Distillation Column Efficiency and Product Purity

For binary distillation with constant relative volatility α = 2.2, the minimum reflux ratio can be found using the Underwood equations or graphically. For a saturated liquid feed (q=1), the minimum reflux ratio is given by: Rmin = (xD - y*)/(y* - x*), where y* and x* are the equilibrium compositions at the feed pinch point. Alternatively, using the McCabe-Thiele method, the pinch occurs at the intersection of the q-line (vertical line at xF = 0.4) and the equilibrium curve. At x = 0.4, the equilibrium y = α x / (1 + (α-1)x) = 2.20.4 / (1 + 1.20.4) = 0.88 / 1.48 = 0.5946. The q-line is x=0.4. The rectifying line passes through (xD, xD) = (0.95, 0.95) and the pinch point (0.4, 0.5946). The slope of the rectifying line at minimum reflux is (0.95 - 0.5946)/(0.95 - 0.4) = 0.3554/0.55 = 0.6462. The slope is also R/(R+1). So Rmin/(Rmin+1) = 0.6462 → Rmin = 0.6462/(1-0.6462) = 0.6462/0.3538 = 1.826. Thus, Rmin ≈ 1.83. If the reflux ratio were set to 1.5 times Rmin, R = 1.51.83 = 2.745. The current reflux ratio is 2.0, which is less than 2.745 and also less than 1.5Rmin. Actually, 2.0 is between Rmin and 1.5Rmin. So the current reflux ratio is higher than minimum but lower than the intended 1.5Rmin. This could explain why the separation is not achieving the desired purity, as the reflux ratio is lower than the design value.

Using the Fenske-Underwood-Gilliland (FUG) method: First, calculate the minimum number of stages (Nmin) from Fenske equation: Nmin = log[(xD/(1-xD))((1-xB)/xB)] / log(α) = log[(0.95/0.05)(0.95/0.05)] / log(2.2) = log[(19)(19)] / log(2.2) = log(361)/0.3424 = 2.557/0.3424 = 7.47. So Nmin ≈ 7.5 stages. Next, calculate Rmin using Underwood equations for q=1: Solve for θ: αxF/(α-θ) + (1-xF)/(1-θ) = 1 - q = 0. So 2.20.4/(2.2-θ) + 0.6/(1-θ) = 0. This gives θ between 1 and 2.2. Solving: 0.88/(2.2-θ) + 0.6/(1-θ) = 0. Multiply: 0.88(1-θ) + 0.6(2.2-θ) = 0 → 0.88 - 0.88θ + 1.32 - 0.6θ = 0 → 2.2 - 1.48θ = 0 → θ = 2.2/1.48 = 1.486. Then Rmin+1 = Σ αxD/(α-θ) for components: For methanol: 2.20.95/(2.2-1.486) = 2.09/0.714 = 2.927; for water: 10.05/(1-1.486) = 0.05/(-0.486) = -0.103. Sum = 2.824. So Rmin = 2.824 - 1 = 1.824, consistent with earlier. Now, at R=2.0, R/(R+1) = 2/3 = 0.6667. Rmin/(Rmin+1) = 1.824/2.824 = 0.6459. The Gilliland correlation: X = (R - Rmin)/(R + 1) = (2.0 - 1.824)/(3.0) = 0.176/3 = 0.0587. Y = (N - Nmin)/(N + 1). From Gilliland chart, for X=0.0587, Y ≈ 0.6 (approximate). So (N - 7.47)/(N+1) = 0.6 → N - 7.47 = 0.6N + 0.6 → 0.4N = 8.07 → N = 20.2 stages. So about 20 theoretical stages are required. The column has only 15 theoretical stages (including reboiler), so it is under-designed for the specified separation at R=2.0. This explains the off-spec products.

The number of theoretical stages from the trays = actual trays × efficiency = 25 × 0.60 = 15 theoretical stages. Including the reboiler (which is typically considered one theoretical stage), the total theoretical stages = 15 + 1 = 16. However, the problem statement says the column has 15 theoretical stages including reboiler, which might already account for efficiency. But to be precise: if 25 actual trays give 15 theoretical stages, plus reboiler gives 16. The required stages at R=2.0 is about 20.2, so the column is short by about 4-5 stages. At the design reflux ratio of 1.5*Rmin = 2.745, the required stages would be less. Using Gilliland with R=2.745: X = (2.745-1.824)/(3.745) = 0.921/3.745 = 0.246. From Gilliland, Y ≈ 0.45. Then (N-7.47)/(N+1)=0.45 → N-7.47=0.45N+0.45 → 0.55N=7.92 → N=14.4 stages. So at R=2.745, only about 14.4 theoretical stages are needed, which is close to the 15 available. Therefore, the column was designed for R=2.745 but is operating at R=2.0, which is insufficient. The most likely cause of poor purities is that the reflux ratio is too low (2.0 instead of 2.745), leading to inadequate separation. Corrective actions: increase reflux ratio to at least 2.745 (preferably higher to provide margin), which may require increasing condenser and reboiler duties. Check if condenser and reboiler can handle the increased load; if not, consider reducing feed rate or improving tray efficiency. Also, verify that the actual reflux ratio is indeed 2.0 and not lower due to measurement error. Additionally, check for fouling, weeping, or flooding that might reduce efficiency. If increasing reflux is not possible, consider adding more trays or increasing tray efficiency through maintenance.