The CRC Handbook of Mechanical Engineering. Chapter 4. Heat and Mass Transfer (776127), страница 54
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From Table 4.7.5, He = CHe= 1710. The liquid phase molardensity can be approximated by the pure water value of c = r/M = 996/18 = 55.3 kmol/m3. The liquidphase diffusion coefficient is obtained from Table 4.7.4 as D12 = vH2O / Sc12 = 0.87 ´ 10–6/420 = 2.07 ´10–9 m2/sec. For negligible relative motion between the droplet and gas, the Sherwood number (see thesection on dimensionless groups) is approximately 2.0, and hence the gas phase mole transfer conductance is Gm1 = 2cD12/D. For the gas phase, the molar density c = P/RT = (1.0133 ´ 105)/(8314)(300) =0.0406 kmol/m3 and D12 = 0.157 ´ 10–4 m2/sec from Table 4.7.2. Thus,G m1 =(2)(0.0406) (0.157 ´ 10 -4 )= 6.37 ´ 10 -4 kmol m 2 sec(0.002)From Equation 4.7.42 with L = R the droplet radius, the mass transfer Biot number is(6.37 ´ 10 )(1710)(0.001) = 9520(55.3) (2.07 ´ 10 )-4Bi m =© 1999 by CRC Press LLC-94-221Heat and Mass TransferThus, even for a small droplet with a relatively large gas-side mole transfer conductance, the absorptionprocess is liquid-side controlled.Diffusion in a Porous CatalystPorous catalysts are used to give a large surface area per unit volume of catalyst surface.
Current practicefor automobile catalytic convertors is to use a ceramic matrix as a support for a thin porous aluminalayer that is impregnated with the catalyst (called a washcoat). A typical matrix has passages of hydraulicdiameter 1 mm, and the washcoat may be about 20 mm thick. Pore sizes are of the order of 1 mm forwhich ordinary and Knudsen diffusion resistances are important. A simple model for diffusion in aporous catalyst isJ1 = - cD1,eff Ñx1 kmol m 2 sec(4.7.43)where the subscript eff denotes an effective diffusivity that accounts for the presence of the solid material.Assuming additive resistances,111=+D1,eff D12,eff D K1,eff(4.7.44)andD12,eff =evD ;t 12D K1,eff =evDt K1,eff(4.7.45)where en is the volume void fraction and t is the tortuosity factor (usually between 4 and 8).
From thekinetic theory of gases the Knudsen diffusion coefficient is12D K1 = 97re (T M1 )m 2 sec(4.7.46)for effective pore radius re in meters and T in kelvins.When a chemical reaction takes place within a porous layer, a concentration gradient is set up, andsurfaces on pores deep within the pellet are exposed to lower reactant concentrations than surfaces nearthe pore openings. For a first-order reaction it is straightforward to obtain the concentration distribution.The results of such an analysis are conveniently given in the form of an effectiveness hp, which is definedas the actual consumption rate of the reactant divided by that for an infinite diffusion coefficient. For alayer of thickness L exposed to reactants on one side, as shown in Figure 4.7.5.tanh bL;hp =bLæ k ¢¢a P öb=ç÷è D1,eff ø12(4.7.47)where k² (m/sec) is the rate constant for a first-order reaction and ap (m–1) is the catalyst area per unitvolume.
Notice that this effectiveness is analogous to the efficiency of a heat transfer fin.For example, consider a 30-mm-thick porous alumina washcoat with a volume void fraction en = 0.8,a tortuosity factor t = 4.0, average pore radius re = 1 mm, and catalytic surface area per unit volume ap= 7.1 ´ 105 cm2/cm3. For carbon monoxide oxidation by copper oxide at 800 K, 1 atm, the rate constantis approximately 4.2 ´ 10–4 m2/sec.
To calculate the effectiveness of the washcoat, we first need tocalculate the effective diffusion coefficient D1,eff:© 1999 by CRC Press LLC4-222Section 4Reactant FlowÆ1,eWashcoatLCeramic MatrixFIGURE 4.7.5 A catalyst layer.D12,eff =ev0.81.06 ´ 10 -4 = 2.12 ´ 10 -5 m 2 secD =t 12 4.0()where D12 is approximated as the CO-air value from Table 4.7.2.D K1,eff =ev0.812D =(97) 1 ´ 10 -6 (800 28) = 1.04 ´ 10 -4 m 2 sect 12 4.0()111=+;D1,eff 2.12 ´ 10 -5 1.04 ´ 10 -4()()( )é 4.2 ´ 10 -4 7.1 ´ 10 5 10 2 ùúb=ê1.76 ´ 10 -5êëúûD1,eff = 1.76 ´ 10 -5 m 2 sec12= 4.2 ´ 10 4 m -1 ;hP =()()bL = 4.2 ´ 10 4 30 ´ 10 -6 = 1.236tanh 1.236= 68.3%1.236In an automobile catalytic convertor, Equation 4.7.47 applies to the catalyst washcoat.
