# Step-by-Step Solution ## (a) Distillation Calculations ### (i) **Minimum Number of Theoretical Stages (\(N_{min}\))** We use the **Fenske Equation** for the minimum number of stages at total reflux: \[ N_{min} = \frac{\log \left[ \frac{(x_{D}/(1-x_{D}))}{(x_{B}/(1-x_{B}))} \right]}{\log \alpha} \] Where: - \(x_D\): mole fraction of more volatile component (Hexane) in distillate - \(x_B\): mole fraction of more volatile component (Hexane) in bottom product - \(\alpha\): relative volatility of Hexane with respect to Octane From the table: - For Hexane: \(x_{f} = .423\), \(x_{D} = .715\), \(x_{B} = .02\), \(\alpha = 2.7\) Substitute into Fenske’s equation: \[ N_{min} = \frac{\log \left[ \frac{.715/(1-.715)}{.02/(1-.02)} \right]}{\log 2.7} \] Calculate numerator: - \(.715/(1-.715) = .715/.285 = 2.5088\) - \(.02/(1-.02) = .02/.98 = .0204\) - Ratio: \(2.5088/.0204 = 122.99\) Calculate denominator: - \(\log 2.7 = .431\) (base 10) So, \[ N_{min} = \frac{\log(122.99)}{.431} \] \[ = \frac{2.089}{.431} \] \[ = 4.85 \] **Final Answer (i):** \[ \boxed{N_{min} = 4.85} \] --- ### (ii) **Minimum Reflux Ratio (\(R_{min}\))** Use the **Underwood Equation** for minimum reflux ratio (for a saturated liquid feed): \[ R_{min} = \sum_{i} \frac{x_{D,i} \alpha_i}{\alpha_i - \theta} - 1 \] Given \(\theta = 2.4\). From the table: - Hexane: \(x_{D,1} = .715\), \(\alpha_1 = 2.7\) - Heptane: \(x_{D,2} = .285\), \(\alpha_2 = 2.2\) - Octane: \(x_{D,3} = \), \(\alpha_3 = 1.\) Plug in: \[ R_{min} = \left[ \frac{.715 \times 2.7}{2.7 - 2.4} + \frac{.285 \times 2.2}{2.2 - 2.4} + \frac{ \times 1.}{1. - 2.4} \right] - 1 \] Calculate each term: - Hexane: \(.715 \times 2.7 = 1.9305\), denominator: \(2.7 - 2.4 = .3\), so \(1.9305/.3 = 6.435\) - Heptane: \(.285 \times 2.2 = .627\), denominator: \(2.2 - 2.4 = -.2\), so \(.627/-.2 = -3.135\) - Octane: zero. Sum: \(6.435 + (-3.135) = 3.3\) \[ R_{min} = 3.3 - 1 = 2.3 \] **Final Answer (ii):** \[ \boxed{R_{min} = 2.3} \] --- ### (iii) **Number of Theoretical Stages (\(N\))** Given: - The feed is a liquid at its boiling point (\(q = 1\)) - Reflux ratio is \(R = 1.1 R_{min} = 1.1 \times 2.3 = 2.53\) Use the **Gilliland Correlation** (graphical or empirical): One empirical form (Eduljee equation): \[ Y = \frac{N - N_{min}}{N + 1} \] \[ X = \frac{R - R_{min}}{R + 1} \] \[ Y = .75 \left( \frac{X^{.566}}{1 + X^{.566} (.75 - .566)} \right) \] But you can use a shortcut for typical design: For \(R = 1.1 R_{min}\): \[ \frac{N - N_{min}}{N + 1} \approx .65 \left(\frac{R - R_{min}}{R + 1}\right) \] Calculate \(X\): \[ X = \frac{2.53 - 2.3}{2.53 + 1} = \frac{.23}{3.53} = .065 \] So, \[ Y = .65 \times .065 = .042 \] Now, \[ \frac{N - 4.85}{N + 1} = .042 \] \[ N - 4.85 = .042(N + 1) \] \[ N - 4.85 = .042N + .042 \] \[ N - .042N = 4.85 + .042 \] \[ .958N = 4.892 \] \[ N = \frac{4.892}{.958} = 5.11 \] **Final Answer (iii):** \[ \boxed{N = 5.1} \] --- ## (b) Weeping in Tray Distillation Columns ### (i) **Describe Weeping and When It Occurs** - **Weeping** is the phenomenon where liquid flows (leaks) through the perforations in a tray instead of being retained on the tray. - It occurs when the vapor flow rate is too low to support (hold up) the liquid on the tray, causing liquid to "weep" down to the tray below. --- ### (ii) **Potential of Weeping for Different Tray Types** - **Sieve Trays:** High potential for weeping at low vapor rates, as holes are always open. - **Valve Trays:** Less susceptible to weeping because valves close at low vapor flow, reducing the chance of liquid passing through. - **Bubble Cap Trays:** Least likely to weep, as caps prevent liquid from passing through unless vapor pushes it up. --- ## **Summary of Answers** ### (a) - (i) \(N_{min} = \boxed{4.85}\) - (ii) \(R_{min} = \boxed{2.3}\) - (iii) \(N = \boxed{5.1}\) ### (b) - (i) Weeping is leakage of liquid through tray holes at low vapor rates. - (ii) Sieve trays are most susceptible, valve trays less so, and bubble cap trays least susceptible to weeping.