Swenson–Walker Crystalliser for Glauber's Salt: 23-Chem-A2 Unit Operations and Separation Processes, December 2014, Q5

Question B2: Swenson–Walker Crystalliser for Glauber's Salt (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. A cooling crystalliser; the hydrate carries 10 waters of crystallisation, so a solids-plus-water balance is needed before the energy balance and area sizing.

Quantity	Value
Product rate $C$ ($\mathrm{Na_2SO_4\cdot10H_2O}$)	0.25 kg/s
Solubility at 300 K / 290 K	40 / 14 kg per 100 kg water
Heat capacity $c_p$ / heat of cryst.	3.8 kJ/kg·K / 230 kJ/kg
$U$ / area per length	0.15 kW/m$^2$K / 3 m$^2$/m
Cooling water 280 → 290 K; liquor 300 → 290 K	countercurrent

Find. the number of 3-m crystalliser sections.

Figure B2 — Countercurrent Swenson–Walker crystalliser. Both terminal temperature differences are 10 K, so the LMTD is exactly 10 K.

Approach. Take a solids/water balance to size the feed and mother-liquor streams, sum the sensible and crystallisation heat loads, size the area from $Q=UA\,\mathrm{LMTD}$, and divide by 3 $\mathrm{m^2/m}$ then by the 3-m section length.

Mass balance (anhydrous basis). The product carries $0.25\times142/322=0.110\ \mathrm{kg/s}$ anhydrous. With feed fraction $40/140$ and mother-liquor fraction $14/114$, the anhydrous balance $F(40/140)=0.110+L(14/114)$ together with $F=C+L$ gives $$F = 0.488\ \mathrm{kg/s},\qquad L = 0.238\ \mathrm{kg/s}.$$
Heat load. Sensible cooling of the feed liquor plus released crystallisation heat: $$Q=F c_p\Delta T + C\,\lambda_c = 0.488(3.8)(10) + 0.25(230)=18.6+57.5=76.1\ \mathrm{kW}.$$ The crystallisation heat dominates.
Log-mean temperature difference. Countercurrent: at the feed end $300-290=10\ \mathrm{K}$; at the product end $290-280=10\ \mathrm{K}$. Equal terminal differences give $$\mathrm{LMTD}=10\ \mathrm{K}.$$
Area and length. From $Q=U A\,\mathrm{LMTD}$, $$A=\frac{76.1}{0.15(10)}=50.7\ \mathrm{m^2}\;\Rightarrow\; \text{length}=\frac{50.7}{3\ \mathrm{m^2/m}}=16.9\ \mathrm{m}.$$
Number of sections. With 3-m sections, $16.9/3=5.6$, rounded up: $$N=\lceil 5.6\rceil = 6\ \text{sections}.$$ 6 crystalliser sections (18 m installed) are required

Quantity	Result
Feed / mother-liquor rate	0.488 / 0.238 kg/s
Sensible / crystallisation load	18.6 kW / 57.5 kW
Total heat load $Q$	76.1 kW
LMTD	10 K
Required area / length	50.7 m$^2$ / 16.9 m
Number of 3-m sections	6

Concept & Theory

Topic. Cooling crystallisation with a hydrate product — combined mass/energy balance and heat-exchanger sizing.

Key relations. Hydrate/anhydrous mass ratio $M_h/M_a$; solids balance $F x_F = C\frac{M_a}{M_h}+L x_L$; heat load $Q=Fc_p\Delta T + C\lambda_c$; sizing $Q=UA\,\mathrm{LMTD}$; sections $=\lceil A/(a\cdot \ell)\rceil$ (Coulson & Richardson Vol. 2, ch. 15).

Why this works. The water of crystallisation locked into the hydrate must be removed from the mother liquor, so the anhydrous balance — not a simple total-solids balance — sets the feed rate. The heat of crystallisation is released as the crystals form and adds to the sensible cooling duty; both are removed by the cooling water through the jacket area.

Common pitfall. Using the hydrate molar mass where the anhydrous solubility applies (or vice versa), and omitting the heat of crystallisation — here it is three times the sensible load, so dropping it under-sizes the unit badly.

Question 5 of 6: Swenson–Walker Crystalliser for Glauber's Salt

Question B2: Swenson–Walker Crystalliser for Glauber's Salt (25 marks)