chemical engineering - ppt - reactor design

1. Design of Fischer Tropsh Reactor
ZUHAIR BINJAWAID
DDP-SP14-BEC-094

2. Multi tubular isothermal reactor

3. Reactor Catalyst
• Catalyst Iron (Fe)
• Low Cost
• Higher Water Gas Shift activity
• Suitable for lower syngas (H2/CO)
• Promoted with Alkali metals. (K2O)
• To obtain high basicity, and to stabilize catalyst.
• Higher alkali level higher the shift to longer
chains.

4. • Particle of pellet Extrudate (short)
• Size 8 mesh (3.04 mm)
• Density 7840 Kg/m3
• Bed void fraction 0.64

5. Requirements
• Reactants Carbon Monoxide , H2
• Temperature 250 C , 523 K
• Pressure 1 atm , 1013.25 Kpa

6. Data Available
• Mass Flow Rate 820112 Kg/hr.
• Volumetric Flow rate 345659 m3/hr
• Moles of CO 29870 Kmol/hr
• Moles of H2 43098 Kmol/hr
• Moles of N2 608 Kmol/hr

7. Component Mole fraction M.W Viscosity(Cp) Density(kg/
CO 0.40 28 0.030 0.082
H2 0.58 2.04 0.014 1.1453
N2 0.02 28.013 0.030 1.1453

8. • Viscosity of Gaseous Mixture 0.074 Cp
• Heat Capacity of Gaseous Mixture (Cp) 16.65 kJ/kg.K
• Density of Gaseous Mixture 2.3726 kg/m3
Reactions:
• 9CO + 15H2→C9H28 + 15H2O
• 15CO + 29H2→C15H28 + 15H2O
• CO + H2O → CO2 + H2

9. Design Calculation
• The performance equation for the fixed bed tubular
reactor is:
•
• Since rate equation is
−rA = K* 𝐶𝑐𝑜
0.7
*𝐶 𝐻2
0.3
• Arrhenius Equation:
K=AeEa/RT

10. • Ea=104 kJ/mol
• R=8.314*10-3 kJ/kmol
• A=8.27*10 8
mol/g.h Atm2
Putting values in Equation
• K= 8.27*10 8
* e-104 / (8.314*10-3*523)
• K= 8.27*10 8
× 4.09 ∗ 10−11
• K= 0.0339

11. • Selecting CO as reference, because the rate of
the reaction depend on it
• Cco=Fco/V
• CCO= FCOo (1-X)/V0 (1+𝜀X)
• CCO= CCOo *
(1−X)
(1+𝜀X)
…… (eq 1)

12. • Parameter evaluation
• 𝜀 = 𝑌𝑐𝑜 * 𝛿
• Where,
• 𝑌𝑐𝑜 = Initial mole fraction of CO
• 𝛿 = change in the no of moles with reaction
• 𝜀 = (0.40)*[(15+1)-(15+29)]
• 𝜀 = -11.2
• CCOo = 𝑌𝑐𝑜 ∗ 𝐶 𝑇 𝑜
= 0.40 *
𝑃𝑜
𝑅𝑇𝑜
= 0.40 *
1013.25
8.314∗523
• CCOo = 0.093

13. • Putting the values of CCOo & 𝜀 in eq 1 :
• CCO= CCOo *
(1−X)
(1+𝜀X)
• CCO= 0.093 *
(1−X)
(1−11.2X)
……. (eq 2)
Similarly,
• 𝐶 𝐻2 = CCOo *
(θ 𝐵−
𝑏
a
X)
(1+𝜀X)
……. (eq 3)

14. • Paramater Evaluation
• CCOo = 0.093
• θ 𝐵 =
𝐹𝐻2
𝐹𝐶𝑂
= 0.12
•
𝑏
a
= 1.93
• Putting all the values in eq 2 :
• 𝐶 𝐻2 = 0.093 *
(0.12−1.93X)
(1−11.2X)
……… (eq 4)

15. • Putting these in rate equation :
−rA = K* 𝐶𝑐𝑜
0.7
*𝐶 𝐻2
0.3
• −rA = 0.0339 0.093 ∗ (1−X)
(1−11.2X)
0.7 0.093 ∗ (0.12−1.93X)
(1−11.2X)
0.3
At 80% conversion rate is given by :
• −rA = 0.0339
0.093 ∗ (1−0.80)
(1−11.2∗0.80)
0.7 0.093 ∗ (0.12−1.93∗0.80)
(1−11.2∗0.80)
0.3
• −rA = 1.411 * 10−04
mol/ g(catalyst).s

16. Levenspiel plot
X −rA
mol/ g(catalyst).s
1/-ra
g(cata).s/mol
0 2.96* 10-4 631
0.1 1.526 *10-3 655
0.2 1.4 *10-3 686
0.3 8.13 *10-4 1228
0.4 5.53 *10-4 1807
0.5 3.99 *10-4 2500
0.6 1.99 *10-4 5018
0.7 1.98 *10-4 5031
0.8 1.41*10-4 7083

17. 0
1000
2000
3000
4000
5000
6000
7000
8000
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Inverse rate law
(1/-ra)
conversion X
LEVENSPEIL PLOT

