BE Civil Engineering (IOE, TU) Sanitary Engineering (IOE, CE 654) Question Paper 2079 Nepal

(a) Flow computations

Average dry weather flow (DWF):

$$Q_{avg} = \frac{\text{Pop} \times \text{lpcd} \times 0.80}{1000} = \frac{45000 \times 135 \times 0.80}{1000} = 4860 \ \text{m}^3/\text{day}$$

Converting to m$^3$/s:

$$Q_{avg} = \frac{4860}{86400} = 0.05625 \ \text{m}^3/\text{s}$$

Infiltration:

$$Q_{inf} = 0.05 \times 12000 = 600 \ \text{m}^3/\text{day} = \frac{600}{86400} = 0.006944 \ \text{m}^3/\text{s}$$

Maximum (peak) design flow including infiltration:

$$Q_{max} = 2.5 \times Q_{avg} + Q_{inf} = 2.5 \times 0.05625 + 0.006944$$ $$Q_{max} = 0.140625 + 0.006944 = 0.14757 \ \text{m}^3/\text{s} \approx \mathbf{0.1476\ m^3/s}$$

Minimum flow (one-third of average, sewage only):

$$Q_{min} = \frac{1}{3} \times 0.05625 = \mathbf{0.01875\ m^3/s}$$

(b) Sewer diameter at half-full

For a circular sewer flowing half full (depth = D/2):

• Flow area: $A = \dfrac{1}{2}\cdot\dfrac{\pi D^2}{4} = \dfrac{\pi D^2}{8}$
• Wetted perimeter: $P = \dfrac{\pi D}{2}$
• Hydraulic radius: $R = \dfrac{A}{P} = \dfrac{\pi D^2/8}{\pi D/2} = \dfrac{D}{4}$

Manning's equation:

$$Q = \frac{1}{n} A R^{2/3} S^{1/2}$$

with $S = 1/600 = 0.0016667$, $S^{1/2} = 0.040825$, $n = 0.013$.

$$Q = \frac{1}{0.013}\cdot\frac{\pi D^2}{8}\cdot\left(\frac{D}{4}\right)^{2/3}\cdot 0.040825$$

Group the constants. $\left(\frac{1}{4}\right)^{2/3} = 0.39685$, and $\frac{\pi}{8} = 0.392699$.

$$Q = \frac{0.040825}{0.013}\times 0.392699 \times 0.39685 \times D^{2}\cdot D^{2/3}$$ $$Q = 3.1404 \times 0.392699 \times 0.39685 \times D^{8/3}$$ $$Q = 0.48937\, D^{8/3}$$

Set $Q = Q_{max} = 0.14757$ m$^3$/s:

$$D^{8/3} = \frac{0.14757}{0.48937} = 0.30155$$ $$D = 0.30155^{3/8} = e^{(3/8)\ln 0.30155} = e^{0.375\times(-1.19884)} = e^{-0.44957} = 0.6379\ \text{m}$$

Adopt the next standard size: D = 700 mm (0.70 m).

Velocity check at half-full (using adopted D = 0.70 m)

$$A = \frac{\pi (0.70)^2}{8} = \frac{\pi \times 0.49}{8} = 0.19242\ \text{m}^2$$ $$R = \frac{D}{4} = \frac{0.70}{4} = 0.175\ \text{m}$$ $$V = \frac{1}{n}R^{2/3}S^{1/2} = \frac{1}{0.013}(0.175)^{2/3}(0.040825)$$

$(0.175)^{2/3} = e^{(2/3)\ln 0.175} = e^{0.6667\times(-1.74297)} = e^{-1.16198} = 0.31285$

$$V = 76.923 \times 0.31285 \times 0.040825 = \mathbf{0.982\ m/s}$$

Since $V = 0.982$ m/s $\geq 0.6$ m/s, the self-cleansing velocity is satisfied at half-full peak flow. (At the half-full condition, velocity equals the full-flowing velocity, so the sewer is self-cleansing throughout the operating range above minimum flow.)

