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 77F, 1 atm, and 60% rh flows at 32.8 ft/s (1970 ft/min),as shown in Figure 7. Find the rate of evaporation, rate of heat transferto the water, and water surface temperature.Solution: Heat transfer to water from a
18、ir supplies the energy requiredto evaporate the water.q = hA(t ts) = hfg= hMA(s )hfgwhereh = convective heat transfer coefficienthM= convective mass transfer coefficientA = 0.328 4.92 2 = 3.23 ft2= surface area (both sides)= evaporation ratets, s= temperature and vapor density at water surfacet, = t
19、emperature and vapor 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
20、Sherwoodnumber Sh, obtained 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
21、equation must be solved 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 somew
22、hat with ts, iteration is required. Start byguessing ts= 57.2F (the dew point of the airstream), giving tf= 67.1F.At these temperatures, values on the right side are found in propertytables or calculated ask = 0.01485 Btu/hftFPr = 0.709Dv= 0.9769 ft2/h = 0.01628 ft2/min from Equation (10) = 0.07435
23、lbm/ft3 = 0.04376 lbm/fthSc = /Dv= 0.6025hfg= 1055.5 Btu/lbm(at 57.2F)= 8.846 104lbm/ft3(from psychrometric chart at 77F, 60% rh)ts= 57.2F (initial guess)Solving yields s= 1.184 103lbm/ft3. The corresponding value ofts= 70.9F. Repeat the process using ts= 70.9F as the initial guess.The result is s=
24、0.977 103lbm/ft3and ts= 64.9F. Continue itera-tions until sconverges to 1.038 103lbm/ft3and ts= 66.9F.To solve for the rates of evaporation and heat transfer, first calculatethe Reynolds number using air properties at tf= (77 + 66.9)/2 = 72.0F.ReL= = 65,871where L = 0.328 ft, the length of the plate
25、 in the direction of flow.Because ReL 500,000, flow is laminar over the entire length of theplate; therefore,Sh = 0.664Re1/2Sc1/3= 144hM= = 7.15 fpm= hMA(s ) = 0.00354 lb/minq = hfg= 3.74 Btu/minThe same value for q would be obtained by calculating the Nusseltnumber and heat transfer coefficient h a
26、nd setting q = hA(t ts).The kind of similarity between heat and mass transfer 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
27、 = 2/3, St = Nu/(Re Pr) is the Stanton number, and jHis theChilton-Colburn 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.Equat
28、ions (35) and (36) are called the Chilton-Colburn j-factoranalogy.The power 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 t
29、o near perfection, is actuallyf /2 from Blasius solution of laminar flow on 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 transfercoe
30、fficient for a flat plate.A wetted-wall column is a vertical tube in which 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 R
31、e. The point spread with variation in /Dvresultsfrom Gillilands finding of 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
32、kL-=hhM-Nu kSh Dv-PrSc-1/3kDv-=PrSc-1/3kDv-ttshfg-uL-0.07435118 200,0.3280.04376-=ShDvL-mmNuRe Pr1n-St Prnf2-=ShRe Sc1n- S tmScnf2-=Dv-0.566.8 2017 ASHRAE HandbookFundamentals Similarly, McAdams (1954) equation for heat transfer in pipescan be expressed asjH= 0.023 Re0.20(38)This is represented by t
33、he dash-dot curve in Figure 9, which fallsbelow the mass transfer data. The curve 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 das
34、h-dot line in Figure 10 represents thedata, it is actually taken from McAdams (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 fric
35、tion.Because the analogies are based on skin friction, normal pressuredrag must 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 su
36、bject to considerable error.In Figure 11, data on evaporation of water into air 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
37、or momentum transfer because total drag has not beenallocated to skin friction 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
38、air is below the airs dew-point temperature, vapor con-densation occurs. Typically, 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 dehu
39、midifier and can be ana-lyzed using the same procedure; however, this is not generally 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
40、 with the Colburn analogy to obtain the mass transfer coef-ficients, but this 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 condi
41、tions with sensible heattransfer. Mass transfer j-factors and friction factors 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 Tra
42、nsfer 6.9behavior. Dry-surface j-factors fall below those obtained underdehumidifying 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
43、becomes thin and more of thetotal flow field is exposed to the droplets. Roughness 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
44、 surface is mainly responsible forbreakdown of the j-factor analogy. The j-factor 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 (19
45、74) related dry sensible j- and f-factors to those for wet-ted dehumidifying surfaces.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.
46、The heat and mass transfer, however, still relate in auseful way by equating jHand 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 mass
47、transfer coefficients could be closely correlated by the following rela-tion:= (0.294 Btu/lbmF)For completely dry air at 70F flowing at a velocity of 31 fps over awet-bulb thermometer of diameter d = 0.300 in., determine the heat andmass transfer coefficients from Figure 10 and compare their ratio w
48、iththe Bedingfield-Drew relation.Solution: For dry air at 70F and standard pressure, = 0.075 lbm/ft3, = 0.044 lbm/hft, k = 0.0149 Btu/hftF, and cp= 0.240 Btu/lbmF.From Equation (10), Dv= 0.973 ft2/h. Therefore,Reda= ud/ = 0.0749 31 3600 0.300/(12 0.044) = 4750Pr = cp/k = 0.240 0.044/0.0149 = 0.709Sc = /Dv= 0.044/(0.0