18. • Area under the curve is found by applying Simpson’s
rule,
• Area under the curve = simpsons rule (0-0.8)
• H= 0.1
• Simpsons rule (0-0.8) = h/3 * [(fo +f8) +4(f1+f3+f5+f7) +
2(f2+f4+f6)]
• = 0.1/3 *[(631+7083)
+4(655+1228+2500+5031) + 2(686+1807+5018)]
• = .033[(7714+37656+15022)]
• Area under the curve = 2013 g.s/mol

19. • Fco = 716883 kg/hr = 199134 g/sec = 7111 mol/sec
• W = 7111 * 2013
=14316312.21 g catalyst
• W = 14316.31 Kg catalyst
AX
A
Ao o
A
dX
W = F *
-r

20. Volume of Catalyst bed
• Volume of Catalyst = Mass/Density
= 14316.31/7840
= 1.82 m3
• Volume of Bed = Vcat/ (1-Void Fraction)
=1.82/ (1-0.64)
=5.05 m3

21. • Dimensions of Tubes
• Tube OD = 1 in = .0254 m
• BWG = 14
• Thickness = .083 in
• Tube ID = 0.834 in =0.0211 m
• Length of tube = 20ft = 6.096 m
• Volume of 1 tube = V =
π
4
D2
L
= 3.14/4 * (.0254)2 * 6.09
= .0030873 m3

22. NO of Tubes
• Nt =
𝑉𝑜𝑙 𝑜𝑓 𝑐𝑎𝑡𝑎𝑙𝑦𝑠𝑡
𝑣𝑜𝑙 𝑜𝑓 1 𝑡𝑢𝑏𝑒
• Nt =
1.82
0 .0030873
• Nt = 589
Surface Area for One Tube
• As = πDoL
= 3.14 * 0.0254 * 6.09
= 0.485 m2

23. Height of Reactor
• Height of the Reactor is equal to the height of
tubes plus 20% allowance
• Height of Tubes = 20 ft = 6.09 m
• Additional Length = 2*.02* 6.09
= 2.439 m
• Height of the reactor = Height of tubes +
Addititional length
= 6.09 + 2.439
= 8.526 m

24. Bundle Diameter
• Db = d°
Nt
K1
1
n1
• Db = 0.0254
589
0.215
1
2.207
• Db = 0.89 m

25. Shell Diameter
• Shell diameter = bundle diameter + clearance
• Ds = Db + Clearence
• Clearence is calculated from following graph :

26. • DS = 0.89 + .097 (from graph)
• Ds = 0.99 m
Volume of Reactor
• V =
π
4
D2L
• V =
3.14
4
∗ 0.992∗ 8.526
• 𝐕 = 𝟔. 𝟓𝟔 𝐦 𝟑

27. Pressure Drop Calculation
∆P
L
=
150 μ G 1−∈ 2
g.ρ D2 ∈3
+1.75
G2 1− ∈
g.ρ D ∈3
• ∆P = Pressure drop along the length of reactor
• L= Length of reactor
• μ = Viscosity of the gaseous feed
• G = Mass velocity
• ρ = Feed density
• D = Diameter of the particle
• ∈ = φ = Porosity
• g = Gravitational constant

28. Parameter evaluation
• L= Length of reactor = 8.53 m = 27.99 ft
• g = Gravitational constant = 4.17 ∗ 108 lbm.ft
lbf.ft2
• G =
mass flow rate
Area of Reactor
• Area of Reactor =
Volume of Reactor
Diameter of Reactor
=
6.56
0.99
= 6.6 m2 = 71 ft2
• G =
820112∗2.2
71
= 22213
lbm
hrft2
• μ = 0.074 cp = 0.17
lb
fthr
• ρ = 2.37
kg
m3 = 0.134
lb
ft3
• Dp = 0.01042 ft
• ∅ = ∈ =0.64

29. ∆P
L
=
150 μ G 1−∈ 2
g.ρ D2 ∈3
+1.75
G2 1− ∈
g.ρ D ∈3
•
∆P
27.99
=
150 ∗0.17∗ 2213∗ 1−0.64 2
4.17∗108∗0.134∗ 0.010422∗0.643 +1.75
22132 1− 0.64
4.17∗108∗0.134∗ 0.1042∗ 0.643
•
∆P
27.99
= 51.65 +0.09
• ∆P = 1443
lb
ft2
• ∆P = 1443
lb
ft2 *
1ft2
144 in2 ∗
1 atm
14.7
lbf
in2
• ∆P = 0.068 atm
• The pressure drop along the length of the reactor is less than the
10% of the operating pressure.

30. Specification Sheet
Reactor
No. required: 01
Design data
Type Fixed Bed Tubular
Position Vertical
Material of construction Carbon steel
Weight of catalyst (kg) 14316.31
Volume of catalyst (m3) 1.82
Diameter of Reactor (m) 0.99
Height of Reactor (m) 9
Volume of reactor (m3) 6.56
Tube diameter (m) 0.0254
No. of Tubes 589

31. • References
[1] A rate equation for fischer tropsh synthesis on iron catalysts
:Robert B.Anderson and Fred S.Kern
[2] Chemical Engineering Design ; Coulson and Richardson vol 6.
[3] Elements of Chemical Reaction Engineering by H.Scott
Foggler.
[4] Process heat transfer by D.Q.Kern.
[5] Fiscer Tropsh Refining by Arno De Clark.
[6] Elementary Principles of chemical processes.