Packed tower problem (gas-liquid absorption)

[SWLABR]

Hi everyone,

I'm having a bit of difficulty with a packed tower problem I'm working on. Not too sure how things work around here so I'll just jump straight in...

The problem is;

A packed column continuously recovers acetone (component A) from air by absorption with water at 60 deg. F. The air contains 3 mole% acetone, where 97% recovery is desired. The gas flow rate is 50 ft^3/min at 60 deg. F, 1 atm. The maximum allowed gas superficial velocity in the column is 2.4 ft/sec.

>> I can assume that in the range of operation, Y* = 1.75X, where Y and X are mole ratios (acetone to pure carrier).

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My question is;

How would I go about determining the minimum water-to-air flow rate ratio? I know I have to start by finding the maximum acetone concentration possible in the aqueous solution but I'm having an absolute mare trying to figure this out.

Any help on this would be greatly appreciated.

 

[CaptainPanic]

First, I am not sure what you mean with the Y*=1.75X. Is that an equilibrium relation? Which phase is Y, which phase is X? One is probably the water phase, one is the air phase. Convention is that X is liquid, Y is gas... so I'll go with that, although it should be specified next time to make sure we're not wasting our time here.

Second, you need to specify whether you are working in countercurrent or co-current operation. Packed columns are mostly operated in countercurrent, but the other is not impossible.

Now, let's assume that the Y*=1.75X equation means: "the molar fraction of acetone in the gas (air) is 1.75 times that of the molar fraction of acetone in water". Let's also assume that we have countercurrent operation (meaning that the gas goes up, water goes down in the column).

This means that the water can theoretically be in equilibrium with the incoming ("fresh") gas which is still full of acetone. After all, the water already traveled all the way through the column so it is pretty much loaded with acetone when it reaches the bottom... and in the last little bit of the tower it finds the fresh air which contains even more acetone than any air encountered in the column so far, and it just absorbs that last bit. You know the air concentration, so you know the air molar fraction. So, you know the maximum molar fraction in water. That is your theoretical minimum.

In reality, you will not reach complete equilibrium, so you must use more water than you just calculated.

Finally, I would like to encourage you to convert all units into SI units when posting. That means: distances in meters, temperatures in either Kelvin or Celcius, time in seconds (although minutes and hours are accepted). I'll immediately admit that I am not certain if this forum has any specific rules about it (I might find myself on an American forum, in which case I should shut my European mouth ).

 

[SWLABR]

Thanks for the reply CaptainPanic, I forgot about posting this here. I have since sorted this problem, but your reply is much appreciated.

It was a countercurrent problem I was dealing with (gas enters through bottom, liquid enters through top). And yes, Y* = 1.75X is as you say "the molar fraction of acetone in the gas (air) is 1.75 times that of the molar fraction of acetone in water".

Sorry about the units, I too am not American (I'm from NZ) but this was an example problem from an American fellow we had as a lecturer so every example we were given, a good deal of time was wasted converting to SI.

Cheers.

 

[mela]

Hi am designing a 50ton methanol plant and need to know what equipment wil i use to remove acetone from the crude methanol

 

[mela]

what equipment will i use to remove acetone from crude methanol

 

[mela]

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[mela]

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