§11.6 Mass transfer at low rates 639 Figure 11.13 Mass diffusion into a semi-infinite stationary medium. with m i = m i,0 for t = 0 (all x) m i = m i,u for x = 0 (t > 0) m i → m i,0 for x→∞ (t > 0) This math problem is identical to that for transient heat conduction into a semi-infinite region (Section 5.6), and its solution is completely analogous to eqn. (5.50): m i −m i,u m i,0 −m i,u = erf   x 2  D im 1 t   The reader can solve all sorts of unsteady mass diffusion problems by direct analogy to the methods of Chapters 4 and 5 when the concen- tration of the diffusing species is low. At higher concentrations of the diffusing species, however, counterdiffusion velocities can be induced, as in Example 11.8. Counterdiffusion may be significant in concentrated metallic alloys, as, for example, during annealing of a butt-welded junc- tion between two dissimilar metals. In those situations, eqn. (11.72)is sometimes modified to use a concentration-dependent, spatially varying interdiffusion coefficient (see [11.6]). 640 An introduction to mass transfer §11.6 Figure 11.14 Concentration boundary layer on a flat plate. Convective mass transfer at low rates Convective mass transfer is analogous to convective heat transfer when two conditions apply: 1. The mass flux normal to the surface, n i,s , must be essentially equal to the diffusional mass flux, j i,s from the surface. In general, this requires that the concentration of the diffusing species, m i , be low. 9 2. The diffusional mass flux must be low enough that it does not affect the imposed velocity field. The first condition ensures that mass flow from the wall is diffusional, as is the heat flow in a convective heat transfer problem. The second condition ensures that the flow field will be the same as for the heat transfer problem. As a concrete example, consider a laminar flat-plate boundary layer in which species i is transferred from the wall to the free stream, as shown in Fig. 11.14. Free stream values, at the edge of the b.l., are labeled with the subscript e, and values at the wall (the s-surface) are labeled with the subscript s. The mass fraction of species i varies from m i,s to m i,e across a concentration boundary layer on the wall. If the mass fraction of species i at the wall, m i,s , is small, then n i,s ≈ j i,s , as we saw earlier in this section. Mass transfer from the wall will be essentially diffusional. This is the first condition. In regard to the second condition, when the concentration difference, m i,s − m i,e , is small, then the diffusional mass flux of species i through the wall, j i,s , will be small compared to the bulk mass flow in the stream- 9 In a few situations, such as catalysis, there is no net mass flow through the wall, and convective transport will be identically zero irrespective of the concentration (see Problems 11.9 and 11.44). §11.6 Mass transfer at low rates 641 wise direction, and it will have little influence on the velocity field. Hence, we would expect that  v is essentially that for the Blasius boundary layer. These two conditions can be combined into a single requirement for low-rate mass transfer, as will be described in Section 11.8. Specifically, low-rate mass transfer can be assumed if B m,i ≡  m i,s −m i,e 1 −m i,s   0.2 condition for low-rate mass transfer (11.74) The quantity B m,i is called the mass transfer driving force. It is writ- ten here in the form that applies when only one species is transferred through the s-surface. The evaporation of water into air is typical exam- ple of single-species transfer: only water vapor crosses the s-surface. The mass transfer coefficient. In convective heat transfer problems, we have found it useful to express the heat flux from a surface, q,as the product of a heat transfer coefficient, h, and a driving force for heat transfer, ∆T . Thus, in the notation of Fig. 11.14, q s = h ( T s −T e ) (1.17) In convective mass transfer problems, we would therefore like to ex- press the diffusional mass flux from a surface, j i,s , as the product of a mass transfer coefficient and the concentration difference between the s-surface and the free stream. Hence, we define