Summary: $Q_{avg}=4860$ m$^3$/day, $Q_{max}=0.1476$ m$^3$/s, $Q_{min}=0.0188$ m$^3$/s; adopt $D = 700$ mm; $V = 0.98$ m/s (OK).

(a) Definitions

BOD (Biochemical Oxygen Demand): the amount of dissolved oxygen consumed by microorganisms in the aerobic biochemical oxidation of organic matter in water over a specified period (typically 5 days at 20°C), expressed in mg/L. It measures the biodegradable organic content.

COD (Chemical Oxygen Demand): the amount of oxygen equivalent consumed in the chemical oxidation of organic matter using a strong oxidant (e.g. K₂Cr₂O₇ in acid). It measures total oxidizable organic matter (biodegradable + non-biodegradable).

BOD/COD ratio: Since COD oxidizes essentially all organics while BOD oxidizes only the biodegradable fraction, the ratio indicates biodegradability. A ratio $> 0.5$ indicates readily biodegradable wastewater (suitable for biological treatment); a ratio $< 0.3$ indicates the wastewater contains a large non-biodegradable/toxic fraction and may not be amenable to conventional biological treatment.

(b) BOD kinetics

(i) Ultimate BOD $L_0$:

$$BOD_t = L_0\left(1 - e^{-k_D t}\right)$$ $$210 = L_0\left(1 - e^{-0.23\times 5}\right) = L_0\left(1 - e^{-1.15}\right) = L_0(1 - 0.31664) = 0.68336\,L_0$$ $$L_0 = \frac{210}{0.68336} = \mathbf{307.3\ mg/L}$$

(ii) BOD remaining after 8 days: the unsatisfied (remaining) BOD is

$$L_t = L_0\,e^{-k_D t} = 307.3 \times e^{-0.23\times 8} = 307.3 \times e^{-1.84}$$ $$e^{-1.84} = 0.15882$$ $$L_t = 307.3 \times 0.15882 = \mathbf{48.8\ mg/L}$$

(Equivalently, BOD exerted in 8 days $= 307.3 - 48.8 = 258.5$ mg/L.)

(iii) Rate constant and 5-day BOD at 27°C:

$$k_{27} = k_{20}\,\theta^{(27-20)} = 0.23 \times 1.047^{7}$$

$1.047^{7} = e^{7\ln 1.047} = e^{7\times 0.045927} = e^{0.32149} = 1.37923$

$$k_{27} = 0.23 \times 1.37923 = \mathbf{0.3172\ day^{-1}}$$

The ultimate BOD $L_0$ is essentially temperature-independent, so using $L_0 = 307.3$ mg/L:

$$BOD_5^{27} = L_0\left(1 - e^{-k_{27}\times 5}\right) = 307.3\left(1 - e^{-0.3172\times 5}\right) = 307.3\left(1 - e^{-1.586}\right)$$

$e^{-1.586} = 0.20479$

$$BOD_5^{27} = 307.3 \times (1 - 0.20479) = 307.3 \times 0.79521 = \mathbf{244.4\ mg/L}$$

The higher temperature accelerates oxidation, so the 5-day BOD rises from 210 mg/L to about 244 mg/L.

(a) Primary settling tank

Surface area from SOR:

$$A = \frac{Q}{SOR} = \frac{18000}{35} = 514.29\ \text{m}^2$$

Diameter:

$$A = \frac{\pi D^2}{4} \Rightarrow D = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4\times 514.29}{\pi}} = \sqrt{654.78} = 25.59\ \text{m}$$

Adopt D = 26 m (provided area $= \frac{\pi}{4}(26)^2 = 530.9$ m$^2$).

Detention time (using provided dimensions, depth 3.0 m):

$$\text{Volume} = A \times H = 530.9 \times 3.0 = 1592.8\ \text{m}^3$$ $$t = \frac{V}{Q} = \frac{1592.8}{18000}\ \text{day} = 0.08849\ \text{day} = 0.08849 \times 24 = \mathbf{2.12\ h}$$

This lies in the usual range of 1.5–2.5 h for primary tanks — OK.

