Question 31.1: The design of aeration systems for aerobic-fermentation proc......

The required $K_{L}a$ is backed out from a material balance on dissolved oxygen (species A) within the well-mixed liquid phase of the fermenter. Recall equation (31-2):

$\dot{V}_{o}(c_{A o}-c_{A})+K_{L}a\cdot V\bigl(c_{A}^{*}-c_{A}\bigr)+R_{A}\,V=0$          (31-2)

$\dot{V}_{o}(c_{A o}-c_{A})+K_{L}a V\bigl(c_{A}^{*}-c_{A}\bigr)+R_{A}\,V=0$

Inserting $R_{A}=-q_{o}c_{X}$ and solving for the required $K_{L}a$ yields

$K_{L}a={\frac{q_{o}c x-{\frac{\dot{V}_{o}}{V}}(c_{A o}-c_{A})}{c_{A}^{*}-c_{A}}}$        (31-4)

The saturation concentration of dissolved oxygen is determined by Henry’s law:

$c_{A}^{*}={\frac{p_{A}}{H}}={\frac{0.21\mathrm{~atm}}{0.826{\frac{{\mathrm{m}}^{3}\cdot{\mathrm{atm}}}{{\mathrm{gmole}}}}}}=0.254{\frac{\mathrm{gmole}\,\mathrm{O}_{2}}{\mathrm{m^{3}}}}$

The partial pressure of oxygen $(p_{A})$ is presumed constant, as the rate of $O_{2}$ transferred to the sparingly soluble liquid is very small in comparison to the molar flow rate of $O_{2}$ in the aeration gas. Finally,

$K_{L}a = \frac{\left({\frac{20\ {\mathrm{gmole}}\,O_{2}}{{\mathrm{kg}}\,\mathrm{cells}\cdot h}}{\frac{5.0\,{\mathrm{kg}}\,\mathrm{cells}}{\mathrm{m}^{3}}} – \frac{1.8\,\mathrm{m^{3}/h}}{3.0\,\mathrm{m^{3}}}(0.010-0.050)\frac{\mathrm{gmole}\,\mathrm{O}_{2}}{\mathrm{m^{3}}}\right)\frac{1\ \mathrm{h}}{3600\ s} }{(0.254-0.050){\frac{\mathrm{gmobte}\,\mathrm{O}_{2}}{\mathrm{m}^{3}}}} = 0.136\ s^{-1}$

The power input to the aerated tank is backed out from the correlation

$(k_{L}a)_{O_{2}}=2\times10^{-3}\left(\frac{P_{g}}{V}\right)^{0.7}\left(u_{g s}\right)^{0.2}$          (30-33)

where $k_{L}a$ has units of $s^{-1},P_{g}/V$  has units of W/m³, and $u_{g s}$ has units of m/s. The superficial velocity of the gas through the empty tank is

$u_{g s}={\frac{4Q_{g}}{\pi d_{T}^{2}}}={\frac{(4)\left({\frac{1.0~m^{3}}{\mathrm{min}}}{\frac{1\,\mathrm{min}}{60\,s}}\right)}{\pi(1.5\,\mathrm{m})^{2}}}=0.0094{\frac{\mathrm{m}}{\mathrm{s}}}$

If the gas is sparingly soluble in the liquid, the interphasemass-transfer process is liquid-phase controlling so that $K_{L}a \cong k_{L}a.$ Therefore,

$0.136=2.0 x 10^{-3}\left({\frac{P_{g}}{V}}\right)^{0.7}(0.0094)^{0.2}$

or

${\frac{P_{g}}{V}}=1572{\frac{\mathrm{W}}{\mathrm{m^{3}}}}$

The total required power input $(P_{g})$ for the 3.0 m³ aerated fermenter is 4716 W.