Water Treatment

Aeration

Aeration is the process of bringing water and air into close contact in order to dissolve oxygen from the air into water and release dissolved gases and volatile substances from the water into the air.

Purposes of aeration

1. To oxidise iron and manganese (Fe$^{2+}$ and Mn$^{2+}$) into insoluble hydroxides or oxides so they can be settled and filtered.
2. To remove dissolved gases such as carbon dioxide, hydrogen sulphide and methane, which cause corrosion and bad taste and smell.
3. To remove taste and odour caused by volatile organic matter, algae products and gases.
4. To raise the dissolved oxygen of deep well or reservoir water and improve its freshness.
5. To reduce free CO$_2$ and increase pH, so less corrosion in pipes occurs.
6. To precipitate some hardness-causing and other dissolved substances by oxidation, and to help with the removal of volatile chemicals.

Methods of aeration

1. Gravity (waterfall) aerators: water falls in thin layers or drops over steps, cascades or inclined planes through the air.

   water in
     \__
        \__   cascade steps
           \__
              \__ -> collecting tank

• Cascade / step aerator: water flows over 3 to 6 steps, each 0.2 to 0.3 m high, with a total fall of about 1 to 3 m. Good for CO$_2$ and iron removal.
• Inclined plane / tray aerator: water trickles over trays of coke, stones or slats (coke trays) 0.3 to 0.5 m apart, in 3 to 5 layers. Coke gives surface for iron oxidation.
• Multiple tray: 3 to 9 trays, 30 to 50 cm apart, 4 to 10 m³/m²·h loading.

2. Spray aerators (fountain): water is pumped through nozzles 2.5 to 4 cm diameter at 70 kPa and sprayed in the air as fine drops (spray height 1 to 3 m). Gives a large contact area and good gas exchange; needs more space and head.

        \ | /     spray
      .  \|/  .   droplets
        ~~~~~~~~~   <- basin (area ~ 0.03 m2 per m3/h)

3. Air diffusion (bubble) aerators: compressed air is bubbled through diffusers (perforated pipes or porous plates) at the bottom of a tank, with 0.3 to 0.75 m³ of air per m³ of water and contact time 10 to 30 min.

   water in -> |~~~~~~~~~~~~~~~| -> out
               |  o  o  o  o   |
               |==diffuser====|<- air

4. Mechanical aerators: rotors, paddles or turbines stir the water surface to draw air in.

5. Pressure aerators / packed tower: air and water are brought into counter-current contact in a tower with packing (a forced draft tower), which gives the highest efficiency for gas removal.

A rapid sand filter (RSF) is a gravity or pressure filter in which water, normally after coagulation and sedimentation, passes down through a sand bed at a high rate (3 to 6 m/h). The bed is cleaned by reversing the flow (backwashing).

Construction

   Inlet (coagulated water)   Wash water trough
        |                      |   |
  ______v______________________v___v____ top
 |  water depth 1.0-1.5 m                |
 |~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~|
 | sand 0.6-0.75 m (ES 0.45-0.7 mm)      |
 |---------------------------------------|
 | gravel 0.45 m (graded, 4 layers)      |
 |----+---+---+---+---+---+---+---+------|
 | laterals with orifices / manifold     |-> filtered
 |_______________________________________|   water out
   wash water in <-    wash waste out ->

• Tank: an open RCC or masonry tank, rectangular in plan with length : width = 1.25 to 1.33, depth about 2.5 to 3.5 m.
• Filter media (sand): 0.6 to 0.75 m depth, effective size 0.45 to 0.7 mm, uniformity coefficient 1.3 to 1.7.
• Base material (gravel): 0.45 m in 4 layers, graded from 2 mm at the top to 50 mm at the bottom, supporting the sand and spreading the flow.
• Under-drainage system: a central manifold with lateral perforated pipes (orifices 6 to 12 mm, area about 0.2 to 0.5 % of bed area). It collects filtered water uniformly and distributes the backwash water. Other types: Wheeler bottom, porous plates, strainers.
• Wash-water troughs above the sand, inlet and outlet pipes with valves, and wash-water and wash-waste pipes.

Working

Coagulated and settled water enters at the top; it flows down through the sand, where particles are removed by straining, sedimentation, and adsorption within the pores (not on a skin like the slow filter). The filtered water leaves via the lateral pipes and manifold to the clear water tank. As impurities collect, the head loss rises from about 0.3 m to the limit of 2.5 to 3.0 m (or the effluent turbidity rises), and the filter must be washed.

Design considerations

• Filtration rate 3000 to 6000 l/h/m² (about 5 m/h) and at least two units (N = 1.22√Q, Q in MLD).
• Area = Q / rate, plus 2 to 4 % of the water for backwashing.
• Depth of water over the sand 1.0 to 1.5 m (prevents negative head), plus 0.5 m free board.
• Terminal head loss 2.5 to 3.0 m.

Operation and maintenance

• Filter runs 24 to 72 h between washes. Record loss of head, turbidity and the rate of filtration; adjust the rate with the rate controller.
• Do not allow negative head (air binding), keep proper water cover.
• Check for mud balls, cracks and sand loss; top up and replace sand every few years. Disinfect the filter after repairs.

Cleaning by backwashing

1. Close the inlet and let water drop to about 0.2 m above the sand.
2. Optionally scour the bed with compressed air (about 0.6 m³/m²·min for 2 to 3 min).
3. Pass clean water upward at 0.6 to 1.0 m/min (600 to 1000 l/min/m²) for 10 to 15 minutes. The bed expands by about 50 %, the sand grains rub together and the dirt is carried out by the wash-water troughs to waste.
4. Stop the wash, close the wash valve, and then reopen the inlet. Run the first filtrate to waste. Backwash uses 2 to 4 % of the filtered water.

Sedimentation with coagulation is the settling of fine suspended and colloidal impurities after chemicals (coagulants such as alum, ferric chloride or ferrous sulphate) have been added, so that the particles join into large, heavy flocs which settle quickly. It is also called chemical or aided sedimentation.

Why it is necessary

• Colloidal particles (0.001 to 1 µm) and very fine silt carry a negative charge and repel one another, so they stay suspended and plain sedimentation cannot remove them. The settling velocity by Stokes' law is extremely small for such sizes.
• It removes turbidity, colour, bacteria and some organic matter, so the filters are loaded less and run longer.
• It makes a much smaller and cheaper tank (higher overflow rates 30 to 40 m³/m²/d, detention 2 to 4 h).

