Question 18.2: A continuous fractionating column is to be designed to separ......
A continuous fractionating column is to be designed to separate 30,000 kg/h of a mixture of 40 percent benzene and 60 percent toluene into an overhead product containing 97 percent benzene and a bottom product containing 98 percent toluene. These percentages are by weight. A reflux ratio of 3.5 mol to 1 mol of product is to be used. The molal latent heats of benzene and toluene are 7360 and 7960 cal/g mol, respectively. Benzene and toluene fonn an ideal system with a relative volatility of about 2.5; the equilibrium curve is shown in Fig. 18.16. The feed has a boiling point of 95°C at a pressure of 1 atm. (a) Calculate the moles of overhead product and bottom product per hour. (b) Determine the number of ideal plates and the position of the feed plate (i) if the feed is liquid and at its boiling point; (ii) if the feed is liquid and at 20°C (specific heat 0.44 cal/g-°C); (iii) if the ,feed is a mixture of two-thirds vapor and one-third liquid. (c) If steam at 20 {\mathrm{lb}_{f}/\mathrm{in}\,^{2}} (1.36 atm) gauge is used for heating, how much steam is required per hour for each of the above three cases, neglecting heat losses and assuming the reflux is a saturated liquid? (d) If cooling water enters the condenser at 25°C and leaves at 40°C, how much cooling water is required, in cubic meters per hour?
(a) The molecular weight of benzene is 78 and that of toluene is 92. The concentra-tions of feed, overhead, and bottoms in mole fraction of benzene are
x_{F}=\frac{\frac{40}{78}}{\frac{49}{78}+\frac{60}{92}}=0.440 x_{D}={\frac{\frac{97}{78}}{\frac{9}{7}}}=0.974
x_{B}={\frac{{\frac{2}{78}}}{{\frac{2}{78}}+{\frac{98}{92}}}}=0.0235
The average molecular weight of the feed is
\frac{100}{\frac{40}{78}+\frac{60}{92}}=85.8
The average heat of vaporization of the feed is
\lambda=0.44(7360)+0.56(7960)=7696\;\mathrm{cal/g\;mol}
The feed rate F is 30,000/85.8 = 350 kg mol/h. By an overall benzene balance, using Eq. (18.5).
\frac{D}{F}=\frac{x_{F}-x_{B}}{x_{D}-x_{B}} (18.5)
D=350{\frac{0.440-0.0235}{0.974-0.0235}}=153.4{\mathrm{~kg~mol/h}}
B=350-153.4=196.6\,\mathrm{Kg\,mol/h}
(b) Next we determine the number of ideal plates and position of the feed plate.
(i) The first step is to plot the equilibrium diagram and on it erect verticals at x_{D},x_{F}, and x_{B}. These should be extended to the diagonal of the diagram. Refer to Fig. 18.16.
The second step is to draw the feed line. Here f = 0, and the feed line is vertical
and is a continuation of line x=x_{F}.\,
The third step is to plot the operating lines. The intercept of the rectifying line on they axis is, from Eq. (18.19), 0.974/(3.5 + I)= 0.216. From the intersection of this operating line and the feed line the stripping line is drawn.
The fourth step is to draw the rectangular steps between the two operating lines and the equilibrium curve. In drawing the steps, the transfer from the rectifying line to the stripping line is at the seventh step. By counting steps it is found that,
besides the reboiler, 11 ideal plates are needed and feed should be introduced on the seventh plate from the top.†
y_{n+1}=\frac{R_D}{R_D+1} x_n+\frac{x_D}{R_D+1} (18.19)
(ii) The latent heat of vaporization of the feed λ is 7696/85.8 = 89.7 cal/g.
Substitution in Eq. (18.24) gives
q=1+{\frac{c_{p L}(T_{b}-T_{F})}{\lambda}} (18.24)
q=1+{\frac{{0.44(95-20)}}{89.7}}=1.37
From Eq. (18.31) the slope of the feed line is -1.37/(1- 1.37) = 3.70. When steps are drawn for this case, as shown in Fig. 18.17, it is found that a reboiler and 10 ideal plates are needed and that the feed should be introduced on the fifth plate.
y=-{\frac{q}{1-q}}\,x+{\frac{x_{F}}{1-q}} (18.31)
(iii) From the definition of q it follows that for this case q={\frac{1}{3}} and the slope of the feed line is -0.5. The solution is shown in Fig. 18.18. It calls for a reboiler and 12 plates, with the feed entering on the seventh plate.
(c) The vapor flow V in the rectifying section, which must be condensed in the condenser, is 4.5 mol per mole of overhead product, or 4.5 × 153.4 = 690 kg mol/h. From Eq. (18.27),
V={\overline{{V}}}+(1-q)F\qquad{\mathrm{and}}\qquad V-{\overline{{V}}}=(1-q)F (18.27)
{\overline{{V}}}=690-350(1-q)
Using the heat of vaporization of toluene rather than that of benzene to be slightly conservative in design, .λ = 7960 cal/g mol. The heat from 1 lb of steam at 20\mathrm{lb}_{f}/\mathrm{in}.^{2} gauge, from Appendix 7, is 939 Btu/lb; hence \lambda_{S}=939/1.8=522\,{\mathrm{cal/g}}. The steam required, from Eq. (18.32), is
{\dot{m}}_{s}=\frac{\bar{V} \lambda}{\lambda_s}. (18.32)
{\dot{m}}_{s}={\frac{7960}{522}}{\overline{{V}}}=15.25[690-350(1-q)]\;{\mathrm{kg/h}}
The results are given in Table 18.2.
(d) The cooling water needed, which is the same in ali cases, is, from Eq. (18.33),
{\dot{m}}_{w}=-{\frac{-{V\lambda}}{T_{2}-T_{1}}}={\frac{V\lambda}{T_{2}-T_{1}}} (18.33)
{\dot{m}}_{w}={\frac{7960\times690}{40-25}}=366,160\mathrm{~kg/h}
The density of water at 25°C (77°F), from Appendix 14, is 62.24\ {\mathrm{lb}}/{\mathrm{ft}}^{3}, or 62.24\times16.{{0}}18=996.3\,{\mathrm{kg/m}}^{3}. The water requirement is 366,160/996.3=367.5\,\mathrm{m}^{3}/\mathrm{h}.
The use of cold feed, case (ii), requires the smaiiest number of plates but the greatest amount of reboiler steam. The total energy requirement for the reboiler and the preheater is about the same for all three cases. The reasons for preheating the feed, in most cases, are to keep the vapor flow rate about the same in both sections of the column and to make use of the energy in a hot-liquid stream such as the bottom product.
| TABLE 18.2 | |||
| Solution to Example 18.2, part (c) | |||
| Case | q | Reboiler steam {\dot{m}}_{s}, kg/h | Number of ideal plates |
| (i) | 1.0 | 10,520 | 11 |
| (ii) | 1.37 | 12,500 | 10 |
| (iii) | 0.333 | 6,960 | 12 |