Wastewater Treatment

A grit chamber is a narrow, long channel or tank placed after the screens (and before the primary sedimentation tank) in which the velocity of sewage is controlled so that heavy inorganic particles (sand, gravel, cinders, eggshell fragments of size about 0.15 mm and above, specific gravity about 2.65) settle, while the lighter organic matter stays in suspension and passes on.

Why a grit chamber is needed

• Grit is abrasive and wears pumps, pipes and sludge-handling equipment.
• It settles in channels, tanks and digesters, reducing their capacity and causing clogging.
• If grit mixes with organic sludge in the primary tank, the sludge becomes heavy and difficult to digest and dewater.
• Removing grit separately gives clean, easily disposable inert material.

Purpose

To remove grit without removing the organic solids, and so protect the downstream mechanical equipment and save tank capacity.

Construction (horizontal-flow type)

 Plan                      Section
 ----------------------    ~~~~~~~~~~~~~~~~~~~~~~~ water level
 inlet -> |           |    -> v = 0.2-0.3 m/s ->
 zone  -> |   channel |  ->  \_ grit settles   _/
 -----------------------      grit storage hopper
          proportional weir / Parshall flume at outlet

• A rectangular channel, normally two or more units in parallel so that one can be cleaned while the other works; inlet and outlet have tapered transitions.
• The outlet has a proportional (Sutro) weir or a Parshall flume to keep the horizontal velocity almost constant at about 0.3 m/s even when the flow varies.
• A hopper at the bottom (or extra depth of 0.2-0.3 m) stores the grit; it is removed by hand-scraping in small plants or by mechanical scrapers, screw conveyors or air-lift in large plants.
• Aerated and vortex-type grit chambers are other variants.

Design criteria

1. Flow-through velocity: about 0.2-0.3 m/s. It must be below the scour velocity of grit, $v_c=\sqrt{\dfrac{8k}{f}(G-1)g\,d}$ ($k=0.04$-$0.06$, $f=0.02$-$0.03$), but high enough to carry organic matter ($v \ge$ about 0.15 m/s).
2. Detention time: 30-60 seconds (usually about 1 minute) at maximum flow.
3. Settling velocity for 0.2 mm grit of specific gravity 2.65 is about 0.02-0.04 m/s (Hazen: $v_s=418(G-1)d^2\dfrac{3T+70}{100}$ mm/s, $d$ in mm).
4. Length $L=\dfrac{v\,H}{v_s}$, increased by 20-50% for inlet and outlet turbulence.
5. Cross-section $A=Q/v$, with depth $H$ chosen (0.6-1.5 m) so that width $B=A/H$ is practical; allow 0.3 m free board and 0.2-0.3 m grit storage depth.
6. Surface loading (overflow rate) about 800-1200 m³/m²/day.
7. Grit removed is about 0.03-0.1 m³ per 1000 m³ of sewage.

Example: $Q=0.2$ m³/s, $v=0.25$ m/s gives $A=0.8$ m²; with $H=0.8$ m, $B=1.0$ m; if $v_s=0.025$ m/s then $L=0.25\times 0.8/0.025 = 8$ m, plus 25% = 10 m.

Biological wastewater treatment uses micro-organisms (mainly bacteria) to convert dissolved and colloidal organic matter into carbon dioxide, water, new cells and stable end products. The organisms may be kept in suspension in the liquid or grown on a fixed surface.

Suspended growth process

The micro-organisms are kept in suspension in the wastewater by mixing or aeration, forming flocs. The mixed liquor is later settled and part of the biomass is recycled.

• Examples: activated sludge process (ASP), aerated lagoons, oxidation ditch, sequencing batch reactor (SBR), anaerobic digesters.
• Needs aeration/mixing energy, a clarifier and sludge return; more sensitive to shock loads; higher operating skill.
• Biomass is measured as MLSS/MLVSS.

Attached (fixed-film) growth process

The micro-organisms grow as a biofilm on a fixed medium (stones, plastic media, discs) over which wastewater flows.

• Examples: trickling filter, rotating biological contactor (RBC), submerged biofilters.
• Simple, low energy, stable with variable loads; needs a medium and a final clarifier to remove sloughed film; larger area.

Principles of biological treatment

1. Food and organisms: organic matter in sewage is food (BOD) for bacteria. They need oxygen (aerobic), nutrients (N, P) and suitable pH (6.5-8.5) and temperature (20-35°C).
2. Aerobic oxidation (energy): $\text{organics} + O_2 \xrightarrow{\text{bacteria}} CO_2 + H_2O + \text{energy}$
3. Synthesis (growth): $\text{organics} + O_2 + N \rightarrow \text{new cells}$
4. Endogenous respiration: when food is short, cells oxidise their own mass: $\text{cells} + O_2 \rightarrow CO_2 + H_2O + NH_3$
5. Colloidal and suspended matter is first adsorbed on the floc/film, then hydrolysed and metabolised.
6. The cells are removed by settling as sludge, which leaves a clear effluent of low BOD.
7. Anaerobic treatment (no oxygen) converts organics to methane and CO₂, used for sludge digestion and strong wastes.
8. The process is controlled by the food-to-micro-organism ratio ($F/M$), mean cell residence time, oxygen supply and temperature.

An oxidation pond (waste stabilisation pond) is a shallow, man-made earthen basin, usually 1-1.5 m deep, in which raw or settled sewage is treated naturally by the combined action of bacteria and algae, using sunlight. It is cheap and suits small towns with plenty of land.

Theory and pollutant removal mechanism

The removal depends on a symbiotic relationship between bacteria and algae:

1. Aerobic bacteria oxidise the organic matter: $\text{organics} + O_2 \rightarrow CO_2 + H_2O + \text{new cells} + NH_3 + PO_4$
2. Algae use sunlight, the CO₂, ammonia and phosphates released by bacteria and produce oxygen by photosynthesis: $CO_2 + H_2O + \text{sunlight} \rightarrow \text{algae} + O_2$
3. This oxygen is used by the bacteria, so no mechanical aeration is needed. Wind action also aerates the surface.
4. Settleable solids go to the bottom, where anaerobic bacteria digest them to CH₄, CO₂ and H₂S. The sludge layer is thin and stabilised.
5. Pathogens die by sunlight (UV), high pH, long detention and predation.

   sunlight
     \ | /
  ---------------------------- water surface
  Algae: CO2 + sunlight -> O2   ^ O2 (and wind)
          |   O2 down   CO2 up  |
  Bacteria (aerobic zone): organics + O2 -> CO2
  - - - - - - - - - - - - - - - - - - - - - -
  Anaerobic zone: sludge -> CH4, CO2, H2S
  ============ bottom (sludge) ==============

Typical data: detention 5-30 days, organic loading 200-400 kg BOD/ha/day, BOD removal 70-90%.

