Heat Exchanger Design Basis

TP - heat exchanger design.ppt*Transport ProcessesOverall heat transfer coefficientFrom previous studies (CPP module):Q=UATLMSome typical U values (all in W/m2K):steam/water:6000 to 18000water/water:850 to 1700steam condenser (water in tubes)1000 to 6000ammonia condenser (water in tubes)800 to 1400alcohol condenser (water in tubes)250 to 700finned tube (air outside, water inside)25 to 50deduce others from charts
TP - heat exchanger design.ppt

TP - heat exchanger design.ppt*Transport ProcessesOverall heat transfer coefficientContains many combinationsMay need to transpose top and bottom fluidsGives rather conservative estimates

TP - heat exchanger design.ppt*Transport ProcessesChoosing right shell-and-tube typeDecision as to TEMA code used depends on fluids used
Shell& tube exchangers
Severe thermal exapansion stresses?
Are bellows allowed?
Is chemical cleaning possible?
High shellside fouling > 0.00035 m2K/W?
High tubeside fouling > 0.00035 m2K/W?
Removable bundle design
Fixed tubesheet
Is interstream leakage allowed?
Are T & P within range for lantern ring?
Is F correction factor < 0.75?
Are there more than 2 passes?
Are F or multi shells allowed?
Frequency of bundle removal
AEL
AEM
BEM
No
No
Yes
BET
AET
No
Yes
Yes
No
No
BEW
AEW
No
Yes
Yes
No
No
BEP
AEP
No
Yes
Yes
No
No
Yes
Yes
No
No
No
AEU AFU
AEU
AFU
No
Yes
No
Yes
Yes
No
No
BES
AES
Is tubeside fouling > 0.00035 m2K/W?
Do we require tube access without disturbing connections?
No
Yes
Yes
Yes
Yes
No
No
Yes
Yes
No
Yes
No
Yes
No
Yes
No
Yes
No
Low
High
Yes

TP - heat exchanger design.ppt*Transport ProcessesLog Mean Temperature Differencee.g.find TLM for both co-current & counter-current flowFluid ATin =120Tout =90CFluid Btin =20tout =80C

TP - heat exchanger design.ppt*Transport ProcessesLog Mean Temperature Differencei.e. less driving force with co-current

TP - heat exchanger design.ppt*Transport ProcessesLog Mean Temperature Differencei.e. more driving force than either of the first twosame value for both co and counter-currentNow make fluid A condensing steam. What happens?

TP - heat exchanger design.ppt*Transport ProcessesLog Mean Temperature DifferenceWhy a log mean temperature difference rather than any other?Consider point along heat exchanger tube:At this point: T = T t d(T) = dT dtalso dQ = -mhCphdT = mcCpcdt(sensible heat change)

TP - heat exchanger design.ppt*Transport ProcessesLog Mean Temperature DifferenceHence:= U(T t).dAassuming constant Cph & Cpc:But

TP - heat exchanger design.ppt*Transport ProcessesLog Mean Temperature DifferenceGiving= UA.TLMCounter-current derivation also true for co-current flowCo-current flow rarely used in practice

TP - heat exchanger design.ppt*Transport ProcessesLog Mean Temperature DifferenceShell & tube exchanger NOT in true counter-current flow if there is more than one tube-side passTLM < TLM for pure counterflowIn this case, calculate TLM as if for counterflow. Multiply by correction factor F to give true value:

TP - heat exchanger design.ppt*Transport ProcessesLog Mean Temperature DifferenceAlternatively, use charts to evaluate F.F should be high (0.75 to 1.0) for efficient operationIf F > 0.75 inachievable, use single tube-side passF then becomes 1

TP - heat exchanger design.ppt*Transport ProcessesDutiesFor sensible heat (i.e. no boiling or condensing)QH = mH CPh (Tin - Tout) QC = mC CPc (tout - tin) For latent heat (boiling and/or condensing)Q = m fgFor perfect balance, QH = QC i.e. heat lost by hot fluid = heat gained by cold fluidIn reality, heat losses always occur

TP - heat exchanger design.ppt*Transport ProcessesFoulingStandard formula for U assumes clean surfacesIn reality, surface fouling increases thermal resistanceExternal fouling layerInternal fouling layer

TP - heat exchanger design.ppt*Transport ProcessesFoulingOccurs for a number of reasonsSlimy film through microbial activity in waterPrecipitation of dissolved saltsReaction of fluid alone (eg. polymerisation of hydrocarbons)Reaction of surface with fluid (eg. corrosion)FreezingSilt

