Question 3 of 18: Liquid Alum Dosing for Phosphorus Removal

Question 3: Liquid Alum Dosing for Phosphorus Removal (10 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. Daily mass and volume of liquid alum required, and the storage tank volume for a 30-day supply.

Approach. Convert the phosphorus mass removed per day to moles, apply the 1.5:1 Al:P molar dosing ratio to get moles of Al (2 mol Al per mole of the alum formula unit), convert to mass of 100%-strength alum, then to mass and volume of the 48%-strength liquid product, and finally multiply by the 30-day storage period.

1. Mass of phosphorus removed per day. $$\Delta P = P_{in}-P_{out} = 8.0-1.0 = 7.0\ \text{mg/L}$$ $$m_P = \Delta P\times Q = 7.0\ \tfrac{\text{mg}}{\text{L}}\times 10{,}000\ \text{m}^3/\text{d} = 70{,}000\ \text{g/d} = 70.0\ \text{kg/d}$$
2. Moles of Al required. With $MW_P=31\ \text{g/mol}$, $$n_P = \frac{70{,}000\ \text{g/d}}{31\ \text{g/mol}} = 2{,}258\ \text{mol/d}, \qquad n_{Al}=1.5\times n_P = 3{,}387\ \text{mol/d}$$
3. 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, so $$n_{\text{alum}}=\frac{3{,}387}{2}=1{,}694\ \text{mol/d}$$ $$m_{\text{alum, 100\%}} = 1{,}694\ \text{mol/d}\times 666.4\ \text{g/mol} = 1{,}128{,}500\ \text{g/d} = 1{,}128.5\ \text{kg/d}$$
4. Mass and volume of the 48% liquid alum product. $$m_{\text{liquid}} = \frac{1{,}128.5\ \text{kg/d}}{0.48} = 2{,}351.1\ \text{kg/d}$$ $$\boxed{V_{\text{liquid alum}} = \frac{2{,}351.1\ \text{kg/d}}{1{,}280\ \text{kg/m}^3} = 1.84\ \text{m}^3/\text{d}}$$
5. Storage tank capacity (30-day supply). $$\boxed{V_{\text{storage}} = 1.84\ \tfrac{\text{m}^3}{\text{d}}\times 30\ \text{d} = 55.1\ \text{m}^3}$$