Commissioning (starting up the pond)

Commissioning means bringing a newly built pond into stable operation.

1. Inspect the embankment, inlet, outlet and lining; remove vegetation and make the pond leak-free.
2. Fill with clean water first (river or tube-well water), to the operating depth, to prevent weed growth, embankment erosion and bad odours; filling with raw sewage directly gives odour and anaerobic conditions.
3. Seed the pond with active sludge or contents of a working pond (about 5-10% of volume) to start the culture.
4. Start loading gradually: begin with a low flow (about 1/3 of design) or low BOD, then raise it over several weeks as algae develop (the water turns green).
5. In cold seasons commission in spring/summer when sun and temperature help algae growth.
6. Check DO, pH (above 7.5 in the day) and colour regularly; if the pond turns grey/black (anaerobic), reduce loading or recirculate effluent.
7. Do not discharge effluent until the pond reaches steady performance.

Data

Flow $Q=1.2\times 10^6$ l/day $=1200$ m³/day; influent BOD (after PST) $=200$ mg/l; surface (hydraulic) loading $=1200$ l/m²/day; organic loading $=250$ g/m³/day.

Step 1: BOD load

$$W = Q\times BOD = 1200\times 200\times 10^{-3} = 240.0\ \text{kg/day}$$

Step 2: Volume (from organic loading)

$$V = \frac{W}{\text{organic loading}} = \frac{240.0}{0.25} = 960.0\ \text{m}^3$$

Step 3: Surface area (from hydraulic loading)

$$A = \frac{Q}{\text{surface loading}} = \frac{1.2\times 10^6}{1200} = 1000.0\ \text{m}^2$$

Step 4: Depth

$$\text{Depth} = \frac{V}{A} = \frac{960.0}{1000.0} = 0.96\ \text{m}$$

(Diameter of the circular filter $D=\sqrt{4A/\pi}=35.7$ m. The calculated depth is smaller than the usual 1.8-3 m of a standard-rate filter; if a depth of 1.8 m is wanted, the area is reduced to 533 m² and the hydraulic loading rises accordingly.)

Step 5: Efficiency by the NRC equation (standard rate, no recirculation, $F=1$)

$$E=\frac{100}{1+0.4432\sqrt{W/V}} = \frac{100}{1+0.4432\sqrt{240.0/960.0}} = 81.86\%$$

Effluent BOD $=200\,(1-81.86/100) = 36.3$ mg/l.

Answer: volume $=960.0$ m³, depth $=0.96$ m (area 1000.0 m²); NRC efficiency $=81.86\%$.

Data: $Q=3\times10^6$ l/day $=3000$ m³/day, settled-sewage BOD$_5$ $=300$ mg/l, surface loading $=3000$ l/m²/day, organic loading $=300$ g/m³/day. (The NRC efficiency is also given below for completeness.)

Step 1: BOD load

$$W=Q\times BOD=3000\times 300\times10^{-3}=900.0\ \text{kg/day}$$

Step 2: Volume (organic loading 300 g/m³/day)

$$V=\frac{W}{\text{organic loading}}=\frac{900.0}{0.300}=3000.0\ \text{m}^3$$

Step 3: Area (hydraulic loading 3000 l/m²/day) and depth

$$A=\frac{Q}{\text{surface loading}}=\frac{3000\times1000}{3000}=1000.0\ \text{m}^2,\qquad \text{Depth}=\frac{V}{A}=\frac{3000.0}{1000.0}=3.00\ \text{m}$$

Diameter of the circular filter: $D=\sqrt{4A/\pi}=35.7$ m.

Step 4: Efficiency by the NRC equation (no recirculation, $F=1$)

$$E=\frac{100}{1+0.4432\sqrt{W/V}}=\frac{100}{1+0.4432\sqrt{900.0/3000.0}}=80.47\%$$

Effluent BOD $=300\,(1-80.47/100)=58.6$ mg/l.

Answer: volume $=3000.0$ m³; depth $=3.00$ m; area $=1000.0$ m²; NRC efficiency $=80.47\%$.

Data and approach

High-rate filter design uses the NRC (National Research Council) equation; $W$ is in kg/day of BOD applied to the filter, $V$ in m³:

$$E = \frac{100}{1+0.4432\sqrt{\dfrac{W}{VF}}},\qquad F=\frac{1+R}{(1+0.1R)^2}$$

Step 1: BOD applied to the filter and required efficiency

$$\begin{aligned} Q &= 5000\ \text{m}^3/\text{day}\\ \text{BOD after primary} &= 250\times(1-0.3) = 175.0\ \text{mg/l}\\ W &= Q\times BOD = 5000\times 175.0\times 10^{-3} = 875.0\ \text{kg/day}\\ E &= \frac{175.0-30}{175.0}\times 100 = 82.86\% \end{aligned}$$

Step 2: Recirculation factor

$$F = \frac{1+1.5}{(1+0.1\times 1.5)^2} = 1.890$$

Step 3: Volume

$$82.86 = \frac{100}{1+0.4432\sqrt{W/(VF)}}\Rightarrow \sqrt{\frac{W}{VF}} = 0.4668 \Rightarrow VF = \frac{875.0}{0.4668^2} = 4015.1$$ $$V = \frac{VF}{F} = \frac{4015.1}{1.890} = 2124.0\ \text{m}^3$$

Step 4: Diameter

Adopt depth $d = 2.0$ m (high-rate filters are 1.8-3 m deep).

$$A = \frac{V}{d} = 1062.0\ \text{m}^2,\qquad D = \sqrt{\frac{4A}{\pi}} = 36.77\ \text{m (adopt 36.8 m)}$$

Step 5: Checks

$$\begin{aligned} \text{Organic loading} &= \frac{W}{V} = 0.41\ \text{kg/m}^3\text{/day}\ (\text{high rate: 0.8-1.6})\\ \text{Hydraulic loading} &= \frac{Q(1+R)}{A} = 11.8\ \text{m}^3/\text{m}^2\text{/day}\ (\text{high rate: 10-40}) \end{aligned}$$

Answer: single-stage high-rate filter of volume about 2124.0 m³, depth 2.0 m, diameter 36.8 m.

Activated sludge process (ASP)

The activated sludge process is an aerobic, suspended-growth biological treatment in which settled sewage is mixed and aerated with a flocculent mass of micro-organisms called activated sludge. The organisms oxidise organic matter, and the mixed liquor is then settled; a part of the settled sludge is returned to the aeration tank to maintain the biomass.

Working

1. Primary treatment (screens, grit chamber, primary sedimentation) removes about 30-35% of the BOD.
2. Aeration tank: the settled sewage is mixed with return activated sludge (RAS). Air (diffused or mechanical) supplies oxygen and keeps the flocs in suspension for 4-8 hours. Bacteria and protozoa in the floc adsorb and oxidise organic matter to CO₂ and water and form new cells.
3. Secondary sedimentation tank: the flocs settle and the clear effluent (BOD 20-30 mg/l) overflows.
4. Sludge return: 25-50% of the settled sludge is recycled to the aeration tank to keep MLSS at about 2000-4000 mg/l. The excess (waste) sludge is sent to thickening and digestion.

