1. Done by: Zainab Al-fadhli Supervised by: Prof: M.Fahim Eng : Yusuf Ismail Kuwait university Engineering and Petroleum college Chemical Engineering Department Plant Design

2. Out lines  cooler: E-100 ( for process 1). E-100 ( for process 2). E-101 E-102 E-103  Separator : V-101 ( for process 1) V-101 ( for process 2)  Compressor : K-100( process1) K-100( process 2)

3.  cooler

4. Cooler design  Objective To cold the stream to required temperature.  Assumptions 1- In our process, Used shell and tube heat exchanger, one shell and two tube passes. 2- The value of the overall heat transfer coefficient was assumed. 3- Assume the outer, the inner diameter and the length of the tube.

5. Design procedure : 1.Heat load,(kW) Q = (m Cp ΔT) hot =(m Cp ΔT) cold 2. Tube side flow, (kg/hr) 3.Log mean Temperature, (˚C)

6. Continue 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 (˚C). 4.Calculate the two dimensionless temperature ratios: R = (Thi - Tho) /(Tco - Tci) S = (Tco - Tci)/(Thi - Tci) 5. From Figure we get Ft by using R and S ratios where : Ft : temperature correction factor

7. Continue 6. Calculate True temperature difference ∆Tm= Ft *∆T lm 7. Choose U from table depending on the type of flows in shell and tube side U: Over all heat transfer coefficient 8. Calculate provisional area :

8. continue 9. Assume inlet tube diameter, outlet tube diameter, and tube length 10. Calculate area of one tube A = 3.14*L*do*10^-3 Where: L: Tube length ( m) do: outlet diameter (mm) 11.Calculate Number of tubes = provisional area / area of one tube

9. Continue 12. 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 constant from table. 13. From Figure determine bundle diametrical clearance Ds = Db + Dc 14. Calculate shell diameter Where: Db : bundle diameter( mm) ( Dc : clearance diameter( mm

10. Continue 15.For Tube side coefficient calculate. Cold stream mean temperature = ( Tci +Tco) /2,(ºc). Tube cross sectional area = (π/4 ) di ^2,(mm2). Tubes per pass = no. of tubes / number of passes. Total flow area = tubes per pass * cross sectional area,(m). Water mass velocity = mass flow rate / total flow area,(Kg/s.m2). 16. To calculate Shell side Coefficient : Baffle spacing (Lb) = 0.2 * Ds, mm) Tube pitch (pt) = 1.25 * do, (mm) Cross flow area (As) = (pt - do)* Ds* Lb / pt,(m2) Mass velocity (Gs) = mass flow rate / cross flow area,(Kg/s.m2)

11. Continue linear velosity (ų ) = mass velosity / density,(m/s). Renolds number (Re) = ρ ų di/ µ. Prantel number (Pr) = Cp µ / κ. (hi di / κ) = jh Re Pr^0.33 * (µ/µwall)^0.14. Assume that the viscosity of the fluid is the same as at the wall,(µ/µwall) = 1 the Then find the value of jh from the figuer. 16. To calculate Shell side Coefficient : Baffle spacing (Lb) = 0.2 * Ds, (mm) Tube pitch (pt) = 1.25 * do, (mm) Cross flow area (As) = (pt - do)* Ds* Lb / pt,(m2) Mass velocity (Gs) = mass flow rate / cross flow area,(Kg/s.m2)

12. Continue 17. Calculate over all coefficient: 1/Uo = (1/ho)+(1/hod)+(doln(do/di)/(2kw))+(do/di)(1/hid)+(do/di)(1/hi) Where: Uo: the overall coefficient, W/m2°C ho: outside fluid film coefficient, W/m2 °C hi : inside fluid film diameter hod : outside dirt coefficient (fouling factor) hid: inside dirt coefficient, W/m2°C kw: thermal conductivity of the tube wall material di : tube inside diameter, m do : tube out side diameter, m

13. 18.Calculate pressure drop for Tube side : ∆Pt = Np(8*jf*(L/di)+2.5)*(ρut2/2) ∆Pt : tube side pressure drop, pa Np : number of tube side passes ut: tube side velocity (m/s) L : length of one tube ( m) jf : From Fig 19. Shell side : ∆P = 8*jf*(Ds/de)*(L/lB)*(ρ*us2/2) Where: ∆P : shell side pressure drop (pa) Ds : shell diameter (m) lB : baffling spacing ( m) jf : From Fig

14. 20. Thickness: (t) = ((Pri)/(SEj-0.6P)) + Cc Where : P: maximum internal pressure, kPa ri: inside radius of shell, m Ej: efficiency of joints as a fraction S: maximum allowable stress, kPa Cc: allowance for corrosion, m 21. Calculate the cost from www.matche.comwww.matche.com

15. Cooler E-100( for process 1) Operating Condition Shell Side 200 Outlet temperature ( o C) 633.8 Inlet temperature ( o C) Tube Side 180 Outlet temperature) o C) 35 Inlet temperature( o C) 220 Number of Tubes 2 Number of Tube Rows 1.154 Shell Diameter (m) 1.082 Tube bundle Diameter (m) 285.456 LMTD ( o C) 3359.58 (Q total (Kw 84.457 Heat Exchanger Area (m^2) 165 U (W/m 2 o C) $52300 Cost ($)

