BE Civil Engineering (IOE, TU) Sanitary Engineering (IOE, CE 654) Question Paper 2078 Nepal

Given: $P = 48{,}000$, water supply $=150$ lpcd, 80% returns as sewage, infiltration $=5000\ \text{L/ha/day}$ over $120$ ha.

(a) Average Dry Weather Flow (DWF)

Sanitary (domestic) wastewater flow:

$$Q_{san} = 48{,}000 \times 150 \times 0.80 = 5{,}760{,}000\ \text{L/day}$$

Infiltration:

$$Q_{inf} = 5000 \times 120 = 600{,}000\ \text{L/day}$$

Total average DWF:

$$Q_{avg} = 5{,}760{,}000 + 600{,}000 = 6{,}360{,}000\ \text{L/day} = 6360\ \text{m}^3/\text{day}$$

In $\text{m}^3/\text{s}$:

$$Q_{avg} = \frac{6360}{86400} = \mathbf{0.0736\ m^3/s}$$

(b) Design peak flow

Peak factor (sanitary flow varies; infiltration taken as steady):

$$PF = \frac{5}{P^{0.2}} = \frac{5}{48^{0.2}}$$

$48^{0.2} = e^{0.2\ln 48} = e^{0.2 \times 3.8712} = e^{0.7742} = 2.169$

$$PF = \frac{5}{2.169} = 2.305$$

Sanitary average flow $= 5{,}760{,}000\ \text{L/day} = 0.0667\ \text{m}^3/\text{s}$. Peak sanitary $= 2.305 \times 0.0667 = 0.1537\ \text{m}^3/\text{s}$.

Add steady infiltration $= 600{,}000/86400 = 0.00694\ \text{m}^3/\text{s}$:

$$Q_{peak} = 0.1537 + 0.00694 = \mathbf{0.1607\ m^3/s}$$

(c) Minimum flow

Minimum sanitary $= \tfrac{1}{3} \times 0.0667 = 0.0222\ \text{m}^3/\text{s}$. Add full infiltration $0.00694\ \text{m}^3/\text{s}$:

$$Q_{min} = 0.0222 + 0.00694 = \mathbf{0.0292\ m^3/s}$$

Why both matter: The peak flow governs the required hydraulic capacity (diameter/slope must carry it without surcharge). The minimum flow governs the self-cleansing check — the sewer must maintain a velocity (typically $\geq 0.6\ \text{m/s}$) at low flow so solids do not settle and form deposits. Designing only for peak risks siltation during lean periods.

Given: $D = 0.400$ m, $S = 1/500 = 0.002$, $n = 0.013$.

(a) Flowing full

Area: $A = \dfrac{\pi D^2}{4} = \dfrac{\pi (0.4)^2}{4} = 0.12566\ \text{m}^2$

Hydraulic radius (full pipe): $R = \dfrac{D}{4} = \dfrac{0.4}{4} = 0.10\ \text{m}$

Manning's velocity:

$$V = \frac{1}{n} R^{2/3} S^{1/2} = \frac{1}{0.013}(0.10)^{2/3}(0.002)^{1/2}$$

$(0.10)^{2/3} = 0.21544$, $(0.002)^{1/2} = 0.044721$

$$V = \frac{1}{0.013} \times 0.21544 \times 0.044721 = 76.923 \times 0.0096348 = \mathbf{0.741\ m/s}$$

Discharge full:

$$Q = A V = 0.12566 \times 0.741 = 0.09312\ \text{m}^3/\text{s} = \mathbf{93.1\ L/s}$$

(b) Half full ($d/D = 0.5$)

From hydraulic elements at $d/D = 0.5$: $v/V = 1.000$, $q/Q = 0.500$.

$$v = 1.000 \times 0.741 = \mathbf{0.741\ m/s}$$ $$q = 0.500 \times 0.09312 = 0.04656\ \text{m}^3/\text{s} = \mathbf{46.6\ L/s}$$

(Consistency check: at half depth the wetted area is exactly half and $R = D/4$ same as full, so $v$ equals full velocity and $q$ is half — confirms the values.)

(c) Self-cleansing check

Velocity at half-full $= 0.741\ \text{m/s} > 0.6\ \text{m/s}$.

The sewer satisfies the self-cleansing requirement at half-full flow. Solids will be transported rather than deposited.

