Super chlorination; overflow rate 5000 L/m2/h

When super-chlorination is recommended

Super-chlorination (a dose far above the demand, followed by de-chlorination) is used:

• during epidemics of cholera, typhoid, hepatitis, etc.;
• when a new main, pipe or tank is commissioned or repaired;
• when the raw water is of highly variable or very poor quality, or the plant is overloaded;
• when the contact time is short (so a high dose compensates);
• for emergency or camp supplies and when pre-treatment is poor. The dose is 3 to 15 mg/l and the excess is removed by sulphur dioxide, sodium thiosulphate or activated carbon.

Advantages of break-point chlorination

• Oxidises ammonia, organic matter and iron/manganese completely, which removes most tastes and odours.
• Kills pathogens and algae far more effectively than combined chlorine, including organisms that resist chloramines.
• Gives a stable free residual which protects the distribution system against recontamination.
• The chlorine demand is fully satisfied and known, so the dose can be controlled accurately.

Numerical: removal in the settling tank

Overflow rate $v_0 = 5000$ l/m²/h $= 5$ m³/m²/h:

$$v_0 = \frac{5}{3600} = 1.389\times10^{-3}\ \text{m/s} = 0.1389\ \text{cm/s}$$

At 20 °C, $\nu = 0.01007$ cm²/s, $S_s=2.65$, $g=981$ cm/s². By Stokes' law, $v_s = \dfrac{g(S_s-1)d^2}{18\nu}$.

Particle 0.06 mm (0.006 cm):

$$v_s = \frac{981\times1.65\times0.006^2}{18\times0.01007} = 0.3215\ \text{cm/s}$$

$Re = 0.19 < 1$ (Stokes' law holds). Since $v_s = 0.3215$ > $v_0=0.1389$ cm/s, the particles are removed 100 %.

Particle 0.03 mm (0.003 cm):

$$v_s = \frac{981\times1.65\times0.003^2}{18\times0.01007} = 0.0804\ \text{cm/s}$$

$Re = 0.02<1$. As $v_s<v_0$, the removal is

$$\frac{v_s}{v_0} = \frac{0.0804}{0.1389} = 0.579 \Rightarrow 57.9\ \%$$

Answer: 0.06 mm particles: 100 % removed; 0.03 mm particles: 57.9 % removed.