Distillation Column 2 Final Revised

SPECIFICATION SHEET

Identification:

Item: Distillation Column 2

No. required = 2

Tray type = Sieve tray

Function: To separate n-butyraldehyde from feed mixture

Operation: Continuous

Design Data:

Parameters Values Parameters Values

Type of Column Tray Column Material of

Construction

Carbon Steel

Tray Type Sieve Tray Hole Diameter 5 mm

No. of Trays 51 Weir Length 1.82 m

Height of Column 22.536 m Pressure Drop/plate 709.263 Pa

Diameter of

Column

2.33 m Tray Thickness 5 mm

Tray spacing 0.45 m Active Area 2.11 m2

Flooding 87 % Reflux Ratio 16.6967

Material Balance @ Distillation 2

N iso-Butyraldehyde

(K)Compound % Comp. Mass(kg/hr) (N)Compound % Comp. Mass(kg/hr)

n-Butyral 79.656 5626.763 n-Butyral 1.3 18.025

iso-Butyral 20.337 1436.62 iso-Butyral 98.7 1368.493

n-Butanol 0.0052 0.370 (O)Compound % Comp. Mass(kg/hr)

Iso-Butanol 0.00087 0.0616 n-Butyral 98.792 5608.738

iso-Butyral 1.20 68.127

n-Butanol 0.0065 0.37

Iso-Butanol 0.00108 0.0616

Distillation Column 2

“Butyraldehyde”

1. Feed Operating line

O n-Butyraldehyde

17.66 kg/hr

1383.05kg/hr

5660.10kg/hr

7063.81kg/hr

L Mixed Butyraldehyde

q = Feed quality (L−LF )

V = Vapor-liquid leaving (Enriching section)L = Vapor-liquid coming (Enriching section)L = Vapor-liquid leaving (Stripping section)V = Vapor-liquid coming (Stripping section)

Feed at Boiling Point = 66.77oC

q = 1

Assumptions:

0.45 m plate spacing 100 mm pressure drop per plate 85% Flooding At Minimum Liquid Rate, 70% turn down Weir Height = 50 mm (assumed according to Coulson’s suggest, see Page 589) Hole Diameter =3 mm (assumed according to Coulson’s suggest, see Page 589) Plate thickness = 5 mm (assumed according to Coulson’s suggest, see Page 590)

Nature of Feed

K=Pvi

PT WHERE Pvi=Vapor Pressure Initial AND PT=Total Pressure

97.92 kmolIso-b=20.7%n-b = 79.3%

19.22 kmolIso-b=98.7%n-b = 1.3%

78.7 kmolIso-b=1.2%n-b = 98.8%

@ 66.36℃

Component Xf K values K Xfn-Butyraldehyde 0.797 0.98144 0.78220768

Iso-Butyraldehyde 0.203 1.07324 0.21786772Total 1.00 1.0000754

Nature of Distillate

@ 64.258℃

Component Xd K values K Xdn-Butyraldehyde 0.013 0.92093 0.0132762

Iso-Butyraldehyde 0.987 1.00112 0.98810544Total 1.00 1.00007753

Nature of Bottom

@ 73.4 ℃

Component Xb K values K Xbn-Butayraldehyde 0.988 0.990485 0.9885036Iso-Butyraldehyde 0.012 1.01383 0.012166

Total 1.00 1.0006696

Vapor Pressure Equation Perry’s CHE HB 8th ED (P in bar, T in Kelvin)

log P=A− BT+C

Component A B Cn-Butyraldehyde 3.59112 952.851 -82.569Iso-Butyraldehyde 3.87395 1060.141 -63.196

Vapor Pressures (P in bar)

Component Feed Top Bottomn-Butyraldehyde 0.8704559 0.70756942 0.95846472

iso-Butyraldehyde 1.236486 1.012942 1.356903

α values

α=(Vapor pressure)Lk(Vapor pressure )Hk

= iso−Butyraldehyden−Butyraldehyde Where α=relative volatility

Component Feed Top Bottomn-Butyraldehyde 1 1 1

iso-Butyraldehyde 1.420504 1.43158 1.4157047

α top=¿1.43158 α bottom=¿1.4157047

α avg=√αtopα bottom=√1.43158(1.4157047)=1.42362

Calculations for the number of plates

Using Fenske Method (Eq 11.58 Coulson & Richardson Chemical Engineering Vol 2, 5 th Ed)

n+1=ln [( Xa

Xb)d( XbXa

)w]

ln (αavg)

n+1=ln [( 0.987

0.013)d( 0.9880.012

)w]

ln(1.42362)

nmin=20.2769≈21 stages

Where a = iso-Butyraldehyde, b = n-Butyraldehyde

Calculating Minimum Reflux Ratio

Using Underwood’s Method (Eq 11.114 Coulson & Richardson Chemical Engineering Vol 2, 5th Ed)

