 4 mm (for corrosive material)

Insulation column = mineral wool 75 mm thick

6.4.3 Design Pressure

In mechanical design, there are two parameters such as temperature and pressure

are important properties in order evaluate the thickness and the stress of material.

Therefore, the safety factor is added as precaution and determined by certain

consideration such as corrosion factor, location and process characteristics.

Generally, design pressure is taken as 5 to 10% above the operating pressure at the

bottom of column to avoid serious operating during minor process upsets. Take

design pressure as 10% above operating pressure,

Design Pressure, Pi = 6.5 x 10 -2 bar x 1.1

= 7.15 x 10 -2 bar

= 7.15 x10 -3 N/mm 2

Noted that this column operate at vacuum condition as the design pressure is at

vacuum pressure.

6.4.4 Design Temperature

The design temperature at which the design stress is evaluated should be taken as

the maximum working temperature of the material, with due allowance for anyuncertainty involved in predicting vessel wall temperature.

Operating Temperature, T = 390.4 K = 153.3 oC

Design Temperature, T = operating temperature ( oC) x 1.1

= 153.3 oC x 1.1

= 168.6 oC



6-65

t =0.885P i Rc

SE−0.1P i ……

t =0.885 7.15 × 10 −2 (1.22 × 10 3 )

101 × 1 −0.1 × 1.5 × 10 −2 + 4 = 4.76 mm

6.4.8 Dead Weight of Vessel, W v

For a steel vessel, W v = 240 C v Dm (H v + 0.8 D m) t

where, W v = total weight of shell, excluding internal fitting such as plates, N

C v = a factor to account for the weight of nozzles, manways and internal

support. (In this case for distillation column ,C v 1.15)

Dm = mean diameter of vessel (D i + t ), mHv = height or length between tangent lines, m

t = wall thickness, m

To get a rough estimate of the weight of this vessel is by using the average

thickness, 4

mm.

Dm = D i + t

= 1.22 + 0.004

= 1.224 m

Hv = 12.6 m

So, W v = 240 (1.15) (1.224) (12.6 + 0.8 (1.224)) (0.004)

= 18.3 kN

6.4.9 Weight of Plate, W p

From Nelson Guide, page 833, Coulson and Richardson’s, Chemical Engineering,Volume 6; take contacting plates, 1.2 kN/m 2.

Plate area = π D2/4



6-66

= π (1.22) 2/4

= 1.17 m2

Weight of plate = 1.2 kN/m 2 x 1.17 m 2

= 1.4 Kn

For 20 plates = 20 x 1.4 kN

= 28.06 kN

6.4.10 Weight of insulation, W i

The mineral wool was choosing as insulation material. By referring to Coulson and

Richardson’s, Chemical Engineering, Volume 6, page 833;

Density, ρ of mineral wool = 130 kg/m 3

Thickness = 50 mm = 0.05 m

Approximate volume of insulation = π x Dm x H v x thickness of insulation

= π (1.224) (12.6) (0.05)

= 2.42 m 3

Weight of insulation, W i = volume of insulation x ρ x g

= 2.42 x 130 x 9.81= 3089.47 N

= 3.089 kN

Double this value to allow fittings, so weight of insulation will be = 17.624 kN

Total weight, W t = W v + W p + W i

= 18.3 + 28.06 +3.089

= 49.499 kN



6-67

6.4.11 Wind Loading

Wind loading will only be important on tall columns installed in the open. Columns

are usually free standing, mounted on skirt supports and not attached to structural

steel work. The wind load is calculated based on location and the weather of

surrounding.

Dynamic wind pressure, P w =(1/2)C dρaUw2

where, P w = wind pressure (load per unit area)

Cd = drag coefficient (shape factor)

ρa = density of air

Uw = wind velocity

Wind speed, U w = 160 km/h

For a smooth cylindrical column, the following semi-empirical equation can be used

to estimate the wind pressure,

P w = 0.05 U w2

where, P w = wind pressure, N/m 2

Uw = wind speed, km/h

P w = 0.05 (160 2)

