Question # 4 . Recently, there has been a blue-green algae bloom in the Rideau c

Question

Question # 4 . Recently, there has been a blue-green algae bloom in the Rideau canal. As the algae bloom needs to be treated as they produce cyanotoxins, you have been asked to verify if using copper sulfate would be a reasonable solution. Copper sulfate, while killing the algae, lyses (breaks opens) the cells which releases the intracellular (stored within the cell) cyanotoxins. These toxins are released at a rate of 0.1 g cyanotoxin per 1 g of algae destroyed. The Rideau Canal has an approximate influent flow rate of 28 m/s, an influent algae concentration of 15 mg/L and the copper sulfate addition removes 87% of the algae. Determine the effluent flowrate eff), and the effluent concentration of (Qef), the efluent concentration of algae(Ce cyanotoxins(Cct.eff). Note: The volumetric flow of copper sulfate solution added is negligible compared to the volumetric flow of the canal. Assume the influent concentration of cyanotoxins is zero. Influent Effluent Copper Sulfate Stream

Explanation / Answer

Qeff = Qinf + Qcoppersulfate

since copper sufate volumetric flow rate can be negleted compared to influent flow

thus Qeff = Qinf = 28 m3/sec

Qalgae, inf = Qinf*Calgae,inf = 28*15*1000 = 420000 mg/s = 0.42 Kg/s

now 87% of this is destroyed by copper sulphate, hence only 100-87 = 13% is left out

thus, Qalgae, efff = 0.42*.13 = 0.0546 kg/s

Calgae, efff = 0.0546 /28/1000 = 1.95x10-6 Kg/L = 1.95 mg/L

now Qcyanotoxins, efff = destroy rate of algae * 0.1 (because toxin formation rate is 0.1 g for 1 gm of alage destroyed)

Qcyanotoxins, efff = 0.42*.87*0.1 = 0.03654 kg/s

Calgae, efff = 0.03654 /28/1000 = 1.305x10-6 Kg/L = 1.305 mg/L

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