Question 3 of 6: Crystallization and Rotary-Filter Entrainment

Question 3: Crystallization and Rotary-Filter Entrainment (Part A — 25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Feed 1000 kg/hr, 60 wt% naphthalene ($M=128.17$) / 40 wt% benzene ($M=78.11$), cooled from 80 to 10 °C. At 10 °C the mixture lies in the “liquid + solid naphthalene” field; from the phase diagram the liquid (filtrate) lies on the naphthalene liquidus at mole fraction benzene $x_{Bz}\approx0.80$, i.e. $x_{naph}\approx0.20$. Measured filtrate rate = 505 kg/hr.

Find. (a) entrainment ratio, kg entrained mother-liquor per kg crystals; (b) the effect of that entrainment on naphthalene recovery and on cake purity, compared with an ideal (entrainment-free) separation.

Approach. Convert the liquidus mole fraction to a mass fraction, close a benzene balance to find the ideal (entrainment-free) liquid and solid rates, then compare the measured filtrate rate against the ideal liquid rate: the shortfall is mother liquor retained (entrained) in the cake, which shifts recovery and purity.

1. Liquidus composition (mole → mass). At 10 °C the filtrate is $x_{naph}=0.20$; on a mass basis $$w_{naph}=\frac{0.20(128.17)}{0.20(128.17)+0.80(78.11)}=\frac{25.6}{88.1}=0.291.$$ So the equilibrium mother liquor is 29.1 wt% naphthalene, 70.9 wt% benzene.
2. Ideal (entrainment-free) split — benzene balance. If the cake were pure crystals, all 400 kg/hr of benzene would report to the liquid: $$L^\ast=\frac{400}{1-0.291}=564\ \text{kg/hr},\qquad S^\ast=1000-564=436\ \text{kg/hr crystals.}$$ The ideal naphthalene recovery would be $S^\ast/600=\boxed{72.6\%}$ at 100% purity.
3. (a) Entrainment from the measured filtrate. Only 505 kg/hr of liquid actually leaves as filtrate, so the balance of the mother liquor is trapped in the cake: $$\dot m_{ent}=L^\ast-505=564-505=59.2\ \text{kg/hr,}\qquad \frac{\dot m_{ent}}{S^\ast}=\frac{59.2}{436}=\boxed{0.136\ \text{kg liquor / kg solids.}}$$
4. (b) Effect on recovery and purity. The cake is now crystals plus entrained liquor: $\text{cake}=1000-505=495$ kg/hr. Naphthalene in the cake is the crystals plus the naphthalene dissolved in the entrained liquor $(0.291\times59.2=17.2$ kg): $$n_{naph}^{cake}=436+17.2=453\ \text{kg/hr}\;\Rightarrow\;\text{recovery}=\frac{453}{600}=\boxed{75.5\%},\quad \text{purity}=\frac{453}{495}=\boxed{91.5\%.}$$

Physically, the entrained mother liquor is a two-edged sword: it carries a little extra naphthalene into the cake, nudging naphthalene recovery up from the ideal 72.6% to 75.5%, but because that liquor is benzene-rich it dilutes the crystals, dropping cake purity from a nominal 100% to 91.5%. Better washing of the cake would recover the trapped benzene and restore purity.