Question 30.6: The “bubbleless” shell-and-tube membrane aeration system sho......

The process analysis and solution strategy will require determination of the overall mass-transfer coefficient K_{L}, and the development of a model to predict the dissolved O_{2} concentration in the flowing water down the length of the tube. The flux of O_{2} from gas to the bulk flowing water is determined by two mass-transfer resistances in series, the diffusion of  O_{2} through the silicone tube wall, and the convective mass transfer of O_{2} to the flowing water inside the tube. If the overall mass-transfer flux is defined based on the liquid phase, then

N_{A}=K_{L}(c_{A L}^{*}-c_{A L})

where c_{A L}^{*}=P_{A}/H,\;\mathrm{and}\;K_{L} is the overall mass-transfer coefficient based on the liquid-phase driving force, defined by

{\frac{1}{K_{L}}}={\frac{1}{k_{L}}}+{\frac{1}{k_{m}}}        (30-40)

with

k_{m}={\frac{H\,D_{A\epsilon}}{S_{m}\,l_{m}}}      (30-41)

In equations (30-40) and (30-41), k_{L} is the convective mass-transfer coefficient for O_{2} mass transfer into the water flowing through the tube, and k_{m} is the membrane permeation coefficient, which describes the diffusion of O_{2} through the silicone tube wall of thickness \ell_{\mathrm{m}}, and includes solubility constant for partitioning of O_{2} gas into the silicone polymer (S_{m}), the effective diffusion coefficient of O_{2}  in the silicone polymer (D_{Ae}), and solubility constant for partitioning of O_{2} gas into the fluid (H). The derivation of equations (30-40) and (30-41), which ignore the radius of curvature of the tube, is left as an exercise for the reader.

To determine k_{L}, the Reynolds number (R_{e}) and Schmidt number (S_{c}) for flow inside the tube must first be estimated. The Reynolds number is

\mathrm{Re}={\frac{\mathrm{v}_{\mathrm{\infty }}\,D}{\nu_{L}}}={\frac{(50\,\mathrm{cm}/s)(1.0\,\mathrm{cm})}{(9.12\times10^{-3}\,\mathrm{cm}^{2}/s)}}=5482

and the Schmidt number (Sc) is

\mathrm{Sc}={\frac{\nu_{L}}{D_{A B}}}={\frac{9.12\times10^{-3}\,\mathrm{cm}^{2}/s}{2.1\times10^{-5}\,\mathrm{cm}^{2}/s}}=434

The process is considered to be in turbulent flow, since {\mathrm{Re}}\gt 2100. Therefore, it is necessary to use the Linton–Sherwood correlation for turbulent flow of liquid flowing inside of a tube, given by equation (30-18):

\mathrm{Sh}={\frac{k_{L}D}{D_{A B}}}=0.023\,\mathrm{Re}^{0.83}\mathrm{Sc}^{1/3}        (30-18)

\mathrm{Sh}=0.023\,\mathrm{Re}^{0.83}\,\mathrm{Sc}^{1/3}=0.023(5482)^{0.83}(434)^{1/3}=221

The convective mass transfer coefficient k_{L} is backed out from the Sherwood number

k_{L}=\frac{\mathrm{Sh}}{D}D_{A B}=\frac{221}{1.0\,\mathrm{cm}}2.1\times10^{-5}\,\mathrm{cm}^{2}/s=4.64\times10^{-3}\,\mathrm{cm}/\mathrm{s}

The membrane permeation constant is

k_{m}={\frac{H D_{A e}}{S_{m}l_{m}}}={\frac{(0.78\,\mathrm{atm}\cdot{\mathrm{m}}^{3}/\mathrm{gmole})(5.0\times10^{-6}\,{\mathrm{cm}}^{2}/{\mathrm{s}})}{(0.029\,\mathrm{atm}\cdot{\mathrm{m}}^{3}/\mathrm{gmole})(0.1\,{\mathrm{cm}})}}=1.35\times10^{-3}\,{\mathrm{cm}}/{\mathrm{s}}

And so, finally, K_{L} is

K_{L}=\frac{k_{L}k_{m}}{k_{L}+k_{m}}=\frac{\left(4.64\times10^{-3}\,\mathrm{cm/s}\right)\left(1.35\times10^{-3}\,\mathrm{cm/s}\right)}{\left(4.64\times10^{-3}\,\mathrm{cm/s}\right)+\left(1.35\times10^{-3}\,\mathrm{cm/s}\right)}=1.04\times10^{-3}\,\mathrm{cm/s}

As an aside, note that as km increases, K_{L} approaches k_{L} and the mass-transfer resistance offered by the membrane becomes small. Increasing k_{m} can be accomplished by decreasing the membrane thickness \ell_{m} or by using a membrane material with a lower S_{m}—i.e., a material that is more soluble for O_{2}.

