Jump to content



Featured Articles

Check out the latest featured articles.

File Library

Check out the latest downloads available in the File Library.

New Article

Product Viscosity vs. Shear

Featured File

Vertical Tank Selection

New Blog Entry

Low Flow in Pipes- posted in Ankur's blog

2

Selecting Property Package For Water And Air Mixture In Aspen Hysys/pl


7 replies to this topic
Share this topic:
| More

#1 Sol.Daray

Sol.Daray

    Brand New Member

  • Members
  • 3 posts

Posted 02 October 2022 - 07:18 AM

Hello Every One,

 

I have a question regarding the property package selection.

I want to produce a PT Envelope for Water and Air Mixture in Aspen HYSYS or Aspen PLUS.

I don't know which property package  should be used.

 

Best Regards



#2 latexman

latexman

    Gold Member

  • Admin
  • 1,428 posts

Posted 02 October 2022 - 08:00 AM

Use the Methods Assistant in Aspen+. It’s in Properties. Water/air interactions are weak so I think you just need to select a method good for your pressures.

#3 Pilesar

Pilesar

    Gold Member

  • Members
  • 1,045 posts

Posted 02 October 2022 - 02:36 PM

Computer simulators allow changing parameters pretty easily. Generally, thermodynamic methods are either based on Liquid Activity Coefficients (e.g. NRTL, Wilson) or based on departures from the gas law using an Equation of State method (e.g. SRK, PR). Sometimes there is a strong interaction in the liquid phase due to component polarities or other causes where liquid activity methods are more suited. Otherwise, equation of state methods better handle the effects of pressure. Both types of methods are 'bendable' and often have extra parameters to allow either to model a system. For your system, I suggest generating your PT envelope and trying several thermodynamic methods to see if it makes any difference. You will not find 'air' in your chemical periodic table. The steady state simulator vendors often include it, but consider whether you should use true components N2, O2, etc, where you might have higher precision.



#4 Sol.Daray

Sol.Daray

    Brand New Member

  • Members
  • 3 posts

Posted 11 October 2022 - 03:09 AM

Computer simulators allow changing parameters pretty easily. Generally, thermodynamic methods are either based on Liquid Activity Coefficients (e.g. NRTL, Wilson) or based on departures from the gas law using an Equation of State method (e.g. SRK, PR). Sometimes there is a strong interaction in the liquid phase due to component polarities or other causes where liquid activity methods are more suited. Otherwise, equation of state methods better handle the effects of pressure. Both types of methods are 'bendable' and often have extra parameters to allow either to model a system. For your system, I suggest generating your PT envelope and trying several thermodynamic methods to see if it makes any difference. You will not find 'air' in your chemical periodic table. The steady state simulator vendors often include it, but consider whether you should use true components N2, O2, etc., where you might have higher precision.

Dear Pliesar,

Many thanks for your reply.

 

The issue is as follow:

On the water steam table, for 150 degree-centigrade, there is 4.758 bar for pure water. It needs a minimum of 4.758 bar pressure to have 150 Celsius of water in the liquid phase.
If we add air to the water and want to maintain the water in the liquid phase, we need more pressure in the vessel due to the air diffusion in the water. I want to know, for example, for a mixture of 1000 kg/hr.  Air and 4000 kg/hr. of Water at 150 Celsius, what is the minimum pressure to have the water in the liquid phase? This pressure and the selected temperature and flow rates fix the air's partial pressure in the gas phase and the solubility of the air in the liquid phase. I want to find the bubble point of this liquid phase and not the bubble point of the whole system (liquid phase=water and part of air transferred from gas phase to liquid water). This pressure will be used to control the Evaporation. I tried NRTL-SRK, SRK with and without Henry Component, SRK -KD. I think I am doing wrong and I don't know how to use Aspen to define the system. Would you please tell me how to do it?
 
Best Regards,
Daray


#5 latexman

latexman

    Gold Member

  • Admin
  • 1,428 posts

Posted 11 October 2022 - 07:12 AM

I quickly tried this in Aspen+ V12.1.  FEED = 5000 kg.hr, 25 C, 4.758 bar_a, in mass fractions water 0.8, N2 0.158, O2 0.042.  FEED into 2 phase flash with specification pressure = 4.758 bar_a and vapor fraction = 0.2.  The LIQUID from this flash was fed to a second 2 phase flash with specification pressure = 4.758 bar_a and vapor fraction = 0 (bubble point).  The second flash was not needed; don't put it in your flowsheet.  The liquid from the first flash is at the bubble point!  Using Base Method NRTL and Henry Components N2 and O2, the temperature of the first (and second) 2 phase flash was about 113 C @ 4.758 bar_a.