the mass transfer coeffi- cient for species i, g m,i (kg/m 2 ·s), as follows: j i,s ≡ g m,i  m i,s −m i,e  (11.75) We expect g m,i , like h, to be determined mainly by the flow field, fluid, and geometry of the problem. The analogy to convective heat transfer. We saw in Sect. 11.5 that the equation of species conservation and the energy equation were quite similar in an incompressible flow. If there are no reactions and no heat generation, then eqns. (11.61) and (6.37) can be written as ∂ρ i ∂t +  v ·∇ρ i =−∇·  j i (11.61) ρc p  ∂T ∂t +  v ·∇T  =−∇·  q (6.37) 642 An introduction to mass transfer §11.6 These conservation equations describe changes in, respectively, the amount of mass or energy per unit volume that results from convection by a given velocity field and from diffusion under either Fick’s or Fourier’s law. We may identify the analogous quantities in these equations. For the capacity of mass or energy per unit volume, we see that dρ i is analogous to ρc p dT (11.76a) or, in terms of the mass fraction, ρdm i is analogous to ρc p dT (11.76b) The flux laws may be rewritten to show the capacities explicitly  j i =−ρD im ∇m i =−D im  ρ∇m i   q =−k∇T =− k ρc p  ρc p ∇T  Hence, we find the analogy of the diffusivities: D im is analogous to k ρc p = α (11.76c) It follows that the Schmidt number and the Prandtl number are directly analogous: Sc = ν D im is analogous to Pr = ν α = µc p k (11.76d) Thus, a high Schmidt number signals a thin concentration boundary layer, just as a high Prandtl number signals a thin thermal boundary layer. Finally, we may write the transfer coefficients in terms of the ca- pacities j i,s = g m,i  m i,s −m i,e  =  g m,i ρ  ρ  m i,s −m i,e  q s = h ( T s −T e ) =  h ρc p  ρc p ( T s −T e ) from which we see that g m,i is analogous to h c p (11.76e) §11.6 Mass transfer at low rates 643 From these comparisons, we conclude that the solution of a heat convec- tion problem becomes the solution of a low-rate mass convection prob- lem upon replacing the variables in the heat transfer problem with the analogous mass transfer variables given by eqns. (11.76). Convective heat transfer coefficients are usually expressed in terms of the Nusselt number as a function of Reynolds and Prandtl number Nu x = hx k = (h/c p )x ρ(k/ρc p ) = fn ( Re x , Pr ) (11.77) For convective mass transfer problems, we expect the same functional de- pendence after we make the substitutions indicated above. Specifically, if we replace h/c p by g m,i , k/ρc p by D i,m , and Pr by Sc, we obtain Nu m,x ≡ g m,i x ρD im = fn ( Re x , Sc ) (11.78) where Nu m,x , the Nusselt number for mass transfer, is defined as indi- cated. Nu m is sometimes called the Sherwood number 10 , Sh. Example 11.10 A napthalene model of a printed circuit board (PCB) is placed in a wind tunnel. The napthalene sublimates slowly as a result of forced convective mass transfer. If the first 5 cm of the napthalene model is a flat plate, calculate the average rate of loss of napthalene from that part of the model. Assume that conditions are isothermal at 303 K and that the air speed is 5 m/s. Also, explain how napthalene sublimation might be used to determine heat transfer coefficients . Solution. Let us first find the mass fraction of napthalene just above the model surface. A relationship for the vapor pressure of napthalene (in mmHg) is log 10 p v = 11.450−3729.3  (T K). At 303 K, this gives p v = 0.1387 mmHg = 18.49 Pa. The mole fraction of napthalene is thus x nap,s = 18.49/101325 = 1.825 × 10 −4 , and with 10 Thomas K. Sherwood (1903–1976) obtained his doctoral degree at M.I.T. under War- ren K. Lewis in 1929 and was a professor of Chemical Engineering there from 1930 to 1969. He served as Dean of Engineering from 1946 to 1952. His research dealt with mass transfer and related industrial processes. Sherwood was also the author of very influential textbooks on mass transfer. 