Foreword Momentum,heatandmasstransportphenomenacanbefoundnearly everywhere innature. Even anearly morning activitysuch asboiling an egg* or makingteaisgovernedbylawswhichwillbetreatedhere.Asolidunder-standing of the principlesof these transport processes is essential for those who apply this science ine.g.chemicaland processengineers. The historyof teaching transport phenomena went from apracticalbut less fundamental approach, via a short period ofa practical and academic approach, tothepresentsophisticatedapproach.Ourexperienceineducationandin industry is that today's abstraction does not appeal to all students and engineers, whofeelthemselveseasilylostinvastliteratureanddifficultmathematics. Hence,our objective inwritingthisbookwasto digestthe enormous amount of newknowledge andpresentitinaformusefulforthosewhoworkaspro-fessional engineers or who study engineering. The present book incorporatesmuchfundamentalknowledge,but wehave alsotriedalwaysto illustratethepracticalapplicationof thetheory.On the otherhand,wehavealsoincludedpracticalinformationandhavenotshied awayfromgivingoneortwousefulempiricalcorrelations,wheretheory wouldhavebeentoodifficult.Thebookisbasedon thetextforacoursein transportphenomenagivenbyW.J.BeekatDelftUniversityfrom1962to 1968.Partsof thelastdrafthavebeenused,togetherwithmostproblems encounteredinthree postgraduate courses. Each chapter ends with anumber ofproblemswhichforman integralpart of the book. We would like to ask the student to solve as many of these problems as possible-this is the best way to absorb and digest the theory. We have given the answers of all problems so that the reader can check hisresults. Where we expected difficulties to arise, we have explained some problems in greater detail. Furthermore,weinvented John,our scientificsleuth,and wehope the reader likes his way of solving problems. The reason for his ability to do so is evident : he hasread the presentbook andhasworked through the problems! *ForidenticaltemperatureequalizationintheeggtheFouriernumbermustbeFo =atf d2 = constant,i.e. for all eggs: boiling timex(mass)-213=constant isvalid.Knowing the optimalbojling time for one species of egg,wecanthuspredictthe boiling time for other eggs, even ostrich eggs. Contents CHAPTER IINTRODUCTIONTOPHYSICALTRANSPORT PHENOMENA1 I.l.Conservation laws2 1.2.Rate of molecular transport processes11 I.3.Microbalances15 1.4.SIunits20 1.5.Dimensional analysis.21 1.6.Problems.28 CHAPTER IIFLOWPHENOMENA ILLLaminar flow37 1.Stationary laminar flowbetween two flathorizontal plates38 2.Flow through ahorizontal circular tube40 3.Flow through a horizontal annulus42 4.Flow caused by moving &urfaces45 5.Flow through pipes with other cross-sections48 6.Non-stationary flow49 7.Problems52 II.2.Turbulent flow.55 1.Turbulent flow in pipes55 2.Pressure drop in straight channels60 3.Pressure drop in pipe systems.64 4.Problems74 II.3.Flow with negligible energy dissipation79 1.Flow of a liquid from an orifice80 2.Flow of g a ~ e sthrough orifices82 3.Flow through,weirs85 4.Problems88 II.4.Flow meters,. _91 1.Venturitube92 2.Orifice plate93 3.Rotameter94 4.Problems95 II.5.Flow aroundobstacles.98 1.General approach.98 2.Spherical particles.101 3.Free fallof droplets104 4.Particles innon-stationary flow105 Vlll 5.Rate of sedimentation of a swarm of particles 6.Cylinders perpendicular to the direction of flow 7.Problems Il.6.Flow through beds of particles I .Fixed bed 106 107 108 112 112 115 116 117 120 120 122 124 124 125 125 127 129 131 137 140 2.Filtration through abed of particles 3.Fluidizedbed 4.Problems II. 7.Stirring and mixing 1.Types of stirrer and flowpatterns 2.Power consumption. 3.Pumping capacity and mixing time 4.Problems II.8.Residence time distribution 1.The F function 2.The E function 3.Simple applications ofF and E functions 4.Continuous flowmodels. 5.Dispersion in fl ow systems 6.Problems CHAPTER IIIHEATTRANSPORT III. I.Stationary heat conduction.145 I.Heat conduction through awall146 2.Heat conduction through cylindrical walls147 3.Heat conduction around a sphere148 4.General approach for the calculationof temperature distribu-tiom1 ~ 5.Temperaturedistributioninacylinderwithuniformheat production150 6.Problems152 III.2.Non-stationary heat conduction156 1.Heat penetration into a semi-infinite medium157 2.Heat penetration into a finite medium161 3.Influence of an outside heat transfer coefficient164 4.Problems167 III. 3.Heat transfer by forced convection in pipes171 I .Heat transfer during laminar flowin pipes171 2.Heat transfer during turbulent flow.174 3.Partial and total heattransfer coefficients176 4.Problems178 III.4.Heat exchangers.182 1.Determinationof mean temperature difference182 2.Height of atransfer unit.186 3.Design of heat exchangers187 4.rroblems189 III.5.Heat transfer by forced convection around obstacles 1.Flow along a fiatplate 2.Heat transfer to falling films 3.Flow around spheres and cylinders 4.Heat transfer in packed beds 5.Heat transfer in fluidizedbeds 6.Problems III.6.Heat transfer during natural convection. 1.Heat transfer during natural convection 2.Problems III. 7.Heat transfer during condensation and boiling 1.Film condensation. 2.Dropwise condensation 3.Boiling. 4.Heat transfer in evaporators 5.Problems III.8.Heat transfer in stirred vessels 1.Problem III.9.Heat transport by radiation. 1.Problems CHAPTER IVMASSTRANSPORT IX 194 194 194 195 197 198 199 201 201 204 208 209 211 212 213 214 215 219 220 223 IV . 1.Stationary diffusion and mass transfer227 1.Stationary diffusion227 2.Mass transfer coefficients231 3.Generalapproachforthecalculationof concentrationdis-tributions233 4.Film theory234 5.Problems236 IV.2.Non-stationary diffusion240 1.Problems242 IV.3.Mass transfer with forced convection246 1.Analogy with heat transfer246 2.Mass transfer 'during laminar flow248 3.Mass transfer during turbulent flow253 4.Problems.- 254 IV.4.Mass exchangers257 l.Thermodynamic equilibrium258 2.Choice of the apparatus259 3.Size determination of the mass exchanger260 4.The concept of theoretical plates263 5.Problems265 IV.5.Mass transfer with chemical reaction270 1.Slow homogeneous first-orderreactions271 2.Fast homogeneous first-order reactions273 3.Homogeneous nth-order reactions275 X 4.Homogeneous second-order reactions 5.Mass transfer withheterogeneous chemical reaction 6.Problems IV.6.Combined heat andmass transport 1.Drying. 2.Problems INDEX 276 280 283 292 292 294 296 CHAPTER I IntroductiontoPhysicalTransport Phenomena During the designing of industrial process plant qualitative and quantitative considerations playarole. On thebasisof qualitative(sometimessemi-quantitative) considerationsa preselection of feasible concepts of processes suitable for carrying out the desired production in an economical way is made. The type of operation, e.g. distillation against extractionor the choice of asolvent, willalsobe fixedby thistypeof reasoning, in which experience and a sound economic feeling play an important role. Assoon asone or two rough concepts of a production unit are selected, the different process steps will be analysed in more detail. This asks for a quantitative appro3.:chwiththeaidof amathematicalmodel.of theunitoperation.The experience that mass, energy and momentum cannot be lost provides the three conservation laws, on which the quantitative analysis of physical and chemical processeswhollyreliesandonwhichtheprocessdesignof aplant isbased. This kind of design, which aims at fixing the main dimensions of a reactor or an apparatusfortheexchangeof mass,momentumandenergyorheat,isthe purposeofthedisciplinesknownas'chemicalengineering'and'chemical reaction engineering'.The basic ideas behind these disciplines are found under theheadings'transportphenomena'and'chemical(reaction)engineering science',whichrely. on deductivescienceand,hence,havetheadvantageof analyticalthoughtbutwhich, becauseofthat,lackthebenefitof induction based on experience when aiming at asynthesis. Qualitative and quant-itative reasoning cannot be separated when setting up aplant, or to put it in another way: no apparatus, howevergood itsprocessdesignmightbe,cancompetewith awell-designed apparatus of a better conception, whichcanbethe device foraprocess designer,or noresearch,howeverbrilliantinconceptionitmightbe,canresultina competitive production plant without having aquantitative basis, whichcan beamotto foraresearch fellow. 2 Examplesof questions,inwhichfeelingandreasoninghavetomatchwell before science is used to some profit, are to be found in the following areas : the potentialpossibilitiesof rawmaterials,intermediateandendproducts,the choiceof materials andespeciallymaterialsof construction, theinfluenceof sidereactionsontheperformanceof subsequentprocessstepsandthecon-siderations on quality and end-use properties of a product. This type of question, although of importance for the integral approach of a design engineer, will not bedealt with in this book, whichwill findits limitations just there. This book treatsthepracticalconsequencesof theconservationlawsforthechemical engineer in an analytical way, trying notto exaggerate scientific nicety where somany other important questionshavetobe raisedandanswered, but also pretending that asolid understanding of the heart of the matter at least solves apart of all questions satisfactorily. The laws of conservation of mass, energy and momentum are introduced in paragraph 1.1.They are extended to phenomena on .amolecular scale inpara-graphs1.2and1.3.Paragraph1.4isconcernedwith .dimensionsof physical quantities,especiallySIunits,whereasparagraph1.5discussesthetechnique of dimensional analysis.We will end this chapter (and most paragraphs of the followingchapters) withsome proposals forexercising and comments on the solution of some of the problems given. After this, three main chapters follow, each of which concentrates on one of the conserved physical quantities: hydrodynamics (mainly momentum transfer) inchapterII, energytransfer(mainlyheattransfer)inchapterIIIandmass transfer in chapter IV. These chapters elaborate the ideas and concepts which are the subject of the following introduction. 