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7/29/2019 Appendix c -Process Equipment Sizing
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APPENDIX C
Calculation for Process Equipment Sizing
1. Catalytic Packed-bed Reactor (R-101)
Design Basic:
During the production of methacrolein as an intermediate to further produce methyl
methacrylate, a catalytic packed-bed reactor (R-101) is being used. Isobutylene (iC4)
with the presence of oxygen (O 2) and steam (H 2O) is fed into the reactor to produce
methacrolein (MAL). Due to the side reactions, carbon monoxide (CO) and carbon
dioxide (CO 2) are formed. The reaction is carried out at 390C and 3 atm.
Multicomponent molybdatecatalysts are preferred in the reaction. The overall conversion
of iC4 is 95.5%. The calculated energy balance shows that the reaction is exothermic.
The general information of the reactor and catalyst can be summarized as below:
Assumptions:
i. The reaction is irreversible
ii. Isothermal reaction
Reactor (R-101):
Type : Catalytic, packed-bed
Orientation : Vertical
Mode of operation : Continuous
Material : Stainless Steel
Temperature : 390 C
Pressure : 2atm
Mass flow rate : 54337.445kg/hr
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Catalyst:
Type :Molybdate catalyst supported by Calcium Nitrate
Bulk density :700 kg/m 3
Catalyst size : 0.04 m pellet
Bed void fraction : 0.5
Chemical Engineering Design
Chemical Reaction:
C4H8+ O 2 C4H6O + H 2O
C4H8+ 4O 2 4CO + 4H 2O
C4H8+ 6O 2 4CO 2 + 4H 2O
Rate Law from journal:
Source: Vyacheslav and Volodymyr (2008).
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Polymath Calculation
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Reactor Sizing
From the polymath tables result, the reactor length needed for 95.5% of iC4 conversion
is 5.73m.Height of reactor with catalyst, H = 5.73m
Diameter of reactor with catalyst, D = 1.8 m
Volume of reactor with catalyst, V c = x R 2x L
= x 0.9 2 x 5.73
= 14.58 m 3
Add extra 5 % of volume at top and bottom of reactor,
Volume of reactor, V = 1.1 x 14.58 m 3 =16.038m 3
Thus, the diameter of the reactor,
And Height of the reactor,
FromPolymath results,
Weight of catalyst, W = 2098.50 kg
Volume of catalyst, Vc = =
= 3.0 m 3
Volumetric flow rate = 38347.15 m3
/hr
Pressure Drop Calculation
Mass flow in reactor, m = 54609.29 kg/hr
= 15.169 kg/s
Void fraction, = 0.5
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Cross sectional area, A = x 0.9 2
= 2.545 m 2
Superficial velocity, G = mass flow / cross sectional area
= 15.169 / 2.545
= 5.96 kg/m 2.s
Inlet gas density, 0 = 0.519 kg/m 3
Catalyst pellet diameter, D p = 0.04 m
Viscosity of gas, = 0.032cP
= 0.000032Pa.s
By using Ergun equation ,
=
= 12047.79
dP = 12047.79 x 5.73
= 69033.83 Pa
= 0.681atm (pressure drop)
P = (2.0 0.681)atm
= 1.3atm (outlet pressure)
Cooling Jacket Design
Since the reaction is exothermic, heat has to be removed from reactor to maintain it at
isothermal. So, cooling water is used to maintain the reactor at isothermal reaction.
Heat release from exothermic reaction, Q = 1.18x10 7 W
Heat should remove to maintain isothermal, Q r = 1.18x107 W
Cooling water inlet into cooling jacket, Tcw in = 25CCooling water outlet from cooling jacket, Tcw out = 80C
Heat capacity of water, Cp = 4.187 kJ/kg.C
Density of water, = 997.042 kg/m 3
Since,
Q = m x Cp x T
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m = Mass flow rate of cooling water, m = = 51.106 kg/sVolumetric flow rate of cooling water, v = = 0.051 m 3/s
Assume the space time of cooling water, = 180 s
Volume of cooling jacket, V j = v x = 0.051 x 180 = 9.18 m 3
Length of cooling jacket, L j = 0.9 x L = 0.9 x 5.73 = 5.157 m
Diameter of cooling jacket, D j = = = 3.28 m
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2. CSTR Reactor (R-102)
Design Basic:
During the production of Methyl Methacrylate, oxidative esterification of Methacrolein is
carry out in a continuous stirred tank reactor.Methacrolein (MAL) with the presence of
oxygen (O 2) and methanol (MeOH) are fed into the reactor to produce Methyl
Methacrylate (MAL). The reaction is carried out at 50C and 3 atm. Palladium catalysts
are preferred in the reaction. The overall conversion of MAL is 99.4%. The calculated
energy balance shows that the reaction is exothermic. The general information of the
reactor and catalyst can be summarized as below:
Reactor (R-102):Type : Continuous Stirred Tank Reactor (CSTR)
Orientation : Vertical
Mode of operation : Continuous
Material : Stainless Steel
Temperature : 50 C
Pressure : 3 atm
Mass flow rate : 5698.772 kg/hr
Catalyst:
Type : Palladium catalyst supported by Calcium Carbonate
Catalyst size : 4 mm pellet
Bed void fraction : 0.6
Chemical Engineering Design
Chemical Reaction:
C4H6O + 0.5O 2 + CH 3OH C5H8O2 + H 2O
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Rate Law from journal:
Source : Li et al. (2004)
Polymath Calculation
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Reactor Sizing
From the polymath tables result, residence time needed for 99.4% of Methacrolein
conversion is 0.585 hr.
Residence time, t = 0.585 hr
Mass flow, m = mass flow rate / residence time
= 5698.772kg/hr / 0.585hr
= 3333.782 kg
Mixture density, = 56.183 kg/m 3
Volume of liquid, = mass flow / density
=
= 59.338 m 3
Thus, the diameter of the reactor, And Height of the reactor, Weight hourly space velocity (WHSV) is defined as the weight of feed flowing per unit
weight of the catalyst per hour.
Weight hourly space velocity = = = 1.709 hr -1
Weight of catalyst, W = =
= 3334.57 kg
Volume of catalyst = =
= 12.8 m 3
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Pressure Drop Calculation
Mass flow in reactor, m = 5698.772 kg/hr
= 1.583 kg/s
Void fraction, = 0.5
Cross sectional area, A = x 1.67 2
= 8.762 m 2
Superficial velocity, G = mass flow / cross sectional area
= 1.583 kg/s / 8.762 m 2
= 0.181 kg/m 2.s
Inlet gas density, 0 = 1.333 kg/m 3
Catalyst pellet diameter, D p = 0.004 m
Viscosity of gas, = 0.016 cP= 0.000016 Pa.s
Impeller Design
For the impeller, based on the 'Rules of Thumb ' (Stanley M. Wales, "Chemical Process
Equipment: Selection and Design", page xvii), the turbine impeller diameter should be
1/3 from the stirred tank diameter, impeller level above bottom = D/3, impeller blade
width = D/15 and four vertical baffles with width = D/10 are used.
Thus, impeller diameter, Impeller height above from the vessel floor
Impeller width, Width of baffles,
Length of impeller blades, Let impellers speed is 4 rps (revolution per second) and the Reynold number isdetermined.
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=
= 396256
(Since high Reynold number, therefore the first assumption made is valid for turbine
impeller selection)
From Figure 10.59, Power correlations for baffled turbine impeller (Sinnott, 1999), at the
curve 5 (where width of impeller/diameter of impeller = 1/8), the power number, N p is
2.7.
Shaft power, = 2.7(378.638)(4) 3(1.023) 5
= 73307.11 W
= 73.31 Kw
Cooling Jacket Design
The reaction temperature in reactor is preferred at 50C, but the resultant temperature of
the reactant mixture is 93.5C. So, extra heats have to be removed to maintain the
temperature of reactor at 50C. Besides, the reaction is exothermic, heat has to be
removed from reactor to maintain it at isothermal. So, cooling water is used to maintain
the reactor at isothermal reaction.
Heat obtained aspen results,
Heat release from exothermic reaction, Q 1 = 4.96 x10 6 W
Heat should remove to maintain 50C, Q 2 = 9.98 x10 4 W
Total heat should remove, Q r = 4.96x10 6 + 9.98x10 4 = 5.06x10 6 W
Cooling water inlet into cooling jacket, Tcw in = 25C
Cooling water outlet from cooling jacket, Tcw out = 35C
Heat capacity of water, Cp = 4.187 kJ/kg.C
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Density of water, = 997.042 kg/m 3
Since,
Q = m x Cp x T
m =
Mass flow rate of cooling water, m = = 120.86 kg/sVolumetric flow rate of cooling water, v = = 0.121 m 3/s
Assume the space time of cooling water, = 180 s
Volume of cooling jacket, V j = v x = 0.121 x 180 = 21.78 m3
Length of cooling jacket, L j = 0.9 x H = 0.9 x 9.21 = 8.29 m
Diameter of cooling jacket, D j = = = 2.94 m
3. Sizing for Absorption Column, AB-101.
Design Basic:
The absorption tower is used to recover the Methacroleinby using water.
Type of Packing:
Many diverse types and shapes of packing can be selected as the internal packing of
absorption tower. Structured packing with a regular geometry such as stacked rings, grids
and arranged structures packing can be used. Random packing such as ring, saddle and
proprietary shapes which are dumped into the column and take up a random arrangement.
The principal of packing is to provide a large surface area between gas and liquid.
Besides, it is used to promote uniform liquid distribution on the packing surface and
promote uniform vapor gas flow across the column cross-section.
Justification:
Different type of absorption tower can be used as the design for scrubber such as tray
column, plate column, spray column, packed column and others. Packed column is
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chosen as the absorption tower because of the appropriate for the corrosive fluids.
Besides, if there is a fouling in the column, it may be cheaper to use packing and replace
the packing. Besides, the pressure drop per equilibrium stage (HETP) can be lower for
packing compare to plates. In this unit, the liquid flows down the column over packing
surface and the gas, counter-currently, up the column.
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Design Calculation:
***From our assumption, we assume that 99.6% of the methacrolein will be
absorbed out with the process water. We assume that all the other gasses other than
methacrolein are insoluble into water.