Processes involved

 Raw    +-------+   +-----------+   +---------+   +--------+
 water->| Rapid |-->|Flocculator|-->|Settling |-->| Filter |
        |  mix  |   |slow stir  |   |  tank   |   +--------+
        +-------+   +-----------+   +---------+
        coagulant     flocs form     flocs settle

1. Coagulation (rapid mixing): the coagulant is added and violently mixed for 30 s to 2 min. The positive metal ions neutralise the negative charge of the colloids (destabilisation), and metal hydroxide forms.

$$Al_2(SO_4)_3 + 3Ca(HCO_3)_2 \rightarrow 2Al(OH)_3\downarrow + 3CaSO_4 + 6CO_2$$

Alkalinity is needed (about 0.5 mg/l per 1 mg/l alum); lime or soda ash is added if the water lacks it. 2. Flocculation (slow mixing): gentle stirring for 20 to 40 min brings the micro-flocs together to form large flocs by inter-particle collision (bridging and sweep floc). 3. Sedimentation: the large flocs settle in the settling tank under gravity; the clarified water is collected at the outlet weir and sludge is withdrawn from the hopper. 4. Filtration follows to remove the remaining fine flocs, then disinfection.

Iron and manganese occur in ground water (and deep reservoirs) as soluble ferrous (Fe$^{2+}$) and manganous (Mn$^{2+}$) bicarbonates. They have to be oxidised to insoluble ferric hydroxide and manganese dioxide, which can then be settled and filtered.

Methods

1. Aeration, sedimentation and filtration (commonest method) Water is aerated (cascade, spray or tray aerator) so it takes oxygen and loses CO$_2$, raising the pH:

$$4Fe(HCO_3)_2 + O_2 + 2H_2O \rightarrow 4Fe(OH)_3\downarrow + 8CO_2$$

The ferric hydroxide floc is settled and then filtered through a rapid sand filter. Fe oxidises easily above pH 7.5; Mn oxidises slowly and needs a pH of about 9.5, so lime may be added.

2. Chemical oxidation Chlorine, potassium permanganate or ozone oxidise Fe and Mn quickly:

$$3Fe^{2+} + MnO_4^- + 7H_2O \rightarrow 3Fe(OH)_3\downarrow + MnO_2\downarrow + 5H^+$$ $$3Mn^{2+} + 2MnO_4^- + 2H_2O \rightarrow 5MnO_2\downarrow + 4H^+$$

Chlorine dose is about 0.64 mg/l per mg/l Fe and 1.3 mg/l per mg/l Mn. The precipitate is removed by coagulation and filtration.

3. Manganese greensand filter: the glauconite sand coated with MnO$_2$ oxidises and filters Fe and Mn, and is regenerated with KMnO$_4$.

4. Lime softening: at pH above 10 Fe and Mn precipitate as hydroxides along with hardness.

5. Ion exchange (zeolite) softener: removes small amounts of Fe and Mn, but the resin gets fouled when oxidised iron is present.

6. Others: coagulation with alum at pH above 8, and in-situ (underground) treatment.

Final treated water should have Fe below 0.3 mg/l and Mn below 0.05 mg/l (WHO limits).

Forms (aspects) of chlorination

1. Plain chlorination: chlorine is added to otherwise clean water, with only a short contact time (30 min or more). Dose 0.5 to 1 mg/l. Used for water that has already been well treated or is already of good quality.
2. Pre-chlorination: chlorine is added before the filters (or before sedimentation) to reduce the load of organisms and algae, control taste and odour, oxidise iron, and help coagulation.
3. Post-chlorination: the normal disinfection step, applied after filtration, just before the water goes to the clear water tank. Dose gives 0.2 to 0.5 mg/l residual.
4. Double chlorination: pre- and post-chlorination used together; also called two-stage chlorination.
5. Break-point chlorination: chlorine is added past the break point so that free residual is present (see the curve).
6. Super-chlorination: the dose is much higher than the demand (3 to 15 mg/l or more), followed by de-chlorination (using sulphur dioxide, sodium thiosulphate, sodium sulphite or activated carbon). Used during epidemics, after repair of mains, with a short contact time or very poor quality water.
7. De-chlorination: removal of the excess chlorine after super-chlorination.
8. Chloramine process: ammonia is added with the chlorine to form chloramines, which give a more lasting, but weaker, residual, used in long distribution systems.

Applications: pre-chlorination in turbid or algae-rich water, post-chlorination in every plant, super-chlorination in emergencies and for new mains, and break-point for water that contains ammonia or taste and odour problems.

Factors affecting chlorination

1. Chlorine dose and demand: dose must exceed the demand (organic matter, ammonia, iron) to leave a residual.
2. Contact time: the longer the contact, the better the kill ($C^n t$ = constant); usually at least 30 min.
3. pH: hypochlorous acid (HOCl) is a much stronger disinfectant than OCl$^-$; at low pH more HOCl is present, so chlorination is best at pH 6 to 7.
4. Temperature: higher temperature speeds up the reaction and increases the kill rate; cold water needs a longer contact time or a higher dose.
5. Turbidity and suspended matter: particles shield bacteria and consume chlorine; clear water is disinfected better.
6. Kind and number of organisms: viruses, cysts and spores are more resistant than bacteria.
7. Nature of chlorine compound: free chlorine is a faster disinfectant than combined (chloramines).
8. Mixing: proper, quick mixing ensures uniform contact.
9. Ammonia, organic nitrogen and reducing substances in the water.

A continuous-flow ideal settling tank removes a particle completely if it settles through the full depth $H$ during the detention time $t$, i.e. $v_s \ge H/t$ (equivalently $v_s\ge$ overflow rate $v_0$). So the detention time for 100 % removal is $t = H/v_s$.

(a) Detention time for 0.02 mm at 25 °C

Take $S_s = 2.65$, $\nu_{25} = 0.00893$ cm²/s, $d=0.002$ cm, $H = 3.5$ m $=350$ cm.

$$v_s=\frac{g(S_s-1)d^2}{18\nu}=\frac{981\times1.65\times0.002^2}{18\times0.00893}=0.0403\ \text{cm/s}$$

$Re = v_sd/\nu = 0.009<1$ (Stokes' law valid).

$$t=\frac{H}{v_s}=\frac{350}{0.0403}=8689\ \text{s} = 2.41\ \text{h}$$

Recommended detention time $\approx 2.4$ h. (Horizontal velocity $=L/t = 2000/8689 = 0.230$ cm/s, well below the scour velocity.)

(b) Removal of 0.01 mm particles at 15 °C in the same tank

The tank has overflow rate $v_0 = H/t=0.0403$ cm/s. At 15 °C, $\nu=0.01139$ cm²/s, $d=0.001$ cm:

$$v_s=\frac{981\times1.65\times0.001^2}{18\times0.01139}=0.00790\ \text{cm/s}\quad (Re=0.0007<1)$$ $$\text{Removal}=\frac{v_s}{v_0}=\frac{0.00790}{0.0403}=0.196\Rightarrow 19.6\ \%$$

Answer: detention time = 2.41 h; removal of 0.01 mm particles at 15 °C = 19.6 %.

