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7-2. Vapor-Liquid Equilibrium

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					             Chemical Engineering Thermodynamics
Engi-3434
             Dr. Charles Xu @ Chemical Engineering, Lakehead University




      7-2. Vapor-Liquid Equilibrium
                                       ⎯ Henry’s Law




                                                                          Air bubbles in water




                               Review of Raoult’s Law
    Two Major Assumptions:

     1. The vapor phase is considered as an ideal gas ⇒
        valid thus only for low to moderate pressure vapors
     2. The liquid phase is considered as an ideal solution ⇒
        valid for the solutions whose comprising species are chemically similar, e.g.,
        solutions of isomers, solutions of similar compounds such as n-hexane / n-
        heptane, ethanol/propanol, benzene/toluene, acetone (CH3COCH3)/acetonitrile
        (CH3CN), etc.

    • Raoult’s Law:
                     y i P = xi Pi sat (i = 1, 2, .... , N)                      (10.1)

            where: P = total pressure of the system = pressure of the vapor or liquid phase.
                    Pi sat = Saturated vapor pressure of species i

            Note: when the temperature T > Tc (critical temp), Roault’s Law is
            not valid since Pi sat does not exist at T > Tc.
                  Limitation of Raoult’s Law
An example:
Consider a system in which air (1) and liquid water (2) is at equilibrium at
25°C and 1 atm, the mole fraction of water vapor (y2) in the gas phase can
be calculated by Roault’s law assuming the dissolved air in water is
negligible, i.e., x1 ≈ 0 or x2 ≈ 1                              sat
                                                                x 2 P2
                                                         y2 =
                                                                    P
From Antoine eq., water’s saturation pressure at 25°C, P2sat =3.166 kPa

                 x 2 P2sat (1.0)(3.166kPa)
          y2 =            =                = 0.0312
                     P        101.33kPa
          y1 = 1 − y 2 = 1 − 0.0312 = 0.9688

However, if we wish to calculate the accurate mole fraction of dissolved air
in the water (x1 = y1P/P1sat?), we CANNOT use Roault’s law, because the
critical temperature of air (TC = 132.2 K) is lower than the temperature
25°C (298.15 K) when the saturated vapor pressure (P1sat) does not exist!




  7.3.2 Simple Models for Vapor/Liquid Equilibrium
            for Solutions ⎯ Henry’s Law
For a species whose critical temperature (TC) is less
than the temperature of application (T) when the
saturated pressure P1sat is meaningless, e.g., for
most small molecule gases (N2, O2, air, H2, CO2,
etc.) in equilibrium with a liquid species, Raoult’s
Law is not valid, and Henry’s law is usually used.

• Henry’s Law:
                                                                William Henry

         y i P = xi H i               (10.4)
                                                                (1775 – 1836),
                                                                English chemist.


 where:
    P = total pressure of the system = pressure of the vapor/liquid phases.
     H i = Henry’s constant of species i, determined by experiment.

 Note: Henry’s law is usually used for a dilute species in the liquid phase.
    7.3.2 Simple Models for Vapor/Liquid Equilibrium
          for Solutions ⎯ Henry’s Law (Cont’d)
         • Examples of Henry’s Constants




          Note: • All measured by experiments
                  • Henry’s constant Hi has the unit of pressure
                  • Henry’s constant of species i, is function of temperature.
   With the Henry’s constant for air in water at 25°C, we can now finish the preceding
   example to calculate the mole fraction of dissolved air in the water (x1),
                                                1bar
                           0.9688 × 101.33kPa
                    y1 P                      100kPa = 1.35 × 10 −5
               x1 =      =
                    H1              72950bar




                                 Example 7.2
Assuming that carbonated water contains only CO2 (1) and H2O
(2), determine the compositions of the vapor and liquid phases in
a sealed can of “soda” at 10°C and 10 bar. Henry’s constant for
CO2 in water at 10°C is about 990 bar.
Solution will be provided at the tutorial #8 class:
  7.4 Vapor/Liquid Equilibrium from K-Value
                Correlations
 The equilibrium ratio Ki for species “i” is defined by

                               yi
                      Ki ≡                                     (10.10)
                               xi
 This ratio is usually called “K-value”, which is a function of temperature,
 total pressure and species. The K-value of species “i”, Ki, suggests the
 tendency of the species to partition itself preferentially between vapor
 and liquid.

