Documents
Resources
Learning Center
Upload
Plans & pricing Sign in
Sign Out

Two-phase behavior

VIEWS: 10 PAGES: 30

									Two-phase behavior
       Behavior in the two
         phase region
                  log Pvp   A 
                                       B
                                     T C



Ln P                       Antoine’s
                           Equation




            Enthalpy
                 Antoine’s Equation
• Antoines equation is used to predict the relationship
  between the T and P in the two phase region. P is
  usually called saturation pressure and T is usually called
  the vaporization temperature or vice-versa
• For propane, the Antoines constants are

• A = 6.80398, B =803.81, C= 246.99
                      (Psat in mmHg)
• Calculate the vapor pressure for propane at 100, 200,
  300 K and the critical point 369.8 K. Compare to the
  values from the chart.
           Antoine’s Equation

Problem 6
The Antoine constants for n-butane are listed in a data base
as A = 6.80896, B = 935.86, and C = 238.73 where Psat is in
mm Hg:
where T is in oC, and P is in mmHg.

Predict the normal boiling point (temperature at which butane
boils at 1 atm pressure.)

Compare the predicted value to the value listed in the critical
properties table given earlier.
             Gibbs Free energy
The Gibbs free energy is especially useful for calculations
in the two phase region. Why?

          ~     ~      ~
         dG   S dT  VdP
 Since dT = 0 and dP = 0, Gliq = Gvap

The Gibbs free energy of the vapor and liquid in a two
phase system are equal. The criteria for equilibrium
between two phases is that G1 = G2.
               Relationship of V and S

        Along each edge of the envelope:
 ~       ~         ~                ~            ~  ~
dGliq  Sliq dT  Vliq dP dGvap  S vap dT  Vvap dP
                   Psat (liquid )  Psat ( vapor)
                    Tvap( liquid )  Tvap( vapor)
                    G( liquid )  G( vapor)
                 ~      ~               ~      ~
           0  (Sliq  S vap )dTvap  (Vliq  Vvap )dPsat
                  ~      ~               ~      ~
            0  (Sliq  S vap )dTvap  (Vliq  Vvap )dPsat

               ~      ~               ~      ~
              (Sliq  S vap )dTvap  (Vliq  Vvap )dPsat

                                ~       ~
                     Psat    ( S vap  S liq )
                              ~        ~
                     Tvap    (Vvap  Vliq )

The way Psat changes with Tvap is directly related to DSvap and DVvap
       Finding the slope of Psat
                vs Tvap



Psat
                          Psat
                                @T1
                          Tvap



              T1
                   Tvap
           Change in Psat with T
                     B                                     B
log Psat  A                                      A
                                                        Tvap  C
                 Tvap  C              Psat  10


                           A B  
                                       
      Psat                 Tvap  C         B
                  ln(10)10            

                                           Tvap  C 
      Tvap                                           2
                         
                                         
                               ~       ~
                    Psat    ( S vap  S liq )
                             ~        ~
                    Tvap    (Vvap  Vliq )
                Numerical (Easier)
• Estimating the slope of Psat vs Tvap at a
  particular T.

        Psat   Psat(T   )  Psat(T   )
              
        Tvap               2

        The smaller the value of , the more accurate the
        estimation.
                     DSvap
• Given the Antoine’s constants, one can
  calculate the way the saturation pressure
  changes with Tvap at any Tvap.
• One can use an EOS to estimate Vvap
  and Vliq at that temperature Tvap.(eg 10-
  24 PR)
• From the difference in Vvap and Vliq one
  can obtain Svap – Sliq (DSvap)
           Shortcut Approximations
• Since Vliq << Vvap, a shortcut is to find
  only Vvap
                          ~       ~
              Psat     ( S vap  S liq )
                             ~
              Tvap          (Vvap )
• Since Vvap ~ RT/P

                                     ~       ~
                 Tvap Psat        ( S vap  S liq )
                               
                 Psat Tvap                 R
              Calculation of DSvap
               from dPsat/dTvap
Problem 7

a. From Antoine’s equation and the P-R equation of state:
Compute the difference in the vapor and liquid specific volumes
    at butane’s normal boiling point and at 25 C.