For 100 % removal a particle must settle the full depth in the detention time: $t=H/v_s$. For smaller particles, the removal is $v_s/v_0$ where $v_0=H/t$.

(a) Detention time for 0.02 mm at 25 °C

$d=0.002$ cm, $S_s=2.65$, $\nu_{25}=0.00893$ cm²/s, $H=300$ cm.

$$v_s=\frac{981\times1.65\times0.002^2}{18\times0.00893}=0.0403\ \text{cm/s}\quad(Re=0.009<1)$$ $$t=\frac{300}{0.0403}=7448\ \text{s}=2.07\ \text{h}$$

(b) Removal of 0.01 mm particles at 10 °C

The tank's overflow rate is $v_0=H/t=0.0403$ cm/s. At 10 °C, $\nu=0.01306$ cm²/s, $d=0.001$ cm:

$$v_s=\frac{981\times1.65\times0.001^2}{18\times0.01306}=0.00689\ \text{cm/s}\quad(Re=0.0005<1)$$ $$\text{Removal}=\frac{0.00689}{0.0403}=0.171\Rightarrow17.1\ \%$$

Answer: detention time = 2.07 h; removal of 0.01 mm particles at 10 °C = 17.1 %.

For complete removal in an ideal continuous-flow tank, $t=H/v_s$; for smaller particles the removal is $v_s/v_0$ with $v_0=H/t$.

(a) Detention time for 0.03 mm at 25 °C

$d=0.003$ cm, $S_s=2.65$, $\nu_{25}=0.00893$ cm²/s, $H=300$ cm.

$$v_s=\frac{981\times1.65\times0.003^2}{18\times0.00893}=0.0906\ \text{cm/s}\quad(Re=0.030<1)$$ $$t=\frac{300}{0.0906}=3310\ \text{s}=55.2\ \text{min}=0.92\ \text{h}$$

(b) Removal of 0.025 mm particles at 20 °C

Overflow rate of the tank: $v_0=H/t=0.0906$ cm/s. At 20 °C, $\nu=0.01007$ cm²/s, $d=0.0025$ cm:

$$v_s=\frac{981\times1.65\times0.0025^2}{18\times0.01007}=0.0558\ \text{cm/s}\quad(Re=0.014<1)$$ $$\text{Removal}=\frac{0.0558}{0.0906}=0.616\Rightarrow61.6\ \%$$

Answer: detention time = 0.92 h (55 min); removal of 0.025 mm particles at 20 °C = 61.6 %.

Treatment processes and the impurities they remove

 Raw -> Screen -> Aerator -> Sedimentation -> Coag/Floc
 water                                          |
 Supply <- Disinfection <- Filter <- Settling <-+

Factors affecting settlement in a sedimentation tank

1. Size of particle: $v_s \propto d^2$; bigger particles settle quicker.
2. Specific gravity of particle: heavier particles settle faster ($v_s \propto S_s-1$).
3. Temperature (viscosity): warmer water has lower viscosity, so settling is faster; $v_s \propto 1/\nu$.
4. Surface overflow rate / tank area: larger surface area for the flow reduces the overflow rate and improves removal.
5. Detention time: the longer, the better the removal.
6. Velocity of flow: should be low (below about 0.3 m/min) so that the settled sludge is not scoured.
7. Tank shape and depth, inlet and outlet arrangement (short-circuiting, density currents, wind).
8. Flocculation: coagulated particles are bigger and settle faster.
9. Hindered settling at high concentration and tank turbulence.

Hydraulic subsidence (settling) value: the overflow rate $v_0=Q/A$ is the minimum settling velocity that gives 100 % removal. A particle of velocity $v_s\ge v_0$ is fully removed; if $v_s<v_0$ only the fraction $v_s/v_0$ is removed.

Effect of temperature and diameter: from Stokes' law, $v_s = \dfrac{g(S_s-1)d^2}{18\nu}$. Settling velocity varies as the square of diameter (doubling the diameter gives four times the speed) and is inversely proportional to the kinematic viscosity, which falls as temperature rises (the viscosity at 10 °C is 1.3 times that at 20 °C, so settling is about 30 % slower in cold water). Tanks are therefore designed for the lowest water temperature.

Theory of settlement of discrete particles

A discrete particle is a particle that keeps its size, shape and weight while settling (e.g. sand, silt). In a quiescent liquid, it first accelerates under gravity, then very quickly reaches a terminal (settling) velocity when the net downward force is balanced by drag.

Stoke's law

For a spherical particle falling freely at constant (terminal) velocity $v_s$, the weight minus buoyancy equals the fluid drag.

$$\begin{aligned} \text{Net weight} &= \frac{\pi d^3}{6}(\rho_s-\rho)\,g \\ \text{Drag (viscous, } Re<1) &= 3\pi\mu d\,v_s \end{aligned}$$

Equating the two:

$$v_s = \frac{g\,(\rho_s-\rho)\,d^2}{18\,\mu} = \frac{g\,(S_s-1)\,d^2}{18\,\nu}$$

where $d$ = particle diameter, $S_s$ = specific gravity of the particle, $\nu=\mu/\rho$ = kinematic viscosity.

Conditions of application in design

• The particle must be spherical, discrete (non-flocculent) and settle freely without being hindered by neighbours.
• The flow around it must be laminar, with Reynolds number $Re=\dfrac{v_s d}{\nu}<1$ (roughly particle diameter below about 0.1 mm for silt of $S_s=2.65$).
• The liquid is at rest or flows at very low velocity (no turbulence, no short-circuiting).
• For larger particles ($1<Re<1000$) the drag coefficient is taken as $C_d=\dfrac{24}{Re}+\dfrac{3}{\sqrt{Re}}+0.34$ and $v_s=\sqrt{\dfrac{4g(S_s-1)d}{3C_d}}$ is found by trial. For $Re>1000$, $v_s=1.8\sqrt{g d (S_s-1)}$ (Newton).

In an ideal rectangular tank the removal depends only on the overflow rate $v_0=Q/A=H/t$: a particle with $v_s\ge v_0$ is entirely removed.

Effect of temperature

Temperature changes the viscosity, and so the settling velocity ($v_s\propto 1/\nu$). The kinematic viscosity at temperature $T$ (°C) can be calculated from

$$\nu = \frac{0.01775}{1+0.0337\,T+0.000221\,T^2}\ \text{cm}^2/\text{s}$$

and for the same particle:

$$\frac{v_{s,T_2}}{v_{s,T_1}} = \frac{\nu_{T_1}}{\nu_{T_2}}$$

Typical values:

Hence the settling velocity at 20 °C is about 1.3 times that at 10 °C. For design, the lowest expected temperature is used, so the tank is also effective in warm weather.