A model to predict dissolved oxygen profile down the length of the tube requires a material balance for O_{2} on a differential volume element of the tube, as shown in Figure 30.13 and as stated below:

All terms are in units of moles O_{2}/time. The source for O_{2} mass transfer is the O_{2} gas, and the sink is the flowing water. The model development requires three primary assumptions: (1) the process is at steady state, (2) the concentration profile of interest is along the z-direction only and represents the local bulk concentration and fluid properties, (3) there is no reaction of O_{2} in the water, and (4) the process is dilute with respect to dissolved O_{2}. Under these assumptions, the differential material balance is

\frac{\pi D^{2}}{4}\mathrm{v}_{\infty }\ c_{A L}\bigg|_{z}+N_{A}\,\pi\,D\,\Delta z-\frac{\pi D^{2}}{4}\mathrm{v}_{\infty }\ c_{A L}\bigg|_{z+\Delta z}+0=0

Rearrangement yields

-\frac{\pi{D}^{2}}{4}{ v}_{\infty}\left(\frac{c_{A L}|_{z+\Delta z}-c_{A L}|_{z}}{\Delta z}\right)+K_{L}\,\pi\,{D}\left(c_{A L}^{*}-c_{A L}\right)=0

When Δz  → 0,

-\frac{dc_{AL}}{dz}+\frac{4K_L}{D_{v∞}}(c^*_{AL}\ -\ c_{AL}) = 0

At constant p_{A},\,c_{A L}^{*} is also constant, and so separation of dependent variable c_{AL} from independent variable z, followed by integration gives

\int_{c_{AL,o}}^{c_{A L}}\frac{-d c_{A L}}{c_{A L}^{*}-c_{A L}}=-\frac{4K_{L}}{v_{\infty }D}\int_{0}^{L}d{z}

or

\mathrm{ln}\biggl(\frac{c_{A L}^{*}-c_{A L,o}}{c_{A L}^{*}-c_{A L}}\biggr)=-\frac{4K_{L}L}{{\mathrm v_{\infty }}D}      (30-42)

c_{A L}^{*}={\frac{p_{A}}{H}}={\frac{1.5\;\mathrm{atm}}{0.78\;\mathrm{atm}\cdot{\mathrm{m}}^{3}/\mathrm{gmole}}}=1.92\;\mathrm{gmole}\;\mathrm{O}_{2}/\mathrm{m}^{3}

and so at 30% of saturation c_{A L} at z=L is 0.3c_{A L}^{*}, or 0.577\ \mathrm{gmole}\ O_{2}/\mathrm{m^{3}}. Therefore, the required length L is

L={\frac{\mathrm{v_{\infty }}D}{4K_{L}}}\ln\left({\frac{c_{A L}^{*}-c_{A L,o}}{c_{A L}^{*}-c_{A L}}}\right)={\frac{(50\operatorname{cm/s})(1.0\operatorname{cm})}{4\cdot{\big(}1.04\times10^{-3}\operatorname{cm/s}\big)}}{\mathrm{ln}}\left({\frac{(1.92-0)\operatorname{gmole}/{\mathrm{m}}^{3}}{(1.92-0.577)\operatorname{gmole}/{\mathrm{m}}^{3}}}\right) = 4296\ cm

or 43 m. Rearrangement of equation (30-42) for the concentration profile gives

c_{A L}(z)=c_{A L}^{*}-\left(c_{A L}^{*}-c_{A L,o}\right)\mathrm{exp}\left(-{\frac{4K_{L Z}}{\mathrm{v}_{\infty D}}}\right)

Two plots of c_{A L}(z) are provided in Figure 30.14, one ending at 50 m, and the second ending at 500 m, to reveal the approach of c_{A L}(z)\ \mathrm{to}\ C_{A L}^{*}.