1、6.1CHAPTER 6MASS TRANSFERMolecular Diffusion 6.1Convection of Mass. 6.5Simultaneous Heat and Mass Transfer Between Water-Wetted Surfaces and Air 6.10Symbols . 6.14ASS transfer by either molecular diffusion or convection isM the transport of one component of a mixture relative to themotion of the mix
2、ture and is the result of a concentration gradient.Mass transfer can occur in liquids and solids as well as gases. Forexample, water on the wetted slats of a cooling tower evaporates intoair in a cooling tower (liquid-to-gas mass transfer), and water vaporfrom a food product transfers to the dry air
3、 as it dries. A piece ofsolid CO2 (dry ice) also gets smaller and smaller over time as theCO2molecules diffuse into air (solid-to-gas mass transfer). A pieceof sugar added to a cup of coffee eventually dissolves and diffusesinto the solution, sweetening the coffee, although the sugar mole-cules are
4、much heavier than the water molecules (solid-to-liquidmass transfer). Air freshener does not just smell where sprayed, butrather the smell spreads throughout the room. The air freshener(matter) moves from an area of high concentration where sprayed toan area of low concentration far away. In an abso
5、rption chiller, low-pressure, low-temperature refrigerant vapor from the evaporatorenters the thermal compressor in the absorber section, where therefrigerant vapor is absorbed by the strong absorbent (concentratedsolution) and dilutes the solution.In air conditioning, water vapor is added or remove
6、d from the airby simultaneous transfer of heat and mass (water vapor) between theairstream and a wetted surface. The wetted surface can be water drop-lets in an air washer, condensate on the surface of a dehumidifyingcoil, a spray of liquid absorbent, or wetted surfaces of an evaporativecondenser. E
7、quipment performance with these phenomena must becalculated carefully because of simultaneous heat and mass transfer.This chapter addresses mass transfer principles and providesmethods of solving a simultaneous heat and mass transfer probleminvolving air and water vapor, emphasizing air-conditioning
8、 pro-cesses. The formulations presented can help analyze performanceof specific equipment. For discussion of performance of coolingcoils, evaporative condensers, cooling towers, and air washers, seeChapters 23, 39, 40, and 41, respectively, of the 2016 ASHRAEHandbookHVAC Systems and Equipment.1. MOL
9、ECULAR DIFFUSIONMost mass transfer problems can be analyzed by considering dif-fusion of a gas into a second gas, a liquid, or a solid. In this chapter,the diffusing or dilute component is designated as component B, andthe other component as component A. For example, when watervapor diffuses into ai
10、r, the water vapor is component B and dry air iscomponent A. Properties with subscripts A or B are local propertiesof that component. Properties without subscripts are local propertiesof the mixture.The primary mechanism of mass diffusion at ordinary tempera-ture and pressure conditions is molecular
11、 diffusion, a result ofdensity gradient. In a binary gas mixture, the presence of a concen-tration gradient causes transport of matter by molecular diffusion;that is, because of random molecular motion, gas B diffuses throughthe mixture of gases A and B in a direction that reduces the concen-tration
12、 gradient.Ficks LawThe basic equation for molecular diffusion is Ficks law. Express-ing the concentration of component B of a binary mixture of com-ponents A and B in terms of the mass fraction B/ or mole fractionCB/C, Ficks law isJB= Dv = JA(1a)(1b)where = A+ Band C = CA+ CB.The minus sign indicate
13、s that the concentration gradient is nega-tive in the direction of diffusion. The proportionality factor Dvis themass diffusivity or the diffusion coefficient. The total massflux and molar flux are due to the average velocity of themixture plus the diffusive flux:(2a)(2b)where v is the mixtures mass
14、 average velocity and v*is the molaraverage velocity.Bird et al. (1960) present an analysis of Equations (1a) and (1b).Equations (1a) and (1b) are equivalent forms of Ficks law. Theequation used depends on the problem and individual preference.This chapter emphasizes mass analysis rather than molar
15、analysis.However, all results can be converted to the molar form using therelation CB B/MB.Ficks Law for Dilute MixturesIn many mass diffusion problems, component B is dilute, with adensity much smaller than the mixtures. In this case, Equation (1a)can be written asJB= Dv(3)when B0.01 or 0.01, espec
16、ially if the Reynolds number islarge.mBBiB-mBmBhMBiB=hMFig. 5 Nomenclature for Convective Mass Transfer from External Surface at Location xWhere Surface Is Impermeable to Gas AhM1A- hmAdAFig. 6 Nomenclature for Convective Mass Transfer from Internal Surface Impermeable to Gas AmBBiBb-mB1 uBAcsAcsuBB
