divinylbenzene equipment design

7/31/2019 divinylbenzene equipment design

1/20

CHAPTER 5EQUIPMENT DESIGNING


2/20

20

5.1 Process Design of Benzene Distillation Column

In Benzene column, a product stream from reactor is coming which consists of benzene,

ethyl benzene & some amount of di-ethyl benzene. Since it is more than two component

case, hence we will perform multicomponent distillation in order to separate benzene and

ethyl benzene from mixture.

DEB 1.5129 KMOLE

EB 0.0824 KMOLE

BENZENE 0.1069 KMOLE

EB 0.0824 KMOLE

BENZENE 0.1069 KMOLE

DEB 1.5129 KMOLE

Here Benzene is light key component and EB is heavy key component. The column is operating at 1 atm.

Now first of all we will calculate bubble point & dew point temperature.

Table 6


3/20

21

BUBBLE POINT CALCULATION:-

Take T=456.82 K (183C)

COMPONENT Pi( mmHg) Ki=Pi/760 Xi Xi*Ki

BENZENE 10911.88464 14.35774295 0 0EB 3151.533924 4.146755164 0 0

DEB 767.8215491 1.010291512 1 1.010292

TOTAL 1.010292

Therefore,bubble point temperature=456.8 K

DEW POINT CALCULATION:-

Take T=388 K(115C)

COMPONENT Pi( mmHg) Ki=Pi/760 Yi Yi/Ki

BENZENE 2074.458326 2.729550429 0.5647 0.206884

EB 413.1979536 0.543681518 0.4352 0.800469

DEB 63.90478783 0.084085247 0 0

TOTAL 1.007353

Therefore,dew point temperature=388 K

CALCULATION OF MINIMUM REFLUX RATIO (Rm):-

For a liquid feed, q = 1

From trial & error

=1.54

COMPONENT Xf Xd Ki iBENZENE 0.0628 0.5647 1.12 5.741

EB 0.0484 0.4352 0.196 1.000

DEB 0.8887 0 0.036 0.1836


4/20

22

Rm+1=1.242

Rm=0.242

Calculation of Operating Reflux Ratio:-

Ro=1.5Rm=0.363

Calculation of number of ideal plates at operating reflux:-

By Gilliland correlation:

(Ro - Rm )/( Ro +1)= 0.0974

(N-Nmin)/(N+1)=0.55 (from perrys engineering handbook 8th

edt. pg-13-26)(a)

Now,Nmin= [ln{(xdi /xb i)/(xd j/xbj )}] / [ln ( avg)]

On calculating,

Nmin=8

Now putting value in equation (a) we get,

N=19

Thus,we need 19 stages ideally.

Flow Rates:

Average molar mass of feed.

Mav = xi Mi

= (78*0.0628) + (106*0.484) + (134*0.8887) = 175.2888 kg/kmol

F = 1.7022 kmol/hr

D = 0.1839 kmol/hr

B = 1.5129 kmol/hr

Now molar flow rates of vapor & liquid at top in enriching section:

L = R * D = 0.363 * 0.1839 = 0.0667 kmol/hr

V = (R+1) * D = (0.363 +1) * 0.1839 = 0.2506 kmol/hr

Molar flow rates of vapor & liquid in stripping section:

L = L + F * q


5/20

23

= 0.0667 + 1.7022 * 1 = 1.7689 kmol/hr

V = F*(q-1) + V = 1.7022*(1-1) + 0.2506 = 0.2506 kmol/h

Calculation of tower diameter:

(a) Tower diameter required at top:

Operating pressure at the top of column = 1 atm = 101.325 kpa

V = 0.2506 kmol/hr

L = 0.0667 kmol/hr

Here total condenser is used,

hence Lw/Vw = L/V = 0.0667/0.2506 = 0.2661m

Density of vapor:

v = (p* Mav)/(R*T) = (49.177*273)/(388*22.414) = 1.5437 kg/m3 ( Mav = xi Mi =49.177)

Density of liquid at top:

l = 879 kg/m3

Liquidvapor flow factor at top:

Flv = (Lw/Vw)*(v/ l)^0.5 = 0.226*(1.5437/879)^0.5 = 0.146

Tray spacing = 0.3 m

Corresponding Cf = 0.06

(from Introduction to Process Engineering & Design by S.B. Thakore & B.I Bhatt, Chapter 6,

Process Design of distillation column, Page no. 448)

Now flooding velocity:

vf = Cf * (/0.02)0.2 * {(l v)/v}^0.5

(where = surface tension of liquid, N/m = ixi)

(assumed)

= 0.06 * (22.267*10-3/0.02)0.2 * {(879-1.5437)/1.5437}^0.5

= 1.126 m/sec.

Now actual velocity:

v = 0.85 * vf = 0.85 * 1.1126


6/20

24

= 0.946 m/sec.