644 An introduction to mass transfer §11.6 eqn. (11.9), the mass fraction is, with M nap = 128.2 kg/kmol, m nap,s = (1.825 ×10 −4 )(128.2) (1.825 ×10 −4 )(128.2) +(1 −1.825 ×10 −4 )(28.96) = 8.074 ×10 −4 The mass fraction of napthalene in the free stream, m nap,s , is zero. With these numbers, we can check to see if the mass transfer rate is low enough to use the analogy of heat and mass transfer, with eqn. (11.74): B m,nap =  8.074 ×10 −4 −0 1 −8.074 ×10 −4  = 8.081 ×10 −4  0.2 The analogy therefore applies. The convective heat transfer coefficient for this situation is that for a flat plate boundary layer. The Reynolds number is Re L = u ∞ L ν = (5)(0.05) 1.867 ×10 −5 = 1.339 ×10 4 where we have used the viscosity of pure air, since the concentration of napthalene is very low. The flow is laminar, so the applicable heat transfer relationship is eqn. (6.68) Nu L = hL k = 0.664 Re 1/2 L Pr 1/3 (6.68) Under the analogy, the Nusselt number for mass transfer is Nu m,L = g m,i L ρD im = 0.664 Re 1/2 L Sc 1/3 The diffusion coefficient for napthalene in air, from Table 11.1,is D nap,air = 0.86×10 −5 m/s, and thus Sc = 1.867×10 −5 /0.86×10 −5 = 2.17. Hence, Nu m,L = 0.664 (1.339 ×10 4 ) 1/2 (2.17) 1/3 = 99.5 and, using the density of pure air, g m,nap = ρD nap,air L Nu m,L = (1.166)(0.86 ×10 −5 ) 0.05 (99.5) = 0.0200 kg/m 2 s §11.6 Mass transfer at low rates 645 The average mass flux from this part of the model is then n nap,s = g m,nap  m nap,s −m nap,e  = (0.0200)(8.074 ×10 −4 −0) = 1.61 ×10 −5 kg/m 2 s = 58.0g/m 2 h Napthalene sublimation can be used to infer heat transfer coeffi- cients by measuring the loss of napthalene from a model over some length of time. Experiments are run at several Reynolds numbers. The lost mass fixes the sublimation rate and the mass transfer coeffi- cient. The mass transfer coefficient is then substituted in the analogy to heat transfer to determine a heat transfer Nusselt number at each Reynolds number. Since the Schmidt number of napthalene is not generally equal to the Prandtl number under the conditions of inter- est, some assumption about the dependence of the Nusselt number on the Prandtl number must usually be introduced. Boundary conditions. When we apply the analogy between heat trans- fer and mass transfer to calculate g m,i , we must consider the boundary condition at the wall. We have dealt with two common types of wall con- dition in the study of heat transfer: uniform temperature and uniform heat flux. The analogous mass transfer wall conditions are uniform con- centration and uniform mass flux. We used the mass transfer analog of the uniform wall temperature solution in the preceding example, since the mass fraction of napthalene was uniform over the entire model. Had the mass flux been uniform at the wall, we would have used the analog of a uniform heat flux solution. Natural convection in mass transfer. In Chapter 8, we saw that the density differences produced by temperature variations can lead to flow and convection in a fluid. Variations in fluid composition can also pro- duce density variations that result in natural convection mass transfer. This type of natural convection flow is still governed by eqn. (8.3), u ∂u ∂x +v ∂u ∂y = (1 −ρ ∞ /ρ)g +ν ∂ 2 u ∂y 2 (8.3) but the species equation is now used in place of the energy equation in determining the variation of density. Rather than solving eqn. (8.3) and 646 An introduction to mass transfer §11.6 the species equation for specific mass transfer problems, we again turn to the analogy between heat and mass transfer. In analyzing natural convection heat transfer, we eliminated ρ from eqn. (8.3) using (1 − ρ ∞ /ρ) = β(T − T ∞ ), and the resulting Grashof and Rayleigh numbers came out in terms of an appropriate β∆T instead of ∆ρ/ρ. These groups could just as well have been written for the heat transfer problem as Gr L = g∆ρL 3 ρν 2 and Ra L = g∆ρL 3 ραν = g∆ρL 3 µα (11.79) although ∆ρ would still have had to have been evaluated from ∆T . With Gr and Pr expressed in terms of density differences instead of temperature differences, the analogy between heat transfer and low-rate mass transfer may be used directly to adapt natural convection heat transfer predictions to natural convection mass transfer. As before, we replace Nu by Nu m and Pr by Sc. But this time we also write Ra L = Gr L Sc = g∆ρL 3 µD 12 (11.80) or calculate Gr L as in eqn. (11.79). The densities must now