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:<math>A_{f}=120\times 0.07 = 8.4\ m^{2}</math>
:<math>A_{f}=120\times 0.07 = 8.4\ m^{2}</math>
So to meet the target particle removal, the forebay will be 8.4 m² in area. This gives the storage volume of 1.26 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 8.4 m² in area. This gives the storage volume of 1.26 m³, which can be returned to the initial equation to determine the minimum cleaning frequency as:
−
:<math>C_{f}=\frac{1.26}{1.7\times 0.8\times 0.6}=1.5\ years</math>
+
:<math>F_{c}=\frac{1.26}{1.7\times 0.8\times 0.6}=1.5\ years</math>
==Gallery==
==Gallery==