Chapter 1 Chemical & Mechanical

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CHAPTER 1

CHEMICAL AND MECHANICAL DESIGN OF REACTOR

1.0 INTRODUCTION

Reactor is a place where reactions take place. It is the heart of a chemical processes. There

are many types of reactor, depending on the type of reaction, kinetics and the total

production. Basically, there are three types of reactor, which are batch reactor, mixed flow

reactor and plug flow reactor.

Usually for lower total production, batch reactor is used, due to low maintenance and

low capital cost. The yield also relatively higher, compared to other two types of reactor.

Many pharmaceutical companies used this type of reactor due to low total production per

year and high dependency to demand.

Second and third type of reactors; mixed flow reactor (MFR), where it is also called

continuous stirred tank reactor (CSTR) and plug flow reactor (PFR). Both are used for higher

production per year. The cost is relatively higher than batch. Mixed flow reactor is about to

mix two or many reactants in a tank, and the reaction will take place in the tank. Agitator may

be used to agitate the mixture, therefore make the reactants react uniformly.

In other hand, plug flow reactor does not need any agitator. Reactants will be

introduced into the reactor, which is in vessel form (cylinder with heads) and the agitation will

occurs naturally. Turbulent flow must be given to the reactants in order to ensure the natural

agitation occurs.

In this process, a CSTR is used since all the reactants, product, and catalyst involved

are in liquid form.


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1.1 CHEMICAL DESIGN OF REACTOR

1.1.1 Determination of Rate of Reaction

The main reaction in this process is between acrylic acid and ethylhexanol producing

ethylhexyl acrylate and water. According to Nowak,1999, the reaction between acrylic acid

and ethylhexanol appears to be second order reaction.

= 1 Since,

=

1

= (1 )So,

= 1 (1 ) (1 )

Where 1 = 0.6537 3 . (Nowak,1999)

= = . 1

=

,

The data collected are;

A =1059.33 kg/m3

B = 836.76 kg/m3

Mw,A = 72.06 kg/kgmol

Mw,B = 130.23 kg/kgmol


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Hence,

=1059.33

72.06 = 14.7 /3

= 836.76130.23 = 6.425 /3

= 1 (1 ) (1 )

= 0.6537(14.7)(1

)(6.425)(1

)

= 61.6923(1 )(1 )

1.1.2 Calculation of Residence Time, of the ReactorBy using Microsoft Excel, the reaction rate is calculated and a graph of reaction rate versus

conversion is plotted.

Table 1.1: Data for the calculation of residence time

X -rA -1/rA

0 61.7455 0.0162

0.1 50.0139 0.0200

0.2 39.5171 0.0253

0.3 30.2553 0.0331

0.4 22.2284 0.0450

0.5 15.4364 0.0648

0.6 9.8793 0.1012

0.7 5.5571 0.1800

0.8 2.4698 0.4049

0.9 0.6175 1.6196


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Figure 1.1: Graph for calculation of volume of reactor

For a CSTR, the area under the black arrow is the residence time. It takes the shape of a

rectangular.

Residence time, = 0.52 x 0.83= 0.432 hr

= 25.92 min

According to Biegler et al, (1997);

=

=1

= 1 =

1

0.432

= 2.31 hr -1

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

1.60

1.80

0 0.2 0.4 0.6 0.8 1

1/rA

X

Reaction Rate, 1/rA vs Concentration, X


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Molar flowrate = 176.221 kgmol/hr

Molar density = 8.4713 kgmol/m3

=

= 176.2218.4713 2.31

= 9.01m3Volume of tank;

= (1 )

Assuming that void fraction is 50%,

= 9.01(1 0.5)

= 18.02m3

1.1.3 Length and Diameter of the Reactor Vessel

Assume;

L

D= 4

D =L

4

V = 4

D2L


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V = D3

D = (V/)1/3

D = (18.02/)1/3

D = 1.790m

D = 5.87ft

L = 4D

= 4 x 1.790

= 7.160m

= 23.49ft

1.2 MECHANICAL DESIGN OF REACTOR

1.2.1 Design Pressure

For safety purpose, the design pressure is 10 % above the operating pressure was chosen.

