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23-Chem-A2 Unit Operations and Separation Processes · December 2017

Question 6 of 6: Design of a Batch Evaporative KCl Crystallizer

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams — 16-CHEM-A2 Unit Operations and Separation Processes, December 2017. 3 hours, open book (one text). Six problems — Part A Unit Operations (A1–A3) and Part B Separation Processes (B1–B3), each worth 25 marks; the rubric asks for at least two problems from each part (only the first two per part are marked). All six are worked below for completeness as a study resource.

Reference texts. McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Geankoplis, Transport Processes and Separation Process Principles (4th ed.); Coulson & Richardson, Chemical Engineering Vol. 2 (particle technology, absorption, drying, crystallization); Treybal, Mass-Transfer Operations (3rd ed.); Perry's Chemical Engineers' Handbook (9th ed.).

Notes on chart / model reads. A1's minor-loss balance uses the Blasius smooth-pipe friction factor $f=0.079\,Re^{-0.25}$ (Fanning) — valid to $Re\approx10^5$, and all three test points fall in $3.8$–$8.9\times10^4$; the 1.5 m is taken as the head lost across the pipe-and-valve test section, because also charging an exit velocity head would make the measured open-valve flow physically impossible (see A1). B1(b) reads the Sherwood/Eckert flooding line supplied with the paper at abscissa $0.061$, giving flooding ordinate $\approx0.17$ (graphical, $0.16$–$0.18$), with the tabulated packing factor $F=160\ \text{ft}^{-1}$ converted to $525\ \text{m}^{-1}$ for the SI form of the ordinate. B3(a) sizes the vessel from its starting liquor volume, allowing for the water evaporated during the batch. Engineering choices are flagged in Check callouts.

Part A — Unit Operations

Question B3: Design of a Batch Evaporative KCl Crystallizer (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.

QuantityValue
Product mass / size $W_p,\ L_p$800 kg / 1.1 mm
Seed size $L_s$80 µm (uniform)
Max slurry density / fill150 kg/m³ / 70%
Growth rate $G_r$$3\times10^{-8}$ m/s
Solubility / soln density / crystal density400 / 1300 / 1900 kg/m³

Find. (a) crystallizer volume; (b) seed mass; (c) batch time; (d) initial and final evaporation rate.

Seeded batch evaporative crystallizer (McCabe ΔL law) water evaporated (vacuum) magma 5.33 m³ (150 kg/m³)70% fill at start → vessel 9.87 m³ seed 80 µmgrow ΔLproduct 1.1 mm
Figure B3 — A fixed number of seeds grows by the same linear increment $\Delta L=G_r t$; mass scales as $L^3$ and deposition/evaporation duty as the surface $L^2$.

Approach. Under McCabe's $\Delta L$ law the crystal number is conserved from seed to product; the maximum-density magma fixes the volume, the cube-size ratio fixes the seed mass, linear growth fixes the time, and the surface growth rate (converted through the solubility) fixes the evaporation duty.

  1. (a) Crystallizer volume. The 70% fill applies at the start, when the vessel holds only saturated liquor; evaporation then shrinks the liquor, so size from the end state and work back. The final magma must hold all 800 kg of crystals at $\le150\ \text{kg/m}^3$: $$V_\text{magma}=\frac{800}{150}=5.33\ \text{m}^3,\quad V_\text{crystals}=\frac{800}{1900}=0.421\ \text{m}^3,\quad V_\text{liq,f}=5.33-0.421=4.91\ \text{m}^3.$$ The KCl deposited (800 less the 0.31 kg of seed found in (b)) came out of saturated solution holding 400 kg KCl/m³, i.e. from $799.7/400=2.00\ \text{m}^3$ of liquor whose $2.00(1300-400)=1799$ kg of water was evaporated. The starting liquor is therefore $$V_0=4.91+2.00=6.91\ \text{m}^3=0.70\,V_\text{vessel}\ \Longrightarrow\ V_\text{vessel}=\frac{6.91}{0.70}=\boxed{9.87\ \text{m}^3}.$$
  2. (b) Mass of seeds. With $N$ and shape factor constant, mass scales as the cube of size, $W\propto N\rho_c L^3$, so $$\frac{W_p}{W_s}=\left(\frac{L_p}{L_s}\right)^3=\left(\frac{1100}{80}\right)^3=13.75^3=2600\ \Longrightarrow\ W_s=\frac{800}{2600}=\boxed{0.308\ \text{kg}\ (308\ \text{g})}.$$
  3. (c) Batch time. The characteristic size grows linearly from seed to product at $G_r$: $$t=\frac{L_p-L_s}{G_r}=\frac{(1.1\times10^{-3})-(80\times10^{-6})}{3\times10^{-8}}=\boxed{3.40\times10^{4}\ \text{s}\approx9.4\ \text{h}}.$$
  4. (d) Evaporation rates. The seed charge fixes the crystal count $N=\dfrac{W_s}{\rho_c L_s^3}=\dfrac{0.308}{1900(80\times10^{-6})^3}=3.16\times10^{8}$. Deposition follows $\dfrac{dW}{dt}=3N\rho_c L^2 G_r$ (growing $\propto L^2$): $$\left.\frac{dW}{dt}\right|_i=3(3.16\times10^{8})(1900)(80\times10^{-6})^2(3\times10^{-8})=3.46\times10^{-4}\ \text{kg/s},$$ $$\left.\frac{dW}{dt}\right|_f=3(3.16\times10^{8})(1900)(1.1\times10^{-3})^2(3\times10^{-8})=6.55\times10^{-2}\ \text{kg/s}.$$ Holding the mother liquor saturated requires evaporating water: it carries $c=\dfrac{400}{1300-400}=0.444$ kg KCl per kg water, so each kg deposited needs $1/c$ kg evaporated: $$\dot m_{\text{evap},i}=\frac{3.46\times10^{-4}}{0.444}\approx\boxed{2.8\ \text{kg/h}},\qquad \dot m_{\text{evap},f}=\frac{6.55\times10^{-2}}{0.444}\approx\boxed{530\ \text{kg/h}}.$$
QuantityValue
(a) Vessel volume9.87 m³
(b) Seed mass $W_s$0.308 kg (308 g)
(c) Batch time $t$3.40×104 s (≈9.4 h)
(d) Evaporation rate (initial / final)≈2.8 / 530 kg/h
Check: (a) takes the 70% fill at the start of the batch. Treating the final 5.33 m³ magma as the 70% fill (7.62 m³) ignores the 2.0 m³ of liquor lost to evaporation and would under-size the vessel. The solubility is read as 400 kg KCl per m³ of saturated solution ($c=0.444$ kg/kg water); reading it per m³ of water ($c=0.40$, closer to handbook KCl data at 40 °C) gives a vessel of about 10.1 m³ and evaporation rates of about 3.1 / 589 kg/h. The evaporation rates integrated over the batch return the same 1799 kg of water as the volume balance in (a).
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