A HEAT TRANSFER TEXTBOOK - THIRD EDITION Episode 3 Part 8 pps

664 An introduction to mass transfer §11.9 Figure 11.20 The wet bulb of a sling psychrometer. perature, is directly related to the amount of water in the surrounding air. 12 The highest ambient air temperatures we normally encounter are fairly low, so the rate of mass transfer should be small. We can test this sug- gestion by computing an upper bound on B m,H 2 O , under conditions that should maximize the evaporation rate: using the highest likely air tem- perature and the lowest humidity. Let us set those values, say, at 120 ◦ F (49 ◦ C) and zero humidity (m H 2 O,e = 0). We know that the vapor pressure on the wet bulb will be less than the saturation pressure at 120 ◦ F, since evaporation will keep the bulb at a lower temperature: x H 2 O,s  p sat (120 ◦ F)/p atm = (11, 671 Pa)/(101, 325 Pa) = 0.115 12 The wet-bulb temperature for air–water systems is very nearly the adiabatic satu- ration temperature of the air–water mixture — the temperature reached by a mixture if it is brought to saturation with water by adding water vapor without adding heat. It is a thermodynamic property of an air–water combination. §11.9 Simultaneous heat and mass transfer 665 so, with eqn. (11.67), m H 2 O,s  0.0750 Thus, our criterion for low-rate mass transfer, eqn. (11.74), is met: B m,H 2 O =  m H 2 O,s −m H 2 O,e 1 −m H 2 O,s   0.0811 Alternatively, in terms of the blowing factor, eqn. (11.104), ln(1 +B m,H 2 O ) B m,H 2 O  0.962 This means that under the worst normal circumstances, the low-rate the- ory 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 is conducted past the u-surface (into the wet bulb), but liquid water flows through it to the surface of the wick where it evaporates. An energy balance on the region between the u and s surfaces gives n H 2 O,s ˆ h H 2 O,s    enthalpy of water vapor leaving − q s    heat convected to the wet bulb = n H 2 O,u ˆ h H 2 O,u    enthalpy of liquid water arriving Since mass is conserved, n H 2 O,s = n H 2 O,u , and because the enthalpy change results from vaporization, ˆ h H 2 O,s − ˆ h H 2 O,u = h fg . Hence, n H 2 O,s h fg   T wet-bulb = h(T e −T wet-bulb ) For low-rate mass transfer, n H 2 O,s  j H 2 O,s , and this equation can be written in terms of the mass transfer coefficient g m,H 2 O  m H 2 O,s −m H 2 O,e  h fg   T wet-bulb = h(T e −T wet-bulb ) (11.107) The heat and mass transfer coefficients depend on the geometry and flow rates of the psychrometer, so it would appear that T wet-bulb should depend on the device used to measure it. The two coefficients are not in- dependent, however, owing to the analogy between heat and mass trans- fer. For forced convection in cross flow, we saw in Chapter 7 that the heat transfer coefficient had the general form hD k = C Re a Pr b 666 An introduction to mass transfer §11.9 where C is a constant, and typical values of a and b are a  1/2 and b  1/3. From the analogy, g m D ρD 12 = C Re a Sc b Dividing the second expression into the first, we find h g m c p D 12 α =  Pr Sc  b Both α/D 12 and Sc/Pr are equal to the Lewis number, Le. Hence, h g m c p = Le 1−b  Le 2/3 (11.108) The Lewis number for air–water systems is about 0.847. Eqn. (11.108) shows that the ratio of h to g m depends primarily on the physical prop- erties of the mixture, rather than the geometry or flow rate. This type of relationship between h and g m was first developed by W. K. Lewis in 1922 for the case in which Le = 1[11.27]. (Lewis’s pri- mary interest was in air–water systems, so the approximation was not too bad.) The more general form, eqn. (11.108), is another Reynolds- Colburn type of analogy, similar to eqn. (6.76). It was given by Chilton and Colburn [11.28] in 1934. Equation (11.107) may now be written as T e −T wet-bulb =  h fg   T wet-bulb c p Le 2/3   m H 2 O,s −m H 2 O,e  (11.109) This expression can be solved iteratively with a steam table to obtain the wet-bulb temperature as a function of the dry-bulb temperature, T e , and the humidity of the ambient air, m H 2 O,e . The psychrometric charts found in engineering handbooks and thermodynamics texts can be generated in this way. We ask the reader to