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Water Trapout In Stripper Column Of Fcc Unit

fccu component trapping in distill water trapout tray

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#1 SANDEEPCH

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Posted 16 June 2018 - 01:18 AM

I am working in FCCU unit where Gas-Con section Pressure is around 15 kg/cm2. In stripper column we have water trapout tray. We usually maintain stripper top temperature around 60o(60 degree Celsius) and bottom temperature is around 125 o(125 degree Celsius). I was wondering why there is need of water trapout tray as at 15 kg/cm2 water boiling point is around 195o? In any case water should go into bottom product. But we find water accumulation in water trapout tray. I could never understand this phenomena. Please explain this.
 
Thanks & Regards !
Sandeep


#2 Bobby Strain

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Posted 16 June 2018 - 10:59 AM

Surely you can apply your own knowledge to understand what's happening in the stripper.

 

Bobby



#3 Saml

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Posted 16 June 2018 - 03:41 PM

Review the VLLE concepts in thermodynamics (vapor-liquid-liquid equilibrium, or two liquid phases in equilibrium with vapor). Also the concepts of  distillation with azeotropes. if you still have doubts, please ask. Let us know if that clarifies what you are looking for. 

 

Smith & Van Ness "Introduction to Chemical Engineering Thermodynamics" has a good description.



#4 SANDEEPCH

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Posted 21 June 2018 - 10:32 AM

@Saml, Thanks for giving me references. I have gone through that topic but not in depth. May be the whole chapter reading may work. Sorry but I could not linked the concept thoroughly with the stripper. But I understand that I need to go in deep to co-relate. I will let you know once it will become very clear to me.

 

Thanks !



#5 Saml

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Posted 21 June 2018 - 06:33 PM

Sandeepch, the basic concept is that when you have two inmiscibles liquid phases, the total vapor pressure is the sum of the vapor pressure of both phases at that temperature. So, inmiscibles phases form a minimum temperature azeotrope. That is why you have water at the top.



#6 SANDEEPCH

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Posted 22 June 2018 - 12:53 AM

@Saml, Thank you very much for simplifying the thing. I will ponder upon it but before that I got confused in basics. Please clarify it if you get time.

 

In steam stripper, steam reduces partial pressure of hydrocarbons. Hence hydrocarbons vaporizes so that partial pressure becomes its vapor pressure. What does it mean that in a mixture liquid will vaporize till that its partial pressure becomes its vapor pressure. Is it always valid? Or this is valid because water and steam is  inmiscibles? 

 

​So for miscible liquids partial pressure of component may not be equal to vapor pressure of component and still be in equilibrium?  Am I correct?

 

Thanks & Regards!  



#7 SANDEEPCH

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Posted 22 June 2018 - 12:43 PM

@Saml, Now I think I have understand the concept. My understandings are outlined below.

 

Since water and hydrocarbon form immiscible liquid mixture (Assume completely immiscible). Corresponding temperature-composition diagram can be given by this.

 

Suppose in stripper column mole fraction of water is very less. At low temperature Water and hydrocarbon both remain in liquid phase and when temperature increases at lower tower then water and hydrocarbon both will vaporise and pressure exerted by both will be equal to their corresponding vapor pressure. when the sum of theses vapor pressure equal to the column pressure there these 3 phases will be in equilibrium. I think at that position water trap-out tray should be there.

Below that tray temperature would be even high. so all the water will remain in vapor phase as given in this diagram and only hdrocarbon will be in liquid phase. (Because mole fraction of hydrocarbon is much high hence all the water will vaporize at equilibrium temp according to this diagram).

 

In other words, water will remain in vapor phase even at much lower temperature than its boiling point. 

 

Thanks & Regards!

Sandeep



#8 Saml

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Posted 22 June 2018 - 12:57 PM

The drawing is missing. However, what you wrote makes sense. 






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