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, Basically, there are three types of reactor, which are batch reactor, mixed m ixed 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 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 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.
1-2
1.1
CHEMICAL CHEMICAL DESIGN OF REACTOR
1.1.1
Determination 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.
− − − =
Since,
1
=
1
(1
=
)
So,
− − − =
Where
1
=
=
= 0.65 0.6537 37
3
1
.
=
,
The data collected are; ρA =1059.33 kg/m 3 ρB = 836.76 kg/m 3
Mw,A = 72.06 kg/kgmol Mw,B = 130.23 kg/kgmol
.
1
(1
)
(Nowak,1999)
(1
)
1-3
Hence,
=
1059.33 = 14.7 72.06
=
836.76 = 6.425 130.23
3
/
3
− − − − − − − − − =
1
(1
)
(1
= 0.6537(14.7)(1 = 61.6923(1
1.1.2
/
)(1
)
)(6.425)(1
)
)
Calculation of Residence Time,
of the Reactor
By 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
1-4
Reaction Rate, 1/r A vs Concentration, X 1.80 1.60 1.40 1.20 A
1.00
r / 1 0.80
0.60 0.40 0.20 0.00 0
0.2
0.4
0.6
0.8
1
X
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 0.432
= 2.31 hr -1
1
1-5
Molar flowrate = 176.221 kgmol/hr Molar density = 8.4713 kgmol/m 3
=
=
176.221 8.4713 2.31
= 9.01m3
Volume of tank;
−Ɛ =
(1
)
Assuming that void fraction is 50%,
− =
9.01 (1 0.5)
= 18.02m 3
1.1.3
Length and Diameter of the Reactor Vessel
Assume;
L =4 D L D= 4
V =
π
4
D2 L
1-6
V = πD3 D = (V/π)1/3 D = (18.02/π)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/mm 2
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 reac tor. T
= 75oC = 348.15 K
1-7
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
1.2.4
= 162.5N/mm2
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 the process 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/mm 2
is the internal pressure
=0.33 N/mm 2
1-8
− =
0.33 1790 2 162.5 0.33
= 1.819mm
Therefore the wall thickness = e + corrosion allowance = 1.819 + 4 = 5.819 mm ≡ 6mm
1.2.7
Heads and Closure
The ends of a cylindrical vessel are closed by head. There are four principal type of closure; flat plates and formed flat heads, hemispherical heads, ellipsoidal heads, and torispherical heads. Hemispherical, ellipsoidal and toripherical can be referred as domed heads. They are formed by pressing or spinning. The diameter of the head will be same with the diameter of the cylinder.
Table 1.2: The Comparison of Head Types Flat
plates
and
formed flat heads
Hemispherical
Ellipsoidal
heads
heads
Torispherical heads
Diagram
covers Uses
for
manways, covers
channel of
Usually used as head closure for high pressure vessels
heat
exchengers
Shape
Strength
Flat
plates
with
flanges Not
a
Domed all
head,
radius
are
same structurally
The
efficient form. Need
and
strongest
Domed
head,
Domed
head,
major to minor knuckle to crown radius ratio =
radius ratio is about
2:1
6/100
can
withstand
can withstand the
can higher pressure
pressure up to 15
1-9
a very thick head to
withstand very
than
withstand
high pressure
torispherical
pressure
high or
large
bar
head
diameter Cheap but can be Price
Cheapest
Highest
Less
than increased with the
hemispherical
increased
of
pressure
All of the minimum thickness of all types of head is calculated to choose the best head for the reactor. The example of calculation for the torispherical head is;
− =
Where
2
+
(
1
= stress concentration factor for torispherical heads = (3 +
= crown radius
= knuckle radius
=
0.2)
= 1.79
= 0.06
= 0.06 × 1.79 = 0.1074
=
1 3+ 4
=
1 1.6 3+ 4 0.1074
4
1-10
= 1.715
− − =
=
2
+
(
0.2)
0.33 1790 1.715 2 1 162.5 + 0.33(1.715
0.2)
= 3.11mm
Therefore the wall thickness = ehead + corrosion allowance = 3.11 + 4 = 7.11mm ≡ 8 mm
Table 1.3: Minimum thickness of different types of closure Heads
Formula Value (mm)
Torispherical
− − − Hemispherical =
4
5
1.2
Ellipsoidal =
2
6
0.2
=
2
+
(
0.2)
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.
