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MANUFACTURING OF
PHOSPHORIC ACID
Presented By
Sagar Mahajan Aniket Mali
Exam No. B120215939 Exam No. B120215940
BE Chemical
Under Guidance of
Prof. H L Kamble
Department of Chemical Engineering
AISSMS College of Engineering, Pune-01
1
1. Introduction
2. Literature Survey
3. Selection of Process
4. Process Description
5. Material Balance
6. Energy Balance
7. Equipment Design
8. Cost Estimation
9. Plant Layout
10. Safety
11. Conclusion
12. References
2
 In this project, we are going to analyze the production of
phosphoric acid by wet process
Phosphoric Acid is made from Phosphate Rock
Figure 1: Structure 3
Food-grade phosphoric acid is used to acidify foods
and beverages such as various colas
Teeth whiteners to eliminate plaque.
As a chemical oxidizing agent for activated carbon
production
As a cleaner by construction trades to remove mineral
deposits, cementations smears, and hard water stains
As a pH adjuster in cosmetics and skin-care products
As a dispersing agent in detergents and leather
treatment
4
5
Sr .
No.
Patent Title Author Patent No Process Raw Material Parameters
1. Process of
manufacturing
phosphoric
acid
Casimer c
Legal,Jr
Pasadena
et al
US2504544
Wet
Process
Sulphuric acid &
phosphate rock
P2O5=33%
Cao=45.79%
Composition:-
Fluorine=3.66%
Moisture=0.70%
Conversion = 98%
2. Method of
manufacturing
wet process
phosphoric
acid
Asalchi
Matsubar
a,Yoshito
Yasutake
US3416887 Wet
Process
Phosphate rock
H2SO4
P2O5=40%
So3=2-2.5%
H2SO4=15-50%
Table 1: Literature Survey
6
Sr .
No.
Paper Title Author Paper no Process Raw Material Parameters
3. Method of
preparing wet
process
phosphate acid
Feng
et al
US7172742
B2
Wet
Process
Decomposing
phosphate rock
in sulphuric
acid
Liq-solid ratio-2.3-2.7
SO3 -0.09g/L
H3PO4-33-39wt%
P2O5-30-35%
Production=80%
4. Phosphoric
Acid
(Dryden’s
Outline of
Chemical
Technology,
3rd Ed.)
M. Gopal
Rao
Page No
150-153
Wet
Process
Phosphate Rock
H2SO4
Phosphate Rock- 2.5 T
H2SO4 – 2.0T
CaSO4 – 2.7 T
Phosphoric Acid:
• Molecular formula : H3PO4
• Molecular weight : 98 gm/mole
• Melting point : 42.4ºC
• Boiling point : 213ºC
• pH: 1.5 (0.1 N aq. sol)
7
Different process are needed because of different
rock and gypsum disposal systems
Two general types of processes are used
Wet Process
Thermal process
8
The processes that use phosphated minerals which
are decomposed with an acid, are known as ‘Wet
Process’
There are 3 Types of Wet Processes:
Nitric
Hydrochloric
Sulphuric
The process using sulphuric acid is most common
and particularly used for fertilizer grade phosphoric
acid
9
Main Reaction:
3CaO + P2O5 Ca3(PO4)2
Ca3(PO4)2 + 3H2SO4 3CaSO4 + 2H3PO4
10
• SO3 + H2O H2SO4
• 2CaO + 2F2 2CaF2 + O2
• CaF2 + H2SO4 + 2H2O 2HF + CaSO4.2H2O
• 6HF + SiO2 H2SiF6 + 2H2O
• CaO + H2O Ca(OH)2
• Ca(OH)2 + H2SO4 CaSO4.2H2O
• Al2O3 + 3H2SO4 Al2(SO4)3 + 3H2O
• Fe2O3 + 3H2SO4 Fe2(SO4)3 + 3H2O
• CO2 + H2O H2CO3
11
Figure 2: PFD 12
Phosphate Rock: Ground to 200Mesh
Reaction Time: 4 – 6 Hr
98% conversion