Volumetric flow rate of liquid at top:

Q = (V* Mav)/v = (0.2506*49.177)/1.5437 = 332.79 m3/hr = 0.0924 m3/sec

Net area required at top:

An = Q/v = 0.0924/0.946 = 0.0973 m2

Let downcomer area Ad,

Ad = 0.12* Ac

(Ac = internal cross-sectional area of tower)

& An = AcAd = Ac0.12*Ac = 0.88 Ac

Ac = An/0.88 = 0.0973/0.88 = 0.11 m2

Inside diameter of column required at top:

Di = {(4*Ac)/}0.5 = 0.375 m

Similarly on calculating diameter at bottom we get Di=0.4m

Checking for weeping:

Minimum vapor velocity through holes to avoid the weeping given by following equation:

vh, min = [{K0.9(25.4dh)}/(v)0.5]

K constant can be obtained from Fig 8.19 (Introduction to Process Engineering & Design by

S.B. Thakore &

B.I Bhatt, Chapter 6, Process Design of distillation column, Page no. 449) or Fig. 4 of

appendix. ,is a function of (hw + how),

where

weir height (hw) = 50 mm

hole diameter(dh) = 5 mm

Plate thickness (t) = 5mm

(a) For enriching section :

height of liquid crest over the weir

how = 750 (Lm/l*lw)

Lm = 0.7*L*Mav


7/20

25

= 0.172 kg/sec

lw = 0.77* Di

= 0.77* 0.375

= 0.2885

Now putting all the values ,

how = 6.075 mm

Thus (hw + how) = 56.075 mm

corresponding K value from graph of K vs(hw + how)

K = 30.2

Thus finally

vh,min = 4.62 m/sec

Actual vapor velocity holes at actual vapor flow rate :

vh,a = (0.7* Qv)/ Ah .(b)

Now Ad = 0.12 * Ac = 0.12 * 0.11 = 0.0132 m2

Active area Aa = Ac2Ad = 0.112(0.12)(0.11) = 0.0836 m2

Thus hole area, Ah= 0.00836 m2

Now putting it in equation (b) we get,

vh,a = 7.74 m/sec

Since vh,a >> vh,min

Thus in enriching section minimum operating rate is well above weep point.

(b) For Stripping section:

height of liquid crest over the weir

how = 750 (Lm/l*lw)

Lm = 0.7*L*Mav

= 0.259 kg/sec

lw = 0.77* Di

= 0.77* 0.400 =0.308


8/20

26

l = 879 kg/m3

Now putting all the values ,

how = 23.19 mm

Thus (hw + how) = 75.19 mm

corresponding K value from graph K vs (hw + how)

K = 30.68

Thus finally

vh,min = 5.8533 m/sec

Actual vapor velocity holes at actual vapor flow rate :

vh,a = (0.7* Qv)/Ah .(c)

Now Ad = 0.12 * Ac = 0.01488 m2

Active area Aa = Ac2Ad = 0.09424 m2

Thus hole area, Ah= 0.009424 m2

Now putting it in equation (c) we get,

vh,a = 6.3 m/sec

Since vh,a >> vh,min

Thus in stripping section minimum operating rate is well above weep point.

Tray pressure drop:

(a) For enriching section:

Dry plate pressure drop:

hd = 51(vh/C)^2*(v/l)..(d)

vh = Qv/Ah = 0.0924/0.00836 = 11.053 m/sec

considering

Plate thickness / hole diameter = 1

Ah/Ap = Ah/Aa =0.1

( Ap is perforated area which is slightly less than active area.)

Thus corresponding C = 0.8422 (from graph of C vs Ah/Ap )


9/20

27

Now putting all the values in equation (d)

hd = 26.58 mm

Maximum height of liquid of crest over the weir:

how = 750 (Lm/l*lw)

= 750{(0.172/0.7)/(867*0.288)}

= 23.36 mm

Residual Pressure drop:

hr = (12.5*103)/l = (12500/867) = 14.41 mm

Total tray pressure drop:

ht = hd + hw + how + hr

= 26.58 + 50 + 23.36 +14.41

= 114.16 mm

(b) For stripping section:

Dry plate pressure drop:

hd = 51(vh/C)^2*(v/l)vh = Qv/Ah = 0.0752/0.009424 = 7.979 m/sec

now,

Plate thickness / Plate Area = 1

Ah/Ap = Ah/Aa =0.1

( Ap is perforated area which is slightly less than active area.)