Filtration

Filtration is the process of passing water through a bed of granular material (usually sand) to remove suspended, colloidal and microbial impurities that sedimentation could not remove.

Theory (mechanisms) of filtration

1. Mechanical straining: particles larger than the pore spaces between sand grains (0.1 to 0.2 mm) are stopped at the top; this effect builds up the schmutzdecke in a slow sand filter, which then strains finer particles.
2. Sedimentation: the pores act like tiny settling basins; fine particles settle on the sand grains' upper surfaces.
3. Adsorption / surface attraction: the sand grain and the sticky film on it attract the oppositely charged colloids and bacteria (electrostatic, van der Waals forces).
4. Biological action: in a slow sand filter the zoogleal film feeds on organic matter and bacteria, oxidising them to harmless compounds, and converting ammonia to nitrates.
5. Electrolytic / chemical action: the ion exchange between sand grains and impurities, and the chemical bonding with coagulant flocs (important for rapid filters).
6. Flocculation in the bed: fine particles collide in the pore passages and combine into bigger ones which are then trapped.

  water in
  ~~~~~~~~~~~~~~~~~
  ..  schmutzdecke     <- straining, biological film
  ::::::::::::::::::
  :::: sand grains ::: <- adsorption, sedimentation
  ::::::::::::::::::
  ooooo gravel ooooooo
  clean water out

Limitations

• Filtration cannot remove dissolved salts, hardness, or dissolved gases, taste and odour.
• Filters need a clean influent: highly turbid water clogs the bed quickly (a slow sand filter needs turbidity below about 30 NTU; a rapid filter needs coagulation first).
• Head loss builds up with time, so cleaning (scraping or backwashing) is frequent and fills the plant with maintenance.
• Viruses and cysts are only partly removed; disinfection is still required.
• Efficiency depends on filtration rate, grain size and temperature; at high rates there is break-through of floc.
• The slow filter needs a large land area, and the rapid filter needs skilled operation, with possible mud balls, cracking and air binding.

Permanent (non-carbonate) hardness is caused by the chlorides, sulphates and nitrates of calcium and magnesium (CaSO$_4$, MgSO$_4$, CaCl$_2$, MgCl$_2$). It is not removed by boiling, because these salts remain dissolved. It is removed by chemical precipitation or ion exchange.

1. Lime-soda process

Soda ash (Na$_2$CO$_3$) is added; it converts the soluble Ca and Mg salts into insoluble carbonate:

$$CaSO_4 + Na_2CO_3 \rightarrow CaCO_3\downarrow + Na_2SO_4$$ $$CaCl_2 + Na_2CO_3 \rightarrow CaCO_3\downarrow + 2NaCl$$

Magnesium salts need lime and soda:

$$MgSO_4 + Ca(OH)_2 \rightarrow Mg(OH)_2\downarrow + CaSO_4$$

and then the CaSO$_4$ formed is removed by soda ash as above. The precipitates are removed by sedimentation and filtration. The remaining hardness is about 30 to 50 mg/l.

2. Zeolite (base-exchange) process

Water passes through a bed of sodium zeolite (Na$_2$Z). Ca and Mg exchange with sodium:

$$CaSO_4 + Na_2Z \rightarrow CaZ + Na_2SO_4$$ $$MgCl_2 + Na_2Z \rightarrow MgZ + 2NaCl$$

When the zeolite is exhausted, it is regenerated with strong brine (NaCl): $CaZ + 2NaCl \rightarrow Na_2Z + CaCl_2$. It gives almost zero hardness water.

3. Demineralisation (ion exchange)

Cation and anion exchange resins remove all dissolved ions.

Break-point chlorination

Break-point chlorination is the addition of chlorine to water in a quantity big enough to oxidise all the chlorine-demanding substances (organic matter, ammonia, etc.) so that a free residual chlorine is left. The dose at which the combined residual falls to its minimum and then starts to rise as free chlorine is the break point.

 Residual
 chlorine
   |      Zone 1  Zone 2   Zone 3   Zone 4
   |      ____    /\                    /
   |     /    \  /  \                 /  free
   |    /      \/    \              /   chlorine
   |   /        a     \           /
   |  /                 \___ ___/
   | /      combined       b break point
   +--------------------------------------
        Chlorine dose added  ->

a = peak of chloramines, b = break point

1. Zone 1 (origin to a): the chlorine first reacts with reducing compounds (Fe$^{2+}$, Mn$^{2+}$, H$_2$S, organic matter) and is used up. There is no residual.
2. Zone 2 (up to peak a): after the demand is met, chlorine reacts with ammonia and amines to form chloramines (NH$_2$Cl, NHCl$_2$). The combined residual rises.
3. Zone 3 (a to break point b): more chlorine oxidises the chloramines to nitrogen gas, N$_2$O and nitrate; so the residual falls. Tastes and odours are worst here.
4. Zone 4 (beyond b): all ammonia and organic matter have been oxidised; the added chlorine remains as free chlorine (HOCl and OCl$^-$) and the residual increases in proportion to the dose.

Effect of pH and temperature on HOCl and OCl$^-$

When chlorine is added to water it hydrolyses:

$$Cl_2 + H_2O \rightleftharpoons HOCl + H^+ + Cl^-$$

The hypochlorous acid is a weak acid and dissociates:

$$HOCl \rightleftharpoons H^+ + OCl^-, \qquad K_i=\frac{[H^+][OCl^-]}{[HOCl]}=2.7\times10^{-8}\ \text{mol/l (at 20 °C)}$$

