1. Designer: Bader Al-Rashedi Supervised by: Prof.M.Fahim ENG: Yousif Ismail

2. Table of content 1- Heat exchanger design (cooler ). 2- Distillation column design. 3- Reactor design. 4- Pump design. 5- compressor design.

3. Introduction A heat exchanger is a device built for efficient heat transfer from one medium to another, whether the media are separated by a solid wall so that they never mix, or the media are in direct contact.

4. Heat Exchanger Design (cooler) Objective (E-103 ):Help to recycle EB back to the start of process. (E-106 ):Decrease Temperature of EB recycled to the start of process. (E-115 ):Decrease Temperature of stream fed to CRV-101-2. (E-101 ):Decrease Temperature of vent gas

5. Assumptions 1-Assume a counter current flow heat exchanger because it provides more effective heat transfer. 2-The value of the overall heat transfer coefficient was assumed in the shell and tube heat exchanger design. 3-Assume the outer, the inner diameter and the length of the tube on both the Shell and tube heat exchanger design.

6. Design Procedures (A) Shell & Tube heat exchanger 1)Calculate Heat Load Heat Load= m*cp*(T2-T1) Where: m= mass flow rate ( Kg/h) cp= Heat capacity ( KJ/kg°C) T2= inlet temperature of shell side°C T1= outlet temperature of shell side°C

7. 2) Calculate Log mean Temperature difference ∆Tlm= ((T1-t2) – (T2-t1))/ln((T1-t2)/ (T2-t1)) where: ∆Tlm: Log mean temperature difference°C. T1 : Inlet shell side fluid temperature°C T2 : Outlet shell side fluid temperature °C t1 : Inlet tube side temperature°C t2: Outlet tube side temperature°C

8. 3) Calculate the two dimensionless temperature ratios: R= (T1-T2)/(t2-t1) S = (t2-t1)/(T1-t1) Where: T1 : Inlet shell side fluid temperature T2 : Outlet shell side fluid temperature t1 : Inlet tube side temperature t2: Outlet tube side temperature

10. 4) From Figure determine Ft by using R and S ratios Where : Ft : temperature correction factor. 5) Calculate True temperature difference ∆Tm= Ft *∆Tlm 6) Choose U from table depending on the type of flows in shell and tube sides. Where: U: Over all heat transfer coefficient.

11. 7) Calculate provisional area 8) Assume inlet tube diameter, outlet tube diameter, and tube length.

12. 9) Calculate area of one tube A = 3.14*L*do*10-3 Where: L: Tube length ( m) do: outlet diameter (mm) 10) Define number of tubes Nt=( Provisional Area / Area of one tube ) 11) Calculate bundle diameter Db=Do(Nt/K1)(1/n1) Where: Db: bundle diameter (mm). Do: tube outer diameter (mm). Nt: number of tubes. K1 and n1 are constants using triangle pitch of 1.25.

13. 12) From Figure determine bundle diametrical clearance.

14. 13) Calculate shell diameter Ds = Db+Dc Where: Db : bundle diameter( mm). Dc : clearance diameter( mm). 14) For Tube side coefficient calculate: - Mean temperature =((t1+t2)/2) - Tube cross sectional area A= (3.14/4)*di^2

15. - Tubes per pass = (Nt/assumed pass) - Total Flow area = (Tubes per pass * A) Where: A : Tube cross sectional area, m2 - mass velocity = (m/ Total Flow area) Where: m : mass Flow rate Tube side, kg/s - Linear velocity ut= (mass velocity/ ρ) Where: ρ : Tube Flow density, kg/m3

16. Tube side Heat Transfer Cofficient: Where: hi : Tube inside coefficient, W/m2°C. t : Mean temperature ( °C). ut : Linear velocity ( m/s). di : Tube inside diameter ( mm).

18. 15) For Shell side coefficient calculate: - baffle spacing = (Ds/5) - Tube pitch = 1.25 * do - As = ((pt – do)*DslB/pt) Where: As : Cross Flow area, m2 pt : tube pitch lB : baffle spacing, m

19. . 16) Calculate over all coefficient 1/Uo= (1/ho)+(doln(do/di)/(2kw))+(do/di)(1/hi) where Uo: overall heat transfer coefficient hi: the tube side heat transfer coefficient ho: the shell side heat transfer coefficient do: outer diameter di: inner diameter Kw: wall thermal conductivity

20. Pressure drop for shell and tube side: Where; Np is the number of passes of tubes. jf is friction factor. ΔP is the pressure drop (Pa).

