John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 95
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Thus, N2,s = 0, and Ns = N1,s = Nvapor,s is simply the evaporation rate of the liquid. The s-surface is just above the surface of theliquid. The mole fraction of the evaporating liquid can be determinedfrom solubility data; for example, if the gas is more-or-less insolublein the liquid, Raoult’s law, eqn. (11.64), may be used. The e-surface isat the mouth of the tube. The gas flow over the top may contain someconcentration of the vapor, although it should generally be near zero.The ratio N1,s /Ns is unity, and the rate of evaporation isx1,e − x1,scD12ln 1 +(11.89)Ns = Nvapor,s =Lx1,s − 1Example 11.13What is the evaporation rate in the Stefan tube if the gas is bubbledup to the liquid surface at some fixed rate, Ngas ?Solution.
Again, N1,s = Nvapor,s is the evaporation rate. However,the total mole flux isNs = Ngas + N1,sSteady mass transfer with counterdiffusion§11.7Thus,Ngas + N1,sx1,e − x1,scD12ln 1 +=Lx1,s − N1,s /(N1,s + Ngas )(11.90)This equation fixes N1,s , but it must be solved iteratively.Once we have found the mole fluxes, we may compute the concentration distribution, x1 (y), using eqn. (11.86):x1 (y) =N1,s + x1,s − N1,s Ns exp(Ns y/cD12 )Ns(11.91)Alternatively, we may eliminate Ns between eqns. (11.86) and (11.87) toobtain the concentration distribution in a form that depends only on theratio N1,s /Ns :y/Lx1,e − N1,s /Nsx1 − N1,s /Ns=(11.92)x1,s − N1,s /Nsx1,s − N1,s /NsExample 11.14Find the concentration distribution of water vapor in a helium–waterStefan tube at 325 K and 1 atm.
The tube is 20 cm in length. Assumethe helium stream at the top of the tube to have a mole fraction ofwater of 0.01.Solution. Let water be species 1 and helium be species 2. Thevapor pressure of the liquid water is approximately the saturationpressure at the water temperature. Using the steam tables, we getpv = 1.341 × 104 Pa and, from eqn.
(11.16),x1,s =1.341 × 104 Pa= 0.1323101, 325 PaWe use eqn. (11.14) to evaluate the mole concentration in the tube:c=101, 325= 0.03750 kmol/m38314.5(325)From eqn. (11.42) we obtain D12 (325 K, 1 atm) = 1.067 × 10−4 m2 /s.Then eqn. (11.89) gives the molar evaporation rate:0.01 − 0.13230.03750(1.067 × 10−4 )ln 1 +N1,s =0.200.1323 − 1−62= 2.638 × 10 kmol/m ·s653An introduction to mass transfer654§11.8This corresponds to a mass evaporation rate:n1,s = 4.754 × 10−5 kg/m2 ·sThe concentration distribution of water vapor [eqn. (11.91)] isx1 (y) = 1 − 0.8677 exp(0.6593y)where y is expressed in meters.Stefan tubes have been widely used to measure mass transfer coefficients, by observing the change in liquid level over a long period of timeand solving eqn. (11.89) for D12 .
These measurements are subject to avariety of experimental errors, however. For example, the latent heat ofvaporization may tend to cool the gas mixture near the interface, causing a temperature gradient in the tube. Vortices near the top of the tube,where it meets the gas stream, may cause additional mixing, and densitygradients may cause buoyant circulation. Additional sources of errorand alternative measurement techniques are described by Marrero andMason [11.7].The problem dealt with in this section can alternatively be solved ona mass basis, assuming a constant value of ρD12 (see Problem 11.33 andProblem 11.34).
The mass-based solution of this problem provides animportant approximation in our analysis of high-rate convective masstransfer in the next section.11.8Mass transfer coefficients at high rates of masstransferIn Section 11.6, we developed an analogy between heat and mass transferthat allowed us to calculate mass transfer coefficients when the rate ofmass transfer was low. This analogy required that the velocity field beunaffected by mass transfer and that the transferred species be dilute.When those conditions are not met, the mass transfer coefficient willbe different than the value given by the analogy.
The difference can beeither an increase or a decrease and can range from a few percent to anorder of magnitude or more, depending upon the concentrations of thediffusing species. In addition to the diffusive transport represented bythe mass transfer coefficient, convective transport can contribute substantially to the total mass flux.§11.8Mass transfer coefficients at high rates of mass transfer655Figure 11.17 The mass concentrationboundary layer.In this section, we model mass convection when the transferred speciesaffects the velocity field and is not necessarily dilute.