The fraction as HOCl is $\dfrac{1}{1+K_i/[H^+]}$. HOCl (neutral, small molecule) is about 80 times more effective as a germicide than OCl$^-$ (charged, cannot enter the cell wall easily).

Effect of pH (computed at 20 °C):

Below pH 6 nearly all is HOCl; at pH 7.5 the two forms are about equal; above pH 8 mainly OCl$^-$. So lower pH (6 to 7) gives better disinfection; higher pH needs a larger dose or contact time.

Effect of temperature: $K_i$ increases with temperature (pKa about 7.8 at 0 °C and 7.5 at 25 °C), so at the same pH 7 the HOCl share is 87 % at about 0 °C and 78 % at about 25 °C. Thus, a higher temperature reduces the HOCl fraction slightly, but at the same time speeds up the kill reaction, which more than makes up for it; so in cold water the contact time must be increased.

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, so the water is free of taste and odour.
• Destroys all pathogens, including those resistant to chloramines, and algae.
• Gives a stable free residual which protects the distribution system against recontamination.
• Chlorine demand is fully known, so the dose can be controlled.

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.

Specific gravity of the particles is not given, so the usual value for silt/sand, $S_s = 2.65$, is assumed. Particles smaller than the cut-off settling velocity $v_0$ are removed in the proportion $v_s/v_0$; those with $v_s \ge v_0$ are removed completely.

Data

• Overflow rate $v_0 = 4000$ l/m²/h $= 4$ m³/m²/h
• $v_0 = \dfrac{4\times100}{3600} = 0.1111$ cm/s
• Temperature 10 °C: $\nu = 0.01306$ cm²/s (viscosity of water at 10 °C), $g = 981$ cm/s²

Settling velocity (Stokes' law)

$$v_s = \frac{g\,(S_s-1)\,d^2}{18\,\nu}$$

(a) d = 0.05 mm = 0.005 cm

$$v_s = \frac{981\times1.65\times0.005^2}{18\times0.01306} = 0.1721\ \text{cm/s}$$

$Re = v_s d/\nu = 0.066 < 1$, so Stokes' law applies. Since $v_s = 0.1721$ > $v_0 = 0.1111$ cm/s, the removal is 100 %.

(b) d = 0.02 mm = 0.002 cm

$$v_s = \frac{981\times1.65\times0.002^2}{18\times0.01306} = 0.0275\ \text{cm/s}$$

$Re = 0.0042 < 1$ (valid). Here $v_s < v_0$:

$$\text{Removal} = \frac{v_s}{v_0} = \frac{0.0275}{0.1111} = 0.248$$

Answer: (a) 0.05 mm particles: 100 % removed; (b) 0.02 mm particles: 24.8 % removed.

Hypochlorous acid (HOCl) is the effective disinfectant; the hypochlorite ion (OCl$^-$) is far weaker and is neglected. The same killing power is obtained at both pH values if the HOCl concentration is kept the same.

$$HOCl \rightleftharpoons H^+ + OCl^-, \qquad K_i=\frac{[H^+][OCl^-]}{[HOCl]}=2.7\times10^{-8}\ \text{mol/l}$$

Fraction of total chlorine present as HOCl:

$$\alpha = \frac{[HOCl]}{[HOCl]+[OCl^-]}=\frac{1}{1+K_i/[H^+]}$$

Step 1: HOCl fraction at the two pH values

• pH 7.0: $[H^+]=10^{-7}$; $K_i/[H^+] = 0.27$; $\alpha_7 = 1/1.27 = 0.7874$
• pH 8.0: $[H^+]=10^{-8}$; $K_i/[H^+] = 2.7$; $\alpha_8 = 1/3.7 = 0.2703$

Step 2: Dose at pH 8.0

HOCl at pH 7 = $0.5\times0.7874 = 0.3937$ mg/l. The same HOCl at pH 8 needs

$$C_8 = \frac{0.5\times0.7874}{0.2703} = 1.457\ \text{mg/l}$$

Step 3: Contact time

With $C^n t=\text{constant}$ and $n=1.5$, the contact time for the changed concentration is

$$t_2 = t_1\left(\frac{C_1}{C_2}\right)^{n} = 10\left(\frac{0.5}{1.457}\right)^{1.5} = 2.01\ \text{min}$$

(Here $C$ is the total chlorine dose applied. The result shows that a higher dose at pH 8 shortens the contact time. If the dose were kept at the pH 7 value, the HOCl would be smaller at pH 8 and the contact time would have to be longer.)

Answer: dose at pH 8 = 1.457 mg/l (about 2.91 times the dose at pH 7); contact time = 2.01 min.

Hypochlorous acid (HOCl) is the effective disinfectant; the hypochlorite ion (OCl$^-$) is far weaker and is neglected. The same killing power is obtained at both pH values if the HOCl concentration is kept the same.

$$HOCl \rightleftharpoons H^+ + OCl^-, \qquad K_i=\frac{[H^+][OCl^-]}{[HOCl]}=2.7\times10^{-8}\ \text{mol/l}$$

Fraction of total chlorine present as HOCl:

$$\alpha = \frac{[HOCl]}{[HOCl]+[OCl^-]}=\frac{1}{1+K_i/[H^+]}$$

Step 1: HOCl fraction at the two pH values

• pH 7.0: $[H^+]=10^{-7}$; $K_i/[H^+] = 0.27$; $\alpha_7 = 1/1.27 = 0.7874$
• pH 8.0: $[H^+]=10^{-8}$; $K_i/[H^+] = 2.7$; $\alpha_8 = 1/3.7 = 0.2703$

Step 2: Dose at pH 8.0

HOCl at pH 7 = $0.7\times0.7874 = 0.5512$ mg/l. The same HOCl at pH 8 needs

$$C_8 = \frac{0.7\times0.7874}{0.2703} = 2.039\ \text{mg/l}$$

Step 3: Contact time

With $C^n t=\text{constant}$ and $n=1.5$, the contact time for the changed concentration is

$$t_2 = t_1\left(\frac{C_1}{C_2}\right)^{n} = 12\left(\frac{0.7}{2.039}\right)^{1.5} = 2.41\ \text{min}$$

(Here $C$ is the total chlorine dose applied. The result shows that a higher dose at pH 8 shortens the contact time. If the dose were kept at the pH 7 value, the HOCl would be smaller at pH 8 and the contact time would have to be longer.)

Answer: dose at pH 8 = 2.039 mg/l (about 2.91 times the dose at pH 7); contact time = 2.41 min.

In an ideal settling tank, particles with $v_s\ge v_0$ are fully removed, and those with $v_s<v_0$ are removed in the ratio $v_s/v_0$, where $v_0=Q/A$ is the overflow rate.

Step 1: Settling velocity of the 0.025 mm particles

$d=0.0025$ cm, $S_s=2.65$, $\nu_{20}=0.01007$ cm²/s.

$$v_s=\frac{981\times1.65\times0.0025^2}{18\times0.01007}=0.0558\ \text{cm/s}\quad(Re=0.014<1)$$

Step 2: Overflow rate

