A Very Simple Design of a Distillation Column
Start with the the feed of the previous illustration, consisting of 50 mol % n-pentane and 50 mol % n-heptane, and recover 95 percent of the n-pentane in the feed in a stream that contains 98 mol % n-pentane. At the top of the column, 1 mole of product will be withdrawn for every 9 moles that are returned to the column, and the distillation column will operate at 1.013 bar pressure.
Basis for the calculations: 100 mol/hr of feed.
The feed contains 50 moles of C_{5}. To meet the recovery target, 50 × 0.95 = 47.5 moles of C_{5} in a stream of 98 percent purity is needed, so the distillate flow rate is D = 47.5/0.98 = 48.47 mol/hr, and the distillate contains 48.47−47.5 = 0.97 moles of C_{7}. Therefore, the bottoms flow rate B = 100−48.47 = 51.53 mol/hr and by a mass balance on C_{5} contains the remaining 2.5 moles of C_{5} and 50 − 0.97 = 49.03 mol/hr of C_{7}. Consequently, the mole fractions of the bottoms product are x_{C 5}=0.0485 \text { and } x_{C 7}=0.9515.
Since the distillate flow rate is 48.47 mol/hr, and 9 moles of liquid are returned to the column for each mole of overhead product, L = 48.47×9 =436.23 mol/hr and L+F = 536.23 mol/hr. Also, the vapor flow in the column V must equal the liquid flow at the top of the column plus the amount of distillate withdrawn, so V = L + D = 436.23 + 48.47 = 484.70 mol/hr.
With the overall column flows now determined, we next consider what happens on each tray. The top tray, which we refer to as tray 1, is shown schematically in Fig. 10.1-9. It has four streams: a vapor stream leaving the tray V_{1}, which is in equilibrium with the liquid stream leaving L_{1} \text {; the vapor stream } V_{2} entering tray 1 from tray 2 below, and an entering liquid stream from the condenser, which we designate as L_{0}. On tray 1 we know that since all the vapor leaving is condensed to product and the returning or reflux liquid, the C_{5} \text { mole fraction of } L_{0} \text { and } V_{1} are 0.98. Also, since L_{1} is in equilibrium with V_{1}, from an equilibrium calculation at P = 1.013 bar, we find that the C_{5} mole fraction of L_{1} is 0.856 and that the equilibrium tray temperature is 313.8 K. Next a mass balance is used to find the C_{5} mole fraction of V_{2} as follows:
y_{C_{5,2}} V+x_{C_{5,0}} L=y_{C_{5,1}} V+x_{C_{5,1}} Ly_{C_{5,2}} \times 484.70+0.98 \times 436.23=0.98 \times 484.70+0.856 \times 436.23
So y_{C_{5,2}}=0.868.
Now since the liquid leaving tray 2 is in equilibrium with the vapor of composition, y_{C_{5,2}}=0.868, a dew point calculation can be done to find the tray temperature and liquid composition. The results are x_{C_{5,2}}=0.458 and the tray temperature is 329.9 K.
Next a mass balance is used to find the vapor composition entering tray 2 from tray 3 below. However, since the liquid composition on this tray is close to the feed composition, the 100 mol/hr should be added to this tray. Therefore, the mass balance is
y_{C_{5,3}} V+x_{C_{5,1}} L+x_{C_{5, F}} F=y_{C_{5,2}} V+x_{C_{5,2}}(L+F)y_{C_{5,3}} \times 484.70+0.856 \times 436.23+0.500 \times 100=0.868 \times 484.70+0.458 \times 536.23
Therefore, y_{C_{5,3}}=0.501.
As the liquid leaving tray 3 is in equilibrium with the vapor of composition, y_{C_{5,3}}=0.501, a dew point calculation is used find the tray temperature and liquid composition. The results are x_{C_{5,3}}=0.137 and the tray temperature is 353.8 K.
Proceeding in this manner, a mass balance is then used to find the vapor composition entering tray 3 from tray 4 below.
y_{C_{5,4}} V+x_{C_{5,2}}(L+F)=y_{C_{5,3}} V+x_{C_{5,3}}(L+F)y_{C_{5,3}} \times 484.70+0.458 \times 536.23=0.501 \times 484.70+0.137 \times 536.23
Therefore, y_{C_{5,4}}=0.146 and from an equilibrium calculation the liquid leaving tray 4 is x_{C_{5,4}}=0.029 and the tray temperature is 366.7 K. As this composition is lower than the specification for the bottoms product of x_{C 5}=0.0485, no further stages are needed in the distillation column. Also, it should be pointed out that while the condenser is used to condense all the vapor, the reboiler vaporizes only part of the liquid. Therefore, it is an equilibrium stage, just as are each of the trays. Consequently, the distillation column we have just “designed” requires a total condenser and 4 “trays” or equilibrium stages, which actually consists of three trays and a reboiler. Also, the feed is to be added to the second tray from the top of the distillation column.