Composition of Gas Stream
component
Gas Inlet Gas Outlet
S8 S10Molar
flowrate(kmol/hr)
Massflowrate(kg/hr)
Molefraction
Molarflowrate
(kmol/hr)
Massflowrate(kg/hr)
Molefraction
Isobutylene 1.437 80.459 0.001 1.436 80.459 0.001Oxygen 279.000 8,928.000 0.144 279.000 8,928.000 0.167Nitrogen 1,387.000 38,863.740 0.715 1,387.000 38,863.740 0.824Water 231.033 4,158.595 0.118 0.000 0.000 0.000Methacrolein 26.710 1,869.744 0.014 0.107 7.479 0.000
CarbonMonoxide 7.562 211.734 0.004 7.562 211.734 0.004CarbonDioxide 7.562 332.724 0.004 7.562 332.724 0.004Total 1,940.304 54,444.996 1.000 1,682.667 48,424.136 1.000
Temperature = 70C
Li uid InletGas Outlet
Liquid Outlet
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Composition of Liquid Stream
component
Liquid Inlet Liquid OutletS9 S11
Molarflowrate
(kmol/hr)
Massflowrate(kg/hr)
Molefraction
Molarflowrate
(kmol/hr)
Massflowrate(kg/hr)
Molefraction
Isobutylene 0.000 0.000 0.000 0.000 0.000 0.000Oxygen 0.000 0.000 0.000 0.000 0.000 0.000Nitrogen 0.000 0.000 0.000 0.000 0.000 0.000Water 586.497 10,556.938 1.000 817.530 14,715.533 0.968Methacrolein 0.000 0.000 0.000 26.603 1,862.265 0.032CarbonMonoxide 0.000 0.000 0.000 0.000 0.000 0.000
CarbonDioxide 0.000 0.000 0.000 0.000 0.000 0.000Total 586.497 10,556.938 1.000 844.133 16,577.798 1.000
Step 1: Physical Properties
From Aspen simulations results, the density of each stream was taken,
; ; Step 2: Selecting Type of Packing
We had selected the Ceramic Intalox Saddles as our packing material. Ceramic packing issuitable for the corrosive liquids. Based on the 'Rules of Thumb ' (Stanley M. Wales,
"Chemical Process Equipment: Selection and Design", page xv), for gas rates more than
2000cfm (1178 m 3/hr), 2inch of packing diameter is preferred. Since our product gas flow
rate is 23661.451m3/hr, we will use 2inch of packing diameter. Therefore the size of the
packing material is 2 inch (51mm). Therefore, from the Table 11.3 (pg 591), (Coulson
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and Richardson, "Chemical Engineering Design", Volume 6), the packing factor, F p =
130m -1, surface area, a = 108 m 2/m3 and bulk density = 609 kg/m 3.
Step 3: Determine the Tower Cross-Sectional Area, and Diameter.
Inlet Gas Flow Rate, V w* = 1940.304 kmol/hr = 0.539 kmol/s
= 54444.996 kg/hr = 15.124 kg/s
Inlet Liquid Flow Rate, L w* = 586.497kmol/hr = 0.163kmol/s
= 10,556.938kg/hr = 2.932kg/s
From Aspen simulation results,
Gas Density at inlet, v Liquid Density at inlet, L Gas Viscosity at inlet = 0.019 cP= 2.1x 10 -5 Ns/m 2
Liquid Viscosity at inlet = 0.913 cP= 9.13 x 10 -4 Ns/m 2
(Based on equation of viscosity of mixture
where x is mole
fraction)
Based on recommended design at subchapter 11.14.4 (pg 602), the design for pressure
drop (mm water per m packing) should be in between 15 to 50. Thus we decided to
choose 40mm H 2O/m packing.
By using the Figure 11.44 (Coulson and Richardson, "Chemical Engineering Design",
Volume 6), at and 40mm H 2O/m packing, thus K 4 = 3.4
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At flooding, K 4 = 7.0
Percentage flooding = Based on the 'Rules of Thumb ' (Stanley M. Wales, "Chemical Process Equipment:
Selection and Design", page xv), packed towers should operate near 70% of the flooding
rate. Thus, this value is satisfactory.
From equation 11.118 (Coulson and Richardson, "Chemical Engineering Design",
Volume 6) with packing factor F p = 130m -1
Vw* = smkg
F
K
L
L p
V LV
.669.2
514.9931013.9
1301.13
)895.0514.993(895.04.3
1.13
)(2
21
1.04
21
1.04
The Column Area Required = 22 667.5/669.2kg/s15.124
m smkg
Diameter = m686.2667.54
Round off to 2.70m
Column Area = 22 726.57.24
m
Ratio diameters of tower and packing = 941.5210517.2
3 m2
Based on the 'Rules of Thumb ' (Stanley M. Wales, "Chemical Process Equipment:Selection and Design", page xv), the ratio diameters of tower and packing should be at
least 15. Thus, this value is satisfactory.
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Step 4: Determine the Number of Stages (Kremser Equation)
Where absorption factor, A = L/KV
K = vapor-liquid equilibrium
L = flowrate of liquid (kmol/h)
V = flowrate of vapor (kmol/h)
y = molar fraction of methacrolein in gas
x = molar fraction of methacrolein in liquid
Thus,
yin = 0.014 ; y out = 0.000
xin = 0.0003 ; x out = 0.032
L = 586.497kmol/hr
V = 1940.304 kmol/hr
From Aspen Plus, the saturated pressure of Methacrolein at 115.90 is
atmThus the vapor-liquid equilibrium constant,
Hence, Number of stages
[ ] [( ) ]
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To estimate the overall stage efficiency,
Where, ;
;
;
Thus,
Hence Step 5: Determine the Tower Height.
Based on typical values for HTU of random packing, (Coulson and Richardson,
"Chemical Engineering Design", Volume 6)
For 2inch of packing materials, the value of height of heat transfer unit, HTU is in
between 0.6 to 1.0m (2 to 3ft). Thus, the HTU of random packing = 0.6 m.
Height of column = HTU Number of stages, N OG
= 0.6 17 = 10.2 m
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Step 6: The Column Internal Features:
Liquid Distributors selection:
For small diameter column, a central open feed-pipe or one fitted with a spray nozzle
may be used.
For larger diameter column, orifice type liquid distributor can be used to ensure good
distribution at all liquid flow rates. Besides, the weir type distributor also can be chosen
for a wider range of liquid rates compare to orifice type.
Therefore, weir type distributor is selected.
Figure 1 : Weir type distributor
Liquid Redistributors selection:Based on chapter 6.1.2 (Ernest J. Henley etc. al, "Separation Process Principles", page
226), for packed column with a height more than 20 ft (6.096m), a liquid redistributors
need to be installed to reduce the channeling.
Liquid redistributors are used to collect liquid that has migrated to the column walls and
redistribute it evenly over the packing. Besides, they will use to prevent any mal-
distribution that has occurred within the packing.
The wall -wiper type re -distributor is selected which is a ring collects liquid from the
column wall and redirects it into the centre packing.
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Figure 2 : "Wall wiper" redistributors
Hold-down plates:
Hold-down plates are used with the ceramic packing to weigh down the top layers and
prevent fluidization. Fluidization occurs when a high gas rates is surged into the column
with incorrect operation, the ceramic packing can break up and the pieces filter down the
column and plug the packing or the packing may be blown out of the column.
Installing packing:
Before installing the packing, the column should be filled with water to ensure truly
random distribution and prevent damage to the packing. Moreover, the height of water
should be kept above the packing at all times.
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4. Sizing for Distillation Column, DC-101
Step 1: Dew Point and Bubble point
Dew point temperature for the top column
1 bar = 0.986923267 atm
At the top column ,P= 1 bar= 1 atm
Trial 1 : Guess dew point, T = 104 C
Component P sat (atm) K i = P sat P y i yi/K i
MAL 2.7938 2.7938 0.97 0.34719737
H2O 1.1506 1.1506 0.03 0K iX i 0.34719737
Trial 2 : Guess dew point T = 59 C
Component P sat (atm) K i = P sat P y i yi/K i
MAL 0.74218 0.74218 0.97 1.30696058
H2O 0.18787 0.18787 0.03 0K iX i 1.30696058
Trial 3 : Guess dew point, T = 68.5 C
Component P sat (atm) K i = Psat P y i yi/K i
MAL 1.0293 1.0293 0.97 0.94238803
H2O 0.29214 0.29214 0.03 0.10269049
yi/Ki 1.04507852
Thus, dew point for the top column = 68.5 C
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Bubble Point Temperature for Top Column
At bottom column pressure, P = 1 bar = 1 atm
Trial 1 : Guess bubble point Temprature = 110 C
Component P sat (atm) K i = P sat P Xi K iX i
MAL 3.2449 3.2449 0.0002 6.1635E-05
H2O 1.4125 1.4125 0.9998 0.70782301
K iX i 0.70788464
Trial 2:Guess bubble point temperature = 101C
Component P sat (atm) K i = Psat P Xi KiXi
MAL 2.587 2.587 0.002 0.005174
H2O 1.0355 1.0355 0.9998 1.0352929
K iX i 1.0404669
Thus, the bubble point temperature for bottom column = 101C
Step 2 : Theoretical Trays
F = W+D
Component
Feed (Kmol/hr) Product
Molar
flow rate
Mole
fraction
top bottom
Molar
flow rate
Mole
fraction
Molar flow
rateMole fraction
MAL 26.6038 0.03153 26.20609 0.99 0.1332 0.000163
H2O 817.153 0.96847 0.264708 0.01 817.53 0.999837
Total 843.7568 1 26.4708 1 817.6632 1
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Fxd = Wx w +Dx d
843.7568 = W+D
W = 843.7568 D
843.7568(0.03153)=(843.7568-D)0.000163+D(0.99)
Therefore:
D = 26.46493803
W = 817.291862
For top operating line operation :
Yn = L nx/V n+Dx d/Vn Yn = 0.714285714(x) + 8.735211876
Bottom operating line operation :
Ym = L m(x)/V m+W(x w)/V m
Ym = 9.82344634 (x) + 0.001437368
Where :
W= 817.2919 ; D=26.46494
Ln = 2.5 x D
Ln = 66.16234507
Vn = L n+D
Vn = 92.62728309
Lm = L n+F
Lm = 909.9191451
Vm = L m-W
Vm = 92.62728309
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Theoretical step= 12.9
Theoretical trays= 12.1=12
Distillation column efficiency is usually in the range 60-90 %
Assuming the distillation efficiency 70 % = 0.7
Nactual=Ntotal/u=17.14285714 trays = 17 trays
Feed inlet at 13 trays
Step 3 : Minimum number of stages
R m/R m+1 = xD-y'/xD-x'
xD = 0.99
x' = 0.9
y' = 0.95
= 0.444444444
R m = 0.8
R = 1.5R m
R = 1.2
Step 4 : Determine Column Height
Assume 2 foot tray spacing spacing 0.6
Extra feed space = 1.5 m
Disangement space = 3m
Skirt height = 1.5 m
Total height = (N-1)(0.6)(0.6096)+1.5+1.5+3
Total height = 11.85216 m
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Step 5: Calculating for Top Section
Step 6 : Relative Molar Mass
Feed (F) = 19.652092
Distillate (D) = 69.5692
Bottom (W) = 18.018484
Step 7 : Calculation for Density
Bottom product :
Liquid Density L = 999.9751 kg/m 3
Vapor density V = 0.549313 kg/m 3
Top Product :
Liquid Density = 848.53 kg/m 3
Vapor Density = 2.290476 kg/m 3
Component
Molecular
weight(kg/kmol)
Feed (MoleFraction)
Distillate (D)Mole fraction
Bottom (W)Mole fraction
liquid density(kg/L)
MAL 70.09 0.031530176 0.99 0.000162903 0.847
H2O 18.01 0.968469824 0.01 0.999837097 1
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Step 8 : Vapor and Liquid Flow
Ln = 66.162345
Vn = 92.627283
Vm = 92.627283
Lm = 909.91915
From Couldson & Richardson, Chemical Enginering can be used to determine the
maximum allowable superficial velocity and hence determine the column area.