Pi = 3 bar x 1.1

= 3.3 bar

= 0.33 N/mm2

1.2.2 Design Temperature

The strength of metal decreased with the increasing of temperature. Therefore, the

maximum allowable design stress will depend on the material temperature. The design

temperature is taken as the maximum working temperature in the reactor.

T = 75oC

= 348.15 K


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1.2.3 Material Used

Stainless steel material is used as the construction because the chemical material involved

in this process is corrosive. Specifically, stainless steel Type 316 is used because it is an

alloy added with molybdenum to improve the corrosion resistance in reducing conditions.From Chemical Engineering Vol 6, Table 13.2, the strength property of this material is:

Design stress, f = 162.5N/mm2

1.2.4 Welded Joint Efficiency

Joint efficiency selected is 1 because the lower joint factor will result in a thicker and heavier

vessel. The joint factor 1 implies that the joint is equally strong as virgin plate.

1.2.5 Corrosion Allowance

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

by corrosion and erosion or scaling. The allowance should be based on experience with the

material of construction under similar service condition to those for the proposed design.

Most design codes and standards specify a minimum allowance of 1.0 mm, but since theprocess involves corrosive material, the corrosion allowance is increased to 4.0 mm.

1.2.6 Minimum Wall Thickness

For cylindrical shell, the minimum thickness required to resist internal pressure is determine

from the following equation;

= 2 where;

Di is the internal diameter = 1.790 m

is the design stress, and =162.5N/mm2

is the internal pressure =0.33 N/mm2


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= 1.715

= 2 + ( 0.2) = 0.33 1790 1.71521162.5 + 0.33(1.715 0.2) = 3.11mmTherefore the wall thickness = ehead + corrosion allowance

= 3.11 + 4

= 7.11mm

8 mm

Table 1.3: Minimum thickness of different types of closure

Heads Hemispherical Ellipsoidal Torispherical

Formula

=

4

1.2

=

2

0.2

=

2

+

(

0.2)

Value(mm)

5 6 8

From the calculation and consideration for choosing the heads, an ellipsoidal head is

chosen for the reactor because the thickness is same with the vessel thickness. So there is

no need to install a conical section (reducers) to make a gradual reduction from one

cylindrical section to another of smaller diameter. Ellipsoidal head is also cheaper than

hemispherical head.


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1.2.8 Total Height of Reactor

Figure 1.2: Major and minor axis of the ellipsoidal dome closure

Major axis, the diameter is equal to internal diameter of the cylinder, which is 1.79 m.

2 = 1.79 = 1.79

2= 0.895

Minor axis, the radius,

, can be calculated by

= 22 = 0.895 = 0.895

2= 0.4475

The height of the dome is equal to the radius at minor axis. Therefore,

Height = 0.4475 m

Total Height of the reactor = height of cylinder + 2 (height of dome)

= 7.160 + 20.4475= 8.051


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1.2.9 Weight Loads

The main sources of load to consider are:

1. Pressure

2. Dead weight of reactor and contents

3. Wind

4. Earthquake (seismic)

5. External loads imposed by piping and attached equipment

Since the plant is situated in Malaysia which is known as an earthquake-free and the reactor

is build on the ground, so the loads cause by earthquake can be neglected.

Dead weight of vessel

For preliminary calculations the approximate weight of a cylindrical vessel with domed end,

and uniform wall thickness can be estimated from the following equation;

=

+ 0.8

10

3

For a steel vessel, the above equation can be reduced to;

= 240 + 0.8Where,

Wv = total weight of the reactor, excluding internal fitting

Cv = 1.15 due to the presence of heating tube

Hv = Length of the cylindrical section, m

t = wall thickness, mm

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

= 1.79 + 6

103

= 1.796


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= 240 1.15 1.7967.160 + 0.81.796(6) = 25568.39