make such calculations in Problem 11.49. The wet-bulb temperature is a helpful concept in many phase-change processes. When a small body without internal heat sources evaporates or sublimes, it cools to a steady “wet-bulb” temperature at which con- vective heating is balanced by latent heat removal. The body will stay at that temperature until the phase-change process is complete. Thus, the wet-bulb temperature appears in the evaporation of water droplets, the sublimation of dry ice, the combustion of fuel sprays, and so on. If the body is massive, however, steady state may not be reached very quickly. §11.9 Simultaneous heat and mass transfer 667 Stagnant film model of heat transfer at high mass transfer rates The multicomponent energy equation. Each species in a mixture car- ries its own enthalpy, ˆ h i . In a flow with mass transfer, different species move with different velocities, so that enthalpy transport by individ- ual species must enter the energy equation along with heat conduction through the fluid mixture. For steady, low-speed flow without internal heat generation or chemical reactions, we may rewrite the energy balance, eqn. (6.36), as −  S (−k∇T)· d  S −  S    i ρ i ˆ h i  v i   ·d  S = 0 where the second term accounts for the enthalpy transport by each species in the mixture. The usual procedure of applying Gauss’s theorem and re- quiring the integrand to vanish identically gives ∇·   −k∇T +  i ρ i ˆ h i  v i   = 0 (11.110) This equation shows that the total energy flux—the sum of heat conduc- tion and enthalpy transport—is conserved in steady flow. 13 The stagnant film model. Let us restrict attention to the transport of a single species, i, across a boundary layer. We again use the stagnant film model for the thermal boundary layer and consider the one-dimensional flow of energy through it (see Fig. 11.21). Equation (11.110) simplifies to d dy  −k dT dy +ρ i ˆ h i v i  = 0 (11.111) From eqn. (11.69) for steady, one-dimensional mass conservation n i = constant in y = n i,s 13 The multicomponent energy equation becomes substantially more complex when kinetic energy, body forces, and thermal or pressure diffusion are taken into account. The complexities are such that most published derivations of the multicomponent energy equation are incorrect, as shown by Mills in 1998 [11.29]. The main source of error has been the assignment of an independent kinetic energy to the ordinary diffusion velocity. This leads to such inconsistencies as a mechanical work term in the thermal energy equation. 668 An introduction to mass transfer §11.9 Figure 11.21 Energy transport in a stagnant film. If we neglect pressure variations and assume a constant specific heat capacity (as in Sect. 6.3), the enthalpy may be written as ˆ h i = c p,i (T −T ref ), and eqn. (11.111) becomes d dy  −k dT dy +n i,s c p,i T  = 0 Integrating twice and applying the boundary conditions T(y = 0) = T s and T(y = δ t ) = T e we obtain the temperature profile of the stagnant film: T −T s T e −T s = exp  n i,s c p,i k y  −1 exp  n i,s c p,i k δ t  −1 (11.112) The temperature distribution may be used to find the heat transfer coefficient according to its definition [eqn. (6.5)]: h ≡ −k dT dy      s T s −T e = n i,s c p,i exp  n i,s c p,i k δ t  −1 (11.113) We define the heat transfer coefficient in the limit of zero mass transfer, h ∗ ,as h ∗ ≡ lim n i,s →0 h = k δ t (11.114) §11.9 Simultaneous heat and mass transfer 669 Substitution of eqn. (11.114) into eqn. (11.113) yields h = n i,s c p,i exp(n i,s c p,i /h ∗ ) −1 (11.115) To use this result, one first calculates the heat transfer coefficient as if there were no mass transfer, using the methods of Chapters 6 through 8. The value obtained is h ∗ , which is then placed in eqn. (11.115) to de- termine h in the presence of mass transfer. Note that h ∗ defines the effective film thickness δ t through eqn. (11.114). Equation (11.115) shows the primary effects of mass transfer on h. When n i,s is large and positive—the blowing case—h becomes smaller than h ∗ . Thus, blowing decreases the heat transfer coefficient, just as it decreases the mass transfer coefficient. Likewise, when n i,s is large and negative—the