John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 93
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Our mathematical problem then becomes∂ 2 mi∂mi= Dim1∂t∂x 2(11.73)Mass transfer at low rates§11.6Figure 11.13 Mass diffusion into asemi-infinite stationary medium.withmi = mi,0fort = 0 (all x)mi = mi,uforx = 0 (t > 0)mi → mi,0forx→ ∞ (t > 0)This math problem is identical to that for transient heat conductioninto a semi-infinite region (Section 5.6), and its solution is completelyanalogous to eqn. (5.50):⎛⎞mi − mi,ux⎠= erf ⎝ 4mi,0 − mi,u2 Dim1 tThe reader can solve all sorts of unsteady mass diffusion problemsby direct analogy to the methods of Chapters 4 and 5 when the concentration of the diffusing species is low.
At higher concentrations of thediffusing species, however, counterdiffusion velocities can be induced,as in Example 11.8. Counterdiffusion may be significant in concentratedmetallic alloys, as, for example, during annealing of a butt-welded junction between two dissimilar metals. In those situations, eqn. (11.72) issometimes modified to use a concentration-dependent, spatially varyinginterdiffusion coefficient (see [11.6]).639An introduction to mass transfer640§11.6Figure 11.14 Concentration boundary layer on a flat plate.Convective mass transfer at low ratesConvective mass transfer is analogous to convective heat transfer whentwo conditions apply:1. The mass flux normal to the surface, ni,s , must be essentially equalto the diffusional mass flux, ji,s from the surface.
In general, thisrequires that the concentration of the diffusing species, mi , be low.92. The diffusional mass flux must be low enough that it does not affectthe imposed velocity field.The first condition ensures that mass flow from the wall is diffusional,as is the heat flow in a convective heat transfer problem. The secondcondition ensures that the flow field will be the same as for the heattransfer problem.As a concrete example, consider a laminar flat-plate boundary layer inwhich species i is transferred from the wall to the free stream, as shownin Fig. 11.14. Free stream values, at the edge of the b.l., are labeled withthe subscript e, and values at the wall (the s-surface) are labeled withthe subscript s.
The mass fraction of species i varies from mi,s to mi,eacross a concentration boundary layer on the wall. If the mass fractionof species i at the wall, mi,s , is small, then ni,s ≈ ji,s , as we saw earlier inthis section. Mass transfer from the wall will be essentially diffusional.This is the first condition.In regard to the second condition, when the concentration difference,mi,s − mi,e , is small, then the diffusional mass flux of species i throughthe wall, ji,s , will be small compared to the bulk mass flow in the stream9In a few situations, such as catalysis, there is no net mass flow through the wall,and convective transport will be identically zero irrespective of the concentration (seeProblems 11.9 and 11.44).Mass transfer at low rates§11.6641wise direction, and it will have little influence on the velocity field.
Hence, is essentially that for the Blasius boundary layer.we would expect that vThese two conditions can be combined into a single requirement forlow-rate mass transfer, as will be described in Section 11.8. Specifically,low-rate mass transfer can be assumed ifBm,i ≡mi,s − mi,e1 − mi,s 0.2condition for low-ratemass transfer(11.74)The quantity Bm,i is called the mass transfer driving force. It is written here in the form that applies when only one species is transferredthrough the s-surface. The evaporation of water into air is typical example of single-species transfer: only water vapor crosses the s-surface.The mass transfer coefficient.
In convective heat transfer problems,we have found it useful to express the heat flux from a surface, q, asthe product of a heat transfer coefficient, h, and a driving force for heattransfer, ∆T . Thus, in the notation of Fig. 11.14,qs = h (Ts − Te )(1.17)In convective mass transfer problems, we would therefore like to express the diffusional mass flux from a surface, ji,s , as the product of amass transfer coefficient and the concentration difference between thes-surface and the free stream.
Hence, we define the mass transfer coefficient for species i, gm,i (kg/m2 ·s), as follows: (11.75)ji,s ≡ gm,i mi,s − mi,eWe expect gm,i , like h, to be determined mainly by the flow field, fluid,and geometry of the problem.The analogy to convective heat transfer.
We saw in Sect. 11.5 thatthe equation of species conservation and the energy equation were quitesimilar in an incompressible flow. If there are no reactions and no heatgeneration, then eqns. (11.61) and (6.37) can be written as∂ρi · ∇ρi = −∇ · ji+v∂t∂T · ∇T = −∇ · q+vρcp∂t(11.61)(6.37)642An introduction to mass transfer§11.6These conservation equations describe changes in, respectively, the amountof mass or energy per unit volume that results from convection by a givenvelocity field and from diffusion under either Fick’s or Fourier’s law.We may identify the analogous quantities in these equations. For thecapacity of mass or energy per unit volume, we see thatdρiis analogous toρcp dT(11.76a)or, in terms of the mass fraction,is analogous toρ dmiρcp dT(11.76b)The flux laws may be rewritten to show the capacities explicitly ji = −ρDim ∇mi = −Dim ρ∇mi = −k∇Tq=−k ρcp ∇TρcpHence, we find the analogy of the diffusivities:Dimis analogous tok=αρcp(11.76c)It follows that the Schmidt number and the Prandtl number are directlyanalogous:Sc =νDimis analogous toPr =µcpν=αk(11.76d)Thus, a high Schmidt number signals a thin concentration boundarylayer, just as a high Prandtl number signals a thin thermal boundarylayer.
