Page 143 - Separation process engineering

(4-3, stage 1)

Assuming that each stage is an equilibrium stage, we know that the liquid and vapor leaving the stage are

in equilibrium. For a binary system, the Gibbs phase rule becomes

Degrees of freedom = C − P + 2 = 2 − 2 + 2 = 2
Since pressure has been set, there is one remaining degree of freedom. Thus for the equilibrium stage the
variables are all functions of a single variable. For the saturated liquid we can write

(4-4a, stage 1)

and for the saturated vapor,

(4-4b, stage 1)

The liquid and vapor mole fractions leaving a stage are also related:

(4-4c, stage 1)

Equations (4-4) for stage 1 represent the equilibrium relationship. Their exact form depends on the
chemical system being separated. Equations (4-1, stage 1) to (4-4c, stage 1) are six equations with six
unknowns: L , V , x , y , H , and h .
1
2
2
1
2
1
Since we have six equations and six unknowns, we can solve for the six unknowns. The exact methods for
doing this are the subject of the remainder of this chapter. For now we will just note that we can solve for
the unknowns and then proceed to the second stage. For the second stage we will use the balance
envelope shown in Figure 4-1B. The mass balances are now

(4-1, stage 2)

(4-2, stage 2)

while the energy balance is

(4-3, stage 2)
The equilibrium relationships are

(4-4, stage 2)

Again we have six equations with six unknowns. The unknowns are now L , V , x , y , H , and h .
2
3
3
2
3
2
We can now proceed to the third stage and utilize the same procedures. After that, we can go to the fourth
stage and then the fifth stage and so forth. For a general stage j (j can be from 1 to f − 1, where f is the
feed stage) in the enriching section, the balance envelope is shown in Figure 4-1C. For this stage the mass