suction case—h becomes very large relative to h ∗ : suc- tion increases the heat transfer coefficient just as it increases the mass transfer coefficient. Condition for the low-rate approximation. When the rate of mass trans- fer is small, we may approximate h by h ∗ , just as we approximated g m by g ∗ m at low mass transfer rates. The approximation h = h ∗ may be tested by considering the ratio n i,s c p,i /h ∗ in eqn. (11.115). For example, if n i,s c p,i /h ∗ = 0.2, then h/h ∗ = 0.90, and h = h ∗ within an error of only 10 percent. This is within the uncertainty to which h ∗ can be pre- dicted in most flows. In gases, if B m,i is small, n i,s c p,i /h ∗ will usually be small as well. Property reference state. In Section 11.8, we calculated g ∗ m,i (and thus g m,i ) at the film temperature and film composition, as though mass transfer were occurring at the mean mixture composition and tempera- ture. We may evaluate h ∗ and g ∗ m,i in the same way when heat and mass transfer occur simultaneously. If composition variations are not large, as in many low-rate problems, it may be adequate to use the freestream composition and film temperature. When large properties variations are present, other schemes may be required [11.30]. 670 An introduction to mass transfer §11.9 Figure 11.22 Transpiration cooling. Energy balances in simultaneous heat and mass transfer Transpiration cooling. To calculate simultaneous heat and mass trans- fer rates, one must generally look at the energy balance below the wall as well as those at the surface and across the boundary layer. Consider, for example, the process of transpiration cooling, shown in Fig. 11.22. Here a wall exposed to high temperature gases is protected by injecting a cooler gas into the flow through a porous section of the surface. A portion of the heat transfer to the wall is taken up in raising the temperature of the transpired gas. Blowing serves to thicken the boundary layer and reduce h, as well. This process is frequently used to cool turbine blades and combustion chamber walls. Let us construct an energy balance for a steady state in which the wall has reached a temperature T s . The enthalpy and heat fluxes are as shown in Fig. 11.22. We take the coolant reservoir to be far enough back from the surface that temperature gradients at the r-surface are negligible and the conductive heat flux, q r , is zero. An energy balance between the r - and u-surfaces gives n i,r ˆ h i.r = n i,u ˆ h i,u −q u (11.116) and between the u- and s-surfaces, n i,u ˆ h i,u −q u = n i,s ˆ h i,s −q s (11.117) §11.9 Simultaneous heat and mass transfer 671 Since there is no change in the enthalpy of the transpired species when it passes out of the wall, ˆ h i,u = ˆ h i,s (11.118) and, because the process is steady, conservation of mass gives n i,r = n i,u = n i,s (11.119) Thus, eqn. (11.117) reduces to q s = q u (11.120) The flux q u is the conductive heat flux into the wall, while q s is the con- vective heat transfer from the gas stream, q s = h(T e −T s ) (11.121) Combining eqns. (11.116) through (11.121), we find n i,s  ˆ h i,s − ˆ h i,r  = h(T e −T s ) (11.122) This equation shows that, at steady state, the heat convection to the wall is absorbed by the enthalpy rise of the transpired gas. Writing the enthalpy as ˆ h i = c p,i (T s −T ref ), we obtain n i,s c p,i (T s −T r ) = h(T e −T s ) (11.123) or T s = hT e +n i,s c p,i T r h +n i,s c p,i (11.124) It is left as an exercise (Problem 11.47) to show that T s = T r +(T e −T r ) exp(−n i,s c p,i /h ∗ ) (11.125) The wall temperature decreases exponentially to T r as the mass flux of the transpired gas increases. Transpiration cooling may be enhanced by injecting a gas with a high specific heat. 672 An introduction to mass transfer §11.9 Sweat Cooling. A common variation on transpiration cooling is sweat cooling, in which a liquid is bled through a porous wall. The liquid is vaporized by convective heat flow to the wall, and the latent heat of vaporization acts as a sink. Figure 11.22 also represents this process. The balances, eqns. (11.116) and (11.117), as well as mass conservation, eqn. (11.119), still apply, but the enthalpies at the