Finally, we may write the transfer coefficients in terms of the capacities gm,iρ mi,s − mi,eji,s = gm,i mi,s − mi,e =ρh=qs = h (Ts − Te )ρcp (Ts − Te )ρcpfrom which we see thatgm,iis analogous tohcp(11.76e)Mass transfer at low rates§11.6643From these comparisons, we conclude that the solution of a heat convection problem becomes the solution of a low-rate mass convection problem upon replacing the variables in the heat transfer problem with theanalogous mass transfer variables given by eqns.
(11.76).Convective heat transfer coefficients are usually expressed in termsof the Nusselt number as a function of Reynolds and Prandtl numberNux =(h/cp )xhx== fn (Rex , Pr)kρ(k/ρcp )(11.77)For convective mass transfer problems, we expect the same functional dependence after we make the substitutions indicated above. Specifically,if we replace h/cp by gm,i , k/ρcp by Di,m , and Pr by Sc, we obtainNum,x ≡gm,i x= fn (Rex , Sc)ρDim(11.78)where Num,x , the Nusselt number for mass transfer, is defined as indicated.
Num is sometimes called the Sherwood number 10 , Sh.Example 11.10A napthalene model of a printed circuit board (PCB) is placed in awind tunnel. The napthalene sublimates slowly as a result of forcedconvective mass transfer. If the first 5 cm of the napthalene modelis a flat plate, calculate the average rate of loss of napthalene fromthat part of the model. Assume that conditions are isothermal at303 K and that the air speed is 5 m/s.
Also, explain how napthalenesublimation might be used to determine heat transfer coefficients .Solution. Let us first find the mass fraction of napthalene justabove the model surface. A relationship for the vaporpressure ofnapthalene (in mmHg) is log10 pv = 11.450 − 3729.3 (T K). At 303 K,this gives pv = 0.1387 mmHg = 18.49 Pa. The mole fraction ofnapthalene is thus xnap,s = 18.49/101325 = 1.825 × 10−4 , and with10Thomas K. Sherwood (1903–1976) obtained his doctoral degree at M.I.T. under Warren K.
Lewis in 1929 and was a professor of Chemical Engineering there from 1930 to1969. He served as Dean of Engineering from 1946 to 1952. His research dealt withmass transfer and related industrial processes. Sherwood was also the author of veryinfluential textbooks on mass transfer.644An introduction to mass transfer§11.6eqn. (11.9), the mass fraction is, with Mnap = 128.2 kg/kmol,mnap,s =(1.825 × 10−4 )(128.2)(1.825 × 10−4 )(128.2) + (1 − 1.825 × 10−4 )(28.96)= 8.074 × 10−4The mass fraction of napthalene in the free stream, mnap,s , is zero.With these numbers, we can check to see if the mass transfer rateis low enough to use the analogy of heat and mass transfer, witheqn. (11.74):8.074 × 10−4 − 0= 8.081 × 10−4 0.2Bm,nap =1 − 8.074 × 10−4The analogy therefore applies.The convective heat transfer coefficient for this situation is thatfor a flat plate boundary layer. The Reynolds number isReL =(5)(0.05)u∞ L== 1.339 × 104ν1.867 × 10−5where we have used the viscosity of pure air, since the concentrationof napthalene is very low.
The flow is laminar, so the applicable heattransfer relationship is eqn. (6.68)NuL =hL1/2= 0.664 ReL Pr1/3k(6.68)Under the analogy, the Nusselt number for mass transfer isNum,L =gm,i L1/2= 0.664 ReL Sc1/3ρDimThe diffusion coefficient for napthalene in air, from Table 11.1, isDnap,air = 0.86 × 10−5 m/s, and thus Sc = 1.867 × 10−5 /0.86 × 10−5 =2.17. Hence,Num,L = 0.664 (1.339 × 104 )1/2 (2.17)1/3 = 99.5and, using the density of pure air,ρDnap,airNum,LL(1.166)(0.86 × 10−5 )(99.5) = 0.0200 kg/m2 s=0.05gm,nap =Mass transfer at low rates§11.6645The average mass flux from this part of the model is then nnap,s = gm,nap mnap,s − mnap,e= (0.0200)(8.074 × 10−4 − 0)= 1.61 × 10−5 kg/m2 s = 58.0 g/m2 hNapthalene sublimation can be used to infer heat transfer coefficients by measuring the loss of napthalene from a model over somelength of time.
Experiments are run at several Reynolds numbers.The lost mass fixes the sublimation rate and the mass transfer coefficient. The mass transfer coefficient is then substituted in the analogyto heat transfer to determine a heat transfer Nusselt number at eachReynolds number. Since the Schmidt number of napthalene is notgenerally equal to the Prandtl number under the conditions of interest, some assumption about the dependence of the Nusselt numberon the Prandtl number must usually be introduced.Boundary conditions.
When we apply the analogy between heat transfer and mass transfer to calculate gm,i , we must consider the boundarycondition at the wall. We have dealt with two common types of wall condition in the study of heat transfer: uniform temperature and uniformheat flux. The analogous mass transfer wall conditions are uniform concentration and uniform mass flux. We used the mass transfer analog ofthe uniform wall temperature solution in the preceding example, sincethe mass fraction of napthalene was uniform over the entire model.
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