Superficial velocity :
lt is plate spacing ranging from 0.5 to 1.5 m
take l t = 1m
Uv = 2.21377 m/s
Column diameter is using equation below:
Vw is maximum vapor flowrate
Vw=4.221128 kg/s
Dc=5.22166 m
Dc= 5 m
Step 9 : Column Area
The column area is calculated from internal column diameter :
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Ac = 19.625 m 2
Ac = 20 m 2
Step 10 : Plate Spacing
In the methyl methacrylate production, the plate spacing was assumed as in the range of
0.5 to 1 m recommended by Coulson and Richardson, Chemical Engineering.
Step 11 : Liquid Flow Arrangement
L = 0.0164279 m3/s
Based on the values of the maximum volumetric flowrates and the column diameter to
figure 11.28 in Chemical Engineering Book, it is considered as a single pass
Step 12: Plate Design
Column Diameter, D (m) 5
Column Area, Ac (m2) 20
Downcomer Area, Ad (m2) 2.4
Net Area, An=Ac-Ad (m2) 17.6
Active Area, Aa=Ac-2Ad (m2) 15.2
Weir Length= 0.76Dc 0.38
From Figure 11.28 According Couldson & Richardson
WhenL = 0.016427886 m
Dc = 5 m
Weir length = 0.38 m
Take :
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For Hole Diameter is 10 mm, Area of One hole is
A1h = 0.00024649 m 2
Number of holes per plate
Nh = total hole area /area of 1 hole
take 10 % from Active area.
Ah = 1.52
Nh = 6166.578766
Step 13 : Downcomer Liquid Backup
The donwcomer area and plate spacing must be such that the level on the liquid and froth
in the down comer is wellbelow the top of the outlet weir on the plate above. If the level
rise above the outlet weir the column will flood
Take
hap = h w-10mm
where
hap = the height of the bottom edge of the apron above the plate
hap = 90 mm = 0.09 m
Area under the apron
Aap = h ap x Iw
Aap = 0.342 m 2
Where A ap is the clearance area under down cover
As the value of A p is less than A d, therefore Equation 11.92 based on
Couldson&Richardson will be used
Weir Height (mm) 100
Hole Diameter (mm) 10
Plate Thickness (mm) 10
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Where :Lwd= liquid flow rate in downcomer kg/s
Am= either the downcomer area Ad of the clearance area under the doncomer A ap ;
whichever is the smaller, m 2
Lwd = 15.5 kg/s
hdc = 0.340989 mm
where :
Lw = weir length ,m
how = weir chest, mm liquid
Lwd = liquid flowrate, kg/s
how = 18.79904 m
At minimum rate, clear liquid depth :
how+hw = 118.799 mm liquid
Residual height ,
Hr = 12500/L
Hr = 12.50031
Dry Plate drop :
Uh= 6.4 kg/s
Co= 0.75
hd= 2.04004 m
for total drop
ht = h d+(h ow+h w)+h r = 133.3394
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Backup in downcomer
h bc = (h ow+hw)+h t =252.47943mm = 0.2524794m
Step 14 : Calculating for Pressure Drop
i) Total Column Pressure Drop
P t = 1109.93 Pa
= 0.01095 atm
ii) Column Pressure Drop
P c = 0.2738 atm
Step 15 : Trial Plate layout
Use cartridge-type construction. Allow 50mm unperforated strip round plate edge; 50
mm wide calming zones
Obtained :
Dc = 5 m
lw = 3.8 m
5. Sizing for Distillation Column, DC-102
D
c = 5 m
50 mm
l w =
3 . 8
m
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Step 1: Dew Point and Bubble point
From Antoine equation, the vapor pressure for each component can be calculated from:
Log 10Psat
= A - C T B
Psat= vapor pressure, torr, T = Temperature in C
For the components, the Antoine coefficients are as follows:
Component A B C
METHA-01 8.072 1574.990 238.870
METHA-02 (LK) 7.067 1204.950 235.350
METHY-01 (HK) 8.384 2191.430 297.058
WATER 8.071 1730.630 233.426
(Ref: Perrys Chemical Engineers 7 th Eddition, pg 13-21)
By applying relation between Raoults Law and Daltons Law:
Ki =
And Rachford Rice expression for determination of dew point and bubble point of liquid
mixture:
Dew point: = 1
Bubble point: K ixi = 1
F eed bubble point temperatu re
By using Antoine equation, the feed bubble point temperature is calculated as trial error.
At feed column, pressure, P = 1.00 atm
Temperature trial: Guess dew point temperature = 99 C
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Component Mole fraction, yi Psat,torr K x*K
METHA-01 0.003 2575.570 3.389 0.010
METHA-02 (LK) 0.003 2904.369 3.822 0.011
METHY-01 (HK) 0.491 710.193 0.934 0.459
WATER 0.503 733.245 0.965 0.486
Total 1.000 1.000
Thus, the feed bubble point temperature is 99 C.
Dew poin t temperatu re for the top column
By using Antoine equation, the dew point temperature is calculated as trial error.
At top column, pressure, P = 1 atm
Temperature trial: Guess dew point temperature = 91 C
Component Mole fraction, yi Psat,torr K y/K
METHA-01 0.186 1985.282 2.612 0.071
METHA-02 (LK) 0.186 2369.782 3.118 0.060
METHY-01 (HK) 0.628 546.146 0.719 0.874Total 1.000 1.000
Thus, the dew point temperature is 91 C
Bubble point temperatu re for bottom column
By using Antoine equation, the bubble point temperature is calculated as trial and error.
At the bottom column pressure, P = 1.02 atm
Temperature trial: Guess bubble point temperature = 102 C
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Component Mole fraction, xi Psat, atm K Kx
METHA-02 0.186 3126.801 4.032 0.751
METHY-01 0.489 781.578 1.008 0.492
WATER 0.511 816.228 1.053 0.538
Total 1.000 1.000
Thus, the bubble point temperature is 102 C
Step 2: Minimum Number of Tray
Temperature C K LK K HK LK/HK
Distillate 91.000 3.118 0.719 4.339Bottom Product 102.000 4.032 1.008 4.001
( LK/HK ) ave = 4.166
Thus, N min = 5.704
Step 3: Reflux Ratio, R
In order to determine the minimum reflux ratio, the value of in equation below is to be
determined by trial and error.
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Since the gap between the dew point and bubble point of distillate and bottom product is
large, therefore average temperature between dew point of distillate and bubble point of
bottom product is used instead.
At average condition (96.5 C, 1.00 atm), the feed is expected to exist as saturated vapor.
Hence, q=1
Thus,
= 1
Trial, = 0.047
Component Mole fraction,
XF
Psat,atm K = K LK * x f
K HK -
METHA-01 0.003 2377.520 3.128 3.630 0.003
METHA-02 0.003 2728.362 3.590 4.166 0.003
METHY-01 0.491 654.980 0.862 1.000 0.515
WATER 0.503 669.577 0.881 1.022 0.528
Total 1.000 1.049
Component Mole
fraction, yi
Psat, atm K y/K = K LK
K HK
i * x Di
i -
METHA-01 0.186 1985.282 2.560 0.073 0.082 0.439
METHA-02 0.186 2369.782 3.056 0.061 0.068 0.598
METHY-01 0.628 546.146 0.704 0.892 1.000 0.659
Total 1.000 1.696
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By using the value of , minimum reflux ratio (R min ) can be determined by underwood
equation
R min = 0.696180
By rule of thumb, R = 1.5 R min = 1.04427
Step 4: Tray Efficiency
The efficiency of tray column could be determined by
Where viscosity of the feed mixture, F can be estimated by using Kern's Equation
With x i= mass fraction of individual component.
Component Mole fraction, x F (cP) x
METHA-01 0.003 0.299 0.001
METHA-02 0.003 0.270 0.001
METHY-01 0.491 0.320 0.157
WATER 0.503 0.355 0.179
Total 1.000 0.311 0.084
Previously, ( HK/LK )ave = 4.166
Eo= 65.743%
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Step 5 : Feed Point Location
Applying Krikbride Equation, the location of feed point calculation from bottom or top of
trays can be determined
B = 51.983 x HK,F = 0.491 x HK, D = 0.628D = 0.826 x LK, F = 0.0029 x LK, B = 0.00006
Meanwhile, N r + N s = N act
By using two relation,
N r = 1 stages
Ns = 5 stages
Step 6 : Actual Trays and Feed Point Location
Thus, the actual tray is equal to
Nactual = 9 plates
Ns = 8 plates
Nr = 1 plates
Hence, the feed has entered at 8th
stages calculated from the bottom of the columnincluding reboiler.
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Step 7. Plate design procedure
a) Column diameter
Sieve plate is chosen because cheap and has good performance.
Balance on condenser
Reflux, R = 1.0443
Component D b (top) kg/h Uap (kg/h) Reflux (kg/h) , kg/m Flow, m /h
METHA-01 4.930 10.077 5.148 791.800 0.006
METHA-02 10.746 21.969 11.222 847.000 0.013
METHY-01 51.900 106.098 0.000 940.000 0.055
WATER 0.000 0.000 0.000 1000.000 0.000
67.576 138.144 16.370 0.074
Balance on reboiler
Component F b kg/h L o (kg/h)
METHA-01 4.930 5.148
METHA-02 10.966 11.222
METHY-01 2595.010 54.198
WATER 478.427 0.000
3089.332 70.568
Component B b, kg/h Flow, m /h
METHA-01 0.000 0.000
METHA-02 0.219 0.000
METHY-01 2543.110 2.705
WATER 478.427 0.478
3021.756 3.184
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Where,
Uap , Kg/h = (R+1)*D b
Reflux,Kg/h = Uap-D b
Flow, m3/h = D b/ Lo, Kg/h = Reflux
Lm, Kg/h = L o+ F b
Diameter Enriching / Top section
P = 1 atm
T = 364.15 K
R = 0.082 L.atm/mol.K
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To find the surface tension the equation is referred based on (p. 567, eq 11.82, Coulson,
1999)
Surface Tension (), F LV =
= 0.00416
Plate spacing from 0.15 m to 1 m are normally used. Close spacing is used with small
diameter columns and where head room is restricted. For column above 1 m diameter,
plate spacing of 0.3 to 0.6 will normally be used and 0.5 m can be taken as an initial
estimate. A larger spacing will be needed between certain plates to accommodate feed
and side stream arrangements and for manways (p.556, Coulson, 1999)
From figure 11-27 (p.567, Coulson, 1999), with plate spacing, ts = 0.25m and F LV .
