Swenson–Walker Crystallizer Sizing: 23-Chem-A2 Unit Operations and Separation Processes, December 2019, Q5

Question B2: Swenson–Walker Crystallizer Sizing (25 points)

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.

Quantity	Value
Product	Na₂SO₄·10H₂O, 0.25 kg/s ($M_\text{hyd}=322.2$, $M_\text{anh}=142.0$)
Solubility (anh.)	40/100 water at 27 °C → 14/100 at 17 °C
$c_p$ liquor / $\Delta H_\text{cryst}$	3.8 kJ/kg·K / 230 kJ/kg
Area density / $U$	3 m²/m / 0.15 kW/m²·K
Cooling water	7 °C → 17 °C (countercurrent)
Section length	3 m each

Find. The number of 3-m crystallizer sections required.

Figure B2 — Both terminal temperature differences equal 10 K, so the LMTD is exactly 10 K. The heat duty is the sum of liquor sensible cooling and the heat of crystallization released.

Approach. Use the hydrate/anhydrous mass balance with the two solubilities to find the feed rate, add the sensible-cooling and crystallization duties, then size the area from $Q=UA\,\Delta T_\text{lm}$ and convert to sections.

Split the crystal product. Anhydrous in crystals $=0.25\times142.0/322.2=0.110$ kg/s; water of crystallization $=0.25-0.110=0.140$ kg/s.
Feed rate from the solubility balance. A water balance and anhydrous balance between the 40/100 feed (27 °C) and the 14/100 mother liquor (17 °C) give feed water $=0.349$ kg/s, so $$F=1.40\times0.349=\boxed{0.488\ \text{kg/s}}\quad(\text{mother liquor }0.238\ \text{kg/s}).$$
Heat duty. Sensible cooling of the feed liquor 27 → 17 °C plus the heat of crystallization released, $$Q=F\,c_p\,\Delta T+\dot m_\text{prod}\Delta H_\text{cryst}=0.488(3.8)(10)+0.25(230)=\boxed{76.0\ \text{kW}}.$$
Log-mean ΔT. Countercurrent, liquor 27 → 17 °C against water 17 ← 7 °C gives $\Delta T=10$ K at both ends, so $\Delta T_\text{lm}=10$ K.
Required area and sections. $$A=\frac{Q}{U\,\Delta T_\text{lm}}=\frac{76.0}{0.15(10)}=50.7\ \text{m}^2\ \Rightarrow\ L=\frac{50.7}{3}=16.9\ \text{m}.$$ At 3 m per section this needs $\lceil16.9/3\rceil=\boxed{6\ \text{sections}}$.

Quantity	Value
Feed rate $F$	0.488 kg/s
Heat duty $Q$	76.0 kW
$\Delta T_\text{lm}$	10 K
Required area	50.7 m² (16.9 m)
Sections (3 m each)	6

Concept & Theory

Topic. Cooling-crystallizer heat and mass balance with a hydrated product.

Key relations. Anhydrous/water split by molar-mass ratio; solubility balance for feed; duty $Q=Fc_p\Delta T+\dot m\Delta H_\text{cryst}$; sizing $Q=UA\Delta T_\text{lm}$.

Why this works. Cooling from 27 to 17 °C shifts the solubility from 40 to 14 kg/100 kg water, and the excess anhydrous salt leaves as the decahydrate — which also removes water of crystallization, concentrating the mother liquor. Both the sensible heat given up by the liquor and the exothermic heat of crystallization must be carried off by the cooling water, and the balanced 10-K terminal differences make the LMTD trivial.

Common pitfall. Reporting crystal mass as anhydrous salt (it is the hydrate, so the water of crystallization must be tracked), or forgetting to add the crystallization heat to the sensible duty.

Source. Coulson & Richardson Vol. 2 ch. 15 (crystallization); McCabe ch. 27; Geankoplis ch. 12.

Question 5 of 6: Swenson–Walker Crystallizer Sizing

Question B2: Swenson–Walker Crystallizer Sizing (25 points)