αAXfAαA−θ

+ aBXfBaB−θ

+ aCXfCαC−θ

+…=1−q

α n−bX f n−bα n−b−θ

+α iso−b X f iso−b

α iso−b−θ=1−q

Assuming θ=1.308783237

1(0.797)1−1.308783237

+1.420504 (0.203)

1.42050−1.308783237=0

∑ axfa−θ

=4.97545x 10−10≅ 0 Accepted

Calculating Rmin

αAXdAαA−θ

+ aBXdBaB−θ

+ aCXdCαC−θ

+…=Rmin+1

α n−bX dn−bαn−b−θ

+α iso−b Xd iso−b

α iso−b−θ=Rmin+1

Rmin+1= 1(0.013)1−1.308783237

+ 1.43158 (0.987)1.43158−1.308783237

=10.46447

Rmax=1.5 Rmin=1.5 (15.2315 )=15.6967

Theoretical Number of Plates

Using Gilliland Correlation - (Eduljee version, Hydro. Proc., Sept. 1975, p. 120)

N−23.7464N+1

=0.75¿

N−31.36N+1

=0.75¿

N=31.7333≈32 plates

Plate Efficiency:

Equation 11.67 Richardson and Coulson’s Chemical Engineering Vol. 6, 4th Edition

Using O’Connell Correlations

Efficiency=(51−32.5 log (μaαa ) )=(51−32.5 log (0.2976804 x1.42362 ) )=63.1178%

Actual Number of Plates

N actual=N theo

Eo

N actual=32

0.631178=50.6988≅ 51 plates

Location of Feed Plate

Equation 11.62 Richardson and Coulson’s Chemical Engineering Vol. 6, 4th Edition

Using Kirkbride’s Equation:

logND

N B=0.206 log [( 78.7251 kmol

hr

19.2272 kmolhr

)( 0.7970.203 )(0.013

0.012 )2

]

N D

N B=1.8314 ; N D=1.8314N B

N D+NB=51

1.8314 N B+NB=51

NB=18.0123≅ 19 th Plate ¿ thebottom

PHYSICAL PROPERTIES:

Assumed: Pressure drop per plate = 100mm

iso-Butyraldehyde = 0.1 x 723 x 9.81 = 709.263 Pa

Column Pressure drop = 709.263 x 46 = 32,626.098 PaPressureBottom = 101325 + 32626.098 = 133,951.098 Pa

Top Bottom

Column top pressure= 101325 PaTemperature= 64.1 °C

𝜌𝑣 = 1.3 kgm3

𝜌𝑙 =793.8 kgm3 (density of the mixture)

Surface tension, σ = 29 x 10 -3 Nm

Liquid viscosity = 0.3095753 cP

MWave = 72.11kgm3

Column Bottom pressure= 133951.098 PaTemperature= 73.4 °C

𝜌𝑣 = 1.67 kgm3

𝜌𝑙 =817 kgm3 (density of the mixture)

MWave = 72.11kgm3

Top:

Ln = R x D =15.6967(19.2272)

= 301.8036 kmolhr

Vn = Ln + D = 301.8036 + 19.2272

= 321.031 kmolhr

Bottom

Lm = Ln + F = 301.8036 + 97.9529

= 399.7565 kmolhr

Vm = Lm – W = 424.4686 - 78.7251 =321.0314

kmolhr

Tray spacing

Tray spacing is selected to minimize entrainment. A large distance between the trays is needed in vacuum columns, where vapour velocities are high and excessive liquid carryover can drastically reduce the efficiencies. Trays are 0.15m to 1.0m. A trial tray spacing of 0.45m was selected.

Figure 11.27 Coulson & Richardson Chemical Engineering Vol. 6, 4th Ed

Estimating column diameter:

FLV Bottom = 8.00734486.4304373 √ 1.67

817= 0.056298

FLV Top = 6.0452946.430493 √ 1.3

793.8 = 0.0380443

Bottom K1 = 0.08

Top K1 = 0.08

Flooding velocity

Equation 11.81 Coulson & Richardson Chemical Engineering Vol. 6, 4th Ed

μf Bottom = 0.08√ 817−1.671.67

= 1.7677 ms

μf Top = 0.08√ 793.8−1.31.3

= 1.9752 ms

Design for 85% flooding at maximum flow rate

Superficial velocity (Qn):