By-Product: Gypsum(CaSO4.2H2O)
13
14
1000 kg/day of Phosphoric Acid
Batch of 8hr
Slurry Feed ratio 1:2.5
Excess H2SO4 =1.4
98% Ca3(PO4)2 conversion
15
COMPOUND CONTRIBUTION (%)
P2O5 32
CaO 49
SiO2 5
F 4
Al2O3 2
Fe2O3 2
CO2 1
SO3 2
Moisture 1
16
% by mass
Table 2: Rock
1
2
3
4 Vent
720.05 Kg Rock
+
1925.13 Kg Water
734.67 Kg H2SO4
Table 3: Material Balance for Reactor
To
Filter
17
3 5
6
Liquid to Sedimentation Tank
Gypsum
+
Waste
Assuming 85%
Filter efficiency
From
Reactor
Table 4: Material Balance for Filter
18
6
7
To
Evaporator
Table 5: Material Balance for Sedimentation Tank
From
Filter
19
8
7
10
Water Vapours
To
Evaporator II
From
Sedimentation Tank
Table 6: Material Balance for Evaporator I
20
11
STEAM
9
483.29 Kg
483.29 Kg
9
13
From Evaporator I
Table 7: Material Balance for Evaporator I
21
14
10
12
483.29 Kg
483.29 Kg
Water Vapours
To
Evaporator III
12
16
Water Vapours
To
Condenser
From Evaporator II
Table 8: Material Balance for Evaporator I
22
17
13
15
444.47 Kg
444.47 Kg
23
Table 9: Properties of Components
24
Ca3(PO4)2 + 3H2SO4 + H2O 3CaSO4.2H2O+ 2H3PO4
Formula ∶ ∆𝐺𝑓𝑅 = ∑(𝑝𝑖 × ∆𝐺𝑓𝑖) 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 − ∑(𝑟𝑖 × ∆𝐺𝑓𝑖) 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡
where ∆𝐺𝑓𝑖 = Gibbs free energy of ith component
𝑝𝑖 = Stoichiometry of Product
• ∆𝐺𝑓𝑅 = 3 × −1797.44 + 2 × −1111.68 − [ 1 × −3884.84 +
3 × −690.06 + (6 × −237.18)]
• ∆𝑮 𝒇𝑹 = -237.58KJ/Kmol
• Reaction is feasible
25
Base Temperature : 298 K
26
Table 10: Total Heat of Reaction for 1 Batch
∴ −∆𝐻 𝑅 = 3448.526 KJ
27
1
2
3
4 Vent
To
Filter
298 K
343 K
298 K
Table 11: Total Heat of Reaction for 1 Batch
Heat to be Added by Jacket = 354055.907 KJ
Steam of 105oC is used at 0.64 Kg/min-batch
28
3 5
6
Liquid to Sedimentation Tank
Gypsum
+
Waste
Assuming 5%
Energy Loss
From
Reactor
343 K
339.5 K
339.5 K
Table 12: Energy Balance for Filter
6
7
To
Pre-Heater
Table 13: Energy Balance for Sedimentation Tank
From
Filter
29
8
339.5 K
337.5 K
337.5 K
Assuming 3%
Energy Loss
First Effect: Wsʎs + WF(tf-t1) = W1ʎ1
Second Effect: W1ʎ1 + (WF-W1)(t1-t2) = W2ʎ2
Second Effect: W2ʎ2 + (WF-W1-W2)(t2-t3) = W3ʎ3
WF-W1-W2-W3 = WP
Steam Supplied: 483.2913 Kg
Vapours Out: 1329.191 Kg
Average Heat Transfer Area Required = 36m2
Steam Economy: 2.75
Table 14: Energy Balance for MEE
30
31
32
We calculated the total volume of input material
Volume of reactor is taken in 10% excess
Diameter, Height and Thickness of reactor
assuming
𝐿
𝐷
=1.5
Various stability checks
Total weight of the reactor
33
Volume of cylinder = 𝜋𝑅2 𝐻
Weight = 𝜋𝐷𝐻𝑡𝑝
Tangential Stress = 𝑓𝑡 =
𝑃 (𝐷 𝑖+𝑡)
2𝑡
Stress due to Internal Pressure = 𝑓1 =
𝑃×𝐷 𝑖
4𝑡
Stress due to Weight = 𝑓2 =
𝑊
𝜋(𝐷 𝑖+𝑡)×𝑡
𝑓𝑎 = 𝑓1 + 𝑓2
𝑓𝑟 = [𝑓1
2
− 𝑓1 𝑓𝑎 + 𝑓𝑎
2
+ 3𝑓2
2
]0.5
34
Table 15: Design of Reactor
35
Calculated the total volume of input material
Volume of tank is taken in 20% excess
Calculations same as for reactor
Diameter, Height and Thickness of tank assuming
𝐿
𝐷
=1.5
36
Table 16: Design of Sludge Separator
We calculated number of tubes required.