= 25.568

Weight of insulation material

Insulation material = Mineral Wool

Density = 130 kg/m3

Thickness = 40 mm

Inner diameter = 1.796 m

Outer diameter = 1.836 m

Length of reactor = 7.160 m

Volume of insulation = Volume by outer diameter Volume by inner diameter

= 18.96 m3 18.14 m3

= 0.82 m3

Total weight = Density x Volume of insulation x Gravitational

acceleration

= 130 kg/m3 x 0.82 m3 x 9.81 m/s2

= 1045.75 N

= 1.06 kN

Weight of insulation wrapper

Wrapper material = Stainless steel 316

Density = 8300 kg/m3

Thickness = 6 mm

Inner diameter = 1.836 m


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Outer diameter = 1.842 m

Length of reactor = 7.160 m

Volume of steel = Volume by outer diameter Volume by inner diameter

= 19.06 m3 18.96 m3

= 0.1 m3

Total weight of steel = Density x Volume of steel x Gravitational Acceleration

= 8300 kg/m3 x 0.1 m3 x 9.81 m/s2

= 8142.3 N

= 8.14 kN

Weight of catalyst

= 9.01m3

= 1840

/m3

= = 1840

m3 9.01m3 9.81 /2

= 162634.10

= 162.634 Weight of fluid

= 18060hr

= 5.017kg/s

= 0.432hr = 1555.2 s


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= 5.017s

1555.2 9.81 /2 = 76541.92 = 76.542

Total weight of the reactor = Weight of vessel + Weight of fluid + Weight of

catalyst + Weight of insulation material and wrapper

= 25.568kN + 76.542kN + 162.634kN + 1.06 + 8.14

= 273.944kN

1.2.10 Analysis of Stresses

Dead weight stress

= ( + t)t

where,

W = total weight

Di = internal diameter

t = thickness of vessel

= 273944N1790 + 66 = 8.09/2


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Bending stresses

= + 2 From Sinnott, 2003, a dynamic wind pressure of 1280 N/m2 can be used in

preliminary design study where it is equivalent to a wind speed of 160 km/h for a column with

a height of 20 m and above. At any site, the wind velocity near the ground will be lower than

that higher up (due to the boundary layer), and in some design methods a lower wind

pressure is used at heights below about 20 m; typically taken as one-half of the pressure

above this height.

Loading per meter;

= Where,

= Wind pressure = 640 N/m2

= Mean diameter

= 640 1.842 = 1178.9 /

Bending moment at bottom tangent line;

= 22 Where,

= Loading per meterH2= The height of concentrated load above the column base

Assume that the height of skirt support to tangent line is 1.5 meter.


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= 1178.9(8.051 + 1.5)22

= 53770.573

= 53770573

The second moment of area of the vessel about the plane of bending, Iv ;

= (4 464 Where,

= + 2 = 1790 + 2(6) = 1802

= (18024 17904)64 = 1.365 10104

So,

= + 2 = 537705731.365 1010 6 + 17902 = 3.55/2


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Resultant longitudinal stress

= + At bottom tangent line, primary stresses are given by the longitudinal and

circumferential stresses due to pressure (internal or external), and . = 2 = 0.33 179626 = 49.39 /2

= 4 = 0.33 179646 = 24.70 /2

= 3.55/2 = 49.39 /2 = 24.70 /2 = 8.09/2 (Compressive, therefore it is negative)

= + = 20.16/2

= = 13.06/2


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As assumed that there is no torsion shear stress, the principal stresses will be

and .

1.2.11 Vessel Support

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

reactor, the design temperature and pressure, the reactor location and arrangement, and the

internal and external fittings and attachments. Supports will impose localised 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. Supports should be

designed to allow easy access to the vessel and fittings for inspection and maintenance.

Skirts support is suitable for the tall, vertical column. In this plant, a straight skirt is

chosen as a support. This is because as they do not impose concentrated loads on the

vessel shell; they are particularly suitable for use with tall columns subject to wind loading

a) Stainless steel has been chosen as the material for straight skirt ( = 90o) with

the design stress = 175 N/mm2 and Young's modulus 200,000 N/mm2 at ambient

temperature

b) Assume height of skirt to tangent line is 1.5 m and the total weight of reactor with

content is 273.944kN

13.06N/mm2

49.39N/mm2

20.16N/mm2

49.39N/mm2

Down-wind Up-wind


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c) Wind loading, Fw is 1178.9 N/m

d) Bending moment, at the base of the skirt

= 5.377 107

e) Take skirt thickness as 10 mm, bending stress in the skirt, ;

=

4

(

+

)

= 2.12 N/mm2

f) The dead weight stress in skirt, w s

=

(

+

)

