John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 96
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Although gm,i depends directly on the rate of∗does not: it is determined only by flow configurationmass transfer, gm,iand physical properties.In a boundary layer, the fluid near the wall is slowed by the no-slipcondition. One way of modeling high-rate mass transfer effects on gm,iis to approximate the boundary layer as a stagnant film—a stationarylayer of fluid with no horizontal gradients in it, as shown in Fig. 11.18.The film thickness, δc , is an effective local concentration boundary layerthickness.The presence of a finite mass transfer rate across the film means thatvertical convection—counterdiffusion effects—will be present. In fact,§11.8Mass transfer coefficients at high rates of mass transferthe stagnant film shown in Fig. 11.18 is identical to the configurationdealt with in the previous section (i.e., Fig. 11.16).
Thus, the solution obtained in the previous section—eqn. (11.88)—also gives the rate of masstransfer across the stagnant film, taking account of vertical convectivetransport.In the present mass-based analysis, it is convenient to use the massbased analog of the mole-based eqn. (11.88). This analog can be shownto be (Problem 11.33)mi,e − mi,sρDimln 1 +ṁ =δcmi,s − ni,s /ṁwhich we may recast in the following, more suggestive formρDim ln(1 + Bm,i )ṁ =Bm,iδcBm,i(11.102)Comparing this equation with eqn. (11.99), we see thatρDim ln(1 + Bm,i )gm,i =δcBm,iand when ṁ approaches zero,∗gm,i= lim gm,i =ṁ →0Bm,i →0ρDimδclim gm,i =(11.103)Hence,gm,i =∗gm,iln(1 + Bm,i )Bm,i(11.104)∗(or δc ) may be found from the solution ofThe appropriate value gm,icorresponding low-rate mass transfer problem, using the analogy of heat∗, in turn, defines the effective conand mass transfer.
(The value of gm,icentration b.l. thickness, δc .)The group [ln(1 + Bm,i )]/Bm,i is called the blowing factor. It accountsthe effect of mass transfer on the velocity field. When Bm,i > 0, we havemass flow away from the wall (or blowing.) In this case, the blowingfactor is always a positive number less than unity, so blowing reducesgm,i . When Bm,i < 0, we have mass flow toward the wall (or suction), andthe blowing factor is always a positive number greater than unity. Thus,659660An introduction to mass transfer§11.8gm,i is increased by suction.
These trends may be better understood ifwe note that wall suction removes the slow fluid at the wall and thins theboundary layer. The thinner b.l. offers less resistance to mass transfer.Likewise, blowing tends to thicken the b.l., increasing the resistance tomass transfer.The stagnant film b.l. model ignores details of the flow in the b.l.and focuses on the balance of mass fluxes across it. It is equally validfor both laminar and turbulent flows. Analogous stagnant film analysesof heat and momentum transport may also be made, as discussed inProblem 11.37.Example 11.16Calculate the mass transfer coefficient for Example 11.15 if the airspeed is 5 m/s, the length of the pan in the flow direction is 20 cm,and the air temperature is 25◦ C. Assume that the air flow does notgenerate waves on the water surface.Solution.
The water surface is essentially a flat plate, as shown inFig. 11.19. To find the appropriate equation for the Nusselt number,we must first compute ReL .The properties are evaluated at the film temperature, Tf = (75 +25)/2 = 50◦ C, and the film composition,mf ,H2 O = (0.050 + 0.277)/2 = 0.164For these conditions, we find the mixture molecular weight from eqn.(11.8) as Mf = 26.34 kg/kmol. Thus, from the ideal gas law,ρf = (101, 325)(26.34) (8314.5)(323.15) = 0.993 kg/m3From Appendix A, we get µair = 1.949×10−5 kg/m·s and µwater vapor =1.062 × 10−5 kg/m·s. Then eqn. (11.54), with xH2 O,f = 0.240 andxair,f = 0.760, yieldsµf = 1.75 × 10−5 kg/m·sso νf = (µ/ρ)f = 1.76 × 10−5 m2 /sWe compute ReL = 5(0.2)/(1.76 × 10−5 ) = 56, 800, so the flow islaminar.The appropriate Nusselt number is obtained from the mass transfer version of eqn. (6.68):1/2Num,L = 0.664 ReL Sc1/3§11.8Mass transfer coefficients at high rates of mass transferFigure 11.19 Evaporation from a tray of water.Equation (11.35) yields DH2 O,air = 2.96 × 10−5 m2 /s, soSc = 1.76/2.96 = 0.595andNum,L = 133Hence,∗gm,H= Num,L (ρDH2 O,air /L) = 0.0195 kg/m2 ·s2OFinally,!"∗gm,H2 O = gm,Hln(1 + Bm,H2 O ) Bm,H2 O2O= 0.0195 ln(1.314)/0.314 = 0.0170 kg/m2 ·sIn this case, the blowing factor is 0.870.