23. 18) Thickness: (t) = ((Pri)/(SEj-0.6P)) + Cc Where : P: maximum internal pressure, kPa ri: inside radius of shell, m(ri=D/2) Ej: efficiency of joints as a fraction S: maximum allowable stress(for carbon steel), kPa Cc: allowance for corrosion, m

24. CoolerEquipment Name Help to recycle EB back to the start of processObjective E-103Equipment Number Bader alrshidiDesigner Shell and tube heat exchanger Type Ethyl benzene oxidation sectionLocation WaterUtility Carbon steelMaterial of Construction miniral woolInsulation 27000Cost ($) Operating Condition Shell Side 65Outlet temperature ( o C)96.988Inlet temperature ( o C) Tube Side 36.48Outlet temperature ( o C)25Inlet temperature ( o C) 31535Number of Tubes8Number of Tube Rows 6.69Shell Diameter (m)6.626Tube bundle Diameter (m) 49.5LMTD ( o C)2.281e+7Q total (Btu/hr) 396.28Heat Exchanger Area (m 2 )52.667U (Btu/hr. o F. ft 2 )

25. CoolerEquipment Name Decrease Temperature of EB recycled to the start of process Objective E-106Equipment Number Bader alrshidiDesigner Shell and tube heat exchanger Type Ethyl benzene oxidation sectionLocation WaterUtility Carbon steelMaterial of Construction miniral woolInsulation 21600Cost ($) Operating Condition Shell Side 25Outlet temperature ( o C)49.868Inlet temperature ( o C) Tube Side 34.8Outlet temperature ( o C)24Inlet temperature ( o C) 729613Number of Tubes8Number of Tube Rows 21473.2Shell Diameter (m)21445.2Tube bundle Diameter (m) 5.18LMTD ( o C)-1.217e+8Q total (Btu/hr) 9168.58Heat Exchanger Area (m 2 )132.6U (Btu/hr. o F. ft 2 )

26. CoolerEquipment Name Decrease Temperature of stream fed to CRV-101-2Objective E-115Equipment Number Bader alrshidiDesigner Shell and tube heat exchanger Type Styrene Production sectionLocation WaterUtility Carbon steelMaterial of Construction miniral woolInsulation 756Cost ($) Operating Condition Shell Side 38Outlet temperature ( o C)394.73Inlet temperature ( o C) Tube Side 200Outlet temperature ( o C)24Inlet temperature ( o C) 729613Number of Tubes8Number of Tube Rows 2.5639Shell Diameter (m)2.5309Tube bundle Diameter (m) 68.65LMTD ( o C)6.868e+5Q total (Btu/hr) 30.18Heat Exchanger Area (m 2 )17.7U (Btu/hr. o F. ft 2 )

27. CoolerEquipment Name Decrease Temperature of vent gasObjective E-101Equipment Number Bader alrshidiDesigner Shell and tube heat exchanger Type Ethyl benzene oxidation sectionLocation WaterUtility Carbon steelMaterial of Construction miniral woolInsulation 21600Cost ($) Operating Condition Shell Side 4Outlet temperature ( o C)49Inlet temperature ( o C) Tube Side 14Outlet temperature ( o C)2Inlet temperature ( o C) 11532.59Number of Tubes8Number of Tube Rows 4.59Shell Diameter (m)4.549Tube bundle Diameter (m) 11.529LMTD ( o C)5e+6Q total (Btu/hr) 353.6 Heat Exchanger Area (m 2 ) 53.275U (Btu/hr. o F. ft 2 )

28. Distillation Column design Objective : T-101-2 : Separates Styrene from MBA & ACP T-102 : Separates propylene &propane from PO & EB.

29. Assumptions 1-Tray spacing= 0.6 m 2-Percent of flooding at maximum flow rate=85% 3- Percent of down comer area of total area=12% 4- The hole area =10% the active area. 5-weir height=50 mm 6-Hole diameter=5 mm 7-Plate thickness=5 mm 8-Turndown percentage=70%

30. For column diameter 1-calculate the liquid –vapors flow factor for top and bottom FLV= Lw / Vw * (ρv / ρl) ½ Where:- LW = liquid mole flow rate in kmol /h Vw = vapors mole flow rate in kmol / h ρv = density of the vapors in kg / m³ ρl = density of the liquid in kg / m³ Main design procedures:

31. 2-from fig 11.27 get constant for the top and the bottom  top k1 and bottom k1 (FLV, tray spacing ) 3-calculate the correction factor for top and the bottom K = (σ / 0.02) ^0.2 * K1 Where:- σ = liquid surface tension in N/m 4-calculate the flooding velocity for top and bottom Uf = K *( (ρl –ρV) / ρV)½ Where:- Uf = flooding vapour velocity in m/s K= constant obtain from fig 11.27 ρl = density of liquid in kg / m³ ρv = density of vapour in kg / m³

32. 5-Assume the flooding percentage is 85% at max flow rate for the top and the bottom UV = 0.85 * Uf 6-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 Volumetric flow rate for top and bottom : V=(Vw*Mwt )/ ρv