Where,

Mx = Maximum bending moment

W = Total weight of vessel with content

Ds = Inside diameter skirt at the base

ts = Skirt thickness

= ( + )= 4.84 N/mm2 (operating)

The resultant stresses in the skirts will be;

() =


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= 2.12 4.84= 2.72 /2

=

+

= 2.12 + 4.84

= 6.96 /2Take joint factor, J = 1

Criteria for design, s (tensile) > fs J sin

-2.81 > 175

s (compressive) > 0.125 E (ts/Ds) sin

5.53 > 140

Both criteria are satisfied, add 2 mm for corrosion, that gives a design thickness

of 12 mm

1.2.12 Anchor Bolts and Base Rings

Scheiman has given a guide rules in selecting the anchor bolt;

1. Bolts smaller than 25 mm cannot be used.

2. Minimum number of bolts is 8

3. Use multiple of 4 bolts.

4. Bolt pitch should not less than 600 mm (2 feet)

= 1 4 Where,

= Area of one bolt at the root of the thread, mm2


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= Number of bolts (8) = Maximum allowable bolt stress, typically 125 N/mm2

= Bending moment at the base, Nm (53770 Nm)

= Bolt circle diameter, m (0.5m)W = Weight of the vessel, N (273944 N)

= 18 1254 537700.5 273944

= 156.216 2So, bolt size of M24 will be used in the design where the root area is 353 mm2

Compressive load on the base ring, Fb

= 42 + = 704.16kN/m

fc is the maximum allowable bearing pressure on the concrete foundation range from 3.5

7.0 N/mm2.

fc = 5 N/mm2 has been chosen.

Minimum width of the base ring;

=

= 704.16 1000/5 1000 /2 = 140.83

From Chemical Engineering Vol 6, for M24:

Lr = 76 mm


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ts = 10 mm

Actual width required, L = Lr + ts + 50mm

= 76 + 10 + 50

= 136mm

= = 704.16/136

= 5.178

/

2

Actual bearing pressure on concrete foundation, fc = 5.178 N/mm2 and fr is typically 140

N/mm2

Base ring thickness;

= 3

= 76 3 5.178140 tb = 25.3mm

26 mm

1.3 COSTING OF REACTOR

Table 1.4: Data from reactor sizing

Reactor

Diameter, D (ft) 5.87

Length, L (ft) 23.49

Material of Construction Stainless steel 316

The costing of reactor is estimated by using the Guthries Modular Method. From

Table 4.11 (Biegler et.al, 1997), if 1 < D < 10ft and 4 < L < 100ft, therefore vertical fabricationvessel can be used to determine the costing.


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Co = $ 1000

Lo = 4.0 ft

Do = 3.0 ft

= 0.81

= 1.05

= 0 0 0

= 1000

23.49

4

0.81

5.87

3

1.05

= $ 8486.49 < $ 200000According to Biegler et al, 1997;

For BC < $ 200 000, MF = 3.18

Vessel pressure = 3.3 bar

= 48.50 psig

Since the operating pressure of the reactor vessel is less than 50 psig, therefore from

Table 4.2 (Biegler et al, 1997), the vessel pressure factor (F P) is equal to 1.00. Since the

material used is stainless steel 316, so the Fm value is 2.25

The material and pressure factor (MPF) for pressure vessel (reactor) is given by (Biegleret.al, 1997):

= = 2.251.00 = 2.25The plant cost index for year 2010 is 586 (Source: CEPCI) and the base cost index

used is based on year 1968 which is 115. (Biegler et.al, 1997)


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=

=586

115

= 5.1The uninstalled cost for the reactor is given by:

= = $ 8486.49 2.25 = $ 19094.60The installation cost can be determined by: (Biegler et.al, 1997)

= 1 = $ 8486.493.18 1 = $ 18400.55Therefore, the total installed cost can be estimated by equation below: (Biegler et.al,1997)

= + 1 = $ 8486.492.25 + 3.18 1 = $ 37595.15Thus, the update bare module cost or equipment estimation cost is given by equation

below: (Biegler et.al, 1997)

= + 1 = 5.1 $ 8486.49 2.25 + 3.18 1

= $ 191735.27

= $ 191735.27 3.0225= RM 579519.86

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Chapter 1 Chemical & Mechanical