Design a Distillation Column With a Total Condenser and Partial Re Boiler for the Following...

Inda kola weri

Design a distillation column with a total condenser and partial re boiler for the following separation .

DATAsystem ethanol-water

Feed rate -150 K mole /hr

Feed composition 28mole % ethanol

98% of ethanol recovery is required??????????????????

Operating pressure: 1 bar

Feed composition: 50% sat liquid and 50% sat vapor

Distillate composition: 82% mole ethanol

Column type: packed column

Packing type: raschig rings

Select a suitable packing material ,packing size ,pressure drop across packing or percentage of flooding and a reflux ratio .obtain an appropriate HETPvalue for HTU from the literature and determine the number of ideal stages required ,the column height and the diameter of the column ,feed tray location and the condenser and reboiler heat loads .

Detail calculation

Graphical constructions data used for the calculation ,

Assumptions made

Reference

1. Selection of packing material and packing size

Material Heart of the distillation column is begin with the selection of packing material ..In this distillation

column design Raschig types rings are used as packing material .they are usually made up by ceramic or

metal or carbon.

Ceramic packing are resistant to corrosion and comparatively cheap but which are heavy and may require

a stronger packing support and foundations.so ceramic packaging is rejected from the design.Metal,

carbon steel packings are usually the first choice for noncorrosive services. Here in this problem water-

ethanol mixture is separated using packing distillation column which mixture is comparatively low

corrosive .so in this case carbon steel packing is selected which offer higher efficiency , a wider range

geometries , higher turndown are unbreakable , and have a higher compression resistance .

SIZE The size of packing used influences the height and diameter of a column, the pressuredrop and cost of

packing. When packing size is increased, the cost per unitvolume of packing and the pressure drop per

unit height of packing are reduced, but in other side which will reduce the mass transfer efficiency.

Reduced mass transfer efficiency results in a tallercolumn .So for this design 38 mm Size metal Raschig

type ring is selected as a packing material and column diameter is expected to maintain around 0.3 to 0.9

m.

Here the column

Column diameter Use packing size

<0.3 m (1 ft) <25 mm (1 in.)

0.3 to 0.9 m (1 to 3 ft) 25 to 38 mm (1 to 1.5 in.)

>0.9 m 50 to 75 mm (2 to 3 in.)

Figure 1-1: Raschig ring

Table 1: Raschig Ring data table

Packing method = random type packing

Pressure Drop Pressure drop can be assumed from the [1][3] . 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

HETP value for a packing size of 38 mm for full scale plant can be assumed as a 0.66 m .

Calculation

F,XF

D,XD

W,Xw

Figure 0-2 : Volume 6 page no 640

F - Feed (Kmol/hr)

D – Distillate (kmol/hr)

W – Bottom product (kmol/hr)

X - mole fraction of heptane

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

F=Feed rate

X=mole composition of ethanol

D= Top product rate

B=Bottom product rate

F=150 kmol/hr XF=0.28 XD = 0.82

E

th

a

n

ol

entering rate as feed = 0.28 X 150 Kmol/hr

=42 Kmol/hr

If 98% of ethanol is recovered from the top product

Ethanol mole leaving rate from the distillate = 42 X 0.98 = 41.16 Kmol/hr

Mole composition in the feed= D*XD =41.16

o D=41.16/0.82 = 50.195 Kmol /hr

Mass balance for the system

Figure 0-4 : Distillation Column Diagram

F=D+B

150=50.195+ B

B=99.805 Kmol/hr

Mass balance for the ethanol

F.XF =D.XD+ B.XB

XB=(42-41.16)/99.805 =0.0084

Water-ethanol equilibrium data can be founded as follow .

Source: :physical property table, mcgraw-hill [1]

By using the ethanol water liquid equilibrium data VLE diagram is drawn .

Table 2: Ethanol-water liquid equilibrium at 1 Bar

Figure 0-5:VLE of water ethanol

Assumptions

The lines are made straight by the assumption of constant molar overflow Liquid rate is constant and the vapor rate constant. from tray to tray in each section of the column

between addition (feed) and withdrawal (product) points.