17、dAcsuB1 Acs AuB dAcsBbmBomBAdA+uBAcs-=mBouBuBuNuShviuMass Transfer 6.7Example 5. Air at 25C, 100 kPa, and 60% rh flows at 10 m/s, as shown inFigure 7. Find the rate of evaporation, rate of heat transfer to the water,and water surface temperature.Solution: Heat transfer to water from air supplies the
18、 energy requiredto evaporate the water.q = hA(t ts) = hfg= hMA(s )hfgwhereh = convective heat transfer coefficienthM= convective mass transfer coefficientA =0.1 1.5 2 = 0.3 m2= surface area (both sides)= evaporation ratets, s= temperature and vapor density at water surfacet, = temperature and vapor
19、density of airstreamThis energy balance can be rearranged to gives = The heat transfer coefficient h is found by first calculating the Nusseltnumber:Nu = 0.664Re1/2Pr1/3for laminar flowNu = 0.037Re4/5Pr1/3for turbulent flowThe mass transfer coefficient hMrequires calculation of Sherwoodnumber Sh, ob
20、tained using the analogy expressed in Equations (33)and (34):Sh = 0.664Re1/2Pr1/3for laminar flowSh = 0.037Re4/5Pr1/3for turbulent flowWith Nu and Sh known,orThis result is valid for both laminar and turbulent flow. Using this resultin the preceding energy balance givess = This equation must be solv
21、ed for s. Then, water surface tempera-ture tsis the saturation temperature corresponding to s. Air propertiesSc, Pr, Dv, and k are evaluated at film temperature tf= (t+ ts)/2, andhfgis evaluated at ts. Because tsappears in the right side and all the airproperties also vary somewhat with ts, iteratio
22、n is required. Start byguessing ts= 14C (the dew point of the airstream), giving tf= 19.5C.At these temperatures, values on the right side are found in propertytables or calculated ask = 0.0257 W/(mK)Pr = 0.709Dv= 2.521 105m2/s from Equation (10) = 1.191 kg/m3 =1.809 105kg/(ms)Sc = /Dv= 0.6025hfg=2.
23、455 106J/kg (at 14C)= 0.01417 kg/m3(from psychrometric chart at 25C, 60% rh)ts= 14C (initial guess)Solving yields s= 0.01896 kg/m3. The corresponding value of ts=21.6C. Repeat the process using ts= 21.6C as the initial guess. Theresult is s= 0.01565 kg/m3and ts= 18.3C. Continue iterations untilsconv
24、erges to 0.01663 kg/m3and ts= 19.4C.To solve for the rates of evaporation and heat transfer, first calculatethe Reynolds number using air properties at tf= (25 + 19.4)/2 = 22.2C.ReL= = 65 837where L = 0.1 m, the length of the plate in the direction of flow.Because ReL 500 000, flow is laminar over t
25、he entire length of theplate; therefore,Sh = 0.664Re1/2Sc1/3= 144hM= = 0.0363 m/s= hMA(s ) = 2.679 105kg/sq = hfg= 65.8 WThe same value for q would be obtained by calculating the Nusseltnumber and heat transfer coefficient h and setting q = hA(t ts).The kind of similarity between heat and mass trans
26、fer that resultsin Equations (31) to (34) can also be shown to exist between heatand momentum transfer. Chilton and Colburn (1934) used this sim-ilarity to relate Nusselt number to friction factor by the analogyjH= (35)where n = 2/3, St = Nu/(Re Pr) is the Stanton number, and jHis theChilton-Colburn
27、 j-factor for heat transfer. Substituting Sh for Nuand Sc for Pr in Equations (33) and (34) gives the Chilton-Colburnj-factor for mass transfer, jD:jD= (36)where Stm= ShPAm/(Re Sc) is the Stanton number for mass transfer.Equations (35) and (36) are called the Chilton-Colburn j-factoranalogy.The powe
28、r of the Chilton-Colburn j-factor analogy is repre-sented in Figures 8 to 11. Figure 8 plots various experimental valuesof jDfrom a flat plate with flow parallel to the plate surface. Thesolid line, which represents the data to near perfection, is actuallyf /2 from Blasius solution of laminar flow o
29、n a flat plate (left-handportion of the solid line) and Goldsteins solution for a turbulentboundary layer (right-hand portion). The right-hand part also rep-resents McAdams (1954) correlation of turbulent flow heat transfercoefficient for a flat plate.A wetted-wall column is a vertical tube in which
30、 a thin liquidfilm adheres to the tube surface and exchanges mass by evaporationor absorption with a gas flowing through the tube. Figure 9 illus-trates typical data on vaporization in wetted-wall columns, plottedas jDversus Re. The point spread with variation in /Dvresultsfrom Gillilands finding of
31、 an exponent of 0.56, not 2/3, represent-ing the effect of the Schmidt number. Gillilands equation can bewritten as follows:jD= 0.023 Re0.17(37)Fig. 7 Water-Saturated Flat Plate in Flowing AirstreammmhhM-ttshfg-hMSh DvL-= hNu kL-=hhM-Nu kSh Dv-PrSc-1/3kDv-=PrSc-1/3kDv-ttshfg-uL-1.191100.11.809 105-=
32、ShDvL-mmNuRe Pr1n-St Prnf2-=ShRe Sc1n- S tmScnf2-=Dv-0.566.8 2017 ASHRAE HandbookFundamentals (SI)Similarly, McAdams (1954) equation for heat transfer in pipescan be expressed asjH= 0.023 Re0.20(38)This is represented by the dash-dot curve in Figure 9, which fallsbelow the mass transfer data. The cu
