We are separating methanol and water. Calculate the internal reflux ratio inside the column, \(\mathrm{L}_{1} / \mathrm{V}_{2}\), for the following cases. The column is at \(101.3 \mathrm{kPa}\). Data are available in Table 2-8 and in Problem 3.E1.

a. Distillate product is \(99.9 \mathrm{~mol} \%\) methanol. External reflux ratio is \(\mathrm{L}_{0} / \mathrm{D}=1.2\). Reflux is cooled to \(40.0^{\circ} \mathrm{C}\) (i.e., it is subcooled).

b. Repeat part a using a saturated liquid reflux. Compare with part a.

Problem 3.E1.

Table 2-8

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E1. 100 kmol/h of a 60 mol% methanol and 40 mol% water feed at 40C is distilled in a column that does not have a reboiler but uses open steam for heating. The column is at 1.0 atm. The steam is pure water vapor (YM = 0) and is a saturated vapor. The distillate product is 99.0 mol% methanol and is a saturated liquid. The bottoms is 2.0 mol% methanol and, because it leaves an equilibrium stage, must be a saturated liquid. The column is adiabatic and has a total condenser. External reflux ratio is L/D = 2.3. Find D, B, Qc, and S. Be careful with units and in selecting basis for energy balance. Data: Amethanol = AHvap = 8.43 kcal/mol = 35.27 kJ/mol (at boiling point) Awater AHvap = 9.72 kcal/mol = 40.656 kJ/mol (at boiling point) Cp,w,liquid 1.0 cal/(g C) = 75.4 J/mol C = CPL,Meoh 75.86 + 0.16837 J/(mol C) Cp,w,vapor 33.46 + 0.006880 T + 0.7604 10-5 T2 -3.593 10-9 T = = x Cp,meoh,vapor 42.93 + 0.08301 T 1.87 x 10-5 T2 - 8.03 10-9 T T is in C, Cp,vapor values in J/mol C Felder et al (2018) and Green and Perry (2008).