However, themass transfer problem also involves a convective process for transport of reactants from the bulk flow.Referring to Figure 4.7.6 there are two mass transfer resistances in series, and the consumption rate ofspecies 1 per unit surface area of the washcoat isJ1,s =- x1,e11+Lh p k ¢¢c G m1kmol m 2 sec(4.7.48)where Gm1 is the mole transfer conductance describing convective transport to the washcoat surface (seethe section on mass and mole transfer conductances). Notice that when Gm1 ! Lhpk²c the reaction rateis controlled by mass transfer from the gas stream to the washcoat surface; when Lhpk²c ! Gm1, thereaction rate is controlled by diffusion within the washcoat and the kinetics of the reaction.© 1999 by CRC Press LLC4-223Heat and Mass TransferFIGURE 4.7.6 Equivalent circuit for mass transfer in an automobile catalytic convertor.Diffusion in a Moving MediumNet mass transfer across a surface results in a velocity component normal to the surface, and an associatedconvective flux in the direction of mass transfer.
This convective flow is called a Stefan flow. The solutionsof a number of mass transfer problems, involving a Stefan flow induced by the mass transfer processitself, follow. When necessary to obtain an analytical result, properties are assumed constant. Thus, useof these results requires evaluation of properties at a suitable reference state.Diffusion with One Component StationaryAs an example, consider the simple heat pipe shown in Figure 4.7.7 with the evaporator and condenserlocated at the ends only (a bad design!). Then, if the working fluid is species 1, and a noncondensablegas is species 2, the concentration distribution isæ 1 - x1 ö æ 1 - x1,e öç1- x ÷ = ç1- x ÷èè1,s ø1,s øz L(4.7.49)and the vapor flux along the heat pipe isN1 =cD12 1 - x1,ekmol m 2 seclnL1 - x1,s(4.7.50)FIGURE 4.7.7 A simple heat pipe with the evaporator and condenser located at its ends.Notice that N2 = 0; that is, the gas is stationary.
The rate of heat flow for a heat pipe of cross-sectionalarea of Ac is Q̇ = N1 M1 hf gAc. Evaluation of the cD product at a reference temperature Tr = (1/2)(Ts +Te) is adequate for most applications. Equation (4.7.50) applies to any situation where a one-dimensionalmodel of mass transport is appropriate.Heterogeneous CombustionAs an example, consider a small carbon particle entrained in a high-temperature airstream, as shown inFigure 4.7.8. The surface reaction is 2C + O2 ® 2CO and there are no reactions in the gas phase.
Thestoichiometric ratio for the reaction is r = 4/3 kg oxygen/kg carbon. The reaction is diffusion controlledat the temperatures under consideration, that is, mO ,s . 0. The mass transfer rate is ns, which we give2© 1999 by CRC Press LLC4-224Section 4the distinctive symbol m˙ ¢¢ since it is usually the desired result of an analysis; in this situation m˙ ¢¢ =nC,u is the combustion rate of carbon, and for a spherical particle of radius R is given bym˙ ¢¢ =rD O2 ,mRé m O ,e - m O , s ùrD O2 ,m22ln ê1 +kg m 2 secú = 0.16043+mRêëúûO 2 ,s(4.7.51)FIGURE 4.7.8 Combusion of a carbon particle in high-temperature air. The surface reaction is 2C + O2 ® 2CO.The carbon particle temperature depends on its radius, and it is required to evaluate the property productrD at an appropriate reference temperature: an energy balance on the particle should be performed bythis purpose.
The resulting particle lifetime t ist=rsolid DO2(1.28 rD O2 ,m)sec(4.7.52)rfor an initial particle diameter of D0. Air properties at an average mean film temperature can be used toevaluate rD O2 ,m .Consider a 10-mm-diameter carbon particle ignited in an airstream at 1500 K and 1 atm. An energybalance on the particle (including radiation to surroundings at 1500 K) shows that the average temperatureof the particle is approximately 2550 K, and, thus, Tr = (1/2)(1500 + 2550) = 2025 K or r . rair = 0.175kg/m3 and D O2 ,m .
D O2 ,air = 4.89 ´ m2/sec (from Table 4.7.1). Then2(1810) (10 ´ 10 -6 )t== 1.65 ´ 10 -3 sec(1.28) (0.175) (4.89 ´ 10 -4 )Droplet EvaporationConsider a small droplet of species 1 entrained in a gas stream, species 2 (Figure 4.7.9). This is asimultaneous heat and mass transfer problem, and the mass transfer rate can be obtained by solvingsimultaneouslym˙ ¢¢ =c p1 (Te - Ts ) öm - m1,s ö k c p1 ærD12 æ2lnç1 + 1,elnç1 +=÷ kg m sec÷Rm1,s - 1 øRh fgèøèm1,s = m1,s (T , P)© 1999 by CRC Press LLC(from vapor - pressure data)(4.7.53a)(4.7.53b)4-225Heat and Mass TransferFIGURE 4.7.9 Evaporation of a droplet.Temperature Ts is the adiabatic vaporization temperature and is essentially the psychrometric wet-bulbtemperature.
Properties can be evaluated at mean film temperature and composition; alternatively, cp1can be set equal to the reference specific heat and all properties evaluated using Hubbard’s 1/3 rule, namely,(m1,r = m1,s + (1 3) m1,e - m1,s)Tr = Ts + (1 3) (Te - Ts )(4.7.54a)(4.7.54b)Droplet CombustionFigure 4.7.10 shows a schematic of a volatile liquid hydrocarbon fuel droplet burning in air at zerogravity. The flame diameter is typically four to six times the droplet diameter. Heat is transferred fromthe flame to the droplet and serves to vaporize the fuel.
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