Question 31.1: The design of aeration systems for aerobic-fermentation proc...
The design of aeration systems for aerobic-fermentation processes is based on gas–liquid mass transfer. Microorganisms growin a liquid suspension and feed on dissolved nutrients such as glucose and mineral salts. Aerobic microorganisms in liquid suspension also require dissolved oxygen for growth. If oxygen is not supplied at a rate sufficient to support cell growth, the cells will die.
In the present process, Aerobacter aerogenes is being cultivated within a continuous flow fermenter of 3 m³ liquid volume (V) and tank diameter \left(d_T\right) of 1.5 m. Fresh nutrient medium containing a trace amount of dissolved \mathrm{O}_2 at concentration 0.01 mole \mathrm{O}_2 / \mathrm{m}^3 enters the fermenter at a flow rate of 1.8 \mathrm{m}^3 / \mathrm{h}. At steady-state conditions, the aerobic fermenter operates at a cell concentration \left(c_X\right) of 5 kg dry mass m³ of liquid culture. The cell concentration is determined by the specific growth rate of the organism and the nutrient composition of the liquid medium, details of which will not be presented here. The liquid cell suspension consumes oxygen proportional to the cell concentration according to the rate equation
R_A=-q_0 \cdot c_Xwhere q_o is the specific oxygen consumption rate of the cells, equal to 20 mole \mathrm{O}_2 / \mathrm{kg} cells · h, which is assumed to be constant. Determine the K_L a value necessary to ensure that the dissolved oxygen concentration in the liquid culture \left(c_A\right) is at least 0.05 mol/m³. Also, determine the power input into a 3 m³ fermenter if the gas flow rate into the fermenter is 1 m³ of air per minute at the process conditions of 298 K and 1 atm. Assume that the bubbles are noncoalescing. At 298 K, Henry’s law constant for dissolution of \mathrm{O}_2 in the liquid nutrient medium is 0.826 atm·m³/mol.
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\left(c_{A o}-c_A\right)+K_L a \cdot V\left(c_A^*-c_A\right)+R_A \cdot V=0 (31-2)
Inserting R_A=-q_o c_X and solving for the required K_L a yields
K_L a=\frac{q_o \cdot c_X-\frac{\dot{V}_o}{V}\left(c_{A o}-c_A\right)}{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{atm} \cdot \mathrm{m}^3}{\mathrm{~mol}}}=0.254 \frac{\mathrm{mol} \mathrm{O}_2}{\mathrm{~m}^3}The partial pressure of oxygen \left(p_A\right) is presumed constant, as the rate of \mathrm{O}_2 transferred to the sparingly soluble liquid is very small in comparison to the molar flow rate of \mathrm{O}_2 in the aeration gas. Finally
K_L a=\frac{\left(20 \frac{\mathrm{mol} \mathrm{O}_2}{\mathrm{~kg} \text { cells } \cdot \mathrm{h}} \cdot 5 \frac{\mathrm{kg} \text { cells }}{\mathrm{m}^3}-\frac{1.8 \mathrm{~m}^3 / \mathrm{h}}{3.0 \mathrm{~m}^3}(0.01-0.05) \frac{\mathrm{mol} \mathrm{O}_2}{\mathrm{~m}^3}\right) \cdot \frac{1 \mathrm{~h}}{3600 \mathrm{~s}}}{(0.254-0.05) \frac{\mathrm{mol} \mathrm{O}_2}{\mathrm{~m}^3}}=0.136 \mathrm{~s}^{-1}In the limiting case of c_A=0 \text { and } c_{A o}=0, the minimum K_L a for \mathrm{O}_2 transfer is defined as
\left(K_L a\right)_{\min }=\frac{q_o \cdot c_X}{c_A^*}=\frac{\left(\frac{\mathrm{mol} \mathrm{O}_2}{\mathrm{~kg} \mathrm{cells} \cdot \mathrm{h}} \cdot 5 \frac{\mathrm{kg} \mathrm{cells}}{\mathrm{m}^3}\right) \frac{1 \mathrm{~h}}{3600 \mathrm{~s}}}{0.254 \frac{\mathrm{mol} \mathrm{O}_2}{\mathrm{~m}^3}}=0.109 \mathrm{~s}^{-1}From the above equation, it is evident that the biological oxygen consumption most strongly determines the required K_L a.
The power input to the aerated tank is backed out from the correlation
\left(k_L a\right)_{\mathrm{O}_2}=2 \times 10^{-3}\left(\frac{P_g}{V}\right)^{0.7}\left(u_{g s}\right)^{0.2} (30-29)
where K_L a has units of \mathrm{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{4 Q_g}{\pi d_T^2}=\frac{(4)\left(1 \frac{\mathrm{m}^3}{\mathrm{~min}} \cdot \frac{1 \mathrm{~min}}{60 \mathrm{~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 interphase mass-transfer process is liquid-phase controlling so that K_L a \cong k_L a. Therefore
0.136=2 \times 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 m³ aerated fermenter is 4716 W.