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Supervised by: Prof. Mohamed Fahim Eng. Yusuf Ismail

Kuwait University College of Engineering & Petroleum Depatment of Chemical Engineering. Propylene Oxide Production by Chlorohydrins Process (Cell- Liqure ). Done by Abdulrahman Habib. Supervised by: Prof. Mohamed Fahim Eng. Yusuf Ismail. Agenda. Distillation column design ( 2 )

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Supervised by: Prof. Mohamed Fahim Eng. Yusuf Ismail

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  1. Kuwait University College of Engineering & Petroleum Depatment of Chemical Engineering Propylene Oxide Production by Chlorohydrins Process (Cell-Liqure) Doneby AbdulrahmanHabib Supervised by: Prof. Mohamed Fahim Eng. Yusuf Ismail

  2. Agenda • Distillation column design ( 2 ) • Packed column ( 2 ) • Heat exchanger design ( 2 Cooler )

  3. Distillation Column Design Objective To separate PO desired product from by-products (PDC DCIPE). • Assumptions • Tray spacing= 0.6 m • Percent of flooding at maximum flow rate=85% • Percent of downcomer area of total area=12% • The hole area =10% the active area. • weir height=50 mm • Hole diameter=5 mm • Plate thickness=5 mm

  4. Main Design Procedures • Specify the properties of outlets streams for both vapor and liquid from HYSYS. • Column Diameter • Where • FLv: liquid-vapor flow factor • Lw: liquid flow rate, kmol/hr • ρL: liquid density,kg/m3 • Vw: vapor flow rate, kmol/hr • ρv :vapor density, kg/m3

  5. Get k1 for both bottom and top from figure 11.27 then use correction factor K = (σ / 0.02) ^0.2 * K1 Where: σ = liquid surface tension in N/m • calculate the flooding velocity for top and bottom Uf = K *( (ρl –ρV) / ρ v)½ Where: Uf = flooding vapour velocity in m/s K= Surface tention correction factor ρl = density of liquid in kg / m³ ρv = density of vapour in kg / m³

  6. Assume the flooding percentage is 85% at max flow rate for the top and the bottom UV = 0.85 * Uf • calculate the net area for the top and the bottom An = V / UV Where: An = net area in m² V = Volumetric flow rate in m³ / s UV = vapour velocity in m/s • Assume as first trail take down comer as 12% of total cross sectional area Ac = An /(1- 0.12) Where: Ac = cross sectional area in m²

  7. Calculate the diameter for the top and the bottom D = ((4 /π) * Ac) ½ • Calculate the column height using the actual number of stage H= (Tray spacing * actual NO. stage ) + D Aa = Ac – 2Ad Ah = 0.1 * Aa Where: Aa = active area in m² Ah = hole area in m²

  8. Check Weeping Where: max Lw: maximum liquid rate, (kg/s). min Lw : minimum liquid rate, (kg/s). max how: mm liquid. min how : mm liquid. • Calculate the actual vapor velocity Calculate the actual vapour velocity = min vapour rate / Ah Uh(min)=[K2-0.90(25.4-dh)]/g0.5

  9. Calculate Pressure Drop: Ht = HD + HW + HOW + HR HD = 51 * (Uh/ C0)² * ρ V / ρL Hr = 12.5E3 / ρL Where: Hd = dry plate drop Uh = min vapour velocity in m/s Hr = residual head Ht = total pressure drop in mm

  10. Downcomer backup

  11. Calculate the residence time TR = (Ad * hb * ρ l) / lwd • Calculate the flooding percentage Flooding percentage = UV / uf * 100 • Calculate the area of the hole A = (3.14 / 4 ) * (dh * 0.001 )² • Calculate number of hole Number of hole = A h / A

  12. Calculate the thickness Where: t: thickness of the separator in (in) P: operating pressure in Pisa ri: radius of the separator in (in) S: is the stress value of carbon steel = 13700 Pisa Ej: joint efficiency (Ej=0.85 for spot examined welding) C0: corrosion allowance = 0.125

  13. Result

  14. Reactor Design (Packed Column) Objective To produce PCH from react C3H6 + Cl2 + H2O PCH + HCl • Assumptions • " 3/4 in " Berl Saddles • Ej = 0.85 • Cc = 0.125 in • Percent of flooding at maximum flow rate=90%