For 93 % removal:

$$v_0=\frac{v_s}{0.93}=\frac{0.0558}{0.93}=0.0600\ \text{cm/s}=51.9\ \text{m}^3/\text{m}^2/\text{day}\ (=2.16\ \text{m}^3/\text{m}^2/\text{h})$$

Surface area $A=11\times5=55$ m², so the tank can treat

$$Q=v_0A=0.000600\times55=0.03301\ \text{m}^3/\text{s}=2852\ \text{m}^3/\text{day}$$

Horizontal velocity $=Q/(B\,H)=0.03301/(5\times3)=0.2200$ cm/s $=0.132$ m/min.

Step 3: Check for 99 % removal of 0.05 mm particles

$d=0.005$ cm:

$$v_s=\frac{981\times1.65\times0.005^2}{18\times0.01007}=0.2232\ \text{cm/s}\quad(Re=0.111<1)$$

$v_s=0.2232$ cm/s is greater than $v_0=0.0600$ cm/s ($v_s/v_0=3.72$), so these particles are removed 100 %, which exceeds 99 %.

Answer: overflow rate = 0.0600 cm/s (51.9 m³/m²/day); yes, the tank is sufficient to remove 99 % of 0.05 mm particles (it removes 100 %).

In an ideal settling tank, particles with $v_s\ge v_0$ are fully removed, and those with $v_s<v_0$ are removed in the ratio $v_s/v_0$, where $v_0=Q/A$ is the overflow rate.

Step 1: Settling velocity of the 0.025 mm particles

$d=0.0025$ cm, $S_s=2.65$, $\nu_{20}=0.01007$ cm²/s.

$$v_s=\frac{981\times1.65\times0.0025^2}{18\times0.01007}=0.0558\ \text{cm/s}\quad(Re=0.014<1)$$

Step 2: Overflow rate

For 95 % removal:

$$v_0=\frac{v_s}{0.95}=\frac{0.0558}{0.95}=0.0587\ \text{cm/s}=50.8\ \text{m}^3/\text{m}^2/\text{day}\ (=2.11\ \text{m}^3/\text{m}^2/\text{h})$$

Surface area $A=12\times5=60$ m², so the tank can treat

$$Q=v_0A=0.000587\times60=0.03525\ \text{m}^3/\text{s}=3046\ \text{m}^3/\text{day}$$

Horizontal velocity $=Q/(B\,H)=0.03525/(5\times3)=0.2350$ cm/s $=0.141$ m/min.

Step 3: Check for 99 % removal of 0.04 mm particles

$d=0.004$ cm:

$$v_s=\frac{981\times1.65\times0.004^2}{18\times0.01007}=0.1429\ \text{cm/s}\quad(Re=0.057<1)$$

$v_s=0.1429$ cm/s is greater than $v_0=0.0587$ cm/s ($v_s/v_0=2.43$), so these particles are removed 100 %, which exceeds 99 %.

Answer: overflow rate = 0.0587 cm/s (50.8 m³/m²/day); yes, the tank is sufficient to remove 99 % of 0.04 mm particles (it removes 100 %).

Data

Population 20 000; peak allowance 112 lpcd; settling velocity $v_s=0.2$ mm/s; $L=2B$; settling (detention) period $t=3$ h.

Step 1: Flow

$$Q=20\,000\times112=2\,240\,000\ \text{l/day}=2240\ \text{m}^3/\text{day}$$

Step 2: Surface area (overflow rate = settling velocity)

$$v_0=0.2\times10^{-3}\times86400=17.28\ \text{m}^3/\text{m}^2/\text{day},\qquad A=\frac{Q}{v_0}=\frac{2240}{17.28}=129.63\ \text{m}^2$$

Step 3: Width and length

$$2B^2=129.63\Rightarrow B=8.05\ \text{m},\qquad L=2B=16.10\ \text{m}$$

Step 4: Depth

$$H=\frac{V}{A}=\frac{Q\,t}{A}=\frac{2240\times3/24}{129.63}=2.16\ \text{m}$$

(Also $H=v_s\,t=0.2\times10^{-3}\times10\,800=2.16$ m.)

Adopt $B=8.1$ m, $L=16.2$ m, effective depth 2.2 m, plus 0.3 m free board and 0.5 m sludge zone, a total of 3.0 m.

Answer: tank about 16.1 m $\times$ 8.1 m in plan and 2.16 m effective depth.

Data

Population 25 000; peak allowance 110 lpcd; $v_s=0.2$ mm/s; $L=2B$; settling period $t=3$ h.

Step 1: Flow

$$Q=25\,000\times110=2\,750\,000\ \text{l/day}=2750\ \text{m}^3/\text{day}$$

Step 2: Surface area

$$v_0=0.2\times10^{-3}\times86400=17.28\ \text{m}^3/\text{m}^2/\text{day},\qquad A=\frac{2750}{17.28}=159.14\ \text{m}^2$$

Step 3: Width and length

$$2B^2=159.14\Rightarrow B=8.92\ \text{m},\qquad L=2B=17.84\ \text{m}$$

Step 4: Depth

$$H=\frac{Q\,t}{A}=\frac{2750\times3/24}{159.14}=2.16\ \text{m}$$

Adopt $B=9.0$ m, $L=18.0$ m, effective depth 2.2 m, free board 0.3 m, sludge zone 0.5 m (total 3.0 m).

Sketch

 PLAN
  +----------------------------------------+
  |inlet|          L = 18.0 m          |outlet|  B = 9.0 m
  +----------------------------------------+
 SECTION
  inlet ->  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~  FB 0.3
            |      H = 2.2 m           |-> outlet
            |__________________________|
            \___ sludge zone 0.5 m ___/ -> drain

Answer: tank 18.0 m $\times$ 9.0 m in plan with 2.2 m effective depth (3.0 m total).

Settling velocity of a discrete spherical particle is first tried with Stokes' law, and the Reynolds number is checked. If $Re>1$ the transition formula must be used.

Data

$d=0.17$ mm $=0.017$ cm, $S_s=2.65$, $\nu=0.00978$ cm²/s (at 21 °C, about 0.978 centistokes), $g=981$ cm/s².

Step 1: Stokes' law (trial)

$$v_s=\frac{g(S_s-1)d^2}{18\nu}=\frac{981\times1.65\times0.017^2}{18\times0.00978}=2.657\ \text{cm/s}$$ $$Re=\frac{v_s d}{\nu}=\frac{2.657\times0.017}{0.00978}=4.62$$

Since $Re>1$, Stokes' law is not valid and the flow is in the transition zone ($1<Re<1000$).

Step 2: Transition zone (Fair and Geyer)

$$C_d=\frac{24}{Re}+\frac{3}{\sqrt{Re}}+0.34,\qquad v_s=\sqrt{\frac{4g(S_s-1)d}{3C_d}}$$

Trial and error:

The values converge to $v_s=2.06$ cm/s, $Re=3.6$, $C_d=8.61$.

Answer: settling velocity $v_s\approx2.06$ cm/s ($20.6$ mm/s), with $Re=3.6$. (Stokes' law would overestimate it as 2.66 cm/s.)

Settling velocity of a discrete spherical particle is first tried with Stokes' law, and the Reynolds number is checked. If $Re>1$ the transition formula must be used.

Data

$d=0.14$ mm $=0.014$ cm, $S_s=2.61$, $\nu=0.00893$ cm²/s (at 25 °C, about 0.893 centistokes), $g=981$ cm/s².

Step 1: Stokes' law (trial)

