1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Psychrometrics - Pure Humidification Process

  1. Sep 23, 2015 #1
    In psychrometrics it is outlined time and again that we cannot achieve a pure humidification process i.e, add only moisture to unsaturated air without a change in its Dry Bulb Temperature(DBT).While pondering over the idea I came up with a possible way of doing it (if it is correct at all) and is explained below:

    Let us take liquid water in a piston cylinder device at 25°C under 1atm or 101.325KPa pressure.We gradually reduce the pressure to 3.17KPa which is the saturation pressure of water at 25°C so that the water is now saturated liquid under prevailing conditions.Now keeping pressure constant at current value we heat the water until it is transformed to saturated vapor at 25°C under 3.17KPa.Now we take an adiabatic chamber with two compartments one filled with unsaturated air at 25°C and 1 atm pressure and the other with this saturated vapor at 25°C and 3.17KPa.Next we remove the partition separating the compartments so that the unsaturated air and saturated water vapor undergo direct mixing.

    Since both air and water vapor are at same temperatures there is no space for sensible heat transfer between them and ideally the temperature of the resulting mixture should be at the same temperature of 25°C.

    A point to be noted lies in the fact that there could be a drop in temperature due to inter molecular Van Der Walls forces but again dry air and water vapor both behave as ideal gases in the range of temperatures in most air conditioning applications(-10°C to 50°C) and the effect of Van Der Walls forces should be negligible.

    So would the idea work as anticipated?
     
  2. jcsd
  3. Sep 25, 2015 #2

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    No. You are expanding dry air at 1 atm. against water vapor at lower pressure (cooling), and compressing water vapor at low pressure (heating).
     
  4. Sep 28, 2015 #3
    After some real brainstorming I feel the need to clarify the following points:

    1. When you say dry air at 1 atm is expanded against water vapor at lower pressure I assume you are neglecting the contribution of the vapor pressure in the air and thus taking the partial pressure of dry air to be 1 atm.

    2. Again when you say dry air at 1atm when expanded against water vapor at a lower pressure results in cooling of dry air,are you assuming dry air undergoing an adiabatic expansion against water vapor?

    In reality as soon as the temperature of dry air would tend to fall a heat transfer from water vapor to dry air would tend to increase its temperature.

    3. When you say water vapor is compressed you are certainly assuming that the specific volume of water vapor decreases.If the moist air was initially unsaturated at the temperature of the water vapor in the other compartment the specific volume of water vapor after mixing should actually increase.It would remain the same if the moist air was saturated at the temperature of the water vapor in the other compartment.Thinking in a practical way the water vapor at the other compartment would combine with the water vapor present in moist air after the partition is removed to occupy the entire volume of the chamber.Since for unsaturated air the specific volume of water vapor in the air is larger than the specific volume of water vapor at the other compartment(before mixing) the specific volume of water vapor after mixing should be larger than the initial specific volume of saturated water vapor existing alone at one of the compartments.Thus, water vapor would actually expand in such a case.

    Then how is water vapor compressed after mixing has been achieved?
     
  5. Sep 28, 2015 #4

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You've established a volume at a pressure of
    and a second volume of air. When the partition is removed, the air expands into the low pressure volume; the two volumes are at thermal, but not mechanical equilibrium at the initiation of the "mixing" process you're describing. The gas/vapor in the lower pressure volume/chamber is compressed.
     
  6. Sep 29, 2015 #5
    Well I am still confused about this.Let's say if there was a movable piston as the partition with stops initially and the stops are removed the vapor at one of the compartments(P=3.17KPa) would get compressed and the piston would move to decrease the specific volume of the vapor as air at a higher pressure(P= 1 atm) expands against the vapor at a lower pressure.

    But if we have direct mixing instead wouldn't the vapor at 3.17Kpa rush to fill the entire volume of the chamber?If it were compressed then its specific volume would have decreased but after the partition is removed vapor should fill the entire volume of the chamber i.e, now vapor is occupying a greater volume and hence specific volume should increase.Of course both dry air and water vapor at the other compartment also fills the chamber and expands likewise.So if the vapor occupies a greater volume after mixing is completed (so that its specific volume increases) how could we say it was compressed in this case?
     
  7. Oct 1, 2015 #6

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Is it going to diffuse as rapidly as dry air expands? I'm hoping to avoid us having to go through construction and evaluation of the whole process step by painful thermodynamic step.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Psychrometrics - Pure Humidification Process
  1. Reversible Processes? (Replies: 10)

  2. Polytropic processes (Replies: 2)

  3. Isobaric processes (Replies: 2)

Loading...