Question

A simple distillation column separates a binary mixture of A and B. The relative volatility of A with respect to B is \( 2 \). The steady-state composition of A in the vapor leaving the 1\textsuperscript{st, 2\textsuperscript{nd}, and 3\textsuperscript{rd} trays in the rectifying section are \( 94\% \), \( 90\% \), and \( 85\% \) (mol\%), respectively. For ideal trays and constant molal overflow, the reflux-to-distillate ratio is:}

• A. 1.9
• B. 2.7
• C. 1.2
• D. 1.1

Hint:

In distillation problems, the reflux ratio can be calculated using the steady-state vapor compositions on successive trays in the rectifying section.

Answer 1

The Correct Option is B

Given: Stream lies on the same side of the operating line. Stream lies on the opposite side of the equilibrium line. Equilibrium Relationships: \[ y_1 = \frac{\alpha x_1}{1 + (\alpha - 1)x_1}, \quad y_2 = \frac{\alpha x_2}{1 + (\alpha - 1)x_2}. \] Substitute the given values: \[ 0.94 = \frac{2x_1}{1 + x_1}, \quad 0.9 = \frac{2x_2}{1 + x_2}. \] Solving for \(x_1\) and \(x_2\): For \(x_1\): \[ x_1 = \frac{0.94}{1.06}. \] \[ x_1 = 0.8868. \] For \(x_2\): \[ x_2 = \frac{0.9}{1.1}. \] \[ x_2 = 0.8182. \] Enriching Section Operating Line: The enriching section operating line can be written as: \[ y = \frac{R}{R+1}x + \frac{x_D}{R+1}. \] Slope of the Operating Line: The slope of the line is: \[ \text{Slope} = \frac{R}{R+1}. \] This slope is equal to the slope between the points \((x_2, y_3)\) and \((x_1, y_2)\): \[ \frac{R}{R+1} = \frac{0.9 - 0.85}{\frac{0.94}{1.06} - \frac{0.9}{1.1}}. \] Simplifying: \[ \frac{R}{R+1} = \frac{0.05}{0.8868 - 0.8182}. \] \[ \frac{R}{R+1} = \frac{0.05}{0.0686}. \] \[ \frac{R}{R+1} = 0.729. \] Solving for \(R\): \[ R = \frac{0.729}{1 - 0.729}. \] \[ R = 2.68. \] Round off: \[ R = 2.7 = \frac{L}{G}. \] Conclusion: The reflux ratio \(R\) is \(2.7\), and it corresponds to \(\frac{L}{G}\).