$$v_s=\frac{g(S_s-1)d^2}{18\nu}=\frac{981\times1.61\times0.014^2}{18\times0.00893}=1.926\ \text{cm/s}$$ $$Re=\frac{v_s d}{\nu}=\frac{1.926\times0.014}{0.00893}=3.02$$

Since $Re>1$, Stokes' law is not valid and the flow is in the transition zone ($1<Re<1000$).

Step 2: Transition zone (Fair and Geyer)

$$C_d=\frac{24}{Re}+\frac{3}{\sqrt{Re}}+0.34,\qquad v_s=\sqrt{\frac{4g(S_s-1)d}{3C_d}}$$

Trial and error:

The values converge to $v_s=1.57$ cm/s, $Re=2.5$, $C_d=12.04$.

Answer: settling velocity $v_s\approx1.57$ cm/s ($15.7$ mm/s), with $Re=2.5$. (Stokes' law would overestimate it as 1.93 cm/s.)

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

$$\frac{H}{v_s}=\frac{L}{v_h}\ \Rightarrow\ v_h=v_s\,\frac{L}{H}$$

Step 1: Settling velocity of the 0.02 mm particle

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

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

Step 2: Horizontal velocity

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

$$v_h=0.03597\times\frac{20}{2.5}=0.2878\ \text{cm/s}=17.27\ \text{cm/min}=0.173\ \text{m/min}$$

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).

For complete removal in an ideal horizontal-flow tank, a particle must settle through depth $H$ while the water flows through length $L$:

$$v_h=v_s\,\frac{L}{H}$$

Step 1: Settling velocity of the 0.02 mm particle

$d=0.002$ cm, $S_s=2.65$, $\nu=0.01$ cm²/s:

$$v_s=\frac{981\times1.65\times0.002^2}{18\times0.01}=0.03597\ \text{cm/s}=0.3597\ \text{mm/s}=21.58\ \text{mm/min}\quad(Re=0.007<1)$$

Step 2: Velocity of flow

$L=30$ m, $H=3$ m, $L/H=10$:

$$v_h=21.58\times10=215.8\ \text{mm/min}$$

(= 0.216 m/min, which is below the usual maximum of 0.3 m/min.)

Answer: velocity of flow $\approx216$ mm/min.

Treatment processes and impurities removed

Factors affecting sedimentation in a tank

1. Size of particle: $v_s \propto d^2$; bigger particles settle quicker.
2. Specific gravity of particle: heavier particles settle faster ($v_s \propto S_s-1$).
3. Temperature (viscosity): warmer water has lower viscosity, so settling is faster; $v_s \propto 1/\nu$.
4. Surface overflow rate / tank area: larger surface area for the flow reduces the overflow rate and improves removal.
5. Detention time: the longer, the better the removal.
6. Velocity of flow: should be low (below about 0.3 m/min) so that the settled sludge is not scoured.
7. Tank shape and depth, inlet and outlet arrangement (short-circuiting, density currents, wind).
8. Flocculation: coagulated particles are bigger and settle faster.
9. Hindered settling at high concentration and tank turbulence.

Why coagulants are used

• Fine silt and colloids (0.001 to 1 µm) have a negative charge and do not settle by themselves in a reasonable time.
• A coagulant such as alum supplies positive ions (Al$^{3+}$) that neutralise the charge, and forms a jelly-like Al(OH)$_3$ floc which enmeshes the particles:

$$Al_2(SO_4)_3 + 3Ca(HCO_3)_2 \rightarrow 2Al(OH)_3\downarrow + 3CaSO_4 + 6CO_2$$

• The large flocs settle quickly, which removes turbidity, colour, bacteria and organic matter, reduces tank size and detention time, and increases the length of filter runs.

Coagulant

A coagulant is a chemical added to water to destabilise the fine suspended and colloidal particles and make them join into large, settleable flocs. Common coagulants are alum (aluminium sulphate), ferrous sulphate (copperas) with lime, ferric chloride, ferric sulphate, sodium aluminate and polyelectrolytes (as coagulant aids).

Alum reaction:

$$Al_2(SO_4)_3\cdot 18H_2O + 3Ca(HCO_3)_2 \rightarrow 2Al(OH)_3\downarrow + 3CaSO_4 + 18H_2O + 6CO_2$$

Factors affecting coagulation

1. Type and dose of coagulant: an under- or over-dose gives poor floc; the optimum is found by the jar test.
2. pH of water: each coagulant works best in a range (alum about pH 6.5 to 8.5; ferrous sulphate with lime pH above 8.5).
3. Alkalinity: alum needs about 0.5 mg/l alkalinity (as CaCO$_3$) for each mg/l of alum.
4. Turbidity and nature of particles: low turbidity needs more coagulant or a coagulant aid, since there are fewer particles to collide.
5. Temperature: low temperature slows the reaction and makes weak floc, increasing the dose and time.
6. Mixing: rapid mixing for 30 s to 2 min ensures instant dispersal; slow stirring then builds floc. Too fast mixing breaks the floc.
7. Detention time in the flocculator (20 to 40 min).
8. Presence of ions and organic matter (e.g. sulphates, colour compounds) alters the dose.
9. Coagulant aids (activated silica, polyelectrolytes, bentonite) help form stronger and heavier floc.

Mixing (rapid mixing) must spread the coagulant quickly and uniformly through the whole of the water within 30 seconds to 2 minutes, so that the charge is neutralised before large flocs begin to form. The usual devices are as follows.

1. Hydraulic mixing (no power)

(a) Mixing channel with baffles (baffled channel): the water flows round the vertical baffles or over and under horizontal baffles; the turbulence at each turn mixes the chemical. Velocity 0.3 to 0.45 m/s.

   -->|  |-->|  |-->|  |-->
   ----+  +--+  +--+  +---->
   (over-and-under baffles)

(b) Hydraulic jump (weir) mixing: the coagulant is dropped at a Parshall flume or a weir where a hydraulic jump forms. Simple and cheap, but the head is lost.

   ~~~~~~\   jump  /~~~~~
   water  \___/\/\___/ coagulant added

