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

Dow RD Dow FSExchanger E-202 E-652Shell Side Steam SteamTubeside CTW CTW

Surf Cond Surf CondNo. Exchangers 1 1Q, mm BTU/HR 286.48 87.000T1, hot in 130 106.35T2, hot out 130 96.35t1, cold in 82 72t2, cold out 106.9 80.6LMTD = 34.05 25.04Area, sf 23600 9,743Tube L, ft. 36 24Tube OD, in. 1.00 1.00No. Tubes 2,504 1,551No. Tube Passes 1 1

Tubesheet Fixed FixedTube Pitch 1.25" 30° Tri 1.25" 30° TriShell ID tubes 70 54.41Shell ID expanded 96 74.62Flux = Q/A = 12,139 8,929U Back Calc'd = 356.55 356.55

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

Input Data

Article Example - Liquid / Liquid Exchanger

Conditions Tubeside Shellside

Flowrate, lb/hr = 307,500 32,800Inlet Temperature, °C = 105 45Outlet Temperature, °C = ? 90Viscosity, Centipoise, Z = 1.7 0.3Specific Heat, btu/h/°F = 0.72 0.9Molecular Wt. = 118 62SG Ref to Water, s = 0.85 0.95Allowable DP, psi = 10 10Maximum Tube Length, ft = 12Minimum Tube Dia. inches = 0.625Inside ID, inches = 0.495Material Construction = cs csThermal Conductivity, metal = 26 26No. Passes, n = 1 1Fouling, h = 1000Therm Cond, btu/hr/sf/(°F/ft.) = 0.102 0.183Baffle Spacing, inches = 5.000

Heat Balances Tubeside: (Wi)(ci)(tH – tL) = (hi)(A)(dTi) Tube walls: ((Wi)(ci)(tH – tL) = (hw)(A)(dTw) Fouling: (Wi)(co)(tH – tL) = (hs)(A)(dTs) Shellside: (Wo)(co)(TH – TL) = (ho)(A)(dTo)

dTi + dTw + dTs + dTo = LMTD = dTM

dTi/dTM + dTw/dTM + dTs/dTM + dTo/dTM = 1

Heat Balances Continued

Tubeside: (Wi)(ci)(tH – tL) / [(hi)(A)(dTM)] +

Tube walls: ((Wi)(ci)(tH – tL) / [(hw)(A)(dTM)] +

Fouling: (Wi)(co)(tH – tL) / [(hs)(A)(dTM)] +

Shellside: (Wo)(co)(TH – TL) / [(ho)(A)(dTM)] +

= 1.0

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 = (2.39E+06) / (307,500) / ((0.72) °C

DT = 6.00 °C

Tubeside Inlet Temp, Ti = 105 °CTubeside Outlet Temp, To = 99 °C

Step 3Calculate LMTD FT fr CLMTD = 1.00

(May Have to Correct if 1-2 Exchanger)105 see CLMTD program90 9915 45

54

CLMTD = 30.45 °C Corrected LMTD = CLMTD = LMTD*FT

Step 4Assume Ud and make First Approximation Of Area

Assume Ud = 250 btu/hr/sf/°F (See Perry 6, Page 10-44)

A = Q / Ud /LMTD = 174.52 sf

Step 5

Calculate No. Tubes for L = 12 ft

Area / Tube = (Do)(L) = 1.9635 sf

No. Tubes = 88.88 No. tubes = A / Area per tube

No tubes per pass = 89 Re = 6.31W/d/cP = 25,908

Step 6

Approximate Shell ID = 1.75(OD)(Nt) 0.47 or (See Perry 6, 11-13)

Shell ID = 9.01 Inches

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

Step 8 Heat Transfer Calcs

Step 1 Tubeside - Use Eqn (1)

Numerical Phys PropFactor, f1 Factor, f2

DTi / DTm = 10.43 (Zi 0.467Mi 0.22/Si 0.89)

DTi / DTm = 10.43 4.27

Step 8 Continued

Work MechanicalFactor, f3 Design Factor, f4

(Wi) 0.2(th-tL)/LMTD (di 0.8/n 0.2/L

0.620 0.0193

Step 8 Continued

dTi/dTM = (f1)(f2)(f3)(f4) dTi/dTM = (10.43)(4.27)(0.62)(0.0193) = 0.5339

Similar Calculations for tube wall, fouling and shell side.

Sum of Products SummaryTrial >>> Case 1 Case 2 Case 3 Case 4Total No. Tubes = 89.00 109 109 121No. Tube Passes 1 1 1 1No. Tubes per pass = 89.00 109 109 121Shell Dia., inches = 9.02 9.92 9.92 10.42Baffle Spacing, inches 5.0 5.0 3.5 3.5Exchanger Area, sq. ft. = 174.8 214.0 214.0 237.6

Product Of FactorsEqn (1) Tubeside 0.5339 0.5127 0.5127 0.5021Eqn (11) Shellside 0.4228 0.3655 0.2951 0.2738Eqn (17) Tube Wall 0.0520 0.0424 0.0424 0.0382Eqn (19) Fouling 0.2497 0.2039 0.2039 0.1837Sum of products, SOP = 1.2584 1.1246 1.0541 0.9978

SOP Message Add Area Add Area Add Area Exchgr OKEqn (21) Tubeside DP = 14.31 9.93 9.93 8.23Eqn (25) Shellside DP = 3.93 3.57 10.42 9.92

End of Presentation

Good luck on your exchanger designs.

If you have any questions call

rhawrela@xcelco.on.ca