= 1280 N/m 2

Mean diameter, including insulation = D c + D c t

= 1.22 + (1.22) (4.76 + 50) x 10 -3

= 1.29 m

Loading per unit length, F w = 1280 N/m 2 x 1.29 m

= 1647 N/m

Bending moment, M x = F w x2 /2

where, x = distance measure from the free end = 1.22 m

Therefore, M x = (1641 N/m) (1.22 m)2

/2= 1221 N.m



6-68

6.4.12 Analysis of stresses

At bottom tangent line:

Pressure stresses: σL = PD i / 4t

Where, P = operating pressure

Di = column diameter

t = thickness

σL = (7.15 x 10 -3) (1.22 x 10 3)/ 4(0.76)

= 2.87 N/mm 2

σh = PD i /2t

= (7.15 x 10 -3) (1.22 x 10 3)/ 2(0.76)

= 5.74 N/mm 2

Dead weight stress,

σw = W v / π (Di + t) t

= 49.499 x 103

/ π (1.22 x 103

+ 0.76) 0.076=16.98 N/mm 2 (compressive stress)

Bending stresses,

Bending moments will be caused by the following loading conditions:

1. The wind loads on tall self-supported vessels

2. Seismic loads on tall column

3. The dead weight and wind loads on piping and equipment

Bending stress can be calculated using this equation:

σb = ±MIv

Di

2+ t

where, M v = the total bending moment at the plate

Iv = the second moment of area of the vessel about the plane of bending



6-69

Iv =π64

Do4 −Di

4

where, D o = (D i + 2t)

= (1220 + 2(0.76))

= 1221.52 mm

Iv =π64

1221.52 4 −1220 4 = 542.9 × 10 6 mm 4

Therefore,

σb = ±1221 × 10 3

542.9 × 10 6

12202

+ 0.76 = ±1.374N/mm 2

The resultant longitudinal stress is,

σz = σL + σw ± σb

σw is compressive and therefore negative,

σz (upwind) = 2.87 - 16.98 + 1.374

= -12.736 N/mm 2

σz (downwind) = 2.87 - 16.98 - 1.374

= - 15.484 N/mm 2

As there is no torsional shear stress, the principle stresses will be σz and σh. The

radial stress σr is negligible.

The greatest difference between the principal stresses will be on the downwind side,

σh - σz (downwind) = 5.74 – (- 15.484)

= 21.224 N/mm 2

Design stress = 130 N/mm2

The value of difference between the principal stresses is well below the maximum

allowable design stress.

Check elastic stability (buckling)

Critical buckling stress, σc = 2 x 10 4 (t/D o)

= 2 x 10 4 (0.76/1220)= 12.45 N/mm 2



6-70

When the column is not under pressure (where the maximum stress occur)

Maximum stress = σw + σh

= 16.98 + 5.74

= 22.72 N/mm 2

6.4.13 Design for the Skirt Support

Type of support = Straight cylindrical skirt, θs = 90 0

Material of construction = stainless steel

Design stress = 130 N/mm 2

Young’s Modulus = 210 kN/mm2 = 210,000 N/mm 2

At this condition of ambient temperature, the maximum dead weight load on the skirt

will occur when the vessel is full of the mixture.

Approximate weight, W approx = ( π /4 x D i2 x H v) x ρL x g

= (π /4 x (1.22)2 x (16.2)) x 902.31 x 9.81

= 167,629 N

= 167.629 kN

Weight of vessel from previous calculation = 49.499 kN

Total weight = W v + W p + W i + W approx

= 18.3 + 28.06 + 3.089 + 167.629

= 217.078 kN

Wind loading from previous calculation = 1641 N/m = 1.641 kN/m

Take skirt support height as 5 m to support the column,

Bending moment at base skirt, M s = F w [ (H v + H skirt )2/2]

= 1.641 [ (16.2 + 5.0) 2/2]

= 368.76 kNm



6-71

Take skirt thickness as same as the thickness of the bottom section of the vessel, t s

= 50 mm.