TP - heat exchanger design.ppt*Transport ProcessesFoulingDynamic problem by natureFouling resistanceTimeCan be held in check byRegular cleaningHigh flow velocitiesLow temperaturesUse of special devices and/or chemical additivesTEMA and others usually quote this assymptotic value

TP - heat exchanger design.ppt*Transport ProcessesFoulingFouling resistances incorporated into formula:Designers assume static Rfo & Rfi. A few examples:FLUIDRf (m2K/W)Seawater & treated boiler water (50C)2 10-4 River water (

TP - heat exchanger design.ppt*Transport ProcessesMechanical considerations of shell-and-tube heat exchanger designTubes held in place by tube sheets with drilled holesHoles align the tubes in square or triangular arrangementDistance between centres of adjacent tubes = tube pitchOuter diameters:16, 20, 25, 30, 38, 50 mm, 2mm thickLengths:1.83, 2.44, 3.66, 4.88, 6.10, 7.32 metres

TP - heat exchanger design.ppt*Transport ProcessesMechanical considerations of shell-and-tube heat exchanger designBaffle spacing:minimum= Ds 5 (but > 5 cm) maximum= 74do0.75 (but < Ds)Baffle cut (segment opening height Ds) = 0.25 to 0.40eg. segmental baffle inside 1.00 m shell25% means segment 25cm high removedSmaller cut leaves smaller holeHigher shell-side film coefficientGreater shell side pressure drop0.25 m

TP - heat exchanger design.ppt*Transport ProcessesFirst design of a shell-and-tube heat exchangerCalculate duty Q (add 10% to include losses and errors)Find appropriate fouling resistancesChoose side for each fluid (based on fouling, corrosion and pressure)Choose type of exchanger from TEMA treeCalculate all temperatures TLM & FKeep things simple to start with; assume 4.88m tubes, do = 20 mm, 2 tube side passes (NP=2)

TP - heat exchanger design.ppt*Transport ProcessesFirst design of a shell-and-tube heat exchangerCool 5.0 kg/s of ethylene glycol from 370 to 330K with cooling water from 283 to 323KEthylene glycol at 350K (average) has following propertiesk = 0.261 W/m.KCp = 2637 J/kg.K = 0.00342 Pa.s = 1079.0 kg/m3Giving Pr = (26370.00342)/0.261 = 34.6Anticipate fouling resistance of Rf = 0.00018 m2K/WDuty is Q = 5.0 2637 (370330) = 527 400 WattsAim to transfer 580 140 W

TP - heat exchanger design.ppt*Transport ProcessesFirst design of a shell-and-tube heat exchangerWater at 303K (average) has following propertiesk = 0.618 W/m.KCp = 4179 J/kg.K = 0.000797 Pa.s = 995.6 kg/m3Giving Pr = (41790.000797)/0.618 = 5.39Anticipate fouling resistance of Rf = 0.0001 m2K/WWater fouls less and is on shell-sideWe need water flowrate3.16 kg/s water on shell-side

TP - heat exchanger design.ppt*Problem we cannot calculate a log meanSolution a log mean is just an average after allWhat is average of 47 and 47?T = 47, F = 0.87Transport ProcessesFirst design of a shell-and-tube heat exchanger

TP - heat exchanger design.ppt*Do we have severe expansion stresses?ie. are the temperatures greatly different to ambient?YesTransport ProcessesFirst design of a shell-and-tube heat exchanger
Shell& tube exchangers
Severe thermal exapansion stresses?
Are bellows allowed?
High shellside fouling > 0.00035 m2K/W?
High tubeside fouling > 0.00035 m2K/W?
Removable bundle design
Fixed tubesheet
Is interstream leakage allowed?
Are T & P within range for lantern ring?
Is F correction factor < 0.75?
Are there more than 2 passes?
Are F or multi shells allowed?
Frequency of bundle removal
AEL
AEM
BEM
No
No
Yes
BET
AET
No
Yes
Yes
No
No
BEW
AEW
No
Yes
Yes
No
No
BEP
AEP
No
Yes
Yes
No
No
Yes
Yes
No
No
No
AEU AFU
AEU
AFU
No
Yes
No
Yes
Yes
No
No
BES
AES
Is tubeside fouling > 0.00035 m2K/W?
Do we require tube access without disturbing connections?
No
Yes
Yes
Yes
Yes
No
No
Yes
Yes
No
Yes
No
Yes
No
Yes
No
Yes
No
Low

Heat Exchanger Design Basis

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