Question 30.5: A holding pond containing a suspension of aerobic microorgan......

At the process temperature of 25 C, the Henry’s law constant for O_{2} gas dissolved in water is H=0.78\mathrm{~atm}\cdot{\mathrm{m}}^{3}/\mathrm{gmole}\mathrm{~O}_{2}, the molecular diffusion coefficient of dissolved O_{2} in water is D_{A B}=2.1\times10^{-9}\,\mathrm{m^{2}/s}\,(A=\mathrm{dissolved\,O_{2}},B=\mathrm{water}), the density of liquid water is \rho_{{B,lid}}=1000\,\mathrm{kg/}{m}^{3}, and the viscosity of liquid water is \mu_{{B,liq}}=825\times10^{-6}\\mathrm{kg/m}\cdot\mathrm{s}. At 25 C and 1.0 atm, the density of air is \rho_{\mathrm{air}}=1.2\ {\mathrm{kg/m}}^{3}.

The process analysis and solution strategy will require determination of the overall volumetric mass-transfer coefficient K_La, and the development of a model to predict the dissolved O_2 concentration within the pond water. From Chapter 28, if the gas–liquid interphase mass-transfer process is liquid film controlling, then K_L ≈ k_L, the liquid film mass-transfer coefficient.

Furthermore, for a dilute process, the physical properties of the liquid mixture approximate the physical properties of liquid water. For 3.0-mm diameter (d_{b}) rising air bubbles, equation (30-15) is used to estimate k_{L} through determination of the Sherwood number (Sh) via the Grashof number (Gr) and Schmidt number (Sc):

\mathrm{Gr}={\frac{d_{b}^{3}\rho_{L}\;g\;(\rho_{L}-\rho_{G})}{\mu_{L}^{2}}}={\frac{(3.0\times10^{-3}\;{\mathrm{m}})^{3}(1000\;\mathrm{kg/m}^{3})(9.8\;\mathrm{m}/s^{2})((1000-1.2)\mathrm{kg/m}^{3})}{(825\times10^{-6}\;\mathrm{kg/m}\cdot s)^{2}}}

=3.88\times10^{5}

\mathrm{Sc}={\frac{\mu}{\rho\,D_{A B}}}={\frac{\mu_{B,\,\mathrm{lid}}}{\rho_{B,\,\mathrm{lid}}D_{A B}}}{\frac{925\times10^{-6}\,\mathrm{kg/m}\cdot{\mathrm{s}}}{(1000\,\mathrm{kg/m}^{3})(2.1\times10^{-9}\,\mathrm{m^{2}/s})}}=441

\mathrm{Sh}={\frac{k_{L}\,d_{b}}{D_{A B}}}=0.42\:\mathrm{Gr}^{1/3}\:\mathrm{Sc}^{1/2}=0.42(3.88\times10^{5})^{1/3}(441)^{1/2}=643

k_{L}={\frac{\mathrm{Sh}\,D_{A B}}{d_{b}}}={\frac{(643)(2.1\times10^{-9}\,\mathrm{m}/s)}{3.0\times10^{-3}\,\mathrm{m}}}=4.5\times10^{-4}\,\mathrm{m}/s

The definition for Sc is based on the liquid phase, since Sh is based on the liquid film surrounding the gas bubble. The interphase mass-transfer area per unit volume of liquid is estimated from the gas holdup (ϕ_{g}) by equation (30-28)

{\frac{A_{i}}{V}}={\frac{V_{g}}{V}}\cdot{\frac{\mathrm{bubble~area}}{\mathrm{bubble~volume}}}=\phi_{g}{\frac{\pi d_{b}^{2}}{\pi d_{b}^{3}/6}}={\frac{6\phi_{g}}{d_{b}}}        (30-28)

a={\frac{6\phi_{g}}{d_{b}}}={\frac{6(0.005\;{\mathrm{m}}^{3}/{\mathrm{m}}^{3})}{3.0\times10^{-3}\,{\mathrm{m}}}}=10\,{\mathrm{m}}^{2}/{\mathrm{m}}^{3}

Consequently,

k_{L}a=(4.5\times10^{-4}\,{\mathrm{m}}/{\mathrm{s}})(10\,{\mathrm{m}}^{2}/{\mathrm{m}}^{3})=4.5\times10^{-3}\,{\mathrm{s}}^{-1}

A model to predict the dissolved oxygen concentration in the pond water requires a material balance on O_{2}, based on the well-mixed liquid phase of the pond:

All terms are in units of moles O_{2}/time . The source for O_{2} mass transfer is the O_{2} in air bubbles, and the sink is the microorganisms that consume O_{2} . In this process, four primary assumptions are made: (1) the liquid phase is well mixed, so dissolved oxygen concentration is uniform within the pond, (2) the total liquid volume is constant, (3) the biological oxygen demand is constant, (4) the process is at a nominal steady state, because the rate of O_{2} input will be balanced by the rate of O_{2} consumption. Under these assumptions, the  material balance on well-mixed liquid phase is

0+N_{A}A_{i}-0+R_{A}V=0

where the R_{A} term is the BOD. If we recall that N_{A}=K_{L}(c_{A L}^{*}-c_{A L}),\;a=A_{i}/V,\;\mathrm{and}\;c_{A L}^{*}=P_{A}/H, then insertion of these relationships into the material balance, and rearrangement, yields

R_{A}=-k_{L}a\left(\frac{P_{A}}{H}-c_{A L}\right)      (30-39)

Equation (30-39) enables estimation of the required R_{A}

R_{A}=(4.5\times10^{-3}\,{\mathrm{s}}^{-1})\left({\frac{0.21\,{\mathrm{at}}\mathrm{m}}{0.78\,\mathrm{at}\mathrm{m}\cdot\mathrm{m}^{3}/\mathrm{gmole}}}-0.05\,{\mathrm{gmole}}/m^{3}\right)=-9.87\times10^{-4}\,{\mathrm{gmole}}/m^{3}\,{\mathrm{s}}

The negative sign on R_{A} indicates that O_{2} is being consumed. The total O_{2} transfer rate is

W_{A}=k_{L}a\left(\frac{P_{A}}{H}-c_{A L}\right)V=-R_{A}\frac{\pi D^{2}H}{4}=-\left(-9.87\times10^{-4}\,\mathrm{gmole}\,\mathrm{O}_{2}/m^{3}{\mathrm{s}}\right)\left(\frac{\pi(15\,\mathrm{m})^{2}(3.0\,\mathrm{m})}{4}\right)

=0.523\ {\mathrm{gmole}}\,\mathrm{O}_{2}/s

The superficial air velocity (u_{gs}) into the pond can be determined by the gas holdup through equation (30-29). In equation (30- 29), the surface tension of water at 25 C, which can be found in many published handbooks, such as the International Critical Tables, is \sigma_{L}=72\times10^{-3}\,\,\mathrm{N/m}. To back out u_{gs}, rearrangement of equation (30-29) yields

Finally, with u_{gs}, known, the volumetric flow rate of air is

Q_{g}=u_{g s}{\frac{\pi D^{2}}{4}}=0.011\,{\mathrm{m/s}}{\frac{\pi(15\,{\mathrm{m}})^{2}}{4}}=1.96\,{\mathrm{m}}^{3}\,{\mathrm{air}}/{\mathrm{s}}

It is left to the reader to verify that the molar flow of O_{2} delivered in volumetric inlet flow rate of air is in large excess to the interphase O_{2} transfer rate into the liquid. Consequently, the O_{2} partial pressure inside the bubble does not measurably decrease.