MASS TRANSFER OPERATION II

CH 4152

DESIGN OF A BERL SADDLES PACKED DISTILLATION COLUMN

NAME : N.G.A.K.K.N. NISSANKA

INDEX NO : 090354 T

DATE OF SUB : 2013-08-01

ContentsINTRODUCTION....................................................................................................................................2

Selection of suitable packing material and size........................................................................................3

Material selection..................................................................................................................................3

Size selection.........................................................................................................................................3

Pressure drop across packing....................................................................................................................4

HETP value...............................................................................................................................................4

Number of ideal stages required...............................................................................................................5

Construction of Q line...........................................................................................................................6

Construct top operating line (TOL) for minimum reflux ratio..............................................................6

Construct top operating line (TOL) for actual reflux ratio....................................................................7

The column height....................................................................................................................................9

Feed tray location......................................................................................................................................9

Diameter of the column.............................................................................................................................9

Flow rates and compositions calculation.............................................................................................10

Diameter for rectifying section............................................................................................................11

Diameter for stripping section.............................................................................................................15

Heat load of condenser............................................................................................................................19

Heat load of re-boiler..............................................................................................................................20

Summary.................................................................................................................................................21

References...............................................................................................................................................22

1

INTRODUCTIONGenerally packed columns are used for distillation, gas absorption and liquid-liquid extraction.

The gas-liquid contact in a packed column is continuous, not stage-wise, as in a plate column. The

liquid flows down in the column over a packing surface and the vapor (or the gas) moves counter-

currently, up the column. The performance of a packed column is very dependent on the maintenance

of good liquid and gas distribution through the packed bed, and this is an important consideration in

packed column design.

I can arrange given data for distillation column as bellow

System : chloroform-benzene

Operating pressure : 1 bar

Feed rate : 48 kmol/hr

Feed condition : 50% sat. liquid and 50% sat. vapour

Feed composition : 45 mol% chloroform

Distillate composition : 85 mol% chloroform

Bottom product composition : 20 mol% chloroform

Column type : packed column

Packing type : berl saddles

As well as we design distillation column with a total condenser and a partial re boiler

I need to find bellow data and calculations for this design

Suitable packing material and packing size

Number of ideal stage required

Column height and diameter

Feed tray location condenser and re boiler heat loads

2

Selection of suitable packing material and size

Material selection There two packing method for packed distillation which random and structured packing. According to

the given data we need to use berl saddles type packing material. Berl saddles type is used in random

packing method. So we have to use random packing method for our distillation process. In berl saddles

type has different type raw material such as ceramic, metal, plastic and carbon. Lot of plastic type

reacts with chloroform without polythene and polypropylene. As well as some pressure and

temperature condition we can ignore plastic type. When we consider the ceramic and plastic both are

good for our distillation column. As well as weight of column increase due to ceramic weight higher

than metal. So metal is good for packing material. But consider the availability in market most berl

saddles type are ceramic. Therefore we use ceramic berl saddle for our packing material.

Material : ceramic berl saddle

Size selectionAccording to market data we can categorize sizes and there data as bellow table

When flow rate increase we need to increase packing size. As well as when size increase efficiency decrease. Our flow rate is got low value. So we can use smaller size packing such as (19 - 25) mm. But low sizes get high weight. So I select packing size as a 25 mm.

Size : 25 mm

3

Packing factorFp

110985240

Pressure drop across packing

We need to assume pressure drop across packing using above table. It can be assumed as a 0.75 inch water /ft packing. That means which is equal to the 62.5 mm water /m packing.

HETP value

According to our packing size we can get HETP value from below table as 0.46 m.

4

Number of ideal stages requiredChloroform-benzene vapour-liquid equilibrium data at 1 bar for 24 data points

temperature(K)mole fraction of chloroform in

liquid phase (X)

mole fraction of chloroform in

vapour phase (Y)353.26 0.0000 0.0000352.67 0.0269 0.0437352.36 0.0417 0.0692349.80 0.1855 0.2581350.87 0.1332 0.1849348.85 0.2372 0.3316347.87 0.2948 0.3980347.04 0.3506 0.4749346.51 0.3921 0.5245345.58 0.4427 0.5714344.57 0.4897 0.6257343.78 0.5347 0.6742342.77 0.5781 0.7164341.95 0.6209 0.7553341.07 0.6654 0.7936340.02 0.7181 0.8257338.87 0.7579 0.8595338.13 0.7954 0.8851337.49 0.8278 0.9057336.86 0.8603 0.9262336.31 0.8892 0.9427335.53 0.9308 0.9648335.07 0.9571 0.9792334.28 1.0000 1.0000

Using above data we can draw Chloroform-benzene vapour-liquid equilibrium curve like below graph.