 Raw    +--------+ +-----+ +---------+ +---------+   Effluent
 sewage |Screen/ | |PST  | |Aeration | |Secondary|-->
 ------>|Grit    |>|     |>|tank+air |>|settling |
        +--------+ +--+--+ +----^----+ +----+----+
                      |        | RAS       |
                      |        +-----------+
              primary sludge     waste sludge -> thickener
                                 -> digester

Advantages

• High BOD removal efficiency (85-95%) and a good quality effluent.
• Needs less land than a trickling filter or oxidation pond.
• No fly or odour nuisance if properly run; no head loss problem.
• Flexible: loading can be adjusted (conventional, extended aeration, step aeration, etc.).
• Capable of nitrification with long sludge age.

Disadvantages

• High capital and operating costs, especially for power for aeration.
• Needs skilled and continuous operation and control (F/M, MLSS, DO, sludge return).
• Sensitive to shock loads and toxic substances.
• Large quantity of sludge is produced; it needs treatment and disposal.
• Sludge bulking (high SVI) and foaming problems may occur.

A grit chamber is designed to settle grit of $d=0.23$ mm and $G=2.65$ at 25°C while organic matter ($G=1.02$) is carried forward. The scour velocity is checked with $k=0.06$, $f=0.03$.

Step 1: Design flow

$$Q = \frac{60\times 10^6}{1000\times 86400} = 0.6944\ \text{m}^3/\text{s}$$

Step 2: Settling velocity of the grit

Hazen's formula (with temperature correction), $d$ in mm:

$$v_s = 418\,(G-1)\,d^2\left(\frac{3T+70}{100}\right)\ \text{mm/s} = 418\times (2.65-1)\times 0.23^2\times\frac{3\times 25+70}{100} = 52.90\ \text{mm/s} = 0.0529\ \text{m/s}$$

Step 3: Horizontal (flow-through) velocity

The velocity must not scour settled grit. Scour (critical) velocity (Camp-Shields):

$$v_c = \sqrt{\frac{8k}{f}(G-1)\,g\,d} = \sqrt{\frac{8\times 0.06}{0.03}(2.65-1)\times 9.81\times 0.23\times 10^{-3}} = 0.244\ \text{m/s}$$

Adopt $v_h = 0.20$ m/s (the scour velocity rounded down; the usual range is 0.2-0.3 m/s). This is high enough to keep the light organic matter moving, so it passes on instead of settling.

Step 4: Cross-section and dimensions

$$\begin{aligned} A_c &= \frac{Q}{v_h} = \frac{0.6944}{0.20} = 3.472\ \text{m}^2\\ \text{Adopt depth } H &= 1.0\ \text{m} \Rightarrow B = \frac{A_c}{H} = 3.47\ \text{m (adopt 3.5 m)}\\ L_{theo} &= \frac{v_h H}{v_s} = \frac{0.20\times 1.0}{0.0529} = 3.78\ \text{m}\\ L &= \max(1.25\,L_{theo},\ 30\,v_h) = \max(4.73,\ 6.00) = 6.00\ \text{m (adopt 6.0 m)} \end{aligned}$$

(25% is added for inlet and outlet turbulence, and the length is not allowed to give less than 30 s detention.)

Step 5: Checks

$$\begin{aligned} \text{Detention time} &= \frac{L}{v_h} = \frac{6.0}{0.20} = 30\ \text{s}\ (\text{within 30-90 s})\\ \text{Surface loading} &= \frac{Q}{B\,L} = \frac{0.6944\times 86400}{3.5\times 6.0} = 2857\ \text{m}^3/\text{m}^2\text{/day} \end{aligned}$$

Total depth $= H + \text{grit storage } 0.25 + \text{free board } 0.30 = 1.55$ m.

 Plan:   inlet ->[ <-- L = 6.0 m --> ]-> outlet (weir)
                 |  B = 3.5 m          |
 Section: water level ~~~~~~~~~~~~~~~~~~~  H = 1.0 m
          grit storage 0.25 m  ______________

Answer: provide a grit chamber (at least two units in parallel, each for the full flow so that one can be cleaned) of size $L\times B\times H = 6.0\ \text{m}\times 3.5\ \text{m}\times 1.0\ \text{m}$ (water depth), total depth about 1.55 m, with flow velocity 0.20 m/s.

The flow is $Q=190$ l/s $=0.19$ m³/s. The particle size is read as 0.21 mm (the "0.21 m" in the question is a typing slip). $k=0.06$, $f=0.03$, SOR (surface overflow rate) $=2$ cm/s $=0.02$ m/s, detention time $=60$ s.

Step 1: Surface area and depth

$$\begin{aligned} Q &= 0.190\ \text{m}^3/\text{s}\\ \text{Surface area } A &= \frac{Q}{SOR} = \frac{0.190}{0.020} = 9.50\ \text{m}^2\\ \text{Volume } V &= Q\,t = 0.190\times 60 = 11.40\ \text{m}^3\\ \text{Depth } H &= \frac{V}{A} = 1.20\ \text{m} \end{aligned}$$

Step 2: Scour velocity for the two kinds of particles

$$v_c=\sqrt{\frac{8k}{f}(G-1)\,g\,d}$$

The horizontal velocity must be above the organic scour velocity (organic matter stays suspended and is carried through) and below the grit scour velocity (settled grit is not re-suspended). Adopt $v_h = 0.20$ m/s, which satisfies 0.081 < 0.20 < 0.233 m/s. (The flow velocity should also be near 0.2-0.3 m/s in practice.)

Step 3: Dimensions

$$\begin{aligned} L &= v_h\,t = 0.20\times 60 = 12.00\ \text{m}\\ B &= \frac{A}{L} = \frac{9.50}{12.00} = 0.79\ \text{m}\\ \text{Check: } A_c &= B\,H = 0.79\times 1.20 = 0.950\ \text{m}^2 \approx \frac{Q}{v_h} = 0.950\ \text{m}^2 \end{aligned}$$

Adopt $L = 12.0$ m, $B = 0.8$ m, water depth $H = 1.20$ m, plus 0.25 m grit storage and 0.30 m free board (total depth about 1.75 m). Provide two units in parallel.

Answer: grit chamber $L\times B\times H = 12.0\times 0.8\times 1.20$ m, flow velocity 0.20 m/s, detention 60 s, surface overflow rate 2 cm/s.

Grit: $d=0.2$ mm, $G=2.65$ at $25^\circ$C; organic matter $G=1.02$; $k=0.06$, $f=0.03$.