Thus corresponding C = 0.8422

Now putting all the values in equation (c)

hd = 23.43 mm

Maximum height of liquid of crest over the weir:

how = 750 (Lm/l*lw)

= 9.74 mm


10/20

28

Residual Pressure drop:

hr = (12.5*103)/l = (12500/879) = 14.22 mm

Total tray pressure drop:

ht = hd + hw + how + hr

= 23.43 + 50 + 9.74 +14.22

= 97.39 mm

Checking of downcomer design:

Type of downcomer: Straight & segmental downcomerarea, Ad = 0.12Ac (for both sections)


hdc = 166 (Lmd/l*Am)

where Lmd = Liquid flow rate through downcomer, kg/sec

= L*Mav

= (0.0667*49.177)/3600

= 0.065 kg/sec

Am = Ad or Aap whichever is smaller

Aap = hap*lw = (hw10)*lw = (50 -10)*0.288 = 0.1152 m2

Ad = 0.0836 m2

Since Aap


11/20

29

(lt + hw)/2 = (300 + 50)/2 = 175 mm

Since hb < (lt + hw)/2

Thus downcomer area & spacing are acceptable.

Checking for residence time :

r = (Ad*hb*l)/Lmd = (0.0836*0.170*867)/0.065 = 189.56 sec

which is greater than 3 sec. thus it is satisfactory.


hdc = 166 (Lmd/l*Am)

where Lmd = Liquid flow rate through downcomer, kg/sec

= L*Mav

= 0.259 kg/sec

Am = Ad or Aap whichever is smaller

Aap = hap*lw = (hw10)*lw = 0.01232 m2

Ad = 0.09424 m2

Since Aap


12/20

30

Checking for residence time :

r = (Ad*hb*l)/Lmd = (0.09424*0.162*879)/(0.259/0.7) = 189.56 sec

which is greater than 3 sec. thus it is satisfactory.

Checking for entrainment:


Vapor velocity based on net area (vn) = Q/ An = 0.0924 / 0.973 = 0.95 m/sec.

% Flooding

Flv = 0.146

From graph of vs Flv

= 0.11

or 11%

Since % > 10% , thus higher efficiency will be obtained.

= (vn/vf)*100 = 85.35%


% Flooding = 85%

Flv = 0.0785

From fig 8.18:

= 0.053

or 5.43%

Since % < 10% , thus higher efficiency will be obtained.


13/20

31

5.2 Mechanical Design of Benzene Distillation Column

Shell Thickness:

Diameter = 400mm (taken for designing)

Operating pressure = 760mmHg

Taking 10% allowance Pd = 684mm Hg

Operating average temperature = 116C

Design temperature = 150 C

Material used is carbon steel

Specific gravity of carbon steel = 7.7

Permissible tensile stress = 950 kg/cm2

(up to 250 C)

Insulation material is mineral wood , 75mm thick , density = 130kg/m3

Shell minimum thickness ts

ts = Pd Di/(2f tJ-Pd) + C take C = 3mm

ts = (0.9x400/2x950x1 -0.9) +3 = 3.20

but for high vessels under external pressure take shell thickness ts= 8mm

Head:-

Torispherical heads

Let thickness = 8mm same as shell

t = P dRc Cs/[2f tJ + Pd(Cs-0.2)]

Cs = [3 + (Rc/Rk)0.5]

Take Rc = 400 mm ; J = 1

On solving Cs = 71.24

Therefore Rk = 5.026x10-3mm

But Rk > 0.06Rc ; therefore Rk = 24mm


14/20

32

Shell Thickness at different Height:-

At a distance X m from the top of shell the stress are

(a) Axial stress:-

fap = Pd Di/4(ts-C) on substituting values

fap = 18 kg/cm2

(b) Stresses due to dead load :

(i) Compressive stress due to weight of shell (fds):-

= 7.7x10 -6X kg/cm2

(ii) Compressive stress due to weight of insulation(fdi):-

Dm = Dins = (0.4 + 0.375)/2 = 0.3875 m

fdi = 3.22x10 -5X kg/cm2

(iii) Compressive stress due to liq. In the column up to height X

fd,liq= 2.134x10 -3X kg/cm2

(iv) stress due to attachement (f d, att)

wt. of attachements = 150 kg/m

therefore fd = 11.46x10 -3X kg/cm2

(c) Stress due to wind load:- fwx

fwx=(1.4PwX2)/(3.14*Dm(ts-C))

wind pressure = 125 kg/m2

= 125x10 -6 kg/mm2

fwx = 1.34x10 -6X2 kg/cm2

Neglecting seisemic load

equating all the stresses to zero

fwx( fdx + fd,liq + fdi + fds )fap = 0

solving for X ;

X = 17.6

Hence thickness taken as 8mm is sufficient as column ht. is 11.7m


15/20

33

Support:-

(a) stress due to dead weight:

Skirt diameter = 400 mm (Ds)

Dead weight attachments = 46000 kg (W)

Fbs=460000/(3.14*(Ds+ts)ts)(a)

(b) stress due to wind load Mw = 0.7PwDoX

fws=4 *Mw/(3.14(Ds+ts)ts).(b)

now,

total compressible stress= (a) + (b)..(c)

also fs(max)=( 0.125*E*ts*cos)/Ds ;cylindrical support so cos=1

E=19.5*10^3 kg/mm2 (d)

