23-Chem-A1 Process Balances and Chemical Thermodynamics · December 2016
Question 1 of 6: Soda-Ash Calciner with Dry Recycle
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2016 — 04-Chem-A1 Process Balances and Chemical Thermodynamics. Three-hour, open-book exam; any non-communicating calculator permitted. Format: two parts — Part A (Q1–Q3) Process Balances, Part B (Q4–Q6) Chemical Thermodynamics; the candidate answers two questions from each part (four constitute a complete paper, equal value). All six questions are solved below for completeness. Property data not printed on the paper (molar masses, critical constants, acentric factors, Rackett Zc) are stated explicitly in each Given block as open-book look-ups.
Reference texts: Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes (4th ed., Wiley) — material balances, combustion/Orsat analysis, humidity; Himmelblau & Riggs, Basic Principles and Calculations in Chemical Engineering (8th ed.) — heats of reaction and Kirchhoff’s law; Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics (8th ed., McGraw-Hill) — fugacity, generalized virial correlations, reaction equilibrium and VLE; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.) and the NIST Chemistry WebBook.
Question 1: Soda-Ash Calciner with Dry Recycle (Part A — equal value)
Given. Wet NaHCO₃ at 2000 kg/h containing 8 wt% free water is blended with a dry Na₂CO₃ recycle stream $R$ to lower the water to 5 wt% before the calciner. Decomposition: $2\,\text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O}$. Molar masses (g/mol): NaHCO₃ 84.01, Na₂CO₃ 105.99, CO₂ 44.01, H₂O 18.02.
Quantity
Value
Wet NaHCO₃ fed
2000 kg/h
— free water in it (8%)
160 kg/h
— dry NaHCO₃ in it
1840 kg/h
Target water after mixing
5 wt%
Recycle stream
dry Na₂CO₃ (no water)
Find. (a) net soda-ash product rate; (b) off-gas rate; (c) CO₂:H₂O mole ratio in the off-gas; (d) recycle rate.
Figure 1 — Wet bicarbonate is blended with dry recycled soda ash to hit 5% moisture, then calcined. The recycle circulates soda ash internally, so the net product is only the freshly formed Na₂CO₃.
Approach. The recycle carries no water, so a single water balance on the mixing tee fixes $R$ independently of the reaction; the dry-bicarbonate reaction then gives the fresh Na₂CO₃, CO₂ and reaction water, and the free moisture leaves with the off-gas. The printed 2000 kg/h is read as the fresh wet (8% water) bicarbonate, as worded; the calciner itself then receives the 3200 kg/h blend.
Recycle rate from a water balance on the mixing tee. The only water is the 160 kg/h in the wet feed; the recycle is dry. Requiring 5% water in the blended stream $(2000+R)$:
$$\frac{160}{2000+R}=0.05 \;\Rightarrow\; 2000+R=3200 \;\Rightarrow\; \boxed{R = 1200\ \text{kg/h}}.$$
This is part (d): the mixing requirement alone fixes the recycle, with no reference to the reaction.
Extent of decomposition. Only the fresh dry NaHCO₃ (1840 kg/h) decomposes; the recycled Na₂CO₃ is already calcined and passes through. Moles of bicarbonate:
$$\dot n_{\text{NaHCO}_3}=\frac{1840}{84.01}=21.90\ \text{kmol/h}\;\Rightarrow\; \dot n_{\text{Na}_2\text{CO}_3}=\dot n_{\text{CO}_2}=\dot n_{\text{H}_2\text{O,rxn}}=\tfrac12(21.90)=10.95\ \text{kmol/h}.$$
Net soda ash produced (part a). The freshly formed Na₂CO₃ is the net product; the 1200 kg/h of recycled Na₂CO₃ merely circulates:
$$\dot m_{\text{Na}_2\text{CO}_3}=10.95\times105.99=\boxed{1161\ \text{kg/h}}.$$
Off-gas rate (part b). The off-gas is CO₂ plus all the water vapour — reaction water plus the 160 kg/h of free moisture driven off:
$$\dot m_{\text{CO}_2}=10.95\times44.01=482.0,\quad \dot m_{\text{H}_2\text{O,off}}=10.95\times18.02+160=357.3\ \text{kg/h},$$
$$\dot m_{\text{off-gas}}=482.0+357.3=\boxed{839\ \text{kg/h}}.$$
CO₂-to-water mole ratio (part c). Water vapour is reaction water (10.95 kmol/h) plus free moisture $(160/18.02=8.88$ kmol/h$)$:
$$\frac{\dot n_{\text{CO}_2}}{\dot n_{\text{H}_2\text{O}}}=\frac{10.95}{10.95+8.88}=\frac{10.95}{19.83}=\boxed{0.552\ \ (\approx 1:1.81)}.$$
The ratio being below 1 : 1 is the tell that free moisture, not just stoichiometric water, is in the off-gas.
An overall check closes the process: in = 2000 kg/h; out = 1161 (net soda ash) + 839 (off-gas) = 2000 kg/h. ✓