Reaction Kinetics:

C6H6 + 4.5O2 K1 C4H2O3 + 2CO2 + 2H2O (R1)

C6H6 + 4.5O2 K2 6C0 + 3H2O (R2)

C4H2O3 + 3O2 K3 4C02 + H2O (R3)

r1 (Kmole/kgcat.sec) = k1CB K1 = 0.0595 (m3/Kgcat.sec)

r2 (Kmole/kgcat.sec) = k2CB K2 = 0.01288 (m3/Kgcat.sec)

r3 (Kmole/kgcat.sec) = k3CMAN K3 = 0.0207 (m3/Kgcat.sec)

Weight of catalyst

On the basis of first reaction

WF 1KC 1 ln 11 X X Where,

Reactant Flow rate, F = 9.836 x 10-3 (kmole/sec) Reaction Constant, K1 = 0.0595 (m3/Kgcat.sec)

Reactant Concentration, C = 0 .000566 (Kmole/m3) Fractional Volume Change, = - 0.091 Conversion (First Rex), XA = 0.73 So,

9.836 x 10 30.0595 x 0 .000566 1 0.091ln 11 0.73 0.091 x 0.73

W = 3670.18 Kgcat

Type and volume of Catalyst

A mixture of V2O5 (70%) and MoO3 (25-30%) having average particle size of 5mm

is used as catalyst in the process

Bulk density of catalyst, c = 1200 Kg/m3

Bed void fraction, = 0.4

Volume of catalyst = 3670.18/1200

= 3.058 m3

Reaction Volume:

Vr = Vcat / (1 - )

= 3.058 / (1 - 0.4)

= 5.097 m3

Tube length and diameter

Assuming tube length of 2.5 m and taking the diameter of tube to prevent deviation from

plug flow assumption. Dt/Dp > 10

Where,

Dt = diameter of tube

Dp = diameter of particle

Tube Dimensions:

Tube outside diameter do = 60.45 mm

Tube inside diameter di = 52.50 mm (Schedule No. 40)

Total number of tubes Nt

!" # 4 . %&' . (

So, = 5.097 / (/4 x 0.052502 x 2.5) = 940

Tube layout

Assuming triangular pitch arrangement then

P = 1.25do

Where P = tube pitch

do = outside tube diameter

So, P = 0.07556 m

Number of tubes at bundle diameter

N+ ,4N- 13 Where, ND = number of tubes at bundle diameter

So, ND = 35

Shell Inside diameter: [ ]1NPD Di +=

So, Shell inside diameter, Di = 2.75 m

Shell Height:

Length of tube = 2.5 m

Leaving 20 % spacing above and below

So height of shell = 2 (0.2 2.5) + 2.5

= 3.5 m

PRESSURE DROP

Tube side pressure drop

Using Eurgen equation.

( )

+

= 1.75GD

1150gD

G

1LP

PcfP

3

= bed void fraction = 0.4

DP = particle diameter = 0.5 cm

f = feed density = .0012 g/cm3

= viscosity of feed = 3.19 x 10-4 g/cm. Sec

gc = 980.67 cm/sec2

L = length = 2.5 m = 250 cm

./012345&4( 6477 81(95&0:, < 311% 6477 3(9= 24010904( 3(9= 4214 Total flow area = Nt . /4. di2

= 940 x /4 x 0.052502 = 2.039 m2

Total mass flow rate = 19.44 kg/sec

So, Interfacial mass velocity = 0.9533 g/cm2.sec

Putting values in above eq. gives

P = 655.32 gm/cm2

And 1033.074 g/cm2 = 1 atm

So P = 0.634 atm

Shell side pressure drop:

Heat duty Q = 2.76 x 107 Kj/hr

Cooling media

Molten salt mixture of Sodium Nitrate (60%) and Potassium Nitrate (40%) is used

to recover heat from the reactor

Specific heat capacity of salt mixture, Cp = 144.40 Kj/kmole-K

Temperature difference, T = 20K

Average molecular weight of salt = 94.7 Kg/Kmole

Molten salt required:

Q = m.Cp.T

2.76 x 107 = m x 144.40 x 20

So molten salt flow rate, m = 9556.78 kmol/hr

Mass flow rate, mw = 251.20 Kg/sec

Shell side flow area:

>? # 4 @AB' %C'D Where, Di = shell inside diameter = 2.72 m

Nt = total number of tubes = 940

do = tube outside diameter= 0.06045 m

Ac = 3.11 m2

Equivalent Diameter:

AEF G4@# 4 AB' #%C' 4 D #%C 12 #AB 12 I Putting values in above formula

AEF = 0.06934 m Shell side mass velocity = salt flow rate / shell side flow area

= 251.20 / 3.11

= 80.68 kg/m2 - sec

Viscosity of molten salt = 1.98 x 10-3 kg/m-sec

Reynolds Number

ee

GDR =

= 2826

Friction factor for tube side, fs = 0.00208

K 3L

Calculations of Heat Transfer Co-efficient

Shell Side

Using Eagle and Ferguson equation

PC 1501 0.0110Q!E.RAFE.' tb = average water temperature of cooling media oF

= 653 oF

De = Equivalent Diameter = 0.2274 ft

Now to calculate V = velocity of cooling salt in fps

Mass velocity, G = 16.54 lb/ft2/s

Salt density, s = 117.36 lb/ft3

Velocity = mass velocity / salt density

= 0.1409 lb/ft

Putting values for shell side heat transfer coefficient

ho = 332.41 Btu/ hr. ft2 oF

Wall Resistance:

ST %B(/%C %B 2UT Kw = Thermal conductivity of wall = 25 Btu/ hr. ft2 oF

Putting the values in above equation:

Rw = 0.00049 ft2-hr-oF/BTU

Tube Side

An equation proposed by LEVA to find heat transfer co-efficient inside the tubes

filled with catalyst particles.

= D

D4.6

0.7

pi

p

e

GD3.5

k

h D

G = tube side mass velocity = 7029.4 lb/hr. ft2

= viscosity of tube side fluid = 0.0772 lb/hr. ft

k = 0.0273 Btu/hr. ft. oF

Dp = diameter of particle = 0.01640 ft

di = Inside diameter of tube = 0.17225 ft

Putting values in above equation

hi = 60.36 Btu/hr. ft2 oF

Inside dirt coefficient:

hid = 500 Btu/hr. ft2 oF

Outside dirt coefficient:

hid = 300 Btu/hr. ft2 oF

Over all H.T. Coefficient:

1VB 1PB 1PBW ST %BPC% %BPBW%C do = tube outside diameter = 2.38 in

di = tube inside diameter = 2.067 in

Ui = overall heat transfer on the basis of inside diameter

By putting the values

Ui = 44.96 Btu/hr. ft2 oF

Area required for Heat Transfer

>214 S1XY&21% ZM[\A. VB Q = 2.62 x 107 Btu/hr

LMTD = 148.5 o F

So

Area required for Heat Transfer = 3916.7 ft2

= 363.818 m2

Area Available for Heat Transfer

Length of tube, Lt = 2.5 m

Inside Diameter of tube, di = 0.05045 m \904( 7Y23451 4214 484&(4](1 #%BM = 940 x x 0.05045 x 2.5

= 372.45 m2

So sufficient area is available for heat transfer