16. Cooler E-100(for process2) Operating Condition Shell Side 1000 Outlet temperature ) o C) 1336 Inlet temperature ( o C( Tube Side 250 Outlet temperature ( o C) 35 Inlet temperature( o C( 98 Number of Tubes 2 Number of Tube Rows 0.818 Shell Diameter (m) 0.750 Tube bundle Diameter (m) 1024.3 LMTD ( o C) 3810.5224 (Q total (Kw 37.63 Heat Exchanger Area (m^2) 100 U (W/m^2 o C) $36900 Cost ($)

17. Cooler E-101(for process2) Operating Condition Shell Side 500 Outlet temperature ( o C( 1000 Inlet temperature ) o C( Tube Side 200 Outlet temperature( o C( 35 Inlet temperature ( o C( 101 Number of Tubes 2 Number of Tube Rows 0.814 Shell Diameter (m) 0.749 Tube bundle Diameter (m) 617.42 LMTD ( o C ( 6686.21 Q total Kw 37.51 Heat Exchanger Area (m^2) 300 U (W/m 2 o C) $36800 Cost ($)

18. Cooler E-102(for process2) Operating Condition Shell Side 250 Outlet temperature ( o C) 500 Inlet temperature ( o C ( Tube Side 150 Outlet temperature ( o C) 35 Inlet temperature ( o C) 112 Number of Tubes 2 Number of Tube Rows 0.862 Shell Diameter (m) 0.797 Tube bundle Diameter (m) 255.144 LMTD ( o C( 3343.106 Q total Kw 43.102 Heat Exchanger Area (m^2) 300 U (W/m 2 o C) $41400 Cost ($)

19. Cooler E-103(for process2) Operating Condition Shell Side 200 Outlet temperature ( o C) 216.7 Inlet temperature ) o C ( Tube Side 120 Outlet temperature ( o C) 35 Inlet temperature(( o C) 46 Number of Tubes 2 Number of Tube Rows 0.589 Shell Diameter (m) 0.5334 Tube bundle Diameter (m)) 127.82 LMT ( o C) 233.319 Q total (Kw) 17.73 Heat Exchanger Area (m^2) 100 U (W/m 2 o C) $11100 Cost ($)

20.  Separator A vapor-liquid separator is a vertical usedin several industrial applications to separate a vapor –liquid mixture.

21. Design procedure 1.calculate the settling velocity of the liquid droplet using the following equation :- Where, Ut = settling velocity in m/s ρl = liquid density in kg/m³ ρv= vapors density in kg/m³ 2- assume there is no demister 3-calculate the actual settling velocity in m/s Ua=0.15Ut.

22. Dv = ( 4Vv/ Π Ua)^(1/2) 4- Calculate the minimum vessel diameter. Where, Dv = minimum vessel diameter in m Vv = gas volumetric flow rate in m³/s Ua= actual settling velocity in m/s 5- assume 10 min hold – up.

23. 6-calculate the volume held in the vessel Volume = VL * hold time VL = Liquid volumetric flow rate in m³/s

24. 7-Calculate the vessel cross sectional area A= Π/4 * Dv² 8-Calculate the liquid depth hv = volume held up/ vessel cross sectional area hv = liquid depth in m 9-calculate the height Ht = hv+ DV/2+DV+DV/2+DV/2

25. 10-Calculate the thickness of the separator using the following equation 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 11- calculate the cost

26. Separator v-101( process1) Operating Condition 515 Operating Pressure (psig) 59.3 Operating Temperature ( o C( Design Considerations 6.9493 Gas Density (kg/m3( 1037.4 Liquid Density (kg/m3) 1 Z factor 0.62201 Viscosity (cp) 858.62 Liquid Flow rate (MMSCFD) 3603.3 Gas Flow rate (MMSCFD) Dimensions 3.5 Height (m) 1.197 Diameter (m) Cost($) 44700

27. Separator v-101( process2) Operating Condition 515 Operating Pressure (psig) 40 Operating Temperature ( o C) Design Considerations 35.203 Gas Density (kg/m3( 1054.3 Liquid Density (kg/m3) 1 Z factor 0.79712 Viscosity (cp) 11506 Liquid Flow rate (MMSCF) 5853.1 Gas Flow rate (MMSCFD) Dimensions 4.5 Height (m) 1.43 Diameter (m) Cost($) 64900

28.  Objective To increase the pressure of the feed from 1 bar to 70 bar Choosing the compressor type.  Choosing the compressor type

29. 1.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)

30. 2. 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:

31. Where, Mw :is the molecular weight of the gas in the stream CP :is the specific heat capacity (Btu/lb◦ F ) 5. Calculate the cost of the compressor from www. Matche. comwww. Matche. com

32. Compressor K-100 ( process 2) Operating Condition 633.8 Outlet Temperature (◦C) 118.2 Inlet Temperature (◦C) 519 Outlet Pressure (psia) 14.7 Inlet Pressure (psia) 1451.9 Power (Hp) 95.55% Efficiency (%)

33. Compressor K-100 (process2 ) Operating Condition 1336 Outlet Temperature (◦C) 343.7 Inlet Temperature (◦C) 1470 Outlet Pressure (psia) 14.7 Inlet Pressure (psia) 2350.844 Power (Hp) 53.988% Efficiency (%)

34. Thank you for listening