Gas containing 0.8 mol% component \(A\) is to be scrubbed with pure water in a packed bed column to reduce the concentration of \(A\) to 0.1 mol% in the exit gas. The inlet gas and water flow rates are 0.1 kmol/s and 3.0 kmol/s, respectively. For the dilute system, both the operating and equilibrium curves are considered linear. If the slope of the equilibrium line is 24, the number of transfer units, based on the gas side, \(N_{OG}\) is __________ (rounded off to 1 decimal place).
Gas containing 0.8 mol% component \(A\) is to be scrubbed with pure water in a packed bed column to reduce the concentration of \(A\) to 0.1 mol% in the exit gas. The inlet gas and water flow rates are 0.1 kmol/s and 3.0 kmol/s, respectively. For the dilute system, both the operating and equilibrium curves are considered linear. If the slope of the equilibrium line is 24, the number of transfer units, based on the gas side, \(N_{OG}\) is __________ (rounded off to 1 decimal place).
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Solution and Explanation
\[ x_{\text{inlet}} = 0.8\% = 0.008 \quad \text{(mole fraction of \(A\) in the inlet gas)} \] \[ x_{\text{outlet}} = 0.1\% = 0.001 \quad \text{(mole fraction of \(A\) in the outlet gas)} \] \[ F = 0.1 \, \text{kmol/s} \quad \text{(inlet gas flow rate)} \] \[ W = 3.0 \, \text{kmol/s} \quad \text{(water flow rate)} \] \[ \text{slope of the equilibrium line} = 24 \]
Step 2: Number of Transfer Units Calculation.The number of transfer units (NTU) based on the gas phase is calculated as: \[ N_{OG} = \frac{\ln\left(\frac{x_{\text{inlet}} - x_{\text{outlet}}}{x_{\text{outlet}}}\right)}{\text{slope of equilibrium line}} \] Substituting the values: \[ N_{OG} = \frac{\ln\left(\frac{0.008 - 0.001}{0.001}\right)}{24} = \frac{\ln(7)}{24} \approx \frac{1.9459}{24} \approx 0.0811 \]
Final Answer: The number of transfer units based on the gas side is \( \boxed{0.0811} \).Top Questions on Absorption
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