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23-Chem-A2 Unit Operations and Separation Processes · May 2016

Question 4 of 6: Cooling Crystallization of Glauber's Salt

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, May 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) and Vol. 1 (heat transfer); McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Geankoplis, Transport Processes and Separation Process Principles (4th ed.); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (8th ed.); Perry's Chemical Engineers' Handbook (9th ed.).

Note on saturation data. Sections B2 and B3 need water saturation temperatures and latent heats that the exam expects from an open-book steam table. These are taken from standard tables (and, for reproducibility). The B1 latent heat of evaporation is not printed in the exam and is taken as 2370 kJ/kg at the cooling range; this and other engineering choices are flagged in Check callouts.

Question B1: Cooling Crystallization of 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. Batch cooling crystallizer.

QuantityValue
Na₂SO₄ charged500 kg
Water charged2500 kg
Vessel (mild steel)750 kg
Cooling range333 K → 283 K (ΔT = 50 K)
Solubility at 283 K8.9 kg anhyd./100 kg water
Crystal phaseNa₂SO₄·10H₂O (M = 322)
Water evaporated2% of 2500 = 50 kg
Heat of solution (291 K)−78.5 MJ/kmol

Find. The total heat $Q$ that must be removed by the coolant during the batch.

Batch cooling crystallizer — energy streams Na₂SO₄ solution 333 K → 283 K Na₂SO₄·10H₂O crystals 50 kg H₂O vapour (+Q abs) Q removed
Figure B1 — The coolant removes sensible heat (solution + vessel) plus the heat of crystallization released, less the latent heat absorbed by the 50 kg of evaporating water.

Approach. First a mass balance on the mother liquor at 283 K (with 50 kg water gone) fixes the yield of decahydrate; then an energy balance sums sensible cooling of solution and steel, plus heat of crystallization released, minus the evaporation latent load.

  1. Decahydrate yield from the solubility balance. Let $C$ be kg of Na₂SO₄·10H₂O formed. It removes $\tfrac{142}{322}C$ of anhydrous salt and $\tfrac{180}{322}C$ of water. With residual water $=2500-50=2450$ kg and solubility 0.089, $$500-\tfrac{142}{322}C=0.089\,(2450-\tfrac{180}{322}C)\;\Rightarrow\;C=\;\boxed{721\ \mathrm{kg}}.$$
  2. Sensible heat of solution + vessel. Cooling the full charge (3000 kg at 3.6 kJ/kg·K) and the 750 kg steel (0.5 kJ/kg·K) over 50 K: $$Q_{sens}=(3000\cdot3.6+750\cdot0.5)(50)=(10800+375)(50)=558.8\ \mathrm{MJ}.$$
  3. Heat of crystallization released. Crystallization is the reverse of dissolution, so it releases 78.5 MJ per kmol of decahydrate. With $721/322=2.24$ kmol, $$Q_{cryst}=2.24(78.5)=175.7\ \mathrm{MJ}\ \text{(released)}.$$
  4. Latent heat absorbed by evaporation. Evaporating 50 kg of water absorbs (at $\lambda\approx2370$ kJ/kg for the cooling range) $$Q_{evap}=50(2370)=118.5\ \mathrm{MJ}\ \text{(absorbed)}.$$
  5. Net heat to be removed. The coolant must take the sensible load and the released crystallization heat, but the evaporation carries some heat off with the vapour: $$Q=Q_{sens}+Q_{cryst}-Q_{evap}=558.8+175.7-118.5=\;\boxed{616\ \mathrm{MJ}}.$$
ContributionValue (MJ)
Decahydrate crystallized721 kg (2.24 kmol)
Sensible (solution + vessel)+558.8
Heat of crystallization released+175.7
Latent heat of evaporation−118.5
Heat to be removed≈ 616
Check (data taken from tables): the heat of solution is quoted at 291 K and applied as the heat of crystallization over the 283–333 K range (its temperature variation is second-order here); the latent heat of the evaporating water, not printed in the exam, is taken as $\lambda\approx2370$ kJ/kg. Omitting the evaporation term (a common exam simplification) gives $Q\approx734$ MJ.