Thermal Design of Heat Exchanger (E-100) Tube side (Cold fluid) Inlet fluid: Glycol to be preheated Outlet fluid: Preheated glycol to absorber Inlet temperature: 67.59o F Outlet temperature: 770F Tavg = 72.2950F Density (by interpolation) =70.58 lb/ ft3 Viscosity = 22.52 cP=54.5 lb/ft-hr Thermal conductivity K= 0.192 Btu/hrft0F Heat capacity Cp =0.5891 Btu/lb0F Flow rate, Wg = 4.88 x 105 lb/hr q = WgCp∆t = (4.88 x 105 x 0.5891 x 9.41) Btu/hr = 2.7x 106 Btu/hr Shell side (Hot fluid) Inlet fluid: Dry gas Outlet fluid: Sales gas Inlet temperature: 99.880 F Outlet temperature: 94.860F Tavg = 97.370F Density (by interpolation) = 3.601 b/ft3 Viscosity = 0.01362 cP=0.0333 lb/ft-hr Thermal conductivity K = 0.023875 Btu/hrft0F Heat capacity Cp = 0.5105 Btu/lb0F Flow rate,Wi = 1.05 x 106 lb/hr q = WiCp∆t = (1.05 x 106 x 0.5105 x 5.02) Btu/hr = 2.7x 106 Btu/hr Tube side calculation LMTD calculation: (First pass counter-current flow) Hot 99.88 (T2) 94.86 (T1) 5.02 Higher Lower Difference LMTD= ¿ ( T 2−t 2 )−(T 1−t 1) ( T 2−t 2 ) ln ( ) T 1−t 1 Cold 67.59 (t1) 77 (t2) 9.41 R= Difference 32.29 17.86 14.43 T 2−T 1 t 2−t 1 S t 2−t 1 T 2−t 1 (LMTD)' = −4.39 ln ( 22.88 /27.27) R= 5.02 9.41 S= 9.41 99.88−67.59 = 25.01oF = 0.53 = 0.2914 From Fig 18 (kern) Page-828 for 1-Shell pass & 2 or more Tube pass heat exchanger FT = 0.95 which is satisfactory.. Corrected LMTD = FT x (LMTD)' = 0.95 25.01 = 24.510F Calculation of heat transfer area and tube numbers no of tubes. Assuming tube length (Lt) is 30 ft. Uassume = 35 Btu/hr ft20F Q 2.7 10 6 3147.heavy organics Overall U = 10-40 Btu/hr ft20F Assume. Re = π × di× µ ×nt = 1100 < 104 . no of tubes.834 in. ID (di) 0. Re = π × di× µ ×nt = 892 < 104 Iteration. Reynolds no. So. Assuming tube length (Lt) is 20 ft. square pitch. 1 4 in. square pitch. Fixed tube plate 1 1in. OD(d0). 1 4 in.2 2nd iteration is started assuming 1 shell pass & 4 tube pass(np). Reynolds no. ID (di) 0. nt = π × d 0×<¿ A req ¿ = 561 Nearest count from table 9(kern) Page-841 is 562 4 × wg× np Now. OD(d0).1 The first iteration is started assuming 1 shell pass & 2 tube pass(np). So. 14 BWG. 14 BWG. Fixed tube plate 1 1in.4 ft 2 U assumee LMTD 35 24.51 A= Iteration.834 in. nt = π × d 0×<¿ A req ¿ = 562 Nearest count from table 9(kern) Page-841 is 574 4 × wg× np Now.Q = Uassume Arequired LMTD For light . Iteration.37 in. Nu = hidi k = 0. square pitch. Re = π × di× µ ×nt = 5647 < 104 Iteration.5 in.027 × Re0. 1 16 in. 16 BWG. So. ID (di) 1. Reynolds no. Fixed tube plate 9 1.25 in.3 3rd iteration is started assuming 1 shell pass & 4 tube pass(np).8 × Cp K 1 3 ( µµw ) × 0. our final tube no 182 & Reynolds no (Re) is 16560 selected shell ID (Ds) is 35 in. Assuming tube length (Lt) is 45 ft. Now. So. 16 BWG. nt = π × d 0×<¿ A req ¿ = 259 Nearest count from table 9(kern) Page-841 is 268 4 × wg× np Now. Assuming tube length (Lt) is 26 ft. ID (di) 1. no of tubes. OD(d0). nt = π × d 0×<¿ A req ¿ = 178 Nearest count from table 9(kern) Page-841 is 182 4 × wg× np Now.12 in. Re = π × di× µ ×nt = 16560 > 104 Therefore. square pitch. OD(d0). no of tubes. 