23-Chem-A2 Unit Operations and Separation Processes · December 2016
Question 4 of 6: Yield from Cooling Crystallization of KNO 3
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — 04-CHEM-A2 Mechanical and Thermal Operations, December 2016. 3 hours, open book. Six problems (Section A Mechanical Operations: A1–A3; Section B Thermal Operations: B1–B3), each 25 marks; candidates attempt at least two from each section (only the first two per section are marked). All six are worked below for completeness.
Reference texts. Coulson & Richardson, Chemical Engineering Vol. 2 (particle technology, sedimentation, fluidization, filtration, crystallization, evaporation, drying) and Vol. 1 (fluid flow, heat transfer); McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Geankoplis, Transport Processes and Separation Process Principles (4th ed.); Perry's Chemical Engineers' Handbook (9th ed.); Kern, Process Heat Transfer.
Note on chart and property data. A1 uses a Fanning friction-factor chart, B1 a KNO3 solubility curve (mole-fraction basis), and B2/B3 standard steam-table / transport-property values. Engineering choices are flagged in Check callouts.
Question B1: Yield from Cooling Crystallization of KNO3(25 marks)
Given. Feed 62.5 wt% KNO3 (anhydrous crystals; no water of crystallization), cooled from 100 °C to 10 °C in a closed crystallizer (no evaporation). Solubility read from the mole-fraction chart: $x_\text{sat}\approx0.035$ at 10 °C (and $\approx0.10$ at 40 °C, $\approx0.165$ at 60 °C). Molar masses $M_\text{KNO3}=101.1$, $M_\text{water}=18.0$.
Find. (a) a general expression for solids yield vs. temperature; (b) the crop at 10 °C for a 22,680 kg charge.
Figure B1 — On cooling, the liquor follows the saturation curve; the KNO₃ that can no longer stay dissolved (difference between feed load and 10 °C solubility) crystallizes out.
Approach. Water is conserved (no evaporation) and the crystals are anhydrous, so on a fixed-water basis the mother liquor at any temperature holds exactly its saturation amount of KNO3; the yield is the feed KNO3 minus what the liquor retains, evaluated through the solubility curve.
Charge composition on a molar basis. For 22,680 kg feed: KNO3 $=0.625(22680)=14{,}175$ kg $=140.2$ kmol; water $=0.375(22680)=8505$ kg $=472.0$ kmol. Feed mole fraction $x_\text{feed}=140.2/612.2=0.229$ — the liquor is unsaturated at 100 °C and begins to deposit only below $\approx79$ °C.
KNO3 retained by the mother liquor at temperature $T$. Water moles $n_W$ are fixed; if the saturated mole fraction is $x_\text{sat}(T)$, the dissolved KNO3 is $$n_\text{ML}=\frac{x_\text{sat}}{1-x_\text{sat}}\,n_W.$$
Yield as a function of temperature. $$\boxed{\;Y(T)=M_\text{KNO3}\Big[n_K-\frac{x_\text{sat}(T)}{1-x_\text{sat}(T)}\,n_W\Big]\;}$$ with $n_K,n_W$ the feed moles of salt and water. The yield is zero at the saturation temperature of the feed ($\approx79$ °C) and grows as $T$ falls and $x_\text{sat}$ drops.
Tabulate the yield along the cooling path. Read $x_\text{sat}$ off the solubility chart; the curve passes through about (0.05, 20 °C), (0.10, 40 °C), (0.15, 55 °C) and (0.20, 70 °C). Applying the boxed expression to the 22,680 kg charge ($n_K=140.2$, $n_W=472.0$ kmol) gives:
$T$ (°C)
$x_\text{sat}$ (chart)
Crystals (kg)
% of feed KNO3
≈79 (onset)
0.229
0
0
70
0.200
2,250
16
60
0.165
4,750
33
50
0.133
6,860
48
40
0.100
8,870
63
30
0.072
10,470
74
20
0.050
11,660
82
10
0.035
12,440
88
Per kilogram of original solution, the yield is these values divided by 22,680. About 84% of the final crop has already formed by the time the liquor reaches 30 °C.
Crop at 10 °C. With $x_\text{sat}=0.035$, the liquor retains $n_\text{ML}=\dfrac{0.035}{0.965}(472.0)=17.1$ kmol, so crystallized KNO3 $=140.2-17.1=123.1$ kmol: $$Y(10^\circ\mathrm C)=123.1(101.1)=\;\boxed{12{,}440\ \mathrm{kg}}.$$ That is 88% of the KNO3 fed (14,175 kg) recovered as solids, or 55% of the original solution mass. The chart can be read to only about ±0.002 in mole fraction near 10 °C, so the crop is good to roughly ±100 kg: a read of 0.037 gives 12,340 kg. The handbook solubility of 20.9 g per 100 g water corresponds to $x=0.036$.