(c) Pipe (in-line) mixing / venturi: the coagulant is injected into a pipe at a throat or a bend.

2. Mechanical mixing

(a) Rapid mix tank with a paddle or propeller (turbine): a square or circular tank with a motor-driven impeller; detention 30 to 60 s, velocity gradient $G=300$ to $1000\ \text{s}^{-1}$.

        [Motor]
          |
   ______|______
  |      |      |  <- water in
  |    --+--    |
  |     paddle  |  -> to flocculator
  |_____________|

(b) Deflecting baffle plate / flash mixer.

3. Compressed-air mixing

Air bubbles through diffusers in the bottom of the tank cause turbulence.

The mixer is followed by the flocculator (slow mix) – baffled channels or paddles at 0.1 to 0.5 m/s and 20 to 40 min – where flocs grow.

A jar test is a laboratory test that finds the optimum coagulant dose (and the best pH) for a given raw water by simulating coagulation, flocculation and sedimentation in a set of beakers.

Apparatus

A jar-test (flocculator) apparatus has six beakers or jars (1 or 2 litres each) with variable-speed paddles driven by a common motor, a pH meter, a turbidimeter and a stop-watch.

Procedure

1. Fill the six jars with equal volumes of raw water and note the temperature, pH, turbidity and alkalinity.
2. Add increasing doses of coagulant solution to the jars (e.g. 5, 10, 15, 20, 25, 30 mg/l), at the same time.
3. Rapid mix at 100 rpm for 1 minute (simulating the rapid mixing basin).
4. Slow mix at 20 to 30 rpm for 15 to 20 minutes (simulating flocculation).
5. Switch off, allow the flocs to settle for 30 minutes (sedimentation) and note the time the first floc forms and its size.
6. Test the supernatant for turbidity, colour and residual coagulant.

Interpretation

The jar that gives the clearest water with the lowest dose, with a quick, big, firm floc is the optimum dose. The test is repeated at different pH (by adding lime or acid) to find the best pH and for different coagulants or coagulant aids.

Uses

• To find the optimum coagulant dose and pH, which changes with the raw water quality.
• To compare coagulants and coagulant aids and to save chemicals.
• To select the mixing and settling time needed for design.

A slow sand filter (SSF) is a gravity filter in which water passes slowly (0.1 to 0.4 m/h) through a fine sand bed. It works by straining, sedimentation, adsorption and especially biological action, and it needs no chemicals.

Construction

   Inlet          supernatant water 1.0-1.5 m
     |      ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
     v      | schmutzdecke (biological film)
  ______    |:::::::::::::::::::::::::::::|
 |      |   | fine sand 0.9-1.0 m         |
 |      |   |-----------------------------|
 |      |   | gravel 0.3-0.45 m (graded)  |
 |______|   |==under-drain pipes (laterals)|
 Free board 0.5 m       |             |
                   main drain -> clear water

• Tank: rectangular RCC or masonry, with watertight floor, L : B about 2:1 to 3:1; area up to about 2000 m² per bed, with a minimum of two units.
• Supernatant water: 1.0 to 1.5 m depth above the sand, plus 0.5 m free board.
• Filter sand: depth 0.9 to 1.2 m, effective size 0.2 to 0.3 mm, uniformity coefficient 2 to 3 (not more than 5), free of clay and organic matter.
• Base material: gravel 0.3 to 0.45 m in 3 or 4 layers from 3 mm (top) to 40 mm (bottom).
• Under-drainage: central main drain with lateral perforated pipes or open-jointed tiles covered with gravel, bed slope 1 in 100.
• Inlet and outlet chambers with valves, weirs, a filter-to-waste valve, and a rate (flow) controller.

Design considerations

• Rate of filtration: 0.1 to 0.4 m/h (100 to 400 l/h/m²).
• Area: $A=Q/\text{rate}$, with one extra unit as standby for cleaning.
• Sand cleaning: when the head loss reaches 0.6 to 1.2 m, remove the top 1 to 3 cm by scraping, and re-sand when the bed depth is below 0.4 to 0.5 m (cleaning every 1 to 3 months).
• Influent turbidity not above 30 NTU (use a roughing filter or sedimentation for turbid water).
• After scraping, the filtrate of the first 1 to 2 days is run to waste until the schmutzdecke is formed again.
• Efficiency: removes 98 to 99 % of bacteria and the turbidity to below 1 NTU; chlorination is still required.

Purposes of aeration

1. To oxidise soluble iron and manganese into insoluble precipitates.
2. To remove dissolved gases such as CO$_2$, H$_2$S and methane, which cause corrosion, taste and odour.
3. To remove taste and odour from volatile and algal substances.
4. To add oxygen to oxygen-deficient waters and make them fresh.
5. To reduce free CO$_2$ and so raise the pH and cut corrosion.

Removal mechanism of iron and manganese

Fe and Mn are present in ground water or reservoir bottom water as soluble ferrous and manganous bicarbonates. Aeration adds oxygen and drives off CO$_2$; the oxidation then makes them insoluble:

$$4Fe(HCO_3)_2 + O_2 + 2H_2O \rightarrow 4Fe(OH)_3\downarrow + 8CO_2$$ $$2Mn(HCO_3)_2 + O_2 \rightarrow 2MnO_2\downarrow + 4CO_2 + 2H_2O\ (\text{slow, pH}>9\text{ or by catalyst})$$

The precipitates are settled out in a tank and removed by filtration. Where aeration is not enough, chemical oxidants (chlorine, KMnO$_4$, ozone), manganese greensand filters, lime softening or ion exchange are used.

Significance of residual chlorine

Residual chlorine is the chlorine left in water after the demand has been satisfied and the required contact time has passed. A free residual of about 0.2 to 0.5 mg/l is usually maintained.

Significance

• It is a safety margin against re-contamination in service reservoirs and distribution pipes.
• Its presence proves that the chlorine demand has been met and the disinfection is complete.
• It is easy to measure (orthotolidine or DPD test) at any point, so it is a quick check of the plant performance.
• It continues to kill bacteria that enter through leaks or back-siphonage.
• Too high a residual gives a taste and may produce trihalomethanes, so it is limited (WHO guideline 5 mg/l; Nepal drinking water standard 0.1 to 0.5 mg/l).

Why high iron and manganese are objectionable

• They give a bitter, metallic taste to water, tea and food.
• They stain clothes, utensils and sanitary fittings (reddish-brown for Fe, black for Mn).
• They support iron bacteria that form slime and clog pipes, and cause red or black water.
• They reduce pipe capacity and foul zeolite softeners, boilers and industrial processes (textile, paper, food).
• Limits: Fe 0.3 mg/l and Mn 0.05 mg/l as per WHO.

Methods of removal

1. Aeration, sedimentation and filtration.
2. Chemical oxidation by chlorine, potassium permanganate or ozone, followed by filtration.
3. Manganese greensand filtration.
4. Lime softening at high pH.
5. Ion exchange.
6. Coagulation at suitable pH.

Methods of aeration

1. Gravity aerators: water falls through the air over cascade steps (3 to 6 steps, each 0.2 to 0.3 m high), through trays of coke or stones (3 to 5 trays at 0.3 to 0.5 m spacing), or over an inclined plane, so that thin sheets and droplets are exposed to air.

      water in
         \__
            \__  steps / trays
               \__ -> collecting basin

2. Spray aerators (fountain): water is forced through nozzles (2.5 to 4 cm) at 70 kPa and spray 1 to 3 m into the air as fine droplets, which gives a very large area for gas exchange.
3. Air diffusion (bubble) aerators: compressed air is passed through porous plates or perforated pipes at the bottom of a tank (0.3 to 0.75 m³ air per m³ water, 10 to 30 min contact).
4. Mechanical aerators: paddles or turbines churn the surface to draw in air.
5. Packed (forced draft) towers: counter-current air and water through packing, for highest gas removal.