1 8 in. Fixed tube plate 7 1. Reynolds no.4 4th iteration is started assuming 1 shell pass & 6 tube pass(np).14 . B = 0. Re = De .So.3 Btu/hrft20F . Gs µs = 465000 So. ho = 184 Btu/hrft2 0F Clean overall heat transfer coefficient (Uc) calculation Uc = ( 1 Ao do−di Ao 1 Ao + Rd + + + Rd) ho Ai 2 Kw Ai hi Ai -1 = 42. as = = 0.03125 ft Shell ID.5 in = 1.5 Ds = 17. Gs = Wi / as = 1.23 x 106 lb/hrft2 4 P ² T d 0 2 / 4 d o Equivalent dia.124 ft Reynolds no. flows area.15625 ft Clearance. Ds = 35 in = 2.55 × Cp K 1 3 ( µµw ) 0. square pitch = 0. Pt = 1 7 8 in.8526 ft2 Mass velocity.36 × Re0. hi = 175.14 × So.74 Btu/hrft2 0F hio = hi x ID OD =151. De = = 0.46 ft (half of the shell ID is selected) Pitch. Nu = hoDe k = 0. C = Pt – do = 0.92 Btu/hrft2 0F Shell side calculation Assumption: 25 % cut segmental baffles Baffles spacing.92 ft ID shell C B PT So. a t= = 182 0. allowable=0. Rd = = 0. Pressure drop calculation Tube side calculation No of tubes flow area per tube No of passes Flow area. Rd is acceptable. Over design calculation: % of over design = Ucalculated −Uassume Uassume = 20.3 Btu/hrft20F Assumed overall Coefficient. design is accepted. Uc = 42.309 ft2 .86 % < 30 % So.00098 Because for natural gas Rd. Ud = 35 Btu/hrft20F Uc U D U c U D Dirt factor.01021 6 = 0.001 (Ref: Mechanical Design of Process System (vol2)) Therefore. Dirt factor calculation Clean overall heat transfer coefficient. 82 ~ 31 From figure 26.56 Psi < 8.8526 ft2 Mass velocity. C = Pt – do = 0. PT = Pf ❑ SG + Pr = 0. Re = 465000 No of baffles.np 2.23 x 106 lb/hrft2 Reynolds no. as = 0. total pressure drop in tube side.5) So.1296 ft2/ft2 s =1(assuming) .di. =1 Pf So.03125 ft Shell ID.92 ft Flows area. Pt = 1 7 8 in.6 104 lb/hrft2 For Ret = 16560 From figure 26.288 ft2/ft2 t Assume. tube side friction factors correlation (Kern) Page-836 Friction factor. tube side friction factors correlation (Kern) Page-836 Friction factor.22 Psi 5.SG.334 f .t 10-13 (2np – 1.Lt . square pitch = 0.Mass velocity.34 Psi = 2. Gt = Wg/at = 1. Gs = 1.G 2 t.7 psi which is acceptable. nb = ¿ B = 30. Shell side calculation Pitch. f = f = 0. frictional pressure drop.22 1010. Ds = 35 in = 2. f = 0. Return loss Pr = 1.15625 ft Clearance. Ds.088 in. of shell: 1 Shell dia.Ds 2. Dp = Wp × 1.De.1 = 1371.P } = 0.s = 9.7 Psi Permissible working pressure.(nb 1) 5. j P Shell thickness. f . Ds = 35 in Working pressure. j = 0. which is acceptable.Ps So. P = 1247 Psi Design pressure.G 2 s.68 mm . f = 11000 Psi Welding efficiency.SG. ts= = 2. Nozzles Take inlet and outlet nozzles as 100mm diameter.175 mm Shell thickness including corrosion allowance = (2. Vent nozzle = 25mm diameter Drain nozzle = 25mm diameter Relief Valve = 50 mm diameter.22 1010.5 Psi. Nozzle thickness = [ P x Di ] / { 2 f J . of pass: 6 Shell side properties: Materials: Stainless Steel No. pressure drop = f .9 P.205 in. Corrosion