Steam Calculation for Methanol Distillat

Cheatsheet Content

Batch Distillation: Methanol-Water System Objective: Calculate steam required for batch distillation of methanol to 99% purity, starting from a feed with 95% methanol, 5% low boiler, and 25% moisture. 1. Input Data and Assumptions Working Volume of Vessel: $10 \text{ kL} = 10 \text{ m}^3$ Feed Composition: Methanol: $95\%$ (on dry basis) Low Boiler (LB): $5\%$ (on dry basis) Moisture Content: $25\%$ (of total feed) Product Purity Desired: $99\%$ Methanol Collection Rate: $500 \text{ kg/hr}$ pure methanol Reflux Rate: $200 \text{ kg/hr}$ Steam Pressure: $3.5 \text{ kg/cm}^2$ gauge (LPS - Low Pressure Steam) Initial Feed Temperature: Room Temperature (assume $25^\circ\text{C}$) Distillation Temperature: Assume boiling point of methanol-water mixture. For simplicity, we'll generally use the boiling point of pure methanol for condensation ($64.7^\circ\text{C}$ at atmospheric pressure) as the product is nearly pure. For reboiler, assume slightly higher due to water. Specific Heat Capacity (Average): Methanol ($C_{p,MeOH}$): $2.5 \text{ kJ/(kg}^\circ\text{C)}$ Water ($C_{p,H_2O}$): $4.18 \text{ kJ/(kg}^\circ\text{C)}$ Vaporization Latent Heat of Methanol ($\lambda_{MeOH}$): $1100 \text{ kJ/kg}$ (at boiling point) Latent Heat of Vaporization of Steam ($\lambda_{steam}$): Approx. $2163 \text{ kJ/kg}$ at $3.5 \text{ kg/cm}^2$ gauge (saturated steam temperature $\approx 148.5^\circ\text{C}$) Density of Feed: Assume average density of methanol-water mixture $\approx 0.9 \text{ kg/L} = 900 \text{ kg/m}^3$ 2. Calculate Total Feed Mass Total Feed Mass = Working Volume $\times$ Density Total Feed Mass = $10 \text{ m}^3 \times 900 \text{ kg/m}^3 = 9000 \text{ kg}$ 3. Calculate Feed Composition (Actual Mass) Let $M_{total}$ be total mass of feed. $M_{water}$ be mass of water, $M_{dry}$ be mass of dry components. Moisture Content = $25\%$ $M_{water} = 0.25 \times M_{total} = 0.25 \times 9000 \text{ kg} = 2250 \text{ kg}$ $M_{dry} = M_{total} - M_{water} = 9000 \text{ kg} - 2250 \text{ kg} = 6750 \text{ kg}$ Dry Basis Composition: Methanol: $95\%$ of $M_{dry}$ $M_{MeOH,feed} = 0.95 \times 6750 \text{ kg} = 6412.5 \text{ kg}$ Low Boiler: $5\%$ of $M_{dry}$ $M_{LB,feed} = 0.05 \times 6750 \text{ kg} = 337.5 \text{ kg}$ Total Feed Composition: Methanol: $6412.5 \text{ kg}$ Low Boiler: $337.5 \text{ kg}$ Water: $2250 \text{ kg}$ Total: $6412.5 + 337.5 + 2250 = 9000 \text{ kg}$ (Checks out) 4. Energy Balance for Distillation The total heat duty ($Q_{total}$) required for distillation consists of: $Q_{sensible}$: Heat to raise the feed temperature to its boiling point. $Q_{vaporization}$: Heat to vaporize the desired components. $Q_{reflux}$: Heat removed by condenser and returned as reflux. 4.1. Sensible Heat ($Q_{sensible}$) Assuming average specific heat capacity for the mixture. Average $C_p$ for feed: $(6412.5 \times 2.5 + 2250 \times 4.18 + 337.5 \times C_{p,LB}) / 9000$ For simplicity, we can use a weighted average or approximate $C_p \approx 3.0 \text{ kJ/(kg}^\circ\text{C)}$ (between methanol and water). Boiling point of methanol-water mixture (initial): Approx. $80^\circ\text{C}$ (varies with composition, but for initial heating, use an average). Temperature difference $\Delta T = 80^\circ\text{C} - 25^\circ\text{C} = 55^\circ\text{C}$ $Q_{sensible} = M_{total} \times C_{p,avg} \times \Delta T$ $Q_{sensible} = 9000 \text{ kg} \times 3.0 \text{ kJ/(kg}^\circ\text{C)} \times 55^\circ\text{C} = 1,485,000 \text{ kJ}$ 4.2. Heat for Vaporization ($Q_{vaporization}$) This is the heat required to vaporize the methanol and low boiler content that will be distilled. We need to distill most of the methanol. Target Methanol Purity: $99\%$ Total Methanol in feed: $6412.5 \text{ kg}$ Assume we distill $95\%$ of the initial methanol to achieve product purity. Methanol to be vaporized $\approx 0.95 \times 6412.5 \text{ kg} = 6091.875 \text{ kg}$ Low Boiler to be vaporized: Assume most of the low boiler will distill with methanol, so $337.5 \text{ kg}$. Total mass to be vaporized ($\approx M_{vap}$) = $6091.875 \text{ kg (MeOH)} + 337.5 \text{ kg (LB)} \approx 6429.375 \text{ kg}$ $Q_{vaporization} = M_{vap} \times \lambda_{MeOH}$ (using methanol's latent heat as dominant) $Q_{vaporization} = 6429.375 \text{ kg} \times 1100 \text{ kJ/kg} = 7,072,312.5 \text{ kJ}$ 4.3. Heat for Reflux ($Q_{reflux}$) The reflux stream is condensed vapor that is returned to the column. The heat removed in the condenser for reflux needs to be supplied by the reboiler to re-vaporize it. Reflux Rate: $200 \text{ kg/hr}$ Collection Rate (Distillate): $500 \text{ kg/hr}$ Vaporization Rate from Reboiler ($V$) = Distillate Rate ($D$) + Reflux Rate ($R$) $V = 500 \text{ kg/hr} + 200 \text{ kg/hr} = 700 \text{ kg/hr}$ This $V$ is the actual amount of vapor generated per hour. Heat required for this hourly vaporization: $Q_{reboiler, hourly} = V \times \lambda_{MeOH}$ $Q_{reboiler, hourly} = 700 \text{ kg/hr} \times 1100 \text{ kJ/kg} = 770,000 \text{ kJ/hr}$ 5. Calculate Total Distillation Time Total Methanol to be collected ($M_{MeOH,prod}$) = $6091.875 \text{ kg}$ (from step 4.2) Collection Rate: $500 \text{ kg/hr}$ Distillation Time = $M_{MeOH,prod} / \text{Collection Rate}$ Distillation Time = $6091.875 \text{ kg} / 500 \text{ kg/hr} \approx 12.18 \text{ hours}$ 6. Total Heat Duty from Reboiler The total heat supplied by the reboiler over the entire batch operation is the sum of sensible heat and the total heat of vaporization for the distillate and reflux over the batch duration. Total Heat of Vaporization over batch = Distillation Time $\times Q_{reboiler, hourly}$ Total Heat of Vaporization = $12.18 \text{ hours} \times 770,000 \text{ kJ/hr} = 9,378,600 \text{ kJ}$ Total Heat Duty ($Q_{total}$) = $Q_{sensible} + \text{Total Heat of Vaporization}$ $Q_{total} = 1,485,000 \text{ kJ} + 9,378,600 \text{ kJ} = 10,863,600 \text{ kJ}$ 7. Calculate Steam Required Steam supplies the latent heat for the process. Latent Heat of Steam ($\lambda_{steam}$): $2163 \text{ kJ/kg}$ (at $3.5 \text{ kg/cm}^2$ gauge) Total Steam Required ($M_{steam}$) = $Q_{total} / \lambda_{steam}$ $M_{steam} = 10,863,600 \text{ kJ} / 2163 \text{ kJ/kg} \approx 5022.47 \text{ kg}$ 8. Calculate Average Hourly Steam Requirement Average Steam Rate = $M_{steam} / \text{Distillation Time}$ Average Steam Rate = $5022.47 \text{ kg} / 12.18 \text{ hours} \approx 412.35 \text{ kg/hr}$ Summary of Results Total Feed Mass: $9000 \text{ kg}$ Total Methanol in Feed: $6412.5 \text{ kg}$ Estimated Distillation Time: $12.18 \text{ hours}$ Total Steam Required for the Batch: $5022.47 \text{ kg}$ Average Steam Flow Rate: $412.35 \text{ kg/hr}$ Important Considerations and Notes This calculation assumes ideal conditions and constant latent heats. Heat losses to the surroundings were not accounted for, which would increase steam consumption. The specific heat and latent heat values are approximate and depend on the exact composition and temperature. The boiling point of the mixture changes during batch distillation, affecting sensible heat and latent heat slightly. The "low boiler" component's properties were assumed similar to methanol for simplicity. This calculation provides an estimate. Actual steam consumption would require more detailed VLE (Vapor-Liquid Equilibrium) data and a rigorous process simulation. A heat-up phase at the beginning of the batch will have higher steam demand than the steady-state distillation phase. The calculated average rate smooths this out.