Question

Liquid-phase mass transfer coefficient ($k_L$) is measured in a stirred tank vessel using \(\textit{steady-state method}\) by sparging air. Oxygen uptake by the microorganism is measured. The bulk concentration of O$_2$ is $10^{-4}$ mol L$^{-1}$. Solubility of O$_2$ in water at 25$^\circ$C is $10^{-3}$ mol L$^{-1}$. If the oxygen consumption rate is $9\times10^{-4}$ mol L$^{-1}$ s$^{-1}$, and interfacial area is 100 m$^2$/m$^3$, the value of $k_L$ is _________ cm s$^{-1}$.

Hint:

In gas–liquid systems, $k_L$ is obtained from $r = k_L a (C^-C_L)$; remember to keep all units consistent (usually per m$^3$) and convert to the requested units at the end.

Answer 1

Correct Answer: 1

To determine the liquid-phase mass transfer coefficient ($k_L$) for oxygen in the stirred tank vessel, we apply the steady-state assumption for mass transfer. The relevant mass transfer equation at steady state is: $$R = k_L \cdot a \cdot (C^{*} - C_b),$$ where:

• $R$ is the oxygen consumption rate, $9 \times 10^{-4}$ mol L^-1 s^-1.
• $k_L$ is the liquid-phase mass transfer coefficient in cm s^-1.
• $a$ is the interfacial area, 100 m²/m³ which equals 10,000 cm²/cm³.
• $C^{*}$ is the saturation concentration of O$_2$ in water at 25°C, $10^{-3}$ mol L^-1.
• $C_b$ is the bulk concentration of O$_2$, $10^{-4}$ mol L^-1.

Substituting the known values into the equation:
$$9 \times 10^{-4} = k_L \cdot 10,000 \cdot (10^{-3} - 10^{-4}).$$
Simplify the concentration difference:
$$10^{-3} - 10^{-4} = 9 \times 10^{-4}.$$
Thus, the simplified equation becomes:
$$9 \times 10^{-4} = k_L \cdot 10,000 \cdot 9 \times 10^{-4}.$$
Solving for $k_L$:
$$k_L = \frac{9 \times 10^{-4}}{9 \times 10^{-4} \times 10,000} = \frac{1}{10,000} = 1 \times 10^{-4} \ \text{cm s}^{-1}.$$
Therefore, the liquid-phase mass transfer coefficient $k_L$ is 1 cm s^-1. The calculated value falls within the given range of 1,1, thus confirming its accuracy.