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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].
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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
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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)
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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)
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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
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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].
= ∆= ( − )
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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
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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:
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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
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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.
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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
Therefore, the pipe used is 25mm pipe.
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
Therefore, the pipe used is 50mm pipe.
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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
Face BoltingDrilling Boss
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
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Table 6.4 Values for bolt and flange of the outlet pipe
Nominal
sized1
FlangeRaised
Face BoltingDrilling Boss
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)
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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)
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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
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= . (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
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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
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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
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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.
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, =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
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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
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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:
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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
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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
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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
Material of construction Austenitic Stainless Steel (18Cr/8Ni,
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
Material of construction Austenitic Stainless Steel (18Cr/8Ni,
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
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 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
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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
S7: D nominal = 200 mm
Table 6.10: Values for bolt and flange of the outlet pipe (S6)
d 1 Flange Raised Face
BoltingDrilling Boss
D b h d 4 f No. d 2 k d 3
219.1 320 20 44 258 3 M16 8 18 280 240
S8: D nominal = 50 mm
Table 6.11: Values for bolt and flange of the outlet pipe (S8)
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
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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.
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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
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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
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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 = .
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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.
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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)
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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
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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
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(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
The corrosion allowance is the additional thickness of metal added to allow
for material lost by corrosion and erosion. Based on table 13.4, Coulson &Richardson, Chemical Engineering, volume 6, page 739, and this minimum
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(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
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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
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(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.
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(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
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≈ 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
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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
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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)
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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
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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)
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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
Density = 0.1167 kg/m 3
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
Density = 801.7 kg/m 3
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.
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Figure 6.7: Typical Standard Flange Design
(Source: R. K. Sinnott, John Metcalfe Coulson, John Francis Richardson, Chemical
Engineering Design. Volume 6)
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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
Total Weight 426.58 kN
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
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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
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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
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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
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= 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
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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
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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
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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
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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
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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
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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
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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.
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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
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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,
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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
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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
where, P s = operating pressure
σ = design stress at working temperature
Therefore, t = 0.0715 (20.2) / (20 (130) + 0.0715)
= 0.0005 mm
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So, thickness of nozzle = corrosion allowance + 0.0012 mm
= 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 ……
where, P s = operating pressure
σ = design stress at working temperature
Therefore, t = 0.00715 (7.39) / (20 (130) + 0.00715)
= 0.0015 mm
So, thickness of nozzle = corrosion allowance + 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.
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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
g = gravitational acceleration, m/s 2
= 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
g = gravitational acceleration, m/s 2
= × 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
Total Weight 8.466 kN
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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