(b) Trickling filter (NRC formula)

BOD entering the filter (after 32% removal in primary tank):

$$BOD_{in} = 250\times(1-0.32) = 250 \times 0.68 = 170\ \text{mg/L}$$

Mass of BOD applied to filter (load):

$$W = \frac{Q \times BOD_{in}}{1000} = \frac{18000 \times 170}{1000} = 3060\ \text{kg/day}$$

Required filter efficiency. Overall plant removal must be 88%, i.e. final effluent BOD $= 250\times(1-0.88) = 30$ mg/L. The primary stage already removed 32%. The trickling filter must remove the remaining BOD relative to its influent (170 mg/L) down to 30 mg/L:

$$E_{filter} = \frac{170 - 30}{170}\times 100 = \frac{140}{170}\times 100 = 82.35\%$$

Solve NRC for volume. With $F = 1$ (no recirculation):

$$E = \frac{100}{1 + 0.4432\sqrt{W/V}}$$ $$82.35 = \frac{100}{1 + 0.4432\sqrt{3060/V}}$$ $$1 + 0.4432\sqrt{3060/V} = \frac{100}{82.35} = 1.21433$$ $$0.4432\sqrt{3060/V} = 0.21433$$ $$\sqrt{3060/V} = \frac{0.21433}{0.4432} = 0.48360$$ $$\frac{3060}{V} = 0.48360^2 = 0.23387$$ $$V = \frac{3060}{0.23387} = \mathbf{13083\ m^3}$$

Filter plan area and diameter (depth = 1.8 m):

$$A_f = \frac{V}{H} = \frac{13083}{1.8} = 7268\ \text{m}^2$$

Use, say, four equal circular filters: $A_f/4 = 1817$ m$^2$ each $\Rightarrow$ $D = \sqrt{4\times1817/\pi} = 48.1$ m each (or several smaller units).

BOD volumetric loading (organic loading):

$$L_{BOD} = \frac{W}{V} = \frac{3060\ \text{kg/day}}{13083\ \text{m}^3} = 0.2339\ \text{kg/m}^3/\text{day} = \mathbf{233.9\ g\,BOD/m^3/day}$$

This loading ($\approx 0.23$ kg/m$^3$/d) is consistent with low-rate trickling filter practice (0.08–0.4 kg/m$^3$/d).

(a) Aeration tank volume

The standard complete-mix design relation linking SRT to volume:

$$\frac{1}{\theta_c} = \frac{Y\,Q\,(S_0 - S)}{V X} - k_d$$

Rearranged for volume:

$$V = \frac{\theta_c\,Y\,Q\,(S_0 - S)}{X\,(1 + k_d\,\theta_c)}$$

Substitute (work in mg/L = g/m$^3$ for concentrations and m$^3$/day for Q):

• $S_0 - S = 160 - 15 = 145$ mg/L
• $1 + k_d\theta_c = 1 + 0.06\times 8 = 1.48$

$$V = \frac{8 \times 0.55 \times 10000 \times 145}{3000 \times 1.48}$$

Numerator $= 8\times0.55 = 4.4$; $4.4\times10000 = 44000$; $44000\times145 = 6{,}380{,}000$. Denominator $= 3000\times1.48 = 4440$.

$$V = \frac{6{,}380{,}000}{4440} = \mathbf{1436.9\ m^3}$$

Hydraulic retention time:

$$\tau = \frac{V}{Q} = \frac{1436.9}{10000}\ \text{day} = 0.14369\ \text{day} = 0.14369\times 24 = \mathbf{3.45\ h}$$

(b) Observed yield and sludge production

Observed yield:

$$Y_{obs} = \frac{Y}{1 + k_d\,\theta_c} = \frac{0.55}{1.48} = 0.37162\ \text{kg VSS/kg BOD}$$

Mass of waste activated sludge (VSS) produced per day:

$$P_x = Y_{obs}\,Q\,(S_0 - S)$$

BOD removed per day $= Q(S_0-S) = \dfrac{10000 \times 145}{1000} = 1450$ kg BOD/day.

