Five hundred kmol/h of a liquid mixture of light alcohols containing, by moles, 40% methanol (M), 35%
Question:
Five hundred kmol/h of a liquid mixture of light alcohols containing, by moles, 40% methanol (M), 35% ethanol (E), 15% isopropanol (IP), and 10% normal propanol (NP) is distilled in a sequence of two distillation columns. The distillate from the first column is 98% pure M with a 96% recovery of M. The distillate from the second column is 92% pure E with a 95% recovery of E from the process feed. Assume no propanols in the distillate from Column C1, no M in the bottoms from Column C2. and no normal propanol in the distillate from Column C2.(a) By material balances. assuming negligible propanols in the distillate from the first column, compute the flow rates in kmol/h of each component in each feed, distillate, and bottoms. Draw a labeled block-flow diagram like Figure. Include the results of the material balances in a table like Table 1.5 and place the table below your block-flow diagram.(b) Compute the mole-percent purity of the propanol mixture leaving as bottoms from the second column in the sequence.(c) If the recovery of ethanol is fixed at 95%, what is the maximum purity that can be achieved for the ethanol in the distillate from the second column?(d) If instead, the purity of the ethanol is fixed at 92%, what is the maximum recovery of ethanol (based on the process feed) that can beachieved?
Step by Step Answer:
a M in distillate from C1 096500040 192 kmolh Total distillate from C1 192098 19592 kmolh E in disti...View the full answer
