The product from a reactor contains 60%w/w A, 25%w/w B and 15%w/w C. The final s
ID: 1817070 • Letter: T
Question
The product from a reactor contains 60%w/w A, 25%w/w B and 15%w/w C. The final sealable product is required to contain 90% A.
The mixture can be purified in a distillation column but under the conditions possible in the column the product produced is 100% A with the bottom product containing no A. It is possible to mix the reactor products and distillation product.
Calculate:
(i)
the amount of reactor product that could by-pass the distillation system to give maximum amount of sealable product at minimum cost.
(ii)
the amount and composition of 'waste ' material.
Explanation / Answer
Original situation/reactor product: 60% of A 25% of B 15% of C Addition of situation 2: 100% A Final/ situation 3: 90% A We need to solve for x as it is our variable amount for the reactor product originally, it contains 0.6 x of A mix with 3x amount of 100% A (situation 2) results in 3.6x amount of 90% A and .4x amount of waste (B and C) The answer for part (i) is : If you need V gallons (for example) per minute of 90% A, then you need x= V/3.6 of reactor product. in the waste composition part (ii): the amount is .4 x = .4 * V/3.6 = V/9 the composition is 25%/40% = 62.5% of B 15%/40% = 37.5% of C
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