$$P_x = 0.37162 \times 1450 = \mathbf{538.8\ kg\,VSS/day}$$

(Check via SRT: solids in tank $= VX = 1436.9 \times 3000 / 1000 = 4310.7$ kg; $P_x = VX/\theta_c = 4310.7/8 = 538.8$ kg/day. ✓)

(c) F/M ratio and volumetric BOD loading

Food-to-microorganism ratio:

$$\frac{F}{M} = \frac{Q\,S_0}{V\,X} = \frac{10000 \times 160}{1436.9 \times 3000} = \frac{1{,}600{,}000}{4{,}310{,}700} = \mathbf{0.371\ kg\,BOD/kg\,MLVSS/day}$$

(This sits in the conventional ASP range of 0.2–0.5.)

Volumetric BOD loading:

$$L_v = \frac{Q\,S_0}{V} = \frac{10000 \times 160 / 1000}{1436.9} = \frac{1600}{1436.9} = \mathbf{1.11\ kg\,BOD/m^3/day}$$

Summary: $V \approx 1437$ m$^3$, $\tau \approx 3.45$ h, $Y_{obs}=0.372$, $P_x \approx 539$ kg VSS/day, F/M $=0.37$, $L_v=1.11$ kg/m$^3$/d.

(a) Generation and collection

(i) Daily mass:

$$M = 0.45 \times 120000 = 54{,}000\ \text{kg/day} = \mathbf{54\ tonnes/day}$$

Loose volume at 250 kg/m$^3$:

$$V_{loose} = \frac{54000}{250} = \mathbf{216\ m^3/day}$$

(ii) Vehicle trips. Mass capacity of one compacting vehicle:

$$m_{veh} = 8\ \text{m}^3 \times 450\ \text{kg/m}^3 = 3600\ \text{kg/trip}$$ $$\text{Trips} = \frac{54000}{3600} = 15\ \text{trips/day}$$

15 vehicle trips per day (no rounding-up needed; exactly 15).

(b) Landfill area for one year

Compacted volume of waste in landfill (600 kg/m$^3$):

$$V_{waste/day} = \frac{54000}{600} = 90\ \text{m}^3/\text{day}$$

Cover soil volume at ratio 1:4 (soil:waste):

$$V_{cover/day} = \frac{1}{4}\times 90 = 22.5\ \text{m}^3/\text{day}$$

Total landfill airspace per day:

$$V_{total/day} = 90 + 22.5 = 112.5\ \text{m}^3/\text{day}$$

Annual volume:

$$V_{year} = 112.5 \times 365 = 41{,}062.5\ \text{m}^3/\text{year}$$

Land area for usable depth 8 m:

$$A = \frac{41062.5}{8} = 5132.8\ \text{m}^2$$

Converting to hectares ($1\ \text{ha} = 10{,}000\ \text{m}^2$):

$$A = \frac{5132.8}{10000} = \mathbf{0.513\ ha/year}$$

(Allowing for side slopes, access roads and buffer, the gross land take would be larger — typically 1.2–1.5× this footprint.)

(c) Functional elements of integrated SWM

1. Waste generation — activities that identify materials as no longer valuable; the starting point. Source reduction here lowers the load on all later stages.
2. On-site handling, storage and segregation — storage of waste at the source and separation of recyclables/organics before collection, which improves recovery efficiency.
3. Collection — gathering waste and transporting it to a transfer station or processing/disposal site; usually the costliest element.
4. Transfer and transport — transfer of waste from small collection vehicles to larger transport units and haul to distant facilities, reducing unit haul cost.
5. Processing and recovery — separation, recycling, composting and energy recovery (e.g. WtE) to reduce volume and recover resources.
6. Disposal — final placement of residual waste, primarily in engineered sanitary landfills with leachate and gas control.

(Any four, explained, earn full marks.)

(a) Sewer appurtenances

Sewer appurtenances are the ancillary structures and fittings provided along a sewerage system for its proper functioning, inspection, cleaning, ventilation and maintenance. Five common appurtenances and their functions:

(Any five accepted.)

