The CRC Handbook of Mechanical Engineering. Chapter 4. Heat and Mass Transfer (776127), страница 53
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For other temperatures use Sc/Sc300 K . (m2/rT)/(m2rT)300 K, where m and r are for water,and T is absolute temperature. For chemically similar solutes of different molecular weights useSc2/Sc1 . (M2/M1)0.4. A table of (m2/rT)/(m2rT)300 K for water follows.T(K)(m2/rT)/(m2/rT)300 K2903003103203303403503603701.661.000.6230.4290.2960.2210.1670.1230.097From Spalding, D.B. 1963. Convective Mass Transfer, McGraw-Hill, New York. With permission.rDmi= Ñ × ji + r˙i¢¢¢Dt(4.7.31)where D/Dt is the substantial derivative operator.If we consider a binary system of species 1 and 2 and introduce Fick’s law, Equation (4.7.24a) intoEquation (4.7.31), then© 1999 by CRC Press LLC4-216Section 4TABLE 4.7.5Diffusion Coefficients in Solids, D = Do exp(–Ea/RT)SystemOxygen-Pyrex glassOxygen-fused silica glassOxygen-titaniumOxygen-titanium alloy (Ti-6Al-4V)Oxygen-zirconiumHydrogen-ironHydrogen-a-titaniumHydrogen-b-titaniumHydrogen-zirconiumHydrogen-Zircaloy–4Deuterium-Pyrex glassDeuterium-fused silica glassHelium-Pyrex glassHelium-fused silica glassHelium-borosilicate glassNeon-borosilicate glassCarbon-FCC ironCarbon-BCC ironDo, m2/secEa, kJ/kmol–84.69 ´ 1043.77 ´ 1042.13 ´ 1052.59 ´ 1057.06 ´ 1055.60 ´ 1035.18 ´ 1042.78 ´ 1044.81 ´ 1046.05 ´ 1054.69 ´ 1043.77 ´ 1042.72 ´ 1042.55 ´ 1042.34 ´ 1043.77 ´ 1041.378 ´ 1058.75 ´ 1046.19 ´ 102.61 ´ 10–95.0 ´ 10–35.82 ´ 10–24.68 ´ 10–57.60 ´ 10–81.80 ´ 10–6l.95 ´ 10–71.09 ´ 10–71.27 ´ 10–56.19 ´ 10–82.61 ´ 10–94.76 ´ 10–85.29 ´ 10–81.94 ´ 10–91.02 ´ 10–102.3 ´ 10–51.1 ´ 10–6Various sources.rDmi= Ñ × (rD12 Ñm1 ) + r˙1¢¢¢Dt(4.7.32)When working on a mass basis we define a stationary medium as one in which the mass average velocityn is zero everywhere.
Substituting in Equation (4.7.32) with no chemical reactions and assuming constantproperties,¶m1= D12 Ñ 2 m1¶t(4.7.33)which is the diffusion equation, and is the mass transfer analog to Fourier’s equation for heat conduction.For steady diffusion, Equation (4.7.33) reduces to Laplace’s equationÑ 2 m1 = 0(4.7.34)Notice that since properties have been assumed constant, any measure of concentration can be used inEquations (4.7.33) and (4.7.34), for example r1, c1, and x1.Diffusion in a Stationary MediumMany problems involving diffusion in a stationary medium are governed by the diffusion equation(Equation 4.7.33). Often solutions may be obtained from their heat conduction analogs. Some importantcases follow.Steady Diffusion through a Plane WallThe mass flow of species 1 across a plane wall of thickness L and cross-sectional area A ism˙ 1 =© 1999 by CRC Press LLCrD12 Am1,u - m1,u¢ kg m 2 secL()(4.7.35)4-217Heat and Mass TransferFIGURE 4.7.3 Steady diffusion across a plane wall.where the u- and u¢-surfaces are shown in Figure 4.7.3.
Solubility data are required to relate the u- andu¢-surface concentrations to s- and s¢-surface concentrations. Alternatively for systems that obey Henry’slaw, a solubility 6 can be defined as the volume of solute gas (at STP of 0°C and 1 atm) dissolved inunit volume when the gas is at a partial pressure of 1 atm. Then, defining permeability P12 as the productD126, the volume flow of species 1 isP AV˙1 = 12 P1,s - P1,s¢ m 3 (STP) secL()(4.7.36)where the partial pressures P 1 are in atmospheres.