1.1.Conservation laws Johnlooked at the still smoking ashes of what had once been theglueandgelatine factory.The fire hadstartedwithan explosioninthebuildingwherebonesweredefattedby extractionwithhexane.Johnrememberedthattheextrac-tion building had a volume of6000 m3 and that the tempera-tureinthebuildingwasalways3rfC higherthanoutside. He knew that per 24 h, 70 ton steam were lost as well as 9 ton hexane. He made a quick calculation and concluded that the steady-state hexanevapourconcentrationintheplantwas well below theexplosionlimit of 12percent by volume and chacsome accident musehaue happened which subsequently ledto the explosion. Physicaltechnologyisbasedonthreeempiricallaws:matter,energyand momentum cannot be lost. The lawof conservation of matter isbased among other things onthe work ofLavoisier,whoprovedthatduringchemicalreactionsnomatter,i.e.no mass (massbeingthemostimportant propertyof matter),islost. Thelawof conservationof matterisnotalwaysvalid:innucleartechnologymatteris transformed into energy but forchemicalor physical technology this exception 3 isof no importance.Itis,of course,possible that matter istransferredfroma desiredformintoan undesiredone(e.g.thedegradationof apolymer,which finallyleads to only C02,H20, etc.). The lawof conservation of energyisbased among other things on the work of Joule,whoprovedthatmechanicalenergyand heatenergyare equivalent. His work finally led to the first law of thermodynamics, which, when formulated foraflowingsystem,isthelawof energyconservationweare looking for.It ishistoricallyremarkablethatittookmorethantwocenturiesbeforethis law, formulated initially for a closed system, was translated into a form in which it could beapplied to flowingsystems. The law of conservation of momentum was finally formulated in its simplest formforasolid body by Newton : if the sum of the forcesacting on abody is different from zero, this difference is (in size and direction) equal to the acceler-ation of thatbody.Together withhis secondlaw, action equals reaction, this formedthebasisfordynamics andhydrodynamics.This timeit did nottake much more thanone centuryto transposetheconcept,originally formulated for arigid body, to the more general case of flow in fluids. These conservation laws play in daily life the same role as the experience that apoundcannotbespenttwiceandthatthedifferenceexistsbetweenthe poundyouowesomebodyandtheone somebodyowesyou.Theeconomic rulesandtheconservationrulesof ourstudyareusedinthesamemanner: balance sheetsaresetupwhichaccountforinflowand outflowandforthe accumulation of the quantity under consideration. Let us denote by Xa certain amount of money, mass, energy or momentum. Then the general law of on which all phenomenological descrip-tions of change in the physical world are based; reads as follows: accumulation of Xin system unit time flow of Xinto systemflow of Xout of systemproduction of X in system ..+.. umt timeumt timeunit t ime (1.1) The system may be acountry, aconcern, afactory,an apparatus, a part of an apparatus (e.g.a tray.),a...pipe or an infinitely small element of volume, etc. This sounds very general and easy, but daily practice proves that we have to develop the qualitative judgement fordefiningthesystemsuchthattheanalysisstays as easy aspossible.To this end, to develop a feelingforthe qualitative aspects ofanalyticalscience,wehavetogothroughmanyaquantitativeexercise. Introductions into adiscipline, suchas thisone,may easilyconfusethe reader if these points are not made clear in the beginning; the subject of our study is thethreelawsof conservation, whichwillappear inmanyformsbecausewe will study many different systems and not because the basic rules are many. If we are only interested inmacroscopic properties like mean concentration in the chosen control volume or the rate of change of the mean temperature in 4 thisvolumewecanchooseamacroscopiccontrolvolume,e.g.acomplete reactor, a complete catalyst particle, etc., for setting up abalance of the desired physicalproperty. If, ontheotherhand,weareinterestedin temperature or concentration distributions we have to start by setting up a microbalance, Le. a balanceoveraninfinitesimallysmallvolumeelement,andtointegratethe differentialequationsobtainedoverthetotal(macroscopic)volume.Both balances willbe treated in d e t a i ~in this and in thefollowing paragraph. Letusnowtrytoformulatethelawof conservationwhichwehavejust defined inwords(equation I. l) inamore preciseway. In order todo so, we need somesymbols : V for the volume of the system (space volume, number of inhabitants, etc.), v, inandcf>v.outfortheingoingand, outgoingvolumetricflowrates(see Figure IJ), rforthe velwnetric production of Xper unit of time and Xforthevolumetric concentration of the physical quantityin study. Cont rol vol ume r--1--------, II I v, out 1 V,r II L-----------J Figure J.lMacrobalance The accumulation per units of volume and time can nowbe denoted by dX/dt ; hence,theaccumulationinthesystemperunitoftimeisgivenbyVdX/ dt. A flowat a rate cf>v,containing a volumetric concentration Xof the considered quantity, represents a flow rate vX of this quantity. Hence, our formulation of the conservation law reads as follows: (1.2) We will now apply this law to the following quantities: money, mass, energy and momentum. Themoney balance Let us start withthe mostcommondaily practice,theconservationlaw for pocket money (the pocket being definedas the system, although some operate isveryunsystematically). ThenumberofpenceinthesystemisgivenbyX ( J = 1),and equation(1.2)can be readasfollows:the accumulationof pence *In order to be completely exactwe have to write this equation as : - u dX- 1-1 vdt=cl>u, inxin- rPu,oulxOUI+Vr ""'here -. ndic.ltes avol ume average and-!a fl ow average. 5 in my pocket pt;r week (whichmay prove to be negative!) equals the difference between the number of pence I have taken in during this weekand thenumber Ihave spentinthis time, increasedbythenumber of penceIproducedinthe meantime. The last contribution sounds somewhatcryptic or even illegal, but anhonestproductionofpencewouldbetochangeothercoinsintopence. Similarlawscanbeexpressedfortheother coins,aswellasfortheoverall contents of the pocket (moneywise).Fromallthese statements itfollowsthat the sum of all the production terms rmust equal zero (expressedas an intrinsic valueandnotasnumbers)or,toputitinanotherway, changing initsown currency never results in a positive gain. To find out how simple these statements might be, try and see what happens to your thinking when Xno longer stands formoney,but formass,energy or momentum! The mass balance The mass balance still looks familiar and comes close to the money balance. Here X= cA,the volumetric concentration of component A in amixture; the units in which cAis measured are kg/m3 * Hence, in a rayon factory, for instance, whereNaOH (in the formof viscose) and HCl (inthe formof the spinning bath) areused, the conservation law for NaOH {A) reads as follows: the accumulation of NaOH on the site in amonth (or any other chosen time unit) is equal to the delivery ofNaOH to the factory in that time, subtracting the amount of NaOH distributed fromthe site inthat unit of time as NaOH and adding the amount ofNaOH produced in the factory duringthattime(aproductionwhichisnegativewhereNaOHisusedasa reactant). A similar mass balance can be set up forthe other reactant, HO (B), as well asfortheproductsNaCl (C) andH20(D).Again,thesumof allindividual production rates must be zero: r A+ r8 + rc+ r0 = 0 (Lavoisier:t.More olten than not, the accountants of afactory are more aware of the implications of the conservation lawsandthe dynamic consequences of varying inflows and out-flowsof afactorythan the engineers in charge of production. 'f#>v,i n l#>v, out vc Figore1.2Massbalanceofastirred vessel The wellstirred continuous flowtank reactor of Figure I.2 gives an example of theapplicationsof themassbalance.By'wellstirred'wemeanthatthe * Fromthe onset, we willacceptkg, m, s,oc asthe basic units for comparing physical phenomena (paragraph1.4). 6 concentration c of acertain compound (e.g. salt) is the same at all. places in the vesselsothatthesaltconcentrationintheeffluentstreamequalsthecon-centration in the vessel.Letus assume that the salt concentration in the vessel at time t= 0 isc0 and that, fromt= 0on, acontinuous stream of pure water (saltconcentrationc= 0) is passedthroughthevessel.Thequestionisthen how does the salt concentration in the vesselchange with time? We can now set up two mass balances, one for the water and one for the salt. The first balance says that, if the liquid volume in the reactor isconstant, the flowrate of liquid out of the reactor must equal the flowrate into the .reactor, v,out= v.in .The second balance says that the decrease in the amount of salt in the vessel (Vc) per unit of time must equal the mass flow rate of salt from the vessel (vc). Thus : d(Vc)= 0 _c dtv or: dec - = dtr where r=Vf4>v is the mean residence time of the fluidin the vessel.Integration between t= 0, c0 and t , c yields: .:_=exp (!) CoT f / T -Figure 1.3Changeof saltconcentrationin stirredvessel ThisfunctionisshowninFigure1.3, whichshowsthatatthetimet='tstill 3""per cent of the initial salt is present in the reactor.For t=i r and ~ rthese values are 65and 22 per cent respectively. Apparently, the liquid inthe reactor has alarge distribution of residence time. 7 Theenergybalance Beforewecanattempttointerpretequation(1.2)intermsofanenergy balance, we must define the meaning of Xas' the volumetric concentration of energy'Er.Thisquantitywillcontaininternalenergyandpotentialenergy. Internalenergyagaincomprisesperceptibleandlatentheatandpressure