23-Chem-A1 Process Balances and Chemical Thermodynamics · May 2013
Question 1 of 7: Crystallization Separation of Sylvinite
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2013 — 04-Chem-A1 Process Balances and Chemical Thermodynamics. Three-hour, open-book exam; any non-communicating calculator permitted. Format: seven questions in three parts — answer one of Q1–Q2 (Part A), one of Q3–Q4 (Part B) and two of Q5–Q7 (Part C); four questions of equal value constitute a complete paper. All seven are solved below for completeness. Where property data are needed they are stated explicitly in each Given block. Units follow the paper (mixed SI and US customary).
Reference texts: Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes (4th ed., Wiley) — material & energy balances; Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics (8th ed., McGraw-Hill) — VLE, residual properties, reaction equilibrium; supporting data from Perry's Chemical Engineers' Handbook (9th ed.) and the NIST Chemistry WebBook.
Question 1: Crystallization Separation of Sylvinite (equal value)
Given. Sylvinite feed $S$ (42.7% NaCl, 57.3% KCl) dissolved in $W=1000$ kg water, cooled to crystallize. Product crystals $C=727$ kg and a mother liquor $M$; compositions (mass%):
Stream
NaCl
KCl
H₂O
Sylvinite $S$
42.7
57.3
0
Water $W$ (1000 kg)
0
0
100
Crystals $C$ (727 kg)
0.5
72.1
27.4
Mother liquor $M$
salts (balance)
50.0
Find. (a) the mass of sylvinite $S$; (b) the full composition of the mother liquor $M$; (c) the percent recovery of KCl into the crystals.
Figure 1 — Cooling crystallizer: sylvinite + water in; KCl-rich crystals and salt-laden mother liquor out. Three species (NaCl, KCl, H₂O) are conserved.
Approach. The water balance alone fixes the mother liquor (its water is known and it is 50% water); the overall balance then gives the feed, and the two salt balances give the mother-liquor composition and the KCl recovery.
Resolve the crystal stream into components. From the crystal composition on the 727 kg product,
$$m_{\text{KCl}}^C = 0.721(727)=524.17,\quad m_{\text{NaCl}}^C = 0.005(727)=3.64,\quad m_{\text{H}_2\text{O}}^C = 0.274(727)=199.20\ \text{kg}.$$
Water balance $\Rightarrow$ mother liquor mass. All 1000 kg of water leaves in either the crystals or the mother liquor, so the water in the mother liquor is $1000-199.20 = 800.80$ kg. Because the mother liquor is 50.0% water,
$$M=\frac{800.80}{0.500}=\boxed{1601.6\ \text{kg}},\qquad \text{salts in }M = 1601.6-800.8 = 800.80\ \text{kg}.$$
Overall balance $\Rightarrow$ sylvinite fed. Total mass in equals total mass out, $S+1000 = C+M$:
$$S = 727 + 1601.6 - 1000 = \boxed{1328.6\ \text{kg of sylvinite}}.$$
As a check, all the sylvinite is salt, and salt out $=527.80\ (\text{crystals}) + 800.80\ (\text{liquor}) = 1328.6$ kg — the salt balance closes exactly.
Salt balances $\Rightarrow$ mother-liquor composition. With the feed known, subtract the salts already in the crystals:
$$m_{\text{NaCl}}^M = 0.427(1328.6)-3.64 = 563.68\ \text{kg},\qquad m_{\text{KCl}}^M = 0.573(1328.6)-524.17 = 237.12\ \text{kg}.$$
Dividing by $M=1601.6$ kg gives mother-liquor mass fractions of 35.19% NaCl, 14.81% KCl and 50.0% water.
KCl recovery. KCl entering (all in the sylvinite) is $0.573(1328.6)=761.29$ kg, of which 524.17 kg reports to the crystals:
$$\text{Recovery}=\frac{524.17}{761.29}=0.6885=\boxed{68.9\%\ \text{of the KCl}}.$$