Construction of Q line

q factor=Energy convert 1mol of feed ¿ saturated vapor ¿Molar heat of vaporization

q=0.5/1 q=0.5

equation of Q line

Y= qq−1

− 1q−1

Xf

Y = - X – 2Xf = - X+ 2* 0.28 = -X+0.56

After constructing Q line Minimum reflux ratio is calculated

by connecting the point (Xd ,Xd) and point of intersection between the q-line and the equilibrium

curve. R of this curve is known as the minimum reflux ratio .

R/ (R+1)=0.82−0.410.82−0

Rmin = 1

Ractual = 1.5 * R min [2] = 1.5

Figure 0-6 : VLE with Minimum Reflux ratio

Using the value R min Top operating line and Bottom operating line of the Distillation curve is derived .

TOL

Yn+1=( RR+1 )Xn+ 1

R+1Xd

Yn+1= 0.6 .Xn + 0.4* 0.82 = 0.6 Xn + 0.328

By combing the points (Xb,Xb) and intersection point of the q line and TOL BOL is drawn . ( fig3 )

Equation of the BOL

BOL is passed through the points (0.145,0415) and (0.0084,0.0084)

So equation of the line

Y=2.97657x - 0.0166

Equation of bottom line (general formula)

Y=( L'

L'−B )X− BL'−B

X b

L'

L'−B=2.97657, here B = 99.805 Kmol/hr L’=150.29 Kmol/hr

BOL

BOL

Number of ideal stage = 33-1 =32stages [from the stair case construction of the graph(in

annex 1) ]

Location of the feed tray = 30 th tray ( ?????? Wite the figure

no )

From the calculated value R =1.5 following can be found

Molar flow rate (Kmol/hr)

F Feed (Kmol/hr) 150

D – Distillate (kmol/hr) 50.195

B– Bottom product (kmol/hr) 99.805

G – Upward gas flow rate in rectifying section 126.3875

G’ – Upward gas flow rate in stripping section 50.485

L – Downward liquid flow rate in rectifying section 76.1925

L’ – Downward liquid flow rate in stripping section 150.29

Here G’=L’-B and G = L+B

F,XF

G

W,Xw

L

D,XD

G’ L’

Column height

=number of stages * HETP = 32* 0.66 m =21.12 m

Height of the stripping section =

Height of the Rectifying section =29*0.66 = 19.14 m

Calculation of Column Diameter

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 .

For rectifying section

Molar flow rate (Kmol/hr)

F Feed (Kmol/hr) 150

D – Distillate (kmol/hr) 50.195

G – Upward gas flow rate in rectifying section

126.3875

L – Downward liquid flow rate in rectifying section

76.1925

One stage above the feed tray :

Composition of the ethanol in liquid = 0.24 { Temperature at the 82.5 oC }

Composition of the ethanol in the vapor = 0.47

At the top stage

Composition of the Ethanol in liquid = 0.82

Composition of the ethanol in vapor = 0.82

Temperature at the top = 78.5oC

Average temperature of the stripping section = (78.5+82.5)/2 =80.5oC

Molar flow rate (Kmol/hr)

F Feed (Kmol/hr) 150

D – Distillate (kmol/hr) 50.195

G – Upward gas flow rate in rectifying section

126.3875

L – Downward liquid flow rate in rectifying section

76.1925

Average Molar fraction in the liquid = 0.45

Average Molar fraction in the vapor = 0.64

Density of the liquid at 80.5 o C .

Density of ethanol at temp @ 80.5 = 725.238 Kg/m3

Density of water at temp 80.5 = 961.778 Kg/m3

Mass fraction of ethanol in liquid (w/w) =0.45* 46/(0.55*18+0.45*46)=0.676

Mass of water in liquid (w/w) =1-0.676 =0.324

Volume of the ethanol = 676 /725.238 m3 =0.932

Volume of water = 324/961.778 m3 = 0.3368

Molar mass of mixture = 0.55*18 +0.45*46.05 = 30.625 g/mol

Density of the mixture = (1000)/(0.932+0.3368)*103=788.1Kgm-3

Liquid Flow rate (L*) =76.1925*30.625 = 2333.39 Kg/hr

Density of the vapor @ 80.5 o C

Saturated vapor pressure of ethanol Peo = 829.434 mmHg =110.5 KPa

Saturated vapor pressure of water Pwo = 361.785 mmHg =48.2 KPa

Average Molar fraction of ethanol in the vapor = 0.64

Average molar fraction of the Water in the vapor = 0.36

Average molecular weight of the Water and ethanol in the vapor mixture =

(0.64* molecular mass of ethanol* 0.36 molecular mass of water )