1-11
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 = 0.895 2
Minor axis, the radius, , can be calculated by
=2
2 = 0.895 =
0.895 = 0.4475 2
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 + 2 0.4475
= 8.051
1-12
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
+ 0.8
Where, 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 = 1.796
−
10
3
1-13
= 240
1.15 1.796 7.160 + 0.8 1.796 (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 m 3 – 18.14 m3 = 0.82 m3 Total weight
= Density x Volume of insulation x Gravitational acceleration = 130 kg/m 3 x 0.82 m 3 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
1-14
Outer diameter
= 1.842 m
Length of reactor
= 7.160 m
Volume of steel
= Volume by outer diameter – Volume by inner diameter = 19.06 m 3 – 18.96 m3 = 0.1 m3
Total weight of steel = Density x Volume of steel x Gravitational Acceleration = 8300 kg/m 3 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
= 162634.10 = 162.634
Weight of fluid
=
18060 hr
= 5.017kg/s
= 0.432hr = 1555.2 s
/
2
1-15
=
5.017 s
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
π =
(
+ t)t
where, W
= total weight
Di
= internal diameter
t
= thickness of vessel
π =
273944N 1790 + 6 6
= 8.09 /
2
1-16
Bending stresses
=±
+
2
From Sinnott, 2003, a dynamic wind pressure of 1280 N/m 2 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/m 2 = Mean diameter
= 640 1.842 = 1178.9 /
Bending moment at bottom tangent line;
2
=
2
Where,
2
H
=
Loading per meter
=
The height of concentrated load above the column base
Assume that the height of skirt support to tangent line is 1.5 meter.
1-17
1178.9(8.051 + 1.5)2 = 2
= 53770.573 = 53770573
The second moment of area of the vessel about the plane of bending, I v ;
− =
(
4
4
64
Where,
− =
+ 2
= 1790 + 2(6) = 1802
=
(18024 17904 ) 64
= 1.365 1010
So,
=±
=±
4
+
2
53770573 1.365 1010
= ±3.55 /
2
6+
1790 2
1-18
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 1796 2 6
2
= 49.39 /
=
4
0.33 1796 4 6
= 24.70 /
2
= ±3.55 /
2
= 49.39 /
2
= 24.70 /
2
2
= 8.09 /
=
(Compressive, therefore it is negative)
+
= 20.16 /
2
=
= 13.06 /
2
.
1-19
As assumed that there is no torsion shear stress, the principal stresses will be
and
. 20.16N/mm 2
13.06N/mm 2
49.39N/mm 2
49.39N/mm2
Down-wind
Up-wind
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 ta ll columns subject to wind loading
a) Stainless steel has been chosen as the material for straight skirt ( = 90o) with the design stress = 175 N/mm 2 and Young's modulus 200,000 N/mm 2 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
1-20
c) Wind loading, F w 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/mm 2
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/mm 2 (operating) The resultant stresses in the skirts will be;
− (
)=
;
1-21
− − = 2.12 =
4.84
2
2.72 / =
+
= 2.12 + 4.84 = 6.96 /
2
Take 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, mm 2
1-22
= Number of bolts (8) = Maximum allowable bolt stress, typically 125 N/mm 2 = Bending moment at the base, Nm (53770 Nm) = Bolt circle diameter, m (0.5m)
W
= Weight of the vessel, N (273944 N)
− =
1 4 53770 8 125 0.5
= 156.216
273944
2
So, bolt size of M24 will be used in the design where the root area is 353 mm 2 Compressive load on the base ring, F b
=
4
2
+
= 704.16kN/m fc is the maximum allowable bearing pressure on the concrete foundation range from 3.5 – 7.0 N/mm 2. 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
1-23
ts
= 10 mm
Actual width required, L = L r + ts + 50mm = 76 + 10 + 50
′ ′ ′
=
=
= 136mm
704.16 / 136
= 5.178 /
2
Actual bearing pressure on concrete foundation, f ’c’ = 5.178 N/mm2 and fr is
typically
140
N/mm2 Base ring thickness;
′ 3
=
= 76
3 5.178 140
t b = 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 Guthrie’s Modular Method. From
Table 4.11 (Biegler et.al, 1997), if 1 < D < 10ft and 4 < L < 100ft, therefore vertical fabrication vessel can be used to determine the costing.
1-24
Co
= $ 1000
Lo
= 4.0 ft
Do
= 3.0 ft
= 0.81
= 1.05
=
0
0
0
23.49 = 1000 × 4
0.81
5.87 3
1.05
= $ 8486.49 < $ 200000
According 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 F m value is 2.25 The material and pressure factor (MPF) for pressure vessel (reactor) is given by (Biegler et.al, 1997):
=
= 2.25 1.00 = 2.25
The 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)
1-25
=
=
586 115
= 5.1
The uninstalled cost for the reactor is given by:
=
= $ 8486.49
2.25 = $ 19094.60
The installation cost can be determined by: (Biegler et.al, 1997)
− − =
1
= $ 8486.49 3.18
1 = $ 18400.55
Therefore, the total installed cost can be estimated by equation below: (Biegler et.al, 1997)
− − =
+
1
= $ 8486.49 2.25 + 3.18
1 = $ 37595.15
Thus, 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
= $ 191735.27
= $ 191735.27 3.0225 = RM 579519.86
1