Given: $Q = 8\ \text{MLD} = 8000\ \text{m}^3/\text{day}$, SOR $= 32\ \text{m}^3/\text{m}^2/\text{day}$, $t = 2.0$ h, $L:B = 4:1$.

(a) Surface area and plan dimensions

Surface area:

$$A = \frac{Q}{\text{SOR}} = \frac{8000}{32} = 250\ \text{m}^2$$

With $L = 4B$: $A = L \times B = 4B^2 = 250$

$$B = \sqrt{250/4} = \sqrt{62.5} = 7.91\ \text{m} \approx \mathbf{8.0\ m}$$ $$L = 4B = 4 \times 7.91 = 31.6\ \text{m} \approx \mathbf{32\ m}$$

Adopt $L = 32$ m, $B = 8$ m (plan area $= 256\ \text{m}^2$, slightly conservative).

(b) Depth and weir loading

Volume required from detention time:

$$\forall = Q \times t = 8000 \times \frac{2.0}{24} = 666.7\ \text{m}^3$$

Depth (using adopted plan area $256\ \text{m}^2$):

$$H = \frac{\forall}{A} = \frac{666.7}{256} = 2.60\ \text{m}$$

Add 0.4 m free board $\Rightarrow$ overall depth $\approx 3.0$ m. Adopt liquid depth $= 2.6$ m.

Weir loading (weir length = width = 8 m):

$$\text{Weir loading} = \frac{Q}{\text{weir length}} = \frac{8000}{8} = 1000\ \text{m}^3/\text{m}/\text{day}$$

This exceeds the permissible $180\ \text{m}^3/\text{m}/\text{day}$. Required weir length:

$$L_{weir} = \frac{8000}{180} = 44.4\ \text{m}$$

A single end weir of 8 m is inadequate; provide multiple/finger (suspended) effluent troughs to give $\geq 44.4$ m of weir crest.

(c) Comment

• Surface loading and detention time are within IS/standard ranges (SOR 30-40 m³/m²/day, $t$ 1.5-2.5 h) — acceptable.
• Depth 2.6-3.0 m is typical for primary tanks — acceptable.
• Single straight weir gives excessive weir loading; must extend weir length using launders so the design is hydraulically acceptable.

Final design: $L = 32$ m, $B = 8$ m, liquid depth $= 2.6$ m, with effluent launders providing $\geq 45$ m of weir crest.

Given: $Q = 6\ \text{MLD} = 6000\ \text{m}^3/\text{day}$, $S_0 = 220\ \text{mg/L}$, $X = 3000\ \text{mg/L}$, HRT $= 6$ h.

(a) Aeration tank volume

$$\forall = Q \times \text{HRT} = 6000 \times \frac{6}{24} = \mathbf{1500\ m^3}$$

(b) F/M ratio and volumetric loading

BOD load applied per day:

$$\text{BOD load} = Q \times S_0 = 6000\ \text{m}^3/\text{day} \times 220\ \text{g/m}^3 = 1{,}320{,}000\ \text{g/day} = 1320\ \text{kg/day}$$

Mass of MLSS in tank:

$$M_X = \forall \times X = 1500\ \text{m}^3 \times 3000\ \text{g/m}^3 = 4{,}500{,}000\ \text{g} = 4500\ \text{kg}$$

F/M ratio:

$$\frac{F}{M} = \frac{Q S_0}{\forall X} = \frac{1320}{4500} = \mathbf{0.293\ kg\ BOD/kg\ MLSS/day}$$

Volumetric (organic) loading:

$$L_v = \frac{Q S_0}{\forall} = \frac{1320}{1500} = \mathbf{0.88\ kg\ BOD/m^3/day}$$

(c) Assessment

$F/M = 0.293$ lies within the conventional range of $0.2$-$0.5$, and the volumetric loading ($\approx 0.88\ \text{kg/m}^3/\text{day}$) is typical of conventional activated sludge ($0.3$-$1.0$). The plant operates in the conventional range.

If F/M is too high (under-loaded biomass relative to food), the microorganisms grow rapidly in a dispersed, non-flocculating manner; filamentous and dispersed growth dominate, the sludge does not flocculate or settle well (bulking sludge, high SVI), leading to solids carry-over in the effluent and loss of the activated-sludge inventory. If F/M is too low (extended aeration), endogenous respiration dominates, giving well-settling but pin-floc sludge.