energy. Pressure energy per unit of volume isnothing but an elaborate expres-sion formerely the pressure inside thevolume.Perceptible and latent heat per unit of volume (U) may be expressed as : (1.3) in whichpisthe specificgravity,cPtheheatcapacityperunitof mass,Tthe temperature,T,areference temperature forcalculatingU, !J.Hthelatentheat per firstorder phase transition (melting,evaporation)perunitmassandLa symboltoindicatethatthecontributionsof allphasetransitionsbetweenT,. and T have to be taken into account. If the process studied does not show phase transition,thereferencetemperatureT,.canbechosenhighenoughtokeep latentheatsoutof theanalysis.If, furthermore,theheatcapacityperunitof volume (pcp) isindependent of temperature,U=pcP(T - T,). The kinetic energyper unitof volume isfpv2,whenv isthevelocityof this volume. If different elements of the volume flowhave different velocities, then the flow average of the squared velocity has to be used for calculating the flow of kinetic energy, " -dxdydz = --" dxdydz- _Y dx dydz- _z dxdydz+ rdxdydz dtdxdydz" or dXd;d;dtp; - -----+r dxdydz (1.17)- = dt Equation (117) is the basic microbalance which can be further evaluated When doing so, wehave to realize that the fluxof Xonthis microscale consistsof a convective transporttermand astatistical transport term.These fluxes inthe n-direction are therefore given by* : dX 4 > ~=-([}or a orv)dn+ v,X(1.18) Thus, if we take II) as a representative of I!J, a or v, we obtain from equation (1.17) with equation (1.18): dXd(vxX)d(vyX) - =- ___;_;.__... dtdxdy (1.19) Inter?reted as the momentum balance, equation 1.18is only valid for Newtonianliquids. We will d;scuss this problem inmore detailimmediately. 17 If weinterpretequation(1.19)asamassbalanceforacomponentAin a mixture of substances A, B, C, etc.(thusX=CA. C8,etc.), weknow that : C A+ Ca + Cc+ ... =p(the specific gravity of that mixture) and rA+ r8+rc +... =0(Lavoisier) Summing up the massbalances forallcomponents wefind: dpd(t.'xP) -=---dtdx (1.20) which is known as the equation of continuity.For a stationary situation, dpjdt equals zero. When pis constant over the flow field, we find furthermore: dvxdv.vdv,. 0-+-+-= dxdydz (1.21) which is avery familiar expression forthe conservation of total mass. If equation (1.19) isto be used as a microbalance for the conservation of heat (or of energy if the transport of heatexceedsthetransportof other formsof energy, as given in equation 1.5),Xstands forpcPTThis microbalance will be further evaluated in paragraph III.l.4, whereas the micromassbalance forone component willbe discussed in more detail in paragraph IV.1.3. We willnow concentrate on the further development of equation (1.17) asa micromomentumbalance.Inordertoobtainarelationwhichisvalidfor liquids of different rheological behaviour (thus Newtonian and non-Newtonian fluids)wewill1 insteadof equation(1.18)usethemoregeneralexpression for the momentum fluxinthe n-direction : (1.22) whichisvalidforallrheologies.Here Xstandsforpv.The momentum pro-ductionterminthen-directionisthesumof thepressureand gravityforces acting inthat direction-onthecontrolvolume: dp rn =- - + pg dnn (1.23) Withtheaboveexpressionsweobtainfromequation(I.17)forthemicro-momentumbalanceinthex-direction : dt (1.24) 18 With the aid of the continuity equation (1.20) this expression can be simplified to : dvxdvxdv.xdvxdT.x.xdTy.xdT:xdp) Pdt=- pv- - pv- - pv- - - - - - - - - +pg{1.25 xdx,dyzdzdxdydzdxx Analogous expressions canbe writtenforthe micromomentumbalance in the y- and z-directions as follows : dvvdv>'dv,dvyd-r"Yd-r,,.dtz,.dp 1 ,6) pdt=- PVx dx- pv, dy- PVzdz- dx- dy- dz- dy +pgY(-and dvzdvzdvzdvzd-rxzd-ryzdtzrdp(l.27) p-- =- pv- - pv- - pv- - - - -- - - - - + pg d txdxYdyzdzdxdydzdzz If, insteadof aCartesianvolume element inrectangular coordinates, wehad considered avolume element in cylindrical coordinates-, asindicated in Figure 1.10, wewouldhaveobtained the followingmicromomentumbalancesinthe r-direction : ' z Figure1.10Volumeelementm cylindrical coordinates p dv,=_p {vdv,+ v6dv,_vi+ v dv,}_~d(rt,)_~d-r,6 + -r80 dt' drrd8rzdzrdrrd8r d-r,%dp ---- +pg dzdrr inthe8-direction : p dv8 =_p {v dv8+ v8dv8+v,v6+ vdv0}_~d(r2-r,9)_~dt89 dt'drrd8rzdzr2 drrd9 (1.28) d - r8~1dp - dz.- ~d8+ pge(1.29) 19 and in the z-direction: dv:{dv:v8 dv:dv:}1d(r't"rz)1 dr8%d-rndp pdt=- pVrdr+-;: d9+ Vzdz-;dr+; d9+dz- dz+ pg. (1.30) These micromomentum balances cannaturallyalso be formulatedin spherical coordinates.Therelationsobtained canbe foundin manyhandbooks. Theabovemomentumbalancesintermsof shear stressesarevalidforall fluids,becausethe shear stresses are independent of the rheological behaviour of the liquid.In the special case of Newtonian liquids -risproportional to the velocity gradient (equation1.13), e.g.: d(pvy) rxy=- vd.x Introducing this expression into the above microbalances weobtain, forrect-angular coordinates, in thex-direction : dvx_{dvxdvxdvx}{d2(pvx)d2(pv,)d2(pvx) } pdt- - PVxdx- l' ydy- v.dz+ vdx2+dy2+dz2 dp -- + pg(1.31) dxx and inthez-direction: dvz __{dv.dvzdvz}{d2(pvz)d2(pvz)d2(pvJ} pdt- pv xdx .+ v>'d y+ v zdz+ vdx2 +d y2 +dz2 - ~ ~+ pgz(1.33) InsphericalcoordinatesNewton' slaw(equation1.13)readsinthez- orr-directions : d(pv6)d(pv) 't"gz=- vdzandrzr=- vdr %.' respectively and inthe 9-direction: (1.34) 20 Introducingthisintothegeneralmicrobalanceintermsof shearstresseswe obtaininthe r-direction : dv,._{dv,.v8dv,.v ~dv,.}{ d( 1d(prv,.)) p-- - pv- +- --- +v- + v- - ___:.____:_ dt,.drrdOrzdzdrrdr +_!_d2(pv,) _~d(pv8)d2(p .v,.)}_ dp r2 d82 ~d8+dz2 dr+ pg,. inthe0-direction : dv6 p-= d t and inthe z-direction: dvz _{dvzv8dvzdvz}{ 1d(d(pvz))1d2(pvz) pdt- - pv,.dr+7dO+Vzdz+v;drrdr+r2 d02 d2(pvz)}dp +dz2- dz+ pgz (1.35) (1.36) (1.37) The micromomentum balances given above form the basis for the calculation of velocity distributions andflowrate- pressuredroprelations during laminar fl ow; we will apply them many times in the next chapter. 1.4.SI units Different systems of units are applied at the moment in different countries in industry,researchand development.The most importantsystems are : the c.g.s.system, based on the centimeter(em),gramme mass (g), second (s) anddegreeCelcius(C).Fortheamountof heatthecalory(cal)isused (derivedunit);* the SI system, based on the metre (m), kilo gramme (kg), second (s) and degree Kelvin(K).Theamountof heat(derivedunit)isexpressedasJoule(J), whichisidenticaltotheunitformechanicalorelectricalenergy(1J = 1 kgm2js2 =1 Nm=1 Ws, where N=Newton, W= Watt) ; the metric system, basedon the metre (m),kilo gramme force (kgf), second (s) and degree Celsius (C). The amount of heat is expressed in kilocalories (kcal) ; thefoot-pound-secondsystem,basedonthefoot(ft),poundmass(lb), second (s)anddegreeFahrenheit (F). Thederivedunitforthe amountof heat istheBritishThermal Unit (BTU) ; Theunitsforelectriccurrentandforluminous intensity willnotbe discussedherebecause they playnoroleinthisbook. 21 theBritishengineeringsystem,basedonthefoot(ft),poundforce(lbf), second (s)and degree Fahrenheit (F). For scientific research the c.g.s.system isused all over the world. Engineers in Anglo-Saxoncountries generallyusetheBritishengineering system; in other countries the metric system isgenerallyused. In the last fewyears the SI system (Systeme International d 'Unites) has been moreandmore applied.ThissystemhasbeenadoptedbytheInternational Organization forStandardization andisrecommendedbyalargenumberof national standard organizations. For that reason we will use SI units throughout this book.In Table I.l a survey of the basic and derived SIunits important for us is given. The reader is advised to use one consistent system of units, preferably the SIunits. Table 1.1Basicand derived SIunits Length Mass Time Temperature Force Quantity Work, energy,quantity of heat Power Pressure Dynamic viscosity Kinematic viscosity Surface energy or tension Enthalpy Heat capacity Heat transfer coefficient Mass transfer coefficient Thermal conductivity SIunit metre kilogramme second degree Kelvin4 Newton Joule Watt m kg s OK Unit symbol N=kgm/s2 J=Nm W=Jj s Nj m2 Nsf m2 m2/s Jj m2 orN/m Jjkg Jj kg C Wjm2C m/s W/ m C 4 Temperature differenceiscommonly expressedin degrees Celsius insteadof degreesKelvin. from one system to the other. In order to facilitate this, Table 1.2 gives a collection of conversion factor$.Table !.3 provides some information givingorientation about the approximate values of some commonproperties of gases, liquids and solids, which can be usefulforrough calculations. 1.5.Dimensional analysis In chemical engineering, relations between parameters are often expressed by means of dimensionless combinations of physical variables.The advantages of this technique are the easy control of dimensional homogeneity of the relation, theconstantsof whicharethenindependentof thesystemof physicalunits applied, andthe factthat thenumber of variables isreduced, whichsimplifies 22 Table 1.2Conversion factors Multiply by MagnitudeExpressed inDivide byIn SJunits Lengthinch (in)0.()254m foot (ft)0305 yard (yd) ().914 rrule1609 Angstrom (A)10- 10 Areain2 645X10-4 m2 ft2 00929 yd2 0836 acre 4047 mile2 259X_106, Volumein3 }64 X10- S m3 ft3 00283 yd3 0765 UK gallon 455X10-3' US gallon 3785xw-3 Time minute (min)60s hour(h)3600 day 864X104 year 316X107 Mass gram 648Xw-s kg ounce (oz) 284xw-2 pound (lb)0454 (cwt)508 ton 1016 Force poundal (pdl)0138N pound force (Jbf) 4-45 dyn w- s kg force (kgf)981 Volumetric Howft3/min 472xw-4 m3/ s UK galj min 7-58xw-s US gal/min 631xw-s Mass flow lb/ min 210xw-6 kgls tonjh0282 Densit ylb/ in3 277X104 kglm3 lbjft3 160 Pressure lbfj in2 6-89X103 N/ m2 lbf/ ft2 479 dynicrn2 01 kgf/ cm2 (=at) 981X104 atm (standard) 1013X105 bar 105 inwater 2-49X102 ftwater 299X103 inHg 339X103 mm Hg (torr)133X102 23 Table 1.2 (com.)Conversion factors Multiply by MagnitudeExpressed inDivide byIn SI units Dynamic viscositylb/ft h 413xw-4 Ns/ m2 lb/ft s149 Poise (P =gjcm s)01 Centipoise (cP) w-3 Kinematic viscosityft2/h 258XJO-S m2/s Stokes (S=cm2 j s) w-4 Centistokes (cS) w-6 Surface tensiondynjcm ( =erg/em 2) w-3 Nj m Temperature differencedegree F(or R)5/9 ocrK) Energy (work, heat)ftlbl0-04211 = Ws=Nm ft lbf136 BTU106X103 CHU1899 hph268X106 erg w-7 kgfm981 kcal419X103 kWh360X106 Power (energy flow)BTU/h 0293 'fl\T CHUjh0528 ftlbf/s136 hp (British)746 hp (metric)736 erg/s w-7 kcal/h1163 calfs 419 Heat fluxBTU/ft2 h 315W/ m2 caljcm2 s419X104 kcal/m2 h 1163 Specific heatBTU/IbF 419X103 JjkgC kcal/kgcC419X103 " .