5 replies to this topic

[Srinivas Rajasekhar]

Dear Members,

We have a Toluene + Menthol feed containing 20%(w/w) of menthol.

Here, the mixture is boiled with 3 kg/cm2 steam through jacket in a vessel at atmospheric pressure.

And when main toulene fraction is removed at 127C. Steam is injected directly into the vessel through a 25NB dip pipe (keeping the jacket steam intact).

 

Here are the observations,

1. The menthol is actually getting boiled @ 141C itself. And its boiling point is 216C. And its latent heat is 87 kcal/kg.

2. How is the menthol getting boiled below its boiling point?

3. Since, steam is losing its latent heat directly to menthol. Is it getting boiled? As enough energy is being given.

 

Kindly, throw some light on the above as I would line to understand the way direct steam injection actually works thermodynamically.

 

[breizh]

Hi ,

To answer to your question , please consider the document attached . More using your favorite search engine ;key word : steam distillation

 

hope this is helping you.

 

Breizh

[Art Montemayor]

rgmohite:

 

What Srinivas is describing (although very inaccurately) is a mixture of MENTHOL (not methyl alcohol - or methanol) + Toluene.

[rgmohite]

@Art Montemayor

 

Thank you for correction

[Art Montemayor]

Srinivas,

 

Kindly, throw some light on what you are doing.  Here are some basic facts about the compounds you are dealing with:

• Toluene has an atmospheric boiling point of 111 ^oC  (232 ^oF)
• Menthol is a waxy, crystalline substance, clear or white in color, which is solid at room temperature and melts at 36-38 ^oC (97-100 ^oF).  It boils at 212 ^oC (414 ^oF) under atmospheric pressure.  It is only slightly soluble in water.  Thermo properties are scarce or non-existent.

Several things are obvious - whether you tell us or not: you are trying to recover the toluene by boiling it out of the mixture and condensing it with an overheads condenser.  The residual menthol will remain in the jacketed vessel that you are heating through a jacket containing saturated steam at 3 kg/cm² (42.67 psi).  You fail to tell us if that is absolute or gauge pressure.  I have to assume that it is GAUGE pressure and the jacket steam is kept at 57.4 psia and 143 ^oC (289 ^oF).

 

You don’t state it, but it seems you are carrying out a Batch Distillation.  Isn’t that correct?  In other words, refer to the attached sketch and confirm this.  You also seem to be sparging in steam (at God only knows what conditions) and using the effect of partial pressures (not any thermodynamics!) as Breizh has explained in his submitted document.

 

Please learn to be accurate and detailed in what you are attempting to explain.  If this is a problem given by your instructors for you to resolve, fine.  But when you describe it to others, you are leaving out a lot of needed basic data.  For example:

• You are going to be able to evaporate (“boil”) out the associated Toluene with the steam pressure you have at 143 ^oC.
• You will also be able to melt the residual menthol left behind with the same steam temperature.
• You will not be able to evaporate the resulting liquid menthol with the same jacket steam.  Therefore you resort to sparging in the steam at conditions that you fail to identify.  This is bad, bad engineering - and even worse communicating.  Therefore it is up to YOU to throw light on the subject, not us.

Some critical comments about your “observations”:

1. You can’t “boil” menthol using the steam you have.  Where did you get the latent heat of menthol?
2. You are not boiling any menthol.
3. No, the menthol is not getting “boiled”.

 

 Toluene + Menthol Mixture.xlsx   12.87KB   14 downloads

[Deepinc]

Dear Shrinivas ,

 

I think with direct steam distillation it is expected that BP of both compounds will come down. Why dont you try vacuum distillation to remove toluene. This will reduce temp and menthol carryover can be avoided.