Question 6 of 6: Batch-Digester House & Recausticizing Mass Balance

Question 6: Batch-Digester House & Recausticizing Mass Balance 20 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.

Batch-digester house mass balance

Given.

Find. (1) white liquor volume, L/batch; (2) black liquor volume, L/batch, and its purpose; (3) brown-stock production rate; (4) recausticizing white-liquor flow (L/min), lime mud (CaCO₃) produced, and the stoichiometric/actual lime (CaO) demand.

Approach. Work one digester's charge from wet-chip volume down to O.D. wood mass, apply the EA% and L/W definitions to get white- and black-liquor volumes, then scale to a house-wide rate. Parts (3) and (4) both require a batch (cook) cycle time, which is not stated in the question; per the exam's own Note 1 ("if doubt exists…submit a clear statement of any assumptions made"), a typical total batch-digester cycle time (fill–cook–blow) of 4.0 hours is assumed. For part (4), the recausticizing white liquor's NaOH/Na₂S/Na₂CO₃ split is backed out from its EA and sulphidity, the causticizing reaction Na₂CO₃+Ca(OH)₂→2NaOH+CaCO₃ (lime slaked in situ, CaO+H₂O→Ca(OH)₂) fixes the lime-mud stoichiometry, and the stated lime availability (the fraction of the reburned/quicklime that is actually active CaO) converts the stoichiometric CaO demand into the ACTUAL raw lime that must be fed to the slaker.

1. Charge mass per digester. Wet-chip volume charged $=0.60\times175=105\ \text{m}^3$, so $$m_{wet}=105\times285=29{,}925\ \text{kg wet wood/batch}\qquad m_{OD}=29{,}925\times(1-0.45)=\boxed{16{,}458.75\ \text{kg O.D. wood/batch}}$$
2. (1) White liquor volume. EA required $=0.180\times16{,}458.75=2962.6\ \text{kg EA}$; at 95 g EA/L: $$V_{WL}=\frac{2962.6\times1000}{95}=\boxed{31{,}185\ \text{L/batch}}\ (31.2\ \text{m}^3/\text{batch})$$
3. (2) Black liquor volume. Total liquor required at L/W=4.1 L per kg O.D. wood: $$V_{total}=4.1\times16{,}458.75=67{,}481\ \text{L/batch}$$ $$V_{BL}=V_{total}-V_{WL}=67{,}481-31{,}185=\boxed{36{,}296\ \text{L/batch}}\ (36.3\ \text{m}^3/\text{batch})$$ The recycled black liquor is added to raise the liquor-to-wood ratio to the level needed for adequate chip saturation, uniform liquor circulation and heat transfer through the charge — not to add fresh cooking chemical — while simultaneously returning residual NaOH/Na₂S left in that black liquor to the cook (chemical economy) instead of drawing an equivalent extra volume of fresh white liquor.
4. (3) Brown-stock production rate. Pulp per batch per digester $=0.54\times16{,}458.75=8887.7\ \text{kg O.D. pulp}$. With the assumed 4.0 h cycle, the house (6 digesters, staggered) discharges $6/4.0=1.5$ batches/h: $$\dot m_{pulp}=8887.7\times1.5=13{,}332\ \text{kg/h}=\boxed{320.0\ \text{O.D. tonnes/day}}\ (\approx356\ \text{ADMT/day at 90\% AD})$$
5. (4a) Recausticizing white-liquor flow. $$\dot V_{WL}=31{,}185\ \text{L/batch}\times1.5\ \text{batch/h}\div60=\boxed{779.6\ \text{L/min}}$$
6. (4b) White-liquor NaOH/Na₂S/Na₂CO₃ split. With EA=NaOH+0.5Na₂S (as Na₂O) and sulphidity S=Na₂S/(NaOH+Na₂S): $$\text{NaOH}=\frac{93}{1+0.5(0.28/0.72)}=\boxed{79.53\ \text{g/L (as Na}_2\text{O)}}$$ Causticizing efficiency CE=NaOH/(NaOH+Na₂CO₃) gives the residual (uncausticized, i.e. causticized-from) carbonate: $$\text{Na}_2\text{CO}_3=\text{NaOH}\left(\frac{1}{CE}-1\right)=79.53(1/0.82-1)=\boxed{17.46\ \text{g/L (as Na}_2\text{O)}}=29.87\ \text{g/L actual Na}_2\text{CO}_3$$
7. (4c) Lime mud and lime stoichiometry. Causticizing reaction $\text{Na}_2\text{CO}_3+\text{Ca(OH)}_2\rightarrow2\text{NaOH}+\text{CaCO}_3$ (1:1:1 molar), lime slaked in situ via $\text{CaO}+\text{H}_2\text{O}\rightarrow\text{Ca(OH)}_2$. At 29.87 g Na₂CO₃/L (M=105.99 g/mol) → 0.2818 mol/L: $$\text{CaO (stoichiometric)}=0.2818\times56.08=15.80\ \text{g/L WL}\qquad \text{CaCO}_3\ \text{(lime mud)}=0.2818\times100.09=28.20\ \text{g/L WL}$$ Scaling by the white-liquor flow (779.6 L/min): $$\dot m_{CaCO_3}=28.20\times779.6/1000=\boxed{21.98\ \text{kg/min}}\ (31.7\ \text{t/day, lime mud to the kiln})$$ $$\dot m_{CaO,stoich}=15.80\times779.6/1000=\boxed{12.32\ \text{kg/min}}$$ Correcting for the stated 88% lime availability (fraction of the reburned lime that is actually active CaO) gives the ACTUAL raw lime the slaker must be fed: $$\dot m_{CaO,actual}=12.32/0.88=\boxed{14.00\ \text{kg/min}}\ (20.2\ \text{t/day})$$