-Latent heatBTU/ lb 233X103 J/ kg kcaljkg419X103 Heat conductivity BTU/ fth OF173 W/ rn oc caljcrn soc419X102 kcaljm h oc1163 Heat transfercoefficientBTU/ft2 h F568 caljcm2 s oc419X104 kcaljm2 hoc1163 24 Table 1.3Physicalpropertiesof some materials AirWater 20C20C p [kg;m3]120998 11[Ns/m2] 17 XJ0-6 w-3 v =('1'// P) [m2/s] 142xw-6 w-6 cp[J/ kg oq103X103 4-19X103 A [W/ m 0C)002506 u[N/ m] 1xw-2 ll)Hz[m2/s] -2 X10-S5xw-9 a =A./ p cP [m2/s] 20XJ0-6 0143X10-6 Pr = vf a=C,!1/A. 07170 Sc =vf O071200 ~ H v[J/kg] 245X106 ~ H m[Jjkg) 335XJOS Gas Law Constant R= 8310 1/ kmol K Avogadro's number N=6023x1026 kmol-1 Gravitational acceleration g =981 m/ s2 Stefan- Boltzmann constant Diffusion c(l) Convection cp0 -4>tr4 constant or, also : 2 = (-11-)4 constant pv,.Dpv,.D (I.43) Since both F1pv;D2 and 11/ pv,.D are dimensionlessof the variables, any relation between these two groups is also dimension;llfy homogeneous.We can therefore conclude that :/ F(17.)(1 ) pv;D2 = fpv,.D= fRe (1.44) HereRe=pv,.D/ 11istheReynoldsnumber,whichcanbeinterpreted asthe ratioof momentumtransport byconvection ("'"' pv,v,.)andbyinternal friction ("' 1]V,./ D). Again,thenumberof variablesinthisproblemof initiall yfivehasbeen reduced to two. Inmost cases the number of dimensionless groups obtained by dimensional analysis equalsthenumberof variables occurring inthe problem minus the numberof independent dimensional equations which can be set up. In flowandmasstransportproblemsthenumber of dimensionalequationsis three (L,T and M) and in heat transport problems four {L, T, Mand tempera-ture). 27 The function fin the result obtained is again unknown and can only be found froma complete analysis of the problem.Sometimes the .result obtained from dimensional analysis can be simplified further by making use of former physical experience. In the case of the sphere, for example, experience tells us that at very lowflowvelocitiestheforceFonthesphereisindependentof thespecific gravityof thefluid. Thusthe relationbetweenthetwodimensionlessgroups mustbe : or F'1 2D2 =-D constant pv,.pv, F -- =constant 17v,D (1.45) This equation represents Stokes's law. That the numerical value of the constant is3ncanonlybefoundbysolvingthemicromomentumbalancesforthis problem. Asalast examplewewill discuss the dropletsformed,forexample, during condensationof watervapourattheundersideof ahorizontal surface.The hanging dropletsgrowuntiltheyreachacriticalvolume~ randfalldown.A dimensionallyhomogeneousrelationfor~ rfollowsfromtheconsideration that ~ - rdepends on the density of the liquid p, the gravitational acceleration g, the surface tension (Jof the liquid and the degree to which the surface iswetted by the liquid. For the great number of cases where the surface is well wetted by the liquid we can write :.. ~ r=f(p, g, a)(1.46) The reader may check that the result is given by: ()3/2 V:: r ~=constant (1.47) Thenumericalvalueof theconstantfollowsonlyfromanexactanalysisor fromexperimentalresults.Analysisyieldshere constant = 666. The problem with initially fourvariables hasbeen reducedto one dimensionlessgroup, the Laplace number which represents the ratio oft he weight of the droplet (-pg ~ r ) and the capillaryforce (-a VtJ If dimensionalanalysisisapplied all relevantvariableshaveto be included intheconsiderations.If wehad,forexample,forgottenthegravitational acceleration inthe analysis of the hanging droplet, wewould not have found a solution.If, on the other hand, we include too many variables, the result becomes unnecessarilycomplicated.Mostprofitisobtainedifbesidesdimensional analysis afragmentary pieceof physical experience isapplied, as illustrated in the aboveexamples. 28 1.6.Problems In the followingthe reader will findanumber of exercises. These problems form an important part of the book a ..1d the reader is asked to solve as many of them as possible. They provide a check on the understanding of the principles discussed,theyillustratethepracticalapplicationoftheseprinciplesand sometimestheyextendthematterpresented.Inthethreechapterswhich follow each paragraph willend with a collection of problems. The answers to these problems are given in order toenable the reader to check his own result. Onafewproblems(markedwithanasterisk)wehavealsoincludedsome comments in order to illustrate an important point or to show that the solution isreally not very difficult if the basic principles are applied in the right way. When solving a problem, the most important thing is to establish the physical mechanismthat governsthe situation studied (e.g.isconvection, conduction and radiation of heat important or can one or two transport mechanisms beneglectedinthisspecialcase?}.Secondly,wehave.todecideoverwhich control volume the balance ismade (e.g.over amicroscopic volume if informa-tion about the temperature distribution is required; over a pipe or heat exchanger oroveratotalreactorwithoveralltransfercoefficientsifonlyinformation about the mean temperature is required). Next, we have to determine whether the situationisin steadystateor not(sometimesasituationcanbemade steady state by assuming the observer to move with the system, e.g. the flowing fluid). Having come so far we are now in aposition to solve the problem with the help of the basic informationprovided inthisbook. This all sounds very difficult-and it is- but the bestwayto solve any difficulties isto try veryhard. 1.A tank is filledwith a liquidhaving aspecific gravity of p= 8 lb/ US gal. What isthe pressure differencebetween the top of the liquid and apoint 4ft below the liquid level? Answer in psi, bar, Njm2,atm and kgfjcm2 Answer :167psi = 0115bar =115x104N/m2 = 0113atm = 0117 kgf(cm2 2.The temperature increase into earth is approx.0025C/m.If the thermal conductivityof earth isi..=186 W j m C,what istheheatlossperunit surfacearea? Answer : =00465 W / m2 *3.11/swater ispressedthroughahorizontalpipe. The pressure difference over the length of the pipe is 2. 105 N/m2. (a)What is the amount of power Anecessary and what is the temperature increase of the water ll. T if there is no beat exchange with the surround-ings? (b)What are4> Aand ll T if thewater flowisdoubled and if thepressure drop is determined by momentum forces (and notby frictionforces)? Answer:(a)cPA=200 W, /:iT=0048C (b)A=1600 w, !J. T= 019 c 29 4.Two spheresof equalweight fallthroughair.What istheratio of their stationary rates of free fallif the ratio of their diameters Dd D2 equals 3? Answer :v Ifv2 =1 5.Show that the power cpAnecessary to rotate astirrer (diameter d)with a speed n (s -1)in afluid of density p and viscosity '1is given by : cpA=r(pnd2' n2d) pn3dsr,g *6.Write down.the conservation laws forakettleof water on the fire: (a)during heating up and (b)during boiling. *7.At the end of aglass capillary (outer diameter 2 mm) single water droplets are formedvery slowly.What isthediameter of thedropletswhichfall fromthecapillary? Answer :d=44 mm 8.Through awell-stirred15m3 tank flows 0.()1m3js of coconut oil. From a certain timeon palm kerneloil is passed throughthe vesselat the same speed.Afterwhat time doesthe effluentoilcontain lessthan1per cent coconut oil? Answer :l=6900 s 9.Water flows at arate of2 mj s through a horizontal pipe of 10 em diameter. A large flat plate is fixed near to the end of the pipe and at right angles to it Findtheforceonthisplateduetothejetof water,assumingthaton reaching the plate the water flowsaway along the surface of the plate. Answer :31-4 N 10.A pulp containing 40 per cent by weight of moisture is fedinto acounter-current drier at.arate of 1000 kg/ h.The pulp enters at atemperature of 20C and leaves at. 70C, the moisture content at the exit being 8 per cent. Air of humidity of 0007 kg water per kg dry air entersthe drier at 120C and the temperature of the air leaving the drier is 80C.What isthe rate of air flowthrough the drier and the humidity of the air leaving the drier? Data :heat losses fromdrier=5x106 J, specific heat of water vapour cP= 2x103 J{kg C,specific heat of drypul pcP= 103 J/kg C, tlH v atsooc =23x106 Jjkg oc Answer:molY =dtY where the symbols have the meaning given in the figure. After partial differentia-tion we can write: d(NX) =NdX + XdN dtdtdt .L 1 =- - = constant fi dN 't'mod/ MolefractionCH30H invapour= Y Nkmole molefraction CH30H =X ott =0, x0, No 33 and these two equations weobtain: dXdNdN N- =-Y- -X =- (Y- X) dtdtdtmol or: dX lPmoldtdN - = -Y - X .NN whichyieldsafter integrationbetweent= 0,x0 and t, x: N=Noexp ( s:,y d: X) Thisequation givestherelationbetweenthetotalnumberof molesinthe evaporator and the methanol content of this residue. We are, however, interested in the average composition of the distillate < Y) which is given by : ft. Y dt-ft..d(NX)- (NX)I'x .. ( Y)=0mol=0=O.Xo mol teNo- NNo- N =-No{xex(JxdX)} =X0 - X) N0-NPx0Y-X. XoThisis the desiredresult ; allthat remains is forusto evaluate graphicallythe integral, as shown in the figure.We findfor the area under the curve : andthus :