Bending stress in the skirt,

=4

+

where, M s = maximum bending moment, evaluated at the base of the skirt (due to

the wind, seismic and eccentric loads)

Ds = inside diameter of the skirt, at the base

ts = skirt thickness

Therefore,

=4 × 368.76 × 10 3 × 10 3

1220 + 4.76 4.76 × 1220= 66 / 2

Dead weight stress in the skirt,

=+

where, W v = total weight of the vessel and contents

Therefore,

=217.078 × 10 3

1220 + 4.76 4.76= 11.8 / 2

=49.499 × 10 3

1220 + 4.76 4.76= 2.7 / 2

Maximum σs (compressive) = σbs + σws = 66 + 11.8

= 77.8 N/mm 2

Maximum σs (tensile) = σbs - σws

= 66 – 11.8

= 54.2 N/mm 2



6-72

Take joint factor, E = 1

Criteria for design,

σs (maximum, tensile) < f s E sin θ

54.2 < 130 (1) sin 90 o

54.2 < 130 N/mm 2

σs (maximum, compressive) < 0.125 E Y ( t s / D s ) sin θ

77.8 > 0.125 x 210 000 x (0.76/1220) sin 90

77.8 > 16.32 N/mm 2

6.4.14 Base Ring and Anchor Bolts Design

Approximate pitch circle diameter, say, 1.25 m

Circumference of bolt circle = 2250 π

Number of bolts required at minimum recommended bolt spacing = 2250 π /600 =

11.78

Closest multiple of 4 = 12 bolts

Take bolt design stress = 125 N/mm 2

Ms = 762.76 kNm

Take W = operating value = 308.716 kN

Ab = 1 [ 4M s /Db – W] / N b f b

where, A b = area of one bolt at the root of the thread, mm 2

Nb = number of bolts

f b = maximum allowable bolt stress, N/mm 2; typically design value 125

N/mm 2 (18,000 psi)

Ms = bending (overturning) moment at the base, Nm

W = weight of the vessel, NDb = bolt circle diameter, m



6-73

So, A b = 1 [ 4 (368.76 x 10 3)/1.25 – (18.3 x 10 3] / 11.78 (125)

= 656 mm 2

Bolt root diameter = √ (656 x 4)/ π = 12.8 mm

Total compressive load on the base ring per unit length

Fb = [ 4M s /π Ds2 + W/ π Ds ]

where, F b = the compressive load on the base ring, Newtons per linear metre

Ds = skirt diameter, m

Therefore, F b = [4(368.76 x 10 3)/π (2.00) 2 + (18.3 x 10 3)/π (2.00)

= 120 kN/m

By assuming that a pressure of 5 N/mm 2 is one of the concrete foundation pad, f c.

Minimum width of the base ring,

Lb = (F b/f c) x (1/10 3)

where, L b = base ring width, mmf c = the maximum allowable bearing pressure on the concrete foundation

pad, which will depend on the mix used, and will typically range from 3.5 to 7 N/mm 2

(500 to 1010 psi)

therefore, L b= 120 x 10 3 /(5 x 10 3)

= 24 mm

Take the skirt bottom diameter as 2 m

Skirt base angle θs = tan -1 (2) = 71.6 / (1/2)

Keep the skirt thickness the same as that calculated for the cylindrical skirt. Highest

stresses will occur at the top of the skirt; where the values will be close to those

calculated for the cylindrical skirt. Sin 71.6 = 0.95, so this term has little effect on the

design criteria.



6-74

Assume bolt circle diameter = 1.25 m

Take number of bolts as 16

Bolt spacing = π x (1.25 x 10 3)/16 = 638 mm, satisfactory

Ab = 1 [ 4M s /Db – W] / N b f b

= 1 [ 4 (761.76 x 10 3)/3.25 – (308.716 x 10 3] / 16 (125)

= 314.42 mm 2

Use M24 bolts (BS 4190:1967) root area = 353 mm 2

Fb = [ 4M s /πDs2 + W/ πDs ]

= [4(761.76 x 10 3)/π (3.00) 2 + (308.716 x 10 3)/π (3.00)

= 140,552.78 N/m

= 140.552 X 103 kN/m

Lb = (F b/f c) x (1/10 3)

= 140,552.78/(5 x 10 3)

= 28.11 mm

This is the minimum width required, actual width will depend on the chair design.