33. 7-Assume as first - trail take down comer as 12% of total cross sectional area for the top and the bottom Ad = An / 0.88 Where: Ad = area of the down comer in m² An = net area in m² 8-calculate the diameter for the top and the bottom D = ((4 /3.14) * Ad) ½ Where: D = Diameter in m Ad = area of the down comer in m² 9- We take the maximum column diameter (top or bottom)

34. 11-calculate the liquid flow pattern Max liquid volumetric flow rate = Lm *MW / ρL * 3600 12-calculate the areas Ac = (3.14 / 4)*D² Where: Ac = total cross sectional area in m² Ad = 0.12 * Ac Where: Ad = area of the down comer in m² 10- we calculate Total height : Total height = (actual no. of stage * tray spacing )+2 Note : From fig 11.28 we have double pass plate

35. An =Ac –Ad Where: An = net area in m² Aa = Ac – 2Ad Where: Aa = active area in m² Ah = 0.1 * Aa Where: Ah = hole area in m²

36. 13- Use fig 11.31 to get L W / D  from 14-Assume weir length hw = 50 mm Hole diameter dh = 5 mm Plate thickness = 5 mm 14-Check weeping h ow max = 750 * (L wd / (L W *ρl)) ^2/3 h ow min = 750 * (L wd / (L W *ρl)) ^2/3 At min rate = hw + h ow min Where:- H ow =Weir liquid crest 15-calculate the weep point Where: U h = min vapor velocity through the hole in m/s dh = hole diameter in m K2 = constant from fig 11.30 (At min rate) U h = k2- 0.9 *(25.4-dh)/ρv½ (A d /A c )*100

37. 16-calculate the actual vapour velocity Calculate the actual vapour velocity = min vapour rate / Ah 17-Calculate plate pressure drop U H = Vv / Ah Where: UH = max. vapor velocity through holes m/s Vv = volumetric flow rate in m³ / s (bottom) Ah = hole area in m² Actual vapor velocity should be >U h

38. HD = 51 * (Uh/ C 0 )² * ρ V / ρ L Where: Hd = dry plate drop Uh = min vapour velocity in m/s Hr = 12.5E3 / ρL Where: Hr = residual head Ht = HD + H W + H OW max + Hr Where: Ht = total pressure drop in mm For C o :from fig 11.34 1)Plate thickness /dh 2)Ah/Ap=Ah/Aa

39. 18-down comer liquid backup Hap = H W – 10 Aap = L W * Hap *0.001 Where: Aap = area of apron Lw = weir length Hdc = 166 * (L Wd max / (ρl * Aap))^2 Backup down comer Hb = Hdc + Ht + h ow max + h w

40. 19-Calculate the residence time tr = (Ad * Hb * ρl) / Lwd max 19-Calculate the flooding percentage Flooding percentage = U V / uf (bottom) * 100 20-Calculate the area of the hole A = (3.14 / 4 ) * (dh * 0.001 )² 21-Calculate number of hole Number of hole = Ah / A U V =vol. flow rate bottom/ An tr should be > 3 s

41. 22-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 23- calculate the cost

42. Distillation Column Equipment Name Separates Styrene from MBA & ACPObjective T-101-2Equipment Number Bader AL-RashidiDesigner TrayType Styrene productionLocation Carbon steelMaterial of Construction miniral wool Insulation 175824Cost ($) Column Flow Rates 1514Recycle (kgmole/hr)405.2Feed (kgmole/hr) 102.3Bottoms (kgmole/hr)302.8Distillate (kgmole/hr) Key Components EBhypreoxideHeavystyreneLight

43. Dimensions 29.2 Height (m)3.85Diameter (m) 5 Reflux Ratio45Number of Trays Sieve tray Type of tray0.6Tray Spacing - Number of Caps/Holes45075Number of Holes Cost 67500 Trays2000Vessel 80000 Reboiler300Condenser Unit

44. Distillation Column Equipment Name Separates propylene &propane from PO & EB.Objective T-102Equipment Number Bader al_RashidiDesigner TrayType Epoxidation of Ethyl benzene HydroperoxideLocation Stainless 304Material of Construction miniral wool Insulation 887600Cost ($) Column Flow Rates 26000 Recycle (kgmole/hr) 2275.6Feed (kgmole/hr) 540 Bottoms (kgmole/hr) 1735.6Distillate (kgmole/hr) Key Components Propylene oxideHeavypropeneLight

45. Dimensions 27.2 Height (m)7.35Diameter (m) 15 Reflux Ratio28Number of Trays Sieve tray Type of tray0.9Tray Spacing - Number of Caps/Holes164430Number of Holes Cost 56000 Trays700000Vessel 50000 Reboiler21600Condenser Unit

