1. Equipment design Done By : : mohammed al-kashan Supervised by: DR. mohamed fahim & Eng. Yusuf ismail

2. Agenda Furnace Furnace Packed bed reactor Packed bed reactor Compressor Compressor absorber absorber

3. Furnace

4. furnace Increasing the temperature of first heat exchanger stream from 265 c ° to 405 c ° on radiation section and increase combined stream to 400 c ° on convection section Furnace contain two main area: -Radiation section - Convection section

5. assumption 1. Fuel: pure methane with amount of 1000 kmol/hr 2. Film heat transfer coefficient (hc) = 0.4 3. Excess air = 100% 4. The tube has staggered pitch arrangement

6. Radiant Section Design qr = saAcpF(Tg4 - Tw4) Where, qr = Radiant heat transfer, Btu/hr s = Stefan-Boltzman constant, 0.173E-8 Btu/ft2-hr-R4 a = Relative effectiveness factor of the tube bank Acp = Cold plane area of the tube bank, ft2 F = Exchange factor Tg = Effective gas temperature in firebox, °R Tw = Average tube wall temperature, °R

7. Alpha calculation Acp = (No. Tubes)*Space*(Eff. Length)  aAcp, where the value of (a) is from graph  aAcp, where the value of (a) is from graph Ar = (W*L*2 + W*H*2 + H*L*2-Exitarea*L ) Aw = Ar - aAcp  Aw /aAcp  Aw /aAcp

9.  (W : H : L) ratio of (width : height : length) Aw = Effective refractory area, ft2 Ar = Total refractory area, ft2 aAcp = Equivalent cold plane area, ft2

10. Gas Emissivity Σ mol% of CO2 and H2O in flue gas= (mole% of O2 + mole% of N2) in product stream beam length=2/3*(volume of furnace)^(1/3) PL=product of the partial pressure of the carbon dioxide (atm.ft) PL= beam length * Σ mol%

12. Exchange Factor (F) From previous graph we got the value of emissivty previous graph we got the value of emissivtyFrom emissivty and (Aw/aAcp) emissivty and (Aw/aAcp) The value of Exchange factor can be calculated

13. Convection Heat Transfer In the Radiant Section qc= hc At (Tg-Tw) At= 2πr * tube length * number of tubes Then, qR= qr + qc

14. Convection Section qc= hc At (Tg-Tw) Outside film heat transfer coefficient, hc: hc = j*Gn*cp(kb/(cp*mb))0.67 Where, J = Colburn heat transfer factor Gn = Mass velocity based on net free area, lb/hr-ft2 cp = Heat capacity, Btu/lb-F kb = Gas thermal conductivity, Btu/hr-ft-F mb = Gas dynamic viscosity, lb/hr-ft

15. Colburn heat transfer factor, j: j=C1*C3*C5(df/do)0.5((Tb+460)/(Ts+460))0.25Where, C1 = Reynolds number correction C3 = Geometry correction C5 = Non-equilateral & row correction df = Outside diameter of fin, in do = Outside diameter of tube, in Tb = Average gas temperature, F Ts = Average fin temperature, F

16. Reynolds number correction, C1: C1 = 0.25*Re-0.35 Where, (Re = Reynolds number) Geometry correction, C3: For segmented fin tubes arranged in,a staggered pattern, C3 = 0.55+0.45*e(-0.35*lf/Sf) If = fin height, Sf = fin spacing, (in inch)

17. Non-equilateral & row correction, C5: C5 = 0.7+(0.70-0.8*e(-0.15*Nr^2))*e(-1.0*Pl/Pt) Nr = Number of tube rows Pl = Longitudinal tube pitch, in Pt = Transverse tube pitch, in

18. Mass velocity (Gn) Gn = Wg/An Where, Wg = Mass gas flow, lb/hr An = Net free area, ft2

19. Net Free Area, An: An = Ad - Ac * Le * Nt Where, Ad = Cross sectional area of box, ft2 Ac = Fin tube cross sectional area/ft, ft2/ft Le = Effective tube length, ft Nt = Number tubes wide Ad = Nt * Le * Pt / 12 Ac = (do + 2 * lf * tf * nf) / 12 tf = fin thickness, in nf = number of fins, fins/in