Actual width required (Figure 13.30),

= L r + t s + 50 mm

= 76 + 52 + 50

= 178 mm

Actual bearing pressure on concrete foundation,

F’c = (140.552 x 103)/(178 x 103) = 0.79 N/mm 2

The minimum thickness is given by,

tb = L r √((3f ’c/f r )

where, L r = the distance from the edge of the skirt to the outer edge of the ring, mm

tb = base ring thickness, mm

f’c = actual bearing pressure on base, N/mm 2

f r = allowable design stress in the ring material, typically 140 N/m2



6-75

tb = L r √((3f ’c/f r )

= 76 √(3) (0.79)/(140)

= 9.9 mm

6.4.15 Design of Stiffness Ring

The plate is supported on rings 75 mm wide and 10 mm deep. The plate spacing is

0.6 m. Take design pressure as 1 bar external or 10 5 N/m 2. The load each ring,

F r = P e Ls

where, P e = external pressure

Ls = spacing between the rings

Therefore, the load per unit length on the ring,

F r = 10 5 N/m 2 x 0.6 m

= 0.6 x 10 5 N/m

Taking Young’s Modulus, E = 210,000 N/mm2

= 2.1 x 10 11 N/m 2

Factor of safety = 6

The second moment of area of the ring to avoid buckling is given by,

P cLs = 24 E I r / Dr 3 x factor of safety

where, I r = second moment of area of the ring cross-section

Dr = diameter of the ring (approximately equal to the shell outside diameter)

0.6 x 105

N/m = 24 (2.1 x 1011

) Ir / 2 x 6Ir = 1.43 x 10 -7 m 4

For a rectangular section, the second moment of area is given by:

I = breath x depth 3 /12

So, I r for the support rings = 10 x (75) 3 x 10 -12 /12

= 3.51 x 10 -7 m 4

And the support ring is of an adequate size to be considered as a stiffening ring,



6-76

L’ / Do = 0.5 / 3.0 = 0.25

Where, L’ = plate spacing Do = internal diameter

Do /t = 2000 / 50 = 40

From Figure 13.13, Coulson and Richardson’s, Chemical Engineering, volume 6,

Kc = 68

From equation 13.53, Coulson & Richardson’s, Chemical Engineering, volume 6,P c = K c x E Y x (t / D o)3

= 68 x 2.1 x 10 11 x (40) 3.

= 2.23 x 10 8 N/m 2.

This is above the maximum design pressure of 2.12 x 10 5 N/m 2. So, design of the

support rings to support the plate is satisfied.

6.4.16 Piping Sizing

By assuming that the flow of the pipe is turbulent flow, therefore to determine

optimum duct diameter is

Optimum duct diameter, d opt = 260 G 0.52 ρ-0.37 (for stainless steel)

where, G = flow rate, kg/s

ρ = density, kg/m 3

For feed stream,

Flow rates, G = 16612 kg/h

= 4.61 kg/s

Density mixture, ρmix = 0.089(1059.33) + 0.001(1158.13) + 0.053(836.76) +

0.847(888.71) +

0.002(724.98) + 0.009(997.99)= 902.96 kg/m 3



6-77

Therefore, d opt = 260 (4.61) 0.52 (902.96) -0.37

= 46.39 mm

Add 4 mm corrosion allowances d op = 50.39 mm

Nozzle thickness, t = P s d opt / 20 σ + P s

where, P s = operating pressure

σ = design stress at working temperature

Therefore, t = 0.0715 (50.39) / (20 (130) + 0.0715)

= 0.0014 mm

So, thickness of nozzle = corrosion allowance + 0.0014 mm

= 4 + 0.0014

= 4.0014 mm

= 4 mm

For top stream,

Flow rates, G = 2855 kg/h

= 0.79 kg/s

Density mixture, ρmix = 0.4274(1059.33) + 0.0026(1158.13) + 0.2579(888.71) +

0.0077(724.98)

+ 0.255(997.99)

= 1297.34 kg/m 3

Therefore, d opt = 260 (0.79) 0.52 (1297.34) -0.37

= 16.2 mm

Add 4 mm corrosion allowances, d op = 20.2 mm

Nozzle thickness, t = P s d opt / 20 σ + P s



Therefore, t = 0.0715 (20.2) / (20 (130) + 0.0715)

= 0.0005 mm



6-78


= 4 + 0.0005

= 4.0005 mm

= 4 mm

For bottom stream,

Flow rates, G = 82.9 kg/h

= 0.023 kg/s

ρmix = 0.0979(1059.33) + 0.0655(1158.13) + 0.8326(888.71) +0.0005(724.98)