Pitch of tube: 75mm (Triangular)
𝐴 =
𝑁×0.866×𝑆 𝑇
2
𝛽
𝛽 = 0.9
Area of Central down-take = 40% of CSA (Tubes)
Diameter of Tube Sheet
Thickness of Calendria
Thickness of Tube Sheet
𝐾 =
𝐸 𝑆×𝑡 𝑆 𝐷 𝑜−𝑡 𝑆
𝐸𝑡×𝑁𝑡×𝑡 𝑡(𝐷𝑡−𝑡 𝑡)
𝐹 =
𝐾
2+3(𝐾)
𝑡𝑡𝑠 = 𝐹 × 𝐷 𝑜 ×
0.25×𝑃
𝑓
37
Area of Drum
𝑅 𝑑 =
𝑉
𝐴
0.0172×
𝜌 𝑙−𝜌 𝑣
𝜌 𝑣
Rd =0.8
Thickness of Vapour Space.
Design for all 3 evaporators will remain same
As the heat transfer area required is equal.
38
Table 17: Tubes and Calendria Specifications
39
40
Table 18: Vapour Space and Head Specifications
41
Figure 2: CFD – Geometry of Evaporator
42
Table 20: Equipment Cost
Final Equipment Cost is 30% excess for supports, pumps, etc.
Therefore, FCE = 1.3 × 2858522.5 = 3716079.3
43
Table 19: Material Cost
Fixed Capital (FCI)
Working Capital (WCI)
TCI = FCI + WCI
= 24773862 +17279693 = ₹ 4,20,53,555
44
Manufacturing Cost
General Expenses
TPC = Manufacturing Cost + General Expenses
= 24773862 +17279693 = ₹ 1,50,93,453
45
Selling Price of H3PO4(SP)= ₹ 80
Operating Time = 330 Days/Year
Capacity of Plant= 1000 Kg/Day
Taxes = 30% of GP
Income = SP × Capacity × Cycles
= 80 × 1000 × 330 = ₹ 2,64,00,000
Gross Profit (GP)= Income – TPC = ₹ 1,13,06,547
Net Profit = GP – Taxes = ₹ 79,14,583
46
Rate of Return:
𝑟 =
Net Profit
Total Capital Investment
× 10
=
7914583
42053555.8
× 100 = 𝟏𝟖. 𝟖𝟐%
Pay-out Period:
T =
Total Capital Investment
Net Profit + Depreciation
= 5.08 𝑌𝑒𝑎𝑟
47
48
After the process flow diagrams are completed and before
detailed piping, structural, and electrical design can begin,
the layout of process units in a plant must be planned
 This layout can play an important part in determining
construction and manufacturing costs
Must be planned carefully with attention being given to
future problems that may arise
49
Operational convenience and accessibility
Economic distribution of utilities and services
Type of buildings and building-code requirements
Health and safety considerations
Waste-disposal requirements
Auxiliary equipment
Space available and space required
Roads and railroads
Possible future expansion
The principal factors to be considered are :-
50
51
Figure 3: Plant Layout
52
Keeping the number of incidences and accidents zero
Having a robust and dynamic safety program
Major Hazard is Fire
Using inherently safe equipment
Carrying out independent audit from competent organizations
Abide by all the Govt. laws and hold sacred all the engineering ethics
Regular training and drill of the employees and workers
Creating safety policies
Developing and monitoring safety programs
53
• Wet process was selected for the production of
Salicylic Acid.
• Production capacity was selected as 1 Ton/Day
after studying supply and demand data.
• From analysing the residence time of the process,
three batches per day were selected.
• Energy balance was done for entire process.
• Design of the equipments was done.
• Costing of equipments was done.
• Plant layout is drawn.
54
55
 R K Sinnott , “Chemical Engineering Design,” 4th ed.,
Elsevier Butterworth-Heinemann (2005)
 J P Holman, “Heat Transfer”, 6th ed., McGraw Hill Book
Company (1986)
 Donald Q Kern, “Process Heat Transfer,” McGraw Hill
Book Company (1988)
 Robert H. Perry, “Perry's Chemical Engineers' Handbook,”
8th ed., McGraw Hill (1934)
 V. V. Mahajani and S. B. Umarji, “Joshi’s Process
Equipment Design”, 5th edition, Trinity Publications.