First, we define themass transfer driving force, which governs the total mass flux from thewall. Then, we relate the mass transfer coefficient at high mass transferrates to that at low mass transfer rates.The mass transfer driving forceFigure 11.17 shows a boundary layer over a wall through which there isa net mass transfer, ns ≡ ṁ , of the various species in the directionnormal to the wall.11 In particular, we will focus on species i. In the freestream, i has a concentration mi,e ; at the wall, it has a concentration mi,s .The mass flux of i leaving the wall is obtained from eqn. (11.21):ni,s = mi,s ṁ + ji,s(11.93)We seek to obtain ṁ in terms of the concentrations mi,s and mi,e .
As before, we define the mass transfer coefficient for species i, gm,i (kg/m2 ·s),asgm,i = ji,s mi,s − mi,e(11.94)Thus, ni,s = mi,s ṁ + gm,i mi,s − mi,e(11.95)The mass transfer coefficient is again based on the diffusive transfer fromthe wall; however, it may now differ considerably from the value for lowrate transport.In this context, we denote the total mass flux through the wall as ṁ , rather thanns , so as to be consistent with other literature on the subject.11An introduction to mass transfer656Equation (11.95) may be rearranged asmi,e − mi,sṁ = gm,imi,s − ni,s /ṁ§11.8(11.96)which express the total mass flux of all species through the wall, ṁ , asthe product of the mass transfer coefficient and a ratio of concentrations.This ratio is called the mass transfer driving force for species i:mi,e − mi,s(11.97)Bm,i ≡mi,s − ni,s /ṁThe ratio of mass fluxes in the denominator is called the mass fractionin the transferred state, denoted as mi,t :mi,t ≡ ni,s /ṁ(11.98)The mass fraction in the transferred state is simply the fraction of thetotal mass flux, ṁ , which is made up of species i.
It is not really a massfraction in the sense of Section 11.2 because it can have any value from−∞ to +∞, depending on the relative magnitudes of ṁ and ni,s . If, forexample, n1,s −n2,s in a binary mixture, then ṁ is very small andboth m1,t and m2,t are very large.Equations (11.96), (11.97), and (11.98) provide a formulation of masstransfer problems in terms of the mass transfer coefficient, gm,i , and thedriving force for mass transfer, Bm,i :ṁ = gm,i Bm,i(11.99)whereBm,i =mi,e − mi,smi,s − mi,t,mi,t = ni,s /ṁ(11.100)These relations are based on an arbitrary species, i.
The mass transfer rate may equally well be calculated using any species in a mixture;one obtains the same result for each. This is well illustrated in a binarymixture for which one may show that (Problem 11.36)gm,1 = gm,2andBm,1 = Bm,2Mass transfer coefficients at high rates of mass transfer§11.8In many situations, only one species is transferred through the wall.If species i is the only one passing through the s-surface, then ni,s = ṁ ,so that mt,i = 1. The mass transfer driving force is simplyBm,i =mi,e − mi,smi,s − 1one speciestransferred(11.101)In all the cases described in Section 11.6, only one species is transferred.Example 11.15A pan of hot water with a surface temperature of 75◦ C is placed inan air stream that has a mass fraction of water equal to 0.05. If theaverage mass transfer coefficient for water over the pan is gm,H2 O =0.0170 kg/m2 ·s and the pan has a surface area of 0.04 m2 , what isthe evaporation rate?Solution.
Only water vapor passes through the liquid surface, sinceair is not strongly absorbed into water under normal conditions. Thus,we use eqn. (11.101) for the mass transfer driving force. Reference toa steam table shows the saturation pressure of water to be 38.58 kPaat 75◦ C, soxH2 O,s = 38.58/101.325 = 0.381Putting this value into eqn. (11.67), we obtainmH2 O,s = 0.277so thatBm,H2 O =0.05 − 0.277= 0.3140.277 − 1.0Thus,ṁH2 O = gm,H2 O Bm,H2 O (0.04 m2 )= (0.0170 kg/m2 ·s)(0.314)(0.04 m2 )= 0.000214 kg/s = 769 g/hr657658An introduction to mass transfer§11.8Figure 11.18 A stagnant film.The effect of mass transfer rates on the mass transfercoefficientWe still face the task of finding the mass transfer coefficient, gm,i .
Themost obvious way to do this would be to apply the same methods we usedto find the heat transfer coefficient in Chapters 6 through 8—solution ofthe momentum and species equations or through correlation of masstransfer data. These approaches are often used, but they are more complicated than the analogous heat transfer problems, owing to the coupling of the flow field and the mass transfer rate.
Simple solutions arenot so readily available for mass transfer problems. We instead employa widely used approximate method that allows us to calculate gm,i fromthe low-rate mass transfer coefficient by applying a correction for theeffect of finite mass transfer rates.To isolate the effect of ṁ on the mass transfer coefficient, we first∗:define the mass transfer coefficient at zero net mass transfer, gm,i∗gm,i≡ lim gm,iṁ →0∗is simply the mass transfer coefficient for low rates thatThe value gm,iwould be obtained from the analogy between heat and mass transfer, asdescribed in Section 11.6.
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