Distillation material balance

[_Greg_]

Homework Statement

Question:

A mixture of 40/60 mole% pentane/heptane is to be separated to give a top product of 95 mole% pentane and a bottom product of 5 mole% pentane.

The column will operate at 1 bar witha reflux ratio of 0.5 kmol/kmol. It will have a total condenser and a partial reboiler. The feed is liquid at its bubble point.

Solution:

Taking a basis of 100 kmol feed:

Kmol-------Feed-----D---------B

pentane----40.0-----37.0----3.1
heptane----60.0-----1.9--- -58.0
total-------100------38.9----61.1

(D = distillate, B = bottoms product)

The Attempt at a Solution

I don't understand these numbers or how there derived??

I would have said, 95% of the 40 kmol pentane in the feed would be in the distillate, which would be 38 kmol.
And 5 % of the 40 kmol would be in the bottom, which is 2 kmol.

What am I doing wrong?

Thanks for looking

 

[KataKoniK]

Dear fellow scientist,

Thank you for your question. It seems like there may be some confusion about how the numbers in the solution were derived. Let me explain the process in more detail.

First, we know that the feed is a mixture of 40 mole% pentane and 60 mole% heptane. This means that out of every 100 moles of feed, 40 moles are pentane and 60 moles are heptane.

Next, we need to determine the composition of the distillate and bottoms products. The problem states that the top product should be 95 mole% pentane and the bottom product should be 5 mole% pentane. This means that out of every 100 moles of top product, 95 moles are pentane and 5 moles are heptane. Similarly, out of every 100 moles of bottom product, 5 moles are pentane and 95 moles are heptane.

Now, we can use these percentages to calculate the actual amounts of pentane and heptane in each product. For the distillate, we know that it is 95% pentane. This means that out of every 100 moles of distillate, 95 moles are pentane. Since we have a total of 40 moles of pentane in the feed, we can calculate that the distillate will contain 95% of 40 moles, which is 38 moles of pentane.

Similarly, for the bottoms product, we know that it is 5% pentane. This means that out of every 100 moles of bottoms product, 5 moles are pentane. Since we have a total of 40 moles of pentane in the feed, we can calculate that the bottoms product will contain 5% of 40 moles, which is 2 moles of pentane.

I hope this helps to clarify the solution for you. If you have any further questions, please do not hesitate to ask. Keep up the great work in your scientific studies!

 

[Mene]

The numbers provided in the solution are based on the material balance equation for distillation, which is:

F = D + B

Where F is the total feed, D is the distillate, and B is the bottoms product. In this case, the total feed is 100 kmol, so the equation becomes:

100 kmol = D + B

The solution then calculates the amount of pentane and heptane in the feed, distillate, and bottoms product based on the given mole percentages and reflux ratio. The numbers may seem confusing, but they are based on the principle of material balance and the given conditions of the distillation column. Your approach of directly calculating the amount of pentane in the distillate and bottoms product is also correct, but it is not taking into account the reflux ratio and the other components in the feed. Both approaches will give the same result, but the solution provided is a more comprehensive and accurate representation of the material balance in the distillation process.