56

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B E Project - Manufacturing of Phosphoric Acid

  • 1. MANUFACTURING OF PHOSPHORIC ACID Presented By Sagar Mahajan Aniket Mali Exam No. B120215939 Exam No. B120215940 BE Chemical Under Guidance of Prof. H L Kamble Department of Chemical Engineering AISSMS College of Engineering, Pune-01 1
  • 2. 1. Introduction 2. Literature Survey 3. Selection of Process 4. Process Description 5. Material Balance 6. Energy Balance 7. Equipment Design 8. Cost Estimation 9. Plant Layout 10. Safety 11. Conclusion 12. References 2
  • 3.  In this project, we are going to analyze the production of phosphoric acid by wet process Phosphoric Acid is made from Phosphate Rock Figure 1: Structure 3
  • 4. Food-grade phosphoric acid is used to acidify foods and beverages such as various colas Teeth whiteners to eliminate plaque. As a chemical oxidizing agent for activated carbon production As a cleaner by construction trades to remove mineral deposits, cementations smears, and hard water stains As a pH adjuster in cosmetics and skin-care products As a dispersing agent in detergents and leather treatment 4
  • 5. 5 Sr . No. Patent Title Author Patent No Process Raw Material Parameters 1. Process of manufacturing phosphoric acid Casimer c Legal,Jr Pasadena et al US2504544 Wet Process Sulphuric acid & phosphate rock P2O5=33% Cao=45.79% Composition:- Fluorine=3.66% Moisture=0.70% Conversion = 98% 2. Method of manufacturing wet process phosphoric acid Asalchi Matsubar a,Yoshito Yasutake US3416887 Wet Process Phosphate rock H2SO4 P2O5=40% So3=2-2.5% H2SO4=15-50% Table 1: Literature Survey
  • 6. 6 Sr . No. Paper Title Author Paper no Process Raw Material Parameters 3. Method of preparing wet process phosphate acid Feng et al US7172742 B2 Wet Process Decomposing phosphate rock in sulphuric acid Liq-solid ratio-2.3-2.7 SO3 -0.09g/L H3PO4-33-39wt% P2O5-30-35% Production=80% 4. Phosphoric Acid (Dryden’s Outline of Chemical Technology, 3rd Ed.) M. Gopal Rao Page No 150-153 Wet Process Phosphate Rock H2SO4 Phosphate Rock- 2.5 T H2SO4 – 2.0T CaSO4 – 2.7 T
  • 7. Phosphoric Acid: • Molecular formula : H3PO4 • Molecular weight : 98 gm/mole • Melting point : 42.4ºC • Boiling point : 213ºC • pH: 1.5 (0.1 N aq. sol) 7
  • 8. Different process are needed because of different rock and gypsum disposal systems Two general types of processes are used Wet Process Thermal process 8
  • 9. The processes that use phosphated minerals which are decomposed with an acid, are known as ‘Wet Process’ There are 3 Types of Wet Processes: Nitric Hydrochloric Sulphuric The process using sulphuric acid is most common and particularly used for fertilizer grade phosphoric acid 9
  • 10. Main Reaction: 3CaO + P2O5 Ca3(PO4)2 Ca3(PO4)2 + 3H2SO4 3CaSO4 + 2H3PO4 10
  • 11. • SO3 + H2O H2SO4 • 2CaO + 2F2 2CaF2 + O2 • CaF2 + H2SO4 + 2H2O 2HF + CaSO4.2H2O • 6HF + SiO2 H2SiF6 + 2H2O • CaO + H2O Ca(OH)2 • Ca(OH)2 + H2SO4 CaSO4.2H2O • Al2O3 + 3H2SO4 Al2(SO4)3 + 3H2O • Fe2O3 + 3H2SO4 Fe2(SO4)3 + 3H2O • CO2 + H2O H2CO3 11
  • 13. Phosphate Rock: Ground to 200Mesh Reaction Time: 4 – 6 Hr 98% conversion By-Product: Gypsum(CaSO4.2H2O) 13
  • 14. 14
  • 15. 1000 kg/day of Phosphoric Acid Batch of 8hr Slurry Feed ratio 1:2.5 Excess H2SO4 =1.4 98% Ca3(PO4)2 conversion 15