dX=117 Y- X Xa x-045- 005 exp (- 117)0435 ( Y)==-- =631per cent 1 - ex p (- 117)069 34 The amount of distillate is found as: N0- N=N0{1- exp (J:., yd:x)}=69 kmol We see that sometimesthe solutionof asimple massbalance can be quite involved. There is, however, a much simpler way to solve this problem approxi-matelybycarrying outthedistillationstepwise. Westartwith100 kmol of a mixture containing a mole fractionof X 0 =0-45methanol.Letus distiloff so much distillate that X drops t ~X 1 =0-35. Apparently,in this range the average mole fractionof methanolinthevapourphasewillbeY0 = 073 andthus we can findthetotal number of moles distilled off froma simplemassbalance as follows: xl =NoXo- (No- N)Yo=0-35=100X045- 073(No- N) N 0 - (N 0 - N)100 - (N 0 - N) WefindN 0 - N=263 kmol,containing192 kmolofmethanol.Nowwe repeatthisexerciseasstatedin thefollowingtableuntil wereachthe desired methanol fractionof X=005 inthe liquid phase. Total numberMole fraction ofMole fraction ofNumber ofkmolNumberofkmol of kmolin potmethanol inpotmethanol in vapourdistilledoffof methanol NXYN 0 - Ndistilledoff Total 100 73-7 56-1 431 315 045- 035 035- 0-25 025 - 0-15 015- 0-05 073 067 0-58 042 263192 17-6118 13076 116 4-9 685435 Thuswefindthat685 kmolweredistilledoff,containing435 kmolof methanol ,i.e.635percent.Thereisaresidueof315 kmol,containing 45- 435=15 kmolof methanol, which is near enoughto the desired5 mol per cent. The reader willrealize that the principal considerations inthis approximate solution are the same as for the exact solution given before. Instead of infinitesi-mal steps dX or dt which made integration necessary, we just used bigger steps which enabled us to obtain an approximate analytical solution after a reasonable amount of additions and multiplications. Problem 12Johnand the burnt-down factory The flowrate of air through the plant v can be found by means of an energy balance (equation!.6).Under steady-state conditions dEtfdt=0 and realizing that the change in heat content of the air pc PL\ Tis much bigger than the changes in the other forms of energy (equation 1.5), we can write : dEt Vdt= 4>vPCp ll.T +A- H= 0 35 Now, no mechanjcaJ energy is added, A=0 and the amount of heat supplied equals ractically zero. The shear force at the wallisthen givenby: ( vx) rw='f'/T (II.24) and the thickness (J h can be estimatedwiththe help of equation (11.22)as soon asthe frictionfactoris known. Withtheaidof thelaminar boundary layerthicknessb1,wecangiveanew mterpretation on the physical meaning of theRe number. The ratio of the tube 58 diameter and thicknessof the laminar boundary layer is found to be : D=Ihw= f Dp(v;;,)=fRe c5h1'/(V),)2112 (II.25) This meansthat forpipe flow at Re=10s (f = 00045) the distancefromthe wallatwhichturbulentmomentumtransportoverridesthetransportby viscosity isc511 4x10-3D. Avelocitydistributionasgivenabove isphysicallyimpossible,butitcan serve as arough modelfor heat and masstransfersin turbulent flow. Inthe caseof heat transferalayerof thickness c5Tisassumed overwhichthe completetemperaturedropoccurs.Outsidethislayertheturbulenteddies dispersethe heat so efficiently that auniform temperature exists in the core of the flow. Within the layer eddies are assumed to be absent. This layer of thickness {JTis called the thermal boundary layer analogous to the hydrodynamic bound-ary layer (thickness oh)treated above. It willbe clear 'that both models donot contradict each other as long asc5Tc511. Thequestionnowarisesastowhetherarelationshipexistsbetweenthe thicknessesofthehydrodynamicandthethermalboundarylayers.Some insight into this problem can be gained from the fact that the eddy diffusivity E increases with the third power of the distance from the pipe wall : E=C1y3 (II.26) Fromthe definitionof o11,thismeansthateddy momentumtransferoverrides viscoustransfer if: pE=pC1y3 11 .orE = C1y3 11/P = v Hence: {II.27) The diffusivityof heatby eddiesisalsodescribedby equation(ll.26),whereas thediffusivityof heatbyconductiononly isgivenby a= J..jpcP(m2/s).Hence, the distancec5Tfromthepipe wallat whichheat transferbyeddiesovertakes conductiveheattransferis givenby areasoningsimilar tothat applied above to be : (II.28) Hence,thesoughtrelationshipbetweenor andohisobtainedfromequations (ll.27)and (11.28) and isfoundtobe: = t=Prt Ora (II.29) where Pr is named the Prandtl number.For gasesPr1 and therefore c51,oT, for viscous liquids Pr1000 and o11 lOoT, and for water at room temperature Pr 7 and c511 2c5r. 59 The reader can imagine that mass transfer between awall and turbulent pipe Howcan be depicted byusing aconcentration boundary layerof thickness be-The ratio between this thickness and the thickness of the hydrodynamic bound-ary layer will then be (see also paragraph IV.3.3): {>(v)t b:=II)= set (11.30) where Scisthe Schmidt number. For gases Sc~1 and ~ ,~l>candforliquids Sc~1000 and b"~lObe.We see that in these cases the thickness of the hydro-dynamicboundarylayerisgreaterthanthoseof therespectivethermaland concentrationboundary layers, b11 >bT, oc. We have .already stated at the beginning of this chapter that it isimpossible to derive theoretically arelation forthe velocity distributionduring turbulent Bow.We can, however, construct arelationbetweenvelocity and the friction factor and see whether we can determine the coefficients in this relation by ex-perimentalresults.Analogouslytothedevdopmentof equation(II.22),we assume that the mean velocity is afunction of the shear stress at the wall, the specific gravity and the kinematic viscosity of the fluid and of the pipe diameter ; so: Via dimensional analysiswefind: ~ = ~ { ~ ~ }= C { ~ ~ r (II.31) which can be rearranged to : rw=f=(C)-2/(l +n>{-v-}2n! p(v)2 2(v)D (11.32) Equation (11.31) indicates that in the turbulent region also the frictionfactor willbeafunctionbf theReynoldsnumberonly.Resultsoffrictionfactor measurements in smooth pipes have been collected and correlated by Blasius as: _4f =0316 Re -0.25 This correlation is discussed in more detail inthe nextsection. Atthemoment wecan use itto determine the constant nin equation (II.32).Via : wefindn =l 2n -- =025 1 +n Repetition of the foregoing procedure forthe flowvelocity in the x-direction atanyplaceryields,if werealizethat vxmustbezeroat 2r=D,insteadof equation(II.32}: (11.33) 60 Thus, maximum flow velocity occurs inthe centre of the tube at r=0 and we can write : _2_ =(l_2r)-+ VmaxD (II.34) In order to find the ratio between mean and maximum velocity we can calculate ( v)from equation (II.34) as : ( v)= _1_JD/2vdr =VmaxJD/2(1 - 2r) -+dr D/ 20XD/ 20D Solving this equation we find: (v)7-- =- = 0875 Vmax8 compared with(v)/ vmax=!found for laminar flowincirculartubes. The velocity distribution (equation II.34) givesarather good descriptionof theactualsituation,as illustratedinFigure11.18,wheremeasuredvaluesof (vfvmu? are plotted against (1- 2r/ D). I >< 0 E "' ' "' 0 i / ~ 0 5l 6 v I 0 5 ,CY 0 0 ~ ,..t c;/ 0\ / 0 _.1-2r/D 'o \ > ~ 'o ' r;\ ~ 0 5 Figure ll.18Measured velocity distribution for turbulentflow I 1.2.2.Pressure drop in straight channels At the end of paragraph 11.1.2 we saw that the pressure drop in a pipe system can only be predicted as a function of the flowrate if the shear force at the wall is known.For laminar flowthis quantity can be predicted directl y because the relationbetween-rxyanddvy/dx(e.g.accordingtoNewton,equation11.1) characteristic of the fluid offers sufficient additional information. For turbulent flow we lack this additional information. Inorder tobeableto predict the flowrate forturbulentflowon thebasis of agivenpressure drop (or, inversely, the pressure drop foragiven flowrate), use is made of graphs and tables of earlier measuring data in which numerical 61 valuescanbefoundofthedimensionlessgroupsthatareimportanttothis problem.On the basis of dimensional considerations these results can then be used for solving every new but analogous problem. We shall deal in succession withthefiowthroughastraightchannel,thefiowthroughfittingsandthe frictionlessfiow. Since in what followsweonly speak of ameanvelocity and not of velocity distributions, the microbalances are of no use to us in this chapter. The quanti-ties which weshall now come across belong to balances over an entire channel or channel system(i.e.theyaremacroscopicquantitiestobeusedinmacro-balances). If the shear stress on the wall of a horizontal straight channel (cross-section A, circumferenceS)isknown,wefindforthestationarystatethemomentum balance (see also paragraph 11.1.2 and Figure Il.4) : L F.x=0=p1A- p2A- twS(x2- x1) and, using equation (II.22), we findforthe pressure drop: S(x2- x 1)Jl 'Jr(11.55) - \. -Thechangeofthecross-sectionofthegasstreamwithpressureisshown 3ChematicallyinFigure 11.29, where: beenplottedagainstpj p1 _ Itcanbeseenthatthe cross-sectionof the jetreachesaminimum at _!!_=(2) K/ (tc- 1) PtK+1 84 p/pl---Figure 11.29Flow of gas through anorifice At thispoint the critical conditions Ac, Po vcand Pcare present. For the critical velocityvcwe findwithequation (II.55): Vc=(2KP1)t =(KPc)t =(KRT)t K+ 1 PtPcM (11.56) which is the velocity of sound at Pc and Pc (for air at 0C, vc=331 mjs). For gases consisting of two atoms, K=14 and consequently pcfp1 = 0527. The expression developed for the mass flow rate (equation JI.55) can be sim-plified for anumber of conditions. (a)p0 ~p1Undertheseconditionsthecompressibilityofgasescanbe neglected and equation (11.55) simplified to: (II. 57) whereA= cross-sectionof jetatp= Po,A 1 =cros5-sectionof orificeand C= contraction factor which is approximately 06 L) Broad- crestedwei r( 0>3h0) Tri angular notch 0 6m /s;ho>0 1 m) 4>u:0 465(L-02h0)J(L:width of weir ) 0 44;-;;g-tong a) (Il.73) The risinggasbubblesthenhave theformof sphericalcaps.Comparison of equations (If.73) and (li.70) written for half asphere shows that for these con-ditions C"'=~ a taRe value of > 12,000 (air in water). This C..., value is compar-able withthat of acircular fiat plate {Ta bleII.4). 11.5.3.Free fallof droplets The fr iction forcesacting on afallingliquid droplet will deform thedroplet fromapurely sphericalto an elonga ted formuntil the droplet breaks up. The surface tension counteracts this deformation andwe can assumeroughly that the droplets will no longer break up if this ratio is equal to or smaller than unity: 1t212 friction forceC.., 4 dP1p8vCd_ 2 ------ - ---..,...- - =wp{JgV~1 surface tensionnadP8v -105 In the dimensionless group we recognize the Weber number, the critical va)ue of which should thenbe of the order of magnitude of: W-pgdpmuvz8 e cr- 19 (Jcw (II.74) (assuming turbulent conditions). Hence free-falling droplets can only be stable for diameters smaller thanor equal to dpmax . Experimental results have shownthat We 22forfreefallandWe13if the droplets are sudden] y exposed to drag forces. 11.5.4.Particles innon-stationary flow Allrelations described in this paragraphso farapply only if the stationary state bas been att ained. If weconsider afree-falling sphere, the non-stationary momentum balance reads : dv Mdt =Mg- CwAtpv2 (II. 75) if we assume Cw to be constant over the whole range of velocities. In the station-ary state(velocityv.J : Mg=CwA!pv; (11.76) isvalid, as we have discussed in the previous section. Thus equation (II. 75)can be simplifiedto : dvv2 M- .=M g - fttf g-dtv2 s Integration between v=0at t=0and v.1 tyields: v 1 +-'V5 = exp(2gt) 1- Vs vs (II. 77) In Figure II.43 vjv5 has beenplotted against the dimensionless factor 2gtfvs. It appears that the velocity is stationary if2gtjv5 >5. For 2gtfvs {2+ exp

