John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 91
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(11.51) to (11.53), although we could simply use tabled datain eqns. (11.54) and (11.55). From Tables 11.2 and 11.3, we obtainSpeciesσ (Å)ε/kB (K)N2O2Ar3.7983.4673.54271.4106.793.3MΩµ28.0232.0039.950.95991.0571.021Substitution of these values into eqn. (11.51) yields626An introduction to mass transfer§11.4Speciesµcalc (kg/m·s)µdata (kg/m·s)N2O2Ar1.767 × 10−52.059 × 10−52.281 × 10−51.80 × 10−52.07 × 10−52.29 × 10−5where we show data from Appendix A (Table A.6) for comparison.
Wethen read cp from Appendix A and use eqn. (11.52) and (11.53) to getthe thermal conductivities of the components:Speciescp (J/kg·K)kcalc (W/m·K)N2O2Ar1041.919.9521.50.025000.025690.01782kdata (W/m·K)0.02600.026150.01787The predictions thus agree with the data to within about 2% for µ andwithin about 4% for k.To compute µm and km , we use eqns. (11.54) and (11.55) and thetabulated values of µ and k. Identifying N2 , O2 , and Ar as species 1,2, and 3, we getφ12 = 0.9894,φ21 = 1.010φ13 = 1.043,φ31 = 0.9445φ23 = 1.058,φ32 = 0.9391and φii = 1.
The sums appearing in the denominators are⎧⎪⎪⎨0.9978 for i = 1$xj φij = 1.008for i = 2⎪⎪⎩0.9435 for i = 3When they are substituted in eqns. (11.54) and (11.55), these valuesgiveµm,calc = 1.861 × 10−5 kg/m·s,µm,data = 1.857 × 10−5 kg/m·skm,calc = 0.02596 W/m·K,km,data = 0.02623 W/m·Kso the mixture values are also predicted within 0.3 and 1.0%, respectively.The equation of species conservation§11.5Finally, we need cpm to compute the Prandtl number of the mix#ture. This is merely the mass weighted average of cp , or i mi cpi ,and it is equal to 1006 J/kg·K. ThenPr = (µcp /k)m = (1.861 × 10−5 )(1006)/0.02596 = 0.721.This is 1% above the tabled value of 0.713.
The reader may wish tocompare these values with those obtained directly using the valuesfor air in Table 11.2 or to explore the effects of neglecting argon inthe preceding calculations.11.5The equation of species conservationConservation of speciesJust as we formed an equation of energy conservation in Chapter 6, wenow form an equation of species conservation that applies to each substance in a mixture.
In addition to accounting for the convection anddiffusion of each species, we must allow the possibility that a speciesmay be created or destroyed by chemical reactions occuring in the bulkmedium (so-called homogeneous reactions). Reactions on surfaces surrounding the medium (heterogeneous reactions) must be accounted forin the boundary conditions.We consider, in the usual way, an arbitrary control volume, R, with aboundary, S, as shown in Fig. 11.8. The control volume is fixed in space,with fluid moving through it.
Species i may accumulate in R, it may travelin and out of R by bulk convection or by diffusion, and it may be createdwithin R by homogeneous reactions. The rate of creation of species i isdenoted as ṙi (kg/m3 ·s); and, because chemical reactions conserve mass,#the net mass creation is ṙ = ṙi = 0. The rate of change of the mass ofspecies i in R is then described by the following balance:ddtρi dR = −Rrate of increaseof i in RS i · dS +nRṙi dR · dS −ṙi dRρi vji · dS + S S R =−rate of convectionof i out of Rdiffusion of iout of Rrate of creationof i in R(11.56)627An introduction to mass transfer628§11.5Figure 11.8 Control volume in afluid-flow and mass-diffusion field.This species conservation statement is identical to our energy conservation statement, eqn. (6.36) on page 293, except that mass of species i hastaken the place of energy and heat.We may convert the surface integrals to volume integrals using Gauss’stheorem [eqn.
(2.8)] and rearrange the result to find: ∂ρi + ∇ · ji − ṙi dR = 0+ ∇ · (ρi v)(11.57)∂tRSince the control volume is selected arbitrarily, the integrand must beidentically zero. Thus, we obtain the general form of the species conservation equation:∂ρi = −∇ · ji + ṙi+ ∇ · (ρi v)∂t(11.58)We may obtain a mass conservation equation for the entire mixture bysumming eqn. (11.58) over all species and applying eqns.