be calculated from the concentrations. Example 11.11 Helium is bled through a porous vertical wall, 40 cm high, into sur- rounding air at a rate of 87.0mg/m 2 ·s. Both the helium and the air are at 300 K, and the environment is at 1 atm. What is the average concentration of helium at the wall, m He,s ? Solution. This is a uniform flux natural convection problem. Here g m,He and ∆ρ depend on m He,s , so the calculation is not as straight- forward as it was for thermally driven natural convection. To begin, let us assume that the concentration of helium at the wall will be small enough that the mass transfer rate is low. Since m He,e = 0, if m He,s  1, then m He,s −m He,e  1 as well. Both conditions for the analogy to heat transfer will be met. The mass flux of helium at the wall, n He,s , is known, and because low rates prevail, n He,s ≈ j He,s = g m,He  m He,s −m He,e  §11.6 Mass transfer at low rates 647 Hence, Nu m,L = g m,He L ρD He,air = n He,s L ρD He,air  m He,s −m He,e  The appropriate Nusselt number is obtained from the mass trans- fer analog of eqn. (8.44b): Nu m,L = 6 5  Ra ∗ L Sc 4 +9 √ Sc +10 Sc  1/5 with Ra ∗ L = Ra L Nu m,L = g ∆ρn He,s L 4 µρD 2 He,air  m He,s −m He,e  The Rayleigh number cannot easily be evaluated without assuming a value of the mass fraction of helium at the wall. As a first guess, we pick m He,s = 0.010. Then the film composition is m He,f = (0.010 +0)/2 = 0.005 From eqn. (11.8) and the ideal gas law, we obtain estimates for the film density (at the film composition) and the wall density ρ f = 1.141 kg/m 3 and ρ s = 1.107 kg/m 3 From eqn. (11.42) the diffusion coefficient is D He,air = 7.119 ×10 −5 m 2 /s. At this low concentration of helium, we expect the film viscosity to be close to that of pure air. From Appendix A, for air at 300 K µ f  µ air = 1.857 ×10 −5 kg/m·s. The corresponding Schmidt number is Sc = (µ f /ρ f )  D He,air = 0.2286. Furthermore, ρ e = ρ air = 1.177 kg/m 3 From these values, Ra ∗ L = 9.806(1.177 −1.107)(87.0 ×10 −6 )(0.40) 4 (1.857 ×10 −5 )(1.141)(7.119 ×10 −5 ) 2 (0.010) = 1.424 ×10 9 648 An introduction to mass transfer §11.7 We may now evaluate the mass transfer Nusselt number Nu m,L = 6  (1.424 ×10 9 )(0.2286)  1/5 5  4 +9 √ 0.2286 +10(0.2286)  1/5 = 37.73 From this we calculate  m He,s −m He,e  = n He,s L ρD He,air Nu m,L = (87.0 ×10 −6 )(0.40) (1.141)(7.119 ×10 −5 )(37.73) = 0.01136 We have already noted that  m He,s −m He,e  = m He,s , so we have ob- tained an average wall concentration 14% higher than our initial guess of 0.010. Using m He,s = 0.01136 as our second guess, we repeat the pre- ceding calculations with revised values of the densities to obtain m He,s = 0.01142 Since this result is within 0.5% of our second guess, a third iteration is not needed. In the preceding example, concentration variations alone gave rise to buoyancy. If both temperature and density vary in a natural convec- tion problem, the appropriate Gr or Ra may be calculated using density differences based on both the local m i and the local T, provided that the Prandtl and Schmidt numbers are approximately equal (that is, if the Lewis number  1). This is usually true in gases. If the Lewis number is far from unity, the analogy between heat and mass transfer breaks down in those natural convection problems that in- volve both heat and mass transfer, because the concentration and ther- mal boundary layers may take on very different thicknesses, complicating the density distributions that drive the velocity field. 11.7 Steady mass transfer with counterdiffusion In 1874, Josef Stefan presented his solution for evaporation from a liquid pool at the bottom of a vertical tube over which a gas flows (Fig. 11.15). This configuration, often called a Stefan tube, is has often been used to [...]