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 At the top of the column, 1 mole of product will be withdrawn for every 9 moles that are returned to the column, and the distillation column will operate at 1.013 bar pressure.

Question

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 At the top of the column, 1 mole of product will be withdrawn for every 9 moles that are returned to the column, and the distillation column will operate at 1.013 bar pressure.

Accepted Answer

Basis for the calculations: 100 mol/hr of feed.

The feed contains 50 moles of [latex]C_{5}[/latex]. To meet the recovery target, 50 × 0.95 = 47.5 moles of [latex]C_{5}[/latex] 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 [latex]C_{7}[/latex]. Therefore, the bottoms flow rate B = 100−48.47 = 51.53 mol/hr and by a mass balance on [latex]C_{5}[/latex] contains the remaining 2.5 moles of [latex]C_{5}[/latex] and 50 − 0.97 = 49.03 mol/hr of [latex]C_{7}[/latex]. Consequently, the mole fractions of the bottoms product are [latex]x_{C 5}=0.0485 ext { and } x_{C 7}=0.9515[/latex].

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Also, the vapor flow in the column V must equal the liquid flow at the top of the column plus the amount of distillate withdrawn, so V = L + D = 436.23 + 48.47 = 484.70 mol/hr.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 [latex]V_{1}[/latex], which is in equilibrium with the liquid stream leaving [latex]L_{1} ext {; the vapor stream } V_{2}[/latex] entering tray 1 from tray 2 below, and an entering liquid stream from the condenser, which we designate as [latex]L_{0}[/latex]. On tray 1 we know that since all the vapor leaving is condensed to product and the returning or reflux liquid, the [latex]C_{5} ext { mole fraction of } L_{0} ext { and } V_{1}[/latex] are 0.98. Also, since [latex]L_{1}[/latex] is in equilibrium with [latex]V_{1}[/latex], from an equilibrium calculation at P = 1.013 bar, we find that the [latex]C_{5}[/latex] mole fraction of [latex]L_{1}[/latex] is 0.856 and that the equilibrium tray temperature is 313.8 K. Next a mass balance is used to find the [latex]C_{5}[/latex] mole fraction of [latex]V_{2}[/latex] as follows:

[latex]y_{C_{5,2}} V+x_{C_{5,0}} L=y_{C_{5,1}} V+x_{C_{5,1}} L[/latex]

[latex]y_{C_{5,2}} imes 484.70+0.98 imes 436.23=0.98 imes 484.70+0.856 imes 436.23[/latex]

So [latex]y_{C_{5,2}}=0.868[/latex].

Now since the liquid leaving tray 2 is in equilibrium with the vapor of composition, [latex]y_{C_{5,2}}=0.868[/latex], a dew point calculation can be done to find the tray temperature and liquid composition. The results are [latex]x_{C_{5,2}}=0.458[/latex] 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

[latex]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)[/latex]

[latex]y_{C_{5,3}} imes 484.70+0.856 imes 436.23+0.500 imes 100=0.868 imes 484.70+0.458 imes 536.23[/latex]

Therefore, [latex]y_{C_{5,3}}=0.501[/latex].

As the liquid leaving tray 3 is in equilibrium with the vapor of composition, [latex]y_{C_{5,3}}=0.501[/latex], a dew point calculation is used find the tray temperature and liquid composition. The results are [latex]x_{C_{5,3}}=0.137[/latex] 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.

[latex]y_{C_{5,4}} V+x_{C_{5,2}}(L+F)=y_{C_{5,3}} V+x_{C_{5,3}}(L+F)[/latex]

[latex]y_{C_{5,3}} imes 484.70+0.458 imes 536.23=0.501 imes 484.70+0.137 imes 536.23[/latex]

Therefore, [latex]y_{C_{5,4}}=0.146[/latex] and from an equilibrium calculation the liquid leaving tray 4 is [latex]x_{C_{5,4}}=0.029[/latex] and the tray temperature is 366.7 K. As this composition is lower than the specification for the bottoms product of [latex]x_{C 5}=0.0485[/latex], 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.

Question 10.1.6: A Very Simple Design of a Distillation Column Start with the...

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