Equipment Design Done by: Sara Saad Al-Quhaim (Group leader)

1. Equipment Design Done by: Sara Saad Al-Quhaim (Group leader)

2. Equipment Design: Distillation Column ( C-201) Distillation Column ( C-202) Cooler (E-103) Compressor ( K-100)

3. Distillation Design: Number of Stages (No. of stages) min = 25 (from HYSYS) Efficiency = 75% (Assumed) Actual stages = (No. of stages) min / Efficiency =33

4. Calculation of actual number of stages ( Short-cut Method): Water ( heavy) a= -7.831 b= 1.7399 c= -2.2505 d= -1.9828 API method: Pr vap= Pc*100*EXP((Tr^-1)*( at+bt^1.5+ct^2.6+dt^5) Y= 18.9621182 α= 1 (relative volatility)

5. Methanol ( light ) a= -8.6413 b= 1.0671 c= -2.3184 d= -1.678 XF= 0.096 XD= 0.996 XB= 0.0000005 API method: Pr vap= Pc*100*EXP((Tr^-1)*( at+bt^1.5+ct^2.6+dt^5) Y= 79.90817981 α= 4.21409565

6. ( Fenske equation) Nm = Log [ XLK/ XHK]d [XLK/ XHK]b ( minimum number of stages) Log LK q=0.98909 (liquid)  [  xf - ] = 1-q

7. = 3.013 by interpolation  [  xd - ] = Rm +1

8. R= 2.8 R/(R+1)= 0.736842105 Rm= 2.50548777 Rm/(Rm+1)= 0.714732994

9. Nm/N= 0.33 N= 32.866 Note: The actual number of stages =33 stages

10. Where: Lw: liquid flow rate, kmol/hr ρL: liquid density,kg/m3 Vw: vapor flow rate, kmol/hr ρv :vapor density, kg/m3 FLv: Liquid-vapor factor Tray spacing= 0.55 ( from hysys) Using figure to find K1

11. uf : flooding vapor velocity, m/s uv: maximum velocity, m/s. x: percentage of flooding at max flowrate.( assume 85”% ) Maximum volumetric flow rate

12. Anet: Net area required, m2 Taking downcomer area as 12 percent of total Cross sectional area ofdowncomer= Column Diameter

13. Column Height Maximum volumetric liquid rate Liquid flow arrangement

14. Total column cross sectional area Ac = π/4 Dc2 Net area available for vapor liquid Active or bubbling area Hole area Ah = 0.1 Aa

15. Ad/Ac*100 Find lw/Dc from Figure Find Weir Length (lw) Minimum liquid rate = 0.7* Max (Lw)

16. Maximum how =750(Lw/(ρL lw))(2/3) Minimum how =750(Lw/(ρL lw))(2/3) Weir crest At minimum rate hw=50mm + how, from Figure 11.30 Find K2

17. Hole diameter =5mm uh (min) =(K2-0.90(25.4-dh))/ρv0.5 Minimum vapor velocity actual minimum vapor velocity = minvapor rate /Ah Hole area Note: actual minimum vapor velocity should be greater than Uh Taking; Plate thickness/hole dia. = 1 Ah/Ap ( perforated area) = Ah/Aa from figure ( orifice cofficient) Co = 0.84

18. Vapor velocity Dry plate drop= hd = 51(uh/Co)2(ρv/ρL) Residual head (hr) = (12.5*103)/ρL Total plate pressure drop (ht) = hd+hw+how+hr

19. Downcomer pressure drop (hap) = hw-10 Area under apron (Aap) = lw *hap*0.001 Head loss in downcomer hdc= hdc = Weir crest Back-up in downcomer (hb) = hw+how+ht+hdc hb less than tray spacing , so tray spacing is acceptable.

20. Down comer backup Downcomer area Residence Time tr: residence time, should be > 3 s Uv = Bottom V / An Percent flooding= From figure, find ψ below 0.1 ( fractional entrainment)

21. Calculate lw/Dc From figure find 

22. Angle subtended by the edge of the plate=180- θc Mean length, unperforated edge strips = Area of unperforated edge strips=0.05*mean length Mean length of calming zone = Area of calming zone =2*mean length of calming zone *0.05 Total area for perforations, Ap =Aa - area of unperforated edge strips - area of calming zone

23. Ah/Ap from figure, find lp/dh satisfactory within 2.5 to 4

24. Number of holes: Area of one hole= Total number of holes = Ah / 1.963E-05 Holes on one plate = total Number of holes/Area of one For condenser: Material Carbon Steel Area of condenser =

25. For reboiler: Material Carbon Steel Area of reboiler= Where: ri = Inside radius of the shell, in P =Maximum allowable internal pressure S = Maximum allowable working stress E = Efficiency of joints Cc = Allowance for corrosion, in

28. Cooler design (heat exchanger)

29. - = Heat load transfer in the hot side, KW. Mass flow rate in Kg/s. Temperature difference of the inlet and outlet.

30. Inlet shell side fluid temperature (oC). Outlet shell side fluid temperature (oC). Inlet tube side temperature (oC). Outlet tube temperature (oC). Measure of temperature efficiency

31. At R and S Temperature correction factor. - Estimate U from table12.1 Provisional area

32. Take: Tube outside diameter(do) =20mm Tube inner diameter(di) =16mm Tube length(L) =4.88m Area of one tube A/ A one tube Number of tubes Bundle diameter K1, n1 depend on number of passes

33. Shell diameter

34. Tube cross sectional area Tube per passes Total flow area Tube mass velocity

35. Tube linear velocity Heat transfer coefficient inside the tube

36. From l/Di and Re Inside coefficient (W/m2oC). Tube side heat transfer factor. Thermal conductivity of stream.

37. Shell baffle spacing Tube pitch Cross flow area Mass velocity

38. Equivalent diameter At 25% baffle cut find jh Pr =Cp/ kf

39. hs = Overall heat transfer coefficient Outside coefficient (fouling factor). Inside coefficient (fouling factor).

40. Jf= friction factor

42. Shell Thickness - Shell thickness (in). Maximum allowable internal pressure (psig). Internal radius of shell before allowance corrosion is added (in). Efficiency of joints. Working stress (psi). Allowance for corrosion (in)

44. CompressorDesign

45. n = Compression factor W = work done (Btu/Ibmol) R = Cp/Cv M= mol flow rate (Ibmol/hr)

46. K = (Mwt*CP)/(Mwt*CP-1.986) Ep = Efficiency of compressor %

48. Thank You