+ 0.0033(997.99)

= 902.31 kg/m 3

Therefore, d opt = 260 (0.023) 0.52 (902.31) -0.37

= 3.39 mm

Add 4 mm corrosion allowances d op = 7.39 mm

Nozzle thickness, t = P s d opt / 20 σ + P s ……



Therefore, t = 0.00715 (7.39) / (20 (130) + 0.00715)

= 0.0015 mm


= 4 + 0.0015

= 4.0015 mm

= 4 mm

6.4.17 Flange Design

The flange class number required for a particular duty will depend on the design

pressure and temperature and the flange material.



6-79

Table 6.14: Summary of Mechanical Design of Distillation Column (T-103)

Operating condition and Material

Construction

Symbol Value Unit

Operating Pressure P 6.5 x 10 -2 bar

Operating Temperature T 117 oC

Design Pressure P o 7.15 x 10 -2 bar

Design temperature T o 168.6 oC

Material Construction Stainless steel

Design Column Dimension

Column Height h c 14.6 m

Shell thickness e 4 mm

Typed of domed Torispherical

Domed thickness e D 0.76 mm

Column Weight

Dead weight of vessel W v 18.3 kN

Weight of plate W p 28.06 kN

Weight of insulation W I 3.089 kN

Total weight W T 49.499 kN

Wind loading F w 1641 kN/mm

Wind pressure P w 1280 N/m 2 Bending moment M x 1221 kNm

Stress Analysis

Longitudinal pressure stress σL 2.87 N/mm

Circumferential pressure stress σH 5.74 N/mm

Dead weight stress σw 16.98 N/mm

Bending stress σB 1.374 N/mm

Upwind stress σz (upwind) -12.736 N/mm

Downwind stress σz(downwind) -15.848 N/mm

Critical buckling stress σc 12.45 N/mm

Vessel support

Type Straight Skirt

Material construction Stainless steel

Skirt thickness t s 0.76 Mm

Skirt diameter D s 2 M

Skirt height h s 5 MSkirt weight W s 167.629 kN



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Insulation

Material construction

Thickness t i 50.76 Mm

Piping sizing

Feed pipe sizing

Flowrate G 4.61 Kg/s

Pipe outer diameter o.d i 50.39 mm

Top pipe sizing

Flowrate G 0.0079 Kg/s

Pipe outer diameter o.d i 16.2 mm

Bottom pipe sizing

Flowrate G 0.023 Kg/sPipe outer diameter o.d i 7.39 mm

6.5 MECHANICAL DESIGN OF HEAT EXCHANGER (E-104)

6.5.1 Design Pressure

Usually, the design pressure is taken above the normal working operation. Thepurpose is to avoid counterfeit operation during minor process upset and safety

condition.

Table 6.15: Design Pressure

Parameter Shell side

Operating Pressure, N/mm 2 0.1

Design pressure, N/mm 2 0.6

6.5.2 Design Temperature

The design temperature at which the design stress is evaluated should be taken as

the maximum working temperature of material with due allowance for any

uncertainty involve in predicting vessel wall temperature. Adding 10% from



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operating temperature to cover the uncertainties in temperature prediction, the

design temperature should be:

Table 6.16: Design Temperature

Parameter Shell side Tube side

Operating temperature, oC 117.2 20

Design temperature, oC 130 22

6.5.3 Exchanger Type

For heat exchanger design, the specification and type of heat exchanger was

discussed in chemical design section. Exchanger with internal floating head is

versatile than other type of heat exchangers. Internal floating head is suitable for

high temperature difference between shell and tubes.

The tubes can be rod from to ends and the bundle are easier to clean and

can be used for fouling liquids. The tube bundle is removable and the floating tube

sheet moves to accommodate differential expansion between shell and tubes.

6.5.4 Material of Construction

Water is fairly corrosive as cooling medium is tending to polymerization. Hence, the

tubes should be constructed by using corrosion resistance materials. Stainless steel

is the most frequently used as corrosion resistance material in chemical industry.