5

Construction of Q line

q factor=Energy convert1mol of feed ¿ saturated vapor ¿Molar heat of vaporization

q= x2x

q = 0.5

Q line equation

Y=( qq−1 )X− 1

q−1X f

X : mole fraction of chloroform in liquid phase

Y : mole fraction of chloroform in vapour phase

Xf : feed composition of chloroform

Y=( 0.50.5−1 )X− 1

0.5−1×0.45

Y = -1 X + 0.90

Construct top operating line (TOL) for minimum reflux ratioAccording to given data value of distillate composition is 0.85. So we can construct line go

through where (0.85, 0.85) and (Q line and equilibrium curve cutting point).

6

Calculate minimum reflux ratio (Rm) Using TOL

Rm

Rm+1=0.85−0.24

0.85=0.7176

Rm = 2.5417

Calculate R actual using Rm value

R actual =Rm × 1.2= 2.5417×1.2 = 3.05 (1.2 is constant vale we can use it between 1.2-1.5)

Construct top operating line (TOL) for actual reflux ratio

Y n+1=( RR+1 )X n+

1R+1

Xd

Xd : distillate composition

Y n+1=( 3.053.05+1 )X n+

13.05+1

×0.85

Y n+1 = 0.7531× Xn + 0.21

Construct Bottom operating line (BOL) using given data

According to given data value of bottom product composition is 0.20. So we can construct line

go through where (0.20, 0.20) and (Q line and TOL curve cutting point).

Q line and TOL curve cutting point = (0.394, 0.501)

Equation for Bottom operating line

y−0.2x−0.2

=0.501−0.20.394−0.2

Y=1.55× X−0.11

Y=( L'

L'−W )X− WL'−W

XW

L' : Downward liquid flow rate in stripping section

W : Bottom product (kmol/hr)

Xw : Bottom product composition of chloroform

Total stages = 14 but it has partial re boiler

7

Number of ideal stages = 13

8

The column height columnheight=number of stages×HETP

columnheight=13×0.46=6m

Feed tray location According to stages counting figure feed location should be between six and seven stages from the top

and six and seven stages from the bottom.

Therefore feed location should be middle of the column

¿ 12×5.98=3m¿ the top∨bottom of the column

Diameter of the column

F : Feed (kmol/hr)

D : Distillate (kmol/hr)

W : Bottom product (kmol/hr)

XF : Feed composition of chloroform

XD : Distillate product composition of chloroform

9

XD

XF

XW

G' L'

XW : bottom product composition of chloroform

G : Upward gas flow rate in rectifying section

G' : Upward gas flow rate in stripping section

L : Downward liquid flow rate in rectifying section

L' : Downward liquid flow rate in stripping section

Flow rates and compositions calculationF : 48kmol/hr

XF : 0.45

XD : 0.85

XW : 0.20

feedingrate of chloroformFCCl4=X F×F=0.45×48=21.6kmol /hr

Mass balance for the system

F = D + W

48 = D + W

Mass balance for the chloroform

F XF = D XD + W XW

21.6 = 0.85 D + 0.2 W

D =18.46 kmol/hr

W = 29.53 kmol/hr

LD

=R

L=R×D=18.46×3.05=56.30kmol /hr

G = L + D = 56.3 + 18.46 = 74.76 kmol/hr

( L'

L'−W )=( L'

L'−29.53 )=1.55bybottom operating lineequation

L' = 83.22 kmol/hr

G' = L' - W = 83.22 - 29.53 = 53.69 kmol/hr

In Rectifying section and Stripping section, from stage to stage composition of G and L are varied, so

it is not possible to get Constant Mass flow rate of vapor and Constant Mass flow rate of liquid. So in

this calculation, it is better to take an average value of the Mass flow rate across the Rectifying Section

and stripping section.

Molecular weight of chloroform = 1395.4 g/mol

Molecular weight of benzene = 827.49 g/mol

10

Diameter for rectifying section I consider one stage above the feed tray and top stage for get average values in rectifying section.

