John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 87
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Figure 11.1 shows a huge cooling tower used to cool the waterleaving power plant condensers or other large heat exchangers. It is essentially an empty shell, at the bottom of which are arrays of cementboards or plastic louvres over which is sprayed the hot water to be cooled.The hot water runs down this packing, and a small portion of it evaporates into cool air that enters the tower from below.
The remaining water,having been cooled by the evaporation, falls to the bottom, where it iscollected and recirculated.The temperature of the air rises as it absorbs the warm vapor and, in599An introduction to mass transfer600§11.2the natural-draft form of cooling tower shown, the upper portion of thetower acts as an enormous chimney through which the warm, moist airbuoys, pulling in cool air at the base.
In a mechanical-draft cooling tower,fans are used to pull air through the packing. Mechanical-draft towersare much shorter and can sometimes be seen on the roofs of buildings(Fig. 11.2).The working mass transfer process in a cooling tower is the evaporation of water into air. The rate of evaporation depends on the temperature and humidity of the incoming air, the feed-water temperature, andthe air-flow characteristics of the tower and the packing.
When the airflow is buoyancy-driven, the flow rates are directly coupled. Thus, masstransfer lies at the core of the complex design of a cooling tower.11.2Mixture compositions and species fluxesThe composition of mixturesA mixture of various chemical species displays its own density, molecularweight, and other overall thermodynamic properties. These propertiesdepend on the types and relative amounts of the component substances,which may vary from point to point in the mixture.
To determine thelocal properties of a mixture, we must identify the local proportion ofeach species composing the mixture.One way to describe the amount of a particular species in a mixture isby the mass of that species per unit volume, known as the partial density.The mass of species i in a small volume of mixture, in kg, divided by thatvolume, in m3 , is the partial density, ρi , for that species, in kg of i perm3 . The composition of the mixture may be describe by stating the partialdensity of each of its components. The mass density of the mixture itself,ρ, is the total mass of all species per unit volume; therefore,ρ=$ρi(11.1)iThe relative amount of species i in the mixture may be described bythe mass of i per unit mass of the mixture, which is simply ρi /ρ.
Thisratio is called the mass fraction, mi :mi ≡ρimass of species i=ρmass of mixture(11.2)Mixture compositions and species fluxes§11.2This definition leads to the following two results:$$mi =ρi /ρ = 1and0 mi 1i(11.3)iThe molar concentration of species i in kmol/m3 , ci , expresses concentration in terms of moles rather than mass. If Mi is the molecularweight of species i in kg/kmol, thenci ≡ρimoles of i=Mivolume(11.4)The molar concentration of the mixture, c, is the total number of molesfor all species per unit volume; thus,$ci .(11.5)c=iThe mole fraction of species i, xi , is the number of moles of i per moleof mixture:xi ≡cimoles of i=cmole of mixture(11.6)Just as for the mass fraction, it follows for mole fraction that$$xi =ci /c = 1and0 xi 1i(11.7)iThe molecular weight of the mixture is the number of kg of mixtureper kmol of mixture: M ≡ ρ/c. By using eqns.
(11.1), (11.4), and (11.6)and (11.5), (11.4), and (11.2), respectively, M may be written in terms ofeither mole or mass fraction$ mi$1=xi Mior(11.8)M=MMiiiMole fraction may be converted to mass fraction using the following relations (derived in Problem 11.1):mi =xi Mixi Mi=#Mk xk Mkandxi =Mmimi /Mi=#Mik mk /Mk(11.9)In some circumstances, such as kinetic theory calculations, one worksdirectly with the number of molecules of i per unit volume.
This numberdensity, Ni , is given byNi = NA ci(11.10)where NA is Avogadro’s number, 6.02214 × 1026 molecules/kmol.601602An introduction to mass transfer§11.2Ideal gasesThe relations we have developed so far involve densities and concentrations that vary in as yet unknown ways with temperature or pressure. Toget a more useful, though more restrictive, set of results, we now combine the preceding relations with the ideal gas law. For any individualcomponent, i, we may write the partial pressure, pi , exterted by i as:pi = ρi Ri T(11.11)In eqn.
(11.11), Ri is the ideal gas constant for species i:Ri ≡R◦Mi(11.12)where R ◦ is the universal gas constant, 8314.472 J/kmol· K. Equation (11.11)may alternatively be written in terms of ci : ◦Rpi = ρi Ri T = (Mi ci )TMi(11.13)= ci R ◦ TEquations (11.5) and (11.13) can be used to relate c to p and Tc=$ici =$ pip= ◦◦TRRTi(11.14)Multiplying the last two parts of eqn. (11.14) by R ◦ T yields Dalton’s lawof partial pressures,1$p=pi(11.15)iFinally, we combine eqns.