+exp (-2gt)'} .VVs1VsVs - = - n-Vs2gt4 18 (II.78) This function is also shown in Figure II.43. lt can be used to calculate (via trial and error, see problem 4)the stationary velocityof freefallfromameasured mean velocity. II 5.5.Rate of sedimentationof a swarmof particles The rate of sedimentation of aswarm of equally large particles in aliquid is lower than that ofone separate particle. In a given system of particles and liquid thisrate of sedimentation (v.Jsappears to depend only onthe fraction4>of the volume which is taken up by the particles. This constant isalso connectedwith theporositye:thisisthevolumefractionofthecontinuousphasesothat e = 1 - qJ.There arevarioustheories abouttherelationbetween(vs)sando or e.A well-usable empirical relation is that of Richardson and Zaki : (vJs=

In this equation isthe rate of settling of one singleparticle of the system question (e= 1) and n avalue. dependenton = and on the rei of the particle diameterand the diameterof the vessel D,inwhich they 107 For dJD1 104 the power numberP0 becomesaconstant.InthisturbulentregionP0 isonlya function of the geometry of the system. At valuesof Re104) (Re104 Impeller PoPo Flat blade turbine six blades. width/ diameter-l/ 5706 four blades, c:litto704-5 Flat blade paddle six blades,widt b/ diameter- 1}6-l/ S703 Propeller,threeblades pitch =diameter 42 03} pitch =two times diameter42 1.() Inclined blade turbine six blades, width/ diameter -l/ 8 7015 Re>JQ4 K, 13 06 13 04-05 08 . 124 I 1.7.3.Pumping capacity and mixing time If weconsider only the biggest velocity component of the liquid flowing from the impeller, the pumping capacity is given by the product of the mean velocity of the liquid leaving the impeller and the area described by the blade tips.For the propeller thisleads to(d- diameter of impeller) : whereas foraturbine we find(W = width of blade): v=iindW If we assume iitobeproportional to thetip speedof the blades (1tnd)wefind for agiven geometry (for turbines W,_d) : (11.91) Inthe turbulent region. Kv isaconstant for agiven impeller geometry.Some valuesof KDare listedin Table TI.6. In many chemicalengineering operations, e.g.the blending of gasoline;the mixing of reactants in a chemical reactor, etc., knowledge of the time necessary toreachhomogeneitymaybeveryimportant.Usuallythemixingtimeis definedasthetimeneededforhomogenizationtothe molecular scale.Since measurements on this scale are beyond experimental capabilities, an investigator isonly ableto measuretheterminal mixing time,beingthetime requiredto attain homogeneityonthescaleof observation.Intheturbulentregion,for many types of impellers the relation (tm=mixing time) : ntm=constant holds, for geometrically similar situations. The mixing time is about four times the circulation time of the vessel contents. Thus, if no dead spacesarepresent: v tm~4 tPc isa good estimation of the mixing time. 1!.7.4.Problems 1.A catalystprecipitationprocesshasbeendevelopedonapilotplant scale (diameter reactor 025 m).It appeared that using a standard flat blade turbine (sixblades,fullybaffledtank),aminimumstirringspeedof 700 rpmwas required to obtain asatisfactory product (pliquid~1300 kglm3) . Ageometrically similar factoryscalereactor(06 mdiameter)has. been designed.Calculatewhichstirring speedshouldbeappliedandwhatthe power consumption would be if we keep constant: (a)the power input perunit volume; (b)the pumping capacity per unit volume; 125 (c)the mixing time; (d)the size of the partic1es that are only just not suspendedinthe liquid Answer:(a)391rpm, 685 W (b)700 rpm, 4000 W (c)700 rpm. 4000 W (d)391rpm, 685w 2.Show that John reached the right conclusion inthe problem described at the beginning of this paragraph. ll.8.Residencetimedistribution Thesergeantonduryin frontof thesupermarket informed Johnthatthem.antheywe,elooking forhadenteredthe supermarkettogetherwithhiswife15 minutes ago. He said tluu they seemed to be doing their shopping forthe next week andthemanlookedrelaxed.John,knowing fromhisown experience thatshoppingfortheweekinthissupermarket cook on the average about 30 niinutes, mode a swift calcula-tion. He decided thathehad a probability of 60 per cent to catch the manif hesearchedthesupermarketdirectlywith his platoon. He therefore ordered them accordingly, but they failedtofindtheman.Atnight,hereadthefollowing chapter and came to the conclusion that the probability ofan event and the actual course ofan evenl are, ash should know, two differentthings. Flow through an apparatus practically always resultsin acertain residence time distribution.Fluid elementsenteringtheapparatus simultaneouslywill generally leave it at different times. This distributionin residence time can be describedquantitativelywithtwodistributionfunctionswhichareclosely related, the F and the E functions. Although residence time distribution isof great importance to the design of processequipment,wewillrestrictourdiscussiontotheprinciple.Amore comprehensivetreatmentofnon-idealflow,includingitsinfluenceonthe designforchemicalreactions,isgivenbyLevenspiel(ChemicalReaction Engineering). Il.8.1.The F function I< 0 ,C = 0 I~0, C = CoSomphng point tracercone. C Figure D.48Measurement of the F curve Considerthecontinuousflowsystem shownin FigureIl.48.From atime t= .oon,wereplacethe feedstreambyafluidcontaining atracermaterial (e.g.colouring agent) at a concentration C0.If we measure the mean concentra-126 T irne1 -Ftgme U.49The F curve tionof thetracermaterialinthe effluentstreamwewillfindthatthetracer concentration (and also < C) / C0,its dimensionless concentration) increases with time,asillustratedinFigure 1149.NowF(t) isdefined asthe volumetricpart of the e:fHuent stream which had aresidence time in the apparatus of less than the timet. Since we know that all untraced fluid in the effluent stream has been in the apparatus for atimet, F(t) is representedby thefraction of tracedfluid ( C)/C0 in theeffluent.Consequently 1- F(t) is thefractionof untracedfluid inthe effluent The Ffunction(and also( C) ) is averaged over asmallvolume taken from the flow (mean cup values).Formeasuring, part of the flowis collected, mixed andsubsequentlyanalysed.If concentration gradients or velocity profilesare present over the cross-section(as isthe case during laminar flow)we must be carefultomeasure the correctly averaged value. The F function bas a number of properties, the most important of which are: F(t}=0fort= 0 (i.e.nofluidremainsinfinitely in the apparatus) or, formulatedmore precisely, F(t)=1forl=oc (i.e. no fluidremains infinitelyin the apparatus) or, formulatedmore precisely. mathematically : lim (1- F(t))= 0 (i.e.allfluidoriginallypresentintheapparatuswillhavelefti tultimately). Funhermore. fromamassbalancewefind: focof/J0(Co - ( C) ) dt =VC0 and using F=C/ C0 weobtain : f::ov (1- f) dt = - ='t' 0cPv (Il.92t 127 t--Figure 11.50Determinat ion of T The mean residence time thus canbe determined fromexperimentally found F(t)curves,asillustratedbyFigureII. 50.A lineatt=t 1 =constant creates the areas A 1 ,A 2 andA 3 such that : rl=At+A3 Because: .T=Lro(1- F) dt =A2+ A3 this line representsthemeanresidencetime r if A1 equalsA2Asacheckon themeasurements,rcan becalculated as-r=Vflf>vfrommeasured valuesof Vandf/J ,. Often, instead of the time r. adimensionlesstime parameter 8=t/ -ris used and we will apply this quantity in all following considerations. It will be evident that forthe F(8) function equation (II.92} becomes: f oCX)(1- F) d8=1 I 1.8.2.TheE function The residence time distribution function of the fluid in the apparatus is given by the Efunction,which therefore representsthe agedistribution of the fluid leaving the vessel. Figure II.S 1 shows atypical E curve. The fraction of material in the exit stream with aresidence time between (Jand (J+d(J is given by E d8, fromwhichfol1owsthat : f oO()E d8= 1 The fraction of effiuentmaterialyounger than 81)which is alsorepresented by theF function,isthen givenby: (11.93) 127 t---Figure 11.50Determination of -r The mean residence time thus canbe determined from experimentally found EU)curves,asillustratedby FigureII.50.Aline att= t 1 = constant creates the areas A 1,A 2 andA 3 suchthat: Because : thisline representsthemean residencetimerif A 1 equals A2Asacheckon themeasurements,tcan becalculatedasr=V /4>.,frommeasured values of ;and tPv Often,instead of the timec, adimensionlesstime parameter 8 =t/ -cis used and we will apply this quantity in all following considerations. It will be evident that for the F(8) function equation (II.92) becomes: f000 (1- F)d8 =1 11.8.2.TheE function The residence