11

According to above graph data we can get below data

DATA Value temperature

F 48 kmol/hr

D 18.46 kmol/hr

G 56.30 kmol/hr

L 74.76 kmol/hr

At one stage above the feed tray

Chloroform Composition in liquid 0.42 346 K

Chloroform Composition in vapour 0.52 346.5 K

At top stage

Chloroform Composition in liquid 0.85 337 K

Chloroform Composition in vapour 0.85 339 K

12

Average temperature for liquid phase ¿

346+3372

+ 346.5+339.52

2=342 K

Average Chloroform Composition in liquid = 0.62

Average Chloroform Composition in vapour = 0.76

Density of the liquid at 342 K

Density of chloroform at temp = 1395.4 Kg/m3

Density of benzene at temp = 827.49 Kg/m3

Mass fraction of chloroform in liquid (w/w) =0.62×119.38

(0.38×78.11+0.62×119.38)=0.7138

Mass fraction of benzene in liquid (w/w) =1-0.7138 = 0.2862

Volume of the chloroform for 1000 kg = 713.8

1395.4=0.5115m3

Volume of benzene for 1000 kg =286.6

827.49=0.3463m3

Molar mass of mixture (M) =(0.38×78.11+0.62×119.38 )=103.7 g/mol

Density of the mixture (ρL) = 1000

0.5115+0.3463=1165.77Kg /m3

Liquid Flow rate (L¿¿W ¿)¿ (L×M) = 74.76×103.7 = 7752.61 Kg/hr

Density of the vapor at 342 K

Saturated vapor pressure of chloroform PC0 = 129.27 KPa

Saturated vapor pressure of benzene PB0 = 70.7 KPa

Average Molar fraction of chloroform XC = 0.76

Average molar fraction of benzene X B = 0.24

Average molecular weight of vapor mixture =(0.24×78.11+0.76×119.38)=109.48g /mol

Vapor flow rate (V ¿¿W ¿)¿ (G×M) = 56.30×109.48= 6163.72 Kg/hr

Raoult’s Law for vapour phase13

PT=PC0 × XC+PB

0 × XB

PT=129.27×0.76+70.7×0.24=115.21KPa

Applying Gas Law for find density

PV=nRT=mRTM

Where

P = absolute pressure (Pa)

V = volume m3

n = mols of gas

T = absolute temperature K

R = universal gas constant 8.314 J/(mol*K)

m = mass

M= molecular weight

ρ = density

Assumption: all gasses behavior is ideal

ρV=PMRT

=115.21×109.488.314×342

=4.44 Kg /m3

Viscosity of the mixture at 342 K

Viscosity of chloroform = 3.7×10−4 Ns/m2

Viscosity of benzene = 3.6×10−4 Ns/m2

Find K4

value

finding

FLV value

and

using

graph

14

According to selected pressure drop value of 62.5 K4 can be found = 1.8

At flooding line K4 is equal to the = 4

So percentage of flooding = √ 1.66

= 63.25 % satisfactory

Packing factor (Fp) = 98

V ¿=( K4 ρV (ρL−ρV )

13.1×F p( μL

ρL)

0.1 )12=( 1.8×4.44 (1165.77−4.44 )

13.1×98×( 3.66×10−4

1165.77 )0.1 )

12=5.68 Kgm−2 s−1

=

columnarea=mass flow rate of vapourV ¿ = 6163.72

3600×5.68=0.3014

columndiameter=d=√ 4×0.3218π

=0.62m

Diameter for stripping section I consider one stage bellow the feed tray and bottom stage for get average values in rectifying section.

15

According to above graph data we can get below data

DATA Value temperature

16

F 48 kmol/hr

W 29.53 kmol/hr

G' 53.69 kmol/hr

L' 83.22 kmol/hr

At one stage bellow the feed tray

Chloroform Composition in liquid 0.39 347 K

Chloroform Composition in vapour 0.50 347.5 K

At bottom stage

Chloroform Composition in liquid 0.20 350.5 K

Chloroform Composition in vapour 0.20 349.5 K

Average temperature for liquid phase ¿

347+350.52

+ 347.5+349.52

2=348.6 K

Average Chloroform Composition in liquid = 0.27

Average Chloroform Composition in vapour = 0.37

Density of the liquid at 348.6 K

Density of chloroform at temp = 1382.04 Kg/m3

Density of benzene at temp = 820.22 Kg/m3

Mass fraction of chloroform in liquid (w/w) =0.27×119.38

(0.73×78.11+0.27×119.38 )=0.3611

Mass fraction of benzene in liquid (w/w) =1-0.3611 = 0.6389

Volume of the chloroform for 1000 kg = 361.1

1382.04=0.2613m3

Volume of benzene for 1000 kg =638.9

820.22=0.7789m3

Molar mass of mixture (M) =(0.73×78.11+0.27×119.38 )=89.25 g /mol

Density of the mixture (ρL) = 1000

0.2613+0.7789=961.35Kg /m3

Liquid Flow rate (L¿¿W ¿)¿ (L'×M) = 83.22×89.25 = 7427.38 Kg/hr

17

Density of the vapor at 348.6 K

Saturated vapor pressure of chloroform PC0 = 157.76 KPa

Saturated vapor pressure of benzene PB0 = 87.62 KPa

Average Molar fraction of chloroform XC = 0.37

Average molar fraction of benzene X B = 0.63

Average molecular weight of vapor mixture =(0.63×78.11+0.37×119.38 )=93.38 g /mol