Substuting values of (d) in equation (c)

fs(max)= 4.875*10^6*ts

At ts=8mm fcompressible


16/20

34

5.3 REACTOR DESIGN (Dehydrogenator):-

Reactor Process design Figure

Reactor Schematic diagram:-

notations:-

11-External shell

12-Inlet header

13-Exit header

14-Feed supply line

16-Tubes

17-Fuel inlet line

18-Fired gas heater

19-Exhaust line

24-Heater nozzle

28-Heat exchanger

INLET (Kmol/hr)

DEB 1.5129 (REACTANT)

STEAM 27.075

OUTLET (Kmol/hr)

DVB 0.6051 (PRODUCT)

DEB 0.6051

HYDROGEN 1.8156

STEAM 27.075

DEB 0.3025


17/20

35

Reaction temperature =600C ( 873 K)

Catalyst: Ferrous Oxide (FeO)

Density of catalyst particle: 4292.008 kg/m3

Catalyst porosity : 0.3

Type of reactor: Fixed bed catalytic reactor

Reaction is endothermic: It is carried out in isothermal manner.Heat is supplied by internal

electric heaters.

(FAo) Inlet of reactant, diethyl benzene to the reactor= 1.512912Kmoles/hr = 203.065

Kg/hr..(from Mass Balance calculations)

(CAo) Intial concentration of reactant diethyl benzene = 0.533 kmoles/m

3

Product Divinyl benzene of desired grade in outlet of reactor=0.6051 Kmoles/hr =78.7916

Kg/hr..(from Mass Balance calculations)

Mass of catalyst:

W/FAO= 1/(-ra)(1)

where,

k=4.3*10^-3 kmole/hr-kg

a=1.8 atm-1

b=39 atm-1

PE=Partial pressure of diethyl benzene at 873K=3.7299 atm

Po=Partial pressure of Oxygen at 873K=28.3676 atm

putting values in equ. ofr we get,

-r=0.0286 Kmole/hr-kg

putting values in equation (1) we get mass of catalyst as

W= 52.898 Kg

Volume of solid catalyst = mass of catalyst/ density of catalyst

=52.898/4292.008


18/20

36

= 0.01232 m3

Porosity of catalyst bed = 0.3

Bulk volume occupied by the catalyst= 0.2326/(1-0.3)

= 0.0176 m3

Now residence time

=5.55 kg/hr(ref:- Journal of catalysis 34,7-12 (1974))

now,

/CAo=V/FAo

On calculation we get,volume of reactor

V=15.611 m3

Diameter and length of reactor:-

L/D= 10.4.(1)( Data from US Patent no: 6781024 B2)

L= length of reactor

D= diameter of reactor

assuming reactor to be cylindrical,

V = (3.14/4)*D^2*L(2)

solving equation (1) & (2) we get,

D = 1.2412 m

L = 12.908 m

Tube design for reactor

Assume Superficial velocity of vapour inlet = 0.55 kg/m2.s

Total cross sectional area of catalyst of tubes= (203.065/3600)/0.55

= 0.0310 m2

MOC of tube = Stainless steel

Tube OD = 50.8mm

Tube ID = 43.28 mm


19/20

37

Total number of tube required

nt = 0.0310/(3.14 /4)(0.04328)^2 =21.08

=22 tubes

Length of tube required

L= net vol. of catalyst/nt (3.14 /4) di^ 2

= ( 0.01232 *4)/(22*3.14)*(0.04328)^2

= 0.3808m

Aav = nt*3.14*do*L

= 22*3.14*0.0508*0.038

= 0.1333 m2

5.4 Mechanical design for Reactor:-

Shell thickness

Material= carbon steel

Number of shells=1

Number of passes=1

Fluid=diethyl benzene & steam

Working pressure=0.2 N/mm2

Design pressure=0.25 N/mm2

Inlet temperature=600C

Outlet temperature=600C

Shell thickness:

t s =PD/(2fJ+P)

= 0.25*1200/((2*87*0.85)+0.52)

= 2.02 mm


20/20

38

Including corrosion allowance.

Use 8 mm thickness.

5.5 Recommended MOC for the process equipments

SR.NO EQUIPMENT MOC1 Flash Drum carbon steel (A285A)

2 Distillation Column D-1 carbon steel (A285A)

3 Distillation Column D-2 SS-316

4 Reactor SS-316

5 Condenser (Hydrogen Separator) SS-316

6 Centrifugal Pump carbon steel (A285A)

7 Pump Seals Viton fluoroelastomerO-ring with carbon face seal

8 Gaskets Teflon resin

9 Piping Carbon steel with teflon

coating

10 Storage Tank black iron

divinylbenzene equipment design

Documents