There are several types of stainless steel that can be divided into 3 classes

according to their microstructure:1. Ferritic: 30% Cr, 0.1% C, no nickel

2. Austenitic: 16 - 26% Cr, 10% Ni

3. Martensitic: 12 – 14% Cr, 0.1 – 1% C, < 2% Ni

To impart corrosion resistance, the chromium content must be above 12%.

Nickel is added to improve the corrosion resistance in non-oxidizing environments.

For tube and shell side, Austenitic stainless steel is also called 316 type or18/8 stainless steel was used. It contains the minimum chromium and nickel that



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give a stable austenitic structure. The carbon content is low enough for heat

treatment and not to be normally needed with this section to prevent weld decay.

They typically have reasonable cryogenic and high temperature strength properties.

It is also highly ductile and tensile strength.

6.5.5 Welded Joint Efficiency and Corrosion Allowance

The strength of welded joint will depend on the type of joint and the quality of the

welding. The soundness of weld is then checked by visual inspection and by

nondestructive testing called radiography. The value of welded joint factor, J can be

assumed as 1.0 which the joint is equally as strong as the virgin steel plate.

The corrosion allowance is additional thickness of metal added to allow for

material lost by corrosion or scaling. For stainless steel, where severe corrosion is

not expected, a minimum corrosion allowance of 3.0 mm is used.

6.5.6 Design Stress

It is necessary to fix a minimum allowance value of stress that can be accepted in

the material of construction. The allowable stress for the material of construction can

be obtained in Table 13.2 from Coulson & Richardson’s, Volume 6 (Refer Appendix E10).

Table 6.17: Design Stress

Material Used Design stress, N/mm 2

Shell: Stainless Steel 304 136 @ 130o

C

6.5.7 Design Criteria

6.5.7.1 Minimum Practical Wall Thickness

To ensure that the vessel is adequately rigid to withstand its own weight and any

incidental loads.



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For cylinder shell the minimum thickness required to resist internal pressure

can be determined as follows:

= ×

2 − (6.5.1)

Where P i = internal Pressure, N/mm 2

Di = Ds= Shell diameter, mm

J = Joint Factor = 1.0

f s = Design stress of shell side, N/mm 2

Shell: Internal diameter, D s = 0.434 m

=

0.6 × 454.7217

2 1 (136) − 0.6 = 0.96

Adding Corrosion allowance: 3 mm

= 0.96 + 3

= . ~ 4.0

6.5.7.2 Head and Closure

Heads closes the ends of cylindrical vessel. For the design, Ellipsoidal heads are

chosen because to save cost since it is more economical. The standard type is with

a major and minor axis ratio 2:1. For this ratio, the following equation can be used:

= ×

2 − 0.2 (6.5.2)

Where P i = internal Pressure, N/mm 2

Di = Ds = Shell diameter, mm

J = Joint Factor = 1.0

f s = Design stress of shell side, N/mm 2

= 0.6 × 434.7560

2 1 (136) −2(0.6)

= 0.957



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Adding Corrosion allowance: 3 mm

= 0.957 + 3

= . ~ 4.0

∴

Same as the wall thickness

6.5.7.3 Baffles

Baffles are used in the shell to direct the fluid flow across tube and increase the fluid

velocity. When the fluid velocity increases, it’s improving the rate of heat transfer.

The assembly of baffles and tubes is hold together by support rods and spacers.

The most commonly used type of baffle is a single-segmental baffle. Baffle

cut used to indicate the dimensions of a segmental baffle. Generally, baffle cut is

20% - 25% will be optimum. The value will give good heat transfer rate without

excessive drop.

With 25% baffle cut:

= – 1 (6.5.3)

= 4.50.1735

– 1

= 24.94~

6.5.7.4 Nozzle (Branches)

Nozzles are used as inlet and outlet stream of the cooler. The nozzles are for

channel side and the shell side of the heat exchanger.

Standard steel pipe are used for the inlet and outlet nozzles. It is important to

avoid flow restrictions at the inlet and outlet nozzles to prevent excessive pressure

drop and flow-induced vibrations of the tubes. Material of construction for nozzles is

same as shell material which is carbon steel.