20. Efficiency Calculation E=(Q/qrls)*100 Q= qr+qc qrls:Heat released by burner,BTU/hr qrls=Wfuel *Lhvfuel

21. Thickness t=(Pri/(SEJ-0.6P))+Cc t : thickness in (inch) p: internal pressure in (psig) ri: inside radius in (inch) S: working stress (psi) EJ: efficiency 0f joint Cc: allowance for corrosion in (inch)

22. Results 7.37E+06 (BTU/hr( qr 9.19E+04 (BTU/hr) qc 7.46E+06 (BTU/hr) QR 2.71E+06 (BTU/hr) Qconv. 1.02E+07 (BTU/hr) Qtotal Glass wall and quartz insulation 0.85290295 (in) thickness 1908300 $ cost

23. Reactor

24. Packed bed reactor A catalytic fixed bed reactor is a cylindrical tube, randomly filled with catalyst particles A catalytic fixed bed reactor is a cylindrical tube, randomly filled with catalyst particles Objective: main reactor of plant and used to produce ethylbenzene Objective: main reactor of plant and used to produce ethylbenzene Reaction involves: Reaction involves: C6H6 + C2H4  C6H5C2H5

25. Assumptions 1. Assume that each bed is reactor to calculate the main reactor dimensions 2. Assume (L/D)=0.135 (for first bed) and Other beds (L/D) = 0.0829 3. Assume the space between each bed=2.75 ft

26. DESIGN OF BACKED BED REACTOR Design equation : Rate low and stoichometry :

27. Arrhenius equation : Arrhenius equation :A=0.69E6 Ea= -6.344E4 R=83.14 Energy equation: Energy equation:

28. Volume of the reactor All the previous equations will solved by polymath simulator. All the previous equations will solved by polymath simulator. The result of the simulation will be the volume of the reactor (1 bed). The result of the simulation will be the volume of the reactor (1 bed). From the volume equation we can get the diameter of the reactor and the height of each bed From the volume equation we can get the diameter of the reactor and the height of each bed

29. Polymath for volume

30. volume of the reactor Equation : Equation : From (L/D) assumption The height of bed and diameter Of The reactor can be calculated

31. Height of reactor From our calculation and assumption the height of the reactor can be calculated. From our calculation and assumption the height of the reactor can be calculated. - H= (space high* no. of spaces) + first bed high + (high of each of fife bed* no. of beds)* (dome high*no. of dome) - Dome high= dimeter/2

32. Area: 2πrh Weight of the catalyst W= V (1- ε ) ρ, ρ = catalyst density W= V (1- ε ) ρ, ρ = catalyst density

33. Results 2.743 m Diameter 10.13 m height 60 m^3 volume 87.295877 m^2 area 2399.76 kgcat Weight of catalyst 0.153 m thickness 1499688 $ cost

34. compressor Objective: To increase the pressure of the feed from 14.5 to 72.52 (psia) Choosing the compressor type.

35. - Calculate the compression factor (n) using the following equation : Where, P1,2 : is the pressure of inlet and outlet respectively (psia) T1,2 : is the temperature of the inlet and outlet respectively (R)

36. Calculate the work done in Btu/lbmol by: Where, R is the ratio of the specific heat capacities (Cp/Cv) 3. Calculate the horse power, Hp using the following equation: Hp=W*M Where, M is the molar flow rate in lbmol/s 4. Calculate the efficiency of the compressor using the following equation:

37. - Where, Mw :is the molecular weight of the gas in the stream Mw :is the molecular weight of the gas in the stream CP :is the specific heat capacity (Btu/lb ◦ F ) CP :is the specific heat capacity (Btu/lb ◦ F )

38. Result 1221 (oC) Inlet Temperature 14.5 (psia) Inlet Pressure 75.96 (%) Efficiency 1318.4 (oC) out Temperature 72.52 (psia) out Pressure 197.913 (hp) Power (Hp) 119100 $ cost