(11.1), (11.17),and (11.22) and the requirement that there be no net creation of mass: $$ ∂ρi = (−∇ · ji + ṙi )+ ∇ · (ρi v)∂tiiso that∂ρ =0+ ∇ · (ρ v)∂t(11.59)The equation of species conservation§11.5This equation applies to any mixture, including those with varying density (see Problem 6.36). = 0Incompressible mixtures. For an incompressible mixture, ∇ · v(see Sect. 6.2 or Problem 11.22), and the second term in eqn. (11.58) maytherefore be rewritten as =v · ∇ρi + ρi ∇ = v · ∇ρi·v∇ · (ρi v) (11.60)=0We may compare the resulting, incompressible species equation to theincompressible energy equation, eqn.
(6.37)∂ρi · ∇ρi = −∇ · ji + ṙi+v∂t∂TDT · ∇T = −∇ · q + q̇=ρcp+vρcpDt∂tDρi=Dt(11.61)(6.37)In these equations: the reaction term, ṙi , is analogous to the heat generation term, q̇; the diffusional mass flux, ji , is analogous to the heat flux,; and dρi is analogous to ρcp dT .qWe can use Fick’s law to eliminate ji in eqn. (11.61).
The resulting equation may be written in different forms, depending upon whatis assumed about the variation of the physical properties. If the product ρDim is independent of (x, y, z)—if it is spatially uniform—theneqn. (11.61) becomesDmi = Dim ∇2 mi + ṙi /ρDt(11.62)where the material derivative, D/Dt, is defined in eqn. (6.38). If, instead,ρ and Dim are both spatially uniform, thenDρi= Dim ∇2 ρi + ṙiDt(11.63)The equation of species conservation and its particular forms mayalso be stated in molar variables, using ci or xi , Ni , and Ji∗ (see Problem 11.24.) Molar analysis sometimes has advantages over mass-basedanalysis, as we discover in Section 11.7.629630An introduction to mass transfer§11.5Figure 11.9 Absorption of ammonia into water.Interfacial boundary conditionsWe are already familiar with the general issue of boundary conditionsfrom our study of the heat equation.
To find a temperature distribution,we specified temperatures or heat fluxes at the boundaries of the domainof interest. Likewise, to find a concentration distribution, we must specify the concentration or flux of species i at the boundaries of the mediumof interest.Temperature and concentration behave differently at interfaces.
Atan interface, temperature is the same in both media as a result of theZeroth Law of Thermodynamics. Concentration, on the other hand, neednot be continuous across an interface, even in a state of thermodynamicequilibrium. Water in a drinking glass, for example, shows discontinouschanges in the concentration of water at both the glass-water interface onthe sides and the air-water interface above. In another example, gaseousammonia is absorbed into water in some types of refrigeration cycles.
Agas mixture containing some particular mass fraction of ammonia willproduce a different mass fraction of ammonia just inside an adjacentbody of water, as shown in Fig. 11.9.To characterize the conditions at an interface, we introduce imaginary surfaces, s and u, very close to either side of the interface. In the§11.5The equation of species conservationammonia absorption process, then, we have a mass fraction mNH3 ,s onthe gas side of the interface and a different mass fraction mNH3 ,u on theliquid side.In many mass transfer problems, we must find the concentration distribution of a species in one medium given only its concentration at theinterface in the adjacent medium. We might wish to find the distribution of ammonia in the body of water knowing only the concentration ofammonia on the gas side of the interface.
We would need to find mNH3 ,ufrom mNH3 ,s and the interfacial temperature and pressure, since mNH3 ,uis the appropriate boundary condition for the species conservation equation in the water.Thus, for the general mass transfer boundary condition, we mustspecify not only the concentration of species i in the medium adjacentto the medium of interest but also the solubility of species i from onemedium to the other.
Although a detailed study of solubility and phaseequilibria is far beyond our scope (see, for example, [11.5, 11.24]), weillustrate these concepts with the following simple solubility relations.Gas-liquid interfaces. For a gas mixture in contact with a liquid mixture,two simplified rules dictate the vapor composition. When the liquid isrich in species i, the partial pressure of species i in the gas phase, pi ,can be characterized approximately with Raoult’s law, which says thatpi = psat,i xifor xi ≈ 1(11.64)where psat,i is the saturation pressure of pure i at the interface temperature and xi is the mole fraction of i in the liquid. When the species i isdilute in the liquid, Henry’s law applies. It says thatpi = H xifor xi 1(11.65)where H is a temperature-dependent empirical constant that is tabulatedin the literature.
Figure 11.10 shows how the vapor pressure varies overa liquid mixture of species i and another species, and it indicates theregions of validity of Raoult’s and Henry’s laws. For example, when xi isnear one, Raoult’s law applies to species i; when xi is near zero, Raoult’slaw applies to the other species.If the vapor pressure were to obey Raoult’s law over the entire range ofliquid composition, we would have what is called an ideal solution.
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