John H. Lienhard IV, John H. Lienhard V. A Heat Transfer Textbook (776116), страница 97
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The Lewis number for air–water systems is about0.847; and, because the concentration of water vapor is generally low, cpcan often be approximated by cpair .This type of relationship between h and gm was first developed byW. K. Lewis in 1922 for the case in which Le = 1 [11.27]. (Lewis’s primary interest was in air–water systems, so the approximation was nottoo bad.) The more general form, eqn. (11.108), is another ReynoldsColburn type of analogy, similar to eqn. (6.76). It was given by Chiltonand Colburn [11.28] in 1934.Equation (11.107) may now be written ashfgTwet-bulb mH2 O,s − mH2 O,e(11.109)Te − Twet-bulb =cpair Le2/3This expression can be solved iteratively with a steam table to obtain thewet-bulb temperature as a function of the dry-bulb temperature, Te , andthe humidity of the ambient air, mH2 O,e .
The psychrometric charts foundin engineering handbooks and thermodynamics texts can be generated inthis way. We ask the reader to make such calculations in Problem 11.49.The wet-bulb temperature is a helpful concept in many phase-changeprocesses. When a small body without internal heat sources evaporatesor sublimes, it cools to a steady “wet-bulb” temperature at which convective heating is balanced by latent heat removal.
The body will stay atthat temperature until the phase-change process is complete. Thus, thewet-bulb temperature appears in the evaporation of water droplets, thesublimation of dry ice, the combustion of fuel sprays, and so on. If thebody is massive, however, steady state may not be reached very quickly.Simultaneous heat and mass transfer§11.9Stagnant film model of heat transfer at high mass transfer ratesThe multicomponent energy equation. Each species in a mixture carries its own enthalpy, ĥi . In a flow with mass transfer, different speciesmove with different velocities, so that enthalpy transport by individual species must enter the energy equation along with heat conductionthrough the fluid mixture. For steady, low-speed flow without internalheat generation or chemical reactions, we may rewrite the energy balance,eqn.
(6.36), as⎛⎞ $i ⎠ · dS = 0− (−k∇T ) · dS − ⎝ ρi ĥi vSSiwhere the second term accounts for the enthalpy transport by each speciesin the mixture. The usual procedure of applying Gauss’s theorem and requiring the integrand to vanish identically gives⎞⎛$i ⎠ = 0ρi ĥi v(11.110)∇ · ⎝−k∇T +iThis equation shows that the total energy flux—the sum of heat conduction and enthalpy transport—is conserved in steady flow.13The stagnant film model. Let us restrict attention to the transport of asingle species, i, across a boundary layer.
We again use the stagnant filmmodel for the thermal boundary layer and consider the one-dimensionalflow of energy through it (see Fig. 11.21). Equation (11.110) simplifies todTd−k+ ρi ĥi vi = 0(11.111)dydyFrom eqn. (11.69) for steady, one-dimensional mass conservationni = constant in y = ni,s13The multicomponent energy equation becomes substantially more complex whenkinetic energy, body forces, and thermal or pressure diffusion are taken into account.The complexities are such that most published derivations of the multicomponentenergy equation are incorrect, as shown by Mills in 1998 [11.29].
The main sourceof error has been the assignment of an independent kinetic energy to the ordinarydiffusion velocity. This leads to such inconsistencies as a mechanical work term in thethermal energy equation.667668An introduction to mass transfer§11.9Figure 11.21 Energy transport in a stagnant film.If we neglect pressure variations and assume a constant specific heatcapacity (as in Sect. 6.3), the enthalpy may be written as ĥi = cp,i (T −Tref ),and eqn.
(11.111) becomesdTd−k+ ni,s cp,i T = 0dydyIntegrating twice and applying the boundary conditionsT (y = 0) = TsandT (y = δt ) = Tewe obtain the temperature profile of the stagnant film:ni,s cp,iexpy −1T − Tsk=ni,s cp,iTe − Tsδt − 1expk(11.112)The temperature distribution may be used to find the heat transfercoefficient according to its definition [eqn. (6.5)]:dT −kdy sn c i,s p,i =(11.113)h≡ni,s cp,iTs − Teexpδt − 1kWe define the heat transfer coefficient in the limit of zero mass transfer,h∗ , ash∗ ≡ lim h =ni,s →0kδt(11.114)Simultaneous heat and mass transfer§11.9Substitution of eqn. (11.114) into eqn.