33、rve f /2, representing friction insmooth tubes, is the upper, solid curve.Data for liquid evaporation from single cylinders into gasstreams flowing transversely to the cylinders axes are shown inFigure 10. Although the dash-dot line in Figure 10 represents thedata, it is actually taken from McAdams
34、(1954) as representative ofa large collection of data on heat transfer to single cylinders placedtransverse to airstreams. To compare these data with friction, it isnecessary to distinguish between total drag and skin friction.Because the analogies are based on skin friction, normal pressuredrag mus
35、t be subtracted from the measured total drag. At Re = 1000,skin friction is 12.6% of the total drag; at Re = 31 600, it is only1.9%. Consequently, values of f /2 at a high Reynolds number,obtained by the difference, are subject to considerable error.In Figure 11, data on evaporation of water into ai
36、r for singlespheres are presented. The solid line, which best represents thesedata, agrees with the dashed line representing McAdams correla-tion for heat transfer to spheres. These results cannot be comparedwith friction or momentum transfer because total drag has not beenallocated to skin friction
37、 and normal pressure drag. Application ofthese data to air/water-contacting devices such as air washers andspray cooling towers is well substantiated.When the temperature of the heat exchanger surface in contactwith moist air is below the airs dew-point temperature, vapor con-densation occurs. Typic
38、ally, air dry-bulb temperature and humidityratio both decrease as air flows through the exchanger. Therefore,sensible and latent heat transfer occur simultaneously. This processis similar to one that occurs in a spray dehumidifier and can be ana-lyzed using the same procedure; however, this is not g
39、enerally done.Cooling coil analysis and design are complicated by the problemof determining transport coefficients h, hM, and f. It would be con-venient if heat transfer and friction data for dry heating coils couldbe used with the Colburn analogy to obtain the mass transfer coef-ficients, but this
40、approach is not always reliable, and Guillory andMcQuiston (1973) and Helmer (1974) show that the analogy is notconsistently true. Figure 12 shows j-factors for a simple parallel-plate exchanger for different surface conditions with sensible heattransfer. Mass transfer j-factors and friction factors
41、 exhibit the sameFig. 8 Mass Transfer from Flat PlateFig. 9 Vaporization and Absorption in Wetted-Wall Columncpk- 0.7Fig. 10 Mass Transfer from Single Cylinders in CrossflowFig. 11 Mass Transfer from Single SpheresMass Transfer 6.9behavior. Dry-surface j-factors fall below those obtained underdehumi
42、difying conditions with the surface wet. At low Reynoldsnumbers, the boundary layer grows quickly; the droplets are sooncovered and have little effect on the flow field. As the Reynoldsnumber increases, the boundary layer becomes thin and more of thetotal flow field is exposed to the droplets. Rough
43、ness caused by thedroplets induces mixing and larger j-factors.The data in Figure 12 cannot be applied to all surfaces, becausethe length of the flow channel is also an important variable. How-ever, water collecting on the surface is mainly responsible forbreakdown of the j-factor analogy. The j-fac
44、tor analogy is approx-imately true when surface conditions are identical. Under someconditions, it is possible to obtain a film of condensate on the sur-face instead of droplets. Guillory and McQuiston (1973) andHelmer (1974) related dry sensible j- and f-factors to those for wet-ted dehumidifying s
45、urfaces.The equality of jH, jD, and f /2 for certain streamlined shapes atlow mass transfer rates has experimental verification. For flow pastbluff objects, jHand jDare much smaller than f /2, based on totalpressure drag. The heat and mass transfer, however, still relate in auseful way by equating j
46、Hand jD.Example 6. Using solid cylinders of volatile solids (e.g., naphthalene, cam-phor, dichlorobenzene) with airflow normal to these cylinders, Beding-field and Drew (1950) found that the ratio between the heat and masstransfer coefficients could be closely correlated by the following rela-tion:=
47、 1230 J/(kgK)For completely dry air at 21C flowing at a velocity of 9.5 m/s over awet-bulb thermometer of diameter d = 7.5 mm, determine the heat andmass transfer coefficients from Figure 10 and compare their ratio withthe Bedingfield-Drew relation.Solution: For dry air at 21C and standard pressure,
48、 = 1.198 kg/m3, = 1.82 105kg/(sm), k = 0.02581 W/(mK), and cp= 1.006 kJ/(kgK). From Equation (10), Dv= 25.13 mm2/s. Therefore,Reda= ud/ = 1.198 9.5 7.5/(1000 1.82 105) = 4690Pr = cp/k = 1.006 1.82 105 1000/0.02581 = 0.709Sc = /Dv= 1.82 105 106/(1.198 25.13) = 0.605From Figure 10 at Reda= 4700, read jH= 0.0089, and jD= 0.010. FromEqu