Bottom, μv = 0.85 x 1.7677 ms = 1.5025

ms

Top, μv = 0.85 x 1.9752 ms = 1.679

ms

Maximum volumetric flow rate (Qv):

Bottom = Vn xMWρv x3600 =

321.031 x72.111.67 x3600

=¿3.85056 m3

s

Top = Vm xMWρv x3600 =

321.031 x72.111.3 x3600

=¿4.9465 m3

s

Net area required:

Bottom = QvQn

=3.850561.5025

=¿2.5626 m2

Top = QvQn

=4.94651.679

=¿2.9461 m2

First trial, downcomer area is taken as 12% of the total downcomer area

Column cross-section area

Bottom = 2.5626

0.88 = 2.91204 m2

Top = 2.94610.88 = 3.34785 m2

Downcomer Area

Bottom = (0.12 x 2.91204) = 0.349445m2

Top = (0.12 x 3.34785) = 0.40174m2

Column diameter

Bottom = 4π √2.91204 = 2.17275 m

Top = 4π √3.34785= 2.3297 m

Column Diameter should be use = 2.33 m

Liquid Flow pattern

Maximum Volume Flow Rate

Top = LnX MWρl X3600

=301.8036 x72.11793.8 x3600

=¿0.0076156 m3

s


Liquid Flow arrangement = f (Dc, Liquid Flow rate) = Cross flow Single pass

Provisional Plate Design

Column Diameter = 2.33 m

Column Area, Ac = 2.91204 m2

Downcomer Area, Ad = 0.40174 m2

Net Area, An = 2.9461 m2

Active Area, = Ac - 2 Ad = 2.91204 – 2(0.40174) = 2.10856 m2

Hole Area, Ah take 10% of Aa as first trial = 0.1 x 2.10856 = 0.2109 m2


From Graph; lwDc

=0.78

Weir Length, lw = 0.78(2.33) = 1.8174 m

Weir Height = 50 mm (assumed according to Coulson’s suggest, see Page 589)

Hole Diameter =5 mm (assumed according to Coulson’s suggest, see Page 589)

Plate thickness = 5 mm (assumed according to Coulson’s suggest, see Page 590)

Check weeping


Maximum Liquid Rate = Lm XMW

3600=399.7565 x72.11

3600=8.0073 kg

s

Minimum Liquid Rate, at 70% turn down = 0.7 x 8.0073 = 5.60514 kgs

maximum how = 750( 8.0073817(1.8174)

)23 = 23.0641 mm liquid

minimum how = 750( 5.60514817(1.8174)

)23 = 18.1832 mm liquid

at minimum rate = hw + how = 50 + 18.1832 = 68.1832 mm


From Graph: K2 = f(hw + how) = 30.75


μh=¿¿

Actual minimum vapour velocity = minimumvapor rate

Ah=0.7 x 3.85056

0.2109=12.7804 m

s

Actual minimum vapour velocity > minimum vapour velocity, No weeping!

Plate pressure drop

Dry plate drop

Maximum vapour velocity through holesμh, max =3.850560.2109

=18.258ms

Plate ThicknessHole Diameter = 1;

Ah

Aa=0.1


From Graph: Co = 0.84


hd=51[ 18.2580.84

]2 1.3

817=38.339mmliquid


hr=12.5x 103

817=15.3mmliquid


ht=hd+(hw+how )+hr=38.339+(50+18.1832 )+15.3=121.822mmliquid

100mm≈121.822;Considerably Acceptable !

Downcomer Liquid Back-up


Downcomer pressure loss

Take hap=hw−10=50−10=40mmliquid

Area under apron, Aap=lw x hap=1.8174 x 0.04=0.0727m2

Since: Ad = 0.40174 m2 > Aap=0.0727m2; So we choose Aap


hdc=166 [ 8.0073448817 (0.0727 )

]2

=2.01698mm


hb=(h¿¿w+how)+ht+hdc=(50+18.1832 )+121.822+2.01698=192.0222mm¿

192.0222mm<12

(plate spacing+weir height )

192.0222mm<1133.7mm; so0.45m plate spacingis acceptable !

Check Residence time


t r=0.40174 (0.1920222)(817)

8.0073448=6.95 sec

7.87 seconds>3 seconds; acceptable !

Check Entrainment

Equation 11.83; Figure 11.29 Coulson & Richardson Chemical Engineering Vol. 6, 4th Ed

μn=Q v

An= 3.85056

2.9461=1.307 m

s

%Flooding=μnμ f

= 1.3071.5025

x100 %=86.99%=87 %

@FlvBottom=0.05894

From the Graph: φ= f (FlvB) = 0.07

0.07>0.1; wellbelow 0.1 ,Process are satisfactory

Trial Layout

Allowance:

50 mm unperforated strip round plate edge

50 mm wide calming zones


2.33

m1.

