Appendix c -Process Equipment Sizing

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


D

c = 5 m

50 mm

l w =

3 . 8

m


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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


33/176

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


34/176

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.


35/176

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%


37/176

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.


38/176

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


39/176

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


40/176

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


41/176

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


42/176

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


= 0.7822 m/s


Uv = U f x 0.80

= 1.3908 m/s

5.0

L

V

W

W

V L


43/176

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


44/176

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)


45/176

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


= 298.2208 m/s


= 54%

Obtained the orifice coefficient Co = 0.77


46/176

The pressure drop through the dry plate can be estimate using eq 11.88 (p.575,

Coulson,1999).

e) Trial Plate Layout


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


47/176

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


48/176

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


49/176

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


50/176

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)


51/176



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.


52/176

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


53/176

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


54/176

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


55/176

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


56/176

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


57/176

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


58/176

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


K' = 0.0412


= 0.6809 m/s


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


59/176

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


60/176


= 0.0527


= 0.7822 m/s


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


61/176

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


62/176

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


= 54%


63/176


From equation 11.88 (p.575, Coulson,1999)

hd = 1063.1457 mm liquid

Base

Dry Pressure Drop


= 56.6988 m/s


= 54%


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


64/176

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


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


65/176

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


66/176

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


67/176

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)


68/176

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


69/176

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


70/176

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


71/176

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


72/176

( )

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


73/176

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;


74/176

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


75/176

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


76/176

= 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


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)


77/176

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


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)


78/176

= 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


= 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


79/176

9. Sizing Heat Exchanger, HE-102


Exchange heat between Stream (from 70C to 80C) and Stream (from 95.31C to

87.92C).




T1 = 95.31C (inlet shell side)





lm = (T 1-t2) - (T 2-t1)

In (T 1-t2)/ (T 2-t1)

=

= 16.58 oC



Volume 6; page 699)

R = T 1 - T 2 S = t 2 - t1

t2 - t1 T1 - t1

= =

= 1.35 = 0.30


80/176


Ft = 0.98

m lm

` = (0.98) (16.58)

= 16.25 C


m

Heat transfer area, A = Q

m

=

= 76.11m 2

(3) Number of Tubes



Tube inside diameter = 12 mmLength of tube, L = 3.66 m

= 0.23 m 2





Tube area

= = 248 tubes


81/176

(4) Shell Diameter




K1 = 0.249

n1 = 2.207


= 0.457m





= 515 mm


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



82/176

= 0.0249




= 0.0979kg/s




= D s / 5

= 257mm



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




p t = 26.51 m 2




83/176


7) Pressure drop

(i) Tube side

av = 926.546kg/m 3

t = 3.922m/s



=

= 189





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


84/176

s = 118.968m/s


= 4627



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



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


85/176



Exchange heat between Stream (from 50C to 80C) and Stream (from 99C to 95.31C).




T1 = 99C (inlet shell side)





lm = (T 1-t2) - (T 2-t1)

In (T 1-t2)/ (T 2-t1)

=

= 30.27 oC



Volume 6; page 699)

R = T 1 - T 2 S = t 2 - t1

t2 - t1 T1 - t1

= =

= 8.13 = 0.07


86/176


Ft = 0.97

m lm

` = (0.97) (30.27)

= 29.36 C


m


m

=

= 41.99m 2

(3) Number of Tubes



Tube inside diameter = 12 mmLength of tube, L = 3.66 m

Area of tube

= 0.23 m 2





Tube area

= = 138 tubes


87/176

(4) Shell Diameter




K1 = 0.249

n1 = 2.207


= 0.35m





= 402 mm



= 97.155 C

Tube cross- 2

4

= 201.152 mm 2

Tube per passes = Number of tube Number of passes

= 69 tubes


= 0.01378


88/176




= 0.04892kg/s




= D s / 5

= 201mm





Baffle cut = 100 mm




p t

= 16.159 m2





89/176

7) Pressure drop

(i) Tube side

av = 920.733kg/m 3

t = 3.5493m/s



= = 181





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


= 490


90/176





= 25.0806 mm

= 0.0250806 m



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


91/176



Exchange heat between Stream (from 56.7C to 80C) and Stream (from 100C to 80C).





T2 = 80C (outlet shell side)

t1 = 56.7C(inlet tube side)



lm = (T 1-t2) - (T 2-t1)

In (T 1-t2)/ (T 2-t1)

=

= 21.68 oC



Volume 6; page 699)

R = T 1 - T 2 S = t 2 - t1

t2 - t1 T1 - t1

= =

= 1.165 = 0.462



92/176

Ft = 0.75

m lm

` = ( 0.75) (21.68)

= 16.20 C


m


m

=

= 110.83m 2

(3) Number of Tubes



Tube inside diameter = 16 mm

Length of tube, L = 4.88 mArea of tube = L

= 0.23 m 2





Tube area

= = 362 tubes


93/176

(4) Shell Diameter




K1 = 0.249

n1 = 2.207


= 0.542m





= 593 mm



= 90 C

Tube cross- 2

4

= 201.152 mm 2


= 180 tubes


= 0.0363


94/176




= 0.05596kg/s




= D s / 5

= 296mm





Baffle cut = 148 mm




p t

= 35.13 m 2





95/176

7) Pressure drop

(i) Tube side

av = 877.386kg/m 3

t = 1.538m/s



= = 11.8





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


= 658


96/176





= 25.0806 mm

= 0.0250806 m



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


97/176



Exchange heat between Stream (from 80C to 297.70C) and Stream (from 390C to

99.50C).







t2 = 297.70C(outlet tube side)


lm = (T 1-t2) - (T 2-t1)

In (T 1-t2)/ (T 2-t1)

=

= 47.23 oC



Volume 6; page 699)

R = T 1 - T 2 S = t 2 - t1

t2 - t1 T1 - t1

= =

= 0.75 = 0.94


98/176


Ft = 3.5

m lm

` = ( 3.5) (47.24)

= 163.31 C


m


m

=

= 110.53m 2

(3) Number of Tubes


Tube diameter,do = 20 mmTube inside diameter = 16 mm

Length of tube, L = 4.88 m

= 0.23 m 2





Tube area

= = 331 tubes


99/176

(4) Shell Diameter




K1 = 0.249

n1 = 2.207


= 0.521m





= 578 mm



= 244.975 C

Tube cross- 2

4

= 201.152 mm 2


= 166 tubes


= 0.0333


100/176




= 14.593kg/s




= D s / 5

= 289mm





Baffle cut = 144 mm




p t

= 33.37 m 2





101/176

7) Pressure drop

(i) Tube side

av = 0.9315kg/m 3

t = 438m/s



= = 246





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



102/176

= = 389From Figure 12.30 (Chemical Eng. Volume 6, page 647), at 25% baffle cuts.




= 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

Documents

Appendix c -Process Equipment Sizing