Step 1: Design flow

$$Q = \frac{50\times 10^6}{1000\times 86400} = 0.5787\ \text{m}^3/\text{s}$$

Step 2: Settling velocity of the grit

Hazen's formula (with temperature correction), $d$ in mm:

$$v_s = 418\,(G-1)\,d^2\left(\frac{3T+70}{100}\right)\ \text{mm/s} = 418\times (2.65-1)\times 0.2^2\times\frac{3\times 25+70}{100} = 40.00\ \text{mm/s} = 0.0400\ \text{m/s}$$

Step 3: Horizontal (flow-through) velocity

The velocity must not scour settled grit. Scour (critical) velocity (Camp-Shields):

$$v_c = \sqrt{\frac{8k}{f}(G-1)\,g\,d} = \sqrt{\frac{8\times 0.06}{0.03}(2.65-1)\times 9.81\times 0.2\times 10^{-3}} = 0.228\ \text{m/s}$$

Adopt $v_h = 0.20$ m/s (the scour velocity rounded down; the usual range is 0.2-0.3 m/s). This is high enough to keep the light organic matter moving, so it passes on instead of settling.

Step 4: Cross-section and dimensions

$$\begin{aligned} A_c &= \frac{Q}{v_h} = \frac{0.5787}{0.20} = 2.894\ \text{m}^2\\ \text{Adopt depth } H &= 1.0\ \text{m} \Rightarrow B = \frac{A_c}{H} = 2.89\ \text{m (adopt 2.9 m)}\\ L_{theo} &= \frac{v_h H}{v_s} = \frac{0.20\times 1.0}{0.0400} = 5.00\ \text{m}\\ L &= \max(1.25\,L_{theo},\ 30\,v_h) = \max(6.25,\ 6.00) = 6.25\ \text{m (adopt 6.5 m)} \end{aligned}$$

(25% is added for inlet and outlet turbulence, and the length is not allowed to give less than 30 s detention.)

Step 5: Checks

$$\begin{aligned} \text{Detention time} &= \frac{L}{v_h} = \frac{6.5}{0.20} = 32\ \text{s}\ (\text{within 30-90 s})\\ \text{Surface loading} &= \frac{Q}{B\,L} = \frac{0.5787\times 86400}{2.9\times 6.5} = 2653\ \text{m}^3/\text{m}^2\text{/day} \end{aligned}$$

Total depth $= H + \text{grit storage } 0.25 + \text{free board } 0.30 = 1.55$ m.

 Plan:   inlet ->[ <-- L = 6.5 m --> ]-> outlet (weir)
                 |  B = 2.9 m          |
 Section: water level ~~~~~~~~~~~~~~~~~~~  H = 1.0 m
          grit storage 0.25 m  ______________

Answer: provide a grit chamber (at least two units in parallel, each for the full flow so that one can be cleaned) of size $L\times B\times H = 6.5\ \text{m}\times 2.9\ \text{m}\times 1.0\ \text{m}$ (water depth), total depth about 1.55 m, with flow velocity 0.20 m/s.

$Q=200$ l/s $=0.2$ m³/s, SOR $=2$ cm/s, detention $=1$ min $=60$ s, particle size 0.21 mm, $G_{inorganic}=2.7$, $k=0.06$, $f=0.03$.

Step 1: Surface area and depth

$$\begin{aligned} Q &= 0.200\ \text{m}^3/\text{s}\\ \text{Surface area } A &= \frac{Q}{SOR} = \frac{0.200}{0.020} = 10.00\ \text{m}^2\\ \text{Volume } V &= Q\,t = 0.200\times 60 = 12.00\ \text{m}^3\\ \text{Depth } H &= \frac{V}{A} = 1.20\ \text{m} \end{aligned}$$

Step 2: Scour velocity for the two kinds of particles

$$v_c=\sqrt{\frac{8k}{f}(G-1)\,g\,d}$$

The horizontal velocity must be above the organic scour velocity (organic matter stays suspended and is carried through) and below the grit scour velocity (settled grit is not re-suspended). Adopt $v_h = 0.20$ m/s, which satisfies 0.081 < 0.20 < 0.237 m/s. (The flow velocity should also be near 0.2-0.3 m/s in practice.)

Step 3: Dimensions

$$\begin{aligned} L &= v_h\,t = 0.20\times 60 = 12.00\ \text{m}\\ B &= \frac{A}{L} = \frac{10.00}{12.00} = 0.83\ \text{m}\\ \text{Check: } A_c &= B\,H = 0.83\times 1.20 = 1.000\ \text{m}^2 \approx \frac{Q}{v_h} = 1.000\ \text{m}^2 \end{aligned}$$

Adopt $L = 12.0$ m, $B = 0.9$ m, water depth $H = 1.20$ m, plus 0.25 m grit storage and 0.30 m free board (total depth about 1.75 m). Provide two units in parallel.

Answer: grit chamber $L\times B\times H = 12.0\times 0.9\times 1.20$ m, flow velocity 0.20 m/s, detention 60 s, surface overflow rate 2 cm/s.

Assumptions: sewage reaching the plant $=0.75\times 180 = 135$ lpcd, designed for the maximum rate ($2.4\times$ average): $Q_{avg}=5500\times 135/1000 = 742.5$ m³/day. Detention 2 h and depth 3 m are adopted (usual design values). Silica grains of $d=0.14$ mm ($G=2.65$) at 20°C.

Step 1: Design flow

$$Q = 5500\times 180\times 0.75\times 2.4/1000 = 1782\ \text{m}^3/\text{day}$$

Step 2: Capacity

Adopt detention period $t = 2.0$ h and side water depth $H = 3.0$ m (typical for plain sedimentation: 2-3 h, 2.5-4 m).

$$V = Q\,t = \frac{1782}{24}\times 2.0 = 148.5\ \text{m}^3,\qquad A = \frac{V}{H} = \frac{148.5}{3.0} = 49.5\ \text{m}^2$$

Step 3: Diameter

Provide one circular tank. $A_1=A$.

$$D = \sqrt{\frac{4A_1}{\pi}} = \sqrt{\frac{4\times 49.5}{\pi}} = 7.94\ \text{m}\ \Rightarrow\ \text{adopt } D = 8.0\ \text{m}$$

Adopt a total depth of $H + 0.5\ (\text{free board}) + 0.5\ (\text{sludge zone}) = 4.0$ m, with a central inlet well and peripheral weir.

