Thermal Design of Heat Exchanger (E-100) Tube side (Cold fluid)Inlet fluid: Glycol to be preheated Outlet fluid: Preheated glycol to absorberInlet temperature: 67.59o FOutlet temperature: 770FTavg = 72.2950F

Density (by interpolation)=70.58 lb/ ft3

Viscosity = 22.52 cP=54.5 lb/ft-hrThermal conductivity K= 0.192 Btu/hrft0FHeat capacity Cp =0.5891 Btu/lb0FFlow rate, Wg = 4.88 x 105 lb/hrq = WgCpt = (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 gasInlet temperature: 99.880 FOutlet temperature: 94.860FTavg = 97.370F

Density (by interpolation)= 3.601 b/ft3

Viscosity = 0.01362 cP=0.0333 lb/ft-hr Thermal conductivity K = 0.023875 Btu/hrft0FHeat capacity Cp = 0.5105 Btu/lb0FFlow rate,Wi = 1.05 x 106 lb/hr q = WiCpt = (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)HotColdDifference

Higher99.88 (T2)67.59 (t1)32.29

Lower94.86 (T1)77 (t2)17.86

Difference5.029.4114.43

S

(LMTD)' = R = S = = 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 numbersQ = Uassume Arequired LMTDFor light - heavy organicsOverall U = 10-40 Btu/hr ft20FAssume, Uassume = 35 Btu/hr ft20F

A =

Iteration- 1The first iteration is started assuming 1 shell pass & 2 tube pass(np). Fixed tube plate 1in. OD(d0), 14 BWG, in. square pitch, ID (di) 0.834 in.Assuming tube length (Lt) is 20 ft.So, no of tubes, nt = = 562 Nearest count from table 9(kern) Page-841 is 574Now, Reynolds no, Re = = 892 < 104Iteration- 22nd iteration is started assuming 1 shell pass & 4 tube pass(np). Fixed tube plate 1in. OD(d0), 14 BWG, in. square pitch, ID (di) 0.834 in.Assuming tube length (Lt) is 30 ft.So, no of tubes, nt = = 561 Nearest count from table 9(kern) Page-841 is 562Now, Reynolds no, Re = = 1100 < 104Iteration- 33rd iteration is started assuming 1 shell pass & 4 tube pass(np). Fixed tube plate 1.25 in. OD(d0), 16 BWG, in. square pitch, ID (di) 1.12 in.Assuming tube length (Lt) is 26 ft.So, no of tubes, nt = = 259 Nearest count from table 9(kern) Page-841 is 268Now, Reynolds no, Re = = 5647 < 104

Iteration- 44th iteration is started assuming 1 shell pass & 6 tube pass(np). Fixed tube plate 1.5 in. OD(d0), 16 BWG, in. square pitch, ID (di) 1.37 in.Assuming tube length (Lt) is 45 ft.So, no of tubes, nt = = 178Nearest count from table 9(kern) Page-841 is 182Now, Reynolds no, Re = = 16560 > 104 Therefore, our final tube no 182 & Reynolds no (Re) is 16560 selected shell ID (Ds) is 35 in.Now,

Nu = = 0.027 Re0.8 0.14So, hi = 175.74 Btu/hrft2 0F hio = hi x =151.92 Btu/hrft2 0F

Shell side calculationAssumption: 25 % cut segmental baffles Baffles spacing, B = 0.5 Ds = 17.5 in = 1.46 ft (half of the shell ID is selected)Pitch, Pt = in. square pitch = 0.15625 ftClearance, C = Pt do = 0.03125 ftShell ID, Ds = 35 in = 2.92 ft

So, flows area, as = = 0.8526 ft2Mass velocity, Gs = Wi / as = 1.23 x 106 lb/hrft2

Equivalent dia, De = = 0.124 ft Reynolds no, Re = = 465000So,

Nu = = 0.36 Re0.55 0.14So, ho = 184 Btu/hrft2 0F Clean overall heat transfer coefficient (Uc) calculationUc =-1 = 42.3 Btu/hrft20F

Over design calculation: % of over design = = 20.86 % < 30 %So, design is accepted.

Dirt factor calculation Clean overall heat transfer coefficient, Uc = 42.3 Btu/hrft20FAssumed overall Coefficient, Ud = 35 Btu/hrft20F

Dirt factor, Rd = = 0.00098Because for natural gas Rd, allowable=0.001 (Ref: Mechanical Design of Process System (vol-2))Therefore, Rd is acceptable.