39. Absorber

40. Absorber Objective : to separate the vent gas and waste from the feed Objective : to separate the vent gas and waste from the feed Material : carbon steel Material : carbon steel

41. Assumption Plate spacing = 0.8 m

42. Absorber design the column diameter: Flv = (Lw / Vw) * ( ρ v / ρ l) ^0.5 Where, Lw : liquid mass flow rate (kg/s) Vw : Vapor mass flow rate (kg/s) Flv : liquid vapor flow factor

43. we assumed try spacing From the figure we get K1

45. - Correction for surface tension: K1 = (surface tention*1E3/20)^.2 * K1 Where, K1: correction for surface tension

46. Flooding vapor velocity : uf = K1(( ρ l- ρ v)/ ρ v) ^.5 Where, uf : flooding vapor velocity (m/s) Design for 85%flooding at maximum flow rate ŭ f = uf*0.85

47. maximum volumetric flow rate = flow rate/density flow rate/density net area required = ( volumetric flow rate / uf )max Take down comer area as 12 % of total area A = A net *0.88 (m2)

48. column diameter =(area*4/π)^.5 Maximum volumetric liquid rate

49. Column height: h =(actual number of stages* tray spacing )+Dmax Where, h: column height (m) Actual number of stage = Efficiency * #of stage

50. Provisional plate design: Where, Dc: column diameter (m) Ac: column area for cylinder = (m2) An: down comer area = 0.12*Ac (m2) Aa: active area= Ac-2Ad (m2) Ah :hole area by taking 10% of Aa

52. - Check weeping Maximum liquid rate= lw*MW (Kg/s) Minimum liquid rate @ 70% turn-down =0.7*max liquid rate (Kg/s) Height of the liquid crest over weir how = 750*(Lw/ρl * lw)^(2/3) in (mm)

53. Assuming, take hole diameter(mm) plate thickness (mm) weir height(hw) (mm) at minimum rate hw + how from figure @ hw + how we get K2

54. Vapor velocity = (K2-0.9*(25.4-dh))/( ρ v ^.5) Where, uh : vapor velocity K2 : constant dh : hole diameter (mm) Actual minimum vapor velocity = minimum vapor rate / Ah

55. Plate pressure drop Maximum vapor velocity through holes = Max volumetric flow rate/Ah From figure For plate thickness/ hole diameter =1, and Ah/Ap = Ah/Aa =0.1 We find Co.

58. hd = 51 *( uh/Co)^2 * (ρv /ρl) hr = (12.5*1000) /ρl ht = hd +(weir length +how )+hr Where, hd: dry plat drop (mm liquid) hr :residual head (mm liquid) ht: total pressure drop (mm liquid)

59. Thickness ( t) = (Pri/ SE-0.6P)+C, (t in (in)) Where, t: thickness (in) p: Internal pressure (psig) ri: Inside radius (in) S: Working stress (psi) Ej: Efficiency 0f joint Cc: Allowance for corrosion (in)

60. Down comer back up Take hap (mm) = hw - 10 Area under apron (m2) =0.6*hap As this less than Ad use Aao(m2) Head loss in the down comer (mm)= hdc = 166*(Lwd/ ρ l * Am)^2 Lwd: liquid flow rate in down comer (kg/s) Am: either Ad, or Aad (the smaller ) (m2) hb (mm) = hw +how +ht +hdc

61. Number of holes: Area of on hole (m2) =Number of holes= hole area/area of one hole hole area/area of one hole

62. weight of the metal di= Internal column diameter (m) do=di+2t (m) Volume of cylinder (di) m3=(3.141*H)*(di/2)^2 Volume of cylinder (do) m3=(3.141*H)*(do/2)^2 Volume of metal m3= volume of cylinder (do)- volume of cylinder (di) Weight (Kg)= volume of metal *7900

63. Result 1.75 m diameter 0.125 m thickness 103500 $ cost 1.444 m Volume of metal 11118.8 kg Weight of metal 7 m Height

64. Thank you