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26.32 Please refer to Example 5 of this chapter. Estimate the total transfer rate of solute A from the mass-transfer system shown in Figure 26.14, in units of gmole/h. As part of this analysis, consider the following steps:
a. Differentiate the analytical solution for cxy with respect to the coordinate of interest, and then develop an equation to determine the local flux at the boundary. For example, at x=0, y=H:
OcOy = N0,y = -DAs Ox
Repeat this approach for the other exposed faces of the device.
b. For each exposed face, integrate the local flux over y to obtain an equation for the average flux. For example, at x=0, from y=0 to y=H:
NA = N0,ydy HJ = 0
EXAMPLE 5
An ethanol-water vapor mixture is being distilled by contact with an ethanol/water liquid solution. The ethanol is removed from the liquid to the vapor phase, and the water is transferred in the opposite direction. The condensation of water vapor provides the energy for vaporization of ethanol. Both components are diffusing through a gas film 0.1 mm thick. The temperature is 368 K and the pressure is 1.013 x 10 Pa. At these conditions, the pure component enthalpies of vaporization of ethanol and water are 840 and 2300 kg, respectively. Develop the flux equation for vaporization of ethanol. Then develop the flux equation for the condensation of water vapor.
Adiabatic wall
We will consider the transfer of a diabatic molecular mass across a gas film of thickness illustrated in Figure 26.13. For ethanol, the differential equation for mass transfer simplifies to Figure 25.13 Adiabatic rectification of an ethanol/water mixture. For a binary gas-phase mixture, Fick's equation is NpOH = -cDEOB-B0 + ybomn(Nn+Mao). We perform an energy balance to relate the flux of ethanol vapor to the flux of water vapor. If the energy used to produce the alcohol vapor is equal to the energy used to produce the water vapor, the energy balance is:
Nno = 1.071NnO
Since the flux of ethanol vapor is opposite in direction to the flux of water, substituting this relationship into Fick's equation, we obtain:
Ng = 1 + 0.07ly
As the flux is in the opposite direction, this equation can be integrated directly to obtain:
Non = Dn-m + 0.0718
Essentially equal, i.e., cmo. Then from the adiabatic energy balance, it is easy to show that -Nncn = No - Thm Npon = cDon-n mom + m(Nc+Nao) reduces to:
NOn = cD-m
which upon integration yields:
So we see that equimolar heats of vaporization result in an equimolar-co adiabatic-distillation process.