(11.113) yieldsh=ni,s cp,iexp(ni,s cp,i /h∗ ) − 1(11.115)To use this result, one first calculates the heat transfer coefficient as ifthere 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 determine h in the presence of mass transfer. Note that h∗ defines theeffective film thickness δt through eqn. (11.114).Equation (11.115) shows the primary effects of mass transfer on h.When ni,s is large and positive—the blowing case—h becomes smallerthan h∗ . Thus, blowing decreases the heat transfer coefficient, just as itdecreases the mass transfer coefficient.
Likewise, when ni,s is large andnegative—the suction case—h becomes very large relative to h∗ : suction increases the heat transfer coefficient just as it increases the masstransfer coefficient.Condition for the low-rate approximation. When the rate of mass transfer is small, we may approximate h by h∗ , just as we approximated gm∗ at low mass transfer rates.
The approximation h = h∗ may beby gmtested by considering the ratio ni,s cp,i /h∗ in eqn. (11.115). For example,if ni,s cp,i /h∗ = 0.2, then h/h∗ = 0.90, and h = h∗ within an error ofonly 10 percent. This is within the uncertainty to which h∗ can be predicted in most flows. In gases, if Bm,i is small, ni,s cp,i /h∗ will usually besmall as well.∗(and thusProperty reference state.
In Section 11.8, we calculated gm,igm,i ) at the film temperature and film composition, as though masstransfer were occurring at the mean mixture composition and tempera∗in the same way when heat and massture. We may evaluate h∗ and gm,itransfer occur simultaneously. If composition variations are not large,as in many low-rate problems, it may be adequate to use the freestreamcomposition and film temperature. When large properties variations arepresent, other schemes may be required [11.30].669670An introduction to mass transfer§11.9Figure 11.22 Transpiration cooling.Energy balances in simultaneous heat and mass transferTranspiration cooling.
To calculate simultaneous heat and mass transfer rates, one must generally look at the energy balance below the wall aswell as those at the surface and across the boundary layer. Consider, forexample, the process of transpiration cooling, shown in Fig. 11.22. Here awall exposed to high temperature gases is protected by injecting a coolergas into the flow through a porous section of the surface.
A portion ofthe heat transfer to the wall is taken up in raising the temperature of thetranspired gas. Blowing serves to thicken the boundary layer and reduceh, as well. This process is frequently used to cool turbine blades andcombustion chamber walls.Let us construct an energy balance for a steady state in which the wallhas reached a temperature Ts . The enthalpy and heat fluxes are as shownin Fig. 11.22. We take the coolant reservoir to be far enough back fromthe surface that temperature gradients at the r -surface are negligible andthe conductive heat flux, qr , is zero. An energy balance between the r and u-surfaces givesni,r ĥi.r = ni,u ĥi,u − qu(11.116)and between the u- and s-surfaces,ni,u ĥi,u − qu = ni,s ĥi,s − qs(11.117)§11.9Simultaneous heat and mass transferSince there is no change in the enthalpy of the transpired species whenit passes out of the wall,ĥi,u = ĥi,s(11.118)and, because the process is steady, conservation of mass givesni,r = ni,u = ni,s(11.119)Thus, eqn.
(11.117) reduces toqs = qu(11.120)The flux qu is the conductive heat flux into the wall, while qs is the convective heat transfer from the gas stream,qs = h(Te − Ts )(11.121)Combining eqns. (11.116) through (11.121), we findni,s ĥi,s − ĥi,r = h(Te − Ts )(11.122)This equation shows that, at steady state, the heat convection to thewall is absorbed by the enthalpy rise of the transpired gas. Writing theenthalpy as ĥi = cp,i (Ts − Tref ), we obtainni,s cp,i (Ts − Tr ) = h(Te − Ts )(11.123)orTs =hTe + ni,s cp,i Trh + ni,s cp,i(11.124)It is left as an exercise (Problem 11.47) to show thatTs = Tr + (Te − Tr ) exp(−ni,s cp,i /h∗ )(11.125)The wall temperature decreases exponentially to Tr as the mass flux ofthe transpired gas increases. Transpiration cooling may be enhanced byinjecting a gas with a high specific heat.671672An introduction to mass transfer§11.9Sweat Cooling. A common variation on transpiration cooling is sweatcooling, in which a liquid is bled through a porous wall.
The liquid isvaporized by convective heat flow to the wall, and the latent heat ofvaporization 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 bythe latent heat of vaporization:ĥi,u + hfg = ĥi,s(11.126)Thus, eqn.
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