Therefore K 1 = 0.075
Correction for surface tension, K' = (FLV/0.02)^0.2)*K1
K' = 0.05477
Velocity of flooding, U f = K'( L - v)/ v)^0.5
= 1.6513 m/s
Assume 80% of flooding in whole velocity
Uv = U f x 0.85
Uv = 1.3211 m/s
Flow rate maximum,
Qv = (V w*BM avg)/( v*3600)
= 0.00741 m 3/s
5.0
L
V
W
W
V L
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Area that needed,
An = Q v/Uv
An = 0.00561 m 2
Downspout taken = 12%
Area of section column
Ac = A n / (1-0.12)
= 0.0637 m 2
Diameter of Enriching/Top section
Dc= (A c*4/)^0.5
= 0.0900 m
Diameter Stripping / Base section
P = 1.02 atm
T = 375.15 K
R = 0.082 L.atm/mol.K
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To find the surface tension the equation is referred based on (p. 567, eq 11.82, Coulson,
1999)
Surface Tension (), F LV =
= 0.3026
Selected plate spacing, t s = 0.25 m
From figure 11-27 (p.567, Coulson, 1999), with plate spacing = 0.25 m and F LV .
Therefore K 1= 0.024
Correction for surface tension, K' = (F LV /0.02)^0.2)*K 1
= 0.0520
Velocity of flooding, U f = K'( L - v)/ v)^0.5
= 0.7822 m/s
Assume 80% of flooding in whole velocity
Uv = U f x 0.80
= 1.3908 m/s
5.0
L
V
W
W
V L
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Flow rate maximum,
Qv = (L w*BM avg)/( v*3600)
= 1.3257 m 3/s
Area that needed,
Area of section column, A n = Q v/Uv
= 0.9532 m 2
Taken downspout = 12 %
Ac = A n/(1-0.12)
= 1.0832 m 2
Diameter of Enriching/Bottom section
Dc= (A c*4/)^0.5
= 1.1747 m
b) Liquid Flow Arrangement
Taken same diameter in top and bottom the feed, D ave = 1.1747 m
Maximum liquid flow rate, Q = (V liquid *M avg)/( L*3600)
= 0.0027 m 3/s
With enter a value Dc and flow rate were fluid to Fig.11.28 (Coulson), note that a single
pass plate can be used as meeting point area cross flow (single pass)
c) Provisional Plate Design
Column diameter, D cavg = 1.1747 m
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Column Area, A c = (Dc/4) ^2
= 1.0832 m 2
Downcomer Area, A d = 12% x A c
= 0.1299 m 2
Net Area, A n = A c Ad
= 0.9532 m 2
Active Area, A a = A c - (2A d)
= 0.8232 m 2
Hole Area, A h = 0.54% x A a
= 0.0044 m 2
Weir length found using Fig 11.31 (Coulson 1999, page 572)
(Ad/Ac) x 100 = 12 %
Obtained l w/Dc = 0.78
Weir Length (l w) = 0.78 x D c ( p.572,Coulson, 1999)
lw = 0.91625m
Taken:
Weir height, h w = 50 mm (p.571, Coulson, 1999)
Hole diameter,d h = 5 mm (p.571,Coulson, 1999)
Plate thickness = 5 mm (p.571, Coulson, 1999)
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d) Plate Pressure Drop
Top
Dry plate Drop
Maximum Vapor Velocity Through holes, Uh = Qv/Ah
= 1.6667 m/s
From (Figure 11.34, Coulson), for plate thickness/ hole diameter = 1, and Ah/Ap Ah/Aa
= 54%
Obtained the orifice coefficient C o = 0.77
The pressure drop through the dry plate can be estimate using eq 11.88 (p.575,
Coulson,1999).
Base
Dry Plate Drop
Maximum Vapor Velocity Through holes, Uh = Qv/Ah
= 298.2208 m/s
From (Figure 11.34, Coulson), for plate thickness/ hole diameter = 1, and Ah/Ap Ah/Aa
= 54%
Obtained the orifice coefficient Co = 0.77
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The pressure drop through the dry plate can be estimate using eq 11.88 (p.575,
Coulson,1999).
e) Trial Plate Layout
Use cartridge-type construction. Allow 50mm unperforated strip round plate edge; 50
mm wide calming zones.
f) Perforated Area
From figure 11.32 (Coulson & Richardson 1999), at l w/D c = 0.78
Obtained the value of, c = 97
Angle subtended at plate edge by unperforated strip = 180 - 97 = 83
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Mean length, unperforated edge strips,l h= (D c -50*10^- 3)*3.14*( c/180)
= 1.903 m
Area of unperforated edge strip = 0.095 m 2
Area of calming zones = 2x (50*10^-3)x(l w-2x50*10^-3)
= 0.082 m 2
Total area available for perforation,
A p = A a - (area of perf.edge strip + area of calm.zone)
= 0.646 m 2
Ah/A p = 0.01
From figure 11.33(p.574, Coulson, 1999), l p/dh = 3.7
The hole pitch (distance between the hole centers) lp should not be less than 2.0 hole
diameters, and the normal range will be 2.5 to 4.0 diameters. Within this range the pitch
can be selected to give the number of active holes required for the total hole area
specified. Square and equilateral triangular patterns are used; triangular is preferred.
(Coulson page 573)
g) Number of Holes
Area of one hole =
= 1.963 x 10 -5 m2
Number of holes = A h/Area of one hole
= 227 holes
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h) Plate Specification
Plate material: Stainless Steel
Downcomer material: Stainless Steel
Plate spacing: 0.3 m
Plate thickness: 5mm
Plate No. : 1
Plate I.D: 1.175 m
Hole size: 5 mmHole Pitch: 19 mm
Active holes: 227 holes
Turn-down = 70 % max rate
i) Tower Height
Height of the column = 1.15 (Nactual) x t s
ts = 0.25 m
Total plate (N act) = 9 plates
Column height = 2.600 m
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Empty space at the top tray = 15 % x column height
= 0.388 m
Empty space at the bottom of the tray = 20% x column height
= 0.518 m
Thus, total height = 3.500 m
j) Shell, Head and Bottom Measurement
i) Shell specification
Design pressure taken = 1 atm
Pop = 14.696 psi
Technical design pressure = 2 x operation pressure
= 29.392 psi
Shell wall thickness (t) calculated by using equation 13.1(Brownell & Young, 1959):
t= ((P*r i)/((f.E)-0.6P)))+c
Where,
Design pressure (P) = 29.392 psi
Inlet diameter (r i) = 40.248 inch
Allowable stress = 1870 psi (table 4, Chap. 13, p.571, Peters)
Joint Efficiency = 85%
Corrosion Allowance, (c) = 0.125 or (1/8) in
Thus, shell wall thickness, t = 0.2103 in
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Taken the standard shell wall thickness = 0.3125 or (5/16) in (Appendix F, Brownell &
Young, 1959)
ii) Specification of head and bottom
Torispherical Dished Head used for operation pressure is 15 bar (Coulson,1989)
Construction material used is Stainless Steel SA-283 Grade C
Head and bottom thickness is calculated by using equation13.1 (Brownell & Young,
1959):
t = ((0.885*P*rc)/((f*E)-(0.1*P)))+c
Where,
Design Pressure (P) = 29.392 psi
Crown radius (r) = ID = 46.278 inch
Allowable Stress (f) = 18700 psi
Joint Efficiency (E) = 85 %
Corrosion Allowance (C) = 0.125 or (1/8) in
Hence, the thickness of Head & Bottom, t = 0.076 inch
Taken the thickness of Head & Bottom standard = 0.1875 in or (3/16) in (Appendix F,
Brownell &Young, 1959)
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6. Sizing for Distillation Column, DC-103
Step 1: Dew Point and Bubble point
From Antoine equation, the vapor pressure for each component can be calculated from:
Log 10Psat = A -C T
B
Psat= vapor pressure, torr, T = Temperature in C
For the components, the Antoine coefficient are as follows:
Component A B CMMA 8.38448 2191.43 297.058H20 8.07131 1730.63 233.426
MAL 7.06691 1204.95 235.35(Ref: Perrys Chemical Engineers 7 th Eddition, pg 13-21)
By applying relation between Raoults Law and Daltons Law:
Ki =
And Rachford Rice expression for determination of dew point and bubble point of liquid
mixture:
Dew point: = 1
Bubble point: K ixi = 1
F eed bubble point temperatu re
By using Antoine equation, the feed bubble point temperature is calculated as trial error.
At feed column, pressure, P = 1.03atm
Temperature trial : Guess dew point temperature = 101.05 C
Component Kmol/hr X f Psat (atm) K (P sat/P) X iK i
MMA 24.4701 0.9574 0.9979 0.9689 0.9276 1.0000H20 1.0838 0.0424 1.0385 1.0082 0.0428 1.0406MAL 0.0063 0.0002 4.0201 3.9031 0.0010 4.0284 25.5602 0.9713
Thus, the feed bubble point temperature is 101.05 C.
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Dew poin t temperatu re for the top column
By using Antoine equation, the dew point temperature is calculated as trial error.
At top column, pressure, P = 1 atm
Temperature trial : Guess dew point temperature = 100.23 C
Component Kmol/hr Y i Psat (atm) K (P sat/P) Y i/K i MMA 0.5094 0.3292 0.9720 0.9720 0.3387 1.0000H20 1.0318 0.6668 1.0083 1.0083 0.6613 1.0374MAL 4.0529 0.0040 3.9393 3.9393 - -
1.5475 1.0010
Thus, the dew point temperature is 100.23 C
Bubble point temperatu re for bottom column
By using Antoine equation, the bubble point temperature is calculated as trial and error.
At the bottom column pressure, P = 1.2 atm
Temperature trial: Guess bubble point temperature = 106.93 C
Thus, the bubble point temperature is 106.93 C
Component Kmol/hr X i Psat (atm) K (Psat/P) X iK i
MMA 24.9606 0.9979 1.1998 0.9998 0.9978 1.0000
H20 0.0520 0.0021 1.2755 1.0629 0.0022 1.0631
MAL 0 0 4.6313 3.8594 0 3.8598
25.0127 1.0000
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Step 2: Minimum Number of Trays
Temperature P LK (atm) P HK (atm) X LK XHK LK /HK
Distillate 100.23 3.9393 0.9720 0.0040 0.3292 4.0528
Bottom Product 106.93 1.2755 1.1999 0.0021 0.9979 1.0630
Thus, Nmin = 2.31
Step 3: Reflux Ratio, R
In order to determine the minimum reflux ratio, the value of in equation below is to be
determined by trial and error.
At feed condition (101.05 C, 1.03atm), the feed is expected to exist as saturated vapor.
