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Chapter 12 Differential Distillation Page 12-1
Chapter 12 Differential Distillation
1. Very Pure Benzene
a) A simple overall mass balance gives:
b) Since the condenser produces equal amounts of product
and reflux, by definition we have:
c) Since the condenser is in equilibrium, we can write:
x
F
= 0.99
x
B
D
x
D
= 0.999
G, y
ℓ
Now we can write a mass balance on the condenser:
d) An overall mass balance on the column, reboiler, and condenser gives:
Chapter 12 Differential Distillation Page 12-2
The vapor entering the column will be in equilibrium with the liquid leaving the reboiler:
We now write a mass balance on the column:
e) The NTU are given by:
2. Nitrogen drying with molecular sieves
y
ℓ
= x
D
= 0.02
x
D
x
D
y
ℓ
= x
D
= 0.02
x
D
x
D
Since the column operates at high reflux, G = L and y = x
in the column. At the top:
o o D
At the bottom:
Chapter 12 Differential Distillation Page 12-3
We can calculate the NTU assuming a linear equilibrium line:
3. Very pure methanol
D
a) Basis: 1 mol fed
Methanol Balance
D0.9980
b) Find NTU from Eq. 12.2-12
Chapter 12 Differential Distillation Page 12-4
c) For staged distillation, from Eq. 13.2-14
Because stages are commonly spaced at 0.6 m,
Remember that stages will normally be much less efficient than this ideal limit.
4. Reducing solvent loss
Assume aromatic loss is small relative to acetone flow, so pseudo-steady state
Chapter 12 Differential Distillation Page 12-5
so
5. Paint shop waste
Calc column concs
iterate by choosing new xB.
6. Distillation with a side stream
a) make-up methanol
so
b), c) operating lines
use RD = 10, not r.
in top of column
below sidestream,
below feed
Chapter 12 Differential Distillation Page 12-7
d) NTU from Eq. 12.4-9
e) feed at value of /HTU where x = 0.1.