interface now differ by the latent heat of vaporization: ˆ h i,u +h fg = ˆ h i,s (11.126) Thus, eqn. (11.120) becomes q s = q u +h fg n i,s and eqn. (11.122) takes the form n i,s  h fg +c p,i f (T s −T r )  = h(T e −T s ) (11.127) where c p,i f is the specific heat of liquid i. Since the latent heat is generally much larger than the sensible heat, a comparison of eqn. (11.127)to eqn. (11.123) exposes the greater efficiency per unit mass flow of sweat cooling relative to transpiration cooling. Thermal radiation. When thermal radiation falls on the surface through which mass is transferred, the additional heat flux must enter the energy balances. For example, suppose that thermal radiation were present dur- ing transpiration cooling. Radiant heat flux, q rad,e , originating above the e-surface would be absorbed below the u-surface. 14 Thus, eqn. (11.116) becomes n i,r ˆ h i,r = n i,u ˆ h i,u −q u −αq rad,e (11.128) where α is the radiation absorptance. Equation (11.117) is unchanged. Similarly, thermal radiation emitted by the wall is taken to originate be- low the u-surface, so eqn. (11.128)isnow n i,r ˆ h i,r = n i,u ˆ h i,u −q u −αq rad,e +q rad,u (11.129) or, in terms of radiosity and irradiation (see Section 10.4) n i,r ˆ h i,r = n i,u ˆ h i,u −q u −(H − B) (11.130) for an opaque surface. 14 Remember that the s- and u-surfaces are fictitious elements of the enthalpy bal- ances at the phase interface. The apparent space between them need be only a few molecules thick. Thermal radiation therefore passes through the u-surface and is ab- sorbed below it. Problems 673 Chemical Reactions. The heat and mass transfer analyses in this sec- tion and Section 11.8 assume that the transferred species undergo no homogeneous reactions. If reactions do occur, the mass balances of Sec- tion 11.8 are invalid, because the mass flux of a reacting species will vary across the region of reaction. Likewise, the energy balance of this section will fail because it does not include the heat of reaction. For heterogeneous reactions, the complications are not so severe. Re- actions at the boundaries release the heat of reaction released between the s- and u-surfaces, altering the boundary conditions. The proper sto- ichiometry of the mole fluxes to and from the surface must be taken into account, and the heat transfer coefficient [eqn. (11.115)] must be modi- fied to account for the transfer of more than one species [11.30]. Problems 11.1 Derive: (a) eqns. (11.8); (b) eqns. (11.9). 11.2 A 1000 liter cylinder at 300 K contains a gaseous mixture com- posed of 0.10 kmol of NH 3 , 0.04 kmol of CO 2 , and 0.06 kmol of He. (a) Find the mass fraction for each species and the pressure in the cylinder. (b) After the cylinder is heated to 600 K, what are the new mole fractions, mass fractions, and molar concen- trations? (c) The cylinder is now compressed isothermally to a volume of 600 liters. What are the molar concentrations, mass fractions, and partial densities? (d) If 0.40 kg of gaseous N 2 is injected into the cylinder while the temperature remains at 600 K, find the mole fractions, mass fractions, and molar con- centrations. [(a) m CO 2 = 0.475; (c) c CO 2 = 0.0667 kmol/m 3 ; (d) x CO 2 = 0.187.] 11.3 Planetary atmospheres show significant variations of temper- ature and pressure in the vertical direction. Observations sug- gest that the atmosphere of Jupiter has the following compo- sition at the tropopause level: number density of H 2 = 5.7 ×10 21 (molecules/m 3 ) number density of He = 7.2 × 10 20 (molecules/m 3 ) number density of CH 4 = 6.5 ×10 18 (molecules/m 3 ) number density of NH 3 = 1.3 ×10 18 (molecules/m 3 ) [...]... 80 00 Pa Air has leaked into the unit and has reached a mass fraction of 0.04 The steam–air mixture is at 45◦ C and is blown downward past the wall at 8 m/s (a) Estimate the rate of condensation on the wall (Hint : The surface of the condensate film is not at the mixture or wall temperature.) (b) Compare the result of part (a) to condensation without air in the steam What do you conclude? 