Hence, q=0
Thus, = 1Trial, = 0.1088
ComponentMole
fraction, X F Psat K = K LK /K HK
* Xf -
MMA 0.9574 0.9689 0.9689 1 1.0742H20 0.0424 1.0385 1.0082 1.0406 0.0473MAL 0.0002 4.0201 3.9031 4.0284 0.0003 1.0000 1.1218
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ComponentMole
fraction,yi
Psat, atm K y/K = K LK /K HK
MMA 0.3292 0.9720 0.9720 0.3387 1 0.380021346H20 0.6668 1.0083 1.0083 0.6613 1.0374 0.738766305MAL 0.0040 3.9393 3.9393 0.0010 4.0529 0.001054903 1.119842554
By assuming 1.12 1, thus = 1.120
By using the value of , minimum reflux ratio (Rmin) can be determined by underwood
equation
R min= 0.120
By rule of thumb, R = 1.2R min = 0.140
Step 4: Tray Efficeincy
The efficiency of tray column could be determined by
Where viscosity of feed mixture, f can be estimated by Kern
With Xi= mass fraction of individual component
Component Mass flow rate (cp) X i X i/ MMA 2550.056 0.2799 0.95735 3.4203H20 19.494 0.291 0.04240 0.1457MAL 0.42 0.2398 0.00025 0.0010 2569.97 3.5671
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Previously, ( HK/LK )ave = 2.1542
Eo= 0.2985
Step 5: Number of Equilibrium Stage
By using Grilland Correlation,
Since N min = 2.3116 stages, R min = 0.1198, R = 0.1438
No of equilibrium stages, N eq or N = 7.43
Actual no. of trays, N act= = 25 trays
No of trays, N tray= N act 1 = 24 trays
Step 6 : Feed Point Location
Applying Krikbride Equation, the location of feed point calculation from bottom or top of
trays can be determined
B = 2499.931 Kg /hrXf,HK = 0.9574 Xb,LK = 0.0021
D = 69.957 Kg /hrXf,LK = 0.0002 Xd,HK = 0.3292Meanwhile, N r + N s = N act
By using two relation,
N r = 15 stages
Ns = 10 stages
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Hence, the feed has entered at 10 stages calculated from the bottom of the column
including reboiler.
7. Plate design procedure
a) Column diameter
Sieve plate is chosen because cheap and has good performance.
Balance around condenser
Reflux, R = 0.1438
ComponentDa(Mole flow rate)
Molecular
weight D b,Kg/hr U ap,Kg/hr Kmol/hr Kg/Kmol
MMA 0.5094 100.12 50.961 58.2898
H20 1.0318 18 18.576 21.2474
MAL 0.0063 70 0.42 0.4804
1.5475 69.957 80.0176
Component Reflux Density,Kg/m3
Flowrate,
m3
/hr MMA 7.3288 853.4229 0.0597
H20 2.6714 959.544 0.0194
MAL 0.0604 746.83 0.0936
10.0606 0.1727
Balance around reboiler
ComponentFa(Mole flow rate)
Molecular weight F b,Kg/hr L o,Kg/hr
Kmol/hr Kg/KmolMMA 24.4701 100.12 2449.9444 7.3288
H20 1.0838 18 19.5085 2.6714MAL 0.0063 70 0.4396 0.0604
25.5602 2469.8925 10.0606
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Component L m, Kg/hr B a, Kmol/hr B b, Kg/hr Flowrate,
m3/hr
MMA 2506.3920 24.9607 2499.0633 2.9283
H20 3.6078 0.0520 0.9364 0.0010
MAL 0.0604 0 0 0
2510.0603 25.0127 2499.9997 2.9293
Where,
Uap, Kg/hr = (R+1)*D b
Reflux, Kg/hr = U ap -D b
Flow, m /hr = D b/
Lo, Kg/hr = Reflux
Lm, Kg/hr = L o + F b
D (Kmol/hr, Kg/hr) = Mole or mass flowrate of each component exist
Diameter Enriching / Top section
Density of vapor (v) = (Pv*BM avg)/(R*Tv)
BM Avg = D b/Da
= 45.2076 Kmol/hr
P = 1 atm
T = 373.23 K
R = 0.082 L.atm/Mol.K
v= 1.4771 Kg/m 3
Density of liquid ( L) = D b/Flowrate
L= 404.9732 Kg/m 3
Reflux = 0.1438
V liquid = R*D a
= 0.2225 Kmol/hr
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Vvapor = V liquid + D a
= 1.7700 Kmol/hr
Refer on (p. 567, eq 11.82, Coulson, 1999)
Surface Tension(), F LV =
= 0.00759
Plate spacings from 0.15 m to 1 m are normally used. Close spacing is used with small
diameter columns and where head room is restricted. For column above 1 m diameter,
plate spacing of 0.3 to 0.6 will normally be used and 0.5 m can be taken as an initial
estimate. A larger spacing will be needed between certain plates to accommodate feed
and side stream arrangements and for manways (p.556, Coulson, 1999)From figure 11-27 (p.567, Coulson, 1999), with plate spacing, ts = 0.3m and FLV,
therefore K1 = 0.05
Correction for surface tension, K' = (FLV/0.02)^0.2)*K1
K' = 0.0412
Velocity of flooding, U f = K'( L - v)/ v)^0.5
= 0.6809 m/s
Assume 85% of flooding in whole velocity
Uv = U f x 0.85
= 0.5787 m/s
Flow rate maximum,
Qv = (V w*BM avg)/( v*3600)
= 0.0150 m 3/s
Area that needed,
An = Q v/Uv
= 0.0260 m 2
Taken downcomer = 65%...
Area of section column
Ac = A n/(1-0.65) = 0.0743 m 2
5.0
L
V
W
W
V L
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Diameter of Enriching/Top section
Dc= (A c*4/)^0.5
= 0.3075 m
Diameter Stripping / Base section
Density of vapor ( v) = (P v*BM avg)/(R*T v)
BM Avg = D b/ D a
= 99.9492 Kmol/hr
P = 1.2 atm
T = 379.93 K
R = 0.082 L.atm/Mol.K v = 3.8499 Kg/m 3
Density of liquid ( L) = B b/Flowrate
L= 853.4583 Kg/m 3
Reflux = 0.1438
V liquid =( R+ 1)*D a
= 1.7700 Kmol/hr
Vvapor = V liquid + B a
= 26.7827 Kmol/hr
Refer on (p. 567, eq 11.82, Coulson, 1999)
Surface Tension(), F LV =
= 1.0163
Selected plate spacing, t s = 0.3 m
From figure 11-27 (p.567, Coulson, 1999), with plate spacing = 0.3 m and FLV, therefore
K1= 0.024
5.0
L
V
W
W
V
L
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Correction for surface tension, K' = (FLV/0.02)^0.2)*K1
= 0.0527
Velocity of flooding, U f = K'( L - v)/ v)^0.5
= 0.7822 m/s
Assume 85% of flooding in whole velocity
Uv = U f x 0.85
= 0.6648 m/s
Flow rate maximum,
Qv = (L w*BM avg)/( v*3600)
= 0.1931 m 3/s
Area that needed,
Area of section column, A n = Q v/Uv
= 0.2905 m 2
Taken downspout = 65%...
Ac = A n/(1-0.65)
= 0.8300 m 2
Diameter of Enriching/Bottom section
Dc= (A c*4/)^0.5
= 1.0280 m
b) Liquid Flow Arrangement
Taken same diameter in top and bottom the feed, D ave = 1.0280 m
Maximum liquid flow rate, Q = (V liquid *BM avg)/( L*3600)
= 0.0009 m 3/s
With enter a value Dc and flow rate were fluid to Fig.11.28 (Coulson), note that a single
pass plate can be used as meeting point area cross flow (single pass)
c) Provisional Plate Design
Column diameter, D cavg = 1.0280 m
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Column Area, A c = (Dc/4)^2
= 0.83 m 2
Downcomer Area, A d = 12%x A c
= 0.0996 m 2
Net Area, A n = A c Ad
= 0.7304 m 2
Active Area, A a = A c - (2A d)
= 0.6308 m 2
Hole Area, A h= 0.54% x A a
= 0.0034 m 2
Weir length found using Fig.11.31 (Coulson, 1999)
(Ad/Ac)x 100 = 12 %Obtained l w/Dc = 0.78
Weir Length (l w) = 0.78 x D c(p.572,Coulson, 1989)
lw = 0.8018 m
Taken:
Weir height, h w= 50 mm (p.571,Coulson, 1999)
Hole diameter,d h = 5 mm (p.571,Coulson, 1999)
Plate thickness = 5 mm (p.571,Coulson, 1999)
d) Check Weeping
Maximum liquid rate, L w = (V liq x BM avg)/3600
= 0.7436 Kg/s
Maximum liquid rate, at 70 per cent turn-down = 0.5205 Kg/s
Where,
how = Weir crest, mm liquid
lw = Weir height, m
Lw = Liquid flowrate, Kg/s
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Maximum h ow = 750(0.7436/(853.4583 x0.6372)^2/3
= 9.24 mm
Minimum h ow = 750(0.52051/(853.4583 x0.6372)^2/3
= 7.28 mm
At minimum rate h w + h ow = 50 + 7.28
= 57.28 mm
From figure 11.30 (Coulson& Richardson, 2005), K 2 = 30.3
Weep point, is the lower limit of the operating range occurs when liquid leakage through
the plate holes become excessive. The vapor value at the weep point is the minimum
value for stable operations, based on Eduljee (1959) correlation,
Minimum design vapor velocity,
K 2 =a constant, dependent on the depth of clear liquid on the plate from figure 11.30
dh = hole diameter, mm
Minimum design vapor velocity, h = 6.0853 m/s
Maximum vapor flowrate, Q v = 0.1931 m 3/s
Actual vapor velocity through holes,
Actual minimum vapor velocity = 39.6891 m/s
So, minimum operating rate is stable far above weep point.
e) Plate Pressure Drop
Top
Dry plate DropMaximum Vapor Velocity Through holes, Uh = Qv/Ah
= 56.6988 m/s
From (Figure 11.34, Coulson), for plate thickness/ hole diameter = 1, and Ah/Ap Ah/Aa
= 54%
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Obtained the orifice coefficient Co = 0.75
From equation 11.88 (p.575, Coulson,1999)
hd = 1063.1457 mm liquid
Base
Dry Pressure Drop
Maximum Vapor Velocity Through holes, Uh = Qv/Ah
= 56.6988 m/s
From (Figure 11.34, Coulson), for plate thickness/ hole diameter = 1, and Ah/Ap Ah/Aa
= 54%
Obtained the orifice coefficient Co = 0.75
From equation 11.88 (p.575, Coulson,1999)
hd = 1314.7977 mm liquid
From figure 11.29 (Coulson & Richardson,2005)
For F lv = 0.00759 and 85 % flooding, = 0.046
= 1114.409 mm liquid
Residual head, hr = 12500/ L,bottom = 14.64629 mm liquid
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Total column pressure drop,
= 1186.335 mm liquid
Total column pressure drop,
= 9932.5 Pa @ 0.0980 atm
Column pressure drop,
= 2.450 atm
f) Trial Plate Layout
Use cartridge-type construction. Allow 50mm unperforated strip round plate edge; 50
mm wide calming zones
`
g) Perforated Area
From figure 11.32 (Coulson & Richardson 1999), at l w/D c = 0.78
Obtained the value of, c = 97
Angle subtended at plate edge by unperforated strip = 180 - 97 = 83
Mean length, unperforated edge strips,l h= (D c-50*10^- 3)*3.14*( c/180)
= 1.4169 m
Area of unperforated edge strip = 0.0708 m 2
D c =
1 . 0 2 8 0
50 mm
l w =
0 . 8 0 1 8 m
5
50 mm
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Area of calming zones = 2x (50*10^-3)x(l w-2x50*10^-3)
= 0.0702 m 2
Total area available for perforation,
A p = A a - (area of perf.edge strip + area of calm.zone)
= 0.4898 m 2
Ah/A p = 0.01
From figure 11.33(p.574,Coulson,1999), l p/dh = 2.8
Hole pitch : Hole diameter (l p/lh), lp should not be less than 2.0 hole
diameter,satisfactory.