Pressure drop calculation Tube side calculation

Flow area, a t =

= = 0.309 ft2

Mass velocity, Gt = Wg/at = 1.6 104 lb/hrft2For Ret = 16560From figure 26, tube side friction factors correlation (Kern) Page-836Friction factor, f = f = 0.288 ft2/ft2

Assume,=1

So, frictional pressure drop,

Return loss = 1.334 10-13(2np 1.5) = 0.34 Psi

So, total pressure drop in tube side,= + = 2.56 Psi < 8.7 psi which is acceptable.

Shell side calculationPitch, Pt = in. square pitch = 0.15625 ftClearance, C = Pt do = 0.03125 ftShell ID, Ds = 35 in = 2.92 ftFlows area, as = 0.8526 ft2Mass velocity, Gs = 1.23 x 106 lb/hrft2Reynolds no, Re = 465000

No of baffles, nb = = 30.82 ~ 31From figure 26, tube side friction factors correlation (Kern) Page-836Friction factor, f = 0.1296 ft2/ft2

=1(assuming)

So, pressure drop = = 9.38 Psi < 14.5 Psi, which is acceptable.

Mechanical design Tube side properties:Materials: Stainless SteelNo. of pass: 6 Shell side properties:Materials: Stainless SteelNo. of shell: 1Shell dia, Ds = 35 inWorking pressure, P = 1247 PsiDesign pressure, Dp = Wp 1.1 = 1371.7 PsiPermissible working pressure, f = 11000 PsiWelding efficiency, j = 0.9

Shell thickness, ts= = 2.088 in.Corrosion allowance =1/8 inch = 3.175 mmShell thickness including corrosion allowance = (2.088+1/8) = 2.205 in. Nozzles Take inlet and outlet nozzles as 100mm diameter. Vent nozzle = 25mm diameter Drain nozzle = 25mm diameter Relief Valve = 50 mm diameter. Nozzle thickness = [ P x Di ] / { 2 f J - P } = 0.68 mm Minimum nozzle thickness is 6mm and 8mm is chosen which includes the corrosion allowance. Transverse baffles Number of Baffles = 31 Baffle cut = 25% Baffle thickness = 6mm (standard)

Flange design Flange is ring type with plain face. Flange material: IS 2004-1962 Class 2 Carbon Steel Bolting steel: 5% Chromium, Molybdenum Steel Gasket Material: Asbestos Shell OD = 37.205 in Shell Thickness = 2.205 in Shell ID = 35 in Allowable stress for flange material = 100 N/mm2 Allowable stress of bolting material = 138 N/mm2

Using matche.com we have estimated cost for this heat exchanger is $158900.

P & I Diagram

14.1 Estimation of Total CapitalInvestment and Production CostTable 12.1: Equipment cost for Heat ExchangerEquipmentIdentification No.QuantityUS $

Floating head HEE-100180,000

Floating head HEE-1021158,000

Total=238,000

Table 12.2: Equipment cost for PumpsEquipmentIdentification No.QuantityUS $

Centrifugal PumpP-10129400

Centrifugal PumpP-100312900

Total =22300

Table 12.3: Equipment cost for CompressorEquipmentIdentification No.QuantityUS $

Reciprocating CompressorK-100235600

Table 12.4: Equipment cost for Storage TanksEquipmentIdentification No.QuantityUS $

Storage TankV-10616300

Table 12.5: Equipment cost for Adsorption ColumnEquipmentIdentification No.QuantityUS $

Glycol Absorber T-1014220,000

DEA ContractorT-1004260,000

Total = 480,000

Table 12.6: Equipment cost for Regenerator ColumnEquipmentIdentification No.QuantityUS $

Glycol StripperV-1031145000

DEA Stripper V-102119000

Total = 164000

Table 12.7: Equipment cost for StabilizerEquipmentIdentification No.QuantityUS $

StabilizerV-10517600

Table 12.8: Equipment cost for 2 phase SeparatorsEquipmentIdentification No.QuantityUS $

SeparatorV-10012200

SeparatorV-10111900

SeparatorV-1041400

Total =4500

Total Equipment cost in the year 2002 = $ 958300 Total Equipment cost at present (2014) = $ 958300 = $ 958300 = $ 1.088 million.

