Question

A flocculation tank used for water treatment has a velocity gradient (G) of 800 s\(^{-1}\). The volume of the tank is 40 m\(^3\). The dynamic viscosity of water is \(9 \times 10^{-4}\) N.s/m\(^2\). The theoretical power required to maintain the given velocity gradient is _______ kW (rounded off to the nearest integer).

Hint:

The theoretical power for a flocculation tank can be calculated from the velocity gradient, dynamic viscosity, and volume of the tank.

Answer 1

Correct Answer: 23

The theoretical power required can be calculated using the formula: \[ P = \frac{G^2 \cdot \mu \cdot V}{\rho} \] where:
- \( G \) is the velocity gradient,
- \( \mu \) is the dynamic viscosity,
- \( V \) is the volume of the tank,
- \( \rho \) is the density of water.
Substituting the given values and solving for the power, we get: \[ P = \frac{(800)^2 \times 9 \times 10^{-4} \times 40}{1000} = 23 \text{ kW}. \] Thus, the power required is \( 23 \, \text{kW} \).