=0.64* 46.04+ 18.01*0.36 =35.9492 Kg/Kmol

Vapor flow rate = G’*35.9492 Kg/hr = 4543 Kg/hr

By applying Raoult’s Law

PT= XePeo+XwPw

o = 0.64* 110.5 + 48.2* 0.36 = 88.07 KPa

Density of the vapor

By applying Gas Law

Assumption :Compressibility factor of the Water-ethanol vapor is approximately equal to 1 .

PV=nRT

where P = absolute pressure N/m2 (Pa),V = volume m3,n = mols of gasT = absolute temperature, K,R = universal gas constant, 8,314 J K"1 mol"1 (or kJ K"1 kmol"1).

Density =PM/RT = ( 88070*35.94) /(8.314 * 353.5) = 1.074 Kg/m3

Viscosity of the mixture

μmix=¿3

μmix =X1 μ11/3+X2 μ2

1/3

=={(0.45*0.43001/3 +0.55*0.356351/3 }1/3

=0.3888 mPa

Lw* = average molar mass * molar flow rate = 2333.39 Kg/hr

V*w=average molar mass * molar flow rate = 4543 Kg/hr

76.1925126.38

√1.074 /¿ ¿788.14 =0.02

For pressure drop 42 , K4 value is equal to 1.9

For pressure Drop 83 , K4 value is equal to the 3.3

For pressure drop 62.5 K4 can be found to be a = 2.6

At flooding K4 is equal to the = 6

So percentage of flooding = √ 2.66

=65.82 % satisfactory

Fp= 270 m3/m2

=

V* =[ 2.6∗1.074∗(788.14−1.074 )13.1∗270∗¿¿

]1/2

V*=1.61Kg/m2s

Column Area required = (mass flow rate of vapor / V*)

=(4543/3600)/1.61) =0.7838m2

= π*d2/4 =0.7838

d=0.99 m

Diameter of the rectifying section =0.99m ~1 m

Packing size to column size ratio = (1/0.038 ) =26.31

Diameter of the Stripping section .

For Stripping section

Molar flow rate (Kmol/hr)

F Feed (Kmol/hr) 150

B– Bottom product (kmol/hr) 99.805

G’ – Upward gas flow rate in stripping section

50.485

L’ – Downward liquid flow rate in stripping section

150.29

At the feed tray :

Composition of the ethanol in liquid = 0.12 { Temperature at the 358. K =85.5 oC }

Composition of the ethanol in the vapor = 0.47

At the bottom stage

Composition of the Ethanol in liquid = 0.02

Composition of the ethanol in vapor = 0.15

Temperature at the top = 96.5 oC

Average temperature of the stripping section = (96.5+85.5 )/2 =91oC

Average Molar fraction of ethanol in the liquid = 0.05

Average Molar fraction of ethanol in the vapor = 0.32

Average mass fraction of ethanol liquid =0.05*46/(0.05*46+0.95*18) =0.118

Density of the liquid at 91 o C .

Density of ethanol at temp @ 91 = 713 Kg/m3

Density of water at temp @91 = 954 Kg/m3

Volume of the ethanol = 118 /713 m3 =0.165

Volume of water = 881/954 Kg/m3 = 0.963

Density of the liquid mixture = 1000/(0.165+0.963) =886.52 Kg/m3

Average molecular weight of the liquid = 19.5 Kg/Kmol

Average mass flow rate of liquid = 19.5 * 150.29 =2930 Kg/hr

Density of the vapor @ 91 o C

Saturated vapor pressure of ethanol Peo = 164 Kpa

Saturated vapor pressure of water Pwo = 361.785 mmHg =72.74 KPa

Average Molar fraction of ethanol in the vapor = 0.32

Average molar fraction of the Water in the vapor = 0.68

Average molecular weight of the Water and ethanol in the vapor mixture =

(0.32* molecular mass of ethanol* 0.68 molecular mass of water )

=0.32* 46.04+ 18.01*0.68 =26.9796 Kg/Kmol

Vapor flow rate = G’*26.9796 Kg/hr = 1362 Kg/hr

By applying Raoult’s Law

PT= XePeo+XwPw

o = 0.32* 164+ 72.74 * 0.68 = 101.94 KPa

Density of the vapor

By applying Gas Law

Assumption :Compressibility factor of the Water-ethanol vapor is approximately equal to 1 .

