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Settling tank 20 m x 2.5 m; velocity for 0.02 mm

Short answer

For complete removal, horizontal velocity v_h = v_s × L/H. Stokes' settling velocity is 0.036 cm/s (Re<1). With L=20 m, H=2.5 m, v_h = 0.288 cm/s (17.3 cm/min).

In a continuous flow settling tank 20 m long and 2.5 m deep, what velocity of water would you recommend for effective removal of 0.02 mm particles at 25°C? Assume specific gravity of particles = 2.65 and kinematic viscosity of water = 0.01 cm2^2/sec.

For complete removal in an ideal horizontal-flow tank, the particle must settle the depth HH in the time the water takes to flow the length LL:

Hvs=Lvh ⇒ vh=vs LH\frac{H}{v_s}=\frac{L}{v_h}\ \Rightarrow\ v_h=v_s\,\frac{L}{H}

Step 1: Settling velocity of the 0.02 mm particle

d=0.002d=0.002 cm, Ss=2.65S_s=2.65, ν=0.01\nu=0.01 cm²/s (given):

vs=981×1.65×0.002218×0.01=0.03597 cm/s(Re=0.007<1)v_s=\frac{981\times1.65\times0.002^2}{18\times0.01}=0.03597\ \text{cm/s}\quad(Re=0.007<1)

Step 2: Horizontal velocity

L=20L=20 m, H=2.5H=2.5 m:

vh=0.03597×202.5=0.2878 cm/s=17.27 cm/min=0.173 m/minv_h=0.03597\times\frac{20}{2.5}=0.2878\ \text{cm/s}=17.27\ \text{cm/min}=0.173\ \text{m/min}

This is below the 0.3 m/min limit for scour, so it is acceptable.

Answer: the velocity of flow should not exceed 0.288 cm/s (17.3 cm/min).

✓ Answer checked by Claude Opus 5.5 (high effort)

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