23-Chem-A1 Process Balances and Chemical Thermodynamics · December 2013
Question 2 of 7: Evaporator–Crystallizer Production of Potassium Sulfate
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 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. Property data needed (densities, molar masses, steam-table and thermochemical values) 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, humidity and gas laws; Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics (8th ed., McGraw-Hill) — excess properties, VLE, residual properties and reaction equilibrium; supporting property data from Perry's Chemical Engineers' Handbook (9th ed.), the NIST/ASME steam tables and the NIST Chemistry WebBook.
Question 2: Evaporator–Crystallizer Production of Potassium Sulfate (equal value)
Given. Fresh feed FF is 18.6 wt% K₂SO₄ (81.4 wt% water). The crystallizer product P is 10 kg solid K₂SO₄ per 1 kg of 40 wt% solution; the recycled filtrate R is a 40 wt% K₂SO₄ solution. Evaporator removes 42.65 % of the water fed to it; the (maximum) evaporation rate is $E = 100$ kg/min.
Stream
K₂SO₄
Water
Fresh feed FF
18.6 %
81.4 %
Recycle R (40 % soln)
40 %
60 %
Product P (per 11 kg: 10 solid + 1 kg 40 % soln)
94.55 %
5.45 %
Evaporated water E
0
100 % (100 kg/min)
Find. (a) solid-K₂SO₄ production rate; (b) fresh-feed rate FF; (c) recycle-to-fresh-feed ratio R/FF.
Approach. The recycle is internal to the overall system, so an overall balance (FF in; E and P out) with one K₂SO₄ balance solves the feed and product; the 42.65 % evaporation datum then fixes the water into the evaporator, and a water balance on the mixing point gives the recycle.
Composition of the product stream P. Per 11 kg of P there are 10 kg solid K₂SO₄ plus 1 kg of 40 % solution (0.4 kg K₂SO₄ + 0.6 kg water):
$$x_{K}^{P} = \frac{10+0.4}{11} = 0.9455,\qquad x_{W}^{P} = \frac{0.6}{11} = 0.0545.$$
Overall balances $\Rightarrow$ P and FF. The recycle never leaves the system, so overall $\text{FF}=E+P$ and the K₂SO₄ balance is $0.186\,\text{FF}=x_K^{P}\,P$. Eliminating FF:
$$P = \frac{0.186\,E}{x_K^{P}-0.186} = \frac{0.186(100)}{0.9455-0.186} = \boxed{24.49\ \text{kg/min}},\qquad \text{FF}=100+24.49=\boxed{124.5\ \text{kg/min}}.$$
The water balance checks: $0.814(124.5)=101.3 = 100 + 0.0545(24.49)$. ✓
Solid-K₂SO₄ production (part a). Ten of every 11 kg of P is solid crystal:
$$\dot m_{\text{solid}} = \tfrac{10}{11}P = \tfrac{10}{11}(24.49) = \boxed{22.3\ \text{kg/min of solid K}_2\text{SO}_4}.$$
Water into the evaporator (from the 42.65 % datum). The evaporator removes 42.65 % of the water fed to it, and that removal is 100 kg/min:
$$\text{water in }F = \frac{E}{0.4265} = \frac{100}{0.4265} = 234.5\ \text{kg/min}.$$
Water balance on the mixing point $\Rightarrow$ recycle (part c). The feed to the evaporator is FF + R; its water is the fresh-feed water plus the recycle water ($R$ is 60 % water):
$$0.814(124.5) + 0.6\,R = 234.5 \;\Rightarrow\; R = \frac{234.5-101.3}{0.6} = \boxed{221.9\ \text{kg/min}},$$
$$\frac{R}{\text{FF}} = \frac{221.9}{124.5} = \boxed{1.78}.$$
Every internal stream then closes: feed to evaporator $=346.4$, to crystallizer $=246.4$, $P+R = 24.49+221.9 = 246.4$ kg/min. ✓