Given: $P = 120{,}000$, generation $= 0.45\ \text{kg/cap/day}$, loose density $= 250\ \text{kg/m}^3$.

(a) Daily generation and loose volume

Mass per day:

$$W = 120{,}000 \times 0.45 = 54{,}000\ \text{kg/day} = \mathbf{54\ tonnes/day}$$

Loose volume:

$$V_{loose} = \frac{54{,}000}{250} = \mathbf{216\ m^3/day}$$

(b) Number of collection vehicles

With a compaction ratio of 2.0, the in-trailer density doubles, so each $5\ \text{m}^3$ trailer carries loose-equivalent volume:

$$5\ \text{m}^3 \times 2.0 = 10\ \text{m}^3 \text{ (loose equivalent) per trip}$$

Effective capacity per vehicle per day ($3$ trips):

$$10 \times 3 = 30\ \text{m}^3/\text{day (loose equivalent)}$$

Number of vehicles:

$$N = \frac{V_{loose}}{30} = \frac{216}{30} = 7.2 \Rightarrow \mathbf{8\ vehicles}$$

(Check by mass: each trailer at compacted density $500\ \text{kg/m}^3$ carries $5 \times 500 = 2500$ kg/trip $= 7500$ kg/day; $54{,}000/7500 = 7.2 \Rightarrow 8$ vehicles. Consistent.)

(c) ISWM hierarchy

The ISWM hierarchy ranks waste-management strategies from most to least preferred:

1. Source reduction / waste prevention (most preferred — avoid generating waste).
2. Reuse of products and materials.
3. Recycling of materials (paper, glass, metal, plastics).
4. Composting / biological treatment of organic fraction.
5. Energy recovery (incineration with energy, waste-to-energy, biogas).
6. Sanitary landfilling / disposal (least preferred).

Sanitary landfilling sits at the bottom of the hierarchy — it is the final disposal option for residual waste that cannot be reduced, reused, recycled, composted, or used for energy recovery. A sanitary landfill uses engineered liners, leachate collection, gas venting and daily soil cover to minimise environmental impact, distinguishing it from open dumping.

BOD vs COD

Always COD $\geq$ BOD for a given sample.

Biodegradability via BOD/COD ratio

$$\frac{\text{BOD}}{\text{COD}} = \frac{180}{400} = \mathbf{0.45}$$

Interpretation:

• Ratio $> 0.5$ $\Rightarrow$ readily biodegradable (suited to biological treatment).
• Ratio $0.3$-$0.5$ $\Rightarrow$ moderately biodegradable.
• Ratio $< 0.3$ $\Rightarrow$ poorly biodegradable; may need physico-chemical treatment.

A ratio of 0.45 indicates the wastewater is reasonably (moderately) biodegradable and is suitable for biological (secondary) treatment.

Sewer appurtenances are the auxiliary structures and fittings provided along a sewer line to ensure its efficient and trouble-free operation, maintenance, inspection, and cleaning.

Four common appurtenances:

1. Manholes — access for inspection, cleaning and maintenance at junctions, bends and changes of grade.
2. Drop manholes — connect a high-level incoming sewer to a lower-level sewer (see below).
3. Catch basins / gully traps — admit surface/storm runoff while trapping grit and preventing odours.
4. Inverted siphons (depressed sewers) — carry sewage under obstructions such as streams, roads or other utilities.

(Others: lamp holes, flushing tanks, vent shafts, grease/oil traps, clean-outs.)

Drop manhole — function and use

A drop manhole is a manhole in which the incoming sewer enters at a level appreciably higher than the outgoing sewer; a vertical drop pipe conveys the flow down to the lower invert.

• Function: to lower the sewage from a higher branch sewer to a lower main sewer without high-velocity splashing, turbulence, and erosion at the manhole base, and to keep workers safe during entry.
• Condition for provision: when the difference in invert levels between the incoming and outgoing sewers exceeds about 0.5-0.6 m (i.e., a steep drop occurs at a junction, typically in hilly terrain or where a lateral meets a deep main). Below this difference, the drop is accommodated by a benched/ramped channel instead.

Given: users $= 30$, flow $= 90\ \text{L/cap/day}$, detention $= 2$ days, sludge allowance $= 40\ \text{L/cap}$, liquid depth $= 1.5$ m, $L:B = 3:1$.

