Where We Go Wrong In Line Sizing - University of Western Ontario
Design of Heat Exchangers Dick Hawrelak Presented to CBE 497 on 31 Oct 00 at UWO Introduction Design using HTRI and based on TEMA Stds TEMA Shell & Head Types, Perry VI, page 11-4 TEMA nomenclature, Perry VI, page 11-6 Liquid / liquid exchanger design example RW Condenser example on CD-ROM TEMA BEM Exchanger
B E M Plant Design, (11) Exchangers Heat Recovery Efficiency Colburn heat transfer method for hi CLMTD Correction Factor, Perry VI, p-10-27 Heat Exchanger Materials
Liquid Liquid Exchanger design example RW Condenser design example Shell Size V1.1 for kettle shell diameter Tube Count Exchanger Comparison Approximate Design Method Tube Count Exchanger Comparison Exchanger Shell Side Tubeside No. Exchangers Q, mm BTU/HR T1, hot in T2, hot out t1, cold in
t2, cold out LMTD = Area, sf Tube L, ft. Tube OD, in. No. Tubes No. Tube Passes Tubesheet Tube Pitch Shell ID tubes Shell ID expanded Flux = Q/A = U Back Calc'd = Dow RD E-202 Steam CTW Surf Cond 1
356.55 Quick Approximate Method Assume Design Ud values, Perry VI, p-10-44. BTU/hr & temperatures from process simulation Assume heating or cooling temperatures Calc LMTD, correct to CLMTD, if required Calc Area = Q / Ud / CLMTD Approx Method Continued
Assume tube od, BWG, tube length, to calc no. tubes (Table 11-2) Assume no. tube passes. Determine shell diameter, Perry VI, Table 11-3 tube count Assume materials & get cost estimate for exchanger Pressure Drop
Exchanger area vs pressure drop. Economics often dictate pressure drop. The designer sets the allowable pressure drops during simulation of process. Confirm pressure drops during exchanger design. Nozzle sizes, baffle spaces, tube dia., tube length, no. tubes per pass all affect pressure drop. Fouling and Overdesign Fouling factors are specified to give the exchanger a cleaning cycle (eg 1 year). In clean hydrocarbon services, a dirt factor
of 0.001 is specified on both sides. The combination of heat transfer coefficients, fouling and material resistance allow calculation of a clean heat transfer coefficient, Uc Over-design Problems Exchanger is designed with a Ud and a corresponding fouled CLMTD. On start-up, the exchanger operates with a Uc and a clean CLMTD. This may result in flow problems for condensing systems. Which steam pressure or refrigerant level should be used?
Temperature Profiles Manual calculations use average in & out temperatures. Subcooling affects LMTD. Partial condenser temperature profiles with inert gases are difficult to model. Good VLE data hard to obtain. Mechanical Design
High RHO-V-SQUARE on inlet shell nozzle can rupture tubes. Impingement plate design not well defined. Tube vibrations with long tube spans. How to join tubes to tubesheet? Maldistribution Shell side maldistribution with small window cuts. Use 20% baffle cuts.
Tube side maldistribution with low tube side pressure drops. Long tubes, small tube diameters. Chinese hat diffusers on tube and shell sides. Acoustics Shell side geometry can cause acoustic vibrations. May require tuning baffles. Entrainment
Minimize entrainment in Kettle refrigeration coolers. See Shell Size V1.2. Entrainment levels often ignored on mass balances. Kettle vapor outlets flow to KO pots in refrigeration compressor design. Expansion Joints. Expansion joints when shell and tubes are different materials.
Expansion joints are a hazard. Expansion joints are fragile. No. flexes per hour usually unknown. Paper clip example. Reboiler Recirculation Problems Low Recirculation due to inert build-up in shell, high tube resistance, low liquid level in column. Low recirculation promotes fouling and unwanted heavies production. Seadrift EO tower explosion due to faulty reboiler design,
Thermosyphon Layout Design of Heat Exchangers Method by Lord, Minton and Slusser, of UCC 26 Jan 70, Chemical Engineering, p-96. Methods suitable for all types of exchangers. Method suitable for spreadsheet analysis. See Liquid Liquid Exchanger and RW Condenser in Plant Design, Exchangers. Alternatively, Process Heat Transfer by Kern
Article Example - Liquid / Liquid Exchanger Input Data Conditions Flowrate, lb/hr = Inlet Temperature, C = Outlet Temperature, C = Viscosity, Centipoise, Z = Specific Heat, btu/h/F = Molecular Wt. = SG Ref to Water, s = Allowable DP, psi = Maximum Tube Length, ft = Minimum Tube Dia. inches = Inside ID, inches = Material Construction = Thermal Conductivity, metal = No. Passes, n = Fouling, h =
(Wi)(ci)(tH tL) / [(hi)(A)(dTM)] + Heat Transfer Coefficients hi = 0.023ciGi/(ciui/ki)^0.67/(DiGi/ui)^0.2 hw = 24kw / (do di) ho = 0.33coGo(0.6)/(couo/ko)^0.67/(DoGo/ko)^0.2 hs = assumed value Arrange Equations Into 4 Factors
For example for dTi/dTM for inside tubes, no phase change, liquid, Nre > 10,000 Numerical factor, f1 = 10.43 Physical Property Factor f2 = (Zi^0.467Mi^0.22)/si^0.89 Work factor f3 = Wi^0.2(tH tL) / dTM Mechanical Design Factor, f4 = di^0.8/n^0.2/L dTi / dTM = (f1)(f2)(f3)(f4) Similarly for hw, ho and hs
Pressure Drops Tubeside pressure drop, psi, Eqn (21) DP = (Zi^0.2/si)(Wi/1000/n)^1.8((L/di)+25)/(5.4di)^3.8 Shellside pressure drop, psi, Eqn (25) DPs = (0.326)/So(Wo/1000)^2(L)/Ps^3/Ds Step 1: Calculate Heat Duty Step 1 Calc Heat Transfer, Q = (m)(cp)(DT) from shellside data. Note Temps in Deg C require 1.8 factor for deg F. Q = = (32,800)(0.9)((90 - 45)(1.8) Q = 2.39E+06 BTU/hr Step 2:
Calc Temp Decrease Hot Liq = Q / m / cp for the tubeside DT = DT = (2.39E+06) / (307,500) / ((0.72) C 6.00 Tubeside Inlet Temp, Ti = Tubeside Outlet Temp, To = C 105 99 C C Step 3 Calculate LMTD
FT fr CLMTD = 1.00 (May Have to Correct if 1-2 Exchanger) see CLMTD program 105 90 15 CLMTD = 99 45 54 30.45 C Corrected LMTD = CLMTD = LMTD*FT
Step 4 Assume Ud and make First Approximation Of Area Assume Ud = A = Q / Ud /LMTD = 250 btu/hr/sf/F 174.52 (See Perry 6, Page 10-44) sf Step 5 Calculate No. Tubes for L = Area / Tube = (Do)(L) = No. Tubes =
No tubes per pass = 88.88 12 ft 1.9635 sf No. tubes = A / Area per tube 89 Re = 6.31W/d/cP = 25,908 Step 6
Approximate Shell ID = 1.75(OD)(Nt)^0.47 Shell ID = 9.01 Inches or (See Perry 6, 11-13) Step 7 Calc Max No Tubes That still gives turbulent Re > 12,600 Nt max = Wi / (2*di*Zi) Nt Max = 183 To Keep Re above 12,600
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