11.52 As part. .. the Damkohler number Explain its significance in catalysis What features dominate the process when Da approaches 0 or ∞? What temperature range characterizes each? 11.45 One typical kind of mass exchanger is a fixed-bed catalytic reactor A flow chamber of length L is packed with a catalyst bed A gas mixture containing some species i to be consumed ˙ by the catalytic reaction flows through the bed at a rate... jsalt (z) from part (b), make a simple estimate of the amount of salt carried upward in one week in a 5 km2 horizontal area of the sea 11.9 In catalysis, one gaseous species reacts with another on a passive surface (the catalyst) to form a gaseous product For example, butane reacts with hydrogen on the surface of a nickel catalyst to form methane and propane This heterogeneous reaction, referred to as... at 25◦ C and 1 atm, with xC2 H4 = 1.75 × 10−5 and H = 11.4 × 1 03 atm 11.26 Use a steam table to estimate (a) the mass fraction of water vapor in air over water at 1 atm and 20◦ C, 50◦ C, 70◦ C, and 90◦ C; (b) the partial pressure of water over a 3 percent-byweight aqueous solution of HCl at 50◦ C; (c) the boiling point at 1 atm of salt water with a mass fraction mNaCl = 0. 18 [(c) TB.P = 101 .8 C.] Problems... with water at 5◦ C and 1 atm, exposed to air at the same conditions, H = 4 .88 × 104 atm; (b) the mole fraction of ammonia in air above an aqueous solution, with xNH3 = 0.05 at 0.9 atm and 40◦ C and H = 1522 mm Hg; (c) the mole fraction of SO2 in an aqueous solution at 15◦ C and 1 atm, if pSO2 = 28. 0 mm Hg and H = 1.42 × 104 mm Hg; and (d) the partial pressure of ethylene over an aqueous solution at 25◦... part of a coating process, a thin film of ethanol is wiped onto a thick flat plate, 0.1 m by 0.1 m The initial thickness of the liquid film is 0.1 mm, and the initial temperature of both the plate and the film is 30 3 K The air above the film moves at 10 m/s and has a temperature of 30 3 K (a) Assume that the plate is a poor conductor, so that heat loss into it can be neglected After a short initial transient,... Mass Transfer in Solids and Fluids Cambridge University Press, Cambridge, 2000 [11.27] W K Lewis The evaporation of a liquid into a gas Mech Engr., 44(7):445–446, 1922 687 688 Chapter 11: An introduction to mass transfer [11. 28] T H Chilton and A P Colburn Mass transfer (absorption) coefficients: Prediction from data on heat transfer and fluid friction Ind Eng Chem., 26:11 83 1 187 , 1 934 [11.29] A F Mills... the heat transfer relation for spheres in a laminar flow, NuD = 2 + 0 .3 Re0.6 Pr1 /3 (Hint : first estimate the D surface temperature using properties for pure air; then correct the properties if necessary.) 11.49 The following data were taken at a weather station over a period of several months: Date Tdry-bulb Twet-bulb 3/ 15 4/21 5/ 13 5 /31 7/4 15.5◦ C 22.0 27 .3 32.7 39 .0 11.0◦ C 16 .8 25 .8 20.0 31 .2... of a reactant per unit surface area The p’s are partial pressures and A, ∆E, n, and m are constants Suppose that n = 1 and m = 0 for the reaction B + C → D Approximate the reaction rate, in terms of mass, as ◦ ˙ rB = A e−∆E/R T ρB,s kg/m2 ·s and find the rate of consumption of B in terms of mB,e and the mass transfer coefficient for the geometry in question (c) The ◦ ∗ ratio Da ≡ A e−∆E/R T /gm is called... moths The latent heat of sublimation and evaporation rate are low enough that the wet-bulb temperature is essentially the ambient temperature Estimate the lifetime of this mothball in a closet with a mean temperature of 20◦ C Use the following data: σ = 6. 18 Å, ε/kB = 561.5 K for C10 H8 , and, for solid naphthalene, ρC10 H8 = 1145 kg/m3 at 20◦ C The vapor pressure (in mmHg) of solid naphthalene near room . balances in simultaneous heat and mass transfer Transpiration cooling. To calculate simultaneous heat and mass trans- fer rates, one must generally look at the energy balance below the wall as well. tempera- ture. We may evaluate h ∗ and g ∗ m,i in the same way when heat and mass transfer occur simultaneously. If composition variations are not large, as in many low-rate problems, it may be adequate to. Thermal radiation therefore passes through the u-surface and is ab- sorbed below it. Problems 6 73 Chemical Reactions. The heat and mass transfer analyses in this sec- tion and Section 11 .8 assume