Alum Dosing for Phosphorus Removal: 18-Env-B9 Environmental Chemistry and Microbiology, May 2013, Q9

Question 9: Alum Dosing for Phosphorus Removal (7 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 data
Quantity	Value
Plant flow, $Q$	12,000 m³/d
Influent TP (50th percentile)	6.0 mg/L
Target effluent TP	1.0 mg/L
Design Al:P molar ratio	1.5 : 1
MW Al / MW P	27 / 31 g/mol
Alum formula, MW	$\text{Al}_2(\text{SO}_4)_3\cdot18\text{H}_2\text{O}$, 666.5 g/mol
Liquid alum strength / density	48% w/w, 1.2 kg/L

Find. (9.1) liquid alum feed rate, m³/d; (9.2) storage volume for a 30-day supply, m³.

Approach. Mass-balance the phosphorus that must be removed, convert to moles of P, apply the 1.5:1 Al:P molar dosing ratio to get moles (then mass) of Al required, convert mass of Al to mass of the alum salt (2 mol Al per mol alum), then to mass and volume of the 48% liquid product, and finally scale by the 30-day storage duration.

Phosphorus to be removed. $$\Delta\text{TP} = 6.0-1.0 = 5.0\ \text{mg/L}, \qquad \dot{m}_P = \Delta\text{TP}\times Q = 5.0\ \tfrac{\text{g}}{\text{m}^3}\times12{,}000\ \tfrac{\text{m}^3}{\text{d}} = 60{,}000\ \text{g/d} = 60.0\ \text{kg/d}$$
Moles of P and required moles/mass of Al. $$n_P = \frac{60{,}000\ \text{g/d}}{31\ \text{g/mol}} = 1935.5\ \text{mol/d}, \qquad n_{Al}=1.5\,n_P = 2903.2\ \text{mol/d}$$ $$m_{Al} = 2903.2\ \text{mol/d}\times27\ \text{g/mol} = 78{,}390\ \text{g/d} = 78.4\ \text{kg/d}$$
Mass of alum salt (100% basis). Each mole of $\text{Al}_2(\text{SO}_4)_3\cdot18\text{H}_2\text{O}$ supplies 2 mol Al, so $$n_{\text{alum}} = n_{Al}/2 = 1451.6\ \text{mol/d}, \qquad m_{\text{alum,100\%}} = 1451.6\times666.5 = 967{,}500\ \text{g/d} = 967.5\ \text{kg/d}$$
9.1 — Liquid alum feed rate. The commercial product is only 48% alum by mass, at density 1.2 kg/L: $$m_{\text{liquid}} = \frac{967.5\ \text{kg/d}}{0.48} = 2015.6\ \text{kg/d}, \qquad V_{\text{liquid}} = \frac{2015.6\ \text{kg/d}}{1.2\ \text{kg/L}} = 1679.7\ \text{L/d}$$ $$\boxed{V_{\text{liquid}} \approx 1.68\ \text{m}^3/\text{d}\ (1680\ \text{L/d})}$$
9.2 — 30-day storage capacity. $$V_{\text{storage}} = 1679.7\ \tfrac{\text{L}}{\text{d}}\times30\ \text{d} = 50{,}390\ \text{L} = \boxed{50.4\ \text{m}^3}$$

Alum dosing — final results
Quantity	Value
P removal rate	60.0 kg/d
Al required	78.4 kg/d
Alum salt required (100% basis)	967.5 kg/d
Liquid alum (48%) feed rate	1.68 m³/d
30-day alum storage capacity	50.4 m³

Concept & Theory

Topic. Chemical phosphorus removal by alum precipitation, and reagent feed-system sizing.

Key relations. $\dot m_P=\Delta\text{TP}\cdot Q$; molar dosing ratio scales moles of Al to moles of P removed; commercial reagent volume $=\dfrac{m_{\text{active}}}{\text{strength}\times\rho}$.

Why this works. $\text{AlPO}_4$ has a very low solubility product, so dosing Al in molar excess of the stoichiometric 1:1 (here 1.5:1, reflecting competing side reactions such as $\text{Al}(\text{OH})_3$ formation and non-ideal mixing) drives the precipitation reaction to the required effluent TP.

Common pitfall. Forgetting the alum salt supplies 2 mol Al per mole of $\text{Al}_2(\text{SO}_4)_3\cdot18\text{H}_2\text{O}$ (an easy factor-of-2 slip), and confusing the 48% liquid product mass/volume with the 100%-basis active-alum mass used in the stoichiometry.

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

Question 9 of 25: Alum Dosing for Phosphorus Removal

Section 1: Chemistry (11 questions, 50 marks)

Question 9: Alum Dosing for Phosphorus Removal (7 marks)