Step 4: Checks

$$\begin{aligned} \text{Surface loading} &= \frac{Q/N}{\pi D^2/4} = \frac{1782}{50.3} = 35.5\ \text{m}^3/\text{m}^2\text{/day}\ (\text{usual 25-40 for primary tanks})\\ \text{Weir loading} &= \frac{Q/N}{\pi D} = \frac{1782}{\pi\times 8.0} = 70.9\ \text{m}^3/\text{m/day}\ (\text{should be below about 300})\\ \text{Actual detention} &= \frac{(\pi D^2/4)H}{Q/N}\times 24 = 2.03\ \text{h} \end{aligned}$$

Check of removal of the 0.14 mm silica: Hazen's formula at 20°C, $v_s=418(G-1)d^2\dfrac{3T+70}{100}=418\times1.65\times0.14^2\times1.3 = 17.57$ mm/s $= 1518$ m/day. The tank overflow rate is only 35.5 m³/m²/day, which is much less than $v_s$, so all silica particles larger than 0.14 mm settle easily; the size is governed by the detention time, not by the settling velocity.

   Section (circular, radial flow)
      inlet well        weir/launder
   ->  |__|~~~~~~~~~~~~~~~~~~~~~|_
       |  \                  /  | H = 3.0 m
       |   \______    ______/   | + sludge zone
       +----------\__/----------+
                  sludge hopper -> sludge pipe
        <---------- D = 8.0 m ---------->

Answer: circular sedimentation tank(s): 1 unit(s) of diameter 8.0 m, side water depth 3.0 m (total depth about 4.0 m), detention 2.0 h, surface loading 35.5 m³/m²/day.

Assumptions: sewage reaching the plant $=0.78\times135 = 105.3$ lpcd, designed for the maximum rate ($2.5\times$ average): $Q_{avg}=80000\times 105.3/1000 = 8424.0$ m³/day. Detention 2 h and depth 3 m are adopted; two tanks are provided. Silica $d=0.12$ mm ($G=2.65$) at 20°C.

Step 1: Design flow

$$Q = 80000\times 135\times 0.78\times 2.5/1000 = 21060\ \text{m}^3/\text{day}$$

Step 2: Capacity

Adopt detention period $t = 2.0$ h and side water depth $H = 3.0$ m (typical for plain sedimentation: 2-3 h, 2.5-4 m).

$$V = Q\,t = \frac{21060}{24}\times 2.0 = 1755.0\ \text{m}^3,\qquad A = \frac{V}{H} = \frac{1755.0}{3.0} = 585.0\ \text{m}^2$$

Step 3: Diameter

Provide 2 circular tanks, each with area $A_1 = A/2$ (one tank can then be shut down for cleaning).

$$D = \sqrt{\frac{4A_1}{\pi}} = \sqrt{\frac{4\times 292.5}{\pi}} = 19.30\ \text{m}\ \Rightarrow\ \text{adopt } D = 19.5\ \text{m}$$

Adopt a total depth of $H + 0.5\ (\text{free board}) + 0.5\ (\text{sludge zone}) = 4.0$ m, with a central inlet well and peripheral weir.

Step 4: Checks

$$\begin{aligned} \text{Surface loading} &= \frac{Q/N}{\pi D^2/4} = \frac{10530}{298.6} = 35.3\ \text{m}^3/\text{m}^2\text{/day}\ (\text{usual 25-40 for primary tanks})\\ \text{Weir loading} &= \frac{Q/N}{\pi D} = \frac{10530}{\pi\times 19.5} = 171.9\ \text{m}^3/\text{m/day}\ (\text{should be below about 300})\\ \text{Actual detention} &= \frac{(\pi D^2/4)H}{Q/N}\times 24 = 2.04\ \text{h} \end{aligned}$$

Check: Hazen's formula at 20°C, $v_s = 418\times1.65\times0.12^2\times1.3 = 12.91$ mm/s $=1116$ m/day, far greater than the overflow rate of the tank, so silica of 0.12 mm and larger is removed.

   Section (circular, radial flow)
      inlet well        weir/launder
   ->  |__|~~~~~~~~~~~~~~~~~~~~~|_
       |  \                  /  | H = 3.0 m
       |   \______    ______/   | + sludge zone
       +----------\__/----------+
                  sludge hopper -> sludge pipe
        <---------- D = 19.5 m ---------->

Answer: circular sedimentation tank(s): 2 unit(s) of diameter 19.5 m, side water depth 3.0 m (total depth about 4.0 m), detention 2.0 h, surface loading 35.3 m³/m²/day.

Data: $Q=4$ MLD $=4000$ m³/day; BOD fed to the first filter $=200$ mg/l; final BOD $\le 30$ mg/l; $E_1=2E_2$; depth $=1.2$ m; $R=2$ for both filters.

Data and approach

Two-stage filter (NRC equations, $W$ in kg/day, $V$ in m³):

$$E_1=\frac{100}{1+0.4432\sqrt{\dfrac{W_1}{V_1F}}},\qquad E_2=\frac{100}{1+\dfrac{0.4432}{1-E_1}\sqrt{\dfrac{W_2}{V_2F}}},\qquad F=\frac{1+R}{(1+0.1R)^2}$$

where $W_2=W_1(1-E_1)$ is the BOD load leaving the first stage ($E_1$ as a fraction in the second equation). Flow $Q=4000$ m³/day.

Step 1: Loads and efficiencies

$$\begin{aligned} BOD\ \text{applied} &= 200\ \text{mg/l},\quad W_1=4000\times 200\times10^{-3}=800.0\ \text{kg/day}\\ \text{Overall filter efficiency } E &= \frac{200-30}{200}=0.8500 \end{aligned}$$

Let $E_2=x$, so $E_1=2x$ (as given). Overall efficiency $E=E_1+E_2(1-E_1)$, i.e. $1-(1-2x)(1-x)=0.8500$.

$$2x^2-3x+0.8500=0\ \Rightarrow\ x=0.3792$$

So $E_2=37.92\%$ and $E_1=75.84\%$.

Step 2: Recirculation factor (same for both stages, $R=2$)

$$F=\frac{1+2}{(1+0.1\times 2)^2}=2.0833$$

Step 3: First stage

$$\sqrt{\frac{W_1}{V_1F}}=\frac{100/E_1-1}{0.4432}=0.7189\ \Rightarrow\ V_1=\frac{800.0}{2.0833\times 0.7189^2}=743.1\ \text{m}^3$$

Step 4: Second stage

$$\begin{aligned} W_2 &= W_1(1-E_1)=800.0\times(1-0.7584)=193.3\ \text{kg/day}\\ \sqrt{\frac{W_2}{V_2F}} &= \frac{(100/E_2-1)(1-E_1)}{0.4432}=0.8926\ \Rightarrow\ V_2=\frac{193.3}{2.0833\times 0.8926^2}=116.5\ \text{m}^3 \end{aligned}$$

Step 5: Diameters (depth $1.2$ m)

$$D_1=\sqrt{\frac{4V_1}{\pi\,d}}=\sqrt{\frac{4\times 743.1}{\pi\times 1.2}}=28.1\ \text{m},\qquad D_2=\sqrt{\frac{4\times 116.5}{\pi\times 1.2}}=11.1\ \text{m}$$

Answer: first-stage filter $D_1\approx 28.1$ m (volume 743.1 m³); second-stage filter $D_2\approx 11.1$ m (volume 116.5 m³).