b. Compute the difference in the entropies of liquid butane and
    butane vapor at each temperature using each of the 3
    methods described in the notes.
(Mathematically or numerically take the derivative of Antoine’s
    equation – pressure wrt temperature. Using this slope,
    calculate the ratio of the entropy to volume change at each
    temperature.)
            Calculation of DHvap
• Since along the two phase region:
                   ~       ~
                  Gliq  Gvap
       ~            ~      ~            ~
       H liq  Tvap Sliq  H vap  Tvap S vap
               ~       ~
               H vap  H liq     ~       ~
                                S vap  S liq
                   Tvap
                      ~       ~
         Psat      ( H vap  H liq )
                        ~      ~
         Tvap     Tvap (Vvap  Vliq )

This can be used to find DHvap at a particular
temperature. One calculates the change in Psat with Tvap
at a particular Tvap, uses an EOS to get the volumes,
then calculates DHvap. This is the best way to get DHvap.
                          Clausius-Clapeyron Eq
                This is an approximate equation based on the fact that
                Vliq<< Vvap and Vvap is ~ RT/P
              ~        ~                           ~       ~
                                                 ( H vap  H liq )
Psat       ( H vap  H liq )            Psat
                                               
                    ~                   Tvap       RTvap
                                                           2

Tvap         Tvap (Vvap )
                                                        Psat


            ~       ~
          ( H vap  H liq )                          ~       ~
                                                   ( H vap  H liq )  1
dPsat
                                        Psat                               1 
                      2
                              dTvap    ln o                               o 
 Psat         RTvap                      Psat             R           Tvap T 
                                                                            vap 
           Clausius-Clapeyron
                     ~       ~
         Psat      ( H vap  H liq )  1     1 
    ln                                   o 
           o
         Psat             R           Tvap T 
                                            vap 




Given Psat at two different Tvaps, DHvap can be computed.
c. Compute the enthalpy difference between the
    saturated vapor and saturated liquid butane at its
    normal boiling point and at 25 C. (How does DHvap
    relate to DSvap?)

d. Using only these two temperatures and the vapor
    pressures corresponding to them, use the Clausius-
    Clapeyron equation to estimate the enthalpy of
    vaporization for butane.
                    Summary
• In the two phase region, there is an explicit
  relationship between P and T. As T
  increases so does the vapor pressure,
  according to Antoine’s equation.
• The relationship between Pvap and Tsat
  can be used to compute the difference in
  the vapor and liquid volumes, entropy, and
  enthalpy.
              Gibbs Phase Rule
The Gibbs phase rule is an equation that tells you now many
properties you are free to designate in a chemical
equilibrium situation.



                 Df  N   2

  How many                                  Number of
  “intensive                                phases in
                          Number of         contact
  properties”             different
  you are free                              equilibrium
                          chemicals or
  to designate            “components”
                     Single Phase

                   Df  N   2
A one component system, with a single phase (vapor, liquid, or
supercritical) Df = 1 – 1 + 2 = 2 (specify 2 intensive variables
and all others are fixed. Giving 1 does not fix the system and
you cannot give 3 (the first 2 fix the third automatically)
                2-phase Region

                 Df  N   2

In the two phase region, one is only free to designate 1
property: Df = 1 – 2 + 2 = 1. Designating the T, fixes the
pressure, liquid and vapor specific volume, enthalpy,
entropy, etc.
Two phase region
         In the two phase region,
         there is only 1
         temperature for each
         pressure.
        2 Components – Two
             phases
You can (must) designate 2 properties (T and
P) this will fix the compositions, enthalpies,
internal energies etc of both phases.
                 Phase Rule in
           Multicomponent Systems
  Example MEEK and Water

                                 MEEK rich phase


                                    Df  N   2
                                 Water rich phase

                                    Df  2  2  2  2
Intensive variables – T, P, xm, ym, Vu,Vl,Su,Sl …
                     3 phases
Question: How many things are you free to specify for 3
phases in equilibrium (V, L, G) if there is only one
component?


Df = 1 – 3 + 2 = 0




 There is one and only one temperature, pressure,
 volume(s) entropies etc where solid, liquid, and
 gaseous water are in equilibrium. This is called the
 “triple point” (0.01C) You have nothing you can pick.
Triple Point
                “Stability Analysis”

In 304, you will be presented with calculational procedures that
will tell you whether or not two phases will form at the conditions
of a solution.
                     Summary
• There is an explicit “rule” that relates the
  number of “degrees of freedom” of
  multiphase and multicomponent systems
  at equilibrium.
• This rule tells one how many of the
  “intensive” variables in a system can be
  independently specified, which keeps
  one’s thoughts organized about systems

								
To top