Vapor flow rate (V ¿¿W ¿)¿ (G'×M) = 53.69×93.38= 5013.57 Kg/hr

Raoult’s Law for vapour phase

PT=PC0 × XC+PB

0 × XB

PT=157.76×0.37+87.62×0.63=113.57KPa

Applying Gas Law for find density

PV=nRT=mRTM

ρV=PMRT

=113.57×93.388.314×348.6

=3.66 Kg /m3

Viscosity of the mixture at 348.6 K

Viscosity of chloroform = 3.57×10−4 Ns/m2

Viscosity of benzene = 3.32×10−4Ns /m2

μmix=(∑ X iμi1 /3 )3

=(0.27×(3.57×10−4)13 +0.73×(3.32×10−4)

13 )

3

=3.39×10−4 Ns/m2

Find K4 value finding FLV value and using graph

FLV=LW

¿

VW¿ √ ρV

ρL

=7427.385013.57 √ 3.66

961.35=0.091

18

According to selected pressure drop value of 62.5 K4 can be found = 1.4

At flooding line K4 is equal to the = 3.2

So percentage of flooding = √ 1.43.2

= 66.14 % satisfactory

Packing factor (Fp) = 98

V ¿=( K4 ρV (ρL−ρV )

13.1×F p( μL

ρL)

0.1 )12=( 1.4×3.66 (961.35−3.66 )

13.1×98×( 3.39×10−4

961.35 )0.1 )

12=4.12 Kgm−2 s−1

=

columnarea=mass flow rate of vapourV ¿ = 5013.57

3600×4.12=0.3380

columndiameter=d=√ 4×0.3380π

=0.656m

19

Heat load of condenser

data chloroform benzene

XD 0.85 0.15

Boiling point Tb 335 k 353.1 K

Critical temperature TC 537 K 562 K

Operating temperature T 339 K 339 K

Latent heat @ boiling point Lv,b 29486.86 J/mol 30462.9 J/mol

Gas flow rate G 56.30 kmol/hr

Latent heat at temperature T

Lv=Lv ,b( T C−TTC−Tb

)0.38

For chloroform

Lv ,C=29486.86( 537−339537−335 )

0.38

=29263.6J /mol

For benzene

Lv , B=30462.9 ( 562−339562−353.1 )

0.38

=31228.45 J /mol

Latent heat of mixture L = Lv ,C× XD ,C+Lv , B×X D, B=29263.6×0.85+31228.45×0.15

29558.33J /mol

Condenser heat load QC¿G×L=29558.33×56.30×10003600

=462.26kW

20

Heat load of re-boiler

Data chloroform benzene

XD 0.85 0.15

Boiling point Tb 335 k 353.1 K

Critical temperature TC 537 K 562 K

Operating temperature T 350.5 K 350.5 K

Latent heat @ boiling point Lv,b 29486.86 J/mol 30462.9 J/mol

Gas flow rate G' 53.69 kmol/hr

Latent heat at temperature T

Lv=Lv ,b( T C−TTC−Tb

)0.38

For chloroform

Lv ,C=29486.86( 537−350.5537−335 )

0.38

=28605.73J /mol

For benzene

Lv , B=30462.9 (562−350.5562−353.1 )

0.38

=30606.42J /mol

Latent heat of mixture L = Lv ,C× XD ,C+Lv , B×X D, B=28605.73×0.2+30606.42×0.8

30206.28J /mol

Condenser heat load QC¿G×L=30206.28×53.69×10003600

=450.49kW

21

Summary

According to all calculations of this report we can summaries all important data in below table

data Value or details

System Chloroform-Benzene

Column type Packed bed

Packing type Berl saddle

Packing material Ceramic

Packing size 25 mm

Pressure drop across packing 62.5 mm water /m

HETP value 0.46 m

Reflux ratio 3.05

Number of ideal stages 13

F feed 48 kmol/hr

D distillate 18.46 kmol/hr

W bottom product 29.53 kmol/hr

G gas flow in rectifying system 56.30 kmol/hr

L liquid flow in rectifying system 74.76 kmol/hr

G' gas flow in stripping system 83.22 kmol/hr

L' liquid flow in stripping system 53.69 kmol/hr

Column height 6 m

Feed tray location Middle point (3 m from bottom)

Diameter of rectifying system 0.62 m

Diameter of stripping system 0.656 m

Heat load of condenser 462.26 kW

Heat load of re-boiler 450.49 kW

22

References

http://ddbonline.ddbst.de/DIPPR105DensityCalculation/DIPPR105CalculationCGI.exe

http://ddbonline.ddbst.de/VogelCalculation/VogelCalculationCGI.exe

http://ddbonline.ddbst.de/AntoineCalculation/AntoineCalculationCGI.exe

http://www.engineeringtoolbox.com/fluids-evaporation-latent-heat-d_147.html

http://www.cheric.org/kdb/kdb/hcvle/showvle.php?vleid=3565 from (http://vle-calc.com/index.html)

Colson & Richadson chemical engineering volume 6

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