(b) Drop manhole

Purpose: A drop manhole is used where a branch (lateral) sewer joins a main sewer at a level considerably higher than the main. Instead of letting the incoming sewage cascade down the full depth of the manhole — which causes splashing, turbulence, erosion of the benching, foul gas release and a hazard to workers — the incoming flow is led down through a vertical drop pipe to the invert level of the outgoing sewer.

Arrangement (description):

  incoming high sewer
        |
        v   (enters manhole wall)
     ====+====   <- inlet, with a TEE/bend
         |       <- vertical drop (down) pipe outside or inside the manhole shaft
         |
         |
     ____v____   <- discharges at the invert of the outgoing main sewer
     -> outgoing sewer (lower level)

The drop pipe carries the bulk flow down; an opening/inspection branch at the original incoming level allows rodding of the inlet sewer. The chamber is otherwise a normal manhole with benching, invert channel and cover.

Typical provision threshold: A drop manhole is generally provided when the difference in invert levels (the drop) between the incoming and outgoing sewers exceeds about 0.5 m to 0.6 m (commonly taken as > 600 mm).

(a) Tank volumes

Liquid (sewage flow) volume for 24 h detention:

$$Q = 50 \times 90 = 4500\ \text{litres/day} = 4.5\ \text{m}^3/\text{day}$$

For a detention period of 24 h, liquid volume:

$$V_{liquid} = Q \times t = 4.5 \times 1\ \text{day} = 4.5\ \text{m}^3$$

Sludge & scum storage volume:

$$V_{sludge} = 0.04 \times 50 = 2.0\ \text{m}^3$$

Total required volume:

$$V_{total} = V_{liquid} + V_{sludge} = 4.5 + 2.0 = \mathbf{6.5\ m^3}$$

(b) Plan dimensions

Required plan area for a liquid depth of 1.5 m:

The total volume is accommodated within the liquid depth (sludge accumulates below the liquid working level over the cleaning interval; design the cross-section for the total volume at the working depth):

$$A = \frac{V_{total}}{\text{liquid depth}} = \frac{6.5}{1.5} = 4.333\ \text{m}^2$$

With $L = 3B$:

$$A = L \times B = 3B \times B = 3B^2 = 4.333$$ $$B^2 = 1.4444 \Rightarrow B = 1.20\ \text{m}$$ $$L = 3 \times 1.20 = 3.60\ \text{m}$$

Adopt B = 1.20 m, L = 3.60 m (provided area $= 4.32$ m$^2$ ≈ required).

Overall depth = liquid depth + free board:

$$D = 1.5 + 0.3 = \mathbf{1.8\ m}$$

Adopted tank: 3.60 m (L) × 1.20 m (B) × 1.8 m (overall depth).

Provided effective (liquid) volume $= 3.60\times1.20\times1.5 = 6.48$ m$^3 \approx 6.5$ m$^3$ ✓

(c) Functions

Functions of a septic tank:

1. Sedimentation — quiescent settling of suspended solids to the bottom as sludge and flotation of grease/oil as a scum layer, clarifying the liquid.
2. Anaerobic digestion — partial decomposition (digestion) of the settled sludge by anaerobic bacteria, reducing sludge volume and stabilizing it; it thereby acts as a combined settling-cum-digestion tank.

Role of the soak pit (soakage pit / leach pit): The clarified effluent from the septic tank still contains dissolved organics and pathogens; it is led to a soak pit where it percolates and infiltrates into the surrounding permeable soil, providing final disposal and further treatment of the effluent through soil filtration and aerobic microbial action in the unsaturated zone. The soak pit thus disposes of the liquid effluent into the ground without surface discharge.

(a) Sludge volumes

The volume of wet sludge is

$$V = \frac{M_s}{\rho_w \cdot S \cdot p_s}$$

where $M_s$ = dry solids mass, $\rho_w = 1000$ kg/m$^3$, $S$ = sludge specific gravity, and $p_s$ = solids fraction $= (1 - \text{moisture fraction})$.

