John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 94
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Hadthe mass flux been uniform at the wall, we would have used the analogof a uniform heat flux solution.Natural convection in mass transfer. In Chapter 8, we saw that thedensity differences produced by temperature variations can lead to flowand convection in a fluid. Variations in fluid composition can also produce density variations that result in natural convection mass transfer.This type of natural convection flow is still governed by eqn. (8.3),u∂u∂2u∂u+v= (1 − ρ∞ /ρ)g + ν∂x∂y∂y 2(8.3)but the species equation is now used in place of the energy equation indetermining the variation of density. Rather than solving eqn.
(8.3) and646An introduction to mass transfer§11.6the species equation for specific mass transfer problems, we again turnto the analogy between heat and mass transfer.In analyzing natural convection heat transfer, we eliminated ρ fromeqn. (8.3) using (1 − ρ∞ /ρ) = β(T − T∞ ), and the resulting Grashof andRayleigh numbers came out in terms of an appropriate β∆T instead of∆ρ/ρ. These groups could just as well have been written for the heattransfer problem asGrL =g∆ρL3ρν 2andRaL =g∆ρL3g∆ρL3=ρανµα(11.79)although ∆ρ would still have had to have been evaluated from ∆T .With Gr and Pr expressed in terms of density differences instead oftemperature differences, the analogy between heat transfer and low-ratemass transfer may be used directly to adapt natural convection heattransfer predictions to natural convection mass transfer.
As before, wereplace Nu by Num and Pr by Sc. But this time we also writeRaL = GrL Sc =g∆ρL3µD12(11.80)or calculate GrL as in eqn. (11.79). The densities must now be calculatedfrom the concentrations.Example 11.11Helium is bled through a porous vertical wall, 40 cm high, into surrounding air at a rate of 87.0 mg/m2 ·s. Both the helium and the airare at 300 K, and the environment is at 1 atm.
What is the averageconcentration of helium at the wall, mHe,s ?Solution. This is a uniform flux natural convection problem. Heregm,He and ∆ρ depend on mHe,s , so the calculation is not as straightforward as it was for thermally driven natural convection.To begin, let us assume that the concentration of helium at the wallwill be small enough that the mass transfer rate is low. Since mHe,e =0, if mHe,s 1, then mHe,s − mHe,e 1 as well.
Both conditions forthe analogy to heat transfer will be met.The mass flux of helium at the wall, nHe,s , is known, and becauselow rates prevail, nHe,s ≈ jHe,s = gm,He mHe,s − mHe,eMass transfer at low rates§11.6Hence,Num,L =ngm,He LL He,s=ρDHe,airρDHe,air mHe,s − mHe,eThe appropriate Nusselt number is obtained from the mass transfer analog of eqn.
(8.44b):Num,L6=5ScRa∗√ L4 + 9 Sc + 10 ScwithRa∗L = RaL Num,L =1/5g∆ρ nHe,s L4 mHe,s − mHe,eµρD2He,airThe Rayleigh number cannot easily be evaluated without assuming avalue of the mass fraction of helium at the wall. As a first guess, wepick mHe,s = 0.010.
Then the film composition ismHe,f = (0.010 + 0)/2 = 0.005From eqn. (11.8) and the ideal gas law, we obtain estimates for thefilm density (at the film composition) and the wall densityρf = 1.141 kg/m3and ρs = 1.107 kg/m3From eqn. (11.42) the diffusion coefficient isDHe,air = 7.119 × 10−5 m2 /s.At this low concentration of helium, we expect the film viscosity tobe close to that of pure air. From Appendix A, for air at 300 Kµf µair = 1.857 × 10−5 kg/m·s.The corresponding Schmidt number is Sc = (µf /ρf ) DHe,air = 0.2286.Furthermore,ρe = ρair = 1.177 kg/m3From these values,Ra∗L =9.806(1.177 − 1.107)(87.0 × 10−6 )(0.40)4(1.857 × 10−5 )(1.141)(7.119 × 10−5 )2 (0.010)= 1.424 × 109647An introduction to mass transfer648§11.7We may now evaluate the mass transfer Nusselt numberNum,L!"1/56 (1.424 × 109 )(0.2286)= !√"1/5 = 37.735 4 + 9 0.2286 + 10(0.2286)From this we calculate mHe,s − mHe,e ==nHe,s LρDHe,air Num,L(87.0 × 10−6 )(0.40)(1.141)(7.119 × 10−5 )(37.73)= 0.01136 We have already noted that mHe,s − mHe,e = mHe,s , so we have obtained an average wall concentration 14% higher than our initial guessof 0.010.Using mHe,s = 0.01136 as our second guess, we repeat the preceding calculations with revised values of the densities to obtainmHe,s = 0.01142Since this result is within 0.5% of our second guess, a third iterationis not needed.In the preceding example, concentration variations alone gave riseto buoyancy.
If both temperature and density vary in a natural convection problem, the appropriate Gr or Ra may be calculated using densitydifferences based on both the local mi and the local T , provided thatthe Prandtl and Schmidt numbers are approximately equal (that is, if theLewis number 1). This is usually true in gases.If the Lewis number is far from unity, the analogy between heat andmass transfer breaks down in those natural convection problems that involve both heat and mass transfer, because the concentration and thermal boundary layers may take on very different thicknesses, complicatingthe density distributions that drive the velocity field.11.7Steady mass transfer with counterdiffusionIn 1874, Josef Stefan presented his solution for evaporation from a liquidpool at the bottom of a vertical tube over which a gas flows (Fig.