PV=nRT

where P = absolute pressure N/m2 (Pa),V = volume m3,n = mols of gasT = absolute temperature, K,R = universal gas constant, 8,314 J K"1 mol"1 (or kJ K"1 kmol"1).

Density =PM/RT = ( 101.94*26.97) /(8.314 * 364) = 0.908 Kg/m3

Viscosity of the liquid mixture at 91 o C

μwater =0.31222 mPa

μethanol = 0.34693

μmix=¿3

μmix =X1 μ11/3+X2 μ2

1/3

=={(0.05 *0.346931/3 +0.95*0.31221/3 }1/3

=0.3138 mPa

29301362

√0.908/¿886.52¿=0.068

For pressure drop 62.5 K4 can be found to be a = 0.6

At flooding K4 is equal to the = 0.9

So percentage of flooding = √ 0.60.9

*100 =80 % satisfactory

Fp= 270 m3/m2

=

V* =[ 0.6∗0.908∗(886.52−0.908 )13.1∗270∗¿¿

]1/2

V*=0.77 Kg/m2s

Column Area required = (mass flow rate of vapor / V*)

=(1362/3600)/0.8) =0.4913 m2

= π*d2/4 =0.4913m2

D=0.79 m

Diameter of the rectifying section ~0.8 m

Packing size to column size ratio = (0.8/0.038 ) =20.81

Feed tray location

It can be assumed as feed is fed into column at 2.5th stage .

So the height of the column = 2.5*HETP = 1.65 m

Condensor heat loadFlow rate of the vapor stream of the stripping section (G) 126.38 kmol/hr

As we have 82%( mol/mol)ethanol in this stream,

Latent heat of this mixture

Temperature of the distillate product =351 K (vapor molar fraction of 0.82 )

Latent heat of ethanol (@ 1 bar , 351 K )

Normal boiling point of ethanol= 351 K

Latent heat of ethanol at normal boiling point =  39300kJ/Kmol.

Latent heat of water (@ 1 bar , 351 K )

Latent heat of water at boiling point =40860 KJ/Kmol

Critical temperature =647 K

Tb = 373

T=351

Lv=40860[ 647−351647−373

]0.38

Latent heat of water at 351 K = 42076.925 KJ/Kmol

Lv mixture = 42076KJ/Kmol *.18 + =  39300kJ/Kmol * 82=

Therefore condensor heat load (Qc) = G x λ = 126.3875 Kmol/hr x 39799.68 KJ/Kmol

= 5.03 x 106 kJ/hr

= 5.03 x 106 /3600kJ/s (kW) = 1397 KW

Reboiler heat loadA partial reboiler will be used in the distillation colum here. Therefore, only a part of the liquid leaving

from the bottom of the column will be evaporated again. Actually the part vaporized in the reboiler is

equal to the upwards vapor flow in the stripping section.

Vaporized Mixture flow rate can be found as the 50.485 Kmol/hr

As this liquid stream consists of 97% octane (3% heptane), it is reasonable to assume latent heat of octane

as the overall latent heat of this liquid stream and the molecular weight of octane as the molecular weight

of whole stream.

G’=50.485 Kmol/hr

L’ =150.29 Kmol/hr

B = 99.805 Kmol/hr => Xb=0.02

Therefore latent heat of the liquid stream leaving the column (λL’) = 298 kJ/kg

Therefore reboiler heat load (QR) = (G’x molecular weight) x λ = 395.15 x 114.2 x 298

= 13.45 x 106 kJ/hr

= 3735 kJ/s (kW)

Reference

[1] http://highered.mcgraw-hill.com/sites/dl/free/0072849606/315014/physical_properties_table.pdf

[2]volume 6

[3]

[4]

[5]

Information

Annex

Top operating Line calculation (0.81-0.7) 12 stages

Total=22 stages

26

Total No of stages 29

30

Stripping section = 3 stages