The document summarizes the design of a 30,000 MTPA maleic anhydride production plant in India. It includes: 1) An introduction describing the importance of maleic anhydride and the aim to design a cost-effective plant using mixed butane as a feedstock. 2) Details of the major process units - feedstock pretreatment, synthesis reactor, product recovery and purification. 3) Evaluation of four process alternatives and selection of the final design incorporating a catalytic partial oxidation reactor, absorber for product recovery, and distillation for purification. 4) Key aspects of the design such as mass balances, equipment sizing for a shell and tube heat exchanger, and the

1. CH-2451 PROJECT WORK
NARENDARAN.K (312211203031)
PRITHIVI RAJ.S (312211203040)
VIVEK KUMAR.R(312211203061)
SUPERVISED BY DR.M.SUBRAMANIAN
Department of Chemical Engineering
Sri Sivasubramaniya Nadar College of Engineering
Kalavakkam – 603 110, Kanchipuram (Dist)
Tamil Nadu, India
02-Jan-2015
DESIGN OF 30,000 MTPA MALEIC ANHYDRIDE PRODUCTION
PLANT

2. CONTENTS:
• Introduction
• Best process route
• Mass balance
• Equipment design
• Saftey
• Economic balance
• Conclusion
• References
02-Jan-2015 M Subramanian

3. INTRODUCTION:
3
Maleic anhydride (MAN) plays an important role in
over half of the global demand production of unsaturated polyester resins.
Thus, the aim of this project is to design a practicable plant which is cost-
effective, profitable and harmless to the environment. The project is to
design Maleic Anhydride production plant by using mixed butane as the
feedstock. The plant could be divided into four main processes which are
feedstock pretreatment, synthesis of MAN, recovery MAN and purification
of MAN

4. PROJECT BACKGROUND:
Uses of Maleic Anhydride:
• synthesis of unsaturated polyester resin
• plasticizers, surface coatings, agrochemicals, lubricants.
0%
10%
20%
30%
40%
50%
North American Japan Middle East Asia Europe/France/Italy
Percentage demand of Maleic Anhydride (2007-2011)
Market in Asia
INDIA: THIRUMALAI CHEMICALS LTD (60,000 TPA)
Thailand: PTT Group Company (15,000 TPA)
Indonesia: Chandra Asri Petrochemical (6000-7000 TPA)

5. Type Annual Requirement
(tonne/year)
Feedstock
Raw butane 72,900
Air 570,000
Product
Iso-butane 24,930
Maleic
Anhydride
36,600
FEED COMPOSITION:

6. PROCESS OVERVIEW:
Feedstock
separation
N-butane
vaporization
Catalytic partial
oxidation with air
Dibutyl phthalate
extracts product
Vacuum distillation
recovers product
from solvent
Storage

7. MAJOR ALTERNATIVES:
Alternativ
e
Isobutane converter? Condense reactor effluent?
1 Yes Yes
2 Yes No
3 No No
4 No Yes

8. Alternative 1: Converter + Condenser:
Mixed
butane
Air
Air
Isobutane
n-butane
Offgas
Offgas
Liquid MA
to solidify
Offgas
Lean solvent
MA rich
MA rich

9. Alternative 2: Converter:
Mixed
butane
Air
Air
Isobutane
n-butane
Offgas
Offgas
Liquid MA
to solidify
Lean solvent

10. Alternative 3: Converter +
Condenser:
Isobutane for sale
Mixed
butane
Air
n-butane
Offgas
Offgas
Liquid MA
to solidify
Lean solvent

11. Alternative 4: Converter +
Condenser:
Mixed
butane
Air
Isobutane for sale
n-butane
Offgas
Offgas
Liquid MA
to solidify
Offgas
Lean solvent
MA rich
MA rich

12. Final Design:
Mixed
butane
Isobutane for sale
Air
Offgas
Lean solvent
Offgas
Liquid MA
to solidify

13. PROCESS FLOW DIAGRAM:
13

14. Feed separation
• Bottoms 98% purity
Reactor
•Conversion: 72.2%
Absorber
•Recovery: 97%
Storage
•Purity:
97.5%
Desorber
•Recovery: 98%