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Tube side nozzles

Table 6.18: Fluid: Chilled water

Equation Inlet OutletMaterial ofconstruction Stainless steel Stainless steel

Temperature, oC 10 20Density, kg/m 999.7 999.7Flow rate, G water , kg/s 21.05 21.05Fluid velocity, m/s 1.23 1.23Flow area, A, m G/( × ) 0.02 0.02Inside diameter pipe,m

4 1/2 0.20 0.20

From Stanley M.Walas, Chemical Process Equipment (Refer Appendix E11)

By taking ID = 7.98 in. the selected tube size nozzle (for inlet and outlet):

Table 6.19: Properties of Steel Pipe for Inlet and Outlet of Chilled Water

Normal pipesize, in.

OD, in Schedule No. ID, inFlow area per

pipe, in 2 8 8.63 40 7.98 50.00

Shell side nozzles

Table 6.20: Fluid: Hydrocarbon Mixture

Inlet OutletMaterial of construction Stainless steel Stainless steelTemperature, oC 117.2 20Density, kg/m 3 802.2 894Flow rate, G HC, kg/s 3.894 3.894Fluid velocity, m/s 0.622 0.622Flow area, A, m 0.01 0.01Inside diameter pipe, m 0.11 0.11

By taking ID = 4.33 in

Table 6.21: Properties of Steel Pipe for Inlet and Outlet of Hydrocarbon Mixture

Normal pipesize, in.

OD, in Schedule No. ID, inFlow area per

pipe, in 2 6 6.625 40 6.065 28.9



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6.5.7.5 Flanged Nozzle

Flanged joint are used for connecting pipes and instrument to vessel, for manholes

cover and for removable vessel head when ease of access is required. Flanged also

used on the vessel body, when it is necessary to divide the vessel into sections for

transport or maintenance. Flanged joints are also used to connect pipe to the

equipments such as pumps and valves.

Flanges dimension must be able to withstand the hydrostatic ends loads and

the bolt loads necessary to ensure tight joint in service. For the design of this heat

exchanger, welding-neck flange are used. It is because welding-neck flanges have a

long tapered hub between the flange ring and the welded joint. This gradual

transition of the section reduces the discontinuity stresses between the flange and

branch. It is also can increase the strength of the flange assembly.

Welding-neck flanges are suitable for extreme service conditions, where

flange are likely to be subjected to temperature, shear and vibration loads. They will

normally be specified for the connections and nozzles on process vessels and

process equipment. The dimensions of welding-neck flanges is chosen base on the

nominal pipe size of the nozzle pipe. All dimension are listed below

From Coulson & Richardson’s, Volume 6 (Refer AppendixE12)

Dimension of selected flanges (BS 4504)

Table 6.22: Design of Flanged Joint

Type NomPipe

PipeOD

Flange Raisedface Bolt Drilling Neck

D B h d 4 f No. d 2 k d 3 h 2 r

Tubeside 203 171.3 268.3 18.1 48.4 205.4 3 M16 8 18 228.3 187.1 12.2 10

Shellside 150 168.3 265 18 44 202 3 M16 8 18 225 184 12 10



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Figure 6.8: Flanged Joint Standard

6.5.7.6 Weight Load

The major sources of dead weight loads are:

Vessel Shell

Tubes

Fluid to fill the vessel

Fluid to fill the tubes

Insulator

1. Vessel Shell Weight

For preliminary design calculation, the approximate weight of cylindrical vessel

with domes ends and uniform wall thickness can be estimated from the following

equation:

= × × × × × + 0.8 × 10 −3 (6.5.4)



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Where C v = a factor to account for the weight of nozzles, manways, internal

supports

Hv = Length (the length of cylindrical section), m

g = gravitational acceleration, 9.81 m/s 2

t = wall thickness, mm

ρm = density of vessel material, kg/m 3

Dm = mean diameter of vessel = (D i + t x 10 -3), m

= 1.08 × × 8000 × (0.434 + 0.004) × 9.81 × 4.5 + 0.8 0.434 4.0× 10 −3

= .

2. The tube weight

= × × 2 −2 × × × (6.5.5)

Where N t = number of tubes

do = outside diameter of tube-side, m

d i = inside diameter of tube-side, m

L = length, m

ρm = density of tube material, kg/m3

g = gravitational acceleration = 9.81 m/s 2

= 130 × × 0.02223 2 −0.01974 2 × 4.5 × 16 × 9.81

= .