The SI units for permeability are m 3(STP)/m2sec(atm/m). Permeability and solubility data are given in Table 4.7.6. For example, considerhelium at 105 Pa contained in a 7056-glass vessel with a 1-mm-thick wall at 680 K. For a surface areaof 0.01 m2, the leakage rate into ambient air is(1.0 ´ 10 )(0.01) 10((0.001)-12V˙ =5)- 0 = 1.0 ´ 10 -6 m 3 (STP) secwhere the value P12 was obtained from Table 4.7.6.In general, mass fractions are discontinuous across phase interfaces. Hence, Equation (4.7.35) cannotbe generalized to a number of walls in series by simply adding diffusion resistances. However, equilibriumpartial pressures P1 are continuous, and for two walls A and B, Equation 4.7.36 becomesV˙1 =P1,s - P1,s¢m 3 (STP) secLALB+P1 A A P1B A(4.7.37)Transient Diffusion in a Semi-Infinite SolidThe typically low diffusion coefficients characterizing solids result in many situations where concentration changes are limited to a thin region near the surface (of thickness dc ~ (D12t)1/2).
Examples includecase-hardening of mild steel and coloring of clear sapphires. Details of the geometry are then unimportant© 1999 by CRC Press LLC4-218Section 4TABLE 4.7.6GasH2Solubility and Permeability of Gases in SolidsSolidVulcanized rubberVulcanized neopreneSilicone rubberNatural rubberPolyethylenePolycarbonateFused silicaNickelHeSilicone rubberNatural rubberPolycarbonateNylon 66TeflonFused silicaPyrex glassO2N2CO2H 2ONeAr7740 glass(94% SiO2 + B2O3 + P2O5,5% Na2O + Li2 + K2O,1% other oxides)7056 glass(90% SiO2 + B2O3 + P2O5,8% Na2O + Li2 + K2O,1% PbO, 5% other oxides)Vulcanized rubberSilicone rubberNatural rubberPolyethylenePolycarbonateSilicone-polycarbonatecopolymer (57% silicone)Ethyl celluloseVulcanized rubberSilicone rubberNatural rubberSilicone-polycarbonatecopolymer (57% silicone)TeflonVulcanized rubberSilicone rubberNatural rubberSilicone-polycarbonatecopolymer (57% silicone)Nylon 66Silicone rubberFused silicaFused silica© 1999 by CRC Press LLCTemperature,K6 (m3(STP)/m3 atm)or 6¢a3002903003003003004008003604403003003003003003008003008004705807206 = 0.0406 = 0.051Permeabilitybm3(STP)/m2 sec (atm/m)0.34 ´0.053 ´4.2 ´0.37 ´0.065 ´0.091 ´10–1010–1010–1010–1010–1010–102.3 ´0.24 ´0.11 ´0.0076 ´0.047 ´10–1010–1010–1010–1010–106¢ .
0 0356¢ . 0.0306¢ = 0.2026¢ = 0.1926¢ . 0.0186¢ . 0.0266¢ . 0.0066¢ . 0.0246 = 0.00846 = 0.00386 = 0.00464.6 ´ 10–131.6 ´ 10–126.4 ´ 10–123906806¢ = 0.00396¢ = 0.00591.2 ´ 10–141.0 ´ 10–123003003003003003006 = 0.070300300300300300300300290300300300310300–1200900–12006 = 0.0356 = 0.0900.15 ´3.8 ´0.18 ´4.2 ´0.011 ´1.2 ´10–1010–1010–1010–1210–1010–100.09 ´0.054 ´1.9 ´0.062 ´0.53 ´10–1010–1010–1210–1010–100.019 ´1.0 ´21 ´1.0 ´7.4 ´10–1010–1010–1010–1010–100.0013 ´ 10–100.91–1.8 ´ 10–106 . 0.0026 .