15. MAIN REACTIONS:

16. MASS BALANCE(spreadsheet):

17. DESIGN OF COOLER(E-201):
Equipment Sizing:
Composition and Properties of Inlet and Outlet:
Stream
Tin (°C) Tout (°C) Tin (K) Tout (K)
Process fluid 256 120 529 393
Stream data of shell side

18. Stream data of tube side
Stream tin (°C) tout (°C) tin (K) tout (K)
Cooling water 30 40 303 313
Cooler duty=3435 kW
R (T1-T2)/(t2-t1)
R 13.6
S (t2-t1)/(T1-t1)
S 0.0442
Calculate correction factor, Ft

19. From the Figure 12.19: temperature correction factor. Retrieved from Chemical Engineering (Volume 6)
Coulson and Richardson

20. The value of Ft = 0.98
• Calculate mean temperature, ∆Tlm
∆Tlm can be calculated from the equation,
Then,
∆Tlm = 136.92 °C
• Calculate actual temperature difference
Ft∆Tlm = 134.19 °C
• Overall heat-transfer coefficient
Typical values of the overall heat-transfer coefficient for various types of heat exchanger are given in Table 12.1
in the Chemical Engineering (Volume 6) Coulson and Richardson book.
Assuming this is cooler
)(
)(
ln
)()(
12
21
1221
tT
tT
tTtT
Tlm





21. U = 300 W/m^2°C
Area of heat exchanger
Provisional area of heat exchanger, A can be obtained through the formula,
lmTUAQ 
lmTU
Q
A


Then,
A = 85.33 m2
Tube Side
From Appendix A.5-2, Transport Processes and Unit
Material Carbon Steel
BWG number 18
Length of tube Lt (m) 2.5
Outer diameter, Dto (mm) 25.4
Inner diameter, Dti (mm) 22.1
Material Thermal Conductivity (W/m.K) 36

22. Heat transfer area of a tube, At:
At = 0.1995 m2
Number of tube, Nt:
𝑁𝑡 =
𝐴
𝐴 𝑡
𝑁𝑡 =
85.33
0.1995
𝑁𝑡 = 428 𝑡𝑢𝑏𝑒𝑠
Number of tubes per pass, Np
𝑁𝑡 =
428
2
𝑁𝑡 = 214 𝑡𝑢𝑏𝑒𝑠
Tube pitch is the distance between tube centre and formulated as
𝑃𝑡 = 1.25 × 𝐷𝑡𝑜
𝑃𝑡 = 1.25 × 25.4
𝑃𝑡 = 31.75 𝑚𝑚
For typical tube arrangements, from Table 12.4, Chemical Engineering (Volume 6) Coulson
and Richardson, for triangular pitch for 2 passes, the constant value as follows:
K1 = 0.249
n = 2.207
tott DLA 

23. The bundle diameter, Db:
𝐷 𝑏 = 𝐷𝑡𝑜 𝑁𝑡 𝐾1
1 𝑛
𝐷 𝑏 = 25.4 428 0.249 1 2.207
𝐷 𝑏 = 0.743 𝑚
Shell internal Diameter, Ds
From Chemical Engineering (Volume 6) Coulson and Richardson, figure 12.10,
For Fixed and U-tube,
Shell-bundle clearance = 87 mm
= 0.087 m
Shell internal Diameter, Ds = Db + shell bundle clearance
Ds = 0.830 m
Tube side Coefficient
Mean temperature (K), Tm:
Tm = (Tcin +Tcout)/2
Tm = 308 K
Tube cross-sectional area, At:
𝐴 𝑡 =
𝜋𝐷𝑡𝑖
2
4
𝐴 𝑡 =
𝜋22.12
4
𝐴 𝑡 = 383.64 𝑚𝑚2