3. Weight of 2-Ethylhexyl Acrylate to fill the vessel

= × 2 4 × × × (6.5.6)

Where D s = diameter of shell-side, m

L = length, m

ρs = density of shell-side, kg/m 3

g = gravitational acceleration, m/s 2

= × 0.434 2 4 × 4.5 × 802.2 × 9.81

= .



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4. Weight of water to fill the tubes

= × × 2 −2 × × × (6.5.7)

Where N t = number of tubes

do = outside diameter of tube-side, m

d i = inside diameter of tube-side, m

L = length, m

ρt = density of tube-side, kg/m 3


= 130 × × 0.02223 2 −0.01974 2 × 4.5× 998.2 × 9.81

= .

5. Weight of insulator

= × × × × × (6.5.8)

Where D s = diameter of shell-side, m

L = length, m

lw = thickness of insulation, m

ρm = density of material insulation, kg/m3


= × 0.434 × 4.5 × 0.025 × 180 × 9.81

= .

Table 6.23: Total Weight of Heat Exchanger

Weight of Sources Values (N)

Vessel Shell Weight 2244.74

The tube weight 30.1462Weight of 2-Ethylhexyl Acrylate to fill 5257.0827

Weight of water to fill the tubes 470.9073

Weight of insulator 60.2943

Total Weight 8063.170

Total Weight with 5% allowance 8466.3290



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6.5.8 Vessel Support

Determination of support a vessel is depending on the size, shape, and weight of

the vessel; the design temperature and pressure; the vessel location and

arrangement; and the internal and external fittings and attachments. Support should

be design to allow easy access to the vessel for inspection and maintenance.

Since cooler in horizontal arrangement, saddle support is chosen as the

support of the cooler. The saddle must be designed to withstand the load imposed

by weight of the vessel and its content. The total weight load of the vessel is 13.9 kN

from calculated above. From the value of the weight and diameter, the dimensions

of saddle are choosing by referring to Appendix B4;

For outer vessel or shell diameter = 0.434 ~ 0.45m (standard construction) for

support only.

Table 6.24: Design of Saddle for Vessel

Shell O.D(mm)

Dimensions (mm) A B C D E F

450 330 410 10 13 130 13

Design of saddle for Vessel Continuation

Dimensions (mm)Bolt size No. of ribs Weight

(kg)G H J100 62 - M.20 1 26



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Figure 6.9: Standard saddle for vessel

6.5.9 Summary of Heat Exchanger Design

Table 6.25: Summary of Mechanical Heat Exchanger Design

Description Specification

Material of constructionShell

Tube

Stainless Steel

Stainless Steel

Shell thickness 4.0 mm

Head Type

Thickness

Ellipsoidal

4.0 mm

Support Type Saddle

Flange Type Welding Neck

(Nominal pressure 6 bar)

Nozzles Tube inlet and outlet

Shell inlet and outlet

200.00 mm

110.00 mm

Baffles Type

Number of baffles

Segmental

25




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6.5.10 Specification sheet of Heat Exchanger (E-104)

Heat Exchanger Data Sheet Equipment No. (Tag): E-104

Descript. : HEATEXCHANGE BETWEENS13 AND S14

Operating Data

Size 0.435 TypeShell and

tube No. of units 1

Shell per unit 1 Surface per Shell 2

Surface per unit 2

Performance of one Unit

SHELL SIDE TUBE SIDE

TOTAL FLUID ENTERINGIN OUT IN OUT

LIQUID (2-EHA) LIQUID LIQUID

WATER LIQUID LIQUID

VISCOSITY LIQUID 0.00043 0.00131

SPECIFIC HEAT 2.336 4.1950THERMALCONDUCTIVITY 0.10500 0.5815

TEMPERATURE 117.2 20 10 20

OPERATING PRESSURE 1 bar

VELOCITY 0.3225 1.060NO. OF PASSES 1 65

PRESSURE DROP 0.1061 0.0972

HEAT EXCHANGED 36.043

Construction of one Shell

DESIGN PRESSURE 6 bar

DESIGN TEMPERATURE 128.92

DUTY 884.1549 kW

TUBESNo. OD: 0.02223 THICKNESS

:0.04 LENGTH: 4.5 PITCH0.028

SHELL I.D 0.435

chapter 6 mechanical design

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