allowance =1/8 inch = 3.088+1/8) = 2.38 Psi < 14. Mechanical design Tube side properties: Materials: Stainless Steel No. Molybdenum Steel Gasket Material: Asbestos Shell OD = 37.Minimum nozzle thickness is 6mm and 8mm is chosen which includes the corrosion allowance. Flange material: IS 2004-1962 Class 2 Carbon Steel Bolting steel: 5% Chromium. . Transverse baffles Number of Baffles = 31 Baffle cut = 25% Baffle thickness = 6mm (standard) Flange design Flange is ring type with plain face.205 in Shell ID = 35 in Allowable stress for flange material = 100 N/mm2 Allowable stress of bolting material = 138 N/mm2 Using matche.205 in Shell Thickness = 2.com we have estimated cost for this heat exchanger is $158900. P & I Diagram . . . 14.1 Estimation of Total Capital . Quantity US $ P-101 2 9400 P-100 3 12900 Total = 22300 Table 12.4: Equipment cost for Storage Tanks Equipment Storage Tank Identification No. Table 12.000 Total= 238.1: Equipment cost for Heat Exchanger Equipment Floating head HE Floating head HE Identification Quantity US $ E-100 1 80.000 E-102 1 158.000 No. Reciprocating Compressor K-100 Quantity US $ 2 35600 Quantity US $ 1 6300 Table 12. V-106 Table 12.2: Equipment cost for Pumps Equipment Centrifugal Pump Centrifugal Pump Identification No.3: Equipment cost for Compressor Equipment Identification No.Investment and Production Cost Table 12.5: Equipment cost for Adsorption Column Equipment Identification No. Quantity US $ . 000 DEA Contractor T-100 4 260. Quantity US $ Separator V-100 1 2200 Separator V-101 1 1900 Separator V-104 1 400 Total = 4500 Total Equipment cost in the year 2002 = $ 958300 Total Equipment cost at present (2014) = $ 958300 . Glycol Stripper V-103 1 145000 DEA Stripper V-102 1 19000 Total = 164000 Quantity US $ 1 7600 Table 12.8: Equipment cost for 2 phase Separators Equipment Identification No.7: Equipment cost for Stabilizer Identification Equipment No.000 Quantity US $ Table 12.Glycol Absorber T-101 4 220.000 Total = 480.6: Equipment cost for Regenerator Column Identification Equipment No. Stabilizer V-105 Table 12. .= $ 958300 443.26 390.088 million.4 = $ 1. 762 Land 6 0.36 Construction & expenses 41 0.32 73.511 Instrumentation and control 20 0.46 488 5.12 Indirect Cost Fixed capital investment (Indirect + Direct cost) Working capital investment (15% of total capital investment) Total Capital Investment Ref: Plant Design and Economics by Peter and Timmerhaus.74 Electrical (installed) 11 0.088 Purchased equipment installation 47 0.12 Building 18 0.22 Piping (installed) 68 0.2 0.9: Estimation of Capital Investment Items Percentage of Purchased Cost Equipment Cost Million $ Purchased equipment 100 1.10: Manufacturing Expenses . page-251 Table 12.1088 Service facilities 70 0.24 Contingency 42 0.81 Engineering and supervision 33 0.196 Yard improvement 10 0.0653 Total direct plant cost 350 3.Table 12.45 Contractor’s fee 22 0.8 561 6. 