5 replies to this topic

[Abm]

I Am currently working in a Instrument air plant and I Would Like To Calculate The Moisture Absorbed in the Bed. Inlet Pressure is 7.2Ksc and Outlet 6.8Ksc. Drier Outlet Dewpoint Analyser Shows 24c. Kindly Guide Me How To Calculate How much Moisure Adsorbed in Bed.

[Art Montemayor]

If you are a student it is wise to note the following for your education:

• If you are using an adsorbent dryer, then the water moisture is ADSORBED - NOT ABSORBED.  There is a BIG difference!
• If you are using sodium hydroxide pellets as the drying agent, then you could use the term absorbed.
• This is the first time I've heard of the pressure unit "Ksc" in over 45 years!  I've not heard it again, nor read it anywhere.  It is considered a "deprecated" form by some of expressing kilograms/cm² .   Why don't you use SI units - after all the money and effort put into establishing it?
• What do you mean by "45c"???  Do you mean +45 degrees Celsius, or +45 ^oC???  If so, that is probably a dew point reading (but at what pressure base?) and should be converted to conventional kg/cubic meter of air (at some indicated temperature and pressure) or lb/ft³ (under the same conditions).  Whoever supplied you with the dryer you are using should have also supplied you with a conversion table.

 

If you are dealing with instrument air drying, you should do your homework in becoming totally aware and knowledgeable with the unit operation and its operations.  This is all treated in Unit Operations courses and texts. 

[Saml]

I was writing this on your duplicates as Art was writing his answer....

 

Water adsorbed:

Assume that inlet is saturated at last separator pressure and temperature. You have the water partial pressure at inlet.

Water partial pressure at outlet is the one that correspond to saturation at Dew Point reference pressure and Dew Point Temperature

Having air flow and water partial pressure at inlet and outlet, the rest is just calculating. 

 

Now, if you want to determine how much water is already in the adsorption beds you need to know the adsorption isotherms for the specific adsorbent you are using. 

 

Notes:

Be sure of what you are measuring. It is Pressure Dew Point or Atmospheric Dew Point?

24°C is a very high dew point for instrument air. Are you sure that is not minus 24°C?

[Abm]

ThankYou For The Response.
Saml It is Pressure Dew Point and the Dewpoint is Minus 24 Degree Celsius.
I Dont Have The Exact Composition of the Drier Inlet Air and thereby Not Having water partial Pressure At the Inlet.

Kindly Guide Me How To Calculate Water partial Pressure in Relation to Dewpoint Reference Pressure and Temperature.

Drier Inlet Pressure is 7.2Kg/cm2
Inlet Temperature is 42 Degree Celsius
Drier Outlet Pressure is 6.8Kg/cm2
Outlet Temperatute is 38 Degree Celsius
And Oulet Dewpoint is Minus 24 Degree Celsius

[Abm]

Dear Montemayor

The Drier Is Comissioned Before 15Years and I Dont Have The Dewpoint Conversion Table. Kindly Share Any Sample Conversion Tables If any For Reference.

Dewpoint Minus 24 Degreecelsius at 6.8Kg/Cm2.

[Saml]

Unless you are in an extremely dry or cold environment (see *), after the last compression stage you will have an aftercooler and a moisture drain. The air going to the dryers is usually saturated at that temperature.

 

So in your case is 42°C the water pressure will be 8200 Pa. (*) If, as an example, the ambient temperature is below 10°C,  or it is 20°C with less than 50% relative humidity, you would not have enough water to condense at 6.8 kg/cm2-g however.

 

I don't have a water/steam table that go down to the ice-vapor equilibrium. So for -24°C exactly you will have to calculate it, it is around 70 Pa. So 99% of the water is removed.

 

You have a reference here as to  a more precise calculation of the vapor pressure for sub-zero temperatures.

http://www.kayelaby..../3_4/3_4_1.html