Question 1 of 6: Cooling-Crystallisation Heat Load (Na₂SO₂·10H₂O)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2019 — 16-Chem-A3 Heat and Mass Transfer. Three-hour, open-book exam (one textbook of the candidate’s choice; any non-communicating calculator). Format: two parts — Part A (Q1–Q3) Heat Transfer and Part B (Q1–Q3) Mass Transfer; at least two questions must be attempted from each part and only the first two in each part are marked, so four questions (each 25 points) constitute a complete paper. All six questions are solved below for completeness. Property values not printed on the paper (molar masses, water latent heat, the dimensionless free-convection peak velocity $f'_{max}$, benzene/toluene physical properties) are stated explicitly in each Given block as open-book look-ups.
Reference texts: Coulson & Richardson (Backhurst, Harker & Richardson), Chemical Engineering, Vol. 1 — Fluid Flow, Heat Transfer and Mass Transfer (6th ed., Butterworth-Heinemann) — the source family for the crystalliser, tube-condenser, Stefan-tube and distillation problems; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (free- and forced-convection correlations, the Ostrach similarity solution); Treybal, Mass-Transfer Operations (3rd ed.) and McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.) — Stefan diffusion, Chilton–Colburn analogy, McCabe–Thiele; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.) and the NIST Chemistry WebBook.
The obscured verb in Part B Q1 is taken as “calculate”, and the Part B Q2 water viscosity at 27 °C is $8.76\times10^{-4}$ Pa·s. One column of the Part A Q2 air-property table, the buoyancy group $g\beta/\nu^{2}$, is internally inconsistent: its printed values are about 40% below what $\rho$, $\mu$ and $\beta=1/T_f$ give. That column is not used; the Grashof number is built from the $\rho$ and $\mu$ columns and $\beta=1/T_f$.
Given. A batch of 500 kg Na₂SO₂ dissolved in 2500 kg water is cooled 50 → 10 °C in a 750 kg steel vessel; 2% of the water evaporates. Glauber salt (Na₂SO₂·10H₂O) crystallises; the mother liquor holds 8.9 kg anhydrous salt per 100 kg free water at 10 °C.
Quantity
Value
Na₂SO₂ charged
500 kg
Water charged
2500 kg
Vessel (mild steel)
750 kg, $c_p$ = 0.5 kJ/kg·K
Solution $c_p$
3.6 kJ/kg·K
Solubility at 10 °C
8.9 kg/100 kg water
Heat of solution
−78.5 kJ/mol (decahydrate)
$M$: Na₂SO₂ / decahydrate / H₂O
142.04 / 322.19 / 18.015 g/mol
Find. The total heat that the coolant must remove between 50 °C and 10 °C.
Figure 1 — Batch cooler-crystalliser. Feed cools 50 → 10 °C; 2% of the water leaves as vapour, Glauber salt crystallises, and the coolant removes sensible + crystallisation heat less the evaporative credit.
Approach. First a mass balance on the anhydrous salt (with the decahydrate’s bound water removed from the free-water pool) fixes the crystal yield; then the coolant duty is the sensible heat of solution + vessel, plus the heat released on crystallisation, minus the latent heat carried off by the evaporating water.
Free water after evaporation. The vapour loss is $0.02\times2500=50$ kg, leaving $2500-50=2450$ kg of water before crystal water is removed.
Crystal yield from a salt balance. Let $C$ = kg of Na₂SO₂·10H₂O. The crystal is $f_s=142.04/322.19=0.4409$ anhydrous salt and $f_w=0.5591$ bound water, so the free water is $(2450-0.5591\,C)$ and the dissolved salt is $0.089\times$(free water). Balancing anhydrous salt:$$500 = 0.4409\,C + 0.089\,(2450-0.5591\,C) \;\Rightarrow\; 281.9 = 0.3911\,C \;\Rightarrow\; \boxed{C = 721\ \text{kg Na}_2\text{SO}_4\!\cdot\!10\text{H}_2\text{O}}.$$Check: crystals hold $0.4409\times721=318$ kg salt, liquor holds $182$ kg in $2450-0.5591\times721=2047$ kg water, i.e. $182/2047=8.9$ per 100 — consistent.
Sensible heat, 50 → 10 °C. Cooling the whole 3000 kg charge plus the 750 kg vessel:$$Q_{sens}=\big[(3000)(3.6)+(750)(0.5)\big]\,(40)=(10800+375)(40)=\boxed{4.47\times10^{5}\ \text{kJ}}.$$
Heat of crystallisation. The moles crystallising are $721/0.32219=2.238\times10^{3}$ mol. Crystallisation is the reverse of dissolution, so it releases $|{-78.5}|=78.5$ kJ/mol:$$Q_{cryst}=2238\times78.5=\boxed{1.76\times10^{5}\ \text{kJ (released)}}.$$
Evaporative credit. The 50 kg of water leaving as vapour carries off its latent heat ($\lambda\approx2430$ kJ/kg at these low temperatures):$$Q_{evap}=50\times2430=1.22\times10^{5}\ \text{kJ (carried away by vapour)}.$$
Coolant duty. Sensible and crystallisation heat must be removed; evaporation removes some for free:$$Q=Q_{sens}+Q_{cryst}-Q_{evap}=447{,}000+176{,}000-122{,}000=\boxed{5.0\times10^{5}\ \text{kJ}\;(\approx501\ \text{MJ})}.$$
Check (sign & property look-ups): Glauber salt dissolves endothermically (it is a textbook cold-pack salt), so its crystallisation is exothermic and adds to the cooling load — the heat of crystallisation is $-(\text{heat of solution})=+78.5$ kJ/mol released. The water latent heat is not printed on the paper; $\lambda\approx2430$ kJ/kg (steam tables, ~30 °C) is used, and the answer changes by only ~2% for any reasonable $\lambda$ between 2380 and 2480 kJ/kg. Sensible heat uses the full charge mass (the standard estimate).