  • 16. COMPOUND CONTRIBUTION (%) P2O5 32 CaO 49 SiO2 5 F 4 Al2O3 2 Fe2O3 2 CO2 1 SO3 2 Moisture 1 16 % by mass Table 2: Rock
  • 17. 1 2 3 4 Vent 720.05 Kg Rock + 1925.13 Kg Water 734.67 Kg H2SO4 Table 3: Material Balance for Reactor To Filter 17
  • 18. 3 5 6 Liquid to Sedimentation Tank Gypsum + Waste Assuming 85% Filter efficiency From Reactor Table 4: Material Balance for Filter 18
  • 19. 6 7 To Evaporator Table 5: Material Balance for Sedimentation Tank From Filter 19 8
  • 20. 7 10 Water Vapours To Evaporator II From Sedimentation Tank Table 6: Material Balance for Evaporator I 20 11 STEAM 9 483.29 Kg 483.29 Kg
  • 21. 9 13 From Evaporator I Table 7: Material Balance for Evaporator I 21 14 10 12 483.29 Kg 483.29 Kg Water Vapours To Evaporator III
  • 22. 12 16 Water Vapours To Condenser From Evaporator II Table 8: Material Balance for Evaporator I 22 17 13 15 444.47 Kg 444.47 Kg
  • 23. 23 Table 9: Properties of Components
  • 24. 24 Ca3(PO4)2 + 3H2SO4 + H2O 3CaSO4.2H2O+ 2H3PO4 Formula ∶ ∆𝐺𝑓𝑅 = ∑(𝑝𝑖 × ∆𝐺𝑓𝑖) 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 − ∑(𝑟𝑖 × ∆𝐺𝑓𝑖) 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 where ∆𝐺𝑓𝑖 = Gibbs free energy of ith component 𝑝𝑖 = Stoichiometry of Product • ∆𝐺𝑓𝑅 = 3 × −1797.44 + 2 × −1111.68 − [ 1 × −3884.84 + 3 × −690.06 + (6 × −237.18)] • ∆𝑮 𝒇𝑹 = -237.58KJ/Kmol • Reaction is feasible
  • 26. 26 Table 10: Total Heat of Reaction for 1 Batch ∴ −∆𝐻 𝑅 = 3448.526 KJ
  • 27. 27 1 2 3 4 Vent To Filter 298 K 343 K 298 K Table 11: Total Heat of Reaction for 1 Batch Heat to be Added by Jacket = 354055.907 KJ Steam of 105oC is used at 0.64 Kg/min-batch
  • 28. 28 3 5 6 Liquid to Sedimentation Tank Gypsum + Waste Assuming 5% Energy Loss From Reactor 343 K 339.5 K 339.5 K Table 12: Energy Balance for Filter
  • 29. 6 7 To Pre-Heater Table 13: Energy Balance for Sedimentation Tank From Filter 29 8 339.5 K 337.5 K 337.5 K Assuming 3% Energy Loss
  • 30. First Effect: Wsʎs + WF(tf-t1) = W1ʎ1 Second Effect: W1ʎ1 + (WF-W1)(t1-t2) = W2ʎ2 Second Effect: W2ʎ2 + (WF-W1-W2)(t2-t3) = W3ʎ3 WF-W1-W2-W3 = WP Steam Supplied: 483.2913 Kg Vapours Out: 1329.191 Kg Average Heat Transfer Area Required = 36m2 Steam Economy: 2.75 Table 14: Energy Balance for MEE 30
  • 31. 31
  • 32. 32 We calculated the total volume of input material Volume of reactor is taken in 10% excess Diameter, Height and Thickness of reactor assuming 𝐿 𝐷 =1.5 Various stability checks Total weight of the reactor
  • 33. 33 Volume of cylinder = 𝜋𝑅2 𝐻 Weight = 𝜋𝐷𝐻𝑡𝑝 Tangential Stress = 𝑓𝑡 = 𝑃 (𝐷 𝑖+𝑡) 2𝑡 Stress due to Internal Pressure = 𝑓1 = 𝑃×𝐷 𝑖 4𝑡 Stress due to Weight = 𝑓2 = 𝑊 𝜋(𝐷 𝑖+𝑡)×𝑡 𝑓𝑎 = 𝑓1 + 𝑓2 𝑓𝑟 = [𝑓1 2 − 𝑓1 𝑓𝑎 + 𝑓𝑎 2 + 3𝑓2 2 ]0.5
  • 34. 34 Table 15: Design of Reactor
  • 35. 35 Calculated the total volume of input material Volume of tank is taken in 20% excess Calculations same as for reactor Diameter, Height and Thickness of tank assuming 𝐿 𝐷 =1.5
  • 36. 36 Table 16: Design of Sludge Separator
  • 37. We calculated number of tubes required. Pitch of tube: 75mm (Triangular) 𝐴 = 𝑁×0.866×𝑆 𝑇 2 𝛽 𝛽 = 0.9 Area of Central down-take = 40% of CSA (Tubes) Diameter of Tube Sheet Thickness of Calendria Thickness of Tube Sheet 𝐾 = 𝐸 𝑆×𝑡 𝑆 𝐷 𝑜−𝑡 𝑆 𝐸𝑡×𝑁𝑡×𝑡 𝑡(𝐷𝑡−𝑡 𝑡) 𝐹 = 𝐾 2+3(𝐾) 𝑡𝑡𝑠 = 𝐹 × 𝐷 𝑜 × 0.25×𝑃 𝑓 37
