Liquid Alum Dosing and Storage for Phosphorus Removal: 18-Env-B9 Environmental Chemistry and Microbiology, May 2016, Q7

Question 7: Liquid Alum Dosing and Storage for Phosphorus Removal (10 marks)

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Given.

Quantity	Value
Influent total P	8.0 mg/L
Allowable effluent P	0.2 mg/L
Al : P dosing ratio	1.5 mol Al / mol P
Plant flow	10,500 m³/d
Alum formula	Al₂(SO₄)₃·18H₂O
Alum strength (liquid)	48% by mass
Liquid alum density	1,280 kg/m³
Storage duration	30 days

Find. Daily liquid-alum dosing rate and the 30-day storage volume.

Approach. Mass-balance the phosphorus actually removed, convert to moles of Al required via the 1.5:1 ratio, convert Al to alum via its formula (2 Al per formula unit), scale the 100%-basis alum mass up through the 48% liquid strength and density to a volumetric feed rate, then multiply by the 30-day storage duration.

Phosphorus removed per day. $\text{mg/L}\times\text{m}^3/\text{d}$ gives g/d directly: $$\Delta P=(8.0-0.2)\ \text{mg/L}\times10{,}500\ \text{m}^3/\text{d}=81{,}900\ \text{g/d}=81.9\ \text{kg/d}$$
Moles of P removed, then moles of Al required (1.5:1). $$n_P=\dfrac{81{,}900}{31}=2{,}642\ \text{mol/d}\qquad n_{Al}=1.5(2{,}642)=3{,}963\ \text{mol/d}$$ Mass of Al: $m_{Al}=3{,}963(27)=107{,}000\ \text{g/d}=107.0\ \text{kg/d}$.
Moles and mass of alum (100% basis). Each mole of $\text{Al}_2(\text{SO}_4)_3\cdot18\text{H}_2\text{O}$ supplies 2 mol Al, and $M=2(27)+3(32+64)+18(18)=54+288+324=666\ \text{g/mol}$: $$n_{\text{alum}}=\dfrac{3{,}963}{2}=1{,}981\ \text{mol/d}\qquad m_{\text{alum,100\%}}=1{,}981(666)=1{,}320{,}000\ \text{g/d}=1{,}320\ \text{kg/d}$$
Liquid alum feed rate (48% strength, 1,280 kg/m³). $$m_{\text{liquid}}=\dfrac{1{,}320}{0.48}=2{,}749\ \text{kg/d}\qquad Q_{\text{alum}}=\dfrac{2{,}749}{1{,}280}=\boxed{2.15\ \text{m}^3/\text{d}}$$
30-day storage volume. $$V_{\text{storage}}=2.15\ \text{m}^3/\text{d}\times30\ \text{d}=\boxed{64.4\ \text{m}^3}$$

Quantity	Value
P removed	81.9 kg/d
Al required	107.0 kg/d
Alum (100% basis)	1,320 kg/d
Liquid alum feed rate	2.15 m³/d
30-day storage volume	64.4 m³

Concept & Theory

Topic. Chemical phosphorus precipitation dosing and bulk-storage sizing.

Key relations. $\Delta P[\text{mg/L}]\times Q[\text{m}^3/\text{d}]=\text{load}\ [\text{g/d}]$; mole ratio scale-up through Al → alum stoichiometry; $V=m/\rho$; storage $=$ daily rate $\times$ days.

Why this works. Al³⁺ precipitates phosphate as AlPO₄ (and Al(OH)₃ co-precipitates additional P and colloids); the 1.5:1 lab-derived ratio accounts for the practical excess over the 1:1 stoichiometric AlPO₄ reaction needed to reach a low effluent target.

Common pitfall. Dosing on the influent (8 mg/L) P rather than the P actually removed (8−0.2 mg/L) — the reagent only needs to precipitate the fraction that must come out of solution to meet the 0.2 mg/L limit.

Source. Metcalf & Eddy, Wastewater Engineering, ch. 6 (chemical phosphorus removal).

Question 7 of 20: Liquid Alum Dosing and Storage for Phosphorus Removal

Section 1: Chemistry (8 questions, 50 marks)

Question 7: Liquid Alum Dosing and Storage for Phosphorus Removal (10 marks)