Fermentation product from a brewery contains 10% ethanol and 90% water. Part of the fermentation product (50,000 kg/h) is pumped to a distillation column on the factory site. Under current operating conditions, a distillate of 45% ethanol and 55% water is produced from the top of the column at a rate of one-tenth that of the feed.What is the composition of the waste "bottoms" from the distillation column.Question 5Select one:

To solve this problem, we need to use the principle of mass balance.

Step 1: Calculate the total amount of ethanol and water in the feed.

The feed contains 10% ethanol and 90% water. So, in 50,000 kg of feed, there are 5,000 kg of ethanol (10% of 50,000 kg) and 45,000 kg of water (90% of 50,000 kg).

Step 2: Calculate the total amount of ethanol and water in the distillate.

The distillate is produced at a rate of one-tenth that of the feed, so it's 5,000 kg/h (one-tenth of 50,000 kg/h). It contains 45% ethanol and 55% water. So, in the distillate, there are 2,250 kg of ethanol (45% of 5,000 kg) and 2,750 kg of water (55% of 5,000 kg).

Step 3: Calculate the total amount of ethanol and water in the waste "bottoms".

The total amount of ethanol in the waste "bottoms" is the total amount of ethanol in the feed minus the total amount of ethanol in the distillate. So, it's 5,000 kg - 2,250 kg = 2,750 kg.

Similarly, the total amount of water in the waste "bottoms" is the total amount of water in the feed minus the total amount of water in the distillate. So, it's 45,000 kg - 2,750 kg = 42,250 kg.

Step 4: Calculate the composition of the waste "bottoms".

The total weight of the waste "bottoms" is the sum of the weight of ethanol and water in it, which is 2,750 kg + 42,250 kg = 45,000 kg.

So, the composition of the waste "bottoms" is 2,750 kg of ethanol and 42,250 kg of water. In percentage terms, it's (2,750/45,000)*100 = 6.11% ethanol and (42,250/45,000)*100 = 93.89% water.