24. Total flow area (m2), AT:
𝐴 𝑇 = 𝑁𝑡. 𝐴𝑡
𝐴 𝑇 = 82098.96 𝑚𝑚2
= 0.821 𝑚2
𝐴 𝑇Physical properties of the tube side fluid are obtained from Appendix A.2-11, Transport
Processes and Unit Operation by, Christie J. Geankoplis:
Water density, ρt (kg/m3
) 996.2600
Viscosity of water, μtL (Ns/m2
) 8.348 x 10-4
Heat capacity, Cp (kJ/kg.K) 4.1798
Thermal conductivity, ktf (W/m.K) 0.6129
Table 1: Physical properties of water

25. Mass flowrate (inside tube), m= 81.27 kg/s
Fluid velocity, vf:
𝑣𝑓 = 𝑚 𝐴 𝑇
𝑣𝑓 = 81.27 0.821 = 98.99 𝑘𝑔 𝑚2
. 𝑠
Linear velocity, u
𝑢 = 𝑣𝑓 𝜌
𝑢 = 98.99 996.26 = 0.09937 𝑚/𝑠
Reynold number, Re
𝑅𝑒 =
𝜌𝑢𝐷𝑡𝑖
𝜇
=
(996.26)(0.09937(0.0221)
8.348 × 10−3
= 26210
Prandtl number, Pr
𝑃𝑟 =
𝜇𝐶 𝑝
𝑘 𝑓
=
(8.348 × 10−3
)(4179.8)
0.6129
= 5.693
Length of tube/Inner diameter (L/Dti) = 2.5/ 0.0221= 113.122
Heat transfer coefficient, jh= 0.0037
Tube side heat transfer coefficient, hi
ℎ𝑖 =
𝑘 𝑓 𝑗ℎ 𝑅𝑒𝑃𝑟0.33
𝐷𝑡𝑖
𝜇
𝜇 𝑤
0.14
= 4848.077 𝑊 𝑚2
K
We assume that viscosity of the fluid is identical at the wall and of the bulk fluid.

26. Tube Side Pressure Drop
Friction factor, jf = 0.0036
Tube side pressure drop can be calculated from the equation below:
Where,
m = 0.25 for laminar flow, Re<2100 ; m = 0.14 for turbulent flow, Re>2100
Np = number of tube side passes
ΔPt= 57.53 Pa
2
]5.2))(/(8[
2
sm
w
tifps
u
DLjNP



 
Shell Side
Fluid density, ρs (kg/m3
)
1.768
Viscosity, μsL (Ns/m2
)
2.073x10-5
Heat capacity, Csp (kJ/kg.K)
1.155
Thermal conductivity, ksf (W/m.K)
0.03081
Physical properties of reactor effluent

27. Shell Side Heat Transfer Coefficient
Shell diameter, Ds = 0.83 m
Tube pitch, Pt = 31.7500 mm = 0.0318 m
Cross flow area, As will be calculated using:
𝐴 𝑠 =
𝑝 𝑡−𝐷𝑡𝑜 𝐷 𝑠 𝑙 𝐵
𝑃𝑡
= 0.0689 m2
Shell side mass velocity, Gs :
𝐺𝑠 =
𝑤𝑠
𝐴 𝑠
=
77110
0.0689
= 310.88 𝑘𝑔 𝑠. 𝑚2
Shell side equivalent diameter, De
𝐷𝑒 =
1.1
𝐷𝑡𝑜
𝑝𝑡
2
− 0.971𝐷𝑡𝑜
2
= 16.526 𝑚𝑚
Mean temperature (K)
Tmean = (Th.in +Th,out)/2
Tmean = 461.0 K
mDlSpacingBuffle B 415.0*2/1, 
mDDiameterBuffle s 8284.00016.0 