... mass basis, assuming a constant value of ρD12 (see Problem 11 .33 and Problem 11 .34 ) The mass-based solution of this problem provides an important approximation in our analysis of high-rate convective mass transfer in the next section 11.8 Mass transfer coefficients at high rates of mass transfer In Section 11.6, we developed an analogy between heat and mass transfer that allowed us to calculate mass transfer. .. and mass transfer Many important engineering mass transfer processes occur simultaneously with heat transfer Cooling towers, dryers, and combustors are just a few examples of equipment that intimately couple heat and mass transfer 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... determines the surface temperature of the burning carbon Simultaneous heat and mass transfer processes may be classified as low-rate or high-rate At low rates of mass transfer, mass transfer has only a negligible influence on the velocity field, and heat transfer rates may be calculated as if mass transfer were not occurring At high rates of mass transfer, the heat transfer coefficient must be corrected for the... species transferred (11.101) In all the cases described in Section 11.6, only one species is transferred Example 11.15 A pan of hot water with a surface temperature of 75 ◦ C is placed in an air stream that has a mass fraction of water equal to 0.05 If the average mass transfer coefficient for water over the pan is gm,H2 O = 0.0 170 kg/m2 ·s and the pan has a surface area of 0.04 m2 , what is the evaporation... possibilities in turn Heat transfer at low rates of mass transfer One very common case of low-rate heat and mass transfer is the evaporation of water into air at low or moderate temperatures An archetypical example of such a process is provided by a sling psychrometer, which is a device used to measure the humidity of air In a sling psychrometer, a wet cloth is wrapped about the bulb of a thermometer, as shown... the heat transfer coefficient in Chapters 6 through 8—solution of the momentum and species equations or through correlation of mass transfer data These approaches are often used, but they are more complicated than the analogous heat transfer problems, owing to the coupling of the flow field and the mass transfer rate Simple solutions are not so readily available for mass transfer problems We instead employ... necessarily dilute First, we define the mass transfer driving force, which governs the total mass flux from the wall Then, we relate the mass transfer coefficient at high mass transfer rates to that at low mass transfer rates The mass transfer driving force Figure 11. 17 shows a boundary layer over a wall through which there is ˙ a net mass transfer, ns ≡ m , of the various species in the direction normal... fluid at the wall and thins the boundary layer The thinner b.l offers less resistance to mass transfer Likewise, blowing tends to thicken the b.l., increasing the resistance to mass 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 valid for both laminar and turbulent flows Analogous stagnant film analyses of heat and... the mass transfer coefficient by about 13% Conditions for low-rate mass transfer When the mass transfer driving force is small enough, the low-rate mass transfer coefficient itself is an adequate approximation to the actual mass transfer coefficient This is because the blowing factor tends toward unity as Bm,i → 0: lim Bm,i →0 ln(1 + Bm,i ) =1 Bm,i Thus, for small values of Bm,i , gm,i ∗ gm,i 661 662 An introduction... mass transfer coefficients when the rate of mass transfer was low This analogy required that the velocity field be unaffected by mass transfer and that the transferred species be dilute When those conditions are not met, the mass transfer coefficient will be different than the value given by the analogy The difference can be either an increase or a decrease and can range from a few percent to an order of magnitude . (8 .3) and 646 An introduction to mass transfer §11.6 the species equation for specific mass transfer problems, we again turn to the analogy between heat and mass transfer. In analyzing natural. mass transfer coefficient at high mass transfer rates to that at low mass transfer rates. The mass transfer driving force Figure 11. 17 shows a boundary layer over a wall through which there is a. convective mass transfer in the next section. 11.8 Mass transfer coefficients at high rates of mass transfer In Section 11.6, we developed an analogy between heat and mass transfer that allowed us to calculate