Step 1 — Liquid (sewage) volume from detention

$$V_{liquid} = 30 \times 90 \times 2 = 5400\ \text{L} = 5.4\ \text{m}^3$$

Step 2 — Sludge/scum storage volume

$$V_{sludge} = 30 \times 40 = 1200\ \text{L} = 1.2\ \text{m}^3$$

Step 3 — Total required volume

$$V = V_{liquid} + V_{sludge} = 5.4 + 1.2 = \mathbf{6.6\ m^3}$$

Step 4 — Surface (plan) area at liquid depth 1.5 m

$$A = \frac{V}{H} = \frac{6.6}{1.5} = 4.4\ \text{m}^2$$

Step 5 — Plan dimensions ($L = 3B$)

$$3B^2 = 4.4 \Rightarrow B = \sqrt{1.467} = 1.21\ \text{m} \approx 1.2\ \text{m}$$ $$L = 3B = 3.63\ \text{m} \approx 3.6\ \text{m}$$

Provided tank: $L = 3.6$ m, $B = 1.2$ m, liquid depth $= 1.5$ m, plus about 0.3 m free board $\Rightarrow$ overall depth $\approx 1.8$ m.

Final answer: Liquid + sludge volume $\approx \mathbf{6.6\ m^3}$; tank $\mathbf{3.6\ m \times 1.2\ m \times 1.5\ m}$ (liquid depth).

Anaerobic sludge digestion is the biological stabilisation of organic sludge by microorganisms in the absence of free oxygen, converting putrescible organic matter into stable end-products (digested sludge, gas and liquor).

Three stages:

1. Hydrolysis (and acidogenesis – solubilisation): Complex insoluble organics (carbohydrates, proteins, fats) are broken down by extracellular enzymes into soluble simple compounds (sugars, amino acids, fatty acids).

2. Acid formation (acetogenesis): Acid-forming bacteria convert the soluble organics into volatile fatty acids (mainly acetic acid), along with $\text{CO}_2$, $\text{H}_2$ and ammonia. pH tends to drop in this phase.

3. Methane formation (methanogenesis): Slow-growing, pH-sensitive methanogenic bacteria convert the volatile acids and $\text{CO}_2/\text{H}_2$ into methane ($\text{CH}_4$) and carbon dioxide. This stage stabilises the sludge.

Principal gas produced: Methane ($\text{CH}_4$) (digester gas is typically ~65% CH₄, ~35% CO₂), a usable fuel (biogas).

Two benefits of digestion:

1. Volume and mass reduction of sludge and destruction of pathogens, giving a stable, less odorous, more easily dewatered sludge.
2. Energy recovery — methane (biogas) can be burned for heating the digester or generating power.

Tertiary (advanced) treatment is the treatment provided after secondary (biological) treatment to remove residual constituents that secondary treatment cannot adequately remove — such as nutrients (nitrogen, phosphorus), residual suspended solids, refractory organics, and pathogens. It produces high-quality effluent suitable for discharge to sensitive waters or for reuse. Examples: filtration, nutrient removal (nitrification-denitrification, P precipitation), activated carbon adsorption, membrane processes, and disinfection.

Chlorination (disinfection): Chlorine (as gas $\text{Cl}_2$, hypochlorite, or chlorine dioxide) is added to the effluent. In water it forms hypochlorous acid (HOCl) and hypochlorite ion (OCl⁻); HOCl is the strong germicidal agent that destroys pathogenic bacteria and viruses by oxidising cell material. A suitable contact time (≈ 30 min) and residual chlorine are maintained to ensure kill. It is cheap and effective but can form disinfection by-products (e.g., chloramines, THMs) and residual chlorine may be toxic to aquatic life (sometimes requiring dechlorination).

Chlorine dose calculation

$Q = 10\ \text{MLD} = 10{,}000\ \text{m}^3/\text{day}$, dose $= 6\ \text{mg/L} = 6\ \text{g/m}^3$.

$$\text{Chlorine required} = Q \times \text{dose} = 10{,}000\ \text{m}^3/\text{day} \times 6\ \text{g/m}^3 = 60{,}000\ \text{g/day}$$ $$= \mathbf{60\ kg/day}$$

One advantage of the separate system: The treatment plant handles only the relatively constant sanitary flow, so it is smaller and more economical to operate, and stormwater can be discharged to natural drainage without treatment.

One advantage of the combined system: Only one set of pipes is laid, which is economical and convenient in congested areas; storm flow also flushes/dilutes the sewage in the sewers.