28. Reynold number, Re
𝑅𝑒 =
𝐺𝑠 𝐷𝑒
𝜇
=
(310.88)(0.016526)
2.073 × 10−5
= 248383
Prandtl number, Pr
𝑃𝑟 =
𝜇𝐶𝑝
𝑘𝑓
=
(2.073 × 10−5
)(1.155)
0.03081
= 0.777
Selecting 25% for baffle cut
From figure 12.29, Chemical Engineering (Volume 6) Coulson and Richardson, the heat
transfer factor is:
Heat Transfer Factor, jh = 0.045
Shell side heat transfer coefficient, hs
hs= 19157.20 W/m2
.K
14.03/1
PrRe







we
hf
s
D
jk
h



29. Overall Heat Transfer Coefficient:
Outside fluid film coefficient, hs, W/m2
.o
C 3036139
Inside fluid film coefficient, hi, W/m2
.o
C 799.3244
Outside dirt coefficient (fouling factor), hod, W/m2
.o
C 6000
Inside dirt coefficient, hid, W/m2
.o
C (from Table 12.2, vol. Six ) 3000
Thermal conductivity of the tube wall material, kw, W/m.o
C 36
Tube inside diameter, Dti, m 0.0221
Tube outside diameter, Dto, m 0.0254
Overall heat transfer coefficient can be calculated by using the formula
So we get,
1/Uo = 0.002319
Uo = 492.153
Percentage difference = |(Ucal-Uass)/Uass|*100% = 64%
iti
to
idti
to
w
titoto
odso hd
d
hd
d
k
ddd
hhU
11
2
)/ln(111


30. Data/requisition sheet for
COOLER E-201
Equipment No.: E-201
OPERATING / MECHANICAL DATA
Type of heat exchanger Shell and Tube Heat Exchanger
Number of units required 1
Duty (W) 3435000
Heat transfer area (m2
) 85.33
Overall heat transfer coefficient
(W/m×°C)
492.153
Type of support Saddle
Insulation None
Parameters Shell Tube
Fluid Process Fluid Cooling Water
Material of construction Carbon steel Carbon Steel
Mass flow rate (kg/h) 77110 292600
Heat transfer coefficient (W/m2
.K) 19157.20 4848.07
Inlet temperature (operating) (°C) 256 30
Outlet temperature (operating) (°C) 120 40
Design temperature (°C) 200 35
Inlet pressure (operating) (kPa) 210 100
Design pressure (kPa) - 140
Bundle diameter (mm) - 743
Outer diameter (mm) 846.526 25.4
Inner diameter (mm) 830.0 22.1
Length (m) - 2.5
Number of tubes - 428
Tube per pass - 214
Total flow area (m2
) - 0.821
Pitch (mm) - 31.75 (triangular)
Baffle cut 0.25 -
Baffle spacing (mm) 415.0 -
Prepared
Checked
Approved
Date Eng, Process Rev By Appr. Date
Service Cool effluent from R-201
Equipment No. E-201
Project No.

31. PLANT LAYOUT:

32. SAFETY AND LOSS PREVENTION
HAZARD AND OPERABILITY STUDIES (HAZOP):
• HAZOP study is a formal procedure of hazard identification and
elimination procedures designed to identify the potential hazards,
safety issues and operability issues especially in process plant.

39. USD 15.7 mil / year
Invest USD 35.4 mil
Payback period 6 years, then
ROR 22%
ECONOMIC BALANCE(spreadsheet):

40. CONCLUSION:
• It has been proven conceptually that the establishment of new production
plant is feasible.
• The plant is to be operated in a continuous mode, which involves the
oxidation of n-butane to form maleic anhydride. The reactor selected for
this process is the fluidized bed reactor. The proposed plant design is
economically justified and is said to be viable for investment based on the
economic potential of the process.
. This project has been extremely helpful in cultivating and enhancing the
knowledge and skills at hand. As final year students ,the experience gained
through out the project has given the opportunity in designing a real
processing plant ,which has improved our understanding in the chemical
engineering field. Not only that ,other skills are also developed in the
process, which includes communication skills, management skills, and most
importantly, teamwork.

41. REFERENCES:
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Exposure Indices. American Conference of Governmental Industrial Hygienists, Cincinnati, OH.
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Netherlands: Taylor & Francis.
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5069687, December 1991.
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