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

Question 2 of 6: Swenson–Walker Copperas Crystalliser

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, May 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, filtration, crystallization, drying, absorption); Perry's Chemical Engineers' Handbook (9th ed.); Badger & Banchero, Introduction to Chemical Engineering.

Notes on chart / table reads. A1 uses the ISO nozzle discharge-coefficient chart ($C_n$ vs $Re$, $\beta$); B1's isotherm concentrations are printed as “g/cm³” but are read as g/L (= kg/m³) so the feed concentration (0.25 kg/m³) sits on the same scale as the data — a units slip in the source; B3's equilibrium $Y$ column carries an implied ×10−4 multiplier (the printed integers 0–202 cannot be literal mol/mol, and the inlet gas $Y_1=0.0177$ coincides exactly with the $X=10$ entry of 177, which fixes the scale). Engineering choices are flagged in Check callouts.

Part A — Unit Operations

Question A2: Swenson–Walker Copperas Crystalliser (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
Crystal production1 ton/hr = 2000 lb/hr FeSO₄·7H₂O
Solution temperatures120°F → 80°F
Cooling-water temperatures60°F → 70°F
Solubility (per 100 free water)140 @ 120°F ; 74 @ 80°F
$U$ / surface per length35 BTU·hr−1ft−2°F−1 / 3.5 ft²/ft
$c_p$ solution / heat of soln.0.70 BTU/(lb·°F) / 28.5 BTU/lb

Find. (a) cooling-water rate in gpm; (b) number of 10-ft crystalliser sections.

Swenson–Walker crystalliser — countercurrent coolingscraped trough + water jacket (U = 35)feed 120°Fslurry 80°FCW in 60°FCW out 70°F1 ton/hr FeSO₄·7H₂O crystals — 3.5 ft² per ft of length, 10 ft/section
Figure A2 — Cooling crystalliser. Warm saturated copperas solution is cooled by jacket water flowing counter-current; the required heat-transfer surface fixes the number of 10-ft sections.

Approach. Close a mass balance on the hydrate (free water is conserved because each crystal carries its own seven waters) to get the feed rate; add sensible cooling and released heat of crystallization for the total duty; then size the water stream from its temperature rise and the surface from $Q=U A\,\Delta T_\text{lm}$.

  1. Mass balance — free water is the tie component. Copperas dissolved per 100 lb free water falls from 140 to 74, so the crystal yield is $(1.40-0.74)=0.66$ lb per lb free water. For 2000 lb/hr crystals, $$W=\frac{2000}{1.40-0.74}=\boxed{3030\ \mathrm{lb/hr\ free\ water}}.$$
  2. Feed and mother-liquor rates. Dissolved copperas in the feed is $1.40W=4242$ lb/hr, so the feed solution is $$F=W+1.40W=3030+4242=7273\ \mathrm{lb/hr},$$ and the mother liquor $L=W+0.74W=5273$ lb/hr (check: $F=C+L=2000+5273=7273$ &checkmark).
  3. Heat duty. Sensible cooling of the feed plus the heat released as crystals form: $$Q_\text{sens}=F\,c_p\,\Delta T=(7273)(0.70)(40)=2.036\times10^{5}\ \mathrm{BTU/hr},$$ $$Q_\text{cryst}=(2000)(28.5)=5.70\times10^{4}\ \mathrm{BTU/hr},$$ $$Q=Q_\text{sens}+Q_\text{cryst}=\boxed{2.61\times10^{5}\ \mathrm{BTU/hr}}.$$
  4. (a) Cooling-water rate. The water rises 10°F with $c_p=1$: $$\dot m_w=\frac{Q}{c_{p,w}\,\Delta T_w}=\frac{2.606\times10^{5}}{(1)(10)}=2.606\times10^{4}\ \mathrm{lb/hr}.$$ At 8.33 lb/gal, $$\dot V_w=\frac{2.606\times10^{4}}{(8.33)(60)}=\boxed{52\ \mathrm{gpm}}.$$
  5. Log-mean temperature difference (counter-current). Hot solution 120→80°F against water 70→60°F: $$\Delta T_\text{lm}=\frac{(120-70)-(80-60)}{\ln\frac{50}{20}}=\frac{30}{\ln 2.5}=32.7\ \mathrm{^\circ F}.$$
  6. (b) Surface area and number of sections. $$A=\frac{Q}{U\,\Delta T_\text{lm}}=\frac{2.606\times10^{5}}{(35)(32.7)}=227\ \mathrm{ft^2}.$$ At 3.5 ft²/ft this is $227/3.5=65$ ft of length; a standard Swenson-Walker section is 10 ft (35 ft²), so $$N=\frac{227}{35}=6.5\ \Rightarrow\ \boxed{7\ \text{sections}}.$$
QuantityValue
Free water $W$ / feed $F$3030 / 7273 lb/hr
Total heat duty $Q$2.61×105 BTU/hr
Cooling water≈52 gpm
$\Delta T_\text{lm}$ / surface $A$32.7°F / 227 ft²
Number of sections7 (10-ft each)
Check: the section count assumes the standard 10-ft Swenson-Walker unit (35 ft² each); the bare surface requirement is 65 ft, which rounds up to 7 sections. If a different section length is specified, divide the 65 ft accordingly.