Data: $Q=5$ MLD $=5000$ m³/day; BOD fed to the first filter $=180$ mg/l; final BOD $\le30$ mg/l; $E_1=2E_2$; depth $=1.2$ m; $R=2$ for both filters.

Data and approach

Two-stage filter (NRC equations, $W$ in kg/day, $V$ in m³):

$$E_1=\frac{100}{1+0.4432\sqrt{\dfrac{W_1}{V_1F}}},\qquad E_2=\frac{100}{1+\dfrac{0.4432}{1-E_1}\sqrt{\dfrac{W_2}{V_2F}}},\qquad F=\frac{1+R}{(1+0.1R)^2}$$

where $W_2=W_1(1-E_1)$ is the BOD load leaving the first stage ($E_1$ as a fraction in the second equation). Flow $Q=5000$ m³/day.

Step 1: Loads and efficiencies

$$\begin{aligned} BOD\ \text{applied} &= 180\ \text{mg/l},\quad W_1=5000\times 180\times10^{-3}=900.0\ \text{kg/day}\\ \text{Overall filter efficiency } E &= \frac{180-30}{180}=0.8333 \end{aligned}$$

Let $E_2=x$, so $E_1=2x$ (as given). Overall efficiency $E=E_1+E_2(1-E_1)$, i.e. $1-(1-2x)(1-x)=0.8333$.

$$2x^2-3x+0.8333=0\ \Rightarrow\ x=0.3681$$

So $E_2=36.81\%$ and $E_1=73.62\%$.

Step 2: Recirculation factor (same for both stages, $R=2$)

$$F=\frac{1+2}{(1+0.1\times 2)^2}=2.0833$$

Step 3: First stage

$$\sqrt{\frac{W_1}{V_1F}}=\frac{100/E_1-1}{0.4432}=0.8083\ \Rightarrow\ V_1=\frac{900.0}{2.0833\times 0.8083^2}=661.1\ \text{m}^3$$

Step 4: Second stage

$$\begin{aligned} W_2 &= W_1(1-E_1)=900.0\times(1-0.7362)=237.4\ \text{kg/day}\\ \sqrt{\frac{W_2}{V_2F}} &= \frac{(100/E_2-1)(1-E_1)}{0.4432}=1.0216\ \Rightarrow\ V_2=\frac{237.4}{2.0833\times 1.0216^2}=109.2\ \text{m}^3 \end{aligned}$$

Step 5: Diameters (depth $1.2$ m)

$$D_1=\sqrt{\frac{4V_1}{\pi\,d}}=\sqrt{\frac{4\times 661.1}{\pi\times 1.2}}=26.5\ \text{m},\qquad D_2=\sqrt{\frac{4\times 109.2}{\pi\times 1.2}}=10.8\ \text{m}$$

Answer: first-stage filter $D_1\approx 26.5$ m (volume 661.1 m³); second-stage filter $D_2\approx 10.8$ m (volume 109.2 m³).

Data and design basis

Conventional ASP (plug flow, diffused air). The aeration tank volume follows from the food-to-micro-organism ratio:

$$\frac{F}{M}=\frac{Q\,S_0}{V\,X}\ \Rightarrow\ V=\frac{Q\,S_0}{(F/M)\,X}$$

Data: settled-sewage $S_0=200$ mg/l, $F/M=0.3$/day, MLSS $=2000$ mg/l. Only the aeration tank is designed.

Step 1: Design flow

$$Q = 50000\times80/1000 = 4000.0\ \text{m}^3/\text{day}$$

Step 2: Volume of aeration tank

$$V = \frac{4000.0\times 200}{0.3\times 2000} = 1333.3\ \text{m}^3$$

(units: $Q$ in m³/day and $S_0$, $X$ in mg/l = g/m³.)

Step 3: Dimensions

$$\begin{aligned} \text{Adopt liquid depth } &= 4.0\ \text{m (usual range 3-4.5 m)}\\ \text{Plan area } A &= \frac{V}{4.0} = 333.3\ \text{m}^2 \end{aligned}$$

Provide 2 tanks (so one can be taken out of service), each with area $166.7$ m² and L : B about 4 : 1:

$$B=\sqrt{\frac{166.7}{4}} = 6.45\ \text{m (adopt 6.5 m)},\qquad L=\frac{166.7}{6.5} = 25.6\ \text{m (adopt 26 m)}$$

Adding 0.5 m free board, the tank depth is 4.5 m. The size of each tank is $26\ \text{m}\times 6.5\ \text{m}\times 4.5\ \text{m}$.

Step 4: Checks

$$\begin{aligned} \text{Hydraulic retention time} &= \frac{V}{Q} = \frac{1333.3}{4000.0}\times 24 = 8.0\ \text{h}\ (\text{usual 4-8 h; accepted})\\ \text{Volumetric loading} &= \frac{Q\,S_0}{V\times 1000} = 0.60\ \text{kg BOD/m}^3\text{/day}\ (\text{usual 0.3-0.6}) \end{aligned}$$

Answer: aeration tank volume $\approx 1333$ m³ (detention 8.0 h, depth 4.0 m); provide 2 tanks, each $26\ \text{m}\times 6.5\ \text{m}$ in plan with liquid depth 4.0 m (tank depth 4.5 m).

Data and design basis

Conventional ASP (plug flow, diffused air). The aeration tank volume follows from the food-to-micro-organism ratio:

$$\frac{F}{M}=\frac{Q\,S_0}{V\,X}\ \Rightarrow\ V=\frac{Q\,S_0}{(F/M)\,X}$$

Data: settled-sewage $S_0=200$ mg/l, $F/M=0.3$/day, MLSS $=2000$ mg/l. Only the aeration tank is designed.

Step 1: Design flow

$$Q = 100000\times96/1000 = 9600.0\ \text{m}^3/\text{day}$$

Step 2: Volume of aeration tank

$$V = \frac{9600.0\times 200}{0.3\times 2000} = 3200.0\ \text{m}^3$$

(units: $Q$ in m³/day and $S_0$, $X$ in mg/l = g/m³.)

Step 3: Dimensions

$$\begin{aligned} \text{Adopt liquid depth } &= 4.0\ \text{m (usual range 3-4.5 m)}\\ \text{Plan area } A &= \frac{V}{4.0} = 800.0\ \text{m}^2 \end{aligned}$$

Provide 2 tanks (so one can be taken out of service), each with area $400.0$ m² and L : B about 4 : 1:

$$B=\sqrt{\frac{400.0}{4}} = 10.00\ \text{m (adopt 10.0 m)},\qquad L=\frac{400.0}{10.0} = 40.0\ \text{m (adopt 40 m)}$$

Adding 0.5 m free board, the tank depth is 4.5 m. The size of each tank is $40\ \text{m}\times 10.0\ \text{m}\times 4.5\ \text{m}$.