• Ki < 1 ⇒ yi < xi:
 Species “i” is a “heavy” constituent in the mixture, exhibiting a higher
 concentration in the liquid phase.

• Ki > 1 ⇒ yi > xi:
 Species “i” is a “light” constituent in the mixture, exhibiting a higher
 concentration in the vapor phase.




  7.4 Vapor/Liquid Equilibrium from K-Value
            Correlations (Cont’d)
Comparing to the preceding Raoult’s law, modified Raoult’s law and Henry’s
law, we have

                       y i Pi sat
                  Ki ≡    =                               (10.11)
                       xi   P

                          yi H i
                  Ki ≡       =                            (10.12)
                          x1   P
 Both Pisat and Hi are dependent on temperature, so that the value of Ki for
 species i depends on temperature and pressure!

 For bubble point calculations:   ∑K x
                                    i
                                            i    i   =1     (by   ∑yi
                                                                        i   = 1)   (10.13)


 For dew point calculations:      ∑yi
                                        i       / K i = 1 (by ∑ xi = 1 )
                                                                    i
                                                                                   (10.14)
                                 Example 7.3
For a mixture of 10 mol% methane, 20 mol% ethane and 70 mol% propane at
temperature T = 50°F, determine: (a) the dew point pressure; and (b) the bubble
point pressure, employing the K-Value method .
Solution:
(a) the dew point pressure
 At the dew point, yi = zi. For the given temperature, the K-values depend on the
 choice of P. Now, we’ll find a P (i.e., the dew point pressure) to satisfy the dew
 point equation (10.14) by trial-and-error.

      ∑i
          y / K = 1 (10.14)
               i   i
                                    i
                                      ∑ xi = 1


(b) the bubble point pressure
At the bubble point, xi = zi.
Now, we’ll find a P (i.e., the
bubble point pressure) to
satisfy the bubble point
equation (10.13) by trial-and-
error.
       ∑   K i xi = 1 (10.13)
           i




     7.5 Application of VLE: Flash Calculation
                       • Flash evaporation/distillation
 When reducing the pressure of a high-pressure liquid to below the bubble
 point pressure, the liquid would “flash” (evaporate) to produce a two-phase
 system of vapor and liquid in equilibrium.


 Flash (or partial) evaporation occurs
 when a saturated liquid stream
 undergoes a reduction in pressure by
 passing through a throttling valve or
 other throttling device to a pressure
 lower than its bubble point pressure.
 The throttling valve or device is
 normally located at the entry into a
 pressure vessel. The vesselis often
 referred to as a flash drum.
7.5 Application of VLE: Flash Calculation (Cont’d)

                                                                , {yi}




          {zi}


                                                               , {xi}
Consider a system containing mixture of non-reacting chemical species
with an overall composition of {zi} subject to flash evaporation.

Let L be the mole fraction of the liquid phase, with mole fraction
concentrations of {xi}, and let V be the mole fraction of the vapor phase,
with mole fraction concentrations of {yi}.




7.5 Application of VLE: Flash Calculation (Cont’d)
Material balance equations:          L +V =1
                                     z i = xi L + y iV (i = 1, 2, ..., N )
Combining the two equations to eliminate L gives:
              z i = xi (1 − V ) + y iV (i = 1, 2 , ..., N )         (10.15)
                                                                y        P sat
Substituting xi = yi/Ki and solving for yi, where: K i ≡ i = i
based on Raoult’s Law.                                  xi   P
                                             zi K i
                                yi =                           (10.16)
                                        1 + V ( K i − 1)

Because   ∑y   i   = 1 , thus   ∑ 1 + V (K
                                          zi K i
                                                          =1   (10.17)
          i                      i             i   − 1)
          Flash calculation procedures:
          (a) find the value of V which satisfies Eq. (10.17);
          (b) solve for yi by Eq. (10.16);
          (c) Find xi by xi = yi / Ki.
                                Example 7.4
The system with 1 mol of mixture of
acetone(1)/acetonitrile(2)/nitromethane(3) at 80°C and 110 kPa has the
overall composition: z1 = 0.45, z2 = 0.35 and z3 = 0.2. Assuming that
Raoult’s law is appropriate to this system, determine L, V, {xi} and {yi}. The
vapor pressures of the pure species at 80°C are: P1sat = 195.75 kPa, P2sat
= 97.84 kPa, P3sat = 50.32 kPa
Solution will be provided at the tutorial #8 class:




                         Example 7.4 (Cont’d)

				
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