Water and Wastewater Treatment and Management · Water Treatment · 1 mark · Numerical answer
A flocculation tank has velocity gradient800 s⁻¹,volume 40 m³,water viscosity9×10⁻⁴ N s/m². Find theoretical mixing power in kW, rounded to an integer.
Use $G=\sqrt{P/(\mu V)}$,so $P=\mu VG^2$. Substitution gives .0009(40)(800²)=23040 W=23.04 kW. Rounded, the answer is **23 kW**. This is theoretical power dissipated into the liquid, without an additional motor-efficiency correction.
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