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5.22: Untitled Page 119

  • Page ID
    18252
  • Chapter 5

    Figure 5.27. Sketch of an equilibrium stage process

    and liquid streams exchange mass inside the equilibrium stage until they are in equilibrium with each other. Equilibrium is determined by a ratio of the molar fractions of each component in the liquid and vapor streams according to ( y )

    Equilibrium relation:

    A 4

    K

    A  1 , 2 , .., N

    A

    ( x )

    A 3

    The streams leaving the stage, S3 (liquid), and S4 (vapor) are in equilibrium with each other and therefore satisfy the above relation. The ratio of the molar flow rates of the output streams is a function of the energy balance within the stage.

    In this problem we assume that the ratio of the liquid output molar flow rate to the vapor output molar flow rate, M

    / M , is given. Assuming that the

    3

    4

    compositions, i.e. the mole fractions of the components and the molar flow rates of the input streams, S1 and S2, are known, and the equilibrium constant for one of the components is given, develop the mass balances for a two component vapor‐liquid equilibrium stage.

    5‐28. A single stage, binary distillation process is illustrated in Figure 5.28. The total molar flow rate entering the unit is M

     and the mole fraction of species A in

    1

    this liquid stream is ( x ) . Heat is supplied in order to generate a vapor stream, A 1

    and the ratio, M

    M

       , depends on the rate at which heat is supplied. At the 2

    3

    vapor‐liquid interface, we can assume local thermodynamic equilibrium (see Eq. 5‐24) in order to express the vapor‐phase mole fraction in terms of the liquid-phase mole fraction according to

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    Two‐Phase Systems & Equilibrium Stages

    219

    x

    Equilibrium relation:

    AB A

    y

    ,

    at the vapor‐liquid interface

    A

    1  x (

     1)

    A

    AB

    Here 

    represents the relative volatility. If the distillation process is slow AB

    enough, one can assume that the vapor and the liquid leaving the distillation unit are in equilibrium; however, at this point in our studies we do not know what is meant by slow enough. In order to proceed with an approximate solution to this problem, we replace the equilibrium relation with a process equilibrium relation given by

     ( x )

    AB

    A 3

    ( y )

    , between the exit streams

    A 2

    1  ( x ) (

    1)

    A 3

    AB

    Use of this relation means that we are treating the system shown in Figure 5.28 as an equilibrium stage. Given a detailed study of mass transfer in a subsequent course, one can make a judgment concerning the conditions that are required in order that this process equilibrium relation be satisfactory. For the present, you are asked to use the above relation to derive an implicit expression for ( y ) in terms A 2

    of ( x ) and examine three special cases: 

     0 ,   0 ,

     1 .

    A 1

    AB

    AB

    Figure 5.28. Single stage binary distillation

    5‐29. A saturated solution of calcium hydroxide enters a boiler as shown in Figure 5.29 and a fraction,  , of the water entering the boiler is vaporized.

    Under these circumstances a portion of the calcium hydroxide precipitates and you would like to determine the mass fraction of this suspended solid calcium hydroxide in the liquid stream leaving the boiler. Assume that no calcium

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    220