46. Reactor Design Objectives:  - CRV -100-2 : Dhydration of MBA to styrene and water.

47. - The limiting reactant is MBA in the reactor(component A) -The conversion is equal to 0.99 in the CRV-100-2 reactor -length(height)=4*daimeter Assumptions:

48. 2. Calculate volumetric flow rate Vo ( ft3/hr) Fo: Molar flow rate of inlet stream(lbmole/hr) ρo: Molal Density of inlet stream(lbmole/ft3) 3-Calculate concentration of component (A) inlet (CAo) CAo=FAo / Vo 4- calculate reaction rate V0=Fo/ρo -ra=k CAo(1-X)

49. 3. Calculate the volume of reactor V= (FAo-FA)/-ra From volume calculate diameter 4. Calculate the length of reactor: length of reactor (L) = 4 * Diameter D=(V/3.14)^1/3

50. 5. Calculate the thickness Where, t is thickness in inch P is internal pressure in psig, r i is the radius of the reactor ri=D/2, in, S is the stress value of carbon steel (S=13700 psia), Ej is the joint efficiency (Ej=0.85 for spot examined welding), Cc is the corrosion allowance (Cc=1/8 in) 6. Calculate cost from www.matche.comwww.matche.com

51. Dhydration reactorEquipment Name Dhydration of MBA to styreneObjective CRV100-2Equipment Number Bader Al-RashidiDesigner CSTRType Styrene productionLocation carbon steelMaterial of Construction Miniral wool Insulation

52. Operating Condition 305.9Volume of Reactor (m 3 )270Operating Temperature ( o C) -Catalyst Type20Operating Pressure (psia) -Catalyst Density (Kg/m 3 )21328Feed Flow Rate (mole/s) - Catalyst Diameter (m)99Conversion (%) 17.4Reactor Height (m)-Weight of Catalyst (Kg) 4.3Reactor Diameter (m)-Number of Beds 0.0069Reactor Thickness (m)-Height of Bed/s (m) 360072Cost ($) - Height of Reactor (m)

53. Pump Design Objective :  p-103 : Increase pressure of ACP & MBA fed to CRV-101-2.  p-104 : Increase pressure of ACP & MBA fed to CRV-101-2 Assumptions:-  Centrifugal pump

54. Pump Design Design procedures:- 1.Calculate the actual head (ft) P1:inlet pressure( lb/ft2 ) P2:outlet pressure( lb/ft2 ) : specific weight (Ib/ft) =mass density(Ib/ft3)*gravity(ft2/sec) 2- calculate Water hourse power: 3- Overall Efficiency:

55. Pump Equipment Name Increase pressure of ACP & MBA fed to CRV-101-2Objective p-103Equipment Number Bader ALRashidiDesigner Centrifugal(8 stages,H)Type Styrene productionLocation Stainless304Material of Construction miniral wool Insulation 41200Cost Operating Condition 83Outlet Temperature ( o C)82Inlet Temperature ( o C) 750Outlet Pressure (psia)400Inlet Pressure (psia) 22.7Power (Hp)75Efficiency (%)

56. Pump Equipment Name Increase pressure of ACP & MBA fed to CRV-101-2Objective p-104Equipment Number Bader ALRashidiDesigner Centrifugal(8 stages,H)Type Styrene productionLocation Stainless304Material of Construction miniral wool Insulation 41200Cost Operating Condition 84Outlet Temperature ( o C)83Inlet Temperature ( o C) 1215Outlet Pressure (psia)750Inlet Pressure (psia) 32.1Power (Hp)95Efficiency (%)

57. Compressor : Objective: K-105 : To increase pressure for recycle EB. Assumptions: type of compressor : Reciprocating

58. Main procedure: 1- ratio of spesific heat of gas at constant pressure to spesific heat of gas at constant volume(K). P1=inlet pressure,Ibf/ft^2 P2=inlet pressure,Ibf/ft^2 T1=inlet temperature, R T2=inlet temperature, R

59. 2- calculate power(ft.Ibf/Ibm) V1:spesific volume of gas at intake condision,ft^3/Ibm

60. 3- calculate horse power(hp) qfm1:cubic feet of gas per minute at intake condition,ft^3/min 4-Ep=effeciency of the compressor

61. Compressor Equipment Name To increase pressure for recycle EBObjective k-105Equipment Number Bader AL-RashidiDesigner ReciprocatingType Ethyl benzene oxidationLocation Carbon steelMaterial of Construction miniral wool Insulation 345600Cost Operating Condition 196.4Outlet Temperature ( o C)65Inlet Temperature ( o C) 58.8Outlet Pressure (psia)1Inlet Pressure (psia) 228Power (Hp)80Efficiency (%)