Equilateral tringular patterns are ued for hole area (p.573, Coulson, 1999)
h) Number of HolesArea of one hole =
= 1.9638E-05 m 2
Number of holes = A h/Area of one hole
= 174 holes
i) Plate Specification
Plate material: Stainless Steel
Downcomer material: Stainless Steel
Plate spacing: 0.3 m
Plate thickness: 5mm
50 mm50 mm
l w =
0 . 8
0 1 8
5 0 m m
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Plate No. : 1
Plate I.D: 1.028 m
Hole size: 5 mm
Hole Pitch: 14 mm
Active holes: 174 holesTurn-down = 70 % max rate
j) Tower Height
Height of the column = (Nactual -1) x t s
ts = 0.3
Total plate (N act) = 25 plates
Column height = 7.2 m
Empty space at the top tray = 15 % x column height
= 1.08 m
Empty space at the bottom of the tray = 20% x column height
= 1.44 m
Thus, total height = 9.72 m
k) Shell, Head and Bottom Measurement
i) Shell specification
Design pressure taken = 1 atm
Pop = 14.696 psi
Technical design pressure = 2 x operation pressure
= 29.392 psi
h ap =40 mm
h w = 50mm
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Shell wall thickness (t) calculated by using equation 13.1(Brownell & Young, 1959):
t= ((P*r i)/((f.E)-0.6P)))+c
Where,
Design pressure (P) = 29.392 psi
Inlet diameter (r i) = 40.4711 in
Allowable stress = 1870 psi (table 4, Chap. 13, p.571, Peters)
Joint Efficiency = 85%
Corrosion Allowance, (c) = 0.125 or (1/8) in
Thus, shell wall thickness, t = 0.8818 in
Taken the standard shell wall thickness = 0.3125 or (5/16)in (Appendix F, Brownell &Young, 1959)
ii) Specification of head and bottom
Torispherical Dished Head used for operation pressure is 15 bar( Coulson,1989)
Construction material used is Stainless Steel SA-283 Grade C
Head and bottom thickness is calculate by using equation13.1 (Brownell & Young,
1959):
t = ((0.885*P*rc)/((f*E)-(0.1*P)))+c
Where,
Design Pressure (P) = 29.392 psi
Crown radius (r)= ID = 40.4711 in
Allowable Stress (f) = 18700 psi
Joint Efficiency (E) = 85 %
Corrosion Allowance (C) = 0.125 or (1/8) in
Hence, the thickness of Head & Bottom, t = 0.1333 in
Taken the thickness of Head & Bottom standard = 0.1875 in or (3/16) in (Appendix F,
Brownell &Young, 1959)
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7. Sizing for Decanter, D-101.
Design Criteria:
Decanters are used to separate liquids where there is a sufficient difference in density
between the liquids for the droplets to settle readily. The concept of the decanter involves
the balancing of liquid heights due to differences in density of the two phases as well as
settling velocity of the heavier phase falling through the lighter or the lighter rising
through the heavier. In an operating decanter there will be three distinct zones or bands:
clear heavy liquid; separating dispersed liquid (the dispersion zone); and clear light
liquid.
Thus, this decanter is designed for continuous operation to separate most of the water
contained in the mixture. However, there are assumptions made for the materials balance
in decanter which are:
i) 96% separation efficiency for water
ii) The bottom weight percent for H2O and MMA is 99.26% and 0.074%
respectively and contains no MAL.
Operating Conditions:
Temperature = 80 oC
Pressure = 1 atm
Design Calculation:
Mass balance in decanter
Component
Mass Flow Rate (kg/hr)
Stream 31 Stream 32(heavy phase)
Stream 33(light phase)
MMA 2568.579 18.522 2550.057MAL 0.420 0.000 0.420H2O 487.710 468.205 19.505
TOTAL 3056.709 486.727 2569.982
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Properties table for materials in decanter
Data obtained :
Average density of light phase, L = 868.741 kg/m
Average density of heavy phase, H = 867.239 kg/m
Average viscosity of light phase, L = 9.844 x 10 - kg/m.s
Average viscosity of heavy phase, H = 1.033 x 10 - kg/m.s
Mass flow rate = 8.491 x 10 - kg/s
Volumetric flow rate = 9.774 x 10 - m/s
Mass flow rate out of light phase, M L = 7.139 x 10 - kg/s
Mass flow rate out of heavy phase, M H = 1.352 x 10 - kg/s
Volumetric flow rate out of light phase, Q L = 8.217 x 10 - m/s
Volumetric flow rate out of heavy phase, Q H = 1.559 x 10 - m/s
Design Step:
i) Determination of dispersed phase
The dispersed phase can be determined using Selker and Sleicher correlation
> 3.3
Thus, heavy phase is always dispersed
ComponentDensity,
(kg/m)Viscosity,
(kg/m.s)
MMA 871.745 8.88 x 10 -6
MAL 766.294 8.90 x 10 -6
H2O 968.184 1.18 x 10 -5
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ii) Settling Velocity of dispersed phased
Stokes Law is used to determined the settling velocity of dispersed phase.
The settling velocity is calculated with an assumed droplet size of 150m which is well below the droplet sizes normally found in decanter feeds
Where (-ve) sign mean the heavy phase rises instead of setting.
iii) Area of Interface
iv) Dimensions
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Diameter is assumed to be half of the height as it is reasonable for cylinder.
Since the diameter is below 8m, so it is most suitable to build vertically.
v) Time required for separation
10% of the height is taken as dispersion band
Residence time of the droplet in the dispersion band:
This is satisfactory since the normal recommended times are within 2 to 5 min.
vi) Droplets size
Velocity of the continuous phase
The entrained droplet size can be determine using
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( )
This value is satisfactory since it below the normal size which is 150 m.
vii) Piping arrangement
To minimize entrainment by the jet of liquid entering the vessel, the inlet velocity for a
decanter should keep below 1 m/s.
Flow rate = volumetric flow rate of light phase+ volumetric flow rate of heavy phase
Flow rate = 9.776 x 10 -4 m3/s
viii) Height of light phase take off
Take the light liquid off-take as at 90% of the vessel height
ix) Height of interface
Take the position of the interface as half-way up to the vessel
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x) Height of heavy phase take off
8. Sizing Heat Exchanger (HE-101)
(1) Design Criteria (1-shell passes 2-tube passes)
Exchange heat between Stream (from 25C to 80C) and Stream (from 87.92C to75.33C).
(2) Heat Transfer Area
Heat load = 210.57kW
For estimation purpose, U = 100 W/m 2.C
T1 = 87.92 C (inlet shell side)
T2 = 75.33C (outlet shell side)
t1 = 25C(inlet tube side)
t2 = 80C(outlet tube side)
From Equation 12.4 (Chem. Eng. Volume 6, page 629), log mean temperature different:
lm = (T 1-t2) - (T 2-t1)
In (T 1-t2)/ (T 2-t1)
=
= 22.93 oC
Log Mean Temperature Different Correction Factor, Ft;
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From Equation 12.6(R) and 12.7(S) for 1 shell - 2 tube pass exchanger (Chem. Eng.
Volume 6; page 699)
R = T 1 - T 2 S = t 2 - t1
t2 - t1 T1 - t1
= =
= 0.47 = 0.20
From Figure 12.19(Chem. Eng. Volume 6, page 657)
Ft = 0.75
m lm
` = (0.75) (22.93)
= 17.20 C
From Equation 12.1 (Chem.Eng. Volume 6, page 612)
m
Heat transfer area, A = Qm
= = 122.42m 2
(3) Number of Tubes
From Table 12.3 (Chem.Eng. Volume 6, page 620)
Tube diameter,do = 20 mm
Tube inside diameter = 12 mm
Length of tube, L = 3.66 m
= 0.23 m 2
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Tube outside diameter 20mm (for most duties; give more compact; cheaper exchangers).
Length 3.66 m (reduced shell diameter; lower cost exchangers). Tube arrangement;
triangular pitch (lower pressure drop; not difficult for mechanical cleaning outside tubes)
Number of tubes required = Heat transfer area
Tube area
= = 400 tubes
(4) Shell Diameter
From Equation 12.3(b) (Chem.Eng. Volume 6, page 623)
Tube bundle diameter Db = do (
From Table 12.4, Triangular pitch; 2passes
K1 = 0.249
n1 = 2.207
Bundle diameter, Db = 567mm
= 0.567m
From Figure 12.10 (Chem.Eng. Volume 6, page 622)
Bundle diameter clearance = 60mm = 0.06 m
(Pull-through floating head-easy to remove the bundle for cleaning process)
Shell Inside Diameter Ds, = 0.627 m
= 627 mm
(5) Tube - side Coefficient
Mean tube-side temperature = = 81.625 C
Tube cross- 2
4
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= 113 mm 2
Tube per passes = Number of tube
Number of passes
= 200 tubes
Total flow area = Tube per passes tube cross-sectional area
= 0.0402
Heat capacity for tube side, Cp = 4.0255 kJ/kg. K
Heat load for HE-101 = 210.57 kW
Mass velocity of tube-side flow, m =
= 0.1669kg/s
(6) Shell-side coefficient
From Figure 12.13(a) (Chem. Eng. Volume 6, page 626)
Choose baffle spacing, b = shell inside diameter / 5
= D s / 5
= 313mm
From Figure 12.5(Chem. Eng. Volume 6, page 526)
Baffle diameter, d b = D s-3/16 in (4.88mm) =627 mm = 0.627 m
The optimum baffle cut is 20 to 25% of baffle disc diameter because its given a good heattransfer rates. So, we choose 25% of baffle disc diameter.
Baffle cut = 157 mm
Tube pitch = triangular pitch, pt = 1.25 do = 25 mm
From Equation 12.21 (Chem. Eng. Volume 6, page 645)
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Cross-flow area, As = (p t - do ) D s b
p t
= 20.41 m 2
Heat Capacity for Shell side, C p = 4.5217kJ/kg. K
Heat load for HE-101 = 210.57 kW
Shell-side Flowrate = 0.847 kg/s
7) Pressure drop
(i) Tube side
av = 936.9kg/m3
t = 4.154m/s
thermal conductivity, k f = 0.001W/m.K
Reynold number, Re = D i
= = 178.17
From Figure 12.24 (Chemical Eng. Volume 6, page 641)
Friction factor, j f = 0.43
Equation 12.20 (Volume 6, page 642) pressure drop of tube side,
Neglecting the viscosity correction term, so
t = Np [8j f (L/d i t2/2)
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= 2(8 x 0.021) = 4093281 kg/m.s 2
= 4093.281 N/m2
(Pa)= 0.04093 atm
(ii) Shell side
av = 966.1415 kg/m 3
s = 17.281m/s
Reynold number, Re = D i
= 421
From Figure 12.30 (Chemical Eng. Volume 6, page 647), at 25% baffle cuts.
