The turbidity of raw water entering a sedimentation tank is \(120\) NTU. After sedimentation, the turbidity is reduced to \(30\) NTU. What is the turbidity removal efficiency of the sedimentation process?
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Removal efficiency:
\[
\boxed{
\eta=
\frac{C_i-C_f}{C_i}\times100
}
\]
where \(C_i\) = initial value and \(C_f\) = final value.
Step 1: Recall the removal efficiency formula. Removal efficiency compares how much a pollutant or parameter drops, using $\text{Efficiency} = \dfrac{\text{Initial} - \text{Final}}{\text{Initial}} \times 100$. Step 2: Substitute the turbidity values. Initial turbidity is 120 NTU and final turbidity is 30 NTU, so efficiency $= \dfrac{120-30}{120}\times100 = \dfrac{90}{120}\times100$. Step 3: Simplify to get the percentage. Working that fraction out gives 75 percent removal. \[ \boxed{75\%} \]
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