Let's solve the step by step as per your requirements. ### Given Data Gas flow rate (\(G\)) = 90 kmol/h Inlet SO₂ mole fraction (\(y_{in}\)) = .10 Outlet SO₂ mole fraction (\(y_{out}\)) = .03 Total pressure (\(P\)) = 1 atm Henry's law (\(y^* = 2.x\)) Overall gas phase mass transfer coefficient (\(K_ya\)) = .16 kmol/(m³·h·atm) Packed height (\(Z\)) = 6 m --- ### 1. Calculate NTU (Number of Transfer Units) NTU for absorption (when the solvent is pure and operates isothermally) is: \[ NTU = \int_{y_{out}}^{y_{in}} \frac{dy}{y - y^*} \] Here, \(y^* = 2.x\), but since the solvent is water (assume \(x \approx \)), \(y^* \approx \). So, the equation simplifies to: \[ NTU = \int_{y_{out}}^{y_{in}} \frac{dy}{y} \] So, \[ NTU = \ln \left(\frac{y_{in}}{y_{out}}\right) = \ln \left(\frac{.10}{.03}\right) = \ln(3.33) \approx 1.204 \] --- ### 2. Calculate HTU (Height of Transfer Unit) HTU is calculated using: \[ HTU = \frac{G}{K_ya \cdot P} \] Convert \(K_ya\) to consistent units: \[ K_ya = .16~\text{kmol}/(\text{m}^3 \cdot \text{h} \cdot \text{atm}) \] \(G = 90~\text{kmol/h}\), but we need per m². Assume cross-sectional area \(A\) is not given, so we will solve per unit area. \[ HTU = \frac{G}{K_ya \cdot P} \] \[ HTU = \frac{90}{.16 \times 1} \] \[ HTU = 562.5~\text{m} \] However, this value is too high; usually, HTU is defined per unit area. If we work on a per m² basis: \[ G' = \text{mol flux} = \frac{G}{A} \] But area is not given, so we typically use the formula as above, assuming per m². --- ### 3. Check Whether Packed Height Is Sufficient \[ \text{Required height} = NTU \times HTU \] \[ = 1.204 \times 562.5 = 677.25~\text{m} \] Given packed height is only 6 m, which is far less than required, so the packed height is **NOT sufficient**. --- ### 4. Calculate Percentage Removal of SO₂ \[ \text{Percentage removal} = \frac{y_{in} - y_{out}}{y_{in}} \times 100\% \] \[ = \frac{.10 - .03}{.10} \times 100 = 70\% \] --- ### 5. Effect of Gas Flow Rate Increase by 25% New gas flow rate \(G_{new} = 1.25 \times 90 = 112.5~\text{kmol/h}\) HTU will increase proportionally: \[ HTU_{new} = \frac{G_{new}}{K_ya \cdot P} = \frac{112.5}{.16 \times 1} = 703.125~\text{m} \] With the same packed height and NTU formula: \[ NTU_{new} = \frac{Z}{HTU_{new}} = \frac{6}{703.125} \approx .0085 \] So, the number of transfer units **decreases** if the gas flow rate increases, meaning less SO₂ will be removed for the same column height. --- ### **Final Answers** NTU = 1.204 HTU = 562.5 m Packed height is **not sufficient** (required: 677.25 m, available: 6 m) Percentage removal of SO₂ = 70% If gas flow increases by 25%, HTU increases to 703.125 m and NTU drops to .0085, so removal efficiency **decreases**.