(Solved): EXAMPLE 10.7-1. Design of an Absorption Tower with a Concentrated Gas Mixture A tower packed with \ ...

EXAMPLE 10.7-1. Design of an Absorption Tower with a Concentrated Gas Mixture A tower packed with \( 25.4-\mathrm{mm} \) ceramic rings is to be designed to absorb \( \mathrm{SO}_{2} \) from air by using pure water at \( 293 \mathrm{~K} \) and \( 1.013 \times 10^{5} \mathrm{~Pa} \) abs pressure. The entering gas contains \( 20 \mathrm{~mol} \% \mathrm{SO}_{2} \) and that leaving \( 2 \mathrm{~mol} \% \). The inert air flow is \( 6.53 \times 10^{-4} \mathrm{~kg} \) mol air/s and the inert water flow is \( 4.20 \times 10^{-2} \mathrm{~kg} \mathrm{~mol} \) water/s. The tower cross-sectional area is \( 0.0929 \mathrm{~m}^{2} \). For dilute \( \mathrm{SO}_{2} \), the film mass-transfer coefficients at \( 293 \mathrm{~K} \) are, for 25.4-mm (1-in.) rings (W1), \( k_{y}^{\prime} a=0.0594 G_{y}^{0.7} G_{x}^{0.25} \quad k_{x}^{\prime} a=0.152 G_{x}^{0.82} \) \( k_{y}^{\prime} a \) \( k_{x}^{\prime} a \) where is \( \mathrm{kg} \mathrm{mol} / \mathrm{s} \cdot \mathrm{m}^{3} \cdot \mathrm{mol} \mathrm{frac} \), is \( \mathrm{kg} \mathrm{mol} / \mathrm{s}^{3} \cdot \mathrm{m}^{3} \mathrm{~mol} \mathrm{frac} \), and \( \mathrm{G}_{\mathrm{x}} \) and \( \mathrm{G}_{\mathrm{y}} \) are \( \mathrm{kg} \) total liquid or gas, respectively, per sec per \( \mathrm{m}^{2} \) tower cross section. Calculate the tower height. Coefficients. Tower Using Overall Using the data for Example 10.7-1, calculate the tower height using the overall mass-transfer \[ K_{y}^{*} a \quad K_{y}^{\prime} a \] Coefficient. [Hint: Calculate at the top of the tower and at the bottom of the tower from the film coefficients. Then use a linear average of the two values for the design. Obtain the values of \( y^{\circ} \) from the operating- and equilibrium-line plot. Numerically or graphically \[ K_{y}^{\prime} a \] integrate Eq. (10.6-19), keeping outside the integral.]