Page 150 - Separation process engineering

B. Explore. Except for some slight changes in the feed temperature and column pressure, this problem
is very similar to Example 3-1. The solution for B and D obtained in that example is still correct. B
= 6667 kg/h, D = 3333 kg/h. Equilibrium data are available in weight fractions in Figure 2-4 and in
mole fraction units in Figure 2-2 and Table 2-1. To use the Lewis method we must have CMO. We
can check this by comparing the latent heat per mole of pure ethanol and pure water. (This checks
the third and most important criterion for CMO. Since the column is well insulated, the first
criterion, adiabatic, will be satisfied.) The latent heats are (Himmelblau, 1974):

The difference of roughly 5% is reasonable particularly since we always use the ratio of L/V or /
. (Using the ratio causes some of the change in L and V to divide out.) Thus we will assume CMO.
Now we must convert flows and compositions to molar units.

C. Plan. First, convert to molar units. Carry out preliminary calculations to determine L/V and / .
Then start at the top, alternating between equilibrium (Figure 2-2 or Eq. (2.B-2)) and the top
operating Eq. (4-12b). Since stage 2 is the feed stage, calculate y from the bottom operating Eq.
3
(4-14).

D. Do It. Preliminary Calculations: To convert to molar units:

Average molecular weight of feed is
= (0.144)(46) + (0.856)(18) =22.03

Feed rate = (10,000 kg/h)/(22.03 kg/kmol) = 453.9 kmol/h

For distillate, the average molecular weight is

which is also the average for the reflux liquid and vapor stream V since they are all the same
composition.
Then D = (3333 kg/h)/35.08 = 95.2 kmol/h and

while

V = L + D = 380.9. Thus,

Because of CMO, L/V is constant in the rectifying section.
Since the feed is a saturated liquid,

where we have converted F to kmol/h. Since a saturated liquid feed does not affect the vapor, = V
= 380.9. Thus,

An internal check on consistency is L/V < 1 and / > 1.

Stage-by-Stage Calculations: At the top of the column, y = x = 0.61. Liquid stream L of
1
1
D
concentration x is in equilibrium with the vapor stream y . From Figure 2-2, x = 0.4. (Note that y >
1
1
1
1