Friction factor, j h = 0.25
Equation 12.22 (Volume 6, page 648),
equivalent diameter, de = 1.27/d o(p t2-0.785d o2 )
= 25.0806 mm
= 0.0250806 m
Equation 12.26 (Volume 6, page 648) pressure drop for shell side,
Neglect Viscosity correction,
s = 8j f (D s/de) (L/ b s2/2)
= 2(8 x 0.025) = 112276kg/m.s 2
= 112.276N/m 2 (Pa)= 0.1123atm
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9. Sizing Heat Exchanger, HE-102
(1) Design Criteria (1-shell passes 2-tube passes)
Exchange heat between Stream (from 70C to 80C) and Stream (from 95.31C to
87.92C).
(2) Heat Transfer Area
Heat load = 123.67kW
For estimation purpose, U = 100 W/m 2.C
T1 = 95.31C (inlet shell side)
T2 = 87.92C (outlet shell side)
t1 = 70C(inlet tube side)
t2 = 80C(outlet tube side)
From Equation 12.4 (Chem. Eng. Volume 6, page 629), log mean temperature different:
lm = (T 1-t2) - (T 2-t1)
In (T 1-t2)/ (T 2-t1)
=
= 16.58 oC
Log Mean Temperature Different Correction Factor, Ft;
From Equation 12.6(R) and 12.7(S) for 1 shell - 2 tube pass exchanger (Chem. Eng.
Volume 6; page 699)
R = T 1 - T 2 S = t 2 - t1
t2 - t1 T1 - t1
= =
= 1.35 = 0.30
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From Figure 12.19(Chem. Eng. Volume 6, page 657)
Ft = 0.98
m lm
` = (0.98) (16.58)
= 16.25 C
From Equation 12.1 (Chem.Eng. Volume 6, page 612)
m
Heat transfer area, A = Q
m
=
= 76.11m 2
(3) Number of Tubes
From Table 12.3 (Chem.Eng. Volume 6, page 620)
Tube diameter,do = 20 mm
Tube inside diameter = 12 mmLength of tube, L = 3.66 m
= 0.23 m 2
Tube outside diameter 20mm (for most duties; give more compact; cheaper exchangers).
Length 3.66 m (reduced shell diameter; lower cost exchangers). Tube arrangement;
triangular pitch (lower pressure drop; not difficult for mechanical cleaning outside tubes)
Number of tubes required = Heat transfer area
Tube area
= = 248 tubes
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(4) Shell Diameter
From Equation 12.3(b) (Chem.Eng. Volume 6, page 623)
Tube bundle diameter Db = do (
From Table 12.4, Triangular pitch; 2passes
K1 = 0.249
n1 = 2.207
Bundle diameter, Db = 457mm
= 0.457m
From Figure 12.10 (Chem.Eng. Volume 6, page 622)
Bundle diameter clearance = 58mm = 0.06 m
(Pull-through floating head-easy to remove the bundle for cleaning process)
Shell Inside Diameter Ds, = 0.515 m
= 515 mm
(5) Tube - side Coefficient
Mean tube-side temperature =
= 91.615 C
Tube cross- 2
4
= 113 mm 2 Tube per passes = Number of tube
Number of passes
= 124 tubes
Total flow area = Tube per passes tube cross-sectional area
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= 0.0249
Heat capacity for tube side, Cp = 4.2079 kJ/kg. K
Heat load for HE-102 = 123.67 kW
Mass velocity of tube-side flow, m =
= 0.0979kg/s
(6) Shell-side coefficient
From Figure 12.13(a) (Chem. Eng. Volume 6, page 626)
Choose baffle spacing, b = shell inside diameter / 5
= D s / 5
= 257mm
From Figure 12.5(Chem. Eng. Volume 6, page 526)
Baffle diameter, d b = D s-3/16 in (4.88mm) =515 mm = 0.515 m
The optimum baffle cut is 20 to 25% of baffle disc diameter because its given a good heat
transfer rates. So, we choose 25% of baffle disc diameter.
Baffle cut = 129 mm
Tube pitch = triangular pitch, pt = 1.25 do = 25 mm
From Equation 12.21 (Chem. Eng. Volume 6, page 645)
Cross-flow area, As = (p t - do ) D s b
p t = 26.51 m 2
Heat Capacity for Shell side, C p = 2.756kJ/kg. K
Heat load for HE-102 = 123.67 kW
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Shell-side Flowrate = 4.487 kg/s
7) Pressure drop
(i) Tube side
av = 926.546kg/m 3
t = 3.922m/s
thermal conductivity, k f = 0.001W/m.K
Reynold number, Re = D i
=
= 189
From Figure 12.24 (Chemical Eng. Volume 6, page 641)
Friction factor, j f = 0.45
Equation 12.20 (Volume 6, page 642) pressure drop of tube side,
Neglecting the viscosity correction term, so
t = Np [8j f (L/d i t2/2)
= 2(8 x 0.021) = 3999282 kg/m.s
2
= 3999.282 N/m 2 (Pa)
= 0.3999 atm
(ii) Shell side
av = 917.67 kg/m 3
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s = 118.968m/s
Reynold number, Re = D i
= 4627
From Figure 12.30 (Chemical Eng. Volume 6, page 647), at 25% baffle cuts.
Friction factor, j h = 0.54
Equation 12.22 (Volume 6, page 648),equivalent diameter, de = 1.27/d o(p t2-0.785d o2 )
= 25.0806 mm
= 0.0250806 m
Equation 12.26 (Volume 6, page 648) pressure drop for shell side,
Neglect Viscosity correction,
s = 8j f (D s/de) (L/ b s2/2)
= 2(8 x 0.025) = 10917244kg/m.s 2
= 10917.244N/m 2 (Pa)
= 10.91724atm
Could be reduced by reducing the baffle pitch.Doubling the pitch halves the shell-side
velocity, which reduces the pressure drop by a factor of approximately (1/4^2)
P s = 10.91724/16
= 0.6823 atm
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10. Sizing Heat Exchanger, HE-103
(1) Design Criteria (1-shell passes 2-tube passes)
Exchange heat between Stream (from 50C to 80C) and Stream (from 99C to 95.31C).
(2) Heat Transfer Area
Heat load = 61.659kW
For estimation purpose, U = 50 W/m 2.C
T1 = 99C (inlet shell side)
T2 = 95.31C (outlet shell side)
t1 = 50C(inlet tube side)
t2 = 80C(outlet tube side)
From Equation 12.4 (Chem. Eng. Volume 6, page 629), log mean temperature different:
lm = (T 1-t2) - (T 2-t1)
In (T 1-t2)/ (T 2-t1)
=
= 30.27 oC
Log Mean Temperature Different Correction Factor, Ft;
From Equation 12.6(R) and 12.7(S) for 1 shell - 2 tube pass exchanger (Chem. Eng.
Volume 6; page 699)
R = T 1 - T 2 S = t 2 - t1
t2 - t1 T1 - t1
= =
= 8.13 = 0.07
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From Figure 12.19(Chem. Eng. Volume 6, page 657)
Ft = 0.97
m lm
` = (0.97) (30.27)
= 29.36 C
From Equation 12.1 (Chem.Eng. Volume 6, page 612)
m
Heat transfer area, A = Q
m
=
= 41.99m 2
(3) Number of Tubes
From Table 12.3 (Chem.Eng. Volume 6, page 620)
Tube diameter,do = 20 mm
Tube inside diameter = 12 mmLength of tube, L = 3.66 m
Area of tube
= 0.23 m 2
Tube outside diameter 20mm (for most duties; give more compact; cheaper exchangers).
Length 3.66 m (reduced shell diameter; lower cost exchangers). Tube arrangement;
triangular pitch (lower pressure drop; not difficult for mechanical cleaning outside tubes)
Number of tubes required = Heat transfer area
Tube area
= = 138 tubes
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(4) Shell Diameter
From Equation 12.3(b) (Chem.Eng. Volume 6, page 623)
Tube bundle diameter Db = do (
From Table 12.4, Triangular pitch; 2passes
K1 = 0.249
n1 = 2.207
Bundle diameter, Db = 350mm
= 0.35m
From Figure 12.10 (Chem.Eng. Volume 6, page 622)
Bundle diameter clearance = 53mm = 0.053 m
(Pull-through floating head-easy to remove the bundle for cleaning process)
Shell Inside Diameter Ds, = 0.402 m
= 402 mm
(5) Tube - side Coefficient
Mean tube-side temperature =
= 97.155 C
Tube cross- 2
4
= 201.152 mm 2
Tube per passes = Number of tube Number of passes
= 69 tubes
Total flow area = Tube per passes tube cross-sectional area
= 0.01378
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Heat capacity for tube side, Cp = 4.20165 kJ/kg. K
Heat load for HE-102 = 61.659 kW
Mass velocity of tube-side flow, m =
= 0.04892kg/s
(6) Shell-side coefficient
From Figure 12.13(a) (Chem. Eng. Volume 6, page 626)
Choose baffle spacing, b = shell inside diameter / 5
= D s / 5
= 201mm
From Figure 12.5(Chem. Eng. Volume 6, page 526)
Baffle diameter, d b = D s-3/16 in (4.88mm) =402 mm = 0.402 m
The optimum baffle cut is 20 to 25% of baffle disc diameter because its given a good heat
transfer rates. So, we choose 25% of baffle disc diameter.
Baffle cut = 100 mm
Tube pitch = triangular pitch, pt = 1.25 do = 25 mm
From Equation 12.21 (Chem. Eng. Volume 6, page 645)
Cross-flow area, As = (p t - do ) D s b
p t
= 16.159 m2
Heat Capacity for Shell side, C p = 2.395kJ/kg. K
Heat load for HE-102 = 61.659 kW
Shell-side Flowrate = 0.8581 kg/s
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7) Pressure drop
(i) Tube side
av = 920.733kg/m 3
t = 3.5493m/s
thermal conductivity, k f = 0.001W/m.K
Reynold number, Re = D i
= = 181
From Figure 12.24 (Chemical Eng. Volume 6, page 641)
Friction factor, j f = 4.5
Equation 12.20 (Volume 6, page 642) pressure drop of tube side,
Neglecting the viscosity correction term, so
t = Np [8j f (L/d i t2/2)
= 2(8 x 0.021) = 35890312 kg/m.s 2 = 35890.31 N/m 2 (Pa)
= 0.35890 atm
(ii) Shell side
av = 908.531 kg/m 3
s = 13.8674m/s
Reynold number, Re = D i
= 490
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From Figure 12.30 (Chemical Eng. Volume 6, page 647), at 25% baffle cuts.