(11.6), (11.13), and (11.15) to obtain a veryuseful relationship between xi and pi :xi =picipi==◦ccR Tp(11.16)in which the last two equalities are restricted to ideal gases.1Dalton’s law (1801) is an empirical principle (not a deduced result) in classicalthermodynamics. It can be deduced from molecular principles, however. We built theappropriate molecular principles into our development when we assumed eqn. (11.11)to be true.
The reason that eqn. (11.11) is true is that ideal gas molecules occupy amixture without influencing one another.Mixture compositions and species fluxes§11.2Example 11.1The most important mixture that we deal with is air. It has the following composition:SpeciesN2O2Artrace gasesMass Fraction0.75560.23150.01289< 0.01Determine xO2 , pO2 , cO2 , and ρO2 for air at 1 atm.Solution. To make these calcuations, we need the molecular weights,which are given in Table 11.2 on page 616.
We can start by checkingthe value of Mair , using the second of eqns. (11.8):−1mN2mO2mArMair =++MN2MO2MAr−10.23150.012890.7556++=28.02 kg/kmol 32.00 kg/kmol 39.95 kg/kmol= 28.97 kg/kmolWe may calculate the mole fraction using the second of eqns. (11.9)xO2 =mO2 M(0.2315)(28.97 kg/kmol)= 0.2095=MO232.00 kg/kmolThe partial pressure of oxygen in air at 1 atm is [eqn. (11.16)]pO2 = xO2 p = (0.2095)(101, 325 Pa) = 2.123 × 104 PaWe may now obtain cO2 from eqn. (11.13):pO2R◦ T= (2.123 × 104 Pa) (300 K)(8314.5 J/kmol·K)c O2 == 0.008510 kmol/m3Finally, eqn. (11.4) gives the partial densityρO2 = cO2 MO2 = (0.008510 kmol/m3 )(32.00 kg/kmol)= 0.2723 kg/m3603604An introduction to mass transfer§11.2Velocities and fluxesEach species in a mixture undergoing a mass transfer process will havei , which can be different for each species inan species-average velocity, vthe mixture, as suggested by Fig.
11.3. We may obtain the mass-average for the entire mixture from the species average velocitiesvelocity,2 v,using the formula$i .=ρi v(11.17)ρviThis equation is essentially a local calculation of the mixture’s net mo as the mixture’s mass flux, n,mentum per unit volume. We refer to ρ vand we call its scalar magnitude ṁ ; each has units of kg/m2 ·s. Likewise,the mass flux of species i isi i = ρi vn(11.18)and, from eqn. (11.17), we see that the mixture’s mass flux equals thesum of all species’ mass fluxes$ i = ρv=n(11.19)niSince each species diffusing through a mixture has some velocity relative to the mixture’s mass-average velocity, the diffusional mass flux, ji ,of a species relative to the mixture’s mean flow may be identified: .i − vji = ρi v(11.20) i , includes both this diffusionalThe total mass flux of the ith species, nmass flux and bulk convection by the mean flow, as is easily shown: i = ρi vi = ρi v + ρi vi − vn + ji= ρi v= + jmn i i convection(11.21)diffusion2 given by eqn.
(11.17) is identical to the fluid velocity,The mass average velocity, v, used in previous chapters. This is apparent if one applies eqn. (11.17) to a “mixu, here because v is the moreture” composed of only one species. We use the symbol vcommon notation in the mass transfer literature.Mixture compositions and species fluxes§11.2605Figure 11.3 Molecules of differentspecies in a mixture moving withdifferent average velocities. The velocityi is the average over all molecules ofvspecies i.Although the convective transport contribution is fully determined assoon as we know the velocity field and partial densities, the causes ofdiffusion need further discussion, which we defer to Section 11.3.Combining eqns.
(11.19) and (11.21), we find that$$$$$i =+++=ji = ρ vji = njinρi vniiiiiHence$ji = 0(11.22)iDiffusional mass fluxes must sum to zero because they are each definedrelative to the mean mass flux.Velocities may also be stated in molar terms. The mole flux of the isi , is ci vi , in kmol/m2 · s. The mixture’s mole flux, N,ith species, Nobtained by summing over all species$$i ==i = c v∗ci v(11.23)NNii ∗ , as shown. The last fluxwhere we define the mole-average velocity, v∗we define is the diffusional mole flux, Ji : ∗i − vJi∗ = ci v(11.24)An introduction to mass transfer606§11.2It may be shown, using these definitions, thati = xi N + J∗Ni(11.25)Substitution of eqn.
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