0.014-219Heat and Mass TransferTABLE 4.7.6GasSolubility and Permeability of Gases in SolidsSolid6 (m3(STP)/m3 atm)or 6¢aTemperature,KPermeabilitybm3(STP)/m2 sec (atm/m)Solubility 6 = volume of solute gas (0°C, 1 atm) dissolved in unit volume of solid when the gas is at 1 atmpartial pressure. Solubility coefficient 6¢ = c1,u /c1,s.bPermeability P12 = D126.From various sources, including Geankoplis, C.J.
1993. Transport Processes and Unit Operations, 3rd ed.,Prentice-Hall; Englewood Cliffs, N.J.; Doremus, R.H. 1973. Glass Science, Wiley, New York; Altemose, V.O.1961. Helium diffusion through glass, J. Appl. Phys., 32, 1309–1316. With permission.aFIGURE 4.7.4 Transient diffusion in a plane slab.and semi-infinite solid model can be used (Figure 4.7.4). For an initial concentration m1,0 and a u-surfaceconcentration suddenly changed to m1,u at time t = 0, the concentration distribution m1(z,t) ism1 - m1,0m1,u - m1,0= erfcz(4.7.38)12(4D12 t )and the dissolution rate isæD öm˙ 1 = j1,u A = rAç 12 ÷è pt ø12(m1,u)- m1,0 kg sec(4.7.39)For example, consider a Pyrex glass slab at 800 K suddenly exposed to helium at 104 Pa. The molarequivalent to Equation (4.7.39) for an assumed constant solid phase molar concentration c is12˙MæD ö1= ç 12 ÷ c1,u - c1,0A è pt ø()From Table 4.7.6, 6¢ = c1,u/c1,s @ 0.024; hence, c1,u = (0.024)(104)/(8314)(800) = 3.61 ´ 10–5 kmol/m3.From Table 4.7.4, D12 = 4.76 ´ 10–8 exp[–(2.72 ´ 104)(103)/(8314)(800)] = 7.97 ´ 10–10 m2/sec.
Hence,12˙Mæ 7.97 ´ 10 -10 ö-5-101=ç÷ 3.61 ´ 10 - 0 = 5.75 ´ 10 t kmol secptA èø(© 1999 by CRC Press LLC)4-220Section 4Transient Diffusion in Slabs, Cylinders, and SpheresTransient heat conduction in slates, cylinders, and spheres with surface convection is dealt with in Section4.1. The analogous mass diffusion problem for the slab –L < z < L is now considered. On a molar basisthe governing differential equation is¶x1¶2 x= D12 21¶t¶z(4.7.40)with initial condition x1 = x1,0 at t = 0. Boundary conditions are ¶x1/¶z = 0 at z = 0, and at the surface z = L,- cD12¶x1¶z(= G m1 y1,s - y1,e)(4.7.41)z= LThe convective boundary condition is of the same form as Newton’s law of cooling, and defines themole transfer conductance Gm1 (kmol/m2sec) (see also the section on mass and mole transfer conductances).
Also, we have followed chemical engineering practice and denoted mole fraction x in the solid(or liquid) phase and y in the liquid (or gas) phase, to emphasize that generally mole fraction is notcontinuous across a phase interface. For example, consider absorption of a sparingly soluble gas into aliquid for which Henry’s law, Equation (4.7.13), applies: then y1,s = Hex1,u.In using heat conduction charts for mass diffusion problems, particular care must be taken with theevaluation of the Biot number. For heat conduction Bi = hcL/k, where k is the solid conductivity.
Formass diffusion the Biot number accounts for the discontinuity in concentration across the phase interface.Using gas absorption into a plane layer of liquid, for example, when Equation (4.7.41) is put into anappropriate dimensionless form, the mass transfer Biot number is seen to beBi m =G m1HeLcD12(4.7.42)For sparingly soluble gases, e.g., O2 or CO2 in water, He, and hence Bim, are very large, and the absorptionprocess is liquid-side controlled; that is, a uniform gas-phase composition can be assumed. Often interfaceequilibrium data are in graphical or tabular form; then an effective Biot number at the concentration ofconcern must be used.For example, consider a 2-mm-diameter droplet of water at 300 K entrained in an air flow at 1 atmpressure containing 1% by volume CO2.
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