time distripution function of the fluid in the apparatus is given by the Efunction, whichthereforerepresents the age distribution of the fluid leaving the vessel.Figure II.51 shows atypical Ecurve. The fraction of material in the exit stream with aresidence time between e and 8+d8 is given byE d8, fromwhich followsthat: Jo:oEd6 = 1 The fraction of efiluent material younger than 81,which is also representedby the Ffunction,isthen givenby: (11.93) 128 G Figure 0.51The Ecurve and the fraction older than 81 is : ra)r8, 1- F(81)= J ~Ed(}= l- J ~E dO 8t0 The E curve can be measured by injecting atracer material into a flow system for avery short time and by measuring its concentration asafunction of time at adownstreampoint.Atypicalplotof theseconcentrations versustime is shown inFigure II.52. The area underthe curve : LCX)( C)dt~L c L\t t -Area ;0:total amount oflf'ljected tracermaterlol Figure D.52Response curve to a tracer injection signal (11.94) representsthetotalamountofuacer substance injected,Q.Inordertofind E(t),this area mustbe unity ; thus all concentration readings must be divided by the totalamount of tracerinjected: ( C)( C)( C) E(r) =s ~Cdt =Q~L ( C)lit (11.95} 129 The mean residence timecan then becalculatedvia : Jco 00 L t( C)Llt t=rE(t} dt~L tE(t} At =~ 00L( C) At (11.96) Since 9=t/ tand E{8)=~ E ( t )we are now able to construct the actual Ecurve in dimensionless parameters. As we will see later, it is often desirable to know the variance of the distribution curve, whichcan be calculated according to : U2 =Jco(8- 1)2E d8~L 82E il8- 1=L t2( C> L\t- 1(II.97) or2L L\t 11.8.3.Simple applications ofFand E functi ons l/.8.3.1.Perfectmixer.Aperfectmixerisavesselinwhichstirringisso effective that the composition of its contents isidentical at allplaces.Naturally the effluent from this vesselhas the same composition as its contents. Keeping inmind that Cout=C a mass balance over this vessel reads : dC A.C0 - A.C=V-o / ~o/vdt (Il98) For the determination of the F curve from timet= 0 on, the tracer concentra-tion of thefeedstreamischangedfromCi n =0toCin= C 0Integration of equation (II.98)therefore yields, via: ---- --! dt JcdCJr l/J oCo - C- oV F=5:._=1- e -e Co (ll.99) For the determination of the E curve at time c =0, atracer is injected into the vesselfora short time.The amount of traceraddedwould,whenwellmixed, giveaconcentrationC0.Integrationofequation(13)forthiscaseyields . (C,n=Ol) : fc dC=- (' lf>a:dt Jc0 CJo V E=C=e - 8 Co (11.100) Ascanbe seen, the resultsobtained agree with equation(Il.93),which can be writtenasE =dF/d8. I / .8.3.2.Plug .flow.Duringplug flowall parts of the fluidmove at the same speedandthereisnoaxialdispersionandnoresidencetimedistribution. 130 Therefore, the F curve is described by : and fortheEcurve: F=0 fort -r E=0for t #T butE= tracer input signal at t=r ! I 1.8.3.3.Laminar flowincircular tube.Here,the residencetimedistribution is caused by the differences in velocity and by diffusion. If we neglectdiffusion, we can writeforany streamline with velocityv and residence time t : =2(1- :,) =i= 9 (11. 101) Becausethe maximum velocity atthe centre istwice themeanvelocity,both the F and the E functions are zero for: v. (v) 20/L).Thereasonforthislimitation isthatforlower velocitiesthediffusion against the direction of flowpresentsaproblemwhich can onlybesolvedbyspecifyingthe flow andthe dispersionmechanismindetailfor the areax001) the maximum occurs atfJ5)the .expressionn!~-,r: e-"fon isvalidandequation (11. 111)becomes : (n>5)(Il.llla) FigureII.56 showsEcurvesaccordingto equations (11.111)and (II.llla) for various numbers of stirred tank reactors in series. The variance of the E curve forn stirredvesselsin series isgiven by : (II.112) 134 H ~ r - - - - - - - - - - - - - - - - - - - - - - - - ; - - - , l 8 Figure ll.56Tanksin seriesmodel,E curves and analogous to the.development of equation (11.110)we find for the Fcurve for n tanks in series : F- 1!'e {1(](nB) 2(n0)3(n9Y' - 1} - -e+n+2! + 3J + ... +(n-1)! (II.113) F curves fordifferentvalues of n are shown in Figure II.57. The analysisof experimentallydeterminedFcurvesismorecomplicated than that of Ecurves. Two cases can be distinguished: Fcurves for n< 50 or Df(vxL>0.01, and Fcurveswith very little dispersion, i.e n>50or D/v;cL< 0.01.Both cases wil1be treated separately. (a)High dispersion :n001 There are twopossible methodsof analysis : (i)ViatheE curve Plot the measuredFvaluesversustimeand determinegraphically at different times the slope dF(t)/ dt = E(t). The values determined forE(t} at the approximate time can then be analysed as described in paragraph II.8.2. 135 10 e--Figure ll.57Tanksinseriesmodel, Fcurves (ii)Via the slopes ofthe F(8) curve at 8 =1 The slope of an F curve at 8 =1 can be shown to be given by: (dF)n" d88= 1~(E)e..,l=(n- 1)! e-n Therelationbetweenthisslopeandthenumberof stirredvesselsin series isshown in Figurell.58. 30.--------------- --- ----, -.. n Figure U.58Tanks in series model, dFfdO at 8 =1 136 To analyse a measured F curve, determine the mean residence time as shown in Figure II. 50. Determine graphically dFfd8 at 8 = l and find the number of stirred vessels in series fromFigure II. 58. (b)Low dispersion:n>50 or D(v:r;Lnlr=R=qnR2 L(fiLl 151 L Figure lll.4Temperature distribution in a .cylinder In order to findthetemperature distribution inthe wire we set up amicro-energy balance over a cylindrical shell ofthickness dr (Figure III.4}. For steady-state conditions, equation(111.1) can be written as: dE, dt =0= 2nr L4>8L.- 21t1' L01) l2 163 (IIL33) (III.33a) (IIL34) (ill.34a) Due tothe factthat forFo>01 the shapes of thetemperature distributions donot changewithtime,theNuvaluesbecomeconstant.The valueof the Nusselt number depends only on the form of the body. For a flat plate we found Nu= 493 and forasphere Nu = 66.For bodies which can be enclosed by a spherewecan estimatetheNunumbersby applyingNUsphcre=a.DtfA.=66 and taking for D1 the diameter of the spherehavingthe same volumeasthe body.For acube (with side D) we find: V=D3 =- D3 Nu=- =-D =66- =53 1taD66 61cube..).Dl6 (exact solution 49).For a cylinder withL=D we find analogously: Nu 1 = 6-6 3{f58 Icy\}3 (exact solution 5.6).Since for Fo >01 the-Nu anaavalues .are constant aheat balance over the total body becomes: d (1) pcP V dt =et.A(T1 - ( T )) (IIL3S) Integrating, with ( T)=T0 at t=0, we find : InT1- (T)=_=_aD _i_t=_Nu T1 - T0 pcPVA.pepVDD (III.36) Thissolutionisidenticaltotherelationfoundwithequation(JII.33a). Bothrelationsshowthataplotoftherelativetemperatureequalization 164 1-0 I ..._--- ....... -- -.. ..._ ...... I -Fo=at/02 FigweJD.8Meantemperatureduring non-stationaryheatcon-ductiond.. ( ~- ( T))/ (T1 - T0)againstFo=! tj D2 shouldyieldastraightlinefor Fo >01.This is indeed the case, as shown by Figure lll.8. Ananalogous graph canbeproduced fortherelative temperature equalizationin thecentre of the body, as shown in Figure 111.9. Since 'H= a(T1 - ( T ))they-axis of the first graph also represents: -1-)JT1 - 4 > ~ ---=--T1 - T0 cr(T1 - T0) (III.38) Figure 111.8canalso be usedto findthe heatflux into the mediumat any time. Il/.2.3.Influence of anoutside heat transfer coefficient InparagraphsUl.2.1andIII2 ?wehaveconsideredcaseswherethebeat transportoutsidethemediumisverybig (au- oo)comparedwiththebeat conduction in the medium. This situation is, for example, present if acold piece ._o I ...-Flatplote Squarecyl inder, L :oo \---\--Cylinder, L "00 Cu be \--\-___.:lr----1\--- C yll nde r, L = 0 \---+---+--- -+----+--Sphere

00 10 20 30 40 5 Fo-=at (D2 FigureIll.9Centretemperatureduring conduction non-stationaryheat 165 of wood (poor heat conductivity)is -placed in asteamatmosphere (good heat transfer) with a saturation -temperature T1 The opposite situationis realized if apiece of metal (goodheat conduction) of temperatureT0 isplaced inair (poor heattransfer)withatemperatU:reT1. In this case the heat transport outside the medium is so slow that temperature equalizationinthemetalispracticallycomplete ;thetemperatureTinside the mediumisthe same inallplacesand the rate of heating or coolingisin-dependent of the heat conductivity of the material.A macroscopic heat balance for this case leads to an exponentially decreasing temperature equalization like equation (111.36),but withthe outside beattransfer coefficient a:" instead of ex: cxuAt ---pcpV (III.38) 166 If both heat transport resistances, the outside (1/ cx11)and the inside (1/ - 4>n- J.A l!:.T- .e.go- skgj --- -\rx ls Ill!1Hd!l.H In 4 hours. therefore, only about 1 kg of water willevaporate. m.4.Heat exchangers Johnhad just pouredout a cupof coffee and filledhispipe when his boss asked to see him for a j ew minutes. ' HeW, John thought. 'the coffee will be cold before I am back.' Remember-ing the pagestocome,Johnputmilk and sugarin hiscoffee andwenttosee his boss. Heat exchangers play an important role in technology, not only for condition-ing liquid ftows but also for attaining a favourable heat economy.On the basis of tbe demands made on the heat exchange (the amount of heatrPH which must be t ransferredperunitof timeandthe availabletemperaturedifference) and of data on the beat resistances, the required exchanger area Aof aheat exchanger can becalculated fromthe formula: l/>8 =U A ~ T (111.55) Usually both U and ..1 Tare dependent on the place in the heat exchanger, so that the formula for the entire heat exchanger should read : (111.56) WehavealreadydiscussedthedeterminationoftheoveraUheattransfer coefficientUforsimple pipe flow(paragraphIll3) and wewillreturn to this subject in later paragraphs (see also Table lll.S).Here, wewill find out how the meantemperaturedifferencecanbedeterminedin anumberofsimplecases and wewill consider the problems which have to be solved when designing heat exchangers. 