Thus, mild blowing hasreduced the mass transfer coefficient by about 13%.Conditions for low-rate mass transfer. When the mass transfer drivingforce is small enough, the low-rate mass transfer coefficient itself is anadequate approximation to the actual mass transfer coefficient. This isbecause the blowing factor tends toward unity as Bm,i → 0:limBm,i →0ln(1 + Bm,i )=1Bm,i∗.Thus, for small values of Bm,i , gm,i gm,i661662An introduction to mass transfer§11.8The calculation of mass transfer proceeds in one of two ways forlow rates of mass transfer, depending upon how the limit of small ṁis reached. The first situation is when the ratio ni,s /ṁ is fixed at anonzero value while ṁ → 0.
This would be the case when only onespecies is transferred, since ni,s /ṁ = 1. Then the mass flux at lowrates is∗Bm,iṁ gm,i(11.105)In this case, convective and diffusive contributions to ni,s are of the sameorder of magnitude, in general. To reach conditions for which the analogyof heat and mass transfer applies, it is also necessary that mi,s 1, sothat convective effects will be negligible, as discussed in Section 11.6.When that condition also applies, and if only one species is transferred,we have∗Bm,iṁ = ni,s gm,imi,e − mi,s∗= gm,imi,s − 1∗(mi,s − mi,e ) gm,iIn the second situation, ni,s remains finite while ṁ → 0.
Then,from eqn. (11.93),∗ni,s ji,s gm,i(mi,s − mi,e )(11.106)The transport in this case is purely diffusive, irrespective of the size ofmi,s . This situation arises is catalysis, where two species flow to a walland react, creating a third species that flows away from the wall. Sincethe reaction conserves mass, the net mass flow through the s-surface iszero, even though ni,s is not (see Problem 11.44).An estimate of the blowing factor can be used to determine whetherBm,i is small enough to justify using the simpler low-rate theory. If, forexample, Bm,i = 0.20, then [ln(1+Bm,i )]/Bm,i = 0.91 and an error of only9 percent is introduced by assuming low rates.
This level of accuracy isadequate for many engineering calculations.§11.911.9Simultaneous heat and mass transferSimultaneous heat and mass transferMany important engineering mass transfer processes occur simultaneously with heat transfer. Cooling towers, dryers, and combustors arejust a few examples of equipment that intimately couple heat and masstransfer.Coupling can arise when temperature-dependent mass transfer processes cause heat to be released or absorbed at a surface.
For example,during evaporation, latent heat is absorbed at a liquid surface when vaporis created. This tends to cool the surface, lowering the vapor pressure andreducing the evaporation rate. Similarly, in the carbon oxidation problem discussed in Example 11.2, heat is released when carbon is oxidized,and the rate of oxidation is a function of temperature. The balance between convective cooling and the rate of reaction determines the surfacetemperature of the burning carbon.Simultaneous heat and mass transfer processes may be classified aslow-rate or high-rate. At low rates of mass transfer, mass transfer hasonly a negligible influence on the velocity field, and heat transfer ratesmay be calculated as if mass transfer were not occurring. At high ratesof mass transfer, the heat transfer coefficient must be corrected for theeffect of counterdiffusion. In this section, we consider these two possibilities in turn.Heat transfer at low rates of mass transferOne very common case of low-rate heat and mass transfer is the evaporation of water into air at low or moderate temperatures.