11.15).This configuration, often called a Stefan tube, is has often been used to§11.7Steady mass transfer with counterdiffusionFigure 11.15 The Stefan tube.measure diffusion coefficients. Vapor leaving the liquid surface diffusesthrough the gas in the tube and is carried away by the gas flow acrosstop of the tube. If the gas stream itself has a low concentration of thevapor, then diffusion is driven by the higher concentration of vapor overthe liquid pool that arises from the vapor pressure of the liquid.A typical Stefan tube is 5 to 10 mm in diameter and 10 to 20 cm long.If the air flow at the top is not too vigorous, and if density variationsin the tube do not give rise to natural convection, then the transport ofvapor from the liquid pool to the top of the tube will be a one-dimensionalupflow.The other gas in the tube is stationary if it is not being absorbed by theliquid (e.g., if it is insoluble in the liquid or if the liquid is saturated withit).
Yet, because there is a concentration gradient of vapor, there mustalso be an opposing concentration gradient of gas and an associated diffusional mass flux of gas, similar to what we found in Example 11.8. Forthe gas in the tube to have a net diffusion flux when it is stationary,there must be an induced upward convective velocity — a counterdiffusion velocity — against which the gas diffuses. As in Example 11.8, thecounterdiffusion velocity can be found in terms of the diffusional massfluxes:v = −jgas ρgas = jvapor ρgas649650An introduction to mass transfer§11.7Figure 11.16 Mass flow across aone-dimensional layer.In this section, we determine the mass transfer rate and concentration profiles in the tube, treating it as the one-dimensional layer shownin Fig.
11.16. The s-surface lies above the liquid and the e-surface liesat the top end of the tube. We allow for the possibility that the counterdiffusion velocity may not be negligible, so that both diffusion andvertical convection may occur. We also allow for the possibility that thegas passes through the liquid surface (N2,s ≠ 0). The results obtainedhere form an important prototype for our subsequent analyses of convective mass transfer at high rates.The solution of the mass transfer problem begins with an appropriateform of the equation of species conservation.
Since the mixture composition varies along the length of the tube, the density may vary as well.If the temperature and pressure are constant, however, the molar concentration of the mixture does not change through the tube [cf. (11.14)].The system is therefore most accurately analyzed using the molar formof species conservation.For one-dimensional steady mass transfer, the mole fluxes N1 and N2have only vertical components and depend only on the vertical coordinate, y. Using eqn.
(11.69), we get, with ni = Mi Ni ,dN1dN2==0dydyso that N1 and N2 are constant throughout the layer. They have s-surfacevalues, N1,s and N2,s , everywhere. These constants will be positive for upward mass flow. (For the orientations in Fig. 11.16, N1,s > 0 and N2,s < 0.)These results are a straightforward consequence of steady-state speciesconservation.Recalling the general expression for Ni , eqn. (11.25), and introducingSteady mass transfer with counterdiffusion§11.7Fick’s law, eqn. (11.34), we writeN1 = x1 N − cD12dx1= N1,sdy(11.81)The term xN1 represents vertical convective transport induced by masstransfer. The total mole flux, N, must also be constant at its s-surfacevalue; by eqn.
(11.23), this isN = N1,s + N2,s = Ns(11.82)Substituting this result into eqn. (11.81), we obtain a differential equationfor x1 :cD12dx1= Ns x1 − N1,sdy(11.83)In this equation, x1 is a function of y, the N’s are constants, and cD12depends on temperature and pressure. If the temperature and pressureare constant, so too is cD12 .
Integration then yields Ns y= ln Ns x1 − N1,s + constantcD12(11.84)We need to fix the constant and the two mole fluxes, N1,s and Ns . Todo this, we apply the boundary conditions at either end of the layer. Thefirst boundary condition is the mole fraction of species 1 at the bottomof the layerx1 = x1,saty =0and it requires thatconstant = − ln(Ns x1,s − N1,s )(11.85)soNs y= lncD12Ns x1 − N1,sNs x1,s − N1,s(11.86)The second boundary condition is the mole fraction at the top of thelayerx1 = x1,eaty =L651An introduction to mass transfer652which yieldsNs L= lncD12x1,e − N1,s /Nsx1,s − N1,s /Ns§11.7(11.87)orx1,e − x1,scD12ln 1 +Ns =Lx1,s − N1,s /Ns(11.88)The last boundary condition is the value of N1,s /Ns .
Since we haveallowed for the possiblity that species 2 passes through the bottom ofthe layer, N1,s /Ns may not equal unity. The ratio depends on the specificproblem at hand, as shown in the two following examples.Example 11.12Find an equation for the evaporation rate of the liquid in the Stefantube described at the beginning of this section.Solution. Species 1 is the evaporating vapor, and species 2 be thestationary gas. Only vapor is transferred through the s-surface, sincethe gas is not significantly absorbed into the already gas-saturatedliquid.
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