82

m

lwDc

=0.78;¿ theGraph ,θ ¿c=99 °

Angle subtended by the edge of the plate = (180 – 99) = 81°

Mean length, unperforated edge strips = (2.09 – 0.05)π ( 81180 )=2.884m

Area of unperforated edge strips = 0.05 x2.884=0.1442m2

Mean length of calming zone, approximately = lw+width of unperforated

¿1.8174+0.05=1.8674m

Area of calming zones = 2 (1.8674 x0.05 )=0.187m2

Total area for perforateions, Ap=2.10856−0.1442−0.187=1.77736m2


Ah

A p= 0.2109

1.77736=0.11866

From the Graph, l pDh

=2.85; satisfactory ,within2.5¿4.0

l pDh

=2.85; l p=2.85 (5mm )=14.55mm, Hole pitch

Area of Single Hole:

α h=π D2

4=π (0.005)2

4=0.000019635m2

Total no. of holes: No. of holes/plate

nT=Ah

α h =

0.21090.000019635 = 10,741.05 ≈10,742holes =

10,74251 = 211 holes

MECHANICAL DESIGN

Design Pressure (Mpa) 0.101325Allowable Stress (Tube Sheet Material) (Mpa) 4.75Nozzle ID, Shell (m) 0.254Joint Effienciency 0.8Corrosion Allowance (m) 0.002

Shell Thickness Calculation

Equation to be found at Chapter 2.3.1 NPTEL – Chemical Engineering – Chemical Engineering Design - II

Ts= pDsfj−0.6 p

+c

=0.101325Mpa(2.33m)

4.75Mpa (0.8 )−0.6(0.101325Mpa)+0.002m=0.06514m=65.14mm

Torrispherical head Calculation

Th= pRiW(2 fj−0.2 p)

+c

For Torrispherical head Ri =Ds


W=14(3+√Riri )

For Torrispherical Head Riri

= 10.06

W=14(3+√ 1

0.06)=1.77

Th= 0.101325Mpa (1.165m)(1.77)(2 ( 4.75Mpa ) (0.85 )−0.2 (0.101325Mpa ))

+0.002m=0.02794m=27.94mm

Channel Cover Diameter and thickness

Channel cover diameter = Shell Outside diameter = 2.39514 m


Tcc=Dc√C1 p10 f

Tcc=2.39514 √0.25(0.101325Mpa)10(4.75Mpa)

=0.00803m=8.03mm

So :

ColumnHeight=(N actual−1)(Plate spacing)+Mechanical Design

¿ (51−1 ) (0.45 )+(0.03597 )=22.536m

Nozzles (Number of Nozzles = 6)

1. Feed Inlet

NozzleDiameter=0.1D=0.1 (2.33m )=233mm≅ 9.17 =254 m

t n=p Dn

2 fj−p+c

f=4.75Mpa

p=0.101325Mpa

c=2mm

j=0.85

t n=0.101325 N

mm2 (254mm )

2(4.75 Nmm2 ) (0.85 )−0.101325 N

mm2

+2mm

t n=6.23mm≅ 0.25 =6.35 m

2. Reflux

Nozzle diameter = 254 mm≅ 10=254 m

t = 6.35 mm

3. Overhead Vapor Oulet gas out

Nozzle diameter = 254 mm≅ 10=254 m

t = 6.35 mm

4. Bottom Out

Nozzle diameter = 152.4 mm≅ 6”

f=4.75Mpa

p=0.133951Mpa

c=2mm

j=0.85

t n=p Dn

2 fj−p+c

t n=0.133951 N

mm2 (152.4mm )

2(4.75 Nmm2 ) (0.85 )−0.133951 N

mm2

+2mm

t n=4.571mm≅ 0.18

5. Reboiler Return

Nozzle diameter = 152.4 mm≅ 6”

t = 4.57 mm

COSTING (Distillation Column 2)

Column Cost Estimation including installation and auxiliaries

(Fig14-84 Perry’s Chemical Engineer’s HB 8th Ed)

Dc=2.39514m=94.297 inch

ColumnHeight=22.536m=73.94 ft

Cost ($ ) per foot height=4000 $

Cost ($ )=4000$ x 73.94 ft=295,748$=Php13,308,661.42

Using Equipment Index Factors; Marshall & Swift/Boeckh, LLC (Marshall & Swift)

Year Average2012 424.11979 222.301

295,748$ (1979 )[ 424.1 (2013 )222.301 (1979 ) ]=564,220.2545$=Php25,389,911.45

Distillation Column 2 Final Revised

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