Changes

:<math>A_{f}=\frac{0.816}{0.15}=5.44\ m^{2}</math>

:<math>A_{f}=\frac{0.816}{0.15}=5.44\ m^{2}</math>

The area required to settle the 1 mm particles is calculated as:

The area required to settle the 1 mm particles is calculated as:

:<math>A_{f}=120\times 0.1 = 12\ m^{2}</math>

:<math>A_{f}=120\times 0.02 = 2.4\ m^{2}</math>

So to meet the target particle removal, the forebay will be 12 m² in area. This gives the storage volume of 1.8 m³, which can be returned to the initial equation to determine the minimum cleaning frequency as:

So to meet the target particle removal, the forebay will be 5.44 m² in area. This gives the storage volume of 1.8 m³, which can be returned to the initial equation to determine the minimum cleaning frequency as:

:<math>C_{f}=\frac{1.8}{1.7\times 0.8\times 0.6}=2.2\ years</math>

:<math>C_{f}=\frac{1.8}{1.7\times 0.8\times 0.6}=2.2\ years</math>