 

You should be able to recreate this.  Play around with the vapor fraction of the 2 phase flash to see sensitivity.  Change physical property methods and repeat.

 

Using component Air, instead of 79% N2 and 21% O2 was problematic in my simulation.  Using an exact temperature (150 C) AND pressure was problematic in my simulation.  The first 2 phase flash created all vapor of the 5000 kg/hr.  I switched from temperature spec to vapor fraction = 0.2 in the first 2 phase flash and the simulation gave results that are about what I think you are looking for.

 

I hope this helps.  I saved the simulation in case you have questions.



#6 Pilesar

Pilesar

    Gold Member

  • Members
  • 1,045 posts

Posted 11 October 2022 - 08:41 AM

Typically in a system of air and water, it is sufficient to calculate how much water will be in the vapor phase. There will always be some! There is water in the air we breathe (humidity) and the same situation will happen in your system. Because the solubility of inert gases in water is low, the amount of oxygen and nitrogen in the liquid phase is usually ignored. It can be predicted rigorously, but those calcs are not super-accurate anyway compared to real-life situations. If pure water is needed after it has been exposed to air, then it must be processed further (deaerated) to remove dissolved gases. The bubble point of the liquid phase is the bubble point of the water-air system. Once you determine the partial pressure of water in the vapor phase, you can look in the saturated steam tables to find the conditions of the liquid phase at that pressure. The dissolved gas in the water should not affect the results much. Your example of 1000 kg air and 4000 kg water is not real-world. No one would add that much air and try to keep all water from evaporating. This system will always have some water in the vapor phase.



#7 Sol.Daray

Sol.Daray

    Brand New Member

  • Members
  • 3 posts

Posted 16 October 2022 - 04:09 AM

Typically in a system of air and water, it is sufficient to calculate how much water will be in the vapor phase. There will always be some! There is water in the air we breathe (humidity) and the same situation will happen in your system. Because the solubility of inert gases in water is low, the amount of oxygen and nitrogen in the liquid phase is usually ignored. It can be predicted rigorously, but those calcs are not super-accurate anyway compared to real-life situations. If pure water is needed after it has been exposed to air, then it must be processed further (deaerated) to remove dissolved gases. The bubble point of the liquid phase is the bubble point of the water-air system. Once you determine the partial pressure of water in the vapor phase, you can look in the saturated steam tables to find the conditions of the liquid phase at that pressure. The dissolved gas in the water should not affect the results much. Your example of 1000 kg air and 4000 kg water is not real-world. No one would add that much air and try to keep all water from evaporating. This system will always have some water in the vapor phase.

Dear Pliesar,

Here, the reaction(oxidation by the Air)  takes place in the liquid phase(almost water).{Contains Sulfide ion (HS-) maximum 1.5%wt .sulfide will convert to thiosulfate , sulfite, sulfate}.The Temperature is fixed(150 to 175 degree centigrade).For these temperatures we have to calculate the minimum pressure in the reactor space to control the vaporization and not  allow the liquid to vaporize. Also, more pressure than this minimum will be required to have a more partial pressure of the oxygen in the vapor space.(more solubility of the oxygen in the liquid phase  will help the reaction conversion).

I will follow the Dear Latexman solution .I hope I can handle this issue( Calculate the minimum (bubble point pressure)  pressures for the relative the temperatures).

 

Best Regards,

Daray



#8 latexman

latexman

    Gold Member

  • Admin
  • 1,428 posts

Posted 16 October 2022 - 05:37 AM

Here, the reaction(oxidation by the Air)  takes place in the liquid phase(almost water).{Contains Sulfide ion (HS-) maximum 1.5%wt .sulfide will convert to thiosulfate , sulfite, sulfate}.

 

Based on this, I would modify my advice in Post # 2 as follows, "Use the Methods Assistant in Aspen+. It’s in Properties. Water/air interactions are weak so I think you just need to select a method good for your pressures, but more importantly now, be sure the method is good for electrolytes.






Similar Topics