111.4.1.Determinacion of mean temperarure difference Letusconsiderasimplemodelof aheatexchanger- twoconcentrictubes (FigureIII.13).Here, forexample,liquid'givesup beattoliquid",either co-currentlyorcountercurrently.IntheformercaseTILshouldinvariablybe ~T{ . In the latter casethe outlet temperature T ~canalso be higherthanT ~ ln the firstinstance aheat balance can be set up over the entire heat exchanger, which shows that in the staionary state just as much heat is supplied as removed. If tneheatexchangewiththesurroundingscanbe neglectedandno heatis T il L Liquid1- _r. ' .P'm' To'--.Noc---!-:;I::::::::::::::::::::::::X::::_.-7,7"!- - L c:----_..:...__----Jil!I1JJI ILiquidI x=ox=L c"r,lf "'m'pto 0-xL0-xL FigureIll.l3Heatexchangerandtemperaturegradientsforco- and countercurrent flow produced in the apparatus, theheat balancewillread: 183 { + countercurrent - cocurrent (lll.57) This relation is valid for any heat exchanger which satisfies the above conditions* and isindispensable for solving a heat exchanger problem. In order to calcula.te the meanT ovei'the entire apparatus in equation (II 1.56), we should first know T' and T" as a functionof the place. For the type of heat exchanger in Figure III.13 we .make a heat balance over a small part dx: -(,.,cP)" dT"= dl/>H Then, according to equation. (III .55),the followingapplies: dn =(T' - T")US dx (III. 58) (III.59) whereS=circumferenceof theexchangearea.Afterintroductionof= T'- T"we findfrom equation (II1.57): l\TL- l\1;,=- H [l+lJ (cf>mcp)'(cJ>mcp'f' Eliminationof the term inbrackets andof dc/>Husing equation(IIL59)finally givesthe following differential equation f o r ~T: d6 T=(A 1i.- .1 T0)S 1). TuHdx (II 1.60) ~ o ra few special casestheintegration can be carried out easily. (a)If Uis independent of xor T (often the case with gases). Then integration ~ f equation (II 1.60)leadsto: !!.TL- ~ T o y=USLinf!..TJll.To =USL(!!.T'hog(IIL61) 102 9 e 7 6 5 4 3 r 2 1 5~ 10 9 8 7 6 5 4 3 2 15 ld!86543215108654321 5 l!. To[oc]---FigureDI.J4Logarithmicmeantemperaturedifferenceasafunctionof ~T, andLlT0 () 0 1-..J (). 7 5. Figure 111.16Correction factor t/Jformultiple pass heatexchangers (one shell,evennumber of tubepasses) asa functionof R and X Jl/.4.2.Height of a transf er unit If weconsider aheatexchangerwithconstant wall temperatureTw wecan set up the following heatbalance : or : n=UATo - ( T )= 4>rnc9(T0- ( T ) ) To -T. In ( T ) - ~ w T0- TwuAUSL ln ( T )- T..,=--mc-P =--mc-P i.e.the meantemperature difference b etweenthe fl uidand walldecreasesbya factor1/ e ( =0-37) over alength Le givenby: L=4>mce eUS (I Il.63) 187 This length, whichischaracteristic of theheattransferprocess,iscalledone HTU, height of a transfer unit', and is used in many books on chemical engineer-mg. We willusethe conceptof HTUs here toshowhowwe canallow forfluid dispersionwhendesigningaheatexchanger.If fluiddispersionoccursina heat exchanger,extraheattransportinthedirectionof flowwillbe thecon-sequence, andthe actual temperature gradients over the length of the exchanger maydifferfromthoseshownintheprevioussection(wheredispersionwas neglected).ThisisillustratedbyFigureU1.17whichshowsthetemperature gradients over the length of aheat exchanger (likeFigure III.13) for cocurrent plug flowand for dispersed flow of liquid.Itwill be clear that in the latter case themeantemperaturedifferencebetweenthetwofluidsissmallerand,con-sequently, alarger surface area (or greater length) of beat exchanger is required forthe transfer of the same amount of heat. We can easily correct for this effect of dispersion by using instead of the height of atransfer unit HTU the HETU, the height equivalent of atransfer unitwhich isgivenby the sum of the HTU and the height of amixing unit HMU: 1 .... HETU=HTU+ HMU r. 'Nodi spersion 0 ,I ........ ...... ....... With di spersi Ofl \ __ \ -L FigureID.17Temperature gradientswithandwithout dispersi ori T ' L (III.64) The height of amixingunit can be determined by the relations given in para-graph11.8.5.Fromwhatisstatedtherewe canconcludethatforturbulent flowandforflowaroundobstac1esHTUHMU(i.e.dispersioncanbe neglected) butthatunder laminar flowconditions acorrection forthe effectof dispersioncanbe necessary. 1 I /.4.3Design of heat exchangers Withtheknowledgenowavailableitispossibletodeterminethemain dimensionsof aheatexchanger.lf theoverallheattransfercoefficient,aB temperaturesand the fluidflowsareknown.we can easily determine the total surfaceoftheexchangerrequired. The task remains to determit}e the number N, diameter D and lengthL of the pipesaswellasthe diameterof theshellDsand the geometric pattern of the r--"' Pressureindex(0/o) N E ' ShellTube -20atm62 D 1..-J 30aim167 d) .2 40atm2914 ... 0. 60atm4423 0 1'-0> Shell materialIndex(%) 304ss9 316ss12 C r ~Mo 5 C r ~Mo4 Ket tleorbellow+ 16% Lengthlndel'.(%) I 2m4 i 1 6m27 24rn!2 30m9 3 6m5 42m2 0 - 3 102.0 Totaloreo[m2] Tubematerial ~ ~ - - - - - - - HastelloyC 50100200 304SS Admlrolity 5Cr ~Mo cs Figure m.t8Costs in1970 of standard carbon steel shell and tube heat exchanger (3/4 in diameter,15/ 16 in pitch,L0/10 atm) inDfl/m2 for allmaterials; cost indexCS 65 per cent/year, all others 45 per centjyear--00 00 189 pipes inthe shell.Forthe solution of thisproblemtwomomentum balances areavailablewhichrestricttheapplicablevelocities,diametersandlengths due to limitsputon pressure drops in the pipes and inthe shell.Furthermore, wehaveaneconomicbalancewhichisusedtofindthecheapestdesignfor agivenproblem.Once wehave -chosenthe type of heat exchangerthesethree balancesdetermineN, LandD,andgeometricconsiderationsthenfurther determine shell diameter, etc. If werealizethat,besjdesthesolutionsof theaboveproblems,questions on constructionmaterialandagreatnumberof constructionaldetailsmust be solved, we can understand why big engineering firmsoften have departments specializedin designing heat exchange equipment. For arough estimationof the costsof aheatexchangerFigureII1.18can beusedThisfigureshows,forinstance,that aheatexchangerof 100m2 with stainless steel316 pipesand aSS 304 shell, suitable for300 psi in the-shell and 600 psiinthe tubes, andwith apipelengtbof10feetwillcost per rn2 surface in1973: 445+ (9+6+ 14+ 9+ 15)%=Oft. 683/m2 Finally,inTableIII.5someheattransfercorrelationsaresummarized, whichare usefulwhen designing heatexchangers. A fewof these relations have alreadybeen discussedinparagraph111.3 ; otherswillbetreatedin following paragraphs. III.4.4.Problems 1.Throughahorizontalsmoothpipe(internaldiameter18 mm)waterof 90Cisforcedup at arateof 08 m/ s.Determinethedistancefromthe entrance at whichthe water begins to boil if the pressure drop is negligible and the pipe wallover its entire length iskept at l10C. Answer :L=15 m 2Aclosedvesselfilledwith 86 1 of waterwhichisstirredthoroughlybas a temperatureofl00Cwhereastheairtemperatureis20C.If thetotal outersurfaceareaof _thevesselisI m2 andthemeanheattransferco-efficientU over this area is25 Wfm2 oc, how long does it take for the tem-perature to fallto50C? Answer :4 h 3.In a double-tube beat exchanger benzene is heated from 20 to 60oC counter-currently with water, which during this process cools down from 88 to 48C. During an overhaul the direction of the water flowis accidentally reversed. Determinethefinalofthebenzene.(Thebeattransferco-efficient isindependent of 6. T.) Answer:52C 190 Table IU.SHeat transfer correlations for heaL exchangers FlowsiluotiollHeattransfercorrelationRtmorks -Flowthrou9flS1r119ht Re"rof which increases in the x-direction. This causes the local heat transfer coefficientsto decrease in thepositivex-direction.For the calculation oftheheattransfercoefficientswecanapplythesametechniqueasused in paragraph 111.3for the calculation of heattransfer coefficients forturbulent flow in pipes. Wefirstcalculatethe thicknessof the hydrodynamic boundarylayer from equation (11.64): Usingourknowledgeabouttheratioof thehydrodynamicandthe thermal boundary layer thickness (equation 11.29): ~= ( ~ ) t=Prt oTa we can now calculate the Nusselt number as: Nu =rtx=_::=.:._Prt=0332~Prt ).{)Tbh-v --; (111.65) This equation is valid for Re=vxf vvincreasesover d:Xby an amount whichis equal to d((v) Wb)(massbalance). Furthermore, on the condensate surface an amount of heat tv! oPd" is released which in the case of laminar flowshould be transportedto the wall by conduction only; so the energy balance willbe : ). l:lH,.p dv = Wdx Now it is assumedthatforeachvalueof xthe -relationbetween4J.,andis given by: 0 .J 3q( v)lJ pg (seeproblem11.1.2)althoughstrictlyspeakingthisequationappliesonlyto laminar vertical filmsof constant thickness.So : cf>v=pgb3W/371,anddcf>v =pgb2Wdb 17 --- - --- - - - - -210 Substitutioninto the energy balance gives: !lT ldx = llH11 p2g03 db 71 After integration over xbetween 0 and Land over b between 0 and OLwe get: J.llTL=AHoplgo1_ 4'7 . The average heat transfer coefficient (a)is defined by: Miup2gWo1. ( ex)AT WL =MlvP4>r:lx=L = 371 From the lasttwoequations we find: 4A. =--3 c5L but in this form the solutionis hardly usable because oLis not an easily measur-able constant.Elimination of bL from the three previous equations gives : ( a )=(}94[AHuPl ).3gJ *(111.77) LYf6T So ( (1.)is inversely proportional to .jL;for this reason horizontal condensers often applied and an increase inthe capacity issought in Wrather than in L.Usuallyhorizontalinternallycooledt ubesareused.Foronehorizontal tube aformulaanalogous to equation (ll1.77) butwitha coefficient of 072 can be derived if instead ofLthe externalpipe diameter Duisused. This relation can be extended to the case where n horizontal pipes are placed under each other. The generalequation forthis is then : [llH &1 p2). 3g]t ( ex)=072nD,J76 T (III.78) The values of ( a: )foundin practiceforlaminar condensate films areon an average20percenthigherthanfollowsfromtheory.FigureIII.23givesa nomograph from which partial heattransfer coefficients for c;:ondensation can be rapidly found. Thevalidity is limitedto those cases wherethe condensate film stillflowslaminarly