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Che 101 Lab Report

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ChE 101 Lab Exercise 4 05 December 2012
Materials Balance (Non-reacting Systems) THR/WRU

1. A 40-mol-% aqueous ethanol solution is to be distilled into two streams: a distillate with 98% ethanol and a side stream with 60%. 98% of the alcohol in the feed is to be recovered in these two products; only 2% goes into the bottoms. If the bottoms product contains 2% ethanol, calculate the amount of the different streams per 1000 kmol of feed stream.

Given: Recovery = 98% of ethanol in feed to distillate and side stream

Feed, F xF = 0.40
Side stream, S xS = 0.60
Distillate, D xD = 0.98
Bottoms, B xB = 0.02

Required: D, S and B

Solution:

2% of the ethanol in the feed goes to the bottoms

Overall material …show more content…

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2. A seed crystal of diameter D (mm) is placed in a solution of dissolved salt, and new crystals are observed to nucleate (form) at a constant rate r (crystals/min). Experiments with seed crystals of different sizes show that the rate of nucleation varies with the seed crystal diameter as

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b. Calculate the crystal nucleation rate in crystals/s corresponding to a crystal diameter of 0.050 inch.

c. Derive a formula for r(crystals/s) in terms of D(inches). Check the formula using the result of part (b).

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