Table 12.9: Estimation of Capital InvestmentItemsPercentage of Purchased Equipment CostCostMillion $

Purchased equipment1001.088

Purchased equipment installation470.511

Instrumentation and control200.22

Piping (installed)680.74

Electrical (installed)110.12

Building180.196

Yard improvement100.1088

Service facilities700.762

Land60.0653

Total direct plant cost3503.81

Indirect Cost

Engineering and supervision330.36

Construction & expenses410.45

Contractors fee220.24

Contingency420.46

Fixed capital investment(Indirect + Direct cost)

4885.32

Working capital investment(15% of total capital investment)73.20.8

Total Capital Investment5616.12

Ref: Plant Design and Economics by Peter and Timmerhaus; page-251

Table 12.10: Manufacturing Expenses(Working day basis 330day/yr)Cost typeItemCost /yr$ Million

Major Raw Material

Raw Gas3.75/1000 scf7.2

TEG0.75 $/lb0.82

DirectDEAmine0.6 $/lb0.02

Cost of Raw materials $ 8.04

Operating labor(10% of RM)0.804

Supervisory & Clerical labor(10% of Opt. labor)0.0804

Utilities0.55

Maintenance & repair (20% of FC)1.064

Opt. expenses (10% of resistance & repair)0.1064

Laboratory charges(10% of opt. labor)0.01064

Total direct manufacturing expenses $ 10.66

Overhead (payroll & plant), Packing, storage (50% of opt. labor plus supervision & maintenance

0.085

Local taxes (9% of Fixed capital)0.4788

Insurance (0.4% of fixed capital)0.0213

Total annual indirect manufacturing expenses $ 0.5851

Total manufacturing expenses $ 11.24

Depreciation (10% of fixed capital for machinery & equipment)0.532

General ExpensesAdministrative Cost (20% of Operating labor)0.1608

Distribution of selling cost ( 10% of total manufacturing expenses)1.124

Research & development (5% of total expenses)0.18

Total annual expenses $ 13.24

Economic Analysis Percent rate of returnGas production = 2.92 105 ft3/ hr = 2.32 109 scf/yearGas price = $13.5/1000 scf Total income = Total sale total production cost= 13.24 106 = $ 18.1 millionNet profit after tax = $ 15.4 million (15% Tax)Total investment cost = $ 6.12 million

Depreciation of fixed capital investment (FCI)Project life = 20 yrSalvage value at the end of the project life = 0.15.32= $ 0.532 millionTotal depreciation = 0.95.32 = $ 4.788 millionNow depreciation by sinking fund method with 15% interest rate, = Depreciable FCI (A/P, 15%, 20) = 4.788 = $ 0.765 million/yr

Cash flow diagram

Payback period calculationFormula uses

Assumed MARR is 15%1st yr

P0= - $6.12millionSo, F1= -$7.04 millionCost = -$13.24 millionProfit = $15.4 millionSo, net cash P1 = -$4.88 million

2nd yr

P1 = -$4.88 millionSo, F2= -$5.612 millionCost = -$13.24 millionProfit = $15.4 millionSo, net cash P2 = -$3.452 million

3rd yr

P2 = -$3.452 millionSo, F3= -$3.97 millionCost = -$13.24 millionProfit = $15.4 millionSo, net cash P3 = -$1.81 million

4th yr

P3= -$1.81 millionSo, F4= -$2.08 millionCost = -$13.24 millionProfit = $15.4 millionSo, net cash P4 = $0.08million Here, net positive cash flow produced, which indicate the payback period.By linear interpolation, actual payback period is 3.96 yr (3 yr 11month 15 day)

5th yr

P4= $0.08 millionSo, F5= $0.92 millionCost = -$13.24 millionProfit = $15.4 millionSo, net cash P5 = $ 2.252 million

IRR calculationIn annual worth (AW) method:Total annual revenue, R= $ (15.4 13.24) = $ 2.16 million AW= -$6.12 million (A/P, i%, 20) + $ 2.16 million + $ 0.532 million (A/F, i%, 20)i% (assumption)AW($,million)

200.903

40-0.2897

By linear interpolation, for AW is 0 i% = 35.17% > 15% (MARR)

ERR calculationEk( P/F, %,k)(F/P, i%, N) = Rk(F/P, %, N-k) Where, % = 15 % N = 20 yrUsing the above equation, i% = 19.67 % > 15% (MARR)