Introduction / Context:
Rapid mixing in water treatment provides intense, short-duration energy to disperse coagulants uniformly. The design power depends on the target velocity gradient G, the fluid viscosity, and the tank volume. This problem applies the standard relationship between G and power dissipation to size the mixer drive.
Given Data / Assumptions:
- Flow, Q = 28,800 m^3/day.
- Detention time, t = 2 minutes = 120 s.
- Velocity gradient, G = 900 s^-1.
- Water density = 1000 kg/m^3; kinematic viscosity ν = 1×10^-6 m^2/s ⇒ dynamic viscosity μ = ρ * ν = 0.001 N·s/m^2.
Concept / Approach:
For rapid mixing, power P relates to velocity gradient by P = μ * G^2 * V, where V is the tank volume. Volume is determined from V = Q * t. Ensure all units are SI for a result in watts (N·m/s).
Step-by-Step Solution:
Convert flow: Q = 28,800 / 86,400 = 0.3333 m^3/s.Compute volume: V = Q * t = 0.3333 * 120 = 40 m^3.Use μ = 0.001 N·s/m^2 and G = 900 s^-1.Compute power: P = μ * G^2 * V = 0.001 * (900)^2 * 40 = 0.001 * 810,000 * 40 = 32,400 W.
Verification / Alternative check:
Express as kW: 32,400 W ≈ 32.4 kW. Compare with typical rapid-mix power densities (≈ 1–10 kW per 100 m^3 at high G, scaling reasonably for 40 m^3).
Why Other Options Are Wrong:
32.4 or 36 W: off by three orders of magnitude due to unit mishandling.324 W: still too low; ignores G^2 scaling.3.24×10^6 W: unrealistically high for a 40 m^3 tank.
Common Pitfalls:
Forgetting to convert day to seconds for Q; using ν instead of μ; omitting the square on G.
Final Answer:
32400
Discussion & Comments