Chapter 20 Transfer between the gas-liquid interfaceDevelop a physical picture of the air-water boundary and how it may control mass transferCalculate mass transfer rates for waterExtend these rates to other compoundsbubbles, oil films, aerosol droplets, etc. which perturb theinterface are not included in this model1From the velocity of a molecule colliding with a surface normal to the direction of movement let us calculate an upper limit of the velocity at which molecules enter the liquid phase from the gas phase.MwRTu21 R = 8.31x107 g/cm2 K-1sec-2 mol-1At 298oC and for a compound of 100g/molu cm x kilometers day136000 5 10 /sec /If only one out of 1000 gas phase molecules ends up in the liquid below the surface the transfer velocity is still 5 kilometers/dayThis is much higher than what is observed and requires additional treatment (1cm/sec)2The Two film modelstagnant air layerFlux = -DC/zstagnant water layerAssumes that concentrations at the boundary layers are constant long enough for a concentration profile to reach steady stateNo chemical reactions in the boundary phasesThere may be different resistivities in the different boundary layersIf we first look at the interface; immediately above is an air concentration, and immediately below a water conc. that are probably in equilibriumair3air (well mixed)water well mixedHenry’s lawWaterKiaw = Ca/w/Cw/a =K’HThe flux at the interface is the same whether is coming from the direction of the water or the airaaa/wawww/awzCCDzCCDFsubstituting for Ca/w from Henry’s lawaaw/aiawawww/awzCCKDzCCDand solving for Cw/a)/zKD)/z(D)C/z(D)C/z(DCaiawawaaaawwww/arecalling the expression for flux in the liquid boundarywww/awzCCDFand substituting we get an expression which includes mass transfer through both boundary layers4iawawiawaawwKCC)KD/z)D/z(F1If the bulk water conc. equals Ca/Kiaw; which means Kiaw = Ca/Cw ie Ca and Cw are in equilibriumthere will be no net fluxIf the bulk water conc. > Ca/Kiaw; ie. Cw > Ca/Kiaw then F will be positive and there will be a net flow from the waterto the air phase and vice versa5iawawiawHaawwKCC)KD/z)D/z(F1moles/(cm2 sec) cm/sec moles/cm3www/awzCCDFwhere this net velocity flux, vtot, is called a mass transfer coef)KD/z)D/z(viawaawwtot1note that zw/Dw has the units of sec/cm or inverse velocity)Kv/)v/(viawawtot1111/vtot= 1/vw + 1/(vaKiaw )6This has the “look and feel” of resistors in parallel1/Rtot= 1/R1 + 1/R2Rtot R1R2if vw is << vaKH vtot ~ vw and transfer will be dominated by the water boundary layer or the resistance will be in the water layerif vw >> vaKH resistance will be in air boundary layer and vtot~ vaKHBoundary layer theory assumptions1. We don’t know much about z thicknesses7as air velocities increase we would expect the thickness of za and zw to decrease, and this is observed in that as volatilization increases so does transfer from the water tothat air phase 2. It is assumed that no reactions take place in the boundary layerWhat is the time needed to traverse the boundary layer compared to reaction times?x = (2Dt)1/2 dropping the 2in the water layer w= zw2/Dw and zw/Dw = 1/vwso w= zw/vwin the air boundary a= za/va8Typical lengths for zw are 5x10-2 to 5x10-3 cmand Dw = 10-5cm2/sec;this gives w= zw2/Dw= (1x10-3)2 cm2 / (1x10-5 cm2/sec) w = 0.1 secand for a= ?? For PAH reacting in sunlight in the air or on particlesdPAH/PAH = -k PAH; k = ~0.02 min-1 r ~ 1/k= 3000 sec3. It is assumed that the bulk concentrations in air and water “above and below” the boundary layers are constant long enough to establish a steady state gradient. What if they are not?9Surface Renewal ModelThis model attempts to describe a continual turnover at the air water interface. Eddies in the water and air immediately above and below the air water interface transport material from the bulk phase to an interface boundary Henry’s law equilibrium at the interface is rapidly re-established Mass transported across the interface does not remainthere and is mixed into the bulk phase on the other side.{Cwwater10Cw/aconst x (Dwt)1/2The concentration in an element just below the interface line will have will have a diffusion length that is some function of (D t)1/2s = const(D t)1/2the concentration in this element Cw - Cw/athe product of this concentration and the diffusion length,s, is the mass/unit area that moves across the boundary or fwfw = const.x (Dw t)1/2 (Cw - Cw/a) = fafa = const.x(Da t)1/2(Ca/w - Ca)const.x (Dw t)1/2 (Cw - Cw/a) = const.x(Da t)1/2(Ca/w - Ca)substituting Kiaw x Cw/a = Ca/w and solving for Cw/a)DKDDCDCCaiawwaawwa/wfw =const.x (Dw t)1/2 (Cw - Cw/a)substituting for Cw/afw = }tDK/{tD/K/CCaiawwHaw11  = fa11fa and fw have the units of mass/area, so if we divide by time, which can be thought of the transfer time across the interface, we have the units mass/(area time) or flux, Fif we call 1/t the surface renewal rate , rfw/t = }rDK/{rD/K/CCaiawwiawaw11  = flux= Fwe can even decouple the renewal rates from different turn over rates in the air and water layersfw/t = }rDK/{rD/K/CCaiawwiawaw11 =FThis looks like the stagnant two film expression iawawiawaawwKCC)KD/z)D/z(F1)KD/z)D/z(viawaawwtot1so by analogy, vtot; with partial transfer velocities vw= (rwDw)1/2 & va=(raDa)1/2 Kiawa. quiescent conditions ----> stagnant film modelb. streams, etc. ------> renewal model12flux will vary D; where  ranges from 0.5 to 113Example: What is the Flux of SO2 to the ocean’s surface?We start with an average atmospheric conc. of 3 g/m3 and the fact that SO2 is rapidly oxidized in the slightly alkaline environments of sea water.Hence what is the Cw value at the water surface?From tables we can obtain values for vw and va.; we will later learn how to calculate these.va(SO2)= 4.4 x 10-3 ms-1vw(SO2)= 9.6 x 10-2 ms-1Kiaw = 0.02vtot = 9x10-5 ms-1F = vtot(0 - 3g/m3/KH) = -0.013 g/m-2s-1 (-sign;air-->water)Total ocean surf= 3.6x1014 m2; Ftot oceans= 1.5x1014g/yearThis is about equal to the annual emissions of SO214How do we obtain air and water transfer velocites?let us 1st use evaporating water as an exampleThe