A solution containing two components (methanol and water) is fed to a distillation column. The distillate is totally condensed and the reflux is returned at the bubble point. The distillate has a 95 weight percent purity of methanol. The residue from the bottom the column contains one weight-percent methanol. If five thousands pounds per hour of the solution, containing 50 percent of A, are fed to the column, calculate composition (mole fraction) and molar low rates of feed, distillate, and bottom products.
Solution:
Let us calculate molecular weight of feed. One pound of feed contains, XF
pounds of A and (one minus XF) pounds of B. The feed contains XF over MA
pound moles of A and (one minus XF)
over MB pound moles of B.
Molecular weight of methanol is 32.04 and that of water is 18.02. With these values molecular weight of feed
is found to be 23.07. Similar approach
could be taken for calculating molecular weights of distillate and residue
(bottoms). Molecular weight of
distillate, MD, and of residue, MW, is found to be 30.83
and 18.1, respectively.
Molar feed rate, F, is equal to the mass flow rate of feed,
F prime, divided by molecular weight of the feed, MF. This value is found to be 216.76 pound-moles
per hour.
Composition of feed (mole fraction), xF, is equal
to the ratio of moles of A to the total number of moles in the feed. This is equal to weight fraction (0.5) time
mass flow rate (5000) over molecular weight (32.04) over (molar feed rate)
216.76. Feed mole fraction is equal to
0.36.
Composition of distillate (mole fraction), xD,
can be calculated as: The distillate
contains XD over MA pound moles of A. So the mole fraction of distillate, xD,
is equal to XD over MA time MD and is found to
be 0.914.
Composition of residue (mole fraction), xW, can
be calculated as: The residue contains
XW over MA pound moles of A. So the mole fraction of residue, xW, is equal to XW
over MA time MW and is found to be 0.00565.
Applying overall mass balance, F, is equal to sum of D and
W. And individual component balance is
given as FxF is equal to the sum of DxD and WxW. Eliminating W from the last equation, D is
equal to F time (xF minus xW) over (xD minus xW). Plugging in the values of F, xF and
xW, one obtains the value of D as 84.51 pound moles. Residue molar flow rate, W, is equal to F
minus D. Its value is found to be
132.25 pound-moles.