  15. Main design procedures First: Calculate Diameter (D) • Determine VG Calculate (∆P = 2 ;Fp= 175 since ¾ Bearl Saddle) Then new capacity parameter is known (from figure 10.6-5) • Determine the mass ratio

  16. Determine GG at 90% Flooding: • GG = 0.9 * VG * ρG(Ib/s.ft2) • Diameter (D): • Area = Feed Gas x (1/Gg) • Diameter = ( Area * π/4 )0.5 ft Second: Calculate Height (HETP) • Determine Gx & Gy • Gy = FG / Area (Ib/hr.ft2) • Gx = FL / Area (Ib/hr.ft2)

  17. Determine the HG & HL HL = HG = • Determine NOG (Y-Y*)m= NOG=

  18. Calculate the KYA • Kya = • Kxa = 1/KYA = • Determine the Height (HETP) Method # 1 Method # 2 HOG = HOG = Height = HOG x NOG

  19. Calculate Thickness (T): Where: t: thickness of the separator in (in) P: operating pressure in Pisa ri: radius of the separator in (in) S: is the stress value of carbon steel = 13700 Pisa Ej: joint efficiency (Ej=0.85 for spot examined welding) C0: corrosion allowance = 0.125

  20. Result

  21. Heat Exchanger Design Objective of ( E-100 ) To decrease the temperature of the stream leaving the reactor and prepare it before interring the next reactor.

  22. Assumptions • Using two shell pass and four or multiple of four tube passes. • Assume the outer, the inner diameter and the length of the tube. • The value of the overall heat transfer coefficient was assumed to be • For = 750 w/m2C.

  23. Main design procedures • Heatload ,(kW) • Q = (m Cp ΔT)hot =(m Cp ΔT)cold • Log mean Temperature, (˚C) ΔT1= Thi-Tco ΔT2= Tho-Tci Where, Thi: inlet hot stream temperature (˚C) Tho: outlet stream temperature (˚C) Tci: inlet cold stream temperature (˚C) Tco: outlet cold temperature

  24. Provisional Area, (m2) • Where: • ΔTm= FtΔTlm • Area of one tube = Lt * do *p • Where: • Outer diameter (do), (mm) • Length of tube (Lt), (mm) • Number of tubes • Nt= provisional area / area of one tube

  25. Bundle diameter • Db = do( Nt / K1) (1/n1) ,mm • Where : • Db: bundle diameter ,mm • Nt : number of tubes • K1 , n1 : constants from table (12.4) using triangular pitch of 1.25 • Shell diameter • Ds = Db + (Db Clearance) ,mm • Where : • we get it from figure (12.10) using split ring floating heat type.

  26. Tube side Coefficient • (hi di / κ) = jh Re Pr0.33 * (µ/µwall)0.14 • Shell side Coefficient • hs = κ * jh *Re *Pr (1/3) / de

  27. Overall heat transfer coefficient • 1/Uo =1/ho + 1/hod + do(ln(do/di))/2kw + do/di * 1/hid + do/di * 1/hi • Where : • Uo : overall coefficient based on outside area of the tube ,w/m^2.C • ho : outside fluid film coefficient, w/m^2.C • hi : inside fluid film coefficient ,w/m^2 • hod : outside dirt coefficient (fouling factor) ,w/m^2.C, from Table (12.2) • hid : inside dirt coefficient (fouling factor),w/m^2.C from Table (12.2) • kw : thermal conductivity of the wall material w/m.Cs for cupronickel • di : tube inside diameter m • do : tube outside diameter m

  28. Pressure drop • Tube side: • ΔP = Np [ 8jf (L/di)(µ/µw)^(-m) +2.5 ] ρυ^2/2,kpa • Where : • ΔP : tube side pressure drop, N/m^2(pa) • Np : number of tube side passes • υ : tube side velocity ,m/s • L : length of one tube , m • jf : tube side friction factor Shell side: ΔPs = 8jf (Ds/de)(L/lb)( ρυ^2/2)(µ/µw)^(-0.14),kpa Where: L : tube length ,m lb : baffle spacing ,m

  29. Shell thickness: • t = (P ri/ S E - 0.6P) + Cc • Where : • t : shell thickness, in • P : internal pressure, psi gage • r i : internal radius of shell, in • E : efficiency of joints • S : working stress, psi (for carbon steel) • Cc : allowance for corrosion, in

  30. Results

  31. Thanks' For Lisiting

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