Step 4: Checks

$$\begin{aligned} \text{Hydraulic retention time} &= \frac{V}{Q} = \frac{3200.0}{9600.0}\times 24 = 8.0\ \text{h}\ (\text{usual 4-8 h; accepted})\\ \text{Volumetric loading} &= \frac{Q\,S_0}{V\times 1000} = 0.60\ \text{kg BOD/m}^3\text{/day}\ (\text{usual 0.3-0.6}) \end{aligned}$$

Answer: aeration tank volume $\approx 3200$ m³ (detention 8.0 h, depth 4.0 m); provide 2 tanks, each $40\ \text{m}\times 10.0\ \text{m}$ in plan with liquid depth 4.0 m (tank depth 4.5 m).

Assumptions (Nepal practice): water supply 135 lpcd, 80% of it reaches the sewer, peak factor 2.5 for a city of this size (so $Q_{avg}=273000\times135\times0.8/1000 = 29484$ m³/day); grit of $d=0.2$ mm, $G=2.65$; $k=0.06$, $f=0.03$; the temperature is 22°C (Dang). Two horizontal-flow units are provided, each able to take the full peak flow.

Step 1: Design flow

$$Q = \frac{273000\times 135\times 0.8\times 2.5}{1000\times 86400} = 0.8531\ \text{m}^3/\text{s}$$

Step 2: Settling velocity of the grit

Hazen's formula (with temperature correction), $d$ in mm:

$$v_s = 418\,(G-1)\,d^2\left(\frac{3T+70}{100}\right)\ \text{mm/s} = 418\times (2.65-1)\times 0.2^2\times\frac{3\times 22+70}{100} = 37.52\ \text{mm/s} = 0.0375\ \text{m/s}$$

Step 3: Horizontal (flow-through) velocity

The velocity must not scour settled grit. Scour (critical) velocity (Camp-Shields):

$$v_c = \sqrt{\frac{8k}{f}(G-1)\,g\,d} = \sqrt{\frac{8\times 0.06}{0.03}(2.65-1)\times 9.81\times 0.2\times 10^{-3}} = 0.228\ \text{m/s}$$

Adopt $v_h = 0.20$ m/s (the scour velocity rounded down; the usual range is 0.2-0.3 m/s). This is high enough to keep the light organic matter moving, so it passes on instead of settling.

Step 4: Cross-section and dimensions

$$\begin{aligned} A_c &= \frac{Q}{v_h} = \frac{0.8531}{0.20} = 4.266\ \text{m}^2\\ \text{Adopt depth } H &= 1.0\ \text{m} \Rightarrow B = \frac{A_c}{H} = 4.27\ \text{m (adopt 4.3 m)}\\ L_{theo} &= \frac{v_h H}{v_s} = \frac{0.20\times 1.0}{0.0375} = 5.33\ \text{m}\\ L &= \max(1.25\,L_{theo},\ 30\,v_h) = \max(6.66,\ 6.00) = 6.66\ \text{m (adopt 7.0 m)} \end{aligned}$$

(25% is added for inlet and outlet turbulence, and the length is not allowed to give less than 30 s detention.)

Step 5: Checks

$$\begin{aligned} \text{Detention time} &= \frac{L}{v_h} = \frac{7.0}{0.20} = 35\ \text{s}\ (\text{within 30-90 s})\\ \text{Surface loading} &= \frac{Q}{B\,L} = \frac{0.8531\times 86400}{4.3\times 7.0} = 2449\ \text{m}^3/\text{m}^2\text{/day} \end{aligned}$$

Total depth $= H + \text{grit storage } 0.25 + \text{free board } 0.30 = 1.55$ m.

 Plan:   inlet ->[ <-- L = 7.0 m --> ]-> outlet (weir)
                 |  B = 4.3 m          |
 Section: water level ~~~~~~~~~~~~~~~~~~~  H = 1.0 m
          grit storage 0.25 m  ______________

Answer: provide a grit chamber (at least two units in parallel, each for the full flow so that one can be cleaned) of size $L\times B\times H = 7.0\ \text{m}\times 4.3\ \text{m}\times 1.0\ \text{m}$ (water depth), total depth about 1.55 m, with flow velocity 0.20 m/s.

Grit particles of $d=0.2$ mm and $G=2.65$ are to be removed from the maximum flow of 10 MLD. The settling velocity ($0.02$ m/s) and the flow velocity ($0.3$ m/s, maintained constant by a proportional weir at the outlet) are given. $k=0.06$ and $f=0.03$ are assumed only for the scour check.

Step 1: Design flow

$$Q = \frac{10\times 10^6}{1000\times 86400} = 0.1157\ \text{m}^3/\text{s}$$

Step 2: Settling velocity of the grit

The settling velocity is given: $v_s = 0.02$ m/s.

Step 3: Horizontal (flow-through) velocity

The velocity must not scour settled grit. Scour (critical) velocity (Camp-Shields):

$$v_c = \sqrt{\frac{8k}{f}(G-1)\,g\,d} = \sqrt{\frac{8\times 0.06}{0.03}(2.65-1)\times 9.81\times 0.2\times 10^{-3}} = 0.228\ \text{m/s}$$

The flow velocity is given: $v_h = 0.3$ m/s. (It is slightly above the Camp-Shields scour value for 0.2 mm grit, but a proportional weir holds it constant and the usual 0.3 m/s limit applies.)

Step 4: Cross-section and dimensions

$$\begin{aligned} A_c &= \frac{Q}{v_h} = \frac{0.1157}{0.30} = 0.386\ \text{m}^2\\ \text{Adopt depth } H &= 0.6\ \text{m} \Rightarrow B = \frac{A_c}{H} = 0.64\ \text{m (adopt 0.7 m)}\\ L_{theo} &= \frac{v_h H}{v_s} = \frac{0.30\times 0.6}{0.0200} = 9.00\ \text{m}\\ L &= \max(1.25\,L_{theo},\ 30\,v_h) = \max(11.25,\ 9.00) = 11.25\ \text{m (adopt 11.5 m)} \end{aligned}$$

(25% is added for inlet and outlet turbulence, and the length is not allowed to give less than 30 s detention.)

Step 5: Checks

$$\begin{aligned} \text{Detention time} &= \frac{L}{v_h} = \frac{11.5}{0.30} = 38\ \text{s}\ (\text{within 30-90 s})\\ \text{Surface loading} &= \frac{Q}{B\,L} = \frac{0.1157\times 86400}{0.7\times 11.5} = 1242\ \text{m}^3/\text{m}^2\text{/day} \end{aligned}$$

Total depth $= H + \text{grit storage } 0.25 + \text{free board } 0.30 = 1.15$ m.

 Plan:   inlet ->[ <-- L = 11.5 m --> ]-> outlet (weir)
                 |  B = 0.7 m          |
 Section: water level ~~~~~~~~~~~~~~~~~~~  H = 0.6 m
          grit storage 0.25 m  ______________

Answer: provide a grit chamber (at least two units in parallel, each for the full flow so that one can be cleaned) of size $L\times B\times H = 11.5\ \text{m}\times 0.7\ \text{m}\times 0.6\ \text{m}$ (water depth), total depth about 1.15 m, with flow velocity 0.30 m/s.

Grit: $d=0.2$ mm, $G=2.65$; organic matter $G=1.2$; $T=25^\circ$C; $k=0.06$, $f=0.03$.

Step 1: Design flow

$$Q = \frac{58\times 10^6}{1000\times 86400} = 0.6713\ \text{m}^3/\text{s}$$

Step 2: Settling velocity of the grit

Hazen's formula (with temperature correction), $d$ in mm:

$$v_s = 418\,(G-1)\,d^2\left(\frac{3T+70}{100}\right)\ \text{mm/s} = 418\times (2.65-1)\times 0.2^2\times\frac{3\times 25+70}{100} = 40.00\ \text{mm/s} = 0.0400\ \text{m/s}$$

Step 3: Horizontal (flow-through) velocity

The velocity must not scour settled grit. Scour (critical) velocity (Camp-Shields):

$$v_c = \sqrt{\frac{8k}{f}(G-1)\,g\,d} = \sqrt{\frac{8\times 0.06}{0.03}(2.65-1)\times 9.81\times 0.2\times 10^{-3}} = 0.228\ \text{m/s}$$

Adopt $v_h = 0.20$ m/s (the scour velocity rounded down; the usual range is 0.2-0.3 m/s). This is high enough to keep the light organic matter moving, so it passes on instead of settling.

Step 4: Cross-section and dimensions

$$\begin{aligned} A_c &= \frac{Q}{v_h} = \frac{0.6713}{0.20} = 3.356\ \text{m}^2\\ \text{Adopt depth } H &= 1.0\ \text{m} \Rightarrow B = \frac{A_c}{H} = 3.36\ \text{m (adopt 3.4 m)}\\ L_{theo} &= \frac{v_h H}{v_s} = \frac{0.20\times 1.0}{0.0400} = 5.00\ \text{m}\\ L &= \max(1.25\,L_{theo},\ 30\,v_h) = \max(6.25,\ 6.00) = 6.25\ \text{m (adopt 6.5 m)} \end{aligned}$$

(25% is added for inlet and outlet turbulence, and the length is not allowed to give less than 30 s detention.)

Step 5: Checks

$$\begin{aligned} \text{Detention time} &= \frac{L}{v_h} = \frac{6.5}{0.20} = 32\ \text{s}\ (\text{within 30-90 s})\\ \text{Surface loading} &= \frac{Q}{B\,L} = \frac{0.6713\times 86400}{3.4\times 6.5} = 2624\ \text{m}^3/\text{m}^2\text{/day} \end{aligned}$$

Total depth $= H + \text{grit storage } 0.25 + \text{free board } 0.30 = 1.55$ m.

 Plan:   inlet ->[ <-- L = 6.5 m --> ]-> outlet (weir)
                 |  B = 3.4 m          |
 Section: water level ~~~~~~~~~~~~~~~~~~~  H = 1.0 m
          grit storage 0.25 m  ______________

Answer: provide a grit chamber (at least two units in parallel, each for the full flow so that one can be cleaned) of size $L\times B\times H = 6.5\ \text{m}\times 3.4\ \text{m}\times 1.0\ \text{m}$ (water depth), total depth about 1.55 m, with flow velocity 0.20 m/s.

Grit: $d=0.2$ mm, $G=2.65$, $T=20^\circ$C. $k=0.06$ and $f=0.03$ are assumed (standard values) for the scour check.

Step 1: Design flow

$$Q = \frac{5\times 10^6}{1000\times 86400} = 0.0579\ \text{m}^3/\text{s}$$

Step 2: Settling velocity of the grit

Hazen's formula (with temperature correction), $d$ in mm:

$$v_s = 418\,(G-1)\,d^2\left(\frac{3T+70}{100}\right)\ \text{mm/s} = 418\times (2.65-1)\times 0.2^2\times\frac{3\times 20+70}{100} = 35.86\ \text{mm/s} = 0.0359\ \text{m/s}$$

Step 3: Horizontal (flow-through) velocity

The velocity must not scour settled grit. Scour (critical) velocity (Camp-Shields):

$$v_c = \sqrt{\frac{8k}{f}(G-1)\,g\,d} = \sqrt{\frac{8\times 0.06}{0.03}(2.65-1)\times 9.81\times 0.2\times 10^{-3}} = 0.228\ \text{m/s}$$

Adopt $v_h = 0.20$ m/s (the scour velocity rounded down; the usual range is 0.2-0.3 m/s). This is high enough to keep the light organic matter moving, so it passes on instead of settling.

Step 4: Cross-section and dimensions

$$\begin{aligned} A_c &= \frac{Q}{v_h} = \frac{0.0579}{0.20} = 0.289\ \text{m}^2\\ \text{Adopt depth } H &= 0.5\ \text{m} \Rightarrow B = \frac{A_c}{H} = 0.58\ \text{m (adopt 0.6 m)}\\ L_{theo} &= \frac{v_h H}{v_s} = \frac{0.20\times 0.5}{0.0359} = 2.79\ \text{m}\\ L &= \max(1.25\,L_{theo},\ 30\,v_h) = \max(3.49,\ 6.00) = 6.00\ \text{m (adopt 6.0 m)} \end{aligned}$$

(25% is added for inlet and outlet turbulence, and the length is not allowed to give less than 30 s detention.)

Step 5: Checks

$$\begin{aligned} \text{Detention time} &= \frac{L}{v_h} = \frac{6.0}{0.20} = 30\ \text{s}\ (\text{within 30-90 s})\\ \text{Surface loading} &= \frac{Q}{B\,L} = \frac{0.0579\times 86400}{0.6\times 6.0} = 1389\ \text{m}^3/\text{m}^2\text{/day} \end{aligned}$$

Total depth $= H + \text{grit storage } 0.25 + \text{free board } 0.30 = 1.05$ m.

 Plan:   inlet ->[ <-- L = 6.0 m --> ]-> outlet (weir)
                 |  B = 0.6 m          |
 Section: water level ~~~~~~~~~~~~~~~~~~~  H = 0.5 m
          grit storage 0.25 m  ______________

Answer: provide a grit chamber (at least two units in parallel, each for the full flow so that one can be cleaned) of size $L\times B\times H = 6.0\ \text{m}\times 0.6\ \text{m}\times 0.5\ \text{m}$ (water depth), total depth about 1.05 m, with flow velocity 0.20 m/s.