Question 5 of 18: Sludge Production With and Without Ferric Chloride, and Lime Requirement

Question 5: Sludge Production With and Without Ferric Chloride, and Lime Requirement (15 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.

Given.

Find. (a) Mass and volume of raw (no-chemical) primary sludge; (b) mass and volume of chemical (FeCl₃-enhanced) sludge; (c) lime (Ca(OH)₂) required, if any, to supplement the natural alkalinity for the FeCl₃ reaction.

Approach. Compute dry-solids capture from each TSS-removal fraction, convert to wet sludge using moisture content and specific gravity; separately balance the FeCl₃–bicarbonate reaction to find the Fe(OH)₃ precipitate mass (added to the chemical sludge) and the alkalinity it consumes, then compare the alkalinity consumed against the 136 mg/L available to decide whether lime supplementation is needed.

1. Raw sludge (no chemical), 60% TSS removal. $$m_{dry}=0.60\times220\ \tfrac{\text{mg}}{\text{L}}\times1{,}000\ \text{m}^3/\text{d} = 132{,}000\ \text{g/d}=132.0\ \text{kg/d}$$ At 94% moisture, solids are 6% of wet mass: $$m_{wet}=\frac{132.0}{1-0.94}=2{,}200\ \text{kg/d}, \qquad \boxed{V_{no\text{-}chem}=\frac{2{,}200}{1.03\times1{,}000}=2.14\ \text{m}^3/\text{d}}$$
2. Fe(OH)₃ precipitate from the FeCl₃ dose. Balancing $2\text{FeCl}_3+3\text{Ca(HCO}_3)_2\to2\text{Fe(OH)}_3+3\text{CaCl}_2+6\text{CO}_2$ (1 mol Fe(OH)₃ per mol FeCl₃), with $MW_{\text{FeCl}_3}=162.2$ and $MW_{\text{Fe(OH)}_3}=106.85\ \text{g/mol}$: $$n_{\text{FeCl}_3}=\frac{40{,}000\ \text{g/d}}{162.2\ \text{g/mol}}=246.6\ \text{mol/d} \;\Rightarrow\; m_{\text{Fe(OH)}_3}=246.6\times106.85=26{,}350\ \text{g/d}=26.35\ \text{kg/d}$$
3. Chemical sludge (with FeCl₃), 85% TSS removal + precipitate. $$m_{TSS}=0.85\times220\times1{,}000=187{,}000\ \text{g/d}=187.0\ \text{kg/d}$$ $$m_{dry,total}=187.0+26.35=213.35\ \text{kg/d}$$ At 92.5% moisture: $$m_{wet}=\frac{213.35}{1-0.925}=2{,}844.7\ \text{kg/d}, \qquad \boxed{V_{chem}=\frac{2{,}844.7}{1.05\times1{,}000}=2.71\ \text{m}^3/\text{d}}$$
4. Alkalinity consumed vs. available — is lime needed? The balanced reaction consumes 3 mol Ca(HCO₃)₂ (each equal to 2 eq $=100$ g CaCO₃-equivalent) per 2 mol FeCl₃: $$\text{alkalinity consumed}=\frac{3}{2}\times246.6\ \text{mol/d}\times100\ \text{g/mol} = 36{,}990\ \text{g/d} = 37.0\ \text{mg/L (as CaCO}_3\text{, over }Q\text{)}$$ Comparing against the 136 mg/L available: $$\text{residual alkalinity}=136-37.0=99.0\ \text{mg/L as CaCO}_3 \;(>0)$$ Since the natural alkalinity comfortably exceeds the FeCl₃ demand (and the 99 mg/L residual is well above the typical 50–70 mg/L minimum reserve used to protect downstream biological pH), $$\boxed{\text{no supplemental lime is required for this FeCl}_3\text{ dose}}$$