  • 38. Area of Drum 𝑅 𝑑 = 𝑉 𝐴 0.0172× 𝜌 𝑙−𝜌 𝑣 𝜌 𝑣 Rd =0.8 Thickness of Vapour Space. Design for all 3 evaporators will remain same As the heat transfer area required is equal. 38
  • 39. Table 17: Tubes and Calendria Specifications 39
  • 40. 40 Table 18: Vapour Space and Head Specifications
  • 41. 41 Figure 2: CFD – Geometry of Evaporator
  • 42. 42
  • 43. Table 20: Equipment Cost Final Equipment Cost is 30% excess for supports, pumps, etc. Therefore, FCE = 1.3 × 2858522.5 = 3716079.3 43 Table 19: Material Cost
  • 44. Fixed Capital (FCI) Working Capital (WCI) TCI = FCI + WCI = 24773862 +17279693 = ₹ 4,20,53,555 44
  • 45. Manufacturing Cost General Expenses TPC = Manufacturing Cost + General Expenses = 24773862 +17279693 = ₹ 1,50,93,453 45
  • 46. Selling Price of H3PO4(SP)= ₹ 80 Operating Time = 330 Days/Year Capacity of Plant= 1000 Kg/Day Taxes = 30% of GP Income = SP × Capacity × Cycles = 80 × 1000 × 330 = ₹ 2,64,00,000 Gross Profit (GP)= Income – TPC = ₹ 1,13,06,547 Net Profit = GP – Taxes = ₹ 79,14,583 46
  • 47. Rate of Return: 𝑟 = Net Profit Total Capital Investment × 10 = 7914583 42053555.8 × 100 = 𝟏𝟖. 𝟖𝟐% Pay-out Period: T = Total Capital Investment Net Profit + Depreciation = 5.08 𝑌𝑒𝑎𝑟 47
  • 48. 48
  • 49. After the process flow diagrams are completed and before detailed piping, structural, and electrical design can begin, the layout of process units in a plant must be planned  This layout can play an important part in determining construction and manufacturing costs Must be planned carefully with attention being given to future problems that may arise 49
  • 50. Operational convenience and accessibility Economic distribution of utilities and services Type of buildings and building-code requirements Health and safety considerations Waste-disposal requirements Auxiliary equipment Space available and space required Roads and railroads Possible future expansion The principal factors to be considered are :- 50
  • 52. 52
  • 53. Keeping the number of incidences and accidents zero Having a robust and dynamic safety program Major Hazard is Fire Using inherently safe equipment Carrying out independent audit from competent organizations Abide by all the Govt. laws and hold sacred all the engineering ethics Regular training and drill of the employees and workers Creating safety policies Developing and monitoring safety programs 53
  • 54. • Wet process was selected for the production of Salicylic Acid. • Production capacity was selected as 1 Ton/Day after studying supply and demand data. • From analysing the residence time of the process, three batches per day were selected. • Energy balance was done for entire process. • Design of the equipments was done. • Costing of equipments was done. • Plant layout is drawn. 54
  • 55. 55  R K Sinnott , “Chemical Engineering Design,” 4th ed., Elsevier Butterworth-Heinemann (2005)  J P Holman, “Heat Transfer”, 6th ed., McGraw Hill Book Company (1986)  Donald Q Kern, “Process Heat Transfer,” McGraw Hill Book Company (1988)  Robert H. Perry, “Perry's Chemical Engineers' Handbook,” 8th ed., McGraw Hill (1934)  V. V. Mahajani and S. B. Umarji, “Joshi’s Process Equipment Design”, 5th edition, Trinity Publications.
  • 56. 56

Editor's Notes

  1. Change gibbs free energy