55 1.124 0.532 0.32 109 scf/year Gas price = $13.18 $ 13.1608 1.4% of fixed capital) Operating labor Supervisory & Clerical labor Utilities Maintenance & repair (20% of FC) Opt.5/1000 scf Total income = Total sale − total production cost 13.064 0.2 0.24 106 = $ 18.92 105 ft3/ hr = 2.24 .1064 0.66 $ 0.0213 0.24 0.75/1000 scf 0.01064 $ 10.82 0.6 $/lb (10% of RM) (10% of Opt.5851 $ 11. labor plus Major Raw Material supervision & Raw Gas maintenance TEG Local taxes DEAmine (9% of Fixed capital) Cost of Raw materials Insurance (0.04 0.5 × 2. labor) 0.32×10 9 = 1000 – 13. expenses (10% of resistance & repair) Laboratory charges (10% of opt.4788 $ 8.804 0. storage (50% of opt.02 0.1 million 7.(Working day basis 330day/yr) Cost type Direct Overhead (payroll & plant).0804 0.085 Cost /yr $ Million 3.75 $/lb 0. labor) Total direct manufacturing expenses Total annual indirect manufacturing expenses Total manufacturing expenses Depreciation (10% of fixed capital for machinery & equipment) General Expenses Administrative Cost (20% of Operating labor) Distribution of selling cost ( 10% of total manufacturing expenses) Research & development (5% of total expenses) Total annual expenses Economic Analysis Percent rate of return Gas production = 2.Item Packing. Net profit after tax = $ 15. 20) = 4. 15%.12 million Depreciation of fixed capital investment (FCI) Project life = 20 yr Salvage value at the end of the project life = 0.32 = $ 4.532 million Total depreciation = 0. = Depreciable FCI (A/P.765 million/yr Cash flow diagram .9×5.4 million (15% Tax) Total investment cost = $ 6.788 = $ 0.788 million Now depreciation by sinking fund method with 15% interest rate.1×5.32= $ 0. 452 million 3rd yr P2 = -$3. F1= -$7.4 million So.24 million Profit = $15. P=R ( 1+i )n −1 i × ( 1+i )n Assumed MARR is 15% 1st yr P0= . net cash P1 = -$4.12million So.612 million Cost = -$13.Payback period calculation Formula uses F=P ( 1+i )n . F2= -$5. net cash P2 = -$3.4 million So.452 million .04 million Cost = -$13.$6.24 million Profit = $15.88 million So.88 million 2nd yr P1 = -$4. which indicate the payback period.96 yr (3 yr 11month 15 day) 5th yr P4= $0.So. By linear interpolation.24) = $ 2. R= $ (15.81 million So.24 million Profit = $15. i%. net cash P4 = $0.81 million 4th yr P3= -$1.252 million IRR calculation In annual worth (AW) method: Total annual revenue. actual payback period is 3.08 million So. i%.08 million Cost = -$13.16 million AW= -$6. F4= -$2. 20) . net cash P3 = -$1. net positive cash flow produced.4 million So. 20) + $ 2.4 million So.08million Here.4 million So.92 million Cost = -$13.4 – 13.532 million (A/F.24 million Profit = $15.97 million Cost = -$13.16 million + $ 0. net cash P5 = $ 2.12 million (A/P. F5= $0.24 million Profit = $15. F3= -$3. i% (assumption) 20 40 AW($. for AW is ‘0’ i% = 35. €%.903 -0. N) = ∑ Rk(F/P.million) 0.2897 By linear interpolation. i%.k)(F/P.17% > 15% (MARR) ERR calculation ∑ Ek( P/F. €%. i% = 19.67 % > 15% (MARR) . €% = 15 % N = 20 yr Using the above equation. N-k) Where.