1. **Problem statement:** A distillation column processes 1000 kg/h of a mixture containing equal mass parts of methanol and water. The bottom stream flow rate is 673 kg/h. The top (overhead) stream contains 96.0 wt% methanol. We need to: (a) Draw and label the flowchart and perform degree-of-freedom analysis. (b) Calculate mass and mole fractions of methanol in bottom stream and molar flow rates of methanol and water in bottom stream. (c) List possible reasons why the actual mole fraction of methanol in bottom stream might be higher than calculated. 2. **Part (a): Flowchart and degree-of-freedom analysis** - Input feed: 1000 kg/h, 50 wt% methanol, 50 wt% water - Top product: flow= $$1000 - 673 = 327\text{ kg/h}$$ - Bottom product: flow=673 kg/h - Oxygen and other species not involved; only methanol (MeOH) and water (H2O) -Degree of freedom analysis: Variables: Mass flows of MeOH and H2O in top and bottom streams, plus total flow rates. Known: Total feed flow and composition, bottom flow, top methanol concentration. Unknowns: mass compositions in bottom stream and molar flow rates. Equations: Total mass balance, methanol mass balance, given stream compositions. Count of equations equals unknown variables, problem is solvable. 3. **Part (b): Mass and mole fractions and molar flow rates in bottom product** - Total feed mass: $F = 1000 \text{ kg/h}$ - Feed methanol mass: $F_{MeOH} = 0.50 \times 1000 = 500 \text{ kg/h}$ - Feed water mass: $F_{H_2O} = 500 \text{ kg/h}$ - Top product mass: $D = 327 \text{ kg/h}$ - Bottom product mass: $B = 673 \text{ kg/h}$ - Methanol mass in top product: $D_{MeOH} = 0.96 \times 327 = 313.92 \text{ kg/h}$ - Water mass in top product: $D_{H_2O} = 327 - 313.92 = 13.08 \text{ kg/h}$ - Methanol mass in bottom product: $$B_{MeOH} = F_{MeOH} - D_{MeOH} = 500 - 313.92 = 186.08 \text{ kg/h}$$ - Water mass in bottom product: $$B_{H_2O} = F_{H_2O} - D_{H_2O} = 500 - 13.08 = 486.92 \text{ kg/h}$$ - Mass fraction methanol in bottom product: $$x_{MeOH, bottom} = \frac{186.08}{673} \approx 0.2765$$ - Mass fraction water in bottom product: $$x_{H_2O, bottom} = \frac{486.92}{673} \approx 0.7235$$ - Convert mass to moles using molar masses: Molar mass MeOH = 32.04 g/mol = 0.03204 kg/mol Molar mass H2O = 18.015 g/mol = 0.018015 kg/mol - Moles methanol in bottom stream: $$n_{MeOH} = \frac{186.08}{0.03204} \approx 5808.9 \text{ mol/h}$$ - Moles water in bottom stream: $$n_{H_2O} = \frac{486.92}{0.018015} \approx 27032.6 \text{ mol/h}$$ - Total moles in bottom stream: $$n_{total} = 5808.9 + 27032.6 = 32841.5 \text{ mol/h}$$ - Mole fraction methanol in bottom stream: $$y_{MeOH, bottom} = \frac{5808.9}{32841.5} \approx 0.177\, (17.7\%)$$ - Mole fraction water in bottom stream: $$y_{H_2O, bottom} = 1 - y_{MeOH, bottom} = 0.823$$ 4. **Part (c): Possible reasons for higher mole fraction methanol in bottom product** - Assumption of steady state may be violated. - Incomplete separation or inefficiencies in the distillation column. - Feed composition might not be exactly 50/50. - Analytical errors in mole fraction measurement in bottom stream. - Presence of entrainment or leaks mixing streams. - Non-ideal behavior or azeotrope effects not considered. - Losses or evaporation of water from bottom stream. - Assumed constant physical properties may be inaccurate.