An archetypicalexample of such a process is provided by a sling psychrometer, which isa device used to measure the humidity of air.In a sling psychrometer, a wet cloth is wrapped about the bulb of athermometer, as shown in Fig. 11.20. This so-called wet-bulb thermometer is mounted, along with a second dry-bulb thermometer, on a swivelhandle, and the pair are “slung” in a rotary motion until they reach steadystate.The wet-bulb thermometer is cooled, as the latent heat of the vaporized water is given up, until it reaches the temperature at which the rateof cooling by evaporation just balances the rate of convective heatingby the warmer air. This temperature, which is called the wet-bulb tem-663664An introduction to mass transfer§11.9Figure 11.20 The wet bulb of a sling psychrometer.perature, is directly related to the amount of water in the surroundingair.12The highest ambient air temperatures we normally encounter are fairlylow, so the rate of mass transfer should be small.
We can test this suggestion by computing an upper bound on Bm,H2 O , under conditions thatshould maximize the evaporation rate: using the highest likely air temperature and the lowest humidity. Let us set those values, say, at 120◦ F(49◦ C) and zero humidity (mH2 O,e = 0).We know that the vapor pressure on the wet bulb will be less than thesaturation pressure at 120◦ F, since evaporation will keep the bulb at alower temperature:xH2 O,s psat (120◦ F)/patm = (11, 671 Pa)/(101, 325 Pa) = 0.11512The wet-bulb temperature for air–water systems is very nearly the adiabatic saturation temperature of the air–water mixture — the temperature reached by a mixtureif it is brought to saturation with water by adding water vapor without adding heat. Itis a thermodynamic property of an air–water combination.Simultaneous heat and mass transfer§11.9so, with eqn.
(11.67),mH2 O,s 0.0750Thus, our criterion for low-rate mass transfer, eqn. (11.74), is met:mH2 O,s − mH2 O,eBm,H2 O = 0.08111 − mH2 O,sAlternatively, in terms of the blowing factor, eqn. (11.104),ln(1 + Bm,H2 O ) 0.962Bm,H2 OThis means that under the worst normal circumstances, the low-rate theory should deviate by only 4 percent from the actual rate of evaporation.We may form an energy balance on the wick by considering the u, s,and e surfaces shown in Fig.
11.20. At the steady temperature, no heat isconducted past the u-surface (into the wet bulb), but liquid water flowsthrough it to the surface of the wick where it evaporates. An energybalance on the region between the u and s surfaces givesnH O,s ĥH O,s − 2 2 enthalpy of watervapor leaving= nH2 O,u ĥH2 O,uheat convectedto the wet bulbenthalpy of liquidwater arrivingqsSince mass is conserved, nH2 O,s = nH2 O,u , and because the enthalpychange results from vaporization, ĥH2 O,s − ĥH2 O,u = hfg . Hence,nH2 O,s hfgTwet-bulb = h(Te − Twet-bulb )For low-rate mass transfer, nH2 O,s jH2 O,s , and this equation can bewritten in terms of the mass transfer coefficient (11.107)gm,H2 O mH2 O,s − mH2 O,e hfgTwet-bulb = h(Te − Twet-bulb )The heat and mass transfer coefficients depend on the geometry andflow rates of the psychrometer, so it would appear that Twet-bulb shoulddepend on the device used to measure it.
The two coefficients are not independent, however, owing to the analogy between heat and mass transfer. For forced convection in cross flow, we saw in Chapter 7 that theheat transfer coefficient had the general formhD= C Rea Prbk665666An introduction to mass transfer§11.9where C is a constant, and typical values of a and b are a 1/2 andb 1/3. From the analogy,gm D= C Rea ScbρD12Dividing the second expression into the first, we find bh D12Pr=gm cp αScBoth α/D12 and Sc/Pr are equal to the Lewis number, Le. Hence,h= Le1−b Le2/3gm cp(11.108)Equation (11.108) shows that the ratio of h to gm depends primarily onthe physical properties of the gas mixture, Le and cp , rather than thegeometry or flow rate.
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