Friction factor, j h = 0.85
Equation 12.22 (Volume 6, page 648),
equivalent diameter, de = 1.27/d o(p t2-0.785d o2 )
= 25.0806 mm
= 0.0250806 m
Equation 12.26 (Volume 6, page 648) pressure drop for shell side,
Neglect Viscosity correction,
s = 8j f (D s/de) (L/ b s2/2)
= 2(8 x 0.025) = 231163kg/m.s 2 = 231.163N/m 2 (Pa)
= 0.23116 atm
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11. Sizing Heat Exchanger, HE-104
(1) Design Criteria (1-shell passes 2-tube passes)
Exchange heat between Stream (from 56.7C to 80C) and Stream (from 100C to 80C).
(2) Heat Transfer Area
Heat load = 538.82kW
For estimation purpose, U = 300 W/m 2.C
T1 = 100C (inlet shell side)
T2 = 80C (outlet shell side)
t1 = 56.7C(inlet tube side)
t2 = 80C(outlet tube side)
From Equation 12.4 (Chem. Eng. Volume 6, page 629), log mean temperature different:
lm = (T 1-t2) - (T 2-t1)
In (T 1-t2)/ (T 2-t1)
=
= 21.68 oC
Log Mean Temperature Different Correction Factor, Ft;
From Equation 12.6(R) and 12.7(S) for 1 shell - 2 tube pass exchanger (Chem. Eng.
Volume 6; page 699)
R = T 1 - T 2 S = t 2 - t1
t2 - t1 T1 - t1
= =
= 1.165 = 0.462
From Figure 12.19(Chem. Eng. Volume 6, page 657)
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Ft = 0.75
m lm
` = ( 0.75) (21.68)
= 16.20 C
From Equation 12.1 (Chem.Eng. Volume 6, page 612)
m
Heat transfer area, A = Q
m
=
= 110.83m 2
(3) Number of Tubes
From Table 12.3 (Chem.Eng. Volume 6, page 620)
Tube diameter,do = 20 mm
Tube inside diameter = 16 mm
Length of tube, L = 4.88 mArea of tube = L
= 0.23 m 2
Tube outside diameter 20mm (for most duties; give more compact; cheaper exchangers).
Length 4.88 m (reduced shell diameter; lower cost exchangers). Tube arrangement;
triangular pitch (lower pressure drop; not difficult for mechanical cleaning outside tubes)
Number of tubes required = Heat transfer area
Tube area
= = 362 tubes
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(4) Shell Diameter
From Equation 12.3(b) (Chem.Eng. Volume 6, page 623)
Tube bundle diameter Db = do (
From Table 12.4, Triangular pitch; 2passes
K1 = 0.249
n1 = 2.207
Bundle diameter, Db = 542mm
= 0.542m
From Figure 12.10 (Chem.Eng. Volume 6, page 622)
Bundle diameter clearance = 51mm = 0.051 m
(Pull-through floating head-easy to remove the bundle for cleaning process)
Shell Inside Diameter Ds, = 0.593 m
= 593 mm
(5) Tube - side Coefficient
Mean tube-side temperature =
= 90 C
Tube cross- 2
4
= 201.152 mm 2
Tube per passes = Number of tube Number of passes
= 180 tubes
Total flow area = Tube per passes tube cross-sectional area
= 0.0363
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Heat capacity for tube side, Cp = 32.096 kJ/kg. K
Heat load for HE-104 = 538.82 kW
Mass velocity of tube-side flow, m =
= 0.05596kg/s
(6) Shell-side coefficient
From Figure 12.13(a) (Chem. Eng. Volume 6, page 626)
Choose baffle spacing, b = shell inside diameter / 5
= D s / 5
= 296mm
From Figure 12.5(Chem. Eng. Volume 6, page 526)
Baffle diameter, d b = D s-3/16 in (4.88mm) =593 mm = 0.593 m
The optimum baffle cut is 20 to 25% of baffle disc diameter because its given a good heat
transfer rates. So, we choose 25% of baffle disc diameter.
Baffle cut = 148 mm
Tube pitch = triangular pitch, pt = 1.25 do = 25 mm
From Equation 12.21 (Chem. Eng. Volume 6, page 645)
Cross-flow area, As = (p t - do ) D s b
p t
= 35.13 m 2
Heat Capacity for Shell side, C p = 1.532kJ/kg. K
Heat load for HE-104 = 538.82 kW
Shell-side Flowrate = 15.094 kg/s
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7) Pressure drop
(i) Tube side
av = 877.386kg/m 3
t = 1.538m/s
thermal conductivity, k f = 0.001W/m.K
Reynold number, Re = D i
= = 11.8
From Figure 12.24 (Chemical Eng. Volume 6, page 641)
Friction factor, j f = 0.75
Equation 12.20 (Volume 6, page 642) pressure drop of tube side,
Neglecting the viscosity correction term, so
P t = Np [8j f (L/d i t2/2)
= 2(8 x 0.021) = 24706194 kg/m.s 2
= 24076.194 N/m 2 (Pa)
= 0.24706 atm
(ii) Shell side
av = 1.5125 kg/m 3
s = 530.268m/s
Reynold number, Re = D i
= 658
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From Figure 12.30 (Chemical Eng. Volume 6, page 647), at 25% baffle cuts.
Friction factor, j h = 0.79
Equation 12.22 (Volume 6, page 648),
equivalent diameter, de = 1.27/d o(p t2-0.785d o2 )
= 25.0806 mm
= 0.0250806 m
Equation 12.26 (Volume 6, page 648) pressure drop for shell side,
Neglect Viscosity correction,
s = 8j f (D s/de) (L/ b s2/2)
= 2(8 x 0.025) = 522980kg/m.s 2
= 522.980N/m 2 (Pa)
= 0.52298atm
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12. Sizing Heat Exchanger, HE-105
(1) Design Criteria (1-shell passes 2-tube passes)
Exchange heat between Stream (from 80C to 297.70C) and Stream (from 390C to
99.50C).
(2) Heat Transfer Area
Heat load = 5035.2kW
For estimation purpose, U = 300 W/m 2.C
T1 = 390C (inlet shell side)
T2 = 99.50C (outlet shell side)
t1 = 80C(inlet tube side)
t2 = 297.70C(outlet tube side)
From Equation 12.4 (Chem. Eng. Volume 6, page 629), log mean temperature different:
lm = (T 1-t2) - (T 2-t1)
In (T 1-t2)/ (T 2-t1)
=
= 47.23 oC
Log Mean Temperature Different Correction Factor, Ft;
From Equation 12.6(R) and 12.7(S) for 1 shell - 2 tube pass exchanger (Chem. Eng.
Volume 6; page 699)
R = T 1 - T 2 S = t 2 - t1
t2 - t1 T1 - t1
= =
= 0.75 = 0.94
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From Figure 12.19(Chem. Eng. Volume 6, page 657)
Ft = 3.5
m lm
` = ( 3.5) (47.24)
= 163.31 C
From Equation 12.1 (Chem.Eng. Volume 6, page 612)
m
Heat transfer area, A = Q
m
=
= 110.53m 2
(3) Number of Tubes
From Table 12.3 (Chem.Eng. Volume 6, page 620)
Tube diameter,do = 20 mmTube inside diameter = 16 mm
Length of tube, L = 4.88 m
= 0.23 m 2
Tube outside diameter 20mm (for most duties; give more compact; cheaper exchangers).
Length 4.88 m (reduced shell diameter; lower cost exchangers). Tube arrangement;
triangular pitch (lower pressure drop; not difficult for mechanical cleaning outside tubes)
Number of tubes required = Heat transfer area
Tube area
= = 331 tubes
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(4) Shell Diameter
From Equation 12.3(b) (Chem.Eng. Volume 6, page 623)
Tube bundle diameter Db = do (
From Table 12.4, Triangular pitch; 2passes
K1 = 0.249
n1 = 2.207
Bundle diameter, Db = 521mm
= 0.521m
From Figure 12.10 (Chem.Eng. Volume 6, page 622)
Bundle diameter clearance = 57mm = 0.057 m
(Pull-through floating head-easy to remove the bundle for cleaning process)
Shell Inside Diameter Ds, = 0.593 m
= 578 mm
(5) Tube - side Coefficient
Mean tube-side temperature =
= 244.975 C
Tube cross- 2
4
= 201.152 mm 2
Tube per passes = Number of tube Number of passes
= 166 tubes
Total flow area = Tube per passes tube cross-sectional area
= 0.0333
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Heat capacity for tube side, Cp = 1.1501 kJ/kg. K
Heat load for HE-104 = 5035.2 kW
Mass velocity of tube-side flow, m =
= 14.593kg/s
(6) Shell-side coefficient
From Figure 12.13(a) (Chem. Eng. Volume 6, page 626)
Choose baffle spacing, b = shell inside diameter / 5
= D s / 5
= 289mm
From Figure 12.5(Chem. Eng. Volume 6, page 526)
Baffle diameter, d b = D s-3/16 in (4.88mm) =578 mm = 0.578 m
The optimum baffle cut is 20 to 25% of baffle disc diameter because its given a good heat
transfer rates. So, we choose 25% of baffle disc diameter.
Baffle cut = 144 mm
Tube pitch = triangular pitch, pt = 1.25 do = 25 mm
From Equation 12.21 (Chem. Eng. Volume 6, page 645)
Cross-flow area, As = (p t - do ) D s b
p t
= 33.37 m 2
Heat Capacity for Shell side, C p = 1.532kJ/kg. K
Heat load for HE-105 = 5035.2 kW
Shell-side Flowrate = 15.094 kg/s
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7) Pressure drop
(i) Tube side
av = 0.9315kg/m 3
t = 438m/s
thermal conductivity, k f = 0.001W/m.K
Reynold number, Re = D i
= = 246
From Figure 12.24 (Chemical Eng. Volume 6, page 641)
Friction factor, j f = 0.3
Equation 12.20 (Volume 6, page 642) pressure drop of tube side,
Neglecting the viscosity correction term, so
t = Np [8j f (L/d i t2/2)
= 2(8 x 0.021) = 299649 kg/m.s 2
= 299.649 N/m 2 (Pa)
= 0.002996 atm
(ii) Shell sideav = 1.1825 kg/m 3
s = 503.62m/s
Reynold number, Re = D i
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= = 389From Figure 12.30 (Chemical Eng. Volume 6, page 647), at 25% baffle cuts.
Friction factor, j h = 0.8
Equation 12.22 (Volume 6, page 648),
equivalent diameter, de = 1.27/d o(p t2-0.785d o2 )
= 25.0806 mm
= 0.0250806 m
Equation 12.26 (Volume 6, page 648) pressure drop for shell side, Neglect Viscosity correction,
s = 8j f (D s/de) (L/ b s2/2)
= 2(8 x 0.025) = 373483kg/m.s 2
= 373.483N/m 2 (Pa)
= 0.37348atm
13. Sizing f
Recommended