(i) Fresh sludge (95% moisture → solids fraction $p_s = 0.05$):

$$V_{fresh} = \frac{2800}{1000 \times 1.02 \times 0.05} = \frac{2800}{51} = \mathbf{54.90\ m^3/day}$$

(ii) Digested sludge (90% moisture → solids fraction $p_s = 0.10$):

$$V_{dig} = \frac{2800}{1000 \times 1.02 \times 0.10} = \frac{2800}{102} = \mathbf{27.45\ m^3/day}$$

Digestion (with dewatering of the bound water) halves the sludge volume from ≈54.9 to ≈27.5 m$^3$/day — a major benefit, since handling cost scales with volume.

(Quick check: a moisture drop from 95% to 90% doubles the solids fraction, so the volume halves — consistent with $54.90/27.45 = 2.0$.)

(b) Sludge digestion

Objectives of sludge digestion:

1. To stabilize the organic matter (reduce putrescibility and odour).
2. To reduce the volume and mass of sludge (by destroying volatile solids and releasing bound water), lowering disposal cost.
3. To reduce pathogens and produce a sludge that can be dewatered and safely disposed/used.
4. To recover energy as biogas (methane) from anaerobic digestion.

Three phases of anaerobic digestion:

1. Hydrolysis / acidogenesis (acid-forming phase) — complex organics are hydrolysed and fermented to volatile fatty acids, alcohols, CO₂ and H₂ by acid-forming bacteria.
2. Acetogenesis — acetogenic bacteria convert the volatile fatty acids to acetic acid, hydrogen and CO₂.
3. Methanogenesis (methane-forming phase) — methanogenic archaea convert acetic acid and H₂/CO₂ to methane (CH₄) and carbon dioxide.

(a) Tertiary / advanced treatment

Tertiary (advanced) treatment is any treatment given to wastewater beyond conventional secondary (biological) treatment to remove specific residual constituents — nutrients (N, P), residual suspended solids, residual organics, pathogens, dissolved solids or specific toxic substances. Common processes: filtration, nutrient removal (nitrification–denitrification, biological/chemical P removal), coagulation, activated carbon adsorption, membrane processes and advanced disinfection.

Two situations where it is necessary:

1. When the effluent is discharged to a sensitive / nutrient-limited water body (e.g. a lake prone to eutrophication) requiring N and P removal.
2. When the effluent is to be reused (irrigation, industrial process water, groundwater recharge) and must meet stringent quality (low SS, pathogens, salinity) standards.

(b) Biological nitrogen removal

Step 1 — Nitrification (aerobic, autotrophic). Ammonia is oxidized to nitrate in two stages by autotrophic bacteria (Nitrosomonas, then Nitrobacter). It requires dissolved oxygen (aerobic conditions) and consumes alkalinity; no external organic carbon is needed (the bacteria fix CO₂).

$$NH_4^+ + 1.5\,O_2 \rightarrow NO_2^- + 2H^+ + H_2O$$ $$NO_2^- + 0.5\,O_2 \rightarrow NO_3^-$$

Overall:

$$NH_4^+ + 2\,O_2 \rightarrow NO_3^- + 2H^+ + H_2O$$

Step 2 — Denitrification (anoxic, heterotrophic). Nitrate is reduced to nitrogen gas by facultative heterotrophs under anoxic conditions (no free dissolved oxygen, but NO₃⁻ present as the electron acceptor). It requires an organic carbon source (e.g. influent BOD, or added methanol) as electron donor.

$$2\,NO_3^- + \text{organic C (e.g. } 1.25\,CH_3OH) \rightarrow N_2\uparrow + CO_2 + H_2O + 2\,OH^-$$

Simplified overall:

$$2\,NO_3^- + 10\,e^- + 12\,H^+ \rightarrow N_2\uparrow + 6\,H_2O$$

The nitrogen gas escapes to the atmosphere, achieving net N removal. Denitrification also recovers about half the alkalinity consumed in nitrification.

(c) Chlorination of effluent

Advantages:

1. Highly effective broad-spectrum disinfectant; inexpensive and well-established.
2. Provides a lasting residual that guards against re-contamination in the receiving/distribution system.

Disadvantage:

• Reacts with organics to form harmful disinfection by-products (e.g. trihalomethanes) and residual chlorine can be toxic to aquatic life, often requiring dechlorination before discharge.

(a) Grit chamber dimensions

Length of channel (so that grit settling is completed within the detention time / flow-through):

$$L = V_h \times t = 0.30\ \text{m/s} \times 60\ \text{s} = \mathbf{18\ m}$$

(Check with settling: the particle must fall the full depth while travelling the length. Required $L = V_h \cdot \frac{H}{v_s}$; we confirm below the depth is consistent.)

Cross-sectional area of flow (from continuity, $Q = A V_h$):

$$A = \frac{Q}{V_h} = \frac{0.30}{0.30} = \mathbf{1.0\ m^2}$$

Depth (width = 1.0 m):

$$H = \frac{A}{\text{width}} = \frac{1.0}{1.0} = \mathbf{1.0\ m}$$

Settling check: time for a grit particle to fall depth $H$ at $v_s = 0.021$ m/s:

$$t_{settle} = \frac{H}{v_s} = \frac{1.0}{0.021} = 47.6\ \text{s} < 60\ \text{s (detention time)} \Rightarrow \text{grit settles. ✓}$$

The required length on the settling criterion $L = V_h\cdot H/v_s = 0.30\times 47.6 = 14.3$ m $< 18$ m provided, so the 18 m chamber comfortably captures the grit.

Adopted grit channel: L = 18 m, width = 1.0 m, depth ≈ 1.0 m (plus free board, say 0.3 m, and extra depth for grit storage).

(b) Control of horizontal velocity

The horizontal velocity is held near 0.3 m/s because:

• It is fast enough to keep lighter organic matter in suspension (so only heavier inorganic grit settles, keeping the grit clean and putrescible organics out of it), and
• It is slow enough to allow the dense grit particles to settle to the bottom.

At higher velocity grit would be scoured/carried over; at lower velocity organic matter would also deposit and putrefy.

Device used: A proportional (Sutro) weir or a parshall flume is provided at the outlet. These control sections make the flow-through (horizontal) velocity nearly constant (≈0.3 m/s) over the full range of varying flows by automatically adjusting the flow depth/area in proportion to the discharge.

(a) Separate vs combined sewerage systems

One advantage of separate system: only the (smaller, more uniform) sanitary flow goes to treatment, so the treatment plant is smaller and cheaper to operate; no overflow of raw sewage during storms.

One advantage of combined system: a single set of sewers is cheaper and simpler to lay (one excavation), and periodic storm flow flushes/scours the sewer, helping self-cleansing.

(b) Factors affecting flows

Dry Weather Flow (DWF) — sanitary sewage flow in dry season; factors:

• Population served and rate of water supply (lpcd).
• Fraction of supplied water reaching the sewer (return factor, typ. 70–80%).
• Type of area (residential / commercial / industrial) and trade/industrial discharges.
• Infiltration of groundwater (depends on water-table, joints, pipe condition).
• Hourly, daily and seasonal variation in usage.

Storm / wet weather flow — runoff entering the system; factors:

• Rainfall intensity, duration and frequency of the design storm.
• Catchment area and its imperviousness (runoff coefficient).
• Time of concentration and slope of the catchment.
• Soil type, antecedent moisture and degree of urbanisation.
• (For combined systems) the amount of illicit/storm inflow connections.

(c) Peak factor in sewer design

Sewers must carry the maximum (peak) flow, not the average, because flow varies through the day (morning/evening peaks). The peak factor = peak flow / average flow is applied to the average DWF to obtain the design (maximum) flow, ensuring the sewer does not surcharge at peak hours.

Variation with population: the peak factor decreases as the contributing population increases, because in a large population the individual usage peaks are spread out in time and tend to average out (statistical smoothing). Thus small upstream sewers use a high peak factor (e.g. 3–4) while large trunk sewers serving big populations use a lower peak factor (e.g. 1.5–2). (Empirically, e.g. Harmon's formula gives peak factor $= 1 + \dfrac{14}{4 + \sqrt{P}}$, with $P$ in thousands, which falls as $P$ rises.)