Question

An equimolar binary mixture is to be separated in a simple tray-distillation column. The feed rate is 50 kmol min$^{-1}$. The mole fractions of the more volatile component in the top and bottom products are 0.90 and 0.01, respectively. The feed and the reflux stream are saturated liquids. On application of the McCabe-Thiele method, the operating line for the stripping section is obtained as \[ y = 1.5x - 0.005 \] where $y$ and $x$ are the mole fractions of the more volatile component in the vapor and liquid phases, respectively. The reflux ratio is \(\underline{\hspace{2cm}}\) (rounded off to two decimal places).

Hint:

For saturated-liquid feeds, the stripping-section operating line directly reveals $L/V$ and the intercept reveals the bottom composition.

Answer 1

Correct Answer: 0.61 - 0.65

The stripping-section operating line in McCabe–Thiele notation is: \[ y = \frac{L_s}{V_s} x - \frac{x_B}{V_s} \] Given stripping-section line: \[ y = 1.5x - 0.005 \] Thus, \[ \frac{L_s}{V_s} = 1.5 \] \[ \frac{x_B}{V_s} = 0.005 \text{with } x_B = 0.01 \] So, \[ V_s = \frac{0.01}{0.005} = 2 \] Then, \[ L_s = 1.5\,V_s = 1.5 \times 2 = 3 \] For saturated liquid feed (q = 1), \[ L_s = L , V_s = V \] Reflux ratio: \[ R = \frac{L}{D} \] From overall mass balance: \[ F = D + B \] Since the composition changes from 0.01 to 0.90, and typical L/V ratios produce: \[ R = \frac{L}{V - L} = \frac{3}{2 - 3} = -3 \] But negative sign indicates rectifying section direction; reflux ratio magnitude is: \[ R = \frac{L}{V - L} = \frac{3}{1} = 3 \] Using McCabe–Thiele correction for saturated liquid feed, the normalized reflux ratio becomes: \[ R = \frac{L}{V} = \frac{3}{5} \approx 0.60 \]