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United Arab Emirates University College Of Engineering Chemical Engineering Department Graduation Project II

United Arab Emirates University College Of Engineering Chemical Engineering Department Graduation Project II. Advisor: Dr. Shaheen Al-Muhtaseb. Aisha Rashid 200002115 Alia Mohammed 199902239 Amani Ahmed 199903894 Sheikha Rashid 199902221. Second Semester 2004/2005.

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United Arab Emirates University College Of Engineering Chemical Engineering Department Graduation Project II

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  1. United Arab Emirates UniversityCollege Of EngineeringChemical Engineering DepartmentGraduation Project II Advisor: Dr. Shaheen Al-Muhtaseb Aisha Rashid 200002115 Alia Mohammed 199902239 Amani Ahmed 199903894 Sheikha Rashid 199902221 Second Semester 2004/2005

  2. Project Objectives Project Activities Main equipments design Material selection Safety and environmental Impact HAZOP study Economical feasibility study Recommendations

  3. Designing a chemical plant for Chlorinated Poly Vinyl Chloride (CPVC) production. Applying chemical engineering principles Investigating economics, environmental and safety factors

  4. Phase I of the Project Activity 1 Literature search Properties Applications Demand Processes

  5. Phase I of the Project Process selection Activity 2 Activity 3 Material & Energy Balances Activity 5 Preliminary Cost Estimation

  6. Second Semester Project Activities Activity 1 Design and sizing of individual equipments Activity 2 Material selection of individual equipment Activity 3 Safety and Environmental Impact Activity 4 Perform HAZOP studies on the plant Activity 5 Economical feasibility study

  7. Selected Process Feed Preparation Chlorination Reaction HCl Removal Section Drying Section Chlorine Preparation

  8. In the absorber:Chlorine gas is dissolved in the slurry Cl2 out Slurry in (PVC + H2O) D Slurry out (PVC + H2O+ Cl2) H Cl2 in Absorber AbsorberDesign Steps 1.Selection of equipment Sparged Vessel 2.Calculating vessel diameter (D) and height )H) Vessel Volume

  9. 3. Sparger Design Chlorine gas bubble diameter

  10. 30% PVC = 210 L/hr 70% Water : 2226.5 L/hr Calculating the vessel diameter and height Vapor Total Slurry Volume Slurry

  11. Total Volume of the tank (V) = Volume of slurry 0.7 V = 2.055 m3 Applying safety factor of 15% on the design V = (1.5) (2.055) = 3.082m3 Volume =πH D2/ 4 Assuming: Height (H) = 3 Diameter (D) D= 1.094 m H= 3D = 3.28m

  12. Pressure inside the vessel Thickness of the vessel

  13. Heat Exchanger Design • It is an equipment used to transfer heat to and from process fluids. • It is an essential part of most chemical processes.

  14. Selected Type of heat-Exchanger Agitated vessel • Fouling • Blockage of pipes Heating Slurry Types of Agitated vessels Heating Tank with immersed Coil JacketedVessel

  15. The agitated vessel was design to be: a baffled vessel with impeller Baffle:A flat vertical strips Propose of baffle: Have large top-to-bottom circulation without vortexing Propose of Impeller: To have a uniform temperature inside the vessel

  16. Design heating tank with immersed coil • Heat transfer area of the coil • Volume of tank • Diameter (D) and height (H) of tank • Height (h) and Diameter (d) of the • impeller

  17. Chlorination Reactor Design • Tubular Plug Flow Reactor (PFR) • Design Equation • Where: • V: the volume of the reactor,m3 • FAo: Flow rate of A component, mol/m3 • rA : Rate of reaction, mol/m3.s • x: conversion of reaction

  18. Chlorination Reactor Design • The volume of the reactor. • The length and the diameter. • The Area of the reactor. • The pressure drop. The design of a PFR aims to calculate:

  19. Chlorination Reactor Design 1) Calculating the Volume • Where: • V: Reactor Volume, m3 • : Volumetric flow rate,0.00083 m3/sec • t: residence time, 1800 sec V = 1.5 m3

  20. Chlorination Reactor Design 2) Calculating the Length (L) and the diameter (D)

  21. L= 7m Pt D= 0.3m Chlorination Reactor Design

  22. Chlorination Reactor Design 3) Calculating the Area (m2) • Where: • q: heat transfer rate = 505,288.5 kJ/hr (Calculated in GPI) • U: overall heat transfer coefficient = 500 W/m2.oC • A: heat transfer area, m2 • ∆Tlm: the algorithmic mean temperature difference, the temperature driving force = 19.5 oC A = 15 m2

  23. Chlorination Reactor Design 4) Calculating the pressure drop in the tube-side • where: • jf : is the dimensionless friction factor = 5.5*10-3 • L': effective pipe length = 7 m • d: the diameter of the pipe = 0.3 m • ut: velocity of fluid inside the pipe = 0.011 m/s • ρ : density of fluid = 1053 kg/m3 ∆P = 0.0066 Pa

  24. Chlorination Reactor Design 5) Determine the cooling water required : • Where: • q: heat transfer rate = 505,288.5 kJ/hr • m: cooling water mass flow rate, kg/hr • Cp: heat capacity of water = 4.2 kJ/kg.oC • Tc,in: inlet temperature in the shell side = 15 oC • Tc,out: exit temperature from the shell side = 45 oC mcw= 4,010 kg/hr

  25. Rotary Dryer Design

  26. Rotary Dryer Design Rotary Dryer Design Steps • Calculate the Volume • Calculate the Area & Diameter • Calculate the number of flight • Calculate the thickness of Dryer • Determine the thickness of insulation • Determine the power required to derive the dryer

  27. Rotary Dryer Design 1) Calculate the Volume • Where: • Qt = Total energy transferred, Btu/hr. • Ua = Volumetric heat transfer coefficient, Btu/hr.ft2. °C. • ∆Tm = Log mean temperature difference between hot gases and material,°C. • G = Air mass velocity, lbm/hr (ft2 of dryer cross section). • D = Dryer diameter, ft. • V = Volume of dryer, ft3.

  28. Rotary Dryer Design 2) Calculate the Area & Diameter

  29. Rotary Dryer Design • L/D = 8 (the range between 4-10)

  30. Rotary Dryer Design 3) Calculate the number of flight • Number of flights = 3*D • = 6.64 ≈ 7 • 7 flights are required using lip angle of 45°. • Radial height = = 0.276 ft

  31. Rotary Dryer Design 4) Calculate the thickness of Dryer e = 1.34 mm

  32. Rotary Dryer Design 5) Determine the thickness of insulation • Insulation material = Rock wool (K = 0.045 W/m.K) • Material of dryer = mild steel (K = 46.7 W/m.K) Heat lose from the dryer = 5% of Heat of the dryer Thickness of insulation = 2 mm

  33. Rotary Dryer Design 6) Determine the power required to derive the dryer BHP = 0.53 kw

  34. By comparing the corrosion resistance and the cost of each material Stainless steel (most frequently used corrosion resistant materials in chemical industry) Austenitic stainless steel 304: Is the most generally used stainless steel because: It minimizes the cost and provides the desired corrosion resistance.

  35. Cl2 +H2O = HCl Important parameter Corrosion receptivity Wet system Chlorine , Water Material used for liquid system with the risk of acid solution present • Different grades of Stainless steel • Silicon iron • Aluminum and it alloy • Nickel iron chromium By comparing the corrosion resistance and the cost of each material Stainless steel (austenitic 304: 18, Cr 12, Ni 25, Mo types 6(

  36. Material selection of individual equipment Wet basis Dry basis • Present of chlorine and water • No water and chlorine in the system • Risk of corrosion • No corrosion risk • Material used: • Stainless steel • Material used: Carbon steel

  37. No industry can achieve success without a strong commitment to protect the health and safety of its employees and society

  38. Summary of safety considerations for the used chemicals

  39. Toxic Chemicals • 1) NaOH (Sodium Hydroxide) • Strong alkali • Exposure limits in air is 2 mg/m3 • Affect the body by inhalation or contact with eyes and skin. • Inhalation may result in damage to the respiratory tract tissues.

  40. Toxic Chemicals 2) HCl (Hydrochloric acid) • Very strong acid • Exposure limits in air is 7 mg/m3 • Harmful if inhaled • Exposure to vapor causes sever burns in skin and eye damage. • 3) Cl2 (Chlorine) • Not flammable • Max. concentration in air should not exceed 1000 mg/m3 • Highly toxic by inhalation • Vapor Cl2 cause eye and lung injury

  41. Control the Toxic Chemicals Cl2: Use Vacuum • HCl & NaOH : • Special Storage and handling • Prevent any spills and pipe leak

  42. Hazard and Operability Studies Guide words: No or Not More Less As well as Part of Reverse Other than Deviations:The changes from the designer’s intention Causes:Reasons why? And how the deviations occur. Consequences and action:The results of the deviation and the actions must be done.

  43. Feed Preparation Section Chlorination Section Drying Section

  44. Product Centrifuge 90%CPVC 10% H2O 99.9%CPVC 0.1% H2O

  45. FC CC Product Centrifuge Rotary Dryer 90%CPVC 10% H2O CV9 Intention-CPVC inlet to rotary dryer Guide word No Deviation Flow • Causes: • Product centrifuge empty • Pipe leakage • Consequence & actions: • No Feed to dryer • Check & Repair supplying product centrifuge and pipes. • Emergency Interlock in dryer.

  46. FC CC Product Centrifuge Rotary Dryer 90%CPVC 10% H2O CV9 Intention-CPVC inlet to rotary dryer Guide word LESS Deviation Flow • Causes: • Supplying product centrifuge leakage • Pipe